/G43/G6F/G6C/G6C/G61/G72/G20 /G42/G65/G61/G72/G69/G6E/G67/G2E/G20 /G32/G37/G2E/G20 /G46._/G6C/G61/G74/G20 /G50/G69/G76/G6F/G74/G20 /G42/G65/G61/G72/G69/G6E/G67/G2E

Chapter 1 :  Introduction   /G6C  1
1Introduction1
Features
1. Definition.
2. Sub-divisions of Theory of
Machines.
3. Fundamental Units.4. Derived Units.5. Systems of Units.6. C.G.S. Units.7. F.P.S. Units.8. M.K.S. Units.9. International System of
Units (S.I. Units).
10. Metre.
11. Kilogram.
12. Second.13. Presentation of Units and
their Values.
14. Rules for S.I. Units.15. Force.16. Resultant Force.17. Scalars and Vectors.18. Representation of V ector
Quantities.

19. Addition of Vectors.20. Subtraction of V ectors.1.1. Definition
The subject Theory of Machines  may be defined as
that branch of Engineering-science, which deals with the studyof relative motion between the various parts of a machine,
and forces which act on them. The knowledge of this subject
is very essential for an engineer in designing the various partsof a machine.
Note:A machine is a device which receives energy in some

available form and utilises it to do some particular type of work.
1.2. Sub-divisions of Theory of Machines
The Theory of Machines may be sub-divided into
the following four branches :
1. Kinematics.  It is that branch of Theory of
Machines which deals with the relative motion between thevarious parts of the machines.
2. Dynamics.  It is that branch of Theory of Machines
which deals with the forces and their effects, while actingupon the machine parts in motion.

3. Kinetics. It is that branch of Theory of Machines
which deals with the inertia forces which arise from the com-bined effect of the mass and motion of the machine parts.
4. Statics. It is that branch of Theory of Machines
which deals with the forces and their effects while the ma-chine parts are at rest. The mass of the parts is assumed to benegligible.

2      /G6C           Theory of Machines
1.3.1.3.1.3.1.3.1.3. Fundamental UnitsFundamental UnitsFundamental UnitsFundamental UnitsFundamental Units
The measurement of
physical quantities is one of the
most important operations inengineering. Every quantity ismeasured in terms of somearbitrary, but internationallyaccepted units, calledfundamental units . All
physical quantities, met withinthis subject, are expressed interms of the following threefundamental quantities :
1.  Length ( L or l ),

2.  Mass ( M or m), and
3. Time ( t).
1.4.1.4.1.4.1.4.1.4. Derived UnitsDerived UnitsDerived UnitsDerived UnitsDerived Units
Some units are expressed in terms of fundamental units known as derived units, e.g., the units
of area, velocity, acceleration, pressure, etc.
1.5.1.5.1.5.1.5.1.5. Systems of UnitsSystems of UnitsSystems of UnitsSystems of UnitsSystems of Units
There are only four systems of units, which are commonly used and universally recognised.

These are known as :
1. C.G.S. units, 2. F.P.S. units, 3. M.K.S. units, and 4. S.I. units.
1.6.1.6.1.6.1.6.1.6. C.G.S. UnitsC.G.S. UnitsC.G.S. UnitsC.G.S. UnitsC.G.S. Units
In this system, the fundamental units of length, mass and time are centimetre , gram and
second  respectively. The C.G.S. units are known as absolute units or physicist's units.
1.7.1.7.1.7.1.7.1.7. FFFFF.P.P.P.P.P.S..S..S..S..S. Units Units Units Units Units

In this system, the fundamental units of length, mass and time are  foot, pound  and second
respectively.
1.8.1.8.1.8.1.8.1.8. M.K.S. UnitsM.K.S. UnitsM.K.S. UnitsM.K.S. UnitsM.K.S. Units
In this system, the fundamental units of length, mass and time are  metre, kilogram and second
respectively. The M.K.S. units are known as gravitational units or engineer's units.

1.9.1.9.1.9.1.9.1.9. InterInterInterInterInter nananananational System of Units (S.I.tional System of Units (S.I.tional System of Units (S.I.tional System of Units (S.I.tional System of Units (S.I.  Units) Units) Units) Units) Units)
The 11th general conference * of weights and measures have recommended a unified and

systematically constituted system of fundamental and derived units for international use. This systemis now being used in many countries. In India, the standards of Weights and Measures Act, 1956 (videwhich we switched over to M.K.S. units) has been revised to recognise all the S.I. units in industryand commerce.

* It is known as General Conference of Weights and Measures (G.C.W.M.). It is an international organisation,of which most of the advanced and developing countries (including India) are members. The conferencehas been entrusted with the task of prescribing definitions for various units of weights and measures, whichare the very basic of science and technology today.
Stopwatch Simple balance

Chapter 1 :  Introduction   /G6C  3
In this system of units, the fundamental units are metre (m), kilogram (kg) and second (s)
respectively. But there is a slight variation in their derived units . The derived units, which will be
used in this book are given below :
Density (mass density) kg/m3
Force N (Newton)Pressure Pa (Pascal) or N/m
2 ( 1 Pa = 1 N/m2)
Work, energy (in Joules) 1 J = 1 N-m
Power (in watts) 1 W = 1 J/s
Absolute viscosity kg/m-sKinematic viscosity m
2/s

2/s
Velocity m/sAcceleration m/s
2
Angular acceleration rad/s2
Frequency (in Hertz) Hz
The international metre, kilogram and second are discussed below :
1.10. Metre
The international metre may be defined as the shortest distance (at 0°C) between the two
parallel lines, engraved upon the polished surface of a platinum-iridium bar, kept at the InternationalBureau of Weights and Measures at Sevres near Paris.
1.11. Kilogram

The international kilogram may be defined as the mass of the platinum-iridium cylinder,
which is also kept at the International Bureau of Weights and Measures at Sevres near Paris.
1.12. Second
The fundamental unit of time for all the three systems, is second, which is 1/24 × 60 × 60
= 1/86 400th of the mean solar day. A solar day may be defined as the interval of time, between the
A man whose mass is 60 kg weighs 588.6 N (60 × 9.81 m/s2) on earth, approximately

96 N (60 × 1.6 m/s2) on moon and zero in space. But mass remains the same everywhere.

4  /G6C   Theory of Machines
instants, at which the sun crosses a meridian on two consecutive days. This value varies slightly
throughout the year. The average of all the solar days, during one year, is called the mean solar day.
1.13. Presentation of Units and their Values
The frequent changes in the present day life are facilitated by an international body known as

International Standard Organisation (ISO) which makes recommendations regarding internationalstandard procedures. The implementation of ISO recommendations, in a country, is assisted by itsorganisation appointed for the purpose. In India, Bureau of Indian Standards (BIS) previously knownas Indian Standards Institution (ISI) has been created for this purpose. We have already discussed thatthe fundamental units inM.K.S. and S.I. units forlength, mass and time is metre,kilogram and second

mass and time is metre,kilogram and second respec-tively. But in actual practice, itis not necessary to express alllengths in metres, all masses inkilograms and all times in sec-onds. We shall, sometimes, usethe convenient units, which aremultiples or divisions of ourbasic units in tens. As a typicalexample, although the metre isthe unit of length, yet a smallerlength of one-thousandth of ametre proves to be more con-venient unit, especially in thedimensioning of drawings. Such convenient units

of drawings. Such convenient units are formed by using a prefix in front of the basicunits to indicate the multiplier. The full list of these prefixes is given in the following table.

Table 1.1. Prefixes used in basic units
Factor by which the uni t Standard form Prefix Abbreviation
is multiplied
1 000 000 000 000 1012tera T
1 000 000 000 109giga G
1 000 000 106mega M
1 000 103kilo k
100 102hecto* h
10 101deca* da
0.1 10–1deci* d
0.01 10–2centi* c
0.001 10–3milli m
0. 000 001 10–6micro µ
0. 000 000 001 10–9nano n
0. 000 000 000 001 10–12pico p
 With rapid development of Information Technology, computers are
playing a major role in analysis, synthesis and design of machines.

*These prefixes are generally becoming obsolete probably due to possible confusion. Moreover, it is becoming
a conventional practice to use only those powers of ten which conform to 103x, where x is a positive or
negative whole number .

Chapter 1 :  Introduction   /G6C  5
1.14. Rules for S.I. Units
The eleventh General Conference of Weights and Measures recommended only the funda-
mental and derived units of S.I. units. But it did not elaborate the rules for the usage of the units. Lateron many scientists and engineers held a number of meetings for the style and usage of S.I. units. Someof the decisions of the meetings are as follows :

1.For numbers having five or more digits, the digits should be placed in groups of three sepa-
rated by spaces * (instead of commas) counting both to the left and right to the decimal point.
2.In a four digit number, ** the space is not required unless the four digit number is used in a
column of numbers with five or more digits.
3.A dash is to be used to separate units that are multiplied together. For example, newton

metre is written as N-m. It should not be confused with mN, which stands for millinewton.
4.Plurals are never used with symbols. For example, metre or metres are written as m.
5.All symbols are written in small letters except the symbols derived from the proper names.
For example, N for newton and W for watt.
6.The units with names of scientists should not start with capital letter when written in full. For
example, 90 newton and not 90 Newton.

At the time of writing this book, the authors sought the advice of various international
authorities, regarding the use of units and their values. Keeping in view the international reputation ofthe authors, as well as international popularity of their books, it was decided to present units *** and
their values as per recommendations of ISO and BIS. It was decided to use :

4500 not 4 500 or 4,50075 890 000 not 75890000 or 7,58,90,0000.012 55 not 0.01255 or .01255
30 × 10
6not 3,00,00,000 or 3 × 107
The above mentioned figures are meant for numerical values only. Now let us discuss about

the units. We know that the fundamental units in S.I. system of units for length, mass and time aremetre, kilogram and second respectively. While expressing these quantities we find it time consum-ing to write the units such as metres, kilograms and seconds, in full, every time we use them. As aresult of this, we find it quite convenient to use some standard abbreviations.
We shall use :
m for metre or metres
km for kilometre or kilometres

kg for kilogram or kilogramst for tonne or tonness for second or secondsmin for minute or minutesN-m for newton × metres ( e.g. work done )
kN-m for kilonewton × metresrev for revolution or revolutions
rad for radian or radians
*In certain countries, comma is still used as the decimal mark.
**In certain countries, a space is used even in a four digit number.

***In some of the question papers of the universities and other examining bodies, standard values are not used.
The authors have tried to avoid such questions in the text of the book. However, at certain places, thequestions with sub-standard values have to be included, keeping in view the merits of the question from thereader’s angle.

6  /G6C   Theory of Machines
1.15. Force
It is an important factor in the field of Engineering science, which may be defined as an agent,
which produces or tends to produce, destroy or tends to destroy motion.
1.16. Resultant Force
If a number of forces P,Q,R etc. are acting simultaneously on a particle, then a single force,
which will produce the same effect as that of all the given forces, is known as a  resultant force . The

forces P,Q,R etc. are called  component forces . The process of finding out the resultant force of the
given component forces, is known as composition of forces .
A resultant force may be found out analytically, graphically or by the following three laws:1. Parallelogram law of forces . It states, “If two forces acting simultaneously on a particle

be represented in magnitude and direction by the two adjacent sides of a parallelogram taken in order,their resultant may be represented in magnitude and direction by the diagonal of the parallelogrampassing through the point.”
2. Triangle law of forces . It states, “If two forces acting simultaneously on a particle be

represented in magnitude and direction by the two sides of a triangle taken in order, their resultantmay be represented in magnitude and direction by the third side of the triangle taken in oppositeorder.”
3. Polygon law of forces . It states, “If a number of forces acting simultaneously on a particle

be represented in magnitude and direction by the sides of a polygon taken in order, their resultant maybe represented in magnitude and direction by the closing side of the polygon taken in opposite order.”
1.17. Scalars and Vectors
1. Scalar quantities are those quantities, which have magnitude only, e.g. mass, time, volume,
density etc.
2. Vector quantities are those quantities which have magnitude as well as direction e.g. velocity,
acceleration, force etc.

3.  Since the vector quantities have both magnitude and direction, therefore, while adding or
subtracting vector quantities, their directions are also taken into account.
1.18. Representation of Vector Quantities
The vector quantities are represented by vectors. A vector is a straight line of a certain length

Chapter 1 :  Introduction   /G6C  7
possessing a starting point and a terminal point at which it carries an arrow head. This vector is cut off
along the vector quantity or drawn parallel to the line of action of the vector quantity, so that thelength of the vector represents the magnitude to some scale. The arrow head of the vector representsthe direction of the vector quantity.
1.19. Addition of Vectors
(a)( b)
Fig. 1.1. Addition of vectors.

(a)( b)
Consider two vector quantities  P and Q, which are required to be added, as shown in Fig.1.1( a).
Take a point A and draw a line AB parallel and equal in magnitude to the vector P. Through B,
draw BC parallel and equal in magnitude to the vector Q. Join AC, which will give the required sum
of the two vectors P and Q, as shown in Fig. 1.1 ( b).
1.20. Subtraction of Vector Quantities

Consider two vector quantities P and Q whose difference is required to be found out as
shown in Fig. 1.2 ( a).
    (a)      ( b)
Fig. 1.2. Subtraction of vectors.
draw BC parallel and equal in magnitude to the vector Q, but in opposite direction.  Join AC, which
gives the required difference of the vectors P and Q, as shown in Fig. 1.2 ( b).

8  /G6C   Theory of Machines
2.1. Introduction
We have discussed in the previous Chapter, that the
subject of Theory of Machines deals with the motion andforces acting on the parts (or links) of a machine. In this chap-ter, we shall first discuss the kinematics of motion i.e. the

relative motion of bodies without consideration of the forcescausing the motion. In other words, kinematics deal with thegeometry of motion and concepts like displacement, velocityand acceleration considered as functions of time.
2.2. Plane Motion
When the motion of a body is confined to only one
plane, the motion is said to be plane motion. The plane mo-
tion may be either rectilinear or curvilinear.
2.3. Rectilinear Motion
It is the simplest type of motion and is along a straight

line path. Such a motion is also known as translatory motion.
2.4. Curvilinear Motion
It is the motion along a curved path. Such a motion,
when confined to one plane, is called plane curvilinear
motion.
When all the particles of a body travel in concentric
circular paths of constant radii (about the axis of rotationperpendicular to the plane of motion) such as a pulley rotating
8Kinematics of
Motion2
Features
1. 1ntroduction.

Motion2
Features
2. Plane Motion.3. Rectilinear Motion.4. Curvilinear Motion.5. Linear Displacement.6. Linear Velocity.7. Linear Acceleration.8. Equations of Linear Motion.9. Graphical Representation of
Displacement with respect toTime.
10. Graphical Representation of
Velocity with respect to Time.
11. Graphical Representation of
Acceleration with respect toTime.
12. Angular Displ acement.
13. Representation of Angular
Displacement by a Vector.

14. Angular Velocity.
15. Angular Acceleration.16. Equations of Angular Motion.17. Relation Between Linear
Motion and Angular Motion.
18. Relation Between Linear and
Angular Quantities ofMotion.
19. Acceleration of a Particle
along a Circular Path.

Chapter 2 : Kinematics of Motion    /G6C  9
about a fixed shaft or a shaft rotating about its
own axis, then the motion is said to be a plane
rotational motion.
Note: The motion of a body, confined to one plane,
may not be either completely rectilinear nor completelyrotational. Such a type of motion is called combinedrectilinear and rotational motion. This motion is dis-
cussed in Chapter 6, Art. 6.1.
2.5. Linear Displacement
It may be defined as the distance moved

by a body with respect to a certain fixed point.The displacement may be along a straight or acurved path. In a reciprocating steam engine, allthe particles on the piston, piston rod and cross-head trace a straight path, whereas all particleson the crank and crank pin trace circular paths,whose centre lies on the axis of the crank shaft. It will be interesting to know, that all the particles onthe connecting rod neither trace a straight path nor a circular one; but trace an oval path, whose

nor a circular one; but trace an oval path, whose radiusof curvature changes from time to time.

The displacement of a body is a vector quantity, as it has both magnitude and direction.
Linear displacement may, therefore, be represented graphically by a straight line.
2.6. Linear Velocity
It may be defined as the rate of
change of linear displacement of a body withrespect to the time. Since velocity is alwaysexpressed in a particular direction, thereforeit is a vector quantity. Mathematically, lin-ear velocity,
v = ds/dt
Notes:  1. If the displacement is along a circular

path, then the direction of linear velocity at any
instant is along the tangent at that point.
2. The speed is the rate of change of linear displacement of a body with respect to the time. Since the
speed is irrespective of its direction, therefore, it is a scalar quantity.
2.7. Linear Acceleration
It may be defined as the rate of change of linear velocity of a body with respect to the time. It
is also a vector quantity. Mathematically, linear acceleration,
2
2dv d ds d sa

2
2dv d ds d sa
dt dt dt dt== = ...dsvdt=/G51
Notes: 1. The linear acceleration may also be expressed as follows:
dv ds dv dvav
dt dt ds ds==×= ×
2. The negative acceleration is also known as deceleration  or retardation.
Spindle
(axis of rotation)
Axis of rotationReference
lineθ
∆θ
θοr

10      /G6C               Theory of Machines
2.8.2.8.2.8.2.8.2.8. Equations of Linear MotionEquations of Linear MotionEquations of Linear MotionEquations of Linear MotionEquations of Linear Motion
The following equations of linear motion are
important from the subject point of view:
1.v = u + a.t 2. s = u.t + 1
2 a.t2
3.v2 = u2 + 2a.s
4.()
2avuvst v t+=× = ×
where u = Initial velocity of the body,
v = Final velocity of the body,
a = Acceleration of the body,

s = Displacement of the body in time t seconds, and
vav = Average velocity of the body during the motion.
Notes:  1.  The above equations apply for uniform
acceleration. If, however, the acceleration is variable,
then it must be expressed as a function of either  t, s
or v and then integrated.
2. In case of vertical motion, the body is
subjected to gravity. Thus  g (acceleration due to grav-
ity) should be substituted for ‘ a’ in the above equa-
tions.

tions.
3. The value of g is taken as + 9.81 m/s2 for
downward motion, and – 9.81 m/s2 for upward mo-
tion of a body.
4. When a body falls freely from a height h,
then its velocity v, with which it will hit the ground is
given by
2. vg h=
2.9.2.9.2.9.2.9.2.9. GraGraGraGraGraphical Reprphical Reprphical Reprphical Reprphical Repr esentaesentaesentaesentaesenta tion oftion oftion oftion oftion of

Displacement with RespectDisplacement with RespectDisplacement with RespectDisplacement with RespectDisplacement with Respect
to to to to to TimeTimeTimeTimeTime
The displacement of a moving body in a given time may be found by means of a graph. Such
a graph is drawn by plotting the displacement as ordinate and the corresponding time as abscissa. We
shall discuss the following two cases :
1. When the body moves with uniform velocity.  When the body moves with uniform velocity,

equal distances are covered in equal intervals of time. By plotting the distances on Y-axis and time on
X-axis, a displacement-time curve ( i.e. s-t curve) is drawn which is a straight line, as shown in Fig. 2.1
(a). The motion of the body is governed by the equation s = u.t, such that
Velocity at instant   1 =  s1/t1
Velocity at instant   2 = s2/t2
Since the velocity is uniform, therefore
3 12
123tans ss
tt t===θ

3 12
123tans ss
tt t===θ
where tan θ is called the slope of s-t curve. In other words, the slope of the s-t curve at any instant
gives the velocity.
t = 0 s
v = 0 m/s
t = 2 s
v = 19.62 m/st = time
v = velocity (downward)
g = 9.81 m/s2 = acceleration
due to gravity
t  = 1 s
v = 9.81 m/s

Chapter 2 : Kinematics of Motion    /G6C  11
2. When the body moves with variable velocity. When the body moves with variable velocity,
unequal distances are covered in equal intervals of time or equal distances are covered in unequal intervalsof time. Thus the displacement-time graph, for such a case, will be a curve, as shown in Fig. 2.1 ( b).
(a) Uniform velocity. ( b) Variable velocity.
Fig. 2.1. Graphical representation of displacement with respect to time.

Consider a point  P on the s-t curve and let this point travels to Q by a small distance δs in a
small interval of time δt. Let the chord joining the points P and Q makes an angle θ with the horizontal.
The average velocity of the moving point during the interval PQ is given by
tan θ = δs/δt . . . (From triangle PQR )
In the limit, when δt approaches to zero, the point Q will tend to approach P and the chord PQ
becomes tangent to the curve at point P. Thus the velocity at  P,

vp = tan θ = ds /dt
where tan θ is the slope of the tangent at P. Thus the slope of the tangent at any instant on the s-t curve
gives the velocity at that instant .
2.10. Graphical Representation of Velocity with Respect to Time
We shall consider the following two cases :
1. When the body moves with uniform velocity.  When the body moves with zero acceleration,
then the body is said to move with a uniformvelocity and the velocity-time curve ( v-t

curve) is represented by a straight line asshown by AB in Fig. 2.2 ( a).
We know that distance covered by a
body in time t second
= Area under the v-t curve AB
= Area of rectangle  OABC
Thus, the distance covered by a
body at any interval of time is given by thearea under the v-t curve.
2. When the body moves with
variable velocity.  When the body moves with

constant acceleration, the body is said to move with variable velocity. In such a case, there is equalvariation of velocity in equal intervals of time and the velocity-time curve will be a straightline AB inclined at an angle θ, as shown in Fig. 2.2 ( b). The equations of motion i.e. v = u + a.t, and
s = u.t +  
1
2a.t2  may be verified from this  v-t curve.

12  /G6C   Theory of Machines
Let u = Initial velocity of a moving body, and
v = Final velocity of a moving body after time t.
Then,Changeinvelocitytan Acceleration ( )TimeBC v uaAC t−θ= = = =
Fig. 2.2. Graphical representation of velocity with respect to time.
Thus, the slope of the v-t curve represents the acceleration of a moving body.
Now tan−=θ = =BC v uaAC torv = u + a.t

Since the distance moved by a body is given by the area under the v-t curve, therefore
distance moved in time ( t),
s = Area OABD = Area OACD + Area ABC
2 11
22.(). .ut v u t ut at=+ − =+ ... (/G51 v – u = a.t )
2.11. Graphical Representation of Acceleration with Respect to Time
(a) Uniform velocity.         ( b) Variable velocity.
Fig. 2.3. Graphical representation of acceleration with respect to time.

1. When the body moves with uniform acceleration.  When the body moves with uniform
acceleration, the acceleration-time curve ( a-t curve) is a straight line, as shown in Fig. 2.3( a). Since
the change in velocity is the product of the acceleration and the time, therefore the area under thea-t curve (i.e. OABC) represents the change in velocity.
2. When the body moves with variable acceleration. When the body moves with variable

acceleration, the a-t curve may have any shape depending upon the values of acceleration at various
instances, as shown in Fig. 2.3( b). Let at any instant of time  t, the acceleration of moving body is a.
Mathematically, a = dv/dtordv = a.dt

Chapter 2 : Kinematics of Motion    /G6C  13
Integrating both sides,
22
11.vt
vtdv adt=∫∫or2
121 . −=∫t
tvv a d t
where v1 and v2 are the velocities of the moving body at time intervals  t1 and t2 respectively.
The right hand side of the above expression represents the area ( PQQ1P1) under the  a-t curve
between the time intervals t1 and t2 . Thus the area under the a-t curve between any two ordinates

represents the change in velocity of the moving body. If the initial and final velocities of the body areu and v, then the above expression may be written as
0.tv u adt−= =∫ Area under a-t curve AB = Area OABC
Example 2.1.  A car starts from rest and
accelerates uniformly to a speed of 72 km. p.h. overa distance of 500 m. Calculate the acceleration andthe time taken to attain the speed.
If a further acceleration raises the speed to

90 km. p.h. in 10 seconds, find this acceleration andthe further distance moved. The brakes are nowapplied to bring the car to rest under uniformretardation in 5 seconds. Find the distance travelledduring braking.
Solution. Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that v
2 = u2 + 2 a.s

2 = u2 + 2 a.s
∴ (20)2 = 0 + 2a × 500 = 1000  aora = (20)2/ 1000 = 0.4 m/s2  Ans.
Time taken by the car to attain the speed
Let t = Time taken by the car to attain the speed.
We know that v = u + a.t
∴ 20 = 0 + 0.4 × tort = 20/0.4 = 50 s  Ans.
Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds.Given : * u = 72 km.p.h. = 20 m/s ; v = 96 km.p.h. = 25 m/s ; t = 10 s

25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m/s
2  Ans.
Distance moved by the car
We know that distance moved by the car,
2211
22. . 20 10 0.5(10) 225msu t a t=+ =×+ × =  Ans.
*It is the final velocity in the first case.

14  /G6C   Theory of Machines
Now consider the motion of the car during the application of brakes for brining it to rest in
5 seconds.
Given : *u = 25 m/s ; v = 0 ; t = 5 s
We know that the distance travelled by the car during braking,
25 05 62.5m22uvst++=× = × =  Ans.
Example 2.2.  The motion of a particle is given by  a = t3 – 3t2 + 5, where a is the acceleration
in m/s2 and t is the time in seconds. The velocity of the particle at t = 1 second is 6.25 m/s, and the

displacement is 8.30 metres. Calculate the displacement and the velocity at t = 2 seconds.
Solution. Given : a = t3 – 3t2 + 5
We know that the acceleration, a = dv/dt. Therefore the above equation may be written as
3235=− +dvttdtor32(35 )=−+dv t t dt
Integrating both sides
43 4
3
1135543 4=− ++=−++tt tvt C t t C ...(i)
where C1 is the first constant of integration. We know that when t = 1 s, v = 6.25 m/s. Therefore
substituting these values of t and v in equation (i),

6.25 = 0.25 – 1 + 5 + C1 = 4.25 + C1orC1 = 2
Now substituting the value of C1 in equation  (i),
34524=−++tvt t ...(ii)
Velocity at t = 2 seconds
Substituting the value of t = 2 s in the above equation,
4
3225 2 2 8 m / s4v=−+ × + =  Ans.
Displacement at t = 2 seconds
We know that the velocity, v = ds/dt, therefore equation (ii) may be written as
44
3352o r 5244ds t tt t ds t t dtdt=−++ = −++ 
54 2
25220 4 2tt tst C=−+ ++ ...(iii)

54 2
where C2 is the second constant of integration. We know that when t = 1 s, s = 8.30 m. Therefore
substituting these values of   t and s in equation (iii),
221158.30 2 4.3
20 4 2CC = −+++ = + orC2 = 4
*It is the final velocity in the second case.

Chapter 2 : Kinematics of Motion    /G6C  15
Substituting the value of C2 in equation (iii),
54 252420 4 2=−++ +tt tst
Substituting the value of t = 2 s, in this equation,
54 2225 22 2 4 15.6m20 4 2×=−+ + × + =s Ans.
Example 2.3.  The velocity of a

train travelling at 100 km/h decreases by10 per cent in the first 40 s after applica-tion of the brakes. Calculate the velocityat the end of a further 80 s assuming that,during the whole period of 120 s,  the re-tardation is proportional to the velocity.
Solution. Given : Velocity in the
beginning ( i.e. when t = 0), v
0 = 100 km/h
Since the velocity decreases by 10

0 = 100 km/h
per cent in the first 40 seconds after theapplication of brakes, therefore velocity at the end of 40 s,
v
40 = 100 × 0.9 = 90 km/h
Let v120 = Velocity at the end of 120 s (or further 80s).
Since the retardation is proportional to the velocity, therefore,
. =− =dvak vdtor .=−dvkd tv
where k is a constant of proportionality, whose value may be determined from the given conditions.
Integrating the above expression,
loge v = – k.t + C ... (i)

where C is the constant of integration. We know that when t = 0, v = 100 km/h. Substituting these
values in equation (i),
loge100 = CorC = 2.3 log 100 = 2.3 × 2 = 4.6
We also know that when  t = 40 s, v = 90 km/h. Substituting these values in equation (i),
loge 90 = – k × 40 + 4.6 ...( /G51 C = 4.6 )
2.3 log 90 = – 40 k + 4.6
or4.6 2.3log90 4.6 2.3 1.95420.002640 40−− ×== =k
Substituting the values of k and C in equation (i),
loge v = – 0.0026 × t + 4.6

or 2.3 log v = – 0.0026 × t + 4.6 ... (ii)
Now substituting the value of t equal to 120 s, in the above equation,
2.3 log v120 = – 0.0026 × 120 + 4.6 = 4.288
or log v120 = 4.288/2.3 = 1.864
∴ v120 = 73.1 km/h  Ans. ... (Taking antilog of 1.864)

16  /G6C   Theory of Machines
Example 2.4.  The acceleration (a) of a slider block and its displacement (s) are related by
the expression, =aks, where k is a constant. The velocity v is in the direction of the displacement
and the velocity and displacement are both zero when time t is zero. Calculate the displacement,
velocity and acceleration as functions of time.
Solution. Given : =aks
We know that acceleration,
dvavds=×    or  dvks vds=× ...dv ds dv dvvdt dt ds ds=×= ×/G51

∴ v × dv = k.s1/2 ds
1/2
0.=∫∫vvdv k s ds or23 / 2
1.
23 / 2=+vk sC ... (i)
where C1 is the first constant of integration whose value is to be determined from the given conditions
of motion. We know that s = 0, when v = 0. Therefore, substituting the values of s and v in equation  (i),
we get C1 = 0.
∴2
3/22.23=vks or3/44
3=×kvs ... (ii)
Displacement, velocity and acceleration as functions of time
We know that3/44
3== ×ds kvsdt... [From equation (ii)]
∴  3/44
3ds kdt

∴  3/44
3ds kdt
s= or3/4 4
3−=ksd s d t
3/4
004
3−=∫∫st ksd s d t
1/4
24
1/4 3=× +sktC ...(iii)
where C2 is the second constant of integration. We know that displacement, s = 0 when t = 0. There-
fore, substituting the values of s and t in equation  (iii), we get C2 = 0.
∴1/44
1/4 3=×skt or 24.
144=kts  Ans.
We know that velocity,
22 3
3.4144 36== × =ds k k tvtdt Ans.24.... Differentiatin g144kt 

and acceleration,22 2


2.336 12==× =dv k k tatdt Ans.23.... Differentiatin g36kt 


Chapter 2 : Kinematics of Motion    /G6C  17
Example 2.5. The cutting stroke of a planing
machine is 500 mm and it is completed in 1 second.The planing table accelerates uniformly during the first125 mm of the stroke, the speed remains constant duringthe next 250 mm of the stroke and retards uniformly duringthe last 125 mm of the stroke. Find the maximum cuttingspeed.
Solution.  Given : s = 500 mm ; t = 1 s ;
s
1 = 125 mm ; s2 = 250 mm ; s3 = 125 mm

s
Fig. 2.4 shows the acceleration-time and veloc-
ity-time graph for the planing table of a planing machine.
Letv = Maximum cutting speed in mm/s.
Average velocity of the table during acceleration
and retardation,
(0 )/2 /2=+ =avvv v
Time of uniform acceleration 1
1125 250s/2avstvv v== =
Time of constant speed,2
2250sstvv==
and time of uniform retardation,3
3125 250s/2avstvv v== =
Fig. 2.4
Since the time taken to complete the stroke is 1 s, therefore

123++=tttt
250 250 2501 ++=vvv or v = 750 mm/s Ans.
2.12. Angular Displacement
It may be defined as the angle described by a particle from one point to another, with respect
to the time. For example, let a line OB has its inclination θ radians to the fixed line OA, as shown in
Planing Machine.

18  /G6C   Theory of Machines
Fig. 2.5. If this line moves from  OB to OC, through an angle δθ during
a short interval of time δt, then δθ is known as the angular
displacement of the line OB.
Since the angular displacement has both magnitude and
direction, therefore it is also a vector quantity.
2.13. Representation of Angular Displacement by
a Vector
In order to completely represent an angular displacement, by a vector, it must fix the follow-
ing three conditions :

1. Direction of the axis of rotation . It is fixed by drawing a line perpendicular to the plane
of rotation, in which the angular displacement takes place. In other words, it is fixed along the axisof rotation.
2. Magnitude of angular displacement . It is fixed by the length of the vector drawn along
the axis of rotation, to some suitable scale.
3. Sense of the angular displacement.  It is fixed by a right hand screw rule. This rule

states that if a screw rotates in a fixed nut in a clockwise direction,  i.e. if the angular displacement
is clockwise and an observer is looking along the axis of rotation, then the arrow head will pointaway from the observer. Similarly, if the angular displacement is anti-clockwise, then the arrowhead will point towards the observer.
2.14. Angular Velocity
It may be defined as the rate of change of angular displacement with respect to time. It is

usually expressed by a Greek letter ω (omega). Mathematically, angular velocity,
/ω= θdd t
Since it has magnitude and direction, therefore, it is a vector quantity. It may be represented
by a vector following the same rule as described in the previous article.
Note : If the direction of the angular displacement is constant, then the rate of change of magnitude of the
angular displacement with respect to time is termed as  angular speed .
2.15. Angular Acceleration

It may be defined as the rate of change of angular velocity with respect to time. It is usually
expressed by a Greek letter α (alpha). Mathematically, angular acceleration,
2
2ωθ θα= = = dd dd
dt dt dt dt...d
dtθ ω=/G51
It is also a vector quantity, but its direction may not be same as that of angular displacement
and angular velocity.
2.16. Equations of Angular Motion
The following equations of angular motion corresponding to linear motion are important

from the subject point of view :
1.0. ω=ω +α t 2. 2
01..2θ=ω + αtt
3.()2 2
02. ω=ω +α θ 4. ()0
2ω+ ωθ=t
where ω0 = Initial angular velocity in rad/s,
ω = Final angular velocity in rad/s,Fig. 2.5. Angular
displacement.

Chapter 2 : Kinematics of Motion    /G6C  19
Fig. 2.6. Motion of a body
along a circular path.t = Time in seconds,
θ = Angular displacement in time  t seconds, and
α = Angular acceleration in rad /s2.
Note : If a body is rotating at the rate of N r.p.m. (revolutions per minute), then its angular velocity,
ω = 2πΝ/ 60 rad/s
2.17. Relation between Linear Motion and Angular Motion
Following are the relations between the linear motion and the angular motion :
Particulars Linear motion Angular motion

Initial velocity u ω0
Final velocity v ω
Constant acceleration a α
Total distance traversed s θ
Formula for final velocity v = u + a.t ω = ω0 + α.t
Formula for distance traversed s = u.t + 1
2a.t2θ = ω0. t + 1
2 α.t2
Formula for final velocity v2 = u2 + 2 a.s ω = (ω0) 2  + 2 α.θ
2.18. Relation between Linear and Angular Quantities of Motion
Consider a body moving along a circular path from A to B as shown in Fig. 2.6.
Let r = Radius of the circular path,

θ = Angular displacement in radians,
s = Linear displacement,
v = Linear velocity,
ω = Angular velocity,
a = Linear acceleration, and
α = Angular acceleration.
From the geometry of the figure, we know that
 s = r . θ
We also know that the linear velocity,
(.).θθ= = =× =ωds d r dvr rdt dt dt
... (i)
and linear acceleration,(. ).ωω== = ×= αdv d r dar rdt dt dt... (ii)

Example 2.6.   A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its
angular acceleration? How many revolutions does the wheel make in attaining the speed of 2000 r.p.m.?
Solution. Given : N0 = 0  or  ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
Angular acceleration
Let α = Angular acceleration in rad/s2.
We know that
ω = ω0 +  α.tor 209.5 = 0 +  α × 20
∴α  =  209.5/20 = 10.475 rad/s2  Ans.

20  /G6C   Theory of Machines
Number of revolutions made by the wheel
We know that the angular distance moved by the wheel during 2000 r.p.m. ( i.e. when
ω = 209.5 rad/s),
()() 0 0 209.5 20209522ω+ ω +θ= = =trad
Since the angular distance moved by the wheel during one revolution is 2 π radians, therefore
number of revolutions made by the wheel,
n = θ/2π = 2095/2 π = 333.4  Ans.
2.19. Acceleration of a Particle along a Circular Path

Consider A and B, the two positions of a particle displaced through an angle δθ in time δt as
shown in Fig. 2.7 ( a).
Let r = Radius of curvature of the circular path,
v = Velocity of the particle at A, and
v + dv = Velocity of the particle at B.
The change of velocity, as the particle moves from A to B may be obtained by drawing the
vector triangle oab, as shown in Fig. 2.7 ( b). In this triangle, oa represents the velocity v and ob

represents the velocity v + dv. The change of velocity in time  δt is represented by ab.
Fig. 2.7. Acceleration of a particle along a circular path.
Now, resolving ab into two components i.e. parallel and perpendicular to oa. Let ac and cb
be the components parallel and  perpendicular to oa respectively.
∴ ac = oc – oa = ob cos δθ  – oa = (v + δv) cos δθ – v
and cb = ob sin δθ = (v + δv) sin δθ
Since the change of velocity of a particle (represented by vector ab) has two mutually

perpendicular components, therefore the acceleration of a particle moving along a circular path hasthe following two components of the acceleration which are perpendicular to each other.
1. Tangential component of the acceleration . The acceleration of a particle at any instant
moving along a circular path in a direction tangential to that instant, is known as tangential componentof acceleration or tangential acceleration.

∴  Tangential component of the acceleration of particle at A or tangential acceleration at  A,
() c o s+δ δθ−==δδtac v v vatt
In the limit, when δt  approaches to zero, then
/.== αtad v d t r ... (i)
2. Normal component of the acceleration . The acceleration of a particle at any instant mov-
ing along a circular path in a direction normal to the tangent at that instant and directed towards thecentre of the circular path ( i.e. in the direction from A to O) is known  as normal component of the

Chapter 2 : Kinematics of Motion    /G6C  21
acceleration or normal acceleration. It is also called radial or centripetal  acceleration.
∴ Normal component of the acceleration of the particle at A or normal (or radial or centrip-
etal) acceleration at A,
()sin+δ θ==δδnvvcbatt
In the limit, when δt approaches to zero, then
2
2..θ=× =ω =×= = ωndv vav v v rdt r r... (ii)
                        []... / , and /dd t v rθ= ω ω =/G51
Since the tangential acceleration ( at) and the normal accelera-

tion (an) of the particle at any instant A are perpendicular to each other,
as shown in Fig. 2.8, therefore total acceleration of the particle ( a) is
equal to the resultant acceleration of at and an.
∴ Total acceleration or resultant acceleration,
()()22=+tnaa a
and its angle of inclination with the tangential acceleration is given by
tan θ = an/at or  θ = tan–1 (an/at)
The total acceleration or resultant acceleration may also be obtained by the vector sum of at
and an.

and an.
Notes : 1. From equations (i) and (ii) we see that the tangential acceleration ( at ) is equal to the rate of change of
the magnitude of the velocity whereas the normal or radial or centripetal acceleration ( an) depends upon its
instantaneous velocity and the radius of curvature of its path.
2. When a particle moves along a straight path, then the radius of curvature is infinitely great. This

means that v2/r is zero. In other words, there will be no normal or radial or centripetal acceleration. Therefore,
the particle has only tangential acceleration  (in the same direction as its velocity and displacement) whose valueis given by
a
t = dv/dt = α.r
3. When a particle moves with a uniform velocity, then dv/dt will be zero. In other words, there will be
no tangential acceleration; but the particle will have only normal or radial or centripetal acceleration, whose
value is given by

an = v2/r  = v.ω = ω2 r
Example 2.7.  A horizontal bar 1.5 metres long and of small cross-section rotates about
vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an intervalof 5 seconds. What is the linear velocity at the beginning and end of the interval ? What are thenormal and tangential components of the acceleration of the mid-point of the bar after 5 secondsafter the acceleration begins ?
Solution. Given : r = 1.5 m ; N

0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ;  t = 5 s
Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r.ω0 = 1.5 × 125.7 = 188.6  m/s Ans.
Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r.ω = 1.5 × 157 = 235.5 m/s Ans.Fig. 2.8. Total acceleration
      of a particle.

22  /G6C   Theory of Machines
Tangential acceleration after 5 seconds
Let α = Constant angular acceleration.
We know that ω = ω0+ α.t
 157 = 125.7 + α × 5 or α = (157 – 125.7)/5 = 6.26 rad/s2
Radius corresponding to the middle point,
r = 1.5/2 = 0.75 m
∴ Tangential acceleration  =  α. r = 6.26 × 0.75 = 4.7 m/s2  Ans.
Radial acceleration after 5 seconds
     Radial acceleration  = ω2 . r = (157)2  0.75 = 18 487 m/s2 Ans.
EXERCISES

EXERCISES
1.A winding drum raises a cage through a height of 120 m. The cage has, at first, an acceleration
of 1.5 m/s2 until the velocity of 9 m/s is reached, after which the velocity is constant until the
cage nears the top, when the final retardation is 6 m/s2. Find the time taken for the cage to reach
the top. [ Ans. 17.1s ]
2.The displacement of a point is given by s = 2t3 + t2 + 6, where s is in metres and t in seconds.

Determine the displacement of the point when the velocity changes from 8.4 m/s to 18 m/s. Find alsothe acceleration at the instant when the velocity of the particle is 30 m/s. [ Ans. 6.95 m ; 27 m/s
2 ]
3.A rotating cam operates a follower which moves in a straight line. The stroke of the follower is 20
mm and takes place in 0.01 second from rest to rest. The motion is made up of uniform acceleration

for 1/4 of the time, uniform velocity for 12 of the time followed by uniform retardation. Find the
maximum velocity reached and the value of acceleration and retardation.
[ Ans. 2.67 m/s ; 1068 m/s2 ; 1068 m/s2 ]
4.A cage descends a mine shaft with an acceleration of 0.5 m/s2. After the cage has travelled 25 metres,

a stone is dropped from the top of the shaft. Determine : 1. the time taken by the stone to hit the cage,and 2. distance travelled by the cage before impact. [ Ans. 2.92 s ; 41.73 m ]
5.The angular displacement of a body is a function of time and is given by equation :
θ = 10 + 3 t + 6 t
2, where t is in seconds.
Determine the angular velocity, displacement and acceleration when t = 5 seconds. State whether or

not it is a case of uniform angular acceleration. [Ans. 63 rad/s ; 175 rad ; 12 rad/s2]
6.A flywheel is making 180 r.p.m. and after 20 seconds it is running at 140 r.p.m. How many revolu-
tions will it make, and what time will elapse before it stops, if the retardation is uniform ?
[ Ans. 135 rev. ; 90 s ]
7.A locomotive is running at a constant speed of 100 km/h. The diameter of driving wheels is 1.8 m. The

stroke of the piston of the steam engine cylinder of the locomotive is 600 mm. Find the centrip-
etal acceleration of the crank pin relative to the engine frame. [ Ans. 288 m/s2 ]
DO YOU KNOW ?
1.Distinguish clearly between speed and velocity. Give examples.
2.What do you understand by the term ‘acceleration’ ? Define positive acceleration and negative accel-
eration.
3.Define ‘angular velocity’ and ‘angular acceleration’. Do they have any relation between them ?

4.How would you find out the linear velocity of a rotating body ?
5.Why the centripetal acceleration is zero, when a particle moves along a straight path ?
6. A particle moving with a uniform velocity has no tangential acceleration. Explain clearly.

Chapter 2 : Kinematics of Motion    /G6C  23
OBJECTIVE TYPE QUESTIONS
1.The unit of linear acceleration is
(a) kg-m ( b)m / s ( c)m / s2(d) rad/s2
2.The angular velocity (in rad/s) of a body rotating at N r.p.m. is
(a)π N/60 ( b)2  π N/60 ( c)π N/120 ( d)π N/180
3.The linear velocity of a body rotating at ω rad/s along a circular path of radius r is given by
(a)ω.r (b)ω/r (c) ω2.r (d)ω2/r
4.When a particle moves along a straight path, then the particle has

(a) tangential acceleration only ( b) centripetal acceleration only
(c) both tangential and centripetal acceleration
5.When a particle moves with a uniform velocity along a circular path, then the particle has
ANSWERS
1.(c) 2.(b) 3.(a) 4.(a) 5.(b)

24  /circle6/circle6/circle6/circle6/circle6   Theory of Machines
24Kinetics of
Motion3
Features
1. Introduction.
2. Newton's Laws of Motion.3. Mass and Weight.4. Momentum.5. Force.6. Absolute and Gravitational
Units of Force.
7. Moment of a Force.8. Couple.9. Centripetal and Centrifugal
Force.
10. Mass Moment of Inertia.
11. Angular Momentum or
Moment of Momentum.
12. Torque.13. Work.14. Power.15. Energy.16. Principle of Conservation of
Energy.

Energy.
17. Impulse and Impulsive Force.18. Principle of Conservation of
Momentum.
19. Energy Lost by Friction
Clutch During Engagement.
20. Torque Required to
Accelerate a Geared System.
21. Collision of Two Bodies.22. Collision of Inelastic Bodies.23. Collision of Elastic Bodies.
24. Loss of Kinetic Energy
During Elastic Impact.3.1. Introduction
In the previous chapter we have discussed the
kinematics of motion, i.e. the motion without considering

the forces causing the motion. Here we shall discuss the
kinetics of motion, i.e. the motion which takes into
consideration the forces or other factors, e.g. mass or weight
of the bodies. The force and motion is governed by the threelaws of motion.
3.2. Newton’s Laws of Motion
Newton has formulated three laws of motion, which

are the basic postulates or assumptions on which the wholesystem of kinetics is based. Like other scientific laws, theseare also justified as the results, so obtained, agree with theactual observations. These three laws of motion are asfollows:
1. Newton’s First Law of Motion . It states,  “Every
body continues in its state of rest or of uniform motion ina straight line, unless acted upon by some external force .”

This is also known as Law of Inertia.
The inertia is that property of a matter, by virtue of
which a body cannot move of itself, nor change the motionimparted to it.

Chapter 3 : Kinetics of Motion    /circle6/circle6/circle6/circle6/circle6  25
2. Newton ’s Second Law of Motion . It states,
“The rate of change of momentum is directly
proportional to the impressed force and takes place in
the same direction in which the force acts .”
3. Newton ’s Third Law of Motion . It states, “To
every action, there is always an equal and oppositereaction. ”
3.3. Mass and Weight
Sometimes much confu-sion and misunder-

standing is created, while using the various systems of units
in the measurements of force and mass. This happensbecause of the lack of clear understanding of thedifference between the mass and the weight. Thefollowing definitions of mass and weight should beclearly understood :
1. Mass . It is the amount of matter contained in a

given body, and does not vary with the change in itsposition on the earth's surface. The mass of a body ismeasured by direct comparison with a standard mass by using a lever balance.
2. Weight . It is the amount of pull, which the earth exerts upon a given body. Since the pull

varies with distance of the body from the centre of the earth, therefore the weight of the body willvary with its position on the earth’s surface (say latitude and elevation). It is thus obvious, that theweight is a force.
The earth’s pull in metric units at sea

level and 45° latitude has been adopted as oneforce unit and named as one kilogram of force.Thus, it is a definite amount of force. But, unfor-tunately, it has the same name as the unit of mass.The weight of a body is measured by the use of aspring balance which indicates the varying ten-sion in the spring as the body is moved from placeto place.
Note:  The confusion in the units of mass and weight

is eliminated, to a great extent, in S.I. units. In this system, the mass is taken in kg and force in newtons.
The relation between the mass ( m) and the weight ( W) of a body is
W = m.g or m = W/g
where W is in newtons, m is in kg and g is acceleration due to gravity.
3.4. Momentum
It is the total motion possessed by a body. Mathematically,
Momentum = Mass × V elocity
Let m = Mass of the body,
u = Initial velocity of the body,
a = Constant acceleration, and

t = Time required (in seconds) to change the velocity from u to v.
The above picture shows space shuttle.
All space vehicles move based onNewton’s third laws.

26  /circle6/circle6/circle6/circle6/circle6   Theory of Machines
Now,      initial momentum = m.u
and  final momentum = m.v
∴    Change of momentum = m.v – m.u
and rate of change of momentum    =  .. ( ).mv mu m v uma
tt−−== ...vuat−= ∵
3.5. Force
It is an important factor in the field of Engineering-science, which may be defined as an
agent, which produces or tends to produce, destroy or tends to destroy motion.

According to Newton’s Second Law of Motion, the applied force or impressed force is
directly proportional to the rate of change of momentum. We have discussed in Art. 3.4, that the rate
of change of momentum
= m.a
where m = Mass of the body, and
a = Acceleration of the body.
∴           Force ,  F ∝ m.a   or  F = k.m.a
where k is a constant of proportionality.
For the sake of convenience, the unit of force adopted is such that it produces a unit
acceleration to a body of unit mass.

∴ F = m.a  = Mass × Acceleration
In S.I. system of units, the unit of force is called newton (briefly written as N). A newton
may be defined as the force while acting upon a mass of one kg produces an acceleration of
1 m/s2 in the direction of which it acts . Thus
         1 N = 1 kg × 1 m/s2 = 1 kg-m/s2
Note: A force equal in magnitude but opposite in direction and collinear with the impressed force producing

the acceleration, is known as inertia force . Mathematically,
Inertia force = –  m.a
3.6. Absolute and Gravitational Units of Force
We have already discussed, that when a body of mass 1 kg is moving with an acceleration of
1 m/s2, the force acting on the body is one newton (briefly written as N). Therefore, when the same
body is moving with an acceleration of 9.81 m/s2, the force acting on the body is 9.81 newtons. But

we denote 1 kg mass, attracted towards the earth with an acceleration of 9.81 m/s2 as 1 kilogram-
force (briefly written as kgf) or 1 kilogram-weight (briefly written as kg-wt). It is thus obvious that
 1 kgf = 1 kg × 9.81 m/s2 = 9.81 kg-m/s2 = 9.81 N  ... (∵ 1 N = 1 kg-m/s2)
The above unit of force i.e. kilogram-force (kgf ) is called  gravitational  or engineer's unit
applied force,  FW, weight (force)
f, friction force
N, normal force

Chapter 3 : Kinetics of Motion    /G6C  27
of force, whereas newton is the absolute or scientific or S.I. unit of force . It is thus obvious, that the
gravitational units are ‘ g’  times the unit of force in the absolute or S.I. units.
It will be interesting to know that the mass of a body in absolute units is numerically equal
to the weight of the same body in gravitational units .
For example, consider a body whose mass, m = 100 kg.

∴ The force, with which it will be attracted towards the centre of the earth,
             F = m.a = m.g = 100 × 9.81 = 981 N
Now, as per definition, we know that the weight of a body is the force, by which it is attracted
towards the centre of the earth. Therefore, weight of the body,
W = 981 N = 981 /9.81 = 100 kgf ... (∵ 1 kgf = 9.81 N)
In brief, the weight of a body of mass m kg at a place where gravitational acceleration is ‘ g’
m/s2 is m.g newtons.
3.7. Moment of a Force

It is the turning effect produced by a force, on the body, on which it acts. The moment of a
force is equal to the product of the force and the perpendicular distance of the point about which themoment is required, and the line of action of the force. Mathematically,
   Moment of a force = F × l
where F  = Force acting on the body, and
l   = Perpendicular distance of the
point and the line of action ofthe force, as shown in Fig. 3.1.
3.8. Couple

The two equal and opposite parallel forces, whose lines of
action are different, form a couple, as shown in Fig. 3.2.
The perpendicular distance ( x) between the lines of action of
two equal and opposite parallel forces ( F) is known as arm of the
couple. The magnitude of the couple ( i.e. moment of a couple) is
the product of one of the forces and the arm of the couple.Mathematically,
Moment of a couple  = F × x

A little consideration will show, that a couple does not produce any translatory motion ( i.e.
motion in a straight line). But, a couple produces a motion of rota-tion of the body, on which it acts.
3.9. Centripetal and Centrifugal Force
Consider a particle of mass m moving with a linear velocity
v in a circular path of radius r.
We have seen in Art. 2.19 that the centripetal acceleration,
  ac = v2/r = ω2.r
and     Force = Mass × Acceleration

∴Centripetal force = Mass × Centripetal acceleration
or                Fc = m.v2/r = m.ω2.rFig. 3.2. Couple.Fig. 3.1. Moment of a force.

28  /G6C   Theory of Machines
This force acts radially inwards and is essential for circular motion.
We have discussed above that the centripetal force acts radially inwards. According to
Newton's Third Law of Motion, action and reaction are equal and opposite. Therefore, the particlemust exert a force radially outwards of equal magnitude. This force is known as centrifugal force
whose magnitude is given by
                     F
c= m.v2/r = m.ω2r
3.10. Mass Moment of Inertia

It has been established since long that a rigid body is
composed of small particles. If the mass of every particle of abody is multiplied by the square of its perpendicular distancefrom a fixed line, then the sum of these quantities(for the wholebody) is known as mass moment of inertia  of the body. It is
denoted by I.
Consider a body of total mass m. Let it is composed of
small particles of masses m
1, m2, m3, m4 etc. If k1, k2, k3, k4 are

the distances of these masses from a fixed line, as shown in Fig.3.3, then the mass moment of inertia of the whole body is givenby
I = m
1 (k1)2 + m2(k2)2 + m3 (k3)2 + m4 (k4)2 +....
If the total mass of body may be assumed to concentrate at one point (known as centre of
mass or centre of gravity), at a distance k from the given axis, such that
    m.k2 = m1(k1)2 + m2(k2)2 + m3(k3)2 + m4 (k4)2 +...
then       I = m.k2

The distance k is called the radius of gyration . It may be defined as the distance, from a
given reference, where the whole mass of body is assumed to be concentrated to give the samevalue of I .
The unit of mass moment of inertia in S.I. units is kg-m
2.
Notes : 1.  If the moment of inertia of a body about an axis through its centre of gravity is known, then the
moment of inertia about any other parallel axis may be obtained by using a parallel axis theorem i.e. moment

of inertia about a parallel axis,
       Ip = IG + m.h2
where        IG = Moment of inertia of a body about an axis through its centre of gravity, and
        h = Distance between two parallel axes.
2. The following are the values of I for simple cases :
(a) The moment of inertia of a thin disc of radius r, about an axis through its centre of gravity and
perpendicular to the plane of the disc is
  I = m.r2/ 2
and moment of inertia about a diameter,
  I = m.r2/4

I = m.r2/4
(b) The moment of inertia of a thin rod of length l, about an axis through its centre of gravity and
perpendicular to its length,
IG = m.l2/12
and moment of inertia about a parallel axis through one end of a rod,
 Ip = m.l2/3Fig. 3.3. Mass moment of inertia.

Chapter 3 : Kinetics of Motion    /G6C  29
3. The moment of inertia of a solid cylinder of radius r and length l, about the longitudinal axis or polar
axis
= m.r2/2
and moment of inertia through its centre perpendicular to longitudinal axis
22
41 2rl=+
3.11. Angular Momentum or Moment of Momentum
Consider a body of total mass m rotating with an angular velocity
of ω rad/s, about the fixed axis O as shown in Fig. 3.4. Since the body

is composed of numerous small particles, therefore let us take one ofthese small particles having a mass dm and at a distance r from the axis
of rotation. Let v is its linear velocity acting tangentially at any instant.
We know that momentum is the product of mass and velocity, thereforemomentum of mass dm
 = dm × v = dm × ω × r
... (/G51 v = ω.r)
and moment of momentum of mass dm about O
= dm × ω × r × r = dm × r2 × ω = Im × ω

where       Im = Mass moment of inertia of mass dm about O = dm × r2
∴ Moment of momentum or angular momentum of the whole body about O
   ..mII=ω = ω∫
where    mI=∫ Mass moment of inertia of the
whole body about O.
Thus we see that the angular momentum or the moment of momentum is the product of mass
moment of inertia ( I ) and the angular velocity ( ω) of the body.
3.12. Torque
It may be defined as the product of

force and the perpendicular distance of its lineof action from the given point or axis. A littleconsideration will show that the torque isequivalent to a couple acting upon a body.
The Newton’s Second Law of
Motion, when applied to rotating bodies,states that the torque is directly proportional
to the rate of change of angular momentum .
Mathematically, Torque,
d /GC9Tdt(. ω )∝
Since I is constant, therefore
.dTI I
dtω=× =α ...d

.dTI I
dtω=α/G51Double length
spannerSame
force
appliedDouble
torqueTorqueFig. 3.4. Angular
momentum.

30  /G6C   Theory of Machines
The unit of torque ( T) in S.I. units is N-m when I is in kg-m2 and α in rad/s2.
3.13. Work
Whenever a force acts on a body and the body undergoes a displacement in the direction of the
force, then work is said to be done. For example, if a force F acting on a body causes a displacement x
of the body in the direction of the force, then
 Work done = Force × Displacement = F × x
If the force varies linearly from zero to a maximum value of F, then
Work done 01

22FxF x+=× = × ×
When a couple or torque ( T ) acting on a body causes the angular displacement ( θ) about an
axis perpendicular to the plane of the couple, then
Work done = Torque × Angular displacement = T.θ
The unit of work depends upon the unit of force and displacement.
In S.I. system of units, the practical unit of work is N-m. It is the work done by a force of 1

newton, when it displaces a body through 1 metre. The work of 1 N-m is known as joule (brieflywritten as J ) such that 1 N-m = 1 J.
Note: While writing the unit of work, it is general practice to put the unit of force first followed by the unit of
displacement ( e.g. N-m).
3.14. Power
It may be defined as the rate of doing work or work done per unit time. Mathematically,
 WorkdonePower = 
Time taken

WorkdonePower = 
Time taken
In S.I. system of units, the unit of power is watt (briefly written as W) which is equal to 1 J/s
or 1 N-m/s. Thus, the power developed by a force of F (in newtons) moving with a velocity v m/s is
F.v watt. Generally a bigger unit of power called kilowatt (briefly written as kW) is used which is
equal to 1000 W.
Notes: 1. If T is the torque transmitted in N-m or J and ω is the angular speed in rad/s, then

Power, P = T.ω = T × 2 π N/60 watts ... ( ∵ ω = 2 π N/60)
where N is the speed in r.p.m.
2. The ratio of power output to power input is known as efficiency of a machine. It is always less than
unity and is represented as percentage. It is denoted by a Greek letter eta ( η). Mathematically,
         Power outputEfficiency, =Powerinputη
3.15. Energy
It may be defined as the capacity to do work. The energy exists in many forms e.g. mechanical,

electrical, chemical, heat, light etc. But we are mainly concerned with mechanical energy.
The mechanical energy is equal to the work done on a body in altering either its position or
its velocity. The following three types of mechanical energies are important from the subject pointof view.

Chapter 3 : Kinetics of Motion    /G6C  31
*     We know that,   v2 – u2 = 2 a.s
Since u = 0 because the body starts from rest, therefore,
v2 = 2 a.s   or   s = v2/2a*1. Potential energy . It is the energy possessed by a body for doing work, by virtue of its
position. For example, a body raised to some height above the ground level possesses potentialenergy because it can do some work by falling on earth’s surface.
Let W = Weight of the body,
m = Mass of the body, and

h = Distance through which the body falls.
Then potential energy,
P.E. = W.h = m.g.h
... (∵ W = m.g)
It may be noted that
(a) When W is in newtons and h in metres, then potential energy will be in N-m.
(b) When m is in kg and  h in metres, then the potential energy will also be in N-m as
discussed below :
We know that potential energy,
2mP.E. . . kg m N m
smgh − == × × =21kg–m1N
s=/G51
2. Strain energy . It is the potential energy

stored by an elastic body when deformed. A com-pressed spring possesses this type of energy, be-cause it can do some work in recovering its originalshape. Thus if a compressed spring of stiffness s
newton per unit deformation ( i.e. extension or com-
pression) is deformed through a distance  x by a load
W, then
      
1
2Strainenergy = Workdone . Wx=
                21
2.sx= ...(∵ W = s × x)

21
In case of a torsional spring of stiffness q N-m per unit angular deformation when twisted
through as angle θ  radians, then
   Strain energy = Work done 21.2q=θ
3. Kinetic energy . It is the energy possessed by a body, for doing work, by virtue of its mass
and velocity of motion. If a body of mass m attains a velocity v from rest in time t, under the
influence of a force F and moves a distance s, then
Work done = F.s = m.a.s ... (∵ F = m.a)

∴ Kinetic energy of the body or the kinetic energy of translation,
          K.E. = m.a.s = m × a ×2
21.22vmva=

32  /G6C   Theory of Machines
It may be noted that when m is in kg and v in m/s, then kinetic energy will be in N-m as
discussed below:
We know that kinetic energy,
2
2
221m k g - mK.E.= = kg× = × m = N-m2 ssm.v21kg-m... 1N
Notes : 1. When a body of mass moment of inertia  I (about a given axis) is rotated about that axis, with an
angular velocity ω, then it possesses some kinetic energy. In this case,
  Kinetic energy of rotation = 21.2Iω

Kinetic energy of rotation = 21.2Iω
2. When a body has both linear and angular motions e.g. in the locomotive driving wheels and
wheels of a moving car, then the total kinetic energy of the body is equal to the sum of kinetic energies of
translation and rotation.
∴Total kinetic energy = 2211..22mv I+ω
Example 3.1.  The flywheel of a steam engine has a radius

of gyration of 1 m and mass 2500 kg. The starting torque of thesteam engine is 1500 N-m and may be assumed constant.Determine: 1. Angular acceleration of the flywheel, and 2. Kinetic
energy of the flyw heel after 10 seconds from the start.
Solution. Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m
1.  Angular acceleration of the flywheel
Let α = Angular acceleration of the flywheel.
We know that mass moment of inertia of the flywheel,
22 2. 2500 1 2500 kg-mIm k==× =
We also know that torque ( T ),

1500 . 2500I=α = × α or21500/2500 0.6 rad/sα= =  Ans.
2.  Kinetic energy of the flywheel after 10 seconds from start
First of all, let us find the angular speed of the flywheel ( ω2 ) after t = 10 seconds from the
start (i.e. ω1 = 0 ).
We know that    ω2 = ω1 + α.t = 0 + 0.6 × 10 = 6 rad/s
∴ Kinetic energy of the flywheel,
22
211( ) 2500 6 45 000J 45 kJ22EI=ω = × × = =  Ans.

Example 3.2.   A winding drum raises a cage of mass 500 kg through a height of 100 metres.
The mass of the winding drum is 250 kg and has an effective radius of 0.5 m and radius of gyrationis 0.35 m. The mass of the rope is 3 kg/m.
The cage has, at first, an acceleration of 1.5 m/s
2 until a velocity of 10 m/s is reached, after
which the velocity is constant until the cage nears the top and the final retardation is 6 m/s2. Find

1. The time taken for the cage to reach the top, 2. The torque which must be applied to the drum at
starting; and  3. The power at the end of acceleration period.
Solution. Given : mC = 500 kg ; s = 100 m ; mD = 250 kg ; r = 0.5 m ; k = 0.35 m,
m = 3 kg/m
Flywheel

Chapter 3 : Kinetics of Motion    /G6C  33
Fig. 3.5
Fig. 3.5 shows the acceleration-time and velocity-time graph for the cage.
1. Time taken for the cage to reach the top
Let t = Time taken for the cage to reach the top = t1 + t2 + t3
where t1 = Time taken for the cage from initial velocity of u1 = 0 to final
       velocity of  v1 = 10 m/s with an acceleration of a1 = 1.5 m/s2,
t2 = Time taken for the cage during constant velocity of v2 = 10 m/s until the
       cage nears the top, andt

cage nears the top, andt
3 = Time taken for the cage from initial velocity of u3 = 10 m/s to final velocity
         of v3 = 0 with a retardation of a3 = 6 m/s2.
We know that   v1 = u1 + a1.t1
10 = 0 + 1.5 t1      or    t1 = 10/1.5 = 6.67 s
and distance moved by the cage during time t1,
11
1110 06.67 33.3522vust++=× =× = m
Similarly, v3 = u3 + a3.t3
 0 = 10 – 6 × t3   or   t3 = 10/6 = 1.67 s
and33
3301 01.67 8.35m22vust++=× = × =

and33
Now, distance travelled during constant velocity of v2 = 10 m/s,
21 3 100 33.35 8.35 58.3m ss ss=− − = − − =
We know that s2 = v2.t2     or    t2 = s2/v2 = 58.3/10 = 5.83 s
∴ Time taken for the cage to reach the top,
  t = t1 + t2 + t3 = 6.67 + 5.83 + 1.67 = 14.17 s Ans.
2. Torque which must be applied to the drum at starting
Let T = Torque which must be applied to the drum at starting = T1 + T2 + T3,

where T1 = Torque to raise the cage and rope at uniform speed,
T2 = Torque to accelerate the cage and rope, and
T3 = Torque to accelerate the drum.

34  /G6C   Theory of Machines
Since the mass of rope, m = 3 kg/m, therefore total mass of the rope for 100 metres,
mR = m.s = 3 × 100 = 300 kg
We know that the force to raise cage and rope at uniform speed,
F1 = (mC + mR) g = (500 + 300) 9.81 = 7850 N
∴ Torque to raise cage and rope at uniform speed,
T1 = F1.r = 7850 × 0.5 = 3925 N-m
Force to accelerate cage and rope,
F2 = (mC + mR) a1 = (500 + 300) 1.5 = 1200 N
∴ Torque to accelerate the cage and rope,
T2 = F2.r = 1200 × 0.5 = 600 N-m

We know that mass moment of inertia of the drum,
     I = mD.k2 = 250 (0.35)2 = 30.6 kg-m2
and angular acceleration of the drum,
2 11.53r a d / s0.5a
rα= = =
∴ Torque to accelerate the drum,
T3 = I.α = 30.6 × 3 = 91.8 N-m
and total torque which must be applied to the drum at starting,
T = T1 + T2 + T3 = 3925 + 600 + 91.8 = 4616.8 N-m Ans.
3.  Power at the end of acceleration period

When the acceleration period is just finishing, the drum torque will be reduced because
there will be s1 = 33.35 m of rope less for lifting. Since the mass of rope is 3 kg/m, therefore mass of
33.35 m rope,
m1 = 3 × 33.35 = 100.05 kg
∴ Reduction of torque,
T4 = (m1.g + m1.a1) r = (100.05 × 9.81 + 100.05 × 1.5) 0.5
    = 565.8 N-m
and angular velocity of drum,
           ω = v/2πr = 10/ 2π × 0.5 = 3.18 rad/s

ω = v/2πr = 10/ 2π × 0.5 = 3.18 rad/s
We know that power = T4.ω = 565.8 × 3.18 = 1799 W = 1.799 kW Ans.
Example 3.3.  A riveting machine is driven by a 4 kW motor. The moment of inertia of the
rotating parts of the machine is equivalent to 140 kg-m2 at the shaft on which the flywheel is mounted.
At the commencement of operation, the flywh eel is making 240 r.p.m. If closing a rivet occupies

1 second and consumes 10 kN-m of energy, find the reduction of speed of the flywheel. What is themaximum rate at which the rivets can be closed ?
Solution :  Given : P = 4 kW = 4000 W ; I = 140 kg-m
2 ; N1 = 240 r.p.m. or  ω1 = 2π ×
240/60 = 25.14 rad/s
Reduction of speed of the flywheel
Let ω2 = Angular speed of the flywheel immediately after closing a rivet.

Chapter 3 : Kinetics of Motion    /G6C  35
Fig. 3.6Since the power of motor is 4000 W, therefore energy supplied by motor in 1 second,
E1 = 4000 N-m ... (∵ 1 W = 1 N-m/s)
We know that energy consumed in closing a rivet in 1 second,
E2 = 10 kN-m = 10 000 N-m
∴ Loss of kinetic energy of the flywheel during the operation,
E = E2 – E1 = 10 000 – 4000 = 6000 N-m
We know that kinetic energy of the flywheel at the commencement of operation
  = 1
2 I (ω1)2 = 1
2× 140 (25.14)2 = 44 240 N-m

= 1
2 I (ω1)2 = 1
∴ Kinetic energy of the flywheel at the end of operation
  = 44 240 – 6000 = 38 240 N-m ... (i)
We also know that kinetic energy of the flywheel at the end of operation
   = 1
2 I (ω2)2 = 1
2 × 140 (ω2)2 = 70 (ω2)2... (ii)
Equating equations (i) and (ii),
70 (ω2)2 = 38 240    or      ( ω2)2 = 38 240/70 = 546.3 and ω = 23.4 rad/s
∴ Reduction of speed
  = ω1 – ω2 = 25.14 – 23.4 = 1.74 rad/s
  = 1.74 × 60/2 π = 16.6 r.p.m.  Ans. ... (∵  ω = 2π N/60)

Maximum rate at which the rivets can be closed
Maximum rate at which the rivets can be closed per minute
Energy supplied by motorpermin 4000 60=2 4
Energyconsumed to close a rivet 10000×== Ans.
Example 3.4.  A wagon of mass 14 tonnes is hauled up an incline of 1 in 20 by a rope which
is parallel to the incline and is being wound round a dr um of 1 m diameter. The drum, in turn, is
driven through a 40 to 1 reduction gear by an electric motor. The frictional resistance to the move-

ment of the wagon is 1.2 kN, and the efficiency of the gear drive is 85 per cent. The bearing frictionat the drum and motor shafts may be neglected. The rotating parts of the drum have a mass of 1.25tonnes with a radius of gyration of 450 mm and the rotating parts on the armature shaft have a massof 110 kg with a radius of gyration of 125 mm.
At a certain instant the wagon is moving up the slope with a velocity of 1.8 m/s and an
acceleration of 0.1 m/s

2. Find the torque on the motor shaft and the power being developed.
Solution. Given : m = 14 t = 14 000 kg ; Slope = 1 in 20 ;
d = 1m or r = 0.5 m ; F = 1.2 kN = 1200 N ; η = 85% = 0.85 ;
m1 = 1.25 t = 1250 kg ; k1 = 450 mm = 0.45 m ; m2 = 110 kg;
k2 = 125 mm = 0.125 m;  v = 1.8 m/s ; a = 0.1 m/s2
Torque on the motor shaft
We know that tension in the rope,
         P1= Forces opposing the motion as shown in   Fig. 3.6.

36  /G6C   Theory of Machines
= Component of the weight down the slope
    + *Inertia force + Frictional  resistance
1.. .20mg ma F=× ++
14 000 9.8114 000 0.1 1200 9467N
20×=+ × + =
∴ Torque on the drum shaft to accelerate load,
T1 = P1.r = 9467 × 0.5 = 4733.5 N-m
I1 = m1 (k1)2 = 1250 (0.45)2 = 253 kg-m2
α = a/r = 0.1/0.5 = 0.2 rad/s
∴ Torque on the drum to accelerate drum shaft,

T2 = I1.α1 = 253 × 0.2 = 50.6 N-m
Since the drum is driven through a 40 to 1 reduction gear and the efficiency of the gear drive
is 85%, therefore
Torque on the armature to accelerate drum and load,
31 211 11( ) (4733.5 50.6) 140.7N-m40 0.85 40 0.85TT T=+ × = + × =
We know that mass moment of inertia of the armature,
I2 = m2 (k2)2 = 110 (0.125)2 = 1.72 kg-m2
and angular acceleration of the armature,
2
20.140 40 8 rad/s0.5a
rα= × = × =

2
rα= × = × =
... (∵ Armature rotates 40 times that of drum)
∴ Torque on the armature to accelerate armature shaft,
T4 = I2.α2 = 1.72 × 8 = 13.76 N-m
and torque on the motor shaft
T = T3 + T4 = 140.7 + 13.76 = 154.46 N-m   Ans.
Power developed by the motor
We know that angular speed of the motor,
1.840 40 144 rad/s0.5v
rω = ×= ×=
∴ Power developed by the motor
= T.ω = 154.46 × 144 = 22 240 W = 22.24 kW Ans.
*Inertia force is equal and opposite to the accelerating force.

Chapter 3 : Kinetics of Motion    /G6C  37
Example 3.5.  A road roller has a total
mass of 12 tonnes. The front roller has a mass of2 tonnes, a radius of gyration of 0.4 m and adiameter of 1.2 m. The rear axle, together with itswheels, has a mass of 2.5 tonnes, a radius ofgyration of 0.6 m and a diameter of 1.5 m.Calculate : 1. Kinetic energy of rotation of the
wheels and axles at a speed of 9 km/h, 2. Total
kinetic energy of road roller, and 3. Braking force

required to bring the roller to rest from 9 km/h in 6m on the level.
Solution. Given : m = 12 t = 12 000 kg ;
m
1 = 2 t = 2000 kg ; k1 = 0.4 m ; d1 = 1.2 m or r1 = 0.6 m ; m2 = 2.5 t = 2500 kg ; k2 = 0.6 m ; d2 = 1.5
m or r2 = 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m
1. Kinetic energy of rotation of the wheels and axles
We know that mass moment of inertia of the front roller,
I1 = m1(k1)2 = 2000 (0.4)2 = 320 kg-m2
and mass moment of inertia of the rear axle together with its wheels,

I2 = m2 (k2)2 = 2500 (0.6)2 = 900 kg -m2
Angular speed of the front roller,
ω1 = v/r1 = 2.5/0.6 = 4.16 rad/s
and angular speed of rear wheels,
ω2 = v/r2 = 2.5/0.75 = 3.3 rad/s
We know that kinetic energy of rotation of the front roller,
22
11 111( ) 320(4.16) 2770 N-m22EI=ω = × =
and kinetic energy of rotation of the rear axle together with its wheels,
22
22 211( ) 900(3.3) 4900 N-m22EI=ω = × =
∴  Total kinetic energy of rotation of the wheels,
12 2770 4900 7670N-m EE E= + =+=  Ans.

2.  Total kinetic energy of road roller
We know that the kinetic energy of motion ( i.e. kinetic energy of translation) of the road roller,
22
311
22. 1 20 0 0( 2 . 5 ) 3 75 0 0N - mEm v== × =
This energy includes the kinetic energy of translation of the wheels also, because the total
mass (m) has been considered.
∴ Total kinetic energy of road roller,
E4  = Kinetic energy of translation + Kinetic energy of rotation

38  /G6C   Theory of Machines
3.  Braking force required to bring the roller to rest
Let   F  = Braking force required to bring the roller to rest, in newtons.
We know that the distance travelled by the road roller,
    s = 6 m ... (Given)
∴ Work done by the braking force
     = F × s = 6 F N-m
This work done must be equal to the total kinetic energy of road roller to bring the roller to
rest, i.e.
   6 F = 45 170    or     F = 45 170/6 = 7528.3 N  Ans.

Example 3.6. A steam engine drop-valve is closed by a spring after the operation of a trip
gear. The stiffness of the spring is s uch that a force of 4 N is required per mm of compression. The
valve is lifted against the spring, and when fully open the compression is 75 mm. When closed thecompression is 30 mm. The mass of the valve is 5 kg and the resistance may be taken as constantand equal to 70 N. Find the time taken to close the valve after the operation of the trip.

Solution. Given : s = 4 N/mm = 4000 N/m ; x
1 = 75 mm = 0.075 m ; x2 = 30 mm = 0.03 m;
m = 5 kg ; R = 70 N
Let x  =  Displacement of the valve (in metres) from its highest position in
time  t seconds.
When the valve is closed, then the value of x
= x1 – x2 = 0.075 – 0.03 = 0.045 m
Since the stiffness of the spring is 4000 N/m ; therefore in any position, the push of the spring
Q = 4000 (0.075 – x ) N
If P is the downward force on the valve, then

P = Q + m.g – R = 4000 (0.075 – x) + 5 × 9.81 – 70 = 279 – 4000 x
 Also             Force, P = Mass × Acceleration
279 – 4000 x = 5 × 2
2dx
dt
or                 2
2279 400056 800 800( 0.07)5dx xxx
dt−== − = − −
Let                     y  = x – 0.07
∴                 22
22800dy dxy
dt dt== −       or     2
2800 0dyy
dt+=
The solution of this differential equation is
 cos 800 sin 800ya tb t=+
0.07 cos 800 sin 800xat bt−= + ... (i)

where a and b are constants to be determined.
Now when t = 0, x = 0, therefore from equation  (i), a = – 0.07
Differentiating equation (i),
800 sin 800 800 cos 800dxatbtdt=− + ... (ii)

Chapter 3 : Kinetics of Motion    /G6C  39
Now when t = 0,dx
dt= 0, therefore from equation (ii), b = 0
Substituting the values of  a and b in equation (i),
x – 0.07 = – 0.07 cos 800 t  or  0.07 (1 cos 800 )xt=−
When  x = 0.045 m, then
0.045 0.07(1 cos 800 ) t =−
or 1 cos 800 0.045/0.07 0.642 t −==   or cos 800 1 0.642 0.358 t=− =
1800 cos (0.358) 69 69 1.2 rad180t− π== ° = × =
∴ 1.2/ 800 1.2/28.3 0.0424s== =t  Ans.
3.16. Principle of Conservation of Energy

It states “The energy can neither be created nor destroyed, though it can be transformed
from one form into any of the forms, in which the energy can exist.”
Note : The loss of energy in any one form is always accompanied by an equivalent increase in another form.

When work is done on a rigid body, the work is converted into kinetic or potential energy or is used in overcom-ing friction. If the body is elastic, some of the work will also be stored as strain energy. Thus we say that the totalenergy possessed by a system of moving bodies is constant at every instant, provided that no energy is rejected
to or received from an external source to the system.
3.17. Impulse and Impulsive Force
The impulse is the product of force and time. Mathematically,

Impulse = F × t
where                            F  = Force, and t = Time.
Now consider a body of mass m. Let a force  F changes its velocity from an initial velocity v1
to a final velocity v2.
We know that the force is equal to the rate of change of linear momentum, therefore
                              21()mv vFt−=  or 21()Ftm v v×= −
i.e.     Impulse = Change of linear momentum

If a force acts for a very short time, it is then known as  impulsive force  or blow. The impulsive
force occurs in collisions, in explosions, in the striking of a nail or a pile by a hammer.
Note: When the two rotating gears with angular velocities ω1 and ω2 mesh each other, then an impulsive
torque acts on the two gears, until they are both rotating at speeds corresponding to their velocity ratio. Theimpulsive torque,
             T.t  = I (ω

T.t  = I (ω
2 – ω1)
3.18. Principle of Conservation of Momentum
It states “ The total momentum of a system of masses (i.e. moving bodies) in any one direc-
tion remains constant, unless acted upon by an external force in that direction .” This principle is
applied to problems on impact, i.e. collision of two bodies. In other words, if two bodies of masses
m1 and m2 with linear velocities v1 and v2 are moving in the same straight line, and they collide and

40  /G6C   Theory of Machines
    Momentum before impact = Momentum after impact
i.e.          11 22 1 2 () mv mv m m v±= +
Notes : 1.  The positive sign is used when the two bodies move in the
same direction after collision. The negative sign is used when they move
in the opposite direction after collision.
2. Consider two rotating bodies of mass moment of inertia I1
and I2 are initially apart from each other and are made to engage as in

the case of a clutch. If they reach a common angular velocity ω, after
slipping has ceased, then
            I1.ω1  ±  I2.ω2 = (I1 + I2) ω
The ± sign depends upon the direction of rotation.
3.19. Energy Lost by Friction Clutch During
Engagement
Consider two collinear shafts A and B connected by
a *friction clutch (plate or disc clutch) as shown in Fig. 3.7.
Let  IA and IB= Mass moment of inertias of the
rotors attached to shafts  A and B
ωA and ωB= Angular speeds of shafts A and B

respectively before engagement  ofclutch, and
ω= Common angular speed of shafts
A and B after engagement of clutch.
By the principle of conservation of momentum,
 I
A.ωA + IB.ωB = (IA + IB) ω
 ∴AA BB
AB..II
IIω+ ωω=
+ ... (i)
Total kinetic energy of the system before engagement,
22
22 AA BB
1A A B B() () 11() ()22 2IIEI Iω+ω=ω +ω =
Kinetic energy of the system after engagement,
2
2 AA BB
2A B A B
AB.. 11() ()22IIEI I I IIIω+ ω=+ ω =+ +
2
AA BB

2
AA BB
AB(. .)
2( )II
IIω+ ω=
+
∴  Loss of kinetic energy during engagement,
22 2
AA BB A A B B
12
AB() () ( . .)
22 ( )ω+ω ω + ω=−= −+II I IEEEII
*Please refer Chapter 10 (Art. 10.32) on Friction.
Fig. 3.7. Friction clutch.

Chapter 3 : Kinetics of Motion    /G6C  41
= 2
AB A B
AB.( )
2( )II
IIω− ω
+ ... (ii)
Notes: 1. If the rotor attached to shaft B is at rest, then ωB= 0. Therefore, common angular speed after engagement,
AA
AB.I
IIωω=
+... [Substituting ωB = 0 in equation (i)] ... (iii)
and loss of kinetic energy,    2
AB A
AB.( )
2( )IIE
IIω=
+... [Substituting ωB = 0 in equation  (ii)] ... (iv)
2. If IB is very small as compared to IA and the rotor B is at rest, then
AA
A
AB.I
IIωω= =ω+... (Neglecting IB)
and2

AA
A
AB.I
and2
BA B11.. .22EI I=ω ω =ω ... [From equations (iii) and (iv)]
    = Energy given to rotor B
Example 3.7.  A haulage rope winds on a drum of radius 500 mm, the free end being

attached to a truck. The truck has a mass of 500 kg and is initially at rest. The drum is equivalent toa mass of 1250 kg with radius of gyration 450 mm. The rim speed of the drum is 0.75 m/s before therope tightens. By considering the change in linear momentum of the truck and in the angular mo-mentum of the drum, find the speed of the truck when the motion becomes steady. Find also theenergy lost to the system.
Solution. Given : r = 500 mm = 0.5 m ; m

1 = 500 kg ; m2 = 1250 kg ; k = 450 mm = 0.45 m ;
u = 0.75 m/s
We know that mass moment of inertia of drum,
 I2 = m2.k2 = 1250 (0.45)2 = 253 kg-m2
Speed of the truck
Let   v = Speed of the truck in m/s, and
 F = Impulse in rope in N-s.
We know that the impulse is equal to the change of linear momentum of the truck. Therefore
    F = m1.v = 500 v N-s
and         moment of impulse = Change in angular momentum  of drum

i.e.                   22 1 2()−×= ω − ω= uvFrI Ir... 21uvuv
rr r− ω− ω= − =/G51
          0.75500 0.5 253
0.5vv−×=     or    250 380 506vv=−
∴         250 v + 506 v = 380      or       v = 380/756 = 0.502 m/s    Ans.
Energy lost to the system
We know that energy lost to the system
= Loss in K.E. of drum – Gain in K.E. of truck
= 22 2
22 1 111() () .22Im v×ω − ω− ×

42  /G6C   Theory of Machines
22
2
21 211.22uvIm v
r−=× −×

22
2
21 (0.75) (0.502) 1253 500(0.502) N-m22 (0.5)−=× −× 

= 94 N-m    Ans.
Example 3.8. The two buffers at one end of a truck each require a force of 0.7 MN/m of

compression and engage with similar buffers on a truck which it overtakes on a straight horizontaltrack. The truck has a mass of 10 tonnes and its initial speed is 1.8 m/s, while the second truck hasmass of 15 tonnes with initial speed 0.6 m/s, in the same direction .
Find : 1. the common velocity when moving together during impact, 2. the kinetic energy
lost to the system, 3. the compression of each buffer to sto re the kinetic energy lost, and 4. the

velocity of each truck on separation if only half of the energy offered in the springs is returned .
Solution. Given : s = 0.7 MN/m = 0.7 × 10
6 N/m ; m = 10 t = 10 × 103 kg ; v1 = 1.8 m/s;
m2 = 15 t = 15 × 103 kg ; v2 = 0.6 m/s
1.  Common velocity when moving together during impact
Let                         v = Common velocity.
We know that momentum before impact = Momentum after impact
i.e.         m1. v1 + m2.v2 = (m1 + m2) v
 10 × 103 × 1.8 + 15 × 103 × 0.6 = (10 × 103 + 15 + 103) v

27 × 103 = 25 × 103 v       or      v = 27 × 103/25 × 103 = 1.08 m/s    Ans.
2.  Kinetic energy lost to the system
Since the kinetic energy lost to the system is the kinetic energy before impact  minus the
kinetic energy after impact, therefore
Kinetic energy lost to the system
                          ()22 2
11 22 1 211 1
22 2mv mv m m v=+ − +
= 32 32 1110 10 (1.8) 15 10 (0.6)
22×× +××
()33 2 110 10 15 10 (1.08)2−× + ×
= 4.35 × 103 N-m = 4.35 kN-m    Ans.

3. Compression of each buffer spring to store kinetic energy lost
Let x= Compression of each buffer spring in metre, and
s= Force required by each buffer spring or stiffness of each spring
= 0.7 MN/m = 0.7 × 106 N/m ... (Given)
Since the strain energy stored in the springs (four in number) is equal to kinetic energy lost
in impact, therefore
        2314 . 4.35 102sx×= ×

Chapter 3 : Kinetics of Motion    /G6C  43
62 3 14 0.7 10 4.35 102x ×× × = ×
or                       62 31.4 10 4.35 10x×= ×
∴        x2 = 4.35 × 103/1.4 × 106 = 3.11 × 10–3
or       x = 0.056 m = 56 mm     Ans.
4. Velocity of each truck on separation
Let       v3 = Velocity of separation for 10 tonnes truck, and
      v4 = Velocity of separation for 15 tonnes truck.
The final kinetic energy after separation is equal to the kinetic energy at the instant of com-

mon velocity plus strain energy stored in the springs. Since it is given that only half of the energy
stored in the springs is returned, therefore
Final kinetic energy after separation
               = Kinetic energy at common velocity +1
2 Energy stored in springs
or         22 2 2
13 24 1 2111 1 1() () ( ) 4 .
222 2 2mv mv m mv s x+= + + × 
32 32 3 3 2 3
34111 110 10 ( ) 15 10 ( ) (10 10 15 10 )(1.08) (4.35 10 )222 2vv ×× +×× = × +× + ×
231... 4 . 4.35 102sx×= ×/G51
          22

          22
3410( ) 15( ) 33.51vv+= ... (i)
We know that initial momentum and final momentum must be equal, i.e.
m1.v3 + m2.v4 = (m1 + m2) v
10 × 103 × v3 + 15 × 103 × v4 = (10 × 103 + 15 × 103) 1.08
10v3 + 15 v4 = 27 ...  (ii)
From equations (i) and (ii),       v3 = 0.6 m/s, and v4 = 1.4 m/s    Ans.
Example 3.9. A mass of 300 kg is allowed to fall vertically through 1 metre on to the top of

a pile of mass 500 kg. Assume that the falling mass and pile remain in contact after impact and thatthe pile is moved 150 mm at each blow. Find, a llowing for the action of gravity after impact 1. The
energy lost in the blow, and 2. The average resistance against the pile.
Solution. Given : m
1 = 300 kg ; s = 1 m ; m2 = 500 kg ; x = 150 mm = 0.15 m
1. Energy lost in the blow
First of all, let us find the velocity of mass m1 with which it hits the pile.

Let v1 = Velocity with which mass m1 hits the pile.
We know that  22
12. vu g s−=
2
110 2 9.81 1 19.62 or 4.43 m/svv−=× ×= =             ... (∵  u = 0 )

44  /G6C   Theory of Machines
Fig. 3.9Again, let v2 =  Velocity of the pile before impact, and
v  =  Common velocity after impact,
We known that momentum before impact
= Momentum after impact
or m1.v1 + m2.v2 = (m1 + m2) v
300 × 4.43 + 500 × 0 = (300 + 500) v
1329 = 800 v
∴ v = 1329/800 = 1.66 m/s
Now, kinetic energy before impact
= Potential energy = m1.g.s
= 300 × 9.81 × 1 = 2943 N-m
and kinetic energy after impact
22
1211( ) (300 500)(1.66) 1102 N-m22mm v=+ = + =
∴  Energy lost in the blow

= 2943 – 1102 = 1841 N-m    Ans.
2. Average resistance against the pile
Let R = Average resistance against the pile in N.
Since the net work done by R, m1 and m2 is equal to the kinetic energy after impact, therefore
(R – m1.g – m2.g) x = Kinetic energy after impact
(R – 300 × 9.81 – 500 × 9.81) 0.15 = 1102
∴ R – 7848 = 1102/0.15 = 7347
or R = 7347 + 7848 = 15 195 N = 15.195 kN    Ans.
Example 3.10. A hammer B suspended from pin C, and

an anvil A suspended from pin D, are just touching each otherat E, when both hang freely as shown in Fig. 3.9. The mass of Bis 0.7 kg and its centre of gravity is 250 mm below C and itsradius of gyration about C is 270 mm. The mass of A is 2.4 kgand its centre of gravity is 175 mm below D and its radius ofgyration about D is 185 mm. The hammer B is rotated 20° to theposition shown dotted and released. Assume that the points ofcontact move horizontally at the instant of impact and that

horizontally at the instant of impact and that theirlocal relative linear velocity of recoil is 0.8 times their relativelinear velocity of impact. Find the angular velocities of hammerand of the anvil immediately after impact .

1 = 0.7 kg ; k1 = 270 mm = 0.27 m ;
m2 = 2.4 kg ;  k2 = 185 mm = 0.185 m
Let ω= Angular velocity of hammer B just before impact, and
h= Distance from release to impact
= Distance of c.g. of mass B below C = 250 mm = 0.25 m ...(Given)Fig. 3.8

Chapter 3 : Kinetics of Motion    /G6C  45
We know that K.E. of hammer B
= Loss of P.E. from relase to impact
2
111.. .2ω=Im g h    or     22
11 11() . .2ω=mk m g h
     22 10.7 (0.27) 0.7 9.81 0.25 (1 cos20 )2×ω = × × − °
0.0255 ω2 = 0.1032
∴                      ω2 =  0.1032 / 0.0255 = 4.05   or  ω = 2.01 rad/s
Let ωA and ωB be the angular velocities of the anvil A and hammer B, in the same direction,
immediately after impact.
∴ Relative linear velocity

= ωA × DL – ωA × CM = ωA × 0.2 – ωB × 0.275
... (DL and CM are taken in metres)
= 0.2 ωA – 0.275 ωB... (i)
But, relative linear velocity
= 0.8 × Relative linear velocity of impact ... (Given)
= 0.8ω × CM = 0.8 × 2.01 × 0.275 = 0.44 ... (ii)
Equating (i) and (ii),
   0.2 ωA – 0.275 ωB = 0.44    or   ωB = 0.727 ωA – 1.6 ... (iii)
Since the linear impulse at E is equal and opposite on A and B, then by moments about D for
A and about C for B, it follows that the ratio

 Decrease in angular momentum of 0.275
Increase in angular momentum of 0.2BC M
AD L==
i.e.BB
AB( ) 0.2751.375
.0 . 2I
Iω−ω==
ω
           2
11 B
2
22 A() ( )1.375
()mk
mkω−ω=
ω      or     2
B
2
A0.7(0.27) (2.01 )1.375
2.4(0.185)−ω=
ω
∴                   2.01– ωB = 2.21 ωA or    ωB = 2.01 – 2.21 ωA... (iv)
From equations (iii) and (iv), we get
        0.727 ωA– 1.6 = 2.01 – 2.21 ωA

0.727 ωA– 1.6 = 2.01 – 2.21 ωA
  0.727 ωA + 2.21 ωA= 2.01 + 1.6     or     ωA = 1.23 rad/s     Ans.
Substituting ωA = 1.23 rad/s in equation (iv),
ωB = 2.01 – 2.21 × 1.23 = – 0.71 rad/s
                              = 0.71 rad/s, in reverse direction     Ans.

46  /G6C   Theory of Machines
Example 3.11. The pendulum of an Izod impact testing machine has a mass of 30 kg. The
centre of gravity of the pendulum is 1 m from the axis of suspension and the striking knife is 150 mmbelow the centre of gravity. The radi us of gyration about the point of suspension is 1.1 m, and about

the centre of gravity is 350 mm. In making a test, the pendulum is released from an angle of 60° tothe vertical. Determine : 1. striking velocity of the pendulum, 2. impulse on the pendulum and
sudden change of axis reaction when a specimen giving an impact value of 54 N-m is broken,3. angle of swing of the pendulum after impact, and 4. average force exerted at the pivot and at the
knife edge if the duration of impact is assumed to be 0.005 second .

Solution. Given : m = 30 kg ;  AG = a = 1 m ; GB = b = 150 mm = 0.15 m ; k
1 = 1.1 m;
k2 = 350 mm = 0.35 m ; θ = 60° ; t = 0.005 s
We know that mass moment of inertia of the pendulum about the point of suspension A,
IA = m (k1)2 = 30 (1.1)2 = 36.3 kg-m2
and mass moment of inertia of the pendulum about centre of
gravity G,
IG = m (k2)2 = 30 (0.35)2
= 3.675 kg-m2
1. Striking velocity of the pendulum
Let v = Striking velocity of the
pendulum, and
ω = Angular velocity of the
pendulum.

pendulum.
Since the potential energy of the pendulum is converted into angular kinetic energy of the
pendulum, therefore,
2
1A1.. .2mgh I=ω
30 × 9.81 (1 – 1 cos 60°) = 2 136.32×ω ... (/G51  h1 = a – a cos 60°)
or 147.15 = 18.15 ω2
∴ω2 = 147.15/18.15 = 8.1   or   ω = 2.85 rad/s
and v = ω × AB = ω (a + b) = 2.85 (1 +  0.15) = 3.28 m/s    Ans.
2.  Impulse on the pendulum
Let F1= Impulse at the pivot A,
F2= Impulse at the knife edge B,

ω= Angular velocity of the pendulum just before the breakage of the
specimen, and
ω1= Angular velocity of the pendulum just after the breakage of the specimen.
Since the loss in angular kinetic energy of the pendulum is equal to the energy used for
breaking the specimen (which is 54 N-m ), therefore
22
A11() 5 42ω− ω =I or22
1136.3(2.85 ) 542×− ω =Fig. 3.10

Chapter 3 : Kinetics of Motion    /G6C  47
∴22
154 2(2.85) 5.125
36.3×ω= − =  or 12.26 rad/sω=
Let vG and vG′ be the linear velocities of G just before and just after the breakage of specimen.
G 2.85 1 2.85m/s vO G=ω× = × =
andG1 2.26 1 2.26m/s vO G′=ω× = × =
We know that  Impulse = Change of linear momentum
F1 + F2 = m (vG – vG′) = 30 (2.85 – 2.26) = 17.7 N ... (i)
Taking moments about G, we get
     Impulsive torque = Change of angular momentum        F
2 × b – F1 × a = IG (ω – ω1)

  F2 × 0.15 – F1 × 1 = 3.675 (2.85 – 2.26) = 2.17 ...  (ii)
From equations (i) and (ii),
F2 = 17.3 N ; and F1 = 0.4 N   Ans.
3. Angle of swing of the pendulum after impact
Let θ = Angle of swing of the pendulum after impact.
Since work done in raising the pendulum is equal to angular kinetic energy of the pendulum,
therefore
  m.g.h1 = 2
A11
2()Iω
          30 × 9.81 (1 – 1 cos θ) = 2 1
236.3(2.26) 92.7×=

 1 – 1 cos θ = 92.7/30 × 9.81 = 0.315 or cos θ = 1 – 0.315 = 0.685
∴  θ  = 46.76°   Ans.
4.  Average force exerted at the pivot and at the knife edge
We know that average force exerted at the pivot
10.480 N0.005F
t== =  Ans.
and average force exerted at the knife edge
217.33460 N
0.005F
t== =  Ans.
Example 3.12. A motor drives a machine through a friction clutch which transmits a torque

of 150 N-m, while slip occurs during engagement. The rotor, for the motor, has a mass of 60 kg, with
radius of gyration 140 mm and the inertia of the machine is equivalent to a mass of 20 kg at the
driving shaft with radius of gyration 80 mm. If the motor is running at 750 r.p.m. and the machineis at rest, find the speed after the engagement of the clutch and the time taken. What will be thekinetic energy lost during the operation ?

48  /G6C   Theory of Machines
Solution. Given : T = 150 N-m ; m1 = 60 kg ; k1 = 140 mm = 0.14 m ; m2 = 20 kg ;
k2 = 80 mm = 0.08 m ; N1 = 750 r.p.m.  or   ω1 = 2 π × 750/60 = 78.55 rad/s ; N2 = 0  or  ω2 = 0
We know that mass moment of inertia of the rotor on motor,
22 2
11 1( ) 60(0.14) 1.176 kg-mIm k== =
and mass moment of inertia of the parts attached to machine,
I2 = m2 (k2)2 = 20 (0.08)2 = 0.128 kg-m2
Speed after the engagement of the clutch and the time taken

Let     ω = Speed after the engagement of the clutch in rad/s,
     t = Time taken in seconds, and
    α = Angular acceleration during the operation in rad/s2.
We know that the impulsive torque = change of angular momentum
 ∴ T.t = I1 (ω1 – ω)o r11() 1.176(78.55 )s150ItTω− ω −ω==      ... (i)
Also T.t = I2 (ω2 – ω)o r22() 0.128s
150It
Tω−ω ×ω== ... (ii)
Equating equations (i) and (ii), ... (/G51 ω2 = 0)
1.176(78.55 )
150−ω 0.128
150ω= or 92.4 – 1.176 ω = 0.128 ω

1.304 ω = 92.4 or ω = 92.4/1.304 = 70.6 rad/s  Ans.
 Substituting the value of ω in equation  (ii),
0.128 70.60.06 s
150t×==  Ans.
Kinetic energy lost during the operation
We know that the kinetic energy lost during the operation,
22
12 1 2 12 1
12 12.( ) . .
2( ) 2( )II IIEII IIω− ω ω==++... (/G51 ω2 = 0)
21.176 0.128 (78.55) 928.8356 N-m2(1.176 0.128) 2.61×== =+ Ans.
3.20. Torque Required to Accelerate a
Geared System

Consider that the two shafts A and B are geared together
as shown in Fig. 3.11. Let the shaft B rotates G times the speed
of shaft A. Therefore, gear ratio,
B
ANGN=
where NA and NB are speeds of shafts A and B (in r.p.m.)
Since the shaft B turns G times the speed of shaft A,
therefore the rate of change of angular speed of shaft B withFig. 3.11. Torque to accelerate a
geared system.

Chapter 3 : Kinetics of Motion    /G6C  49
respect to time ( i.e. angular acceleration of shaft B, αB ) must be equal to G times the rate of change
of angular speed of shaft A with respect to time ( i.e. angular acceleration of shaft A, αA ).
∴αB = G.αA...(i)
Let IA and IB= Mass moment of inertia of the masses attached to shafts A and B
∴ Torque required on shaft A to accelerate itself only,
TA = IA.αA
and torque required on shaft B to accelerate itself only,

TB = IB.αB = G.IB.αA... [From equation (i)]  ... (ii)
In order to provide a torque TB on the shaft B, the torque applied to shaft A must be G × TB.
Therefore, torque applied to shaft A in order to accelerate shaft B,
TAB = G.TB = G2.IB.αA... [From equation (ii)] ... (iii)
∴ Total torque which must be applied to shaft A in order to accelerate the geared system,
T = TA + TAB = IA.αA + G2.IB.αA
= (IA + G2.IB) αA = I.αA... (iv)

where I = IA + G2. IB and may be regarded as equivalent mass moment of inertia of geared system
referred to shaft A.
Let the torque T required to accelerate the geared system, as shown in Fig. 3.11, is applied
by means of a force F which acts tangentially to a drum or pulley of radius r.
∴ T = F × r = I.αA... (v)
We know that the tangential acceleration of the drum,
a = αA.r     or     αA = a/r
∴ ()2
AB.aaFrI I G Irr×=× = +2
AB ...( . )II G I=+/G51

or          ()2
AB2..eaFI G I a m
r=+= ... (vi)
where  ()2
AB21.emI G I
r=+  and may be regarded as equivalent mass of the system referred to the
line of action of the accelerating force F.
Notes : 1.  If η is the efficiency of the gearing between the two shafts A and B, then the torque applied to shaft
A in order to accelerate shaft B,
2
BA
AB..GITα=η
and the total torque applied to shaft A in order to accelerate the geared system,
22
BA B

22
BA B
AA BA A A A A.. ...GI GITT T I I Iα=+ =α + = + α = α ηη
where2
B
A.=+ηGIII , and may be regarded as the equivalent mass moment of inertia of the geared system
2. If the number of shafts (say A to X ) are geared together in series, then the equivalent mass moment
of inertia referred to shaft A is given by,

50  /G6C   Theory of Machines
2
Axx
xGIII=+η∑
where Gx = Ratio of speed of shaft X to the speed of shaft A,
Ix = Mass moment of inertia of mass attached to shaft  X, and
ηx = Overall efficiency of the gearing from shaft A to shaft X.
3. If each pair of gear wheels is assu med to have t he same efficiency  η and there are m gear pairs
through which the power is transmitted from shaft A to shaft X, then the overall efficiency from shaft A to X is
given by,
                       ηx = ηm

given by,
                       ηx = ηm
4. The total kinetic energy of the geared system,
                    2
A1K.E. = ( )2ωI
where      I =  Equivalent mass moment of inertia of the geared system referred to shaft A, and
                        ωA = Angular speed of shaft  A.
Example 3.13. A mass M of 75 kg is hung from a

rope wrapped round a drum of effective radius of 0.3 metre,which is keyed to shaft A. The shaft A is geared to shaft Bwhich runs at 6 times the speed of shaft A. The total massmoment of inertia of the masses attached to shaft A is 100kg-m
2 and that of shaft B is 5 kg-m2.
Find the acceleration of mass M if 1. it is al-
lowed to fall freely, and 2. when the efficiency of the

gearing system is 90%. The configuration of the systemis shown in Fig. 3.12.
Solution. Given : M = 75 kg ; r = 0.3 m ; N
B = 6 NA
or G = NB/NA = 6 ; IA = 100 kg-m2 ; IB = 5 kg-m2; η = 90% = 0.9
Let  a = Acceleration of the mass M, in m/s2.
1. When it is allowed to fall freely
We know that equivalent mass of the geared system referred to the circumference of the
drum (or the line of action of the accelerating mass M ),
() ()22
AB2211. 100 6 5 3111kg
(0.3)emI G I
r=+= + × =

r=+= + × =
and total equivalent mass to be accelerated,
 Me = me + M = 3111 + 75 = 3186 kg
∴ Force required to accelerate this equivalent mass ( Me)
= Me.a = 3186 a N ... (i)
and the accelerating force provided by the pull of gravity on the mass  M suspended from the rope
= M.g = 75 × 9.81 = 736 N ... (ii)
3186 a = 736 or a = 736/3186 = 0.231 m/s2 Ans.
2. When the efficiency of the gearing system is 90%

We know that the equivalent mass of the geared system referred to the circumference of the
drum,Fig. 3.12

Chapter 3 : Kinetics of Motion    /G6C  51
Fig. 3.13()2 2
B
A22. 11 6 5100 3333 kg0.9 0.3eGImI
r  ×=+ = += η 
Me = me + M = 3333 + 75 = 3408 kg
= Me.a = 3408 a N ... (iii)
and accelerating force provided by the pull of gravity on the mass  M suspended from the rope
= M.g = 75 × 9.81 = 736 N ... (iv)
Now equating equations  (iii) and (iv),

3408 a = 736    or    a = 736/3408 = 0.216 m/s2 Ans.
Example. 3.14. The motor shaft A exerts a con-
stant torque of 100 N-m and is geared to shaft B asshown in Fig. 3.13. The moments of inertia of the partsattached to the motor shaft A is 2 kg-m
2 and that of the
parts attached to other shaft B is 32 kg-m2.
Find the gear ratio which gives the maximum
angular acceleration of shaft B and the correspondingangular acceleration of each shaft.

Solution. Given : T = 100 N-m ; I
A = 2 kg-m2 ;
IB = 32 kg-m2
Gear ratio which gives the maximum acceleration
Let G = Gear ratio which gives the maximum
      acceleration.
αA = Angular acceleration of shaft A, and
αB = Angular acceleration of shaft B.
 We know that αA = G.αB                                        ... (i)
∴ Torque required on motor shaft A to accelerate rotating
parts on it,
TA = IA. αA= IA.G. αB
... [From equation (i)]

and torque required on motor shaft A to accelerate rotating parts on shaft B,
BB
AB.ITGα=
Assuming that there is no resisting torque and the torque exerted on the motor shaft A is
utilised to overcome the inertia of the geared system.
∴2
BB A B
AA BA B B....II G ITT T I GGGα+=+ = α + = α 
or B 2
AB.
.GT
IG Iα=
+... (ii)
Parallel shaft gear motor.

52  /G6C   Theory of Machines
For maximum angular acceleration of B, differentiate with respect to G and equate to zero, i.e.
2
AB B.
.0o r 0GTd
IG I d
dG dG
+ α ==
2
AB A
22
AB(. ) .( 2 )0
(. )IG I T G TI G
IG I+− ×=
+or22
AB A.2 . 0IG I GI+− =
∴2
BA.IG I= orB
A3242IGI===  Ans.
Angular acceleration of each shaft
Substituting the value of G in equation  (ii),
2
B 24 1006.25 rad/s
24 3 2×α= =
×+Ans.
and2
AB. 4 6.25 25 rad/sGα=α=× = Ans.

×+Ans.
and2
Example 3.15.  A motor vehicle of total mass 1500 kg has road wheels of 600 mm effective
diameter. The effective moment of ine rtia of the four road wheels and of the rear axle together is
8 kg-m2 while that of the engine and flywheels is 1 kg-m2. The transmission efficiency is 85%
and a tractive resistance at a speed of 24 km/h is 300 N. The total available engine torque is200 N-m. Determine  :

1. Gear ratio, engine to back axle, to provide maximum acceleration on an upgrade whose
sine is 0.25, when travelling at 24 km/h,
2. The value of this maximum acceleration, and
3. The speed and power of the engine under these conditions .
Solution. Given : m = 1500 kg ; d = 600 mm = 0.6 m   or    r = 0.3 m ; I
A = 8 kg-m2 ;
IB = 1 kg-m2; η = 85% = 0.85 ; v = 24 km/h ; F = 300 N ; TB = 200 N-m ; sin θ = 0.25
1. Gear ratio, engine to back axle, to provide maximum acceleration

Let G = Gear ratio, engine to back axle, to provide maximum acceleration.
∴ Torque at road wheels,
WB 0.85 200 170 N-m TG T G G=η× × = × × =
and available tangential force at road wheels,
W170567 N
0.3T GPG
r== =
Let the vehicle travels up the gradient a distance of s metre while its speed changes from u
to  v m/s.
We know that work done by the tangential force P
= Change of linear K.E. of vehicle + Change of angular K.E. of road

wheels and axle + Change of angular K.E. of engine and flywheel +Work done in raising vehicle + Work done in overcoming tractiveresistance

Chapter 3 : Kinetics of Motion    /G6C  53
or22 22 2 22
A2 1 B 2 111 1
22 2( )() . . () . . . s i n . Ps m v u I I G m g s F s×= − + ω− ω+ η ω− ω+ θ +
or2 22
AB
22..(. . s i n)2 η −−θ − = + +II G vusP m g F m
rr
... (Substituting ω1 = u/r, and ω2 = v/r)
 22 2
228 1 0.85(567 1500 9.81 0.25 300) 15002 0.3 0.3vu GsG−× ×−× ×−= ++ 
22
2(567 3980) (1590 9.44 )2vusG G−−= + ... (i)
We know that linear acceleration,
22
2567 3980
21590 9.44vu Gas G−−==
+... [From equation (i)] ... (ii)

For maximum acceleration, differentiate equation (ii) with respect to G and equate to zero,
i.e.
0da
dG=
   2
22(1590 9.44 ) (567 3980)(9.44 2 )0
(1590 9.44 )GG G
G+− − ×=
+
or   901 530 + 5352 G2 – 10 705 G2 + 75 142 G = 0
      G2 – 14 G – 168.4 = 0
∴214 (14) 4 168.4 14 29.521.7522G±+ × ±== =  or 22 Ans.
... (Taking + ve sign)
2. Value of maximum acceleration
Substituting the value of G = 22 in equation  (ii), maximum acceleration,
2567 22 39801.38m/s

1590 9.44(22)maxa×−==
+ Ans.
3. Speed and power of the engine
Let ω = Speed of the engine in rad/s.
We know that the speed of the road wheels,
v = 24 km/h = 6.67 m/s ... (Given)
∴ Angular speed of the road wheels
6.6722.23 rad/s0.3v
r== =

54  /G6C   Theory of Machines
Since the speed of the engine is G times the speed of the road wheels, therefore
22.23 22 22.23 489rad/sGω = ×= ×=  Ans.
We know that power of the engine
   = TB.ω = 200 × 489 = 97 800 W = 97.8 kW Ans.
Example 3.16. A super charged road racing automobile has an engine capable of giving

an output torque of 1 kN-m, this torque being reasonably constant over a speed range from100 km/h to 275 km/h in top gear. The road wheels are of 0.9 m effective di ameter, and the back axle

ratio is 3.3 to 1. When travelling at a steady speed of 170 km/h in top gear on a level road, thepower absorbed is 50 kW. The vehicle has a mass of 1000 kg, the four road wheels each has mass of40 kg and a radius of gyration of 0.25 m. The moment of inertia of the engine and all parts forwardof the differential is 6 kg-m
2.
Assuming that the resistance caused by windage and road drag varies as the square of the

speed, determine the time taken for the speed to rise from 100 km/h to 275 km/h in top gear at fullthrottle on an upgrade of 1 in 30.
Solution. Given : T
B = 1 kN-m = 1000 N-m ; v1 = 100 km/h = 27.8 m/s ; v2 = 275 km/h =
76.4m/s ; d = 0.9 m or r = 0.45 m ; G = 3.3 ; v = 170 km/h = 47.2 m/s ; P = 50 kW = 50 × 103 W ;
M = 1000 kg ; m = 40 kg ; k = 0.25 m ; IB = 1 kg-m2
We know that moment of inertia of four road wheels,
IA = 4 × m.k2 = 4 × 40 (0.25)2 = 10 kg -m2

Let F  = Resistance caused by windage and road drag in newtons.
∴ Power absorbed by the automobile at a steady speed ( P),
50 × 103 = F.v = F × 47.2     or     F = 50 × 103/47.2 = 1060 N
Since the resistance caused by windage and road drag ( F) varies as the square of the speed
(v), therefore
F = k.v2   or  k = F/v2 = 1060/(47.2)2 = 0.476
∴  F = 0.476 v2 N
We know that the torque at road wheels,
WE 3.3 1000 3300N-m TG T=×= × =

W
T33007333N
0.45TF
r== =
Since the gradient is 1 in 30, therefore proceeding in the same way as discussed in the
previous example, we get the linear acceleration,
()2
22
AB
22 22. 1000 9.817333 0.47630 30
.1 0 1 3 . 31000
(0.45) 0.45TMgFF vdvadt II GM
rr×−− − −
== =
×++ ++
= 6.65 – 0.43 × 10–3 v2 – 0.3
∴32 326.65 0.43 10 0.3 6.35 0.43 10dv dvdt
vv−−==
−× − −×

Chapter 3 : Kinetics of Motion    /G6C  55
Let326.35 0.43 10dvdt
v−=
−×∫∫
3
22 21023250.4314 768 (121.5)dv dv
vv==
−−∫∫
∴12325 121.5log2 121.5 121.5evtCv+=+×−... (i)
221... lo g2edv a v
aa v av += − −∫/G51
where C1 is the constant of integration. We know that when t = 0, v1 = 27.8 m/s.
∴ 12325 121.5 27.80l o g2 121.5 121.5 27.8e C+=+×−... (Substituting v = v1)
11149.39.6 log 9.6 log 1.6
93.7ee CC =+ = +
∴19.6log 1.6 9.6 0.47 4.5e C=− =− × =−

Now the expression  (i) may be written as
2325 121.5log 4.52 121.5 121.5evtv+=−×−
When v2 = 76.4 m/s, the time taken for the speed to rise
2325 121.5 76.4 197.9log 4.5 9.6 log 4.52 121.5 121.5 76.4 45.1ee+=− = −×−
9.6log 4.38 4.5 9.6 1.48 4.5 9.7se=− = × − =  Ans.
Example 3.17. An electric motor drives a machine through a speed reducing gear of ratio
9:1. The motor armature, with its shaft and gear wheel, has moment of inertia 0.6 kg-m2. The

rotating part of the driven machine has moment of inertia 45 kg-m2. The driven machine has resist-
ing torque of 100 N-m and the effic iency of reduction gear is 95%. Find
1. The power which the motor must develop to drive the machine at a uniform speed of 160
r.p.m.,
2. The time required for the speed of the machin e to increase from zero to 60 r.p.m., when the
torque developed on the motor armature in starting from rest is 30 N-m, and

3. If the gear ratio were altered so as to give the machine the greatest possible angular
acceleration in starting from rest, what would then be the gear ratio ? The starting torque of themotor is 30 N-m as before.
Solution.  Given : G = 9; I
A = 0.6 kg-m2; IB = 45 kg-m2; TB = 100 N-m;
η = 95% = 0.95; N = 160 r.p.m. ; N1 = 0 ; N2 = 60 r.p.m. ; TA = 30 N-m
A motor driving a machine is shown in Fig. 3.14.

56  /G6C   Theory of Machines
1. Power which the motor must develop
We know that the power which the motor must develop,
B2 . 2 160 100W60 60 0.95NTPππ × ×==×η ×
= 1764 W = 1.764 kW Ans.
2. Time required for the speed of the machine to increase from
    zero to 60 r.p.m .
Let          t   =   Time required for the speed of  the
     machine to increase from zero to 60 r.p.m.
αA= Angular acceleration of motor, and
αB= Angular acceleration of machine.

Since the speed of motor A is G times the speed of machine B, therefore
 αA=G.αB = 9 αB
We know that torque developed on motor armature,
TA= 30 N-m ... (Given)
Due to the torque ( TA) and efficiency of gearing ( η), the torque transmitted to machine B,
TB1 = G.TA.η = 9 × 30 × 0.95 = 256.5 N-m
We know that resisting torque on machine B,
TB = 100 N-m ... (Given)
∴ Net torque on machine B
= TB1 – TB = 256.5 – 100 = 156.5 N-m ... (i)

We know that total torque to be applied to machine B in order to accelerate the geared system
= Torque required on B to accelerate B only + Torque required on B
to accelerate A
= IB.αB + G.TA.η = IB.αB + G.IA.αA.η ... (∵ TA = IA.αA)
= IB.αB + G2. IA .αB.η ...(∵ αA= G.αB)
= 45 αB + 92 × 0.6 × αB × 0.95 = 45 αB + 46.2 αB
= 91.2 αB... (ii)
αB = 156.5/91.2 = 1.7 rad/s2

We are given that initial angular speed, ω1 = 0, and final angular speed,
2
2226 06.28 rad/s60 60Nπ π×ω= = =... (∵  N2 = 60 r.p.m.)
We know that ω2 = ω1 + αB.t
6.28 = 0 + 1.7  t = 1.7 t       or      t = 6.28/1.7 = 3.7 s  Ans.
3. Gear ratio for maximum angular acceleration of the machine
Let G1 = Gear ratio for maximum angular acceleration of the machine.Fig. 3.14

Chapter 3 : Kinetics of Motion    /G6C  57
We know that net torque on machine B
= TB1 – TB = G1.TA.η – TB = G1 × 30 × 0.95 – 100
= 27.5 G1 – 100 ... (iii)
We also know that total torque required to be applied to machine B in order to accelerate the
geared system
= IB.αB + (G1)2 αB.IA.η
= 45 × αB + (G1)2 αB × 0.6 × 0.95 = αB [45 + 0.57 ( G1)2] ...(iv)
From equations  (iii) and (iv),
1
B 2
127.5 100
45 0.57( )G
G−α=
+

1
B 2
127.5 100
45 0.57( )G
G−α=
+
For maximum angular acceleration, differentiate the above expression and equate to zero, i.e.
B
10d
dGα=
or2
111
2
1[45 0.57( ) ](27.5) (27.5 100) (2 0.57 )0
[45 0.57( ) ]GGG
G+− − ×=
+
1237.5 + 15.675 ( G1)2 – 31.34 ( G1)2 + 114 G1 = 0
15.675 (G1)2 – 114 G1 – 1237.5 = 0
(G1)2 – 7.27 G1 – 78.95 = 0
∴2
17.27 (7.27) 4 78.95 7.27 19.213.23522G±+ × ±== =  Ans.

Example 3.18. A hoisting gear, with a 1.5 m diameter drum, operates two cages by ropes
passing from the drum over two guide pulleys of  1 m diameter. One cage (loaded) rises while the
other (empty) descends. The drum is driven by a motor through double reduction gearing. Theparticulars of the various parts are as follows :
S.No. Part Maximum Mass (kg) Radius of Frictional
Speed (r.p.m.) gyration (mm) resistance
1.Motor 900 200 90 –
2.Intermediate gear 275 375 225 150 N-m

3.Drum and shaft 50 2250 600 1125 N-m
4.Guide pulley ( each) – 200 450 150 N-m
5.Rising rope and cage – 1150 – 500 N
6. Falling rope and cage – 650 – 300 N
Determine the total motor torque necessary to produce a cage an acceleration of 0.9 m/s
2.
Solution. Given : d = 1.5 m or r = 0.750 m ; d1 = 1 m ; NM = 900 r.p.m ; N1 = 275 r.p.m. ;
ND = 50 r.p.m ; mM = 200 kg; kM = 90 mm = 0.09 m ; mI = 375 kg ; kI = 225 mm = 0.225 m ;

MD = 2250 kg ;  kD = 600 mm = 0.6 m ; mP = 200 kg ; kP = 450 mm = 0.45 m ; m1 = 1150 kg ;
m2 = 650 kg ; FI = 150 N-m ; FD = 1125 N-m ; FP = 150 N-m ; F1 = 500 N ;  F2 = 350 N ; a = 0.9 m/s2

58  /G6C   Theory of Machines
Fig. 3.15
We know that speed of guide pulley ( P),
PD
11.550 75 r.p.m.1dNNd=× = ×=
Gear ratio for the intermediate gear and motor,
 11 M/ 275/900 0.306GN N== =
Gear ratio for the drum and motor,
2D M/ 50/900 0.055GNN== =
Gear ratio for the guide pulley and motor,
3P M/7 5 / 9 0 0 0 . 0 8 3GN N== =
Mass moment of inertia of the motor,
222
MM M( ) 200(0.09) 1.62 kg-m Im k== =
Mass moment of inertia of the intermediate gear,
222
II I( ) 375(0.225) 18.98 kg-m== =Im k

222
Mass moment of inertia of the drum and shaft,
22 2
DD D( ) 2250(0.6) 810 kg-mIm k== =
Mass moment of inertia of the guide pulley,
222
PP P( ) 200(0.45) 40.5 kg-mIm k== =
2
D/ 0.9/0.75 1.2 rad/sarα= = =

Chapter 3 : Kinetics of Motion    /G6C  59
*The bodies, which rebound after impact are called elastic bodies and the bodies which does not rebound at
all after its impact are called inelastic bodies.Since the speed of the drum ( ND) is 0.055 times the speed of motor ( NM), therefore angular
acceleration of the drum ( αD),
1.2 = 0.055 αMorαM = 1.2/0.055 = 21.8 rad/s2
We know that the equivalent mass moment of inertia of the system ( i.e. motor, intermediate

gear shaft and wheel,drum and two guide pulleys) referred to motor M,
I = IM + (G1)2 II + (G2)2 ID + 2 (G3)2 IP
  = 1.62 + (0.306)2 18.98 + (0.055)2 810 + 2 (0.083)2 40.5
  = 1.62 + 1.78 + 2.45 + 0.56 = 6.41 kg-m2
∴  Torque at motor to accelerate the system,
T1 = I.αM = 6.41 × 21.8 = 139.7 N-m
and torque at motor to overcome friction at intermediate gear, drum and two guide pulleys,
T2 = GI.FI + G2.FD + 2 G3.FP
     = 0.306 × 150 + 0.055 × 1125 + 2 × 0.83 × 150 N-m

= 45.9 + 61.8 + 25 = 132.7 N-m
Now for the rising rope and cage as shown in Fig. 3.15, tension in the rope between the
pulley and drum,
Q1= Weight of rising rope and cage + Force to accelerate rising rope
and cage (inertia force) + Frictional resistance
=m1.g + m1.a + F1 = 1150 × 9.81 + 1150 × 0.9 + 500
= 12 816 N
Similarly for the falling rope and cage, as shown in Fig. 3.15, tension in the rope between
the pulley and drum,
Q2= Weight of falling rope and cage – Force to accelerate falling

rope and cage (inertia force) – Frictional resistance
=m2.g – m2.a – F2 = 650 × 9.81 – 650 × 0.9 – 350 = 5441 N
∴  Torque at drum, TD = (Q1 – Q2) r = (12 816 – 5441) 0.75 = 5531 N-m
and torque at motor to raise and lower cages and ropes and to overcome frictional resistance
T3 = G2 × TD = 0.055 × 5531 = 304 N-m
∴ Total motor torque required,
T = T1 + T2 + T3 = 139.7 + 132.7 + 304 = 576.4 N-m Ans.
3.21. Collision of Two Bodies

Consider the impact between two bodies which move with different velocities along the
same straight line. It is assumed that the point of the impact lies on the line joining the centers ofgravity of the two bodies. The behaviour of these colliding bodies during the complete period ofimpact will depend upon the properties of the materials of which they are made. The material of thetwo bodies may be *perfectly elastic or perfectly inelastic.

In either case, the first effect of impact is approximately the same. The parts of each body
adjacent to the point of impact is deformed and the deformation will continue until the centre ofgravity of the two bodies are moving with the same velocity. Assuming that there are no externalforces acting on the system, the total momentum must remain constant.

60  /G6C   Theory of Machines
3.22. Collision of Inelastic Bodies
When two *inelastic bodies A and B, as shown in Fig. 3.16 ( a), moving with different ve-
locities, collide with each other as shown in Fig. 3.16 ( b), the two bodies will remain together after
impact and will move together with a common velocity.
Let m1 = Mass of first body A.
m2 = Mass of second body B.
u1 and u2 = Velocities of bodies A and B respectively before impact, and
v = Common velocity of bodies A and B after impact.

(a) Before impact. ( b) After impact.
Fig. 3.16. Collision of inelastic bodies.
A little consideration will show that the impact will take place only, if u1 is greater than u2.
Now according to principle of conservation of momentum,
Momentum before impact = Momentum after impact
  m1.u1 + m2.u2 = (m1 + m2) v
∴11 22
12..mu muvmm+=+... (i)
The loss of kinetic energy during impact may be obtained by finding out the kinetic energy

of the two bodies before and after impact. The difference between the two kinetic energies of thesystem gives the loss of kinetic energy during impact.
We know that the kinetic energy of the first body, before impact
2
111()2=mu
and kinetic energy of the second body, before impact
2
221()2=mu
∴ Total kinetic energy of the system before impact,
22
11 1 2 21 1() ()2 2=+Em u m u
When the two bodies move with the same velocity v after impact, then
Kinetic energy of the system after impact,
2

2
21 21()2=+Em m v
∴ Loss of kinetic energy during impact,
22 2
L12 1 1 2 2 12111.. ( )222EEE m u m u m m v=−= + − +Point of impact
*The impact between two lead spheres or two clay spheres is approximately  an inelastic impact.

Chapter 3 : Kinetics of Motion    /G6C  61
2
22 11 22
11 21 1 2
12.. 111.. ( )222mu mumu mu m mmm+=+− + +
. . . [From equation (i)]
2
22 11 22
11 22
12(. .) 11..22 2 ( )mu mumu mumm+=+−+
22 2
1 2 11 22 11 22
121() ( ) ( )2( )m m mu mu mu mumm =+ + ⋅ + ⋅ − ⋅ + ⋅+
. . . [Multiplying the numerator and denominator by ( m1 + m2)]
22 2 2 22
11 122 1 2 1 22
121.. .. ..2( )mu m mu mmu mumm =+ + + +
22 22
11 22 1 2 1 2.. 2mu mu m m u u  −−− 
22
122 121 1 2 1 2

22
121.. .. 22( )mmu mm u m mu umm =+ −+
22 2 12 12
12 1 2 1 2
12 12..2. ( )2( ) 2( )mm mmuu u u uumm mm =+ − = −++
This *loss of kinetic energy is used for doing the work in deforming the two bodies and is

absorbed in overcoming internal friction of the material. Since there will be no strain energy storedup in the material due to elastic deformation, therefore the bodies cannot regain its original shape.Hence the two bodies will adhere together and will move with reduced kinetic energy after impact.The reduction of kinetic energy appears as heat energy because of the work done in overcoming theinternal friction during deformation.
3.23. Collision of Elastic Bodies

When two elastic bodies, as shown in Fig. 3.17 ( a), collide with each other, they suffer a
change of form. When the bodies first touch, the pressure between them is zero. For a short timethereafter, the bodies continue to approach each other and the pressure exerted by one body over theother body increases. Thus the two bodies are compressed and deformed at the surface of contactdue to their mutual pressures.
Fig. 3.17.  Collision of elastic bodies.

If one of the bodies is fixed then the other will momentarily come to rest and then rebound.
However, if both the bodies are free to move, then each body will momentarily come to rest relativeto the other. At this instant, the pressure between the two bodies becomes maximum and thedeformation is also a maximum. At this stage the two bodies move with a **common velocity, as
shown in Fig. 3.17 ( b).

**This common velocity ( v) may be calculated as discussed in the previous article.*According to principle of conservation of energy, the energy cannot be lost.

62  /G6C   Theory of Machines
The work done in deforming the two bodies is stored up as strain energy. Since no energy is
absorbed in overcoming internal friction, therefore there will be no conversion of kinetic energy intoheat energy. Thus immediately after the instant at which the two bodies move with same velocity, thebodies begin to regain their original shape.This process of regaining the original shape is calledrestitution .

The strain energy thus stored is reconverted into kinetic energy and the two bodies
ultimately separates as shown in Fig. 3.17 ( c). In this case, the change of momentum of each body
during the second phase of impact ( i.e. when the bodies are separating) is exactly equal to the
change of momentum during the first phase of impact ( i.e. when the bodies are approaching or
colliding).
Let m
1 = Mass of the first body,
 u1 = Velocity of the first body before impact,

u1 = Velocity of the first body before impact,
 v1 = Velocity of the first body after impact,
             m2, u2 and v2 = Corresponding values for the second body, and
  v = Common velocity of the two bodies at the instant when
        compression has just ended.
∴ Change of momentum of first body during the second phase of impact
     = m1 (v1 – v)
and change of momentum of the same body during first phase of impact
     = m1 (v – u1)

= m1 (v – u1)
∴ m1 (v1 – v) = m1 (v – u1)      or     v1 = 2 v – u1... (i)
Similarly, for the second body, change of momentum of the second body during second
phase of impact
     = m2 (v2 – v)
and change of momentum of the second body during first phase of impact
    = m2 (v – u2)
∴  m2 (v2 – v) = m2 (v – u2)o rv2 = 2v – u2... (ii)
Subtracting equation (ii) from equation  (i), we get
v1 – v2 = (u2 – u1) = – (u1 – u2)  ... (iii)

Therefore, we see that the relative velocity of the two bodies after impact is equal and
opposite to the relative velocity of the two bodies before impact. Due to the fact that physical bodiesare not perfectly elastic, the relative velocity of two bodies after impact is always less than therelative velocity before impact. The ratio of the former to the latter is called coefficient of restitu-

tion and is represented by e. Mathematically, coefficient of restitution,
12
12Relativevelocityafterimpact
Relativevelocitybeforeimpact ( )vveuu−==−−
12
21vv
uu−=−or21
12vv
uu−
−
The value of e = 0, for the perfectly inelastic bodies and e = 1 for perfectly elastic bodies. In
case the bodies are neither perfectly inelastic nor perfectly elastic, then the value of  e lies between 0
and 1.

Chapter 3 : Kinetics of Motion    /G6C  63
The final velocities of the colliding bodies after impact may be calculated as discussed below:
Since the change of velocity of each body during the second phase of impact is e times the
change of velocity during first phase of impact, therefore for the first body,
v1 – v = e (v – u1)   or     v1 = v (1 + e) – e.u1...(iv)
Similarly for the second body,
v2 – v = e (v – u2)o rv2 = v (1 + e) – e.u2 ...(v)

When e = 1, the above equations  (iv) and (v) reduced to equations  (i) and (ii).
Notes : 1. The time taken by the bodies in compression, after the instant of collision, is called the  time of
compression or compression period .
2. The period of time from the end of the compression stage to the instant when the bodies separate ( i.e.
the time for which the restitution takes place) is called  time of restitution or restitution period .

3. The sum of compression period and the restitution  period is called  period of collision or period of
impact.
4. The velocities of the two bodies at the end of restitution period will be different from their common
velocity at the end of the compression period.
3.24. Loss of Kinetic Energy During Elastic Impact
Consider two bodies 1 and 2 having an elastic impact as shown in Fig. 3.17.
Let m1 = Mass of the first body,

v1 = Velocity of the first body after impact,
m2, u2 and v2 = Corresponding values for the second body,
e = Coefficient of restitution, and
EL= Loss of kinetic energy during impact.
   2
111.2=mu
Similarly, kinetic energy of the second body, before impact
   2
221.2=mu
∴  Total kinetic energy of the two bodies, before impact,
22
11 1 2 211.22=+ ⋅Em u m u    . . . (i)

22
Similarly, total kinetic energy of the two bodies, after impact
22
21 12 211.22=+ ⋅Em vm v               . . .  (ii)
∴  Loss of kinetic energy during impact,
           22 2 2
12 1 1 2 2 1 1 2 211 11.. .22 22LEEE m u m u m v m v  =−= + − + ⋅     

64  /G6C   Theory of Machines
   () ()22 22
11 22 11 221.. .2mu mu mv mv=+ − + ⋅
Multiplying the numerator and denominator by ( m1 + m2),
()()() ()()22 22
121 1 2 2 12 1 1 2 2
121.
2LEm m m u m u m m m v m v
mm =+ ⋅ + ⋅ − + + ⋅+
      ()()22 2 2 22
11 122 121 22
121.. . .2mu mmu mmu mumm =⋅ + ⋅ + ++
()22 2 2 22
11 122 121 22mv mmv mmv mv  −⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅
(){}22 22 2 2
11 22 121 2
121()
2mu mu mmu u
mm =⋅ + ⋅ + ⋅ ++
{}22 22 2 2
11 22 1 21 2 ( mv mv mmv v  −⋅ + ⋅ + ⋅ +

()() ( ) ( ) (){}22
1 1 22 12 1 2 121 2 12 1 2
121.. 2 . 22=+ − + − ++mu mu mmuu mm u u mmuumm
() ( ) ( ) ( ){}22
11 22 1 2 1 2 1 2 1 2 1 2 12..2 . 2−+− + − +mv mv mmvv mm v v mmvv
()() ( ){}
() ( ){}22
1 1 22 121 2
12
22
11 22 121 21.. .2
.. .mu m u mm u umm
mv mv mmv v=+ + −+
−++ −
We know that in an elastic impact,
Total momentum before impact = Total momentum after impact
i.e. m1.u1 + m2.u2 = m1.v1 + m2.v2

or ( m1.u1 + m2.u2 )2 =  (m1.v1 + m2.v2 )2   ... (Squaring both sides)
∴∴∴∴∴ Loss of kinetic energy due to impact,
()22
L 121 2 121 2
121.( ) .( )
2Em m u u m m v v
mm =− − −+
Substituting  v1 – v2 = e (u1 – u2) in the above equation,
()22 2
L1 2 1 2 1 2 1 2
121.( ) ..( )
2Em m u u m m e u u
 ()22 12
12
12.() ( 1 )2=− −+mmuu emm
Notes : 1. The loss of kinetic energy may be found out by calculating the kinetic energy of the system before

impact, and then by subtracting from it the kinetic energy of the system after impact.

Chapter 3 : Kinetics of Motion    /G6C  65
2.  For perfectly inelastic bodies, e = 0, therefore
2 12
L1 2
12.()2( )=−+mmEu umm  . . . (same as before)
3.  For perfectly elastic bodies, e = l, therefore EL = 0.
4. If weights (instead of masses) of the two bodies are given, then the same may be used in all the
relations.
Example 3.19. A sphere of mass 50 kg moving at 3 m/s overtakes and collides with another

sphere of mass 25 kg moving at 1.5 m/s in the same direction. Find the velocities of the two massesafter impact and loss of kinetic energy during impact in the following cases :
1. When the impact is inelastic,  2. When the impact
is elastic, and 3. When coefficient of restitution is 0.6 .
1 = 50 kg ; u1 = 3 m/s ;
m2 = 25 kg ; u2 = 1.5 m/s
1.  When the impact is inelastic
In case of inelastic impact, the two spheres adhere

after impact and move with a common velocity. We knowthat common velocity after impact,
11 22
12.. 50 3 25 1.52.5m s50 25+ ×+ ×== =++mu muvmm Ans.
and loss of kinetic energy during impact,
22 12
L1 2
12.5 0 2 5( ) (3 1.5) N-m2( ) 2(50 25)mmEu umm×=− =−++
=  18.75 N-m  Ans.
2. When the impact is elastic
Let v1 = Velocity of the first sphere immediately after impact, and
v2 = Velocity of the second sphere immediately after impact.

We know that when the impact is elastic, the common velocity of the two spheres is the same
i.e. common velocity, v = 2.5 m/s.
∴ v1 = 2v – u1 = 2 × 2.5 – 3 = 2 m/s  Ans.
and v2 = 2v – u2 = 2 × 2.5 – 1.5 = 3.5 m/s  Ans.
We know that during elastic impact, there is no loss of kinetic energy, i.e.  EL = 0 Ans.
3.  When the coefficient of restitution, e = 0.6
We know that    v1 = (1 + e) v – e.u1 = (1 + 0.6) 2.5 – 0.6 × 3 = 2.2 m/s Ans.

and       v2 = (1 + e) v – e.u2 = (1 + 0.6) 2.5 – 0.6 × 1.5 = 3.1 m/s  Ans.
Loss of kinetic energy during impact,
22 12
L1 2
12.() ( 1 )2( )mmEu u emm=− −+
22 50 25(3 1.5) (1 0.6 ) 12N-m2(50 25)×=− − =+Ans.

66  /G6C   Theory of Machines
Example 3.20. A loaded railway wagon has a mass of 15 tonnes and moves along a level
track at 20 km/h. It over takes and collides with an empty wagon of mass 5 tonnes, which is movingalong the same track at 12 km/h. If the each wagon is fitted with two buffer springs of stiffness 1000kN/m, find the maximum deflection of each spring during impact and the speeds of the wagonsimmediately after impact ends .

If the coefficient of re stitution for the buffer springs is 0.5, how would the final speeds be
affected and what amount of energy will be dissipated during impact  ?
1 = 15 t = 15 000 kg ; u1 = 20 km/h = 5.55 m/s ; m2 = 5 t = 5000 kg ;
u2 = 12 km/h = 3.33 m/s ; s = 1000 kN/m = 1 × 106 N/m ; e = 0.5
During impact when both the wagons are moving at the same speed ( v) after impact, the
magnitude of the common speed ( v) is given by
11 22

11 22
12. . 15 000 5.55 5000 3.335m/s15 000 5000mu muvmm+× + ×== =++ Ans.
Maximum deflection of each spring
Let x = Maximum deflection of each buffer spring during impact, and
s = Stiffness of the spring = 1000 kN/m = 1 × 106 N/m          ... (Given)
∴ Strain energy stored in one spring
26 2 3 211. 1 10 500 10 N-m22sx x x== × × × = ×
Since the four buffer springs (two in each wagon) are strained, therefore total strain energy
stored in the springs

= 4 × 500 × 103 x2 = 2 × 106 x2 N-m       ... (i)
Difference in kinetic energies before impact and during impact
22 12
12
12. 15 000 5000( ) (5.55 3.33) N-m2( ) 2(15 000 5000)mmuumm×=− = −++
= 9240 N-m ... (ii)
The difference between the kinetic energy
before impact and kinetic energy during impact isabsorbed by the buffer springs. Thus neglecting alllosses, it must be equal to strain energy stored in thesprings.
2 × 10
6 x2 = 9240

2 × 10
6 x2 = 9240
or x2 = 9240/2 × 106 = 0.00 462
∴ x = 0.068 m = 68 mm   Ans.
Speeds of the wagons immediately after impact ends
Immediately after impact ends, let v1 and v2 be the speeds of the loaded wagon and empty
wagon respectively.
We know that        v1 = 2v – u1 = 2 × 5 – 5.55 = 4.45 m/s  Ans.
and v2 = 2v – u2 = 2 × 5 – 3.33 = 6.67 m/s  Ans.
When the coefficient of restitution, e = 0.5 is taken into account, then
 v1 = (1 + e)v – e.u1 = (1+ 0.5) 5 – 0.5 × 5.55 = 4.725 m/s  Ans.

Chapter 3 : Kinetics of Motion    /G6C  67
and                 v2 = (1 + e)v – e.u2 = (1 + 0.5)5 – 0.5 × 3.33 = 5.635 m/s  Ans.
Amount of energy dissipated during impact
We know that amount of energy dissipated during impact,
22 2 12
L1 2
12.( ) (1 ) 9240 (1 0.5 )N-m2( )mmEu u emm=− − = −+
= 9240 × 0.75 = 6930 N-m  Ans.
Example 3.21.  Fig. 3.18 shows a flywheel A connected through a torsionally flexible spring

to one element C of a dog clutch. The other element D of the clutch is free to slide on the shaft butit must revolve with the shaft to which the flywheel B is keyed .
The moment of inertia of A and B are 22.5 kg-m
2
and 67.5 kg-m2 and the torsional stiffness of the spring is
225 N-m per radian. When the flywheel A is revolving at 150r.p.m. and the flywheel B is at rest, the dog clutch is suddenlyengaged. Neglecting all losses, find : 1. strain energy stored

in the spring,  2. the maximum twist of the spring, and 3. the
speed of flywheel when the spring regains its initial unstrainedcondition .
Solution. Given : I
A = 22.5 kg-m2 ; IB = 67.5 kg-m2 ; q = 225 N-m/rad ; NA = 150 r.p.m. or
ωA = 2π × 150/60 = 15.71 rad/s
Immediately after the clutch is engaged, the element C of the clutch comes to rest
momentarily. But the rotating flywheel A starts to wind up the spring, thus causing equal and oppo-

site torques to act on flywheels A and B. The magnitude of the torque increases continuously until
the speeds of flywheels A and B are equal. During this interval, the strain energy is stored in the
spring. Beyond this, the spring starts to unwind and the strain energy stored in the spring is recon-verted into kinetic energy of the flywheels.
Since there is no external torque acting on the system, therefore the angular momentum will

remain constant. Let ω be the angular speed of both the flywheels at the instant their speeds are equal.
∴ (I
A + IB) ω = IA . ωAorAA
AB. 22.5 15.713.93rad/s22.5 67.5I
IIω×ω= = =++
Kinetic energy of the system at this instant ( i.e. when speeds are equal),
22
2A B11( ) (22.5 67.5) (3.93) 695 N-m22EI I=+ ω = + =
and the initial kinetic energy of the flywheel A,
22
1A A11( ) 22.5 (15.71) 2776N-m22EI=ω = × =
1.  Strain energy stored in the spring
We know that strain energy stored in the spring

= E1 – E2 = 2776 – 695 = 2081 N-m Ans.
2.  Maximum twist of the spring
Let θ = Maximum twist of the spring in radians, and
q = Torsional stiffness of spring = 225 N-m/rad           ...(Given)Fig. 3.18

68  /G6C   Theory of Machines
We know that the strain energy,
        22 2112081 . 225 112.522q=θ = ×θ = θ
∴            θ2 = 2081/112.5 = 18.5
or θ = 4.3 rad = 4.3 × 180/ π = 246.3°  Ans.
3.  Speed of each flywheel when the spring regains its initial unstrained condition
Let  NA1 and NB1 be the speeds of the flywheels  A and B respectively, when the spring
regains its initial unstrained condition. We know that
A1 A A60 60 3.932 – 2 2 15022NN N Nω×  ==− = −   ππ  

= 75 – 150 = – 75 r.p.m.
 SimilarlyB1 B2 7 507 5 r . p . m .NN N=−= − =B...( 0)N=/G51
From above we see that when the spring regains its initial unstrained condition, the flywheel
A will revolve at 75 r.p.m. in the opposite direction to its initial motion and the flywheel B will
revolve at 75 r.p.m. in same direction as the initial motion of flywheel A. Ans.
EXERCISES
1.A flywheel fitted on the crank shaft of a steam engine has a mass of 1 tonne and a radius of gyration

0.4 m. If the starting torque of the engine is 650 J which may be assumed constant, find 1. Angularacceleration of the flywheel, and 2. Kinetic energy of the flywheel after 10 seconds from the start.
[Ans. 4.06 rad/s
2 ; 131.87 kN-m]
2.A load of mass 230 kg is lifted by means of a rope which is wound several times round a drum and

which then supports a balance mass of 140 kg. As the load rises, the balance mass falls. The drum hasa diameter of 1.2 m and a radius of gyration of 530 mm and its mass is 70 kg. The frictional resistanceto the movement of the load is 110 N, and that to the movement of the balance mass 90 N. Thefrictional torque on the drum shaft is 80 N-m.
Find the torque required on the drum, and also the power required, at the instant when the load has an

upward velocity of 2.5 m/s and an upward acceleration of 1.2 m/s
2.
[Ans. 916.2 N-m ; 4.32 kW]
3.A riveting machine is driven by a 3.5 kW motor. The moment of inertia of the rotating parts of the
machine is equivalent to 67.5 kg-m2 at the shaft on which the flywheel is mounted. At the commence-

ment of an operation, the flywheel is making 240 r.p.m. If closing a rivet occupies 1 second andcorresponds to an expenditure of 9 kN-m of energy, find the reduction of speed of the flywheel. Whatis the maximum rate at which rivets can be closed ? [Ans. 33.2 r.p.m. ; 24 per min ]
4.The drum of a goods hoist has a mass of 900 kg. It has an effective diameter of 1.5 m and a radius of

gyration of 0.6 m. The loaded cage has a mass of 550 kg and its frictional resistance in the verticalline of travel is 270 N. A maximum acceleration of 0.9 m/s
2 is required. Determine : 1. The necessary
driving torque on the drum, 2. The tension in the rope during acceleration, and 3. The power devel-oped at a steady speed of 3.6 m/s. [Ans. 4.64 kN-m ; 6.16 kN ; 22.3 kW]
5.A valve operating in a vertical direction is opened by a cam and closed by a spring and when fully

open the valve is in its lowest position. The mass of the valve is 4 kg and its travel is 12.5 mm and theconstant frictional resistance to the motion of the valve is 10 N. The stiffness of the spring is 9.6 N/mm and the initial compression when the valve is closed is 35 mm. Determine 1. the time takento close the valve from its fully open position, and 2. the velocity of the valve at the moment ofimpact. [Ans. 0.0161 s ; 1.4755 m/s]

Chapter 3 : Kinetics of Motion    /G6C  69
6.A railway truck of mass 20 tonnes, moving at 6.5 km/h is brought to rest by a buffer stop. The
buffer exerts a force of 22.5 kN initially and this force increases uniformly by 60 kN for each 1 mcompression of the buffer. Neglecting any loss of energy at impact, find the maximum compressionof the buffer and the time required for the truck to be brought to rest. [Ans. 0.73 m ; 0.707 s]

7.A cage of mass 2500 kg is raised and lowered by a winding drum of 1.5 m diameter. A brake drum

is attached to the winding drum and the combined mass of the drums is 1000 kg and their radius ofgyration is 1.2 m. The maximum speed of descent is 6 m/s and when descending at this speed, thebrake must be capable of stopping the load in 6 m. Find 1. the tension of the rope during stoppingat the above rate, 2. the friction torque necessary at the brake, neglecting the inertia of the rope, and3. In a descent of 30 m, the load starts from rest and falls freely until its speed is 6 m/s. The

and falls freely until its speed is 6 m/s. The brakeis then applied and the speed is kept constant at 6 m/s until the load is 10 m from the bottom. Thebrake is then tightened so as to give uniform retardation, and the load is brought to rest at thebottom. Find the total time of descent. [Ans. 32 kN ; 29.78 kN-m ; 7.27 s]

8.A mass of 275 kg is allowed to fall vertically through 0.9 m on to the top of a pile of mass 450 kg.
Assuming that the falling mass and the pile remain in contact after impact and that the pile ismoved 150 mm at each blow, find allowing for the action of gravity after impact, 1. The energy lostin the blow, and 2. The average resistance against the pile. [Ans. 13.3 kN ; 1.5 kN-m]
9.Fig. 3.19 shows a hammer of mass 6 kg and pivoted at A. It falls against a wedge of mass 1 kg

which is driven forward 6 mm, by the impact into a heavy rigid block. The resistance to the wedge
varies uniformly with the distance through which it moves, varying zero to  R newtons.
Fig. 3.19 Fig. 3.20
Neglecting the small amount by which the hammer rises after passing through the vertical through
A and assuming that the hammer does not rebound, find the value of R. [Ans. 8.38 kN]
10.Fig. 3.20 shows a tilt hammer, hinged at O, with its head A resting on top of the pile B. The

hammer, including the arm OA, has a mass of 25 kg. Its centre of gravity G is 400 mm horizontally
from O and its radius of gyration about an axis through G parallel to the axis of the pin O is 75 mm.

The pile has a mass of 135 kg. The hammer is raised through 45° to the position shown in dottedlines, and released. On striking the pile, there is no rebound. Find the angular velocity of thehammer immediately before impact and the linear velocity of the pile immediately after impact.Neglect any impulsive resistance offered by the earth into which the pile is being driven.
[Ans. 5.8 rad/s, 0.343 m/s]

11.The tail board of a lorry is 1.5 m long and 0.75 m high. It is hinged along the bottom edge to the
floor of the lorry. Chains are attached to the top co rners of the board and to the sides of the lorry

so that when the board is in a horizontal position the chains are parallel and inclined at 45° to thehorizontal. A tension spring is inserted in each chain so as to reduce the shock and these are adjustedto prevent the board from dropping below the horizontal. Each spring exerts a force of 60 N/mm ofextension.
Find the greatest force in each spring and the resultant force at the hinges when the board falls

freely from the vertical position. Assume that the tail board is a uniform body of mass 30 kg.
[Ans. 3636 N ; 9327 N]

70  /G6C   Theory of Machines
12.A motor drives a machine through a friction clutch which transmits 150 N-m while slip
occurs during engagement. For the motor, the rotor has a mass of 60 kg with radius of gyration 140mm and the inertia of the machine is equivalent to a mass of 20 kg with radius of gyration 80 mm.If the motor is running at 750 r.p.m. and the machine is at rest, find the speed after engaging theclutch and the time taken. [Ans. 70.87 rad/s ; 0.06 s]

13.A shaft carrying a rotor of moment of inertia 10 kg-m
2 revolves at a speed of 600 r.p.m. and is
engaged by means of a friction clutch to another shaft on the same axis having a moment of inertiaof 15 kg-m
2. If the second shaft is initially at rest, find 1. the final speed of rotation of the two
shafts together after slipping has ceased, 2. the time of slip if the torque is constant at 250 N-mduring slipping, and 3. the kinetic energy lost during the operation.

[Ans. 25.136 rad/s ; 1.5 s ; 11.85 kN-m]
14.A self-propelled truck of total mass 25 tonnes and wheel diameter 750 mm runs on a track for

which the resistance is 180 N per tonne. The engine develops 60 kW at its maximum speed of 2400r.p.m. and drives the axle through a gear box. Determine : 1. the time to reach full speed from reston the level if the gear reduction ratio is 10 to 1. Assume the engine torque to be constant and agearing efficiency of 94 per cent, and 2. the gear ratio required to give an acceleration of 0.15 m/s
2
on an up gradient of 1 in 70 assuming a gearing efficiency of 90 per cent. [Ans. 157 s ; 20.5]

15.A motor vehicle of mass 1000 kg has road wheels of 600 mm rolling diameter. The total moment of
inertia of all four road wheels together with the half shafts is 10 kg-m2, while that of the engine and
clutch is 1 kg-m2. The engine torque is 150 N-m, the transmission efficiency is 90 per cent and the

tractive resistance is constant at 500 N. Determine 1. Gear ratio between the engine and the roadwheels to give maximum acceleration on an upgrade of 1 in 20, and 2. The value of this maximumacceleration. [Ans. 13 ; 1.74 m/s
2]
16.In a mine hoist a loaded cage is raised and an empty cage is lowered by means of a single rope. This

rope passes from one cage, over a guide pulley of 1.2 m effective diameter, on to the winding drumof 2.4 m effective diameter, and then over a second guide pulley, also of 1.2 m effective diameter, tothe other cage. The drum is driven by an electric motor through a double reduction gear.
Determine the motor torque required, at an instant when the loaded cage has an upward accelera-
tion of 0.6 m/s
2, given the following data :
S.No. Part Maximum speed Mass Radius of Fricti onal

(r.p.m.)( kg)gyration (mm)resistance
1 Motor and pinion N 500 150     –
2. Intermediate gear shaft5N600 225 45 N-m
and attached wheel
3. Drum and attached gear20N3000 900 1500 N-m
4. Guide pulley, each – 125 450 30 N-m
5. Rising rope and cage – 10 000 – 2500 N
6. Falling rope and cage – 5000 – 1500 N
[Ans. 4003.46 N-m]
1.State Newton’s three laws of motion.
2.What do you understand by mass moment of inertia ? Explain clearly.

3.What is energy ? Explain the various forms of mechanical energies.
4.State the law of conservation of momentum.
5.Show that for a relatively small rotor being started from rest with a large rotor, the energy lost in the
clutch is approximately equal to that given to the rotor.

Chapter 3 : Kinetics of Motion    /G6C  71
6.Prove the relation for the torque required in order to accelerate a geared system.
7.Discuss the phenomenon of collision of elastic bodies.
8.Define the term ‘coefficient of restitution’.
1.The force which acts along the radius of a circle and directed ...... the centre of the circle is known
as centripetal force.
(a) away from ( b) towards
2.The unit of mass moment of inertia in S.I. units is

(a)m4(b) kgf-m-s2(c) kg-m2(d)N - m
3.Joule is a unit of
(a) force ( b)w o r k ( c) power ( d) none of these
4.The energy possessed by a body, for doing work by virtue of its position, is called
(a) potential energy ( b) kinetic energy
(c) electrical energy ( d) chemical energy
5.When a body of mass moment of inertia I (about a given axis) is rotated about that axis with an
angular velocity, then the kinetic energy of rotation is
(a) 0.5 I.ω (b)I.ω (c) 0.5 I.ω2(d)I.ω2

6.The wheels of a moving car possess
(a) potential energy only
(b) kinetic energy of translation only
(c) kinetic energy of rotation only
(d) kinetic energy of translation and rotation both.
7.The bodies which rebound after impact are called
(a) inelastic bodies ( b) elastic bodies
8.The coefficient of restitution for inelastic bodies is
(a)  zero ( b) between zero and one
(c) one ( d) more than one
9.Which of the following statement is correct ?

(a) The kinetic energy of a body during impact remains constant.
(b) The kinetic energy of a body before impact is equal to the kinetic energy of a body after impact.
(c) The kinetic energy of a body before impact is less than the kinetic energy of a body after
impact.
(d) The kinetic energy of a body before impact is more than the kinetic energy of a body after
impact.

impact.
10.A body of mass m moving with a constant velocity v strikes another body of same mass m moving
with same velocity but in opposite direction. The common velocity of both the bodies after collisionis
(a)v (b)2  v (c)4  v (d)8  v
ANSWERS
1.(b) 2.(c) 3.(b) 4.(a) 5.(c)
6.(d) 7.(b) 8.(a) 9.(d) 10.(b)

72  /G6C   Theory of Machines
Simple
Harmonic
Motion4
Features
2. Velocity and Acceleration of
a Particle Moving withSimple Harmonic Motion.
3. Differential Equation of
Simple Harmonic Motion.
4. Terms Used in Simple
Harmonic Motion.
5. Simple Pendulum.6. Laws of Simple Pendulum.7. Closely-coiled Helical
Spring.
8. Compound Pendulum.
9. Centre of Percussion.
10. Bifilar Suspension.
11. Trifilar Suspension
(Torsional Pendulum).4.1. Introduction
Consider a particle

moving round the circumfer-ence of a circle in ananticlockwise direction, witha constant angular velocity,as shown in Fig. 4.1. Let P
be the position of the particleat any instant and N be the
projection of P on the diam-
eter XX′ of the circle.
It will be noticed that when
the point P moves round the
circumference of the circle from X to
Y, N moves from X to O, when P
moves from Y to X ′, N moves from O
to X ′. Similarly when P moves from

X ′ to Y ′, N moves from X ′ to O and
finally when P moves from Y ′ to X, N
moves from O to X. Hence, as P
completes one revolution, the point N
completes one vibration about theFig. 4.1. Simple harmonic
motion.
A clock pendulumexecutes SimpleHarmonic Motion.
72

Chapter 4 : Simple Harmonic Motion    /G6C  73
point O. This to and fro motion of N is known as simple
harmonic motion  (briefly written as S.H.M.).
4.2. Velocity and Acceleration of a
Particle Moving with SimpleHarmonic Motion
Consider a particle, moving round the circumfer-
ence of a circle of radius r, with a uniform angular velocity
ω rad/s, as shown in Fig. 4.2. Let P be any position of the
particle after t seconds and θ be the angle turned by the
particle in t seconds. We know that
θ = ω.t

θ = ω.t
If N is the projection of P on the diameter X X ′,
then displacement of N from its mean position O is
              x  = r.cos θ = r.cos ω.t             ... (i)
The velocity of N is the component of the velocity
of P parallel to XX ′,  i.e.
22
Nsin . sinvv r r x=θ = ω θ = ω −                  ... (ii)
22... . , and sinvr r N Pr x=ω θ= = −/G51
Fig. 4.2. Velocity and acceleration of a particle.

A little consideration will show that velocity is maximum, when x = 0, i.e. when N passes
through O i.e., its mean position.
∴ vmax = ω.r
We also know that the acceleration of P is the centripetal acceleration whose magnitude is
ω2.r. The acceleration of N is the component of the acceleration of P parallel to XX ′ and is directed
towards the centre O, i.e.,
22
N.c o s .ar x=ω θ=ω ... (c o s )xr=θ/G51  ...(iii)
The acceleration is maximum when x = r i.e. when P is at X or  X′.
∴ amax = ω2.r

It will also be noticed from equation (iii) that when x = 0, the acceleration is zero i.e. N passes
through O. In other words, the acceleration is zero at the mean position. Thus we see from equation
(iii) that the acceleration of N is proportional to its displacement from its mean position O, and it is
Movements of a ship up and down in
a vertical plane about transverse axis
(called Pitching) and about longitude
(called rolling) are in Simple

74  /G6C   Theory of Machines
*always directed towards the centre O; so that the motion of N is simple harmonic.
In general, a body is said to move or vibrate with simple harmonic motion, if it satisfies the
following two conditions :
1. Its acceleration is always directed towards the centre, known as point of reference or
mean position  ;
2. Its acceleration is proportional to the distance from that point .
4.3. Differential Equation of Simple Harmonic Motion

We have discussed in the previous article that the displacement of N from its mean position O is
x = r.cos θ = r.cos ωt ... (i)
Differentiating equation  (i), we have velocity of N,
N.s i ndxvr tdt== ω ω ... (ii)
Again differentiating equation (ii), we have acceleration of N,
2
22
N2. . cos . cos .dxar t r t x
dt== − ω ω ω = − ω ω = − ω ... (iii)
... (/G51 r cos ωt = x)
or22
20dxx
dt+ω =
This is the standard differential equation for simple harmonic motion of a particle. The

solution of this differential equation is
x = A cos ω t + B sin ω t ... (iv)
where A and B are constants to be determined by the initial conditions of the motion.
In Fig. 4.2, when t = 0, x = r i.e. when points P and N lie at X, we have from equation (iv), A = r
Differentiating equation (iv),
.. s i n .c o sdxAt Btdt=− ω ω + ω ω
When 0, 0,dxtdt==  therefore, from the above equation, B = 0. Now the equation  (iv) becomes
x = r cos ω t . . . [Same as equation (i)]

The equations (ii) and (iii) may be written as
N.s i n .c o s ( / 2 )dxvr t r tdt== − ω ω = ω ω + π
and222
N2.c o s .c o s ( )dxar t r t
dt== − ω ω = ω ω + π
These equations show that the velocity leads the displacement by 90° and acceleration leads
the displacement by 180°.
*The negative sign shows that the direction of  acceleration is opposite to the direction in which x increases,
i.e. the acceleration is always directed towards the point O.

Chapter 4 : Simple Harmonic Motion    /G6C  75
4.4. Terms Used in Simple Harmonic Motion
The following terms, commonly used in simple harmonic motion, are important from the
subject point of view.
1.  Amplitude . It is the maximum displacement of a body from its mean position. In Fig. 4.2,
OX or OX ′ is the amplitude of the particle P. The amplitude is always equal to the radius of the
circle.
2. Periodic time . It is the time taken for one complete revolution of the particle.

∴       Periodic time, tp = 2π/ω seconds
We know that the acceleration,
a = ω2.xor2oraa
xxω= ω =
∴pDisplacement 22 2 secondsAccelerationxtaπ== π= πω
It is thus obvious, that the periodic time is independent of amplitude.
3. Frequency . It is the number of cycles per second and is the reciprocal of time period, tp.
∴Frequency, 11Hz22pantxω===ππ
Notes : 1.  In S.I. units, the unit of frequency is hertz (briefly written as Hz) which is equal to one cycle per
second.

second.
2. When the particle moves with angular simple harmonic motion, then the periodic time,
pAngulardisplacement22 sAngularaccelerationtθ=π =πα
and frequency, 1Hz2nα=πθ
Example 4.1. The piston of a steam engine moves with simple harmonic motion. The crank
rotates at 120 r.p.m. with a stroke of 2 metres. Find the velocity and acceleration of the piston, when
it is at a distance of 0.75 metre from the centre .

Solution. Given : N = 120 r.p.m.   or    ω = 2π × 120/60 = 4 π rad/s ; 2r = 2 m   or  r = 1 m;
x = 0.75 m
Velocity of the piston
We know that velocity of the piston,
22 24 1 (0.75) 8.31m/s vr x=ω − = π − =  Ans.
Acceleration of the piston
We also know that acceleration of the piston,
a = ω2.x = (4π)2 0.75 = 118.46 m/s2 Ans.
Example 4.2.  A point moves with simple harmonic motion. When this point is 0.75 metre

from the mid path, its velocity is 11 m/s and w hen 2 metres from the centre of its path its velocity is
3 m/s. Find its angular velocity, periodic time and its maximum acceleration .

76  /G6C   Theory of Machines
Solution. Given : When x = 0.75 m, v = 11 m/s ; when x = 2 m, v = 3 m/s
Angular velocity
Let ω = Angular velocity of the particle, and
r = Amplitude of the particle.
We know that velocity of the point when it is 0.75 m from the mid path ( v),
22 2 211 (0.75)rx r=ω − =ω − . . . (i)
Similarly, velocity of the point when it is 2 m from the centre ( v),
2232r=ω − . . . (ii)
Dividing equation (i) by equation  (ii),
22 22
22 22(0.75) (0.75) 11
322rr
rrω− −==
ω− −

22 22
322rr
rrω− −==
ω− −
Squaring both sides,
2
20.5625 121
9 4r
r−=
−
121 r2 – 484 = 9 r2 – 5.06 or 112 r2 = 478.94
∴ r2 = 478.94/ 112 = 4.276 or r = 2.07 m
Substituting the value of r in equation  (i),
2211 (2.07) (0.75) 1.93=ω − = ω
∴ω  = 11/1.93 = 5.7 rad/s Ans.
Periodic time
We know that periodic time,
tp = 2π/ω = 2π/5.7 = 1.1 s Ans.
Maximum acceleration
We know that maximum acceleration,
amax = ω2.r = (5.7)2 2.07 = 67.25 m/s2 Ans.
4.5. Simple Pendulum

A simple pendulum, in its simplest form, consists
of heavy bob suspended at the end of a light inextensibleand flexible string. The other end of the string is fixed atO, as shown in Fig. 4.3.
Let L= Length of the string,
m= Mass of the bob in kg,
W= Weight of the bob in newtons
= m.g, and
θ= Angle through which the string
is displaced.
Fig 4.3. Simple pendulum.

Chapter 4 : Simple Harmonic Motion    /G6C  77
When the bob is at A, the pendulum is in equilibrium position. If the bob is brought to  B or C
and released, it will start oscillating between the two positions B and C, with A as the mean position.
It has been observed that if the angle θ is very small (less than 4° ), the bob will have simple harmonic
motion. Now, the couple tending to restore the bob to the equilibrium position or restoring torque,
T = m.g sin θ × L

Since angle θ is very small, therefore sin θ = θ radians.
∴ T = m.g.L. θ
We know that the mass moment of inertia of the bob about an axis through the point of
suspension,
I = mass × (length)2 = m.L2
∴ Angular acceleration of the string,
2... .or
.mgL gTL
IL gmLθθ θα= = = =α
i.e.          Angular displacement
Angular accelerationL
g=
We know that the periodic time,
Displacement22AccelerationpLtg=π =π ... (i)
and frequency of oscillation,
11
2pgntL==π... (ii)

11
From above we see that the periodic time and the frequency of oscillation of a simple
pendulum depends only upon its length and acceleration due to gravity. The mass of the bob has noeffect on it.
Notes : 1.  The motion of the bob from one extremity to the other ( i.e. from B to C or C to B) is known as  beat
or swing. Thus one beat = 1
2 oscillation.
∴ Periodic time for one beat = Lgπ/

2. A pendulum, which executes one beat per second ( i.e. one complete oscillation in two seconds) is
known as a second’s  pen dulum.
4.6. Laws of Simple Pendulum
The following laws of a simple pendulum are important from the subject point of view :
1. Law of isochronism . It states, “The time period ( tp) of a simple pendulum does not depend
upon its amplitude of vibration and remains the same, provided the angular amplitude ( θ) does not
exceed 4°.”

2. Law of mass . It states, “The time period ( tp) of a simple pendulum does not depend upon
the mass of the body suspended at the free end of the string.”
3. Law of length . It states, “The time period ( tp) of a simple pendulum is directly propor-
tional to L, where L is the length of the string.”
4. Law of gravity . It states, “The time period ( tp) of a simple pendulum is inversely propor-
tional to g, where g is the acceleration due to gravity.”

78  /G6C   Theory of Machines
Fig. 4.4. Closely-coiled
helical spring.
*The differential equation for the motion of the spring is
22
22..o r –dx dx s xms xm dt dt=− = ... ()2Heres
mω=
The – ve sign indicates that the restoring force s.x is opposite to the direction of disturbing force.Note: The above laws of a simple pendulum are true from the equation of the periodic time i.e.
2/ptL g =π
4.7. Closely-coiled Helical Spring
Consider a closely-coiled helical spring, whose upper end is

fixed, as shown in Fig. 4.4. Let a body be attached to the lower end.Let AA be the equilibrium position of the spring, after the mass is
attached. If the spring is stretched up to BB and then released, the
mass will move up and down with simple harmonic motion.
Let m= Mass of the body in kg,
W= Weight of the body in newtons = m.g,
x= Displacement of the load below equilib-
rium position in metres,
s= Stiffnes of the spring in N/m i.e. restoring

force per unit displacement from the equi-librium position,
a= Acceleration of the body in m/s
2.
We know that the deflection of the spring,
.mg
sδ= ... (i)
Then disturbing force = m.a
and                   restoring force = s.x ... (ii)
m.a = s.x*    or   =xm
as

Chapter 4 : Simple Harmonic Motion    /G6C  79
Displacement22Accelerationpxta=π =π
22m
sgδ=π =π ... mg
sδ=/G51
and  frequency, 11 1
22pg sntm== =ππ δ
Note: If the mass of the spring ( m1) is also taken into consideration, then the periodic time,
1/32pmmts+=π seconds,
and                     frequency,
11Hz2/ 3snmm=π+
Example 4.3.  A helical spring, of negligible mass, and which is found to extend 0.25 mm

under a mass of 1.5 kg, is made to support a mass of 60 kg. The spring and the mass system isdisplaced vertically through 12.5 mm and released. Determine the frequency of natural vibration ofthe system. Find also the velocity of the mass, when it is 5 mm below its rest position.
Solution. Given : m = 60 kg ; r = 12.5 mm = 0.0125 m ; x = 5 mm = 0.005 m
Since a mass of 1.5 kg extends the spring by 0.25 mm, therefore a mass of 60 kg will extend
the spring by an amount,
0.2560 10mm = 0.01m1.5δ= × =

Frequency of the system
We know that frequency of the system,
11 9 . 8 14.982 2 0.01gn== =πδ πHz Ans.
Velocity of the mass
Let v = Linear velocity of the mass.
We know that angular velocity,
ω*9.81* 31.320.01g== =δ rad/s
and
22 2 231.32 (0.0125) (0.005) vr x=ω − = − = 0.36 m/s Ans.
4.8. Compound Pendulum
When a rigid body is suspended vertically, and it oscillates

with a small amplitude under the action of the force of gravity, thebody is known as compound pendulum , as shown in Fig. 4.5.
Let m= Mass of the pendulum in kg,
W= Weight of the pendulum in
newtons = m.g,
Fig. 4.5. Compound pendulum.
*We know that periodic time,
tp = 2π/ωorω = 2π/tp = 2π × n = 2π × 4.98 = 31.3 rad/s ...( ∵ n = 1/tp)

80  /G6C   Theory of Machines
kG= Radius of gyration
about an axisthrough the centreof gravity G and
perpendicular tothe plane ofmotion, and
h= Distance of point
of suspension O
from the centre ofgravity G of the
body.
If the pendulum is given a small
angular displacement θ, then the couple
tending to restore the pendulum to theequilibrium position OA,
T = mg sin θ × h = mgh sin θ
Since θ is very small, therefore sub-
stituting sin θ = θ radians, we get
T = mgh θ

T = mgh θ
Now, the mass moment of inertia about the axis of suspension O,
()22 2
GG. II m hm kh=+ = + . . . (By parallel axis theorem)
∴ Angular acceleration of the pendulum,
22 22
GG()mgh ghT
Imk h k hθθα= = =
++= constant × θ
We see that the angular acceleration is directly proportional to angular displacement,
therefore the pendulum executes simple harmonic motion.
∴22
G
.kh
gh+θ=α
Displacement22Accelerationptθ=π =πα
22

22
G2.kh
gh+=π ... (i)
and frequency of oscillation, 22
G. 11
2pghnt kh==π+... (ii)

Chapter 4 : Simple Harmonic Motion    /G6C  81
Notes : 1.  Comparing this equation with equation (ii) of simple pendulum, we see that the equivalent length of
a simple pendulum, which gives the same frequency as compound pendulum, is
22 2
GGkhkLhhh+== +
2. Since the equivalent length of simple pendulum ( L) depends upon the distance between the point of
suspension and the centre of gravity ( G), therefore L can be changed by changing the position of point of suspen-

sion. This will, obviously, change the periodic time of a compound pendulum. The periodic time will be minimumif L is minimum. For L to be minimum, the differentiation of L with respect to h must be equal to zero, i.e.
2
G0o r 0k dL dhdh dh h=+ = 

∴2
G
210k
h−+= or    kG = h
Thus the periodic time of a compound pendulum is minimum when the distance between the point of
suspension and the centre of gravity is equal to the radius of gyration of the body about its centre of gravity.

∴ Minimum periodic time of a compound pendulum,
G
()22pm i nktg=π . . . [Substituting h = kG in equation  (i)]
4.9. Centre of Percussion
The centre of oscillation is sometimes termed as  cen-
tre of percussion . It is defined as that point at which a blow
may be struck on a suspended body so that the reaction at thesupport is zero.
Consider the case of a compound pendulum suspended
at O as shown in Fig. 4.6. Suppose the pendulum is at rest in

the vertical position, and a blow is struck at a distance  L from
the centre of suspension. Let the magnitude of blow is F new-
tons. A little consideration will show that this blow will havethe following two effects on the body :
1. A force (F) acting at C will produce a linear motion
with an acceleration a, such that
F = m.a ... (i)
where m is the mass of the body.
2. A couple with moment equal to ( F × l ) which will tend to produce a motion of rotation in

the clockwise direction about the centre of gravity G. Let this turning moment ( F × l) produce an
angular acceleration ( α), such that
F × l = I
G × α ... (ii)
where IG is the moment of inertia of the body about an axis passing through G and parallel to the axis
of rotation.
From equation  (i)    a = F/m ... (iii)
and from equation  (ii),
G.Fl
Iα=Fig. 4.6. Centre of percusssion.

82  /G6C   Theory of Machines
Now corresponding linear acceleration of O,
       0 2G G.. ..
.Flh FlhahImk=α. = =                              ... (iv)
    2
GG(. )Im k=/G51
where kG is the radius of gyration of the body about the centre of
gravity G.
Since there is no reaction at the support when the body is
struck at the centre of percussion, therefore a should be equal to a0.
Equating equations (iii) and (iv),
2
G..
.FF l h
mmk=
or2
2 G
G.,a n dkkl h lh==      ... (v)

mmk=
or2
2 G
We know that the equivalent length of a simple pendulum,
       22
GGkhkLh l hhh+== + = +                        ... (vi)
From equations (v) and (vi), it follows that
1. The centre of percussion is below the centre of gravity
and at a distance 2
G/.kh
2. The distance between the centre of suspension and the centre of percussion is equal to the
equivalent length of a simple pendulum.
Note: We know that mass moment of inertia of the body about O,

22 2 2
OG G O .o r. . . I I mh mk mk mh=+ = +
∴222 2
OG .( ) kkhl h hh l hO G O C= + = + =+ =× ... 2
G(. )kl h=/G51
It is thus obvious that the centre of suspension ( O) and the centre of percussion ( C) are inter-changeable.
In other words, the periodic time and frequency of oscillation will be same, whether the body is suspended at the
point of suspension or at the centre of percussion.
Example 4.4.  A uniform thin rod, as shown in Fig. 4.7, has a mass of 1 kg and carries a

concentrated mass of 2.5 kg at B. The rod is hinged at A and is maintained in the horizontal positionby a spring of stiffness 1.8 kN/m at C.
Find the frequency of oscillation, neglecting t he effect of the mass of the spring .
Fig. 4.7
A pendulum clock designed
by Galileo. Galileo was the
first to deisgn a  clock based
on the relationship between
gravitational force ( g), length
of the pendulum ( l) and time
of oscillation ( t).

Chapter 4 : Simple Harmonic Motion    /G6C  83
Solution. Given : m = 1 kg ; m1 = 2.5 kg ; s = 1.8 kN/m = 1.8 × 103 N/m
We know that total length of rod,
l = 300 + 300 = 600 mm = 0.6 m
∴ Mass moment of inertia of the system about A,
IA= Mass moment of inertia of 1 kg about A + Mass moment of interia of
2.5 kg about A
2 222
11(0.6) .. 2.5 (0.6) 1.0233mlml =+= + = kg-m2
If the rod is given a small angular displacement  θ and then released, the extension of the spring,
δ = 0.3 sin θ = 0.3θ m

. . . ( ∵ θ is very small, therefore substituting sin θ = θ )
∴  Restoring force = s.δ = 1.8 × 103 × 0.3 θ = 540 θ N
and restoring torque about A  = 540 θ × 0.3 = 162 θ N-m ... (i)
We know that disturbing torque about A
=  IA × α = 1.02α N-m ... (ii)
1.02 α = 162 θorα/θ = 162/1.02 = 159
We know that frequency of oscillation,
1115922nα==πθ π= 2.01 Hz Ans.

Example 4.5.  A small flywheel of mass 85 kg is suspended in a vertical plane as a compound
pendulum. The distance of centre of gravity from the knife edge support is 100 mm and the flywheelmakes 100 oscillations in 145 seconds. Find the moment of inertia of the flywheel through the centreof gravity.
Solution. Given : m = 85 kg ; h = 100 mm = 0.1 m
Since the flywheel makes 100 oscillations in 145 seconds, therefore frequency of oscillation,
n = 100/145 = 0.69 Hz

Let L = Equivalent length of simple pendulum, and
k
G = Radius of gyration through C.G.
We know that frequency of oscillation ( n),
1 1 9.81 0.50.6922g
LL L== =ππ
∴ L= 0.5/0.69 = 0.7246  or   L = 0.525 m
We also know that equivalent length of simple pendulum ( L),
22 2 2
GG G (0.1)0.525 0.10.1 0.1kk khh+=+ =+=
22 2
G0.525 0.1 (0.1) 0.0425 mk=× −=

84      /G6C               Theory of Machines
and moment of inertia of the flywheel through the centre of gravity,
I = 2
G.mk = 85 × 0.0425 = 3.6 kg-m2 Ans.
Example 4.6.  The connecting rod of an oil engine has a mass of 60 kg, the distance between
the bearing centres is 1 metre. The diameter of the big end bearing is 120 mm and of the small end

bearing is 75 mm. When suspended vertically with a knife-edge through the small end, it makes 100oscillations in 190 seconds and with knife-edge through the big end it makes 100 oscillations in 165seconds. Find the moment of inertia of the rod in kg-m
2 and the distance of C.G. from the small end
centre.
Solution.  Given : m = 60 kg ;  h1 + h2 = 1 m ; d2* = 102 mm; d1* = 75 mm
Moment of inertia of the rod
First of all, let us find the radius of gyration of the connecting

rod about the centre of gravity ( i.e. kG).
Let h1 and h2= Distance of centre of gravity from
the small and big end centres respec-tively,
L
1 and L2= Equivalent length of simple
perdulum when the axis of oscilla-tion coincides with the small and bigend centres respectively
When the axis of oscillation coincides with the small end cen-
tre, then frequency of oscillation,
n
1 = 100/190 = 0.526 Hz
When the axis of oscillation coincides with the big end centre, the frequency of oscillation,

n2 = 100/165 = 0.606 Hz
We know that for a simple pendulum,
1
11Hz2gnL=π
∴ 1 22
19.810.9m
(2 ) (2 0.526)gL
n== =
ππ ×
Similarly2 22
29.810.67 m
(2 ) (2 0.606)gL
n== =
ππ ×
We know that    22
22 G1
1G 1 1 1
1()or . ( )khLk L h hh+== − ... (i)
Similarly22
G2 2 2 .( ) kL hh=− ... (ii)
From equations (i) and (ii), we have
L1.h1 – (h1)2 = L2.h2 – (h2)2
* Superfluous data.Connecting rod

Chapter 4 : Simple Harmonic Motion    /G6C  85
0.9 × h1 – (h1)2 = 0.67 (1 – h1) – (1 – h1)2. . . (/G51 h1 + h2 = 1 m)
= 0.67 – 0.67 h1 – 1 – (h1)2 + 2h1
  0.9 h1 + 0.67 h1 – 2 h1 = – 0.33     or      – 0.43 h1 = – 0.33
∴ h1 = 0.33/0.43 = 0.767 m
Substituting the value of h1 in equation (i), we have
22 2
G0.9 0.767 (0.767) 0.69 0.59 0.1mk= × − =−=
We know that mass moment of inertia of the rod,
 2
G.Im k== 60 × 0.1 =  6 kg-m2 Ans.
Distance of C.G. from the small end centre

We have calculated above that the distance of C.G. from the small end centre,
h1 = 0.767 m  Ans.
Example 4.7.  A uniform slender rod 1.2 m long is fitted with a transverse pair of knife-
edges, so that it can swing in a vertical plane as a compound pendulum. The position of the knifeedges is variable. Find the time of swing of the rod, if 1. the knife edges are 50 mm from one end of

the rod, and 2. the knife edges are so placed that the time of swing is minimum.
In case (1) find also the maximum angular velocity and the maximum angular acceleration
of the rod if it swings through 3° on either side of the vertical .
Solution. Given : l = 1.2 m ; θ = 3° = 3 × π /180 = 0.052 rad
1. Time of swing of the rod when knife edges are 50 mm
Since the distance between knife edges from one end of the rod is 50 mm = 0.05 m, therefore
distance between the knife edge and C.G. of the rod,

1.20.05 0.55m2h=− =
We know that radius of gyration of the rod about C.G.,
k*
G1.20.35m
12 12l===
∴ Time of swing of the rod,
        22 22
G (0.35) (0.55)22.9 . 8 1 0 . 5 5pkhtgh+ +=π =π×
= 1.76 s Ans.
2. Minimum time of swing
We know that minimum time of swing,
G
()2 20 . 3 52 2 1.68s9.81pm i nktg×=π =π =  Ans.
*We know that mass moment of inertia of the rod about an axis through C.G.
I = m. l2/12
Also I = m.k2or k2 = I/m = m.l2/12 × m = l2/12       or k = 12l

86  /G6C   Theory of Machines
Maximum angular velocity
In case (1), the angular velocity,
ω = 2π/tp = 2π/1.76 = 3.57 rad/s
We know that maximum angular velocity,
ωmax = ω.θ = 3.57 × 0.052 = 0.1856 rad/s Ans.
Maximum angular acceleration
We know that maximum angular acceleration,
αmax = ω2.θ = (3.57)2 × 0.052 = 0.663 rad/s2 Ans.
Example 4.8.  The pendulum of an Izod impact testing machine has a mass of 30 kg. Its

centre of gravity is 1.05 m from the axis of suspension and the striking knife is 150 mm below the
centre of gravity. The time for 20 small free oscillations is 43.5 seconds. In making a test the pendu-lum is released from an angle of 60° to the vertical. Determine :
1. the position of the centre of percussion relative to the striking knife and the striking veloc-
ity of the pendulum, and 2. the impulse on the pendulum and the sudden change of axis reaction

when a specimen giving an impact value of 55 N-m is broken.
Solution. Given : m = 30 kg ; OG = h = 1.05 m ; AG = 0.15 m
Since the time for 20 small free oscillations is 43.5 s, therefore frequency of oscillation,
200.46 Hz43.5n==
1. The position of centre of percussion relative to the striking knife and the striking velocity of the
pendulum
Let L= Equivalent length of simple pendulum,
kG= Radius of gyration of the pendulum about the centre of gravity, and

kO= Radius of gyration of the pendulum about O.
We know that the frequency of oscillation,
1
2gnL=π
or229.81
(2 ) (2 0.46)gL
n==
ππ ×
       = 1.174 m
∴ Distance of centre of percussion ( C) from
the centre of gravity ( G),
CG = OC – OG = L – OG
     = 1.174 – 1.05 = 0.124 m
and distance of centre of percussion ( C) from knife edge A,
AC = AG – CG = 0.15 – 0.124 = 0.026 m Ans.
We know that22
O( ) . 1.174 1.05 1.233 mkh l h L h=+ = = × =

A little consideration will show that the potential energy of the pendulum is converted into
kinetic energy of the pendulum before it strikes the test piece. Let v and ω be the linear and angular
velocity of the pendulum before it strikes the test piece.Fig. 4.8

Chapter 4 : Simple Harmonic Motion    /G6C  87
∴ m.g.h1 = 22 2
O11.. .22mv mk=ω ...  (∵ v = kO.ω)
30 × 9.81 × 1.05 (1 – cos 60°) = 2 130 1.2332×× ω or 154.5 = 18.5 ω2
∴ω2 = 154.5/18.5 = 8.35 or ω = 2.9 rad/s
∴  Velocity of striking = ω × OA = 2.9 (1.05 + 0.15) = 3.48 m/s Ans.
2. Impulse on the pendulum and sudden change of axis reaction
It is given that the impact value of the specimen ( i.e. the energy used for breaking the specimen)

is 55 N-m. Let ω1 be the angular velocity of the pendulum immediately after impact. We know that
Loss of kinetic energy  = 22 2 22
1O 111() . () 5 522Im kω− ω = ω− ω = N-m
∴   22
1130 1.233 (2.9 ) 552×× − ω =
18.5 (8.41 – 2
1ω) = 55 or2
1ω = 8.41 – 55/18.5 = 5.44
∴ω1 = 2.33 rad/s
Let P and Q be the impulses at the knife edge A and at the pivot O respectively as shown in
Fig. 4.8.
∴ P + Q = Change of linear momentum
= m.h (ω – ω1) = 30 × 1.05 (2.9 – 2.33) = 17.95 ... (i)
Taking moments about G,

0.15 P – 1.05 Q = Change of angular momentum
= 22 2
G1 O 1.( ) ( ) ( )mk m k hω−ω = − ω−ω
= 30 (1.233 – 1.052) (2.9 – 2.33) = 2.27 ... (ii)
From equations  (i) and (ii),
P = 17.6 N-s; and Q = 0.35 N-s Ans.
∴ Change in axis reaction when pendulum is vertical
= Change in centrifugal force
                                    22 2 2
1( ) 30 (2.9 2.33 )mh=ω − ω= − 1.05 = 94 N Ans.
4.10. Bifilar Suspension

The moment of inertia of a body may be determined experimentally by an apparatus called
bifilar suspension . The body whose moment of inertia is to be determined (say AB) is suspended by
two long parallel flexible strings as shown in Fig. 4.9. When the body is twisted through a small angleθ about a vertical axis through the centre of gravity G, it will vibrate with simple harmonic motion in
a horizontal plane.
Let m= Mass of the body,

k
G= Radius of gyration about an axis through the centre of gravity,

88  /G6C   Theory of Machines
I= Mass moment of inertia of the
body about a vertical axis
through 2
G.,Gm k=
 l=Length of each string,
x= Distance of A from G (i.e. AG),
y= Distance of B from G (i.e. BG),
θ= Small angular displacement of
the body from the equilibriumposition in the horizontal plane,
φ
A and φB= Corresponding angular dis-
placements of the strings, and
α= Angular acceleration towards
the equilibrium position.
When the body is stationary, the tension in the strings are
given by

given by
AB.. ..,a n dmgy mgxTTxy xy==++    ...(Taking moments about B and A respectively,)
When the body is displaced from its equilibrium position in a horizontal plane through a
small angle θ, then the angular displacements of the strings are given by
AA′ = φA.l = x.θ; and BB′ = φB.l = y.θ
∴ AB. .;a n dy x
llθ θφ= φ=
Component of tension TA in the horizontal plane, acting normal to A′B′ at  A′ as shown in Fig. 4.9
              = AA.. ......()mgy mgxyxTxy l l xyθ θφ= × =++

Component of tension TB in the horizontal plane, acting normal to A′ B′ at B′ as shown in Fig. 4.9
              = BB.. . .....()mgx y mgxyTxy l l xyθθφ= × =++
These components of tensions TA and TB are equal and opposite in direction, which gives rise
to a couple. The couple or torque applied to each string to restore the body to its initial equilibriumposition, i.e. restoring torque
AA BB.. ..Tx Ty=φ+φ
.... ....()()mgxy mgxyxylx y lθθ=+ =+... (i)
and accelerating (or disturbing) torque
2

2
G.. .Im k=α = α ... (ii)
                     2
G......mgxymklθ=α    or   2
G.
..kl
gxyθ=αFig. 4.9. Bifilar suspension.

Chapter 4 : Simple Harmonic Motion    /G6C  89
i.e.  2
G. Angular displacement
Angulara cceleration . .kl
gxy=
2
G. Angular displacement22Angular acceleration . .pkltgxy=π =π
G2..lkgxy=π
and frequency,  
G.. 11
2pgxyntk l==π
Note : The bifilar suspension is usually used for finding the moment of inertia of a connecting rod of an engine.
In this case, the wires are attached at equal distances from the centre of gravity of the connecting rod ( i.e. x = y)

so that the tension in each wire is same.
Example 4.9.  A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by
two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centreof gravity. If the rod makes 20 oscillations in 40 seconds, find the radius of gyration and the massmoment of inertia of the rod about a vertical axis through the centre of gravity.
Solution. Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m

Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation,
n = 20/40 = 0.5 Hz
Radius of gyration of the connecting rod
Let k
G = Radius of gyration of the connecting rod.
GG9.81 0.12 0.12 ..1 1 0.05350.52 2 1.25gxy
kl k k××== =ππ
∴ kG = 0.0535/0.5 = 0.107 m = 107 mm Ans.
Mass moment of inertia of the connecting rod
We know that mass moment of inertia,
I = m (kG)2 = 1.5 (0.107)2 = 0.017 kg-m2 Ans.

4.11. Trifilar Suspension (Torsional Pendulum)
It is also used to find the moment of inertia of a body experi-
mentally. The body (say a disc or flywheel) whose moment of inertiais to be determined is suspended by three long flexible wires A, B
and C, as shown in Fig. 4.10. When the body is twisted about its axis
through a small angle θ and then released, it will oscillate with simple
harmonic motion.

k
G= Radius of gyration about an axis
through c.g.,
I= Mass moment of inertia of the disc about an axis through O and per-
pendicular to it = m.k2,Fig. 4.10.  Trifilar suspension.

90  /G6C   Theory of Machines
l= Length of each wire,
r= Distance of each wire from the axis of the disc,
θ= Small angular displacement of the disc,
φ= Corresponding angular displacement of the wires, and
α= Angular acceleration towards the equilibrium position.
Then, for small displacements,
r. θ = l. φorφ = r.θ/l
Since the three wires are attached symmetrically with respect to the axis, therefore the ten-
sion in each wire will be one-third of the weight of the body.

∴ Tension in each wire  = m.g/3
Component of the tension in each wire perpendicular to  r
.. s i n .. ...
33 3mg mg mgr
lφφ θ== = . . . ( ∵ φ is a small angle, and φ = r.θ/l)
∴ Torque applied to each wire to restore the body to its initial equilibrium position i.e.
restoring torque
           2... .. .
33mgr mgrrllθθ=× =
Total restoring torque applied to three wires,
            22.. . .. .33mgr mgrTllθθ=× = ... (i)
We know that disturbing torque
            = I.α = 2
G..mkα ... (ii)

= I.α = 2
Equating equations  (i) and (ii),
            2
2
G.. ...mgrmklθ=α or       2
G
2.
.lk
grθ=α
i.e.  2
G
2. Angular displacement
Angular acceleration .lk
gr=
2
GG
2.2 Angular displacement22Angulara cceleration .plk k ltrg grπ=π =π =
and frequency, 
G1
2pgrntk l==π
Example 4.10. In order to find the radius of gyration of a car, it is suspended with its axis

vertical from three parallel wires 2.5 metres long. The wires are attached to the rim at points spaced
120° apart and at equal distances 250 mm from the axis.
It is found that the wheel makes 50 torsional oscillations of small amplitude about its axis in
170 seconds. Find the radius of gyration of the wheel.
Solution. Given : l = 2.5 m; r = 250 mm = 0.25 m;
Since the wheel makes 50 torsional oscillations in 170 seconds, therefore frequency of
oscillation,
n = 50/170 = 5/17 Hz

Let kG = Radius of gyration of the wheel

Chapter 4 : Simple Harmonic Motion    /G6C  91
GG G5 0.25 9.81 0.079
17 2 2 2.5rg
kl k k== =ππ
∴ kG = 0.079 × 17/5 = 0.268 m = 268 mm Ans.
Example 4.11.  A connecting rod of mass 5.5 kg is placed on a horizontal platform whose

mass is 1.5 kg. It is suspended by three equal wires, each 1.25 m long, from a rigid support. The wiresare equally spaced round the circumference of a circle of 125 mm radius. When the c.g. of theconnecting rod coincides with the axis of the circle, the platform makes 10 angular oscillations in 30seconds. Determine the mass moment of inertia about an axis through its c.g.
1 = 5.5 kg ; m2 = 1.5 kg ; l = 1.25 m ; r = 125 mm = 0.125 m

Since the platform makes 10 angular oscillations in 30 s, therefore frequency of oscillation,
n = 10/30 = 1/3 Hz
Let kG = Radius of gyration about an axis through the c.g.
We know that frequency of oscillation ( n)
GG G1 0.125 9.81 0.056
32 2 1 . 2 5gr
kl k k== =ππ
∴ kG = 0.056 × 3 = 0.168 m
and mass moment of inertia about an axis through its c.g.,
22 2 2
G1 2 G. ( ) (5.5 1.5) (0.168) kg-mIm k mmk== + = +
= 0.198 kg-m2 Ans.
EXERCISES

EXERCISES
1.A particle, moving with simple harmonic motion, performs 10 complete oscillations per minute and
its speed, when at a distance of 80 mm from the centre of oscillation is 3/5 of the maximum speed.
Find the amplitude, the maximum acceleration and the speed of the particle, when it is 60 mm fromthe centre of the oscillation. [Ans. 100 mm ; 109.6 mm/s
2 ; 83.76 mm/s]
2.A piston, moving with a simple harmonic motion, has a velocity of 8 m/s, when it is 1 metre from the

centre position and a velocity of 4 m/s, when it is 2 metres from the centre. Find : 1. Amplitude, 2.Periodic time, 3. Maximum velocity, and 4. Maximum acceleration.
[Ans. 2.236 m ; 1.571 s ; 8.94 m/s ; 35.77 m/s
2]
3.The plunger of a reciprocating pump is driven by a crank of radius 250 mm rotating at 12.5 rad/s.
Assuming simple harmonic motion, determine the maximum velocity and maximum acceleration ofthe plunger. [Ans. 3.125 m/s ; 39.1 m/s
2]

2]
4.A part of a machine of mass 4.54 kg has a reciprocating motion which is simple harmonic in character.
It makes 200 complete oscillations in 1 minute. Find : 1. the accelerating force upon it and its velocitywhen it is 75 mm, from midstroke ; 2. the maximum accelerating force, and 3. the maximum velocityif its total stroke is 225 mm i.e. if the amplitude of vibration is 112.5 mm.
[Ans. 149.5 N ; 1.76 m/s ; 224 N ; 2.36 m/s]

5.A helical spring of negligible mass is required to support a mass of 50 kg. The stiffness of the spring
is 60 kN/m. The spring and the mass system is displaced vertically by 20 mm below the equilibriumposition and then released. Find : 1. the frequency of natural vibration of the system ; 2. the velocityand acceleration of the mass when it is 10 mm below the rest position.
[Ans. 5.5 Hz ; 0.6 m/s ; 11.95 m/s
2]

2]
6.A spring of stiffness 2 kN/m is suspended vertically and two equal masses of 4 kg each are attached to
the lower end. One of these masses is suddenly removed and the system oscillates. Determine : 1. theamplitude of vibration, 2. the frequency of vibration, 3. the velocity and acceleration of the mass when

92  /G6C   Theory of Machines
passing through half amplitude position, and 4. kinetic energy of the vibration in joules.
 [Ans. 0.019 62 m ; 3.56 Hz ; 0.38 m/s , 4.9 m/s2 ; 0.385 J]
7.A vertical helical spring having a stiffness of 1540 N/m is clamped at its upper end and carries a mass

of 20 kg attached to the lower end. The mass is displaced vertically through a distance of 120 mm andreleased. Find : 1. Frequency of oscillation ; 2. Maximum velocity reached ; 3. Maximum accelera-tion; and 4. Maximum value of the inertia force on the mass.
[Ans. 1.396 Hz ; 1.053 m/s ; 9.24 m/s
2 ; 184.8 N]
8.A small flywheel having mass 90 kg is suspended in a vertical plane as a compound pendulum. The

distance of centre of gravity from the knife edge support is 250 mm and the flywheel makes 50oscillations in 64 seconds. Find the moment of inertia of the flywheel about an axis through the centreof gravity. [Ans. 3.6 kg-m
2]
9.The connecting rod of a petrol engine has a mass 12 kg. In order to find its moment of inertia it is

suspended from a horizontal edge, which passes through small end and coincides with the small endcentre. It is made to swing in a vertical plane, such that it makes 100 oscillations in 96 seconds. If thepoint of suspension of the connecting rod is 170 mm from its c.g., find : 1.  radius of gyration about anaxis through its c.g., 2. moment of inertia about an axis through its c.g., and 3. length of the equivalent
simple pendulum. [Ans. 101 mm ; 0.1224 kg-m
2 ; 0.23 m]

2 ; 0.23 m]
10.A connecting rod of mass 40 kg is suspended vertically as a compound pendulum. The distance between

the bearing centres is 800 mm. The time for 60 oscillations is found to be 92.5 seconds when the axis ofoscillation coincides with the small end centre and 88.4 seconds when it coincides with the big endcentre. Find the distance of the centre of gravity from the small end centre, and the moment of inertia ofthe rod about an axis through the centre of gravity. [Ans. 0.442 m ; 2.6 kg-m
2]
11.The following data were obtained from an experiment to find the moment of inertia of a pulley by

bifilar suspension :Mass of the pulley = 12 kg ; Length of strings = 3 m ; Distance of strings on either side of centre ofgravity = 150 mm ; Time for 20 oscillations about the vertical axis through c.g. = 46.8 secondsCalculate the moment of inertia of the pulley about the axis of rotation.
[Ans. 0.1226 kg-m
2]
12.In order to find the moment of inertia of a flywheel, it is suspended in the horizontal plane by three

wires of length 1.8 m equally spaced around a circle of 185 mm diameter. The time for 25 oscillationsin a horizontal plane about a vertical axis through the centre of flywheel is 54 s. Find the radius ofgyration and the moment of inertia of the flywheel if it has a mass of 50 kg.
[Ans. 74.2 mm; 0.275 kg-m
2]
1.Explain the meaning of S.H.M. and give an example of S.H.M.
2.Define the terms amplitude, periodic time, and frequency as applied to S.H.M.

3.Show that when a particle moves with simple harmonic motion, its time for a complete
oscillation is independent of the amplitude of its motion.
4.Derive an expression for the period of oscillation of a mass when attached to a helical spring.
5.What is a simple pendulum ? Under what conditions its motion is regarded as simple harmonic?
6.Prove the formula for the frequency of oscillation of a compound pendulum. What is the length of a

simple pendulum which gives the same frequency as compound pendulum ?
7.Show that the minimum periodic time of a compound pendulum is
G
()22pm i nktg=π
where kG is the radius of gyration about the centre of gravity.
8.What do you understand by centre of percussion ? Prove that it lies below the centre of gravity of the
body and at a distance 2
G/kh, where kG is the radius of gyration about c.g. and h is the distance
between the centre of suspension and centre of gravity.

Chapter 4 : Simple Harmonic Motion    /G6C  93
9.Describe the method of finding the moment of inertia of a connecting rod by means of bifilar suspen-
sion. Derive the relations for the periodic time and frequency of oscillation.
10.What is a torsional pendulum ? Show that periodic time of a torsional pendulum is
G2
pkltrgπ=
where                   kG = Radius of gyration,
l = Length of each wire, and
r = Distance of each wire from the axis of the disc.

1.The periodic time ( tp) is given by
(a)ω/2  π (b)2  π/ω (c)2  π × ω (d)π/ω
2.The velocity of a particle moving with simple harmonic motion is . . . . at the mean position.
(a) zero ( b) minimum ( c) maximum
3.The velocity of a particle ( v) moving with simple harmonic motion, at any instant is given by
(a)22rxω− (b)22xrω− (c)222rxω− (d)222xrω−
4.The maximum acceleration of a particle moving with simple harmonic motion is
(a)ω (b)ω.r (c)ω2.r (d)ω2/r

5.The frequency of oscillation for the simple pendulum is
(a)1
2L
gπ(b)1
2g
Lπ(c) 2L
gπ (d)2g
Lπ
6.When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the
force of gravity, the body is known as(a) simple pendulum ( b) torsional pendulum
(c) compound pendulum ( d) second’s pendulum
7.The frequency of oscillation of a compound pendulum is
(a)
22
G.1
2gh
khπ+(b)22
G1
2.kh
gh+
π(c)22
G.2gh
khπ
+(d)22
G2.kh
gh+π

G1
2.kh
gh+
π(c)22
G.2gh
khπ
+(d)22
G2.kh
gh+π
where     kG  =  Radius of gyration about the centroidal axis, and
      h  =    Distance between the point of suspension and centre of gravity of the body.
8.The equivalent length of a simple pendulum which gives the same frequency as the compound pendulum
is
(a)22
Gh
kh+ (b)22
Gkh
h+
(c)2
22
Gh
kh+(d)22
G
2kh
h+
9.The centre of percussion is below the centre of gravity of the body and is at a distance equal to
(a)h/kG(b)h.kG(c)h2/kG(d)2
G/kh

G/kh
10.The frequency of oscillation of a torsional pendulum is
(a)G2kg
rlπ(b)
G2gr
klπ(c)G2kl
rgπ
(d)
G2rl
kgπ
ANSWERS
1.(b) 2.(c) 3.(a) 4.(c) 5.(b)
6.(c) 7.(a) 8.(b) 9.(d) 10.(b)

94  /G6C   Theory of Machines
5.1. Introduction
We have already discussed that a machine is a de-
vice which receives energy and transforms it into some use-ful work. A machine consists of a number of parts or bodies.In this chapter, we shall study the mechanisms of the variousparts or bodies from which the machine is assembled. This isdone by making one of the parts as fixed, and the relativemotion of other parts is determined with respect to the fixedpart.
5.2. Kinematic Link or Element

Each part of a machine, which moves relative to some
other part, is known as a kinematic link  (or simply link) or
element. A link may consist of several parts, which are rig-

idly fastened together, so that they do not move relative toone another. For example, in a reciprocating steam engine,as shown in Fig. 5.1, piston, piston rod and crosshead consti-tute one link ; connecting rod with big and small end bear-ings constitute a second link ; crank, crank shaft and flywheela third link and the cylinder, engine frame and main bearingsa fourth link.
94Simple
Mechanisms5
Features
1.  Introduction.

94Simple
Features
2. Kinematic Link or Element.3. Types of Links.4. Structure.5. Difference Between a
Machine and a Structure.
6. Kinematic Pair.7. Types of Constrained
Motions.
8. Classification of Kinematic
Pairs.
9. Kinematic Chain.
10. Types of Joints in a Chain.
11. Mechanism.
12. Number of Degrees of
Freedom for PlaneMechanisms.
13. Application of Kutzbach
Criterion to PlaneMechanisms.
14.  Grubler's Criterion for
Plane Mechanisms.

15. Inversion of Mechanism.16. Types of Kinematic Chains.17. Four Bar Chain or Quadric
Cycle Chain.
18. Inversions of Four Bar
Chain.
19.  Single Slider Crank Chain.
20. Inversions of Single Slider
Crank Chain.
21. Double Slider Crank Chain.22. Inversions of Double Slider

Chapter 5 : Simple Mechanisms    /G6C  95
Fig. 5.1. Reciprocating steam engine.
A link or element need not to be a
rigid body, but it must be a  resistant body . A
body is said to be a resistant body if it iscapable of transmitting the required forceswith negligible deformation. Thus a linkshould have the following two characteristics:
1. It should have relative motion, and
2. It must be a resistant body.
5.3. Types of Links
In order to transmit motion, the driver

and the follower may be connected by the following three types of links :
1. Rigid link . A rigid link is one which does not undergo any deformation while transmitting
motion. Strictly speaking, rigid links do not exist. However, as the deformation of a connecting rod,crank etc. of a reciprocating steam engine is not appreciable, they can be considered as rigid links.

2. Flexible link . A flexible link is one which is partly deformed in a manner not to affect the
transmission of motion. For example, belts, ropes, chains and wires are flexible links and transmittensile forces only.
3. Fluid link . A fluid link is one which is formed by having a fluid in a receptacle and the
motion is transmitted through the fluid by pressure or compression only, as in the case of hydraulicpresses, jacks and brakes.
5.4. Structure

It is an assemblage of a number of resistant bodies (known as members) having no relative
motion between them and meant for carrying loads having straining action. A railway bridge, a rooftruss, machine frames etc., are the examples of a structure.
5.5. Difference Between a Machine and a Structure
The following differences between a machine and a structure are important from the subject
point of view :

1. The parts of a machine move relative to one another, whereas the members of a structure
do not move relative to one another.
2. A machine transforms the available energy into some useful work, whereas in a structure
no energy is transformed into useful work.
3. The links of a machine may transmit both power and motion, while the members of a
structure transmit forces only.Piston and piston rod of an IC engine.

96  /G6C   Theory of Machines
5.6. Kinematic Pair
The two links or elements of a machine, when in contact with each other, are said to form a
pair. If the relative motion between them is completely or successfully constrained ( i.e. in a definite
direction), the pair is known as kinematic pair .
First of all, let us discuss the various types of constrained motions.
5.7. Types of Constrained Motions

Following are the three types of constrained motions :1. Completely constrained motion . When the motion between a pair is limited to a definite
direction irrespective of the direction of force applied, then the motion is said to be a completelyconstrained motion. For example, the piston and cylinder (in a steam engine) form a pair and themotion of the piston is limited to a definite direction ( i.e. it will only reciprocate) relative to the

cylinder irrespective of the direction of motion of the crank, as shown in Fig. 5.1.
Fig. 5.2. Square bar in a square hole. Fig. 5.3. Shaft with collars in a circular hole.
The motion of a square bar in a square hole, as shown in Fig. 5.2, and the motion of a shaft
with collars at each end in a circular hole, as shown in Fig. 5.3, are also examples of completelyconstrained motion.
2. Incompletely constrained motion . When the motion between a pair can take place in more

than one direction, then the motion is called an incompletely constrained motion. The change in thedirection of impressed force may alter the direction of relative motion between the pair. A circular baror shaft in a circular hole, as shown in Fig. 5.4, is an example of an incompletely constrained motionas it may either rotate or slide in a hole. These both motions have no relationship with the other.
Fig. 5.4. Shaft in a circular hole. Fig. 5.5. Shaft in a foot step bearing.

3. Successfully constrained motion. When the motion between the elements, forming a pair,is

such that the constrained motion is not completed by itself, but by some other means, then the motionis said to be successfully constrained motion. Consider a shaft in a foot-step bearing as shown in Fig.5.5. The shaft may rotate in a bearing or it may move upwards. This is a case of incompletely con-strained motion. But if the load is placed on the shaft to prevent axial upward movement of the shaft,then the motion of the pair is said to be successfully constrained motion. The motion of an

Chapter 5 : Simple Mechanisms    /G6C  97
valve (these are kept on their seat by a spring) and the piston reciprocating inside an engine cylinder
are also the examples of successfully constrained motion.
5.8. Classification of Kinematic Pairs
The kinematic pairs may be classified according to the following considerations :
1. According to the type of relative motion between the elements. The kinematic pairs ac-

cording to type of relative motion between the elements may be classified as discussed below:
(a) Sliding pair.  When the two elements of a pair are connected in such a way that one can

only slide relative to the other, the pair is known as a sliding pair. The piston and cylinder, cross-headand guides of a reciprocating steam engine, ram and its guides in shaper, tail stock on the lathe bedetc. are the examples of a sliding pair. A little consideration will show, that a sliding pair has acompletely constrained motion.
(b)  Turning pair . When the two elements of a pair are connected in such a way that one can

only turn or revolve about a fixed axis of another link, the pair is known as turning pair. A shaft withcollars at both ends fitted into a circular hole, the crankshaft in a journal bearing in an engine, lathespindle supported in head stock, cycle wheels turning over their axles etc. are the examples of aturning pair. A turning pair also has a completely constrained motion.
(c)  Rolling pair . When the two elements of a pair are connected in such a way that one rolls

over another fixed link, the pair is known as rolling pair. Ball and roller bearings are examples ofrolling pair.
(d) Screw pair. When the two elements of a pair are connected in such a way that one element
can turn about the other by screw threads, the pair is known as screw pair. The lead screw of a lathewith nut, and bolt with a nut are examples of a screw pair.
(e) Spherical pair . When the two elements of a pair are connected in such a way that one

element (with spherical shape) turns or swivels about the other fixed element, the pair formed iscalled a spherical pair. The ball and socket joint, attachment of a car mirror, pen stand etc., are theexamples of a spherical pair.
2.  According to the type of contact between the elements . The kinematic pairs according to
the type of contact between the elements may be classified as discussed below :
(a) Lower pair . When the two elements of a pair have a surface contact when relative motion

takes place and the surface of one element slides over the surface of the other, the pair formed isknown as lower pair. It will be seen that sliding pairs, turning pairs and screw pairs form lower pairs.
(b) Higher pair . When the two elements of a pair have a line or point contact when relative

motion takes place and the motion between the two elements is partly turning and partly sliding,thenthe pair is known as higher pair. A pair of friction discs, toothed gearing, belt and rope drives, ball androller bearings and cam and follower are the examples of higher pairs.
3. According to the type of closure . The kinematic pairs according to the type of closure
between the elements may be classified as discussed below :

(a) Self closed pair . When the two elements of a pair are connected together mechanically in
such a way that only required kind of relative motion occurs, it is then known as self closed pair. Thelower pairs are self closed pair.
(b) Force - closed pair . When the two elements of a pair are not connected mechanically but

are kept in contact by the action of external forces, the pair is said to be a force-closed pair. The camand follower is an example of force closed pair, as it is kept in contact by the forces exerted by springand gravity.

98  /G6C   Theory of Machines
Fig. 5.6. Arrangement of three links.5.9. Kinematic Chain
When the kinematic pairs are
coupled in such a way that the last linkis joined to the first link to transmitdefinite motion ( i.e. completely or
successfully constrained motion), it iscalled a kinematic chain.  In other

words, a kinematic chain may be de-fined as a combination of kinematicpairs, joined in such a way that eachlink forms a part of two pairs and therelative motion between the links orelements is completely or successfullyconstrained. For example, the crank-shaft of an engine forms a kinematicpair with the bearings which are fixedin a pair, the connecting rod with thecrank forms a second kinematic pair,the piston with the connecting rod forms a third pair and the piston with the cylinder forms a

pair and the piston with the cylinder forms a fourthpair. The total combination of these links is a kinematic chain.

If each link is assumed to form two pairs with two adjacent links, then the relation between
the number of pairs ( p ) forming a kinematic chain and the number of links ( l ) may be expressed in
the form of an equation :
l = 2p – 4 . . . (i)
Since in a kinematic chain each link forms a part of two pairs, therefore there will be as many
links as the number of pairs.
Another relation between the number of links ( l) and the number of joints ( j) which

constitute a kinematic chain is given by the expression :
322jl=− ...(ii)
The equations (i) and (ii) are applicable only to kinematic chains, in which lower pairs are
used. These equations may also be applied to kinematic chains, in which higher pairs are used. In thatcase each higher pair may be taken as equivalent to two lower pairs with an additional element or link.
Let us apply the above equations to the following cases to determine whether each of them is
a kinematic chain or not.

1. Consider the arrangement of three links AB, BC and CA with pin joints at A, B and C as
shown in Fig. 5.6. In this case,
Number of links, l = 3
Number of pairs, p = 3
and       number of joints,  j = 3
From equation (i),                l = 2p – 4
or 3=  2 × 3 – 4 = 2i.e. L.H.S. > R.H.S.
Now from equation (ii),
322jl=− or333 2 2 . 52=× −=
Lawn-mover is a combination of kinematic links.

Chapter 5 : Simple Mechanisms    /G6C  99
i.e. L.H.S.  > R.H.S.
Since the arrangement of three links, as shown in Fig. 5.6, does not satisfy the equations (i)
and (ii) and the left hand side is greater than the right hand side, therefore it is not a kinematic chain
and hence no relative motion is possible. Such type of chain is called  locked chain  and forms a rigid
frame or structure which is used in bridges and trusses.

2. Consider the arrangement of four links AB, BC, CD and DA as shown in Fig. 5.7. In this case
l = 4, p = 4, and j = 4
From equation (i), l = 2 p – 4
4 = 2 × 4 – 4 = 4
i.e. L.H.S. = R.H.S.
From equation (ii),322jl=−
344 2 42=× − =
i.e.                       L.H.S. = R.H.S.
Since the arrangement of four links, as shown in Fig. 5.7, satisfy the equations (i) and (ii),
therefore it is a kinematic chain of one degree of freedom .

A chain in which a single link such as AD in Fig. 5.7 is sufficient to define the position of all
other links, it is then called a kinematic chain of one degree of freedom.
A little consideration will show that in Fig. 5.7, if a definite displacement (say θ) is given to
the link AD, keeping the link AB fixed, then the resulting displacements of the remaining two links BC
and CD are also perfectly definite. Thus we see that in a four bar chain, the relative motion is com-

pletely constrained. Hence it may be called as a constrained kinematic chain , and it is the basis of all
machines.
3. Consider an arrangement of five links, as shown in Fig. 5.8. In this case,
l = 5, p = 5, and j = 5
From equation  (i),
l = 2 p – 4 or 5 = 2 × 5 – 4 = 6
i.e. L.H.S. < R.H.S.
From equation (ii),
322jl=− or  355 2 5 . 52=×−=
Since the arrangement of five links, as shown in Fig. 5.8 does not satisfy the equations and

left hand side is less than right hand side, therefore it is not a kinematic chain. Such a type of chain iscalled unconstrained chain  i.e. the relative motion is not completely constrained. This type of chain
is of little practical importance.
4. Consider an arrangement of six links, as shown in Fig. 5.9. This chain is formed by adding
two more links in such a way that these two links form a pair with the existing links as well as formthemselves a pair. In this case

l = 6, p = 5, and   j = 7Fig. 5.7. Arrangement of four links.
Fig. 5.8. Arrangement of five links.

100  /G6C   Theory of Machines
Fig. 5.11.  Kinematic chain having
binary and ternary joints.From equation  (i),
l = 2 p – 4       or 6 = 2 × 5 – 4 = 6
322jl=− or376 2 72=× −=
Since the arrangement of six links, as shown in Fig.
5.9, satisfies the equations ( i.e. left hand side is equal to right
hand side), therefore it is a kinematic chain.
Note : A chain having more than four links is known as  compound kinematic chain.

5.10. Types of Joints in a Chain
The following types of joints are usually found in a chain :
1. Binary joint . When two links are joined at the same connection, the joint is known as
binary joint. For example, a chain as shown in Fig. 5.10, has four links and four binary joins at A, B,
C and D.
In order to determine the nature of chain, i.e. whether

the chain is a locked chain (or structure) or kinematic chainor unconstrained chain, the following relation between thenumber of links and the number of binary joints, as given byA.W. Klein, may be used :
3222hjl+= −                            ... (i)
where j = Number of binary joints,
h = Number of higher pairs, and
l = Number of links.
When h = 0, the equation (i), may be written as
322jl=− . . . (ii)
Applying this equation to a chain, as shown in Fig. 5.10, where l = 4 and j = 4, we have

34 4242=×−=
Since the left hand side is equal to the right hand side, therefore the chain is a kinematic chain
or constrained chain.
2. Ternary joint .  When three links are joined at the
same connection, the joint is known as ternary joint. It is equiva-lent to two binary joints as one of the three links joined carrythe pin for the other two links. For example, a chain, as shownin Fig. 5.11, has six links. It has three binary joints at A, B and

D and two ternary joints at C and E. Since one ternary joint is
equivalent to two binary joints, therefore equivalent binary jointsin a chain, as shown in Fig. 5.11, are 3 + 2 × 2 = 7
Let us now determine whether this chain is a kinematic
chain or not. We know that l = 6 and j = 7, therefore fromFig. 5.10. Kinematic chain with all
binary joints.Fig. 5.9. Arrangement of six links.

Chapter 5 : Simple Mechanisms    /G6C  101
equation (ii),
322jl=−
or37 6272=×−=
Since left hand side is equal to right hand side, therefore the chain, as shown in Fig. 5.11, is
a kinematic chain or constrained chain.
3. Quaternary joint.  When four links are joined at the same connection, the joint is called a
quaternary joint. It is equivalent to three binary joints. In general, when l number of links are joined
at the same connection, the joint is equivalent to ( l – 1) binary joints.

For example consider a chain having eleven links, as shown in Fig. 5.12 ( a). It has one binary
joint at D, four ternary joints at A, B, E and F, and two quaternary joints at C and G. Since one
quaternary joint is equivalent to three binary joints and one ternary joint is equal to two binary joints,therefore total number of binary joints in a chain, as shown in Fig. 5.12 ( a), are
(a) Looked chain having binary, ternary ( b) Kinematic chain having binary

and quaternary joints. and ternary joints.
Fig. 5.12
1 + 4 × 2 + 2 × 3 = 15
Let us now determine whether the chain, as shown in Fig. 5.12 ( a), is a kinematic chain or
not. We know that l = 11 and j = 15. We know that,
32,2jl=− or315 11 2 14.5,2=×−=  i.e., L.H.S. > R.H.S.
Since the left hand side is greater than right hand side, therefore the chain, as shown in Fig.
5.12 (a) , is not a kinematic chain. We have discussed in Art 5.9 , that such a type of chain is called

locked chain and forms a rigid frame or structure.
If the link CG is removed, as shown in Fig. 5.12 ( b), it has ten links and has one binary joint
at D and six ternary joints at A, B, C, E, F and G.
Therefore total number of binary joints are 1 + 2 × 6 = 13. We know that
32,2jl=− or313 10 2 132=× −= , i.e. L.H.S. = R.H.S.
Since left hand side is equal to right hand side, therefore the chain, as shown in Fig. 5.12 ( b),
is a kinematic chain or constrained chain.
5.11. Mechanism

When one of the links of a kinematic chain is fixed, the chain is known as mechanism . It may
be used for transmitting or transforming motion e.g. engine indicators, typewriter etc.

102  /G6C   Theory of Machines
*The differential of an automobile requires that the angular velocity of two elements be fixed in order to
know the velocity of the remaining elements. The differential mechanism is thus said to have two degreesof freedom. Many computing mechanisms have two or more degrees of freedom.A mechanism with four links is known as  simple mechanism , and the mechanism with more
than four links is known as compound mechanism . When a mechanism is required to transmit power

or to do some particular type of work, it then becomes a machine. In such cases, the various links or
elements have to be designed to withstand the forces (both static and kinetic) safely.
A little consideration will show that a mechanism may be regarded as a machine in which
each part is reduced to the simplest form to transmit the required motion.
5.12. Number of Degrees of Freedom for Plane Mechanisms

In the design or analysis of a mechanism, one of the most important concern is the number of
degrees of freedom (also called movability) of the mechanism. It is defined as the number of inputparameters (usually pair variables) which must be independently controlled in order to bring themechanism into a useful engineering purpose. It is possible to determine the number of degrees offreedom of a mechanism directly from the number of links and the number and types of joints whichit includes.

Consider a four bar chain, as shown in Fig. 5.13 ( a). A little consideration will show that only
one variable such as θ is needed to define the relative positions of all the links. In other words, we say
that the number of degrees of freedom of a four bar chain is one. Now, let us consider a five bar chain,as shown in Fig. 5.13 ( b). In this case two variables such as θ
1 and θ2 are needed to define completely

the relative positions of all the links. Thus, we say that the number of degrees of freedom is * two.
In order to develop the relationship in general, consider two links AB and CD in a plane
motion as shown in Fig. 5.14 ( a).
Fig. 5.14. Links in a plane motion.
The link AB with co-ordinate system OXY is taken as the reference link (or fixed link). The

position of point P on the moving link CD can be completely specified by the three variables, i.e. the(a) Four bar chain. ( b) Five bar chain.
Fig. 5.13

Chapter 5 : Simple Mechanisms    /G6C  103
co-ordinates of the point P denoted by x and y and the inclination θ of the link CD with X-axis or link
AB. In other words, we can say that each link of a mechanism has three degrees of freedom before it
is connected to any other link. But when the link CD is connected to the link AB by a turning pair at
A, as shown in Fig. 5.14 ( b), the position of link CD is now determined by a single variable θ and thus
has one degree of freedom.

From above, we see that when a link is connected to a fixed link by a turning pair ( i.e. lower
pair), two degrees of freedom are destroyed. This may be clearly understood from Fig. 5.15, in whichthe resulting four bar mechanism has one degree of freedom ( i.e. n = 1 ).
Fig. 5.15. Four bar mechanism.
Now let us consider a plane mechanism with l number of links. Since in a mechanism, one of

the links is to be fixed, therefore the number of movable links will be ( l – 1) and thus the total number
of degrees of freedom will be 3 ( l – 1) before they are connected to any other link. In general, a
mechanism with l number of links connected by j number of binary joints or lower pairs ( i.e. single
degree of freedom pairs) and h number of higher pairs ( i.e. two degree of freedom pairs), then the
number of degrees of freedom of a mechanism is given by
n = 3 (l – 1) – 2j – h ... (i)

This equation is called Kutzbach criterion for the movability of a mechanism having plane
motion.
If there are no two degree of freedom pairs ( i.e. higher pairs), then h = 0. Substituting
h = 0 in equation ( i), we have
n = 3 (l – 1) – 2j ... (ii)
5.13. Application of Kutzbach Criterion to Plane Mechanisms
We have discussed in the previous article that Kutzbach criterion for determining the number
of degrees of freedom or movability ( n) of a plane mechanism is

n = 3 (l – 1) – 2j – h
Fig. 5.16.  Plane mechanisms.
The number of degrees of freedom or movability ( n) for some simple mechanisms having no
higher pair ( i.e. h = 0), as shown in Fig. 5.16, are determined as follows :

104  /G6C   Theory of Machines
1. The mechanism, as shown in Fig. 5.16 ( a), has three links and three binary joints, i.e.
l = 3 and j = 3.
∴ n = 3 (3 – 1) – 2 × 3 = 0
2. The mechanism, as shown in Fig. 5.16 ( b), has four links and four binary joints, i.e.
l = 4 and j = 4.
∴ n = 3 (4 – 1) – 2 × 4 = 1
3. The mechanism, as shown in Fig. 5.16 ( c), has five links and five binary joints, i.e.
l = 5, and j = 5.
∴ n = 3 (5 – 1) – 2 × 5 = 2

4. The mechanism, as shown in Fig. 5.16 ( d), has five links and six equivalent binary joints
(because there are two binary joints at B and D, and two ternary joints at A and C), i.e.
l = 5 and j = 6.
∴ n = 3 (5 – 1) – 2 × 6 = 0
5. The mechanism, as shown in Fig. 5.16 ( e), has six links and eight equivalent binary joints
(because there are four ternary joints at A, B, C and D), i.e. l = 6 and j = 8.
∴ n = 3 (6 – 1) – 2 × 8 = – 1

(a)When n = 0, then the mechanism forms a structure and no relative motion between the
links is possible, as shown in Fig. 5.16 ( a) and (d).
(b)When n = 1, then the mechanism can be driven by a single input motion, as shown in Fig.
5.16 (b).
(c)When n = 2, then two separate input motions are necessary to produce constrained
motion for the mechanism, as shown in Fig. 5.16 ( c).

(d)When n =  – 1 or less, then there are redundant constraints in the chain and it forms a
statically indeterminate structure, as shown in Fig. 5.16 ( e).
The application of Kutzbach’s criterion applied to mechanisms with a higher pair or two
degree of freedom joints is shown in Fig. 5.17.
Fig. 5.17. Mechanism with a higher pair.
In Fig. 5.17 ( a), there are three links, two binary joints and one higher pair, i.e. l = 3, j = 2 and h = 1.
  ∴ n = 3 (3 – 1) – 2 × 2 – 1 = 1

∴ n = 3 (3 – 1) – 2 × 2 – 1 = 1
In Fig. 5.17 ( b), there are four links, three binary joints and one higher pair, i.e. l = 4,
j = 3 and h = 1
∴ n = 3 (4 – 1) – 2 × 3 – 1 = 2
Here it has been assumed that the slipping is possible between the links ( i.e. between the
wheel and the fixed link). However if the friction at the contact is high enough to prevent slipping, thejoint will be counted as one degree of freedom pair, because only one relative motion will be possiblebetween the links.

Chapter 5 : Simple Mechanisms    /G6C  105
5.14. Grubler’s Criterion for Plane Mechanisms
The Grubler’s criterion applies to mechanisms with only single degree of freedom joints
where the overall movability of the mechanism is unity. Substituting n = 1 and h = 0 in Kutzbach
equation, we have
1 = 3 (l – 1) – 2 jor 3l – 2j – 4 = 0
This equation is known as the Grubler's criterion for plane mechanisms with constrained
motion.

motion.
A little consideration will show that a plane mechanism with a movability of 1 and only
single degree of freedom joints can not have odd number of links. The simplest possible machanismsof this type are a four bar mechanism and a slider-crank mechanism in which l = 4 and j = 4.
5.15. Inversion of Mechanism
We have already discussed that when one of links is fixed in a kinematic chain, it is called a

mechanism. So we can obtain as many mechanisms as the number of links in a kinematic chain byfixing, in turn, different links in a kinematic chain. This method of obtaining different mechanisms byfixing different links in a kinematic chain, is known as inversion of the mechanism .
It may be noted that the relative motions between the various links is not changed in any

manner through the process of inversion, but their absolute motions (those measured with respect tothe fixed link) may be changed drastically.
Note: The part of a mechanism which initially moves with respect to the frame or fixed link is called driver and
that part of the mechanism to which motion is transmitted is called  follower. Most of the mechanisms are

reversible, so that same link can play the role of a driver and follower at different times. For example, in areciprocating steam engine, the piston is the driver and flywheel is a follower while in a reciprocating air
compressor, the flywheel is a driver.
5.16. Types of Kinematic Chains
The most important kinematic chains are those which consist of four lower pairs, each pair

being a sliding pair or a turning pair. The following three types of kinematic chains with four lowerpairs are important from the subject point of view :
1. Four bar chain or quadric cyclic chain,
2. Single slider crank chain, and
3. Double slider crank chain.
These kinematic chains are discussed, in detail, in the following articles.
5.17. Four Bar Chain or Quadric Cycle Chain
We have already discussed that the kinematic chain is a combination of four or more

kinematic pairs, such that the relative motion between the links or elements is completely constrained.The simplest and the basic kinematic chain is a four bar chain or quad-ric cycle chain, as shown in Fig. 5.18. It consists of four links, each ofthem forms a turning pair at A, B, C and D. The four links may be of
different lengths. According to  Grashof’s law  for a four bar mecha-

nism, the sum of the shortest and longest link lengths should not begreater than the sum of the remaining two link lengths if there is to becontinuous relative motion between the two links.
A very important consideration in designing a mechanism is to
ensure that the input crank makes a complete revolution relative to the
Fig. 5.18.  Four bar chain.

106  /G6C   Theory of Machines
*Refer Chapter 9, Art. 9.6other links. The mechanism in which no link makes a complete revolution will not be useful. In a four
bar chain, one of the links, in particular the shortest link, will make a complete revolution relative tothe other three links, if it satisfies the Grashof’s law. Such a link is known as crank or driver. In Fig.
5.18, AD (link 4 ) is a crank. The link BC (link 2) which makes a partial rotation or oscillates is known

as lever or rocker or follower and the link CD (link 3) which connects the crank and lever is called
connecting rod or coupler. The fixed link AB (link 1) is known as  frame of the mechanism.
When the crank (link 4) is the driver, the mechanism is transforming rotary motion into
oscillating motion.
5.18. Inversions of Four Bar Chain
Though there are many inversions of the four bar
chain, yet the following are important from the subjectpoint of view :
1. Beam engine (crank and lever mechanism ).

A part of the mechanism of a beam engine (also known ascrank and lever mechanism) which consists of four links,is shown in Fig. 5.19. In this mechanism, when the crankrotates about the fixed centre A, the lever oscillates about
a fixed centre D. The end E of the lever CDE is
connected to a piston rod which reciprocates due to therotation of the crank. In other words, the purpose of thismechanism is to convert rotary motion into reciprocatingmotion.

Fig. 5.20.  Coupling rod of a locomotive.
 Fig. 5.19.  Beam engine.
2. Coupling rod of a locomotive (Double crank mechanism ). The mechanism of a coupling
rod of a locomotive (also known as double crank mechanism) which consists of four links, is shownin Fig. 5.20.
In this mechanism, the links AD and BC (having equal length) act as cranks and are con-
nected to the respective wheels. The link CD acts as a coupling rod and the link AB is fixed in order

to maintain a constant centre to centre distance between them. This mechanism is meant for transmit-ting rotary motion from one wheel to the other wheel.
3. Watt’s indicator mechanism (Double lever mechanism ). A *Watt’s indicator mechanism
(also known as Watt's straight line mechanism or double lever mechanism) which consists of four

Chapter 5 : Simple Mechanisms    /G6C  107
links, is shown in Fig. 5.21. The four links are : fixed link
at A, link AC, link CE and link BFD. It may be noted that
BF and FD form one link because these two parts have no
relative motion between them. The links CE and BFD act
as levers. The displacement of the link BFD is directly
proportional to the pressure of gas or steam which acts onthe indicator plunger. On any small displacement of themechanism, the tracing point E at the end of the link CE

traces out approximately a straight line.
The initial position of the mechanism is shown in
Fig. 5.21 by full lines whereas the dotted lines show theposition of the mechanism when the gas or steam pressureacts on the indicator plunger.
5.19. Single Slider Crank Chain
A single slider crank chain is a modification of the basic four bar chain. It consist of one

sliding pair and three turning pairs. It is,usually, found in reciprocating steam engine mechanism.This type of mechanism converts rotary motion into reciprocating motion and vice versa.
In a single slider crank chain, as shown in Fig. 5.22, the links 1 and 2, links 2 and 3, and links
3 and 4 form three turning pairs while the links 4 and 1 form a sliding pair.
Fig. 5.22. Single slider crank chain.
The link 1 corresponds to the frame of the engine, which is fixed. The link 2 corresponds to

the crank ; link 3 corresponds to the connecting rod and link 4 corresponds to cross-head. As thecrank rotates, the cross-head reciprocates in the guides and thus the piston reciprocates in thecylinder.
5.20. Inversions of Single Slider Crank Chain
We have seen in the previous article that a single slider crank chain is a four-link mechanism.

We know that by fixing, in turn, different links in a kinematic chain, an inversion is obtained andwe can obtain as many mechanisms as the links in a kinematic chain. It is thus obvious, that fourinversions of a single slider crank chain are possible. These inversions are found in the followingmechanisms.
1. Pendulum pump or Bull engine . In this mechanism, the inversion is obtained by fixing the

cylinder or link 4 ( i.e. sliding pair), as shown in Fig. 5.23. In this case, when the crank (link 2) rotates,
the connecting rod (link 3) oscillates about a pin pivoted to the fixed link 4 at A and the piston
attached to the piston rod (link 1) reciprocates. The duplex pump which is used to supply feed waterto boilers have two pistons attached to link 1, as shown in Fig. 5.23.Fig. 5.21.  Watt’s indicator mechanism.

108       /G6C               Theory of Machines
Fig. 5.23.  Pendulum pump. Fig. 5.24.  Oscillating cylinder engine.
2. Oscillating cylinder engine . The ar-
rangement of oscillating cylinder engine mecha-

nism, as shown in Fig. 5.24, is used to convertreciprocating motion into rotary motion. In thismechanism, the link 3 forming the turning pair isfixed. The link 3 corresponds to the connectingrod of a reciprocating steam engine mechanism.When the crank (link 2) rotates, the piston at-tached to piston rod (link 1) reciprocates and thecylinder (link 4) oscillates about a pin pivoted tothe fixed link at A.
3. Rotary internal combustion engine

or Gnome engine . Sometimes back, rotary in-
ternal combustion engines were used in aviation.But now-a-days gas turbines are used in its place.It consists of seven cylinders in one plane andall revolves about fixed centre D, as shown in
Fig. 5.25, while the crank (link 2) is fixed. Inthis mechanism, when the connecting rod (link4) rotates, the piston (link 3) reciprocates insidethe cylinders forming link 1.
Fig. 5.25.  Rotary internal combustion engine.

4. Crank and slotted lever quick return motion mechanism . This mechanism is mostly used
in shaping machines, slotting machines and in rotary internal combustion engines.
Rotary engine

Chapter 5 : Simple Mechanisms    /G6C  109
In this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as shown in Fig.
5.26. The link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crankCB revolves with uniform angular speed about the fixed centre C. A sliding block attached to the crank
pin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point A. A

short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the
line of stroke R
1R2. The line of stroke of the ram ( i.e. R1R2) is perpendicular to AC produced.
Fig. 5.26.  Crank and slotted lever quick return motion mechanism.
In the extreme positions, AP1 and AP2 are tangential to the circle
and the cutting tool is at the end of the stroke. The forward or cuttingstroke occurs when the crank rotates from the position CB
1 to CB2 (or

1 to CB2 (or
through an angle β) in the clockwise direction. The return stroke occurs
when the crank rotates from the position CB2 to CB1 (or through angle α)
in the clockwise direction. Since the crank has uniform angular speed,therefore,
Time of cutting stroke 360orTime of return stroke 360°−α ββ==α° − β α
Since the tool travels a distance of R1 R2 during cutting and return
stroke, therefore travel of the tool or length of stroke
 = R1R2 = P1P2  = 2P1Q = 2AP1 sin ∠ P1 AQ

= R1R2 = P1P2  = 2P1Q = 2AP1 sin ∠ P1 AQ
 = ()12s i n 9 0 2c o s22AP APαα°− =1.....( )AP AP=/G51
  12CBAPAC=× ... 1cos2CB
ACα= /G51
  2CBAPAC=× ... 1()CB CB=/G51
Note: From Fig. 5.26, we see that the angle β made by the forward or cutting stroke is greater than the angle α
described by the return stroke. Since the crank rotates with uniform angular speed, therefore the return stroke is
completed within shorter time. Thus it is called quick return motion mechanism.
The Shaping Machine

110  /G6C   Theory of Machines
5. Whitworth quick return motion mechanism. This mechanism is mostly used in shaping
and slotting machines. In this mechanism, the link CD (link 2) forming the turning pair is fixed, as
shown in Fig. 5.27. The link 2 corresponds to a crank in a reciprocating steam engine. The drivingcrank CA (link 3) rotates at a uniform angular speed. The slider (link 4) attached to the crank pin at A

slides along the slotted bar PA (link 1) which oscillates at a pivoted point D. The connecting rod PR
carries the ram at R to which a cutting tool is fixed. The motion of the tool is constrained along the
line RD produced, i.e. along a line passing through D and perpendicular to CD.
Fig. 5.27. Whitworth quick return motion mechanism.
When the driving crank CA moves from the position CA1 to CA2 (or the link DP from the

position DP1 to DP2) through an angle α in the clockwise direction, the tool moves from the left hand
end of its stroke to the right hand end through a distance 2 PD.
Now when the driving crank moves from the position CA2 to CA1 (or the link DP from DP2 to
DP1 ) through an angle β in the clockwise direction, the tool moves back from right hand end of its
stroke to the left hand end.
A little consideration will show that the time taken during the left to right movement of the

ram (i.e. during forward or cutting stroke) will be equal to the time taken by the driving crank to move
from CA1 to CA2. Similarly, the time taken during the right to left movement of the ram (or during the
idle or return stroke) will be equal to the time taken by the driving crank to move from CA2 to CA1.
Since the crank link CA rotates at uniform angular velocity therefore time taken during the

cutting stroke (or forward stroke) is more than the time taken during the return stroke. In other words,the mean speed of the ram during cutting stroke is less than the mean speed during the return stroke.The ratio between the time taken during the cutting and return strokes is given by
Time of cutting stroke 360orTime of return stroke 360°−β αα==β° − α β
Note.In order to find the length of effective stroke R1 R2, mark P1 R1 = P2 R2 = PR. The length of effective
stroke is also equal to 2 PD.

Example 5.1. A crank and slotted lever mechanism used in a shaper has a centre distance of
300 mm between the centre of oscillation of the slotted lever and the centre of rotation of the crank.The radius of the crank is 120 mm. Find the ratio of the time of cutting to the time of return stroke .
Solution. Given : AC = 300 mm ; CB
1 = 120 mm
The extreme positions of the crank are shown in Fig. 5.28. We know that

Chapter 5 : Simple Mechanisms    /G6C  111
Fig. 5.29Fig. 5.28    1 sin sin(90 /2)CAB∠= ° − α
    11200.4300CB
AC===
∴     190 /2CAB∠= ° − α
    = sin–1 0.4 = 23.6°
or   α/2 = 90° – 23.6° = 66.4°
and    α = 2 × 66.4 = 132.8°
Time of cutting stroke 360 132.8 360
Time of return stroke 132.8°− ° °−α==α°= 1.72 Ans.
Example 5.2.  In a crank and slotted lever quick return motion mechanism, the distance

between the fixed centres is 240 mm and the length of the driving crank is 120 mm. Find the inclina-tion of the slotted bar with the vertical in the extreme position and the time ratio of cutting stroke tothe return stroke.
If the length of the slotted bar is 450 mm, find the length of the stroke if the line of stroke
passes through the extreme positions of the free end of the lever .
Solution. Given : AC = 240 mm ; CB
1 = 120 mm ; AP1 = 450 mm
Inclination of the slotted bar with the vertical

Let ∠CAB1 = Inclination of the slotted bar with the vertical.
The extreme positions of the crank are
shown in Fig. 5.29. We know that
() 1 sin sin 902CABα∠= ° −
                     11200.5240BC
AC===
∴ 1902CABα∠= ° −
1sin 0.5 30−== °  Ans.
Time ratio of cutting stroke to the return stroke
We know that                90° – α/ 2 = 30°
∴                     α / 2 = 90° – 30° = 60°
or  α = 2 × 60° = 120°
∴Time of cutting stroke 360 360 120

Time of return stroke 120°−α °− °==α° = 2 Ans.

112  /G6C   Theory of Machines
Length of the stroke
We know that length of the stroke,
R1R2 = P1P2 =  2 P1Q = 2 AP1 sin (90° – α/2)
= 2 × 450 sin (90°– 60°) = 900 × 0.5 = 450 mm Ans.
Example 5.3.  Fig. 5.30 shows the lay out of a quick return mechanism of the oscillating link

type, for a special purpose machine. The driving crank BC is 30 mm long and time ratio of theworking stroke to the return stroke is to be 1.7. If the length of the working stroke of R is 120 mm,determine the dimensions of AC and AP.
Solution.  Given : BC = 30 mm ; R
1R2 = 120 mm ; Time ratio of working stroke to the return
stroke = 1.7
                            Fig. 5.30                                                               Fig. 5.31
Time of working stroke 360

Time of return stroke−α=αor3601.7−α=α
∴α  = 133.3° or α / 2 = 66.65°
The extreme positions of the crank are shown in Fig. 5.31. From right angled triangle AB1C,
we find that
sin (90° – α/2) = 1BC
ACor1
sin(90 /2) cos /2BC BCAC==°−α α
... (∵ B1C = BC)
∴30 3075.7cos66.65 0.3963AC== =°mm Ans.
We know that length of stroke,
R1R2 = P1P2 = 2P1Q = 2 AP1 sin (90° – α/ 2) = 2 AP1 cos α/2
120 = 2 AP cos 66.65° = 0.7926 AP ... (∵ AP1 = AP)

∴ AP = 120/ 0.7926 = 151.4 mm Ans.
Example 5.4.  In a Whitworth quick return motion mechanism, as shown in Fig. 5.32, the
distance between the fixed centers is 50 mm and the length of the driving crank is 75 mm. The lengthof the slotted lever is 150 mm and the length of the connecting rod is 135 mm. Find the ratio of thetime of cutting stroke to the time of re turn stroke and also the effective stroke .

Chapter 5 : Simple Mechanisms    /G6C  113
Solution.  Given : CD = 50 mm ; CA = 75 mm ; PA = 150 mm ; PR = 135 mm
Fig. 5.32 Fig. 5.33
The extreme positions of the driving crank are shown in Fig. 5.33. From the geometry of the
figure,
250cos /2 0.66775CD
CAβ= = = ... (/G51CA2 = CA)
∴ /2 48.2β= ° orβ = 96.4°
Ratio of the time of cutting stroke to the time of return stroke
Time of cutting stroke 360 360 96.42.735Time of return stroke 96.4−β −== =β Ans.
Length of effective stroke

In order to find the length of effective stroke ( i.e. R1R2), draw the space diagram of the
mechanism to some suitable scale, as shown in Fig. 5.33. Mark P1R2 = P2R2 = PR. Therefore by
measurement we find that,
Length of effective stroke = R1R2 = 87.5 mm  Ans.
5.21. Double Slider Crank Chain
A kinematic chain which consists of two turning pairs and two sliding pairs is known as

double slider crank chain , as shown in Fig. 5.34. We see that the link 2 and link 1 form one turning
pair and link 2 and link 3 form the second turning pair. The link 3 and link 4 form one sliding pair andlink 1 and link 4 form the second sliding pair.
5.22.  Inversions of Double Slider Crank Chain
The following three inversions of a double slider crank chain are important from the subject

1. Elliptical trammels . It is an instrument used for drawing ellipses. This inversion is obtained
by fixing the slotted plate (link 4), as shown in Fig. 5.34. The fixed plate or link 4 has two straightgrooves cut in it, at right angles to each other. The link 1 and link 3, are known as sliders and form slidingpairs with link 4. The link AB (link 2) is a bar which forms turning pair with links 1 and 3.

When the links 1 and 3 slide along their respective grooves, any point on the link 2 such as
P traces out an ellipse on the surface of link 4, as shown in Fig. 5.34 ( a). A little consideration will
show that AP and BP are the semi-major axis and semi-minor axis of the ellipse respectively. This can
be proved as follows :

114  /G6C   Theory of Machines
(a)( b)
Fig. 5.34.  Elliptical trammels.
Let us take OX and OY as horizontal and vertical axes and let the link BA is inclined at an
angle θ with the horizontal, as shown in Fig. 5.34 ( b). Now the co-ordinates of the point P on the link
BA will be
x = PQ = AP  cos θ; and y = PR = BP  sin θ
or cos ; and siny x
AP BP=θ =θ
Squaring and adding,
     2 222
22cos sin 1
() ()y x
AP BP+ = θ+ θ=

2 222
22cos sin 1
() ()y x
This is the equation of an ellipse. Hence the path traced by point P is an ellipse whose semi-
major axis is  AP and semi-minor axis is BP.
Note : If P is the mid-point of link BA, then AP = BP. The above equation can be written as
                        22
221
() ()xy
AP AP+=           or      x2 + y2 = (AP)2
This is the equation of a circle whose radius is AP. Hence if P is the mid-point of link BA, it will trace
a circle.

a circle.
2. Scotch yoke mechanism . This mechanism is used for converting rotary motion into a
reciprocating motion. The inversion is obtained by fixing either the link 1 or link 3. In Fig. 5.35, link1 is fixed. In this mechanism, when the link 2 (whichcorresponds to crank) rotates about  B as centre, the link
4 (which corresponds to a frame) reciprocates. The fixedlink 1 guides the frame.
3. Oldham’s coupling . An oldham's coupling is

used for connecting two parallel shafts whose axes areat a small distance apart. The shafts are coupled in sucha way that if one shaft rotates, the other shaft also rotatesat the same speed. This inversion is obtained by fixingthe link 2, as shown in Fig. 5.36 ( a). The shafts to be
connected have two flanges (link 1 and link 3) rigidlyfastened at their ends by forging.
Fig. 5.35.  Scotch yoke mechanism.

Chapter 5 : Simple Mechanisms    /G6C  115
The link 1 and link 3 form turning pairs with link 2. These flanges have diametrical slots cut
in their inner faces, as shown in Fig. 5.36 ( b). The intermediate piece (link 4) which is a circular disc,
have two tongues ( i.e. diametrical projections) T1 and T2 on each face at right angles to each other, as
shown in Fig. 5.36 ( c). The tongues on the link 4 closely fit into the slots in the two flanges (link 1 and

link 3). The link 4 can slide or reciprocate in the slots in the flanges.
(a)( b)( c)
Fig. 5.36.  Oldham’s coupling.
When the driving shaft A is rotated, the flange C (link 1) causes the intermediate piece (link
4) to rotate at the same angle through which the flange has rotated, and it further rotates the flange D
(link 3) at the same angle and thus the shaft B rotates. Hence links 1, 3 and 4 have the same angular

velocity at every instant. A little consideration will show, that there is a sliding motion between thelink 4 and each of the other links 1 and 3.
If the distance between the axes of the shafts is constant, the centre of intermediate piece will
describe a circle of radius equal to the distance between the axes of the two shafts. Therefore, themaximum sliding speed of each tongue along its slot is equal to the peripheral velocity of the centreof the disc along its circular path.

Let ω = Angular velocity of each shaft in rad/s, and
r = Distance between the axes of the shafts in metres.
∴ Maximum sliding speed of each tongue (in m/s),
v = ω.r
EXERCISES
1. In a crank and slotted lever quick return mechanism, the distance between the fixed centres is 150
mm and the driving crank is 75 mm long. Determine the ratio of the time taken on the cutting and
return strokes. [Ans. 2]

2.In a crank and slotted lever quick return motion mechanism, the distance between the fixed centres O
and C is 200 mm. The driving crank CP is 75 mm long. The pin Q on the slotted lever, 360 mm from
the fulcrum O, is connected by a link QR 100 mm long, to a pin R on the ram. The line of stroke of R
is perpendicular to OC and intersects OC produced at a point 150 mm from C. Determine the ratio of
times taken on the cutting and return strokes. [Ans. 1.647]

3.In a crank and slotted lever quick return mechanism, as shown in Fig. 5.37, the driving crank length is
75 mm. The distance between the fixed centres is 200 mm and the length of the slotted lever is 500mm. Find the ratio of the times taken on the cutting and idle strokes. Determine the effective stroke
also. [Ans. 1.67 ; 380 mm]

116  /G6C   Theory of Machines
All dimensions in mm
Fig. 5.37
4.The Whitworth quick return motion mechanism has the driving crank 150 mm long. The distance
between fixed centres is 100 mm. The line of stroke of the ram passes through the centre of rotation of
the slotted lever whose free end is connected to the ram by a connecting link. Find the ratio of time ofcutting to time of return. [Ans. 2.735]
5.A Whitworth quick return motion mechanism, as shown in Fig. 5.38, has the following

particulars :
Fig. 5.38
Length of stroke = 150 mm ; Driving crank length = 40 mm; Time of cutting stroke=2Timeof returnstroke
Find the lengths of CD and PD. Also determine the angles α and β.
[Hint : Length of stroke = R1R2 = P1P2 = 2PD] [Ans. 20 mm, 75 mm; 240°, 120°]
1.Explain the term kinematic link. Give the classification of kinematic link.
2.What is a machine ? Giving example, differentiate between a machine and a structure.

3.Write notes on complete and incomplete constraints in lower and higher pairs, illustrating your answer
with neat sketches.
4.Explain different kinds of kinematic pairs giving example for each one of them.
5.Explain the terms : 1. Lower pair, 2. Higher pair, 3. Kinematic chain, and 4. Inversion.
6.In what way a mechanism differ from a machine ?
7. What is the significance of degrees of freedom of a kinematic chain when it functions as a mecha-
nism? Give examples.

Chapter 5 : Simple Mechanisms    /G6C  117
8.Determine the mobility (degrees of freedom) of the mechanism shown in Fig. 5.39 ( a) and (b) using
Kutzbach mobility criterion and classify them.
Fig. 5.39
9.Explain Grubler’s criterion for determining degree of freedom for mechanisms. Using Grubler’s cri-
terion for plane mechanism, prove that the minimum number of binary links in a constrained mecha-
nism with simple hinges is four.

10.Sketch and explain the various inversions of a slider crank chain.
11.Sketch and describe the four bar chain mechanism. Why it is considered to be the basic chain?
12.Show that slider crank mechanism is a modification of the basic four bar mechanism.
13.Sketch slider crank chain and its various inversions, stating actual machines in which these are used in
practice.

practice.
14.Sketch and describe the working of two different types of quick return mechanisms. Give examples of
their applications. Derive an expression for the ratio of times taken in forward and return stroke forone of these mechanisms.
15.Sketch and explain any two inversions of a double slider crank chain.
16.Identify the kinematic chains to which the following mechanisms belong :
1. Steam engine mechanism ; 2. Beam engine ; 3. Whitworth quick return motion mechanism;

4. Elliptical trammels.
1.In a reciprocating steam engine, which of the following forms a kinematic link ?
(a) cylinder and piston ( b) piston rod and connecting rod
(c) crank shaft and flywheel ( d) flywheel and engine frame
2.The motion of a piston in the cylinder of a steam engine is an example of
(a) completely constrained motion ( b) incompletely constrained motion
(c) successfully constrained motion ( d) none of these

3.The motion transmitted between the teeth of gears in mesh is
(a) sliding ( b) rolling
(c) may be rolling or sliding depending upon the shape of teeth
(d) partly sliding and partly rolling
4.The cam and follower without a spring forms a
(a) lower pair ( b) higher pair
(c) self closed pair ( d) force closed pair
5.A ball and a socket joint forms a
(a) turning pair ( b) rolling pair ( c) sliding pair ( d) spherical pair
6.The lead screw of a lathe with nut forms a

(a) sliding pair ( b) rolling pair ( c) screw pair ( d) turning pair
7.When the elements of the pair are kept in contact by the action of external forces, the pair is said to be a

118  /G6C   Theory of Machines
8.Which of the following is a turning pair ?
(a) Piston and cylinder of a reciprocating steam engine
(b) Shaft with collars at both ends fitted in a circular hole
(c) Lead screw of a lathe with nut
(d) Ball and socket joint
9.A combination of kinematic pairs, joined in such a way that the relative motion between the links is
completely constrained, is called a
(a) structure ( b) mechanism
(c) kinematic chain ( d) inversion

10.The relation between the number of pairs ( p) forming a kinematic chain and the number of links ( l) is
(a)l = 2p – 2 ( b)l = 2p – 3 ( c)l = 2p – 4 ( d)l = 2p – 5
11.The relation between the number of links ( l) and the number of binary joints ( j) for a kinematic chain
having constrained motion is given by 32.2jl=−  If the left hand side of this equation is greater than
right hand side, then the chain is
(a) locked chain ( b) completely constrained chain

(c) successfully constrained chain ( d) incompletely constrained chain
12.In a kinematic chain, a quaternary joint is equivalent to
(a) one binary joint ( b) two binary joints ( c) three binary joints ( d)  four binary joints
13.If n links are connected at the same joint, the joint is equivalent to
(a)(n – 1) binary joints ( b)(n – 2) binary joints ( c)( 2n – 1) binary joints ( d) none of these

14.In a 4 – bar linkage, if the lengths of shortest, longest and the other two links are denoted by s, l, p and
q, then it would result in Grashof’s linkage provided that
(a)l + p < s + q (b)l + s < p + q (c)l + p = s + q (d) none of these
15.A kinematic chain is known as a mechanism when
(a) none of the links is fixed ( b) one of the links is fixed
(c) two of the links are fixed ( d) all of the links are fixed

16.The Grubler’s criterion for determining the degrees of freedom ( n) of a mechanism having plane motion
is
(a)n = (l – 1) – j (b)n = 2 (l – 1) – 2j(c)n = 3 (l – 1) – 2j(d)n = 4 (l – 1) – 3j
where l = Number of links, and j = Number of binary joints.
17.The mechanism forms a structure, when the number of degrees of freedom ( n) is equal to
(a)0 ( b)1 ( c)2 ( d)–  1
18.In a four bar chain or quadric cycle chain

(a) each of the four pairs is a turning pair ( b) one is a turning pair and three are sliding pairs
(c) three are turning pairs and one is sliding pair ( d) each of the four pairs is a sliding pair.
19.Which of the following is an inversion of single slider crank chain ?
(a) Beam engine ( b) Watt’s indicator mechanism
(c) Elliptical trammels ( d) Whitworth quick return motion mechanism

20.Which of the following is an inversion of double slider crank chain ?
(a) Coupling rod of a locomotive ( b) Pendulum pump
(c) Elliptical trammels ( d) Oscillating cylinder engine
ANSWERS
1.(c) 2.(a) 3.(d) 4.(c) 5.(d)
6.(c) 7.(d) 8.(b) 9.(c) 10.(c)
11.(a) 12.(c) 13.(a) 14.(b) 15.(b)
16.(c) 17.(a) 18.(a) 19.(d) 20.(c)

119Velocity inVelocity inVelocity inVelocity inVelocity in
MechanismsMechanismsMechanismsMechanismsMechanisms
(Instantaneous Centre Method)(Instantaneous Centre Method)(Instantaneous Centre Method)(Instantaneous Centre Method)(Instantaneous Centre Method)/G36
FeaturesFeaturesFeaturesFeaturesFeatures
/G31/G2E /G31/G6E/G74/G72/G6F/G64/G75/G63/G74/G69/G6F/G6E/G2E

120  /G6C   Theory of Machines
Fig. 6.2. Instantaneous
centre of rotation.A1 B1. Such a motion of link AB to A1
B1 is an example of combined motion
of rotation and translation, it beingimmaterial whether the motion ofrotation takes first, or the motion oftranslation.
In actual practice, the motion
of link AB is so gradual that it is
difficult to see the two separatemotions. But we see the two separatemotions, though the point B moves
faster than the point A. Thus, this

combined motion of rotation andtranslation of the link AB may be assumed to be a motion of pure rotation about some centre I, known
as the instantaneous centre of rotation (also called centro or virtual centre ). The position of
instantaneous centre may be located as discussed below:
Since the points A and B of the link has moved to A
1 and B1
respectively under the motion of rotation (as assumed above), there-

fore the position of the centre of rotation must lie on the intersection ofthe right bisectors of chords A A
1 and B B1. Let these bisectors intersect
at I as shown in Fig. 6.2, which is the instantaneous centre of rotation or
virtual centre of the link AB.
From above, we see that the position of the link AB goes on
changing, therefore the centre about which the motion is assumed totake place ( i.e. the instantaneous centre of rotation) also goes on chang-

ing. Thus the instantaneous centre of a moving body may be defined asthat centre which goes on changing from one instant to another . The
locus of all such instantaneous centres is known as centrode. A line
drawn through an instantaneous centre and perpendicular to the planeof motion is called instantaneous axis . The locus of this axis is known as axode.
6.2. Space and Body Centrodes
A rigid body in plane motion relative to a second rigid body, supposed fixed in space, may be

assumed to be rotating about an instantaneous centre atthat particular moment. In other words, the instantaneouscentre is a point in the body which may be consideredfixed at any particular moment. The locus of theinstantaneous centre in space during a definite motion ofthe body is called the  space centrode and the locus of the
instantaneous centre relative to the body itself is calledthe body centrode . These two centrodes have the

instantaneous centre as a common point at any instant andduring the motion of the body, the body centrode rollswithout slipping over the space centrode.
Let I
1 and I2 be the instantaneous centres for the
two different positions A1 B1 and A2 B2 of the link A1 B1
after executing a plane motion as shown in Fig. 6.3. Similarly, if the number of positions of the link
A1 B1 are considered and a curve is drawn passing through these instantaneous centres ( I1, I2....), then

the curve so obtained is called the space centrode.
Mechanisms on a steam automobile engine.
Fig. 6.3. Space and body centrode.

Chapter 6 :  Velocity in Mechanisms   /G6C  121
Now consider a point C1 to be attached to the body or link A1 B1 and moves with it in such
a way that C1 coincides with  I1 when the body is in position A1 B1. Let C2 be the position of the
point C1 when the link A1 B1 occupies the position A2 B2. A little consideration will show that the
point C2 will coincide with I2 (when the link is in position A2 B2) only if triangles A1 B1 C1 and
A2 B2 C2 are identical.

∴                 A1 C2 = A2 I2           and                 B1 C2 = B2 I2
In the similar way, the number of positions of the point C1 can be obtained for different
positions of the link A1B1. The curve drawn through these points ( C1, C2....) is called the body
centrode.
6.3. Methods for Determining the Velocity of a Point on a Link
Though there are many methods for determining the velocity of any point on a link in a

mechanism whose direction of motion ( i.e. path) and velocity of some other point on the same link
is known in magnitude and direction, yet the following two methods are important from the subjectpoint of view.
1.  Instantaneous centre method,  and   2.  Relative velocity method.
The instantaneous centre method is convenient and easy to apply in simple mechanisms,

whereas the relative velocity method may be used to any configuration diagram. We shall discuss therelative velocity method in the next chapter.
6.4. Velocity of a Point on a Link by
Instantaneous Centre Method
The instantaneous centre method of analysing the motion

in a mechanism is based upon the concept (as discussed in Art.6.1) that any displacement of a body (or a rigid link) havingmotion in one plane, can be considered as a pure rotationalmotion of a rigid link as a whole about some centre, known asinstantaneous centre or virtual centre of rotation.
Consider two points A and B on a rigid link. Let v
A and
vB be the velocities of points A and B, whose directions are given
by angles α and β as shown in Fig. 6.4. If vA is known in

magnitude and direction and vB in
direction only, then the magnitude ofv
B may be determined by the
instantaneous centre method asdiscussed below :
Draw AI and  BI perpendicu-
lars to the directions v
A and vB respec-
tively. Let these lines intersect at I,
which is known as instantaneous cen-tre or virtual centre of the link. Thecomplete rigid link is to rotate or turnabout the centre I.
Since A and B are the points

on a rigid link, therefore there cannotbe any relative motion between themalong the line AB.Fig. 6.4. Velocity of a point on
a link.
Robots use various mechanisms to perform jobs.

122  /G6C   Theory of Machines
Now resolving the velocities along AB,
 vA cos α=vB cos β
orA
Bv
v=cos sin(90 – )
cos sin(90 – )β° β=α° α...(i)
Applying Lami’s theorem to triangle ABI,
                    sin(90 – ) sin(90 – )AI BI=°β °α
or                                sin(90 – )
sin(90 – )AI
BI°β=°α...(ii)
From equation  (i) and (ii),
                                    A
BvA I
vB I=            or                ABvv
AI BI== ω ...(iii)

where                             ω  = Angular velocity of the rigid link.
If C is any other point on the link, then
                                   C AB v vv
AI BI CI== ...(iv)
From the above equation, we see that
1.  If vA is known in magnitude and direction and vB in direction only, then velocity of point
B or any other point C lying on the same link may be determined in magnitude and direction.

2.  The magnitude of velocities of the points on a rigid link is inversely proportional to the
distances from the points to the instantaneous centre and is perpendicular to the line joining the pointto the instantaneous centre.
6.5. Properties of the Instantaneous Centre
The following properties of the instantaneous centre are important from the subject point of
view :
1.  A rigid link rotates instantaneously relative to another link at the instantaneous centre for

the configuration of the mechanism considered.
2.  The two rigid links have no linear velocity relative to each other at the instantaneous
centre. At this point ( i.e. instantaneous centre), the two rigid links have the same linear velocity
relative to the third rigid link. In other words, the velocity of the instantaneous centre relative to anythird rigid link will be same whether the instantaneous centre is regarded as a point on the first rigidlink or on the second rigid link.

6.6. Number of Instantaneous Centres in a
Mechanism
The number of instantaneous centres in a constrained
kinematic chain is equal to the number of possible combina-tions of two links. The number of pairs of links or the numberof instantaneous centres is the number of combinations of n
links taken two at a time. Mathematically, number of instanta-neous centres,
                    
(– 1 ),2nnN=  where n = Number of links.
Four bar mechanisms.Bar 2
BaseGround 1RevolutesBar 123
4
Ground 2Bar 3

Chapter 6 :  Velocity in Mechanisms   /G6C  123
6.7. Types of Instantaneous Centres
The instantaneous centres for a mechanism are
of the following three types :
1. Fixed instantaneous centres,  2. Permanent
instantaneous centres, and  3. Neither fixed nor per-
manent instantaneous centres.
The first two types i.e. fixed and permanent
instantaneous centres are together known as primary
instantaneous centres  and the third type is known as
secondary instantaneous centres .

Consider a four bar mechanism ABCD as
shown in Fig. 6.5. The number of instantaneous cen-tres (N) in a four bar mechanism is given by
                                      ( – 1) 4(4 – 1)622nnN=== ... (∵ n = 4)
The instantaneous centres  I12 and I14 are called the fixed instantaneous centres  as they re-
main in the same place for all configurations of the mechanism. The instantaneous centres I23 and I34

are the permanent instantaneous centres  as they move when the mechanism moves, but the joints
are of permanent nature. The instantaneous centres I13 and I24 are neither fixed nor permanent
instantaneous centres  as they vary with the configuration of the mechanism.
Note:  The instantaneous centre of two links such as link 1 and link 2 is usually denoted by I12 and so on. It is
read as I one two and not I twelve.
6.8. Location of Instantaneous Centres

The following rules may be used in locating the instantaneous centres in a mechanism :
1.  When the two links are connected by a pin joint (or pivot joint), the instantaneous centreFig. 6.5. Types of instantaneous centres.
Computer disk drive mechanisms.
Note : This picture is given as additional information and is not a direct example of the current chapter.Track selector
mechanism
The read/write head
is guided by informa-
tion stored on the disk
itself

itself
The hard disk is
coated with a
magnetic materialsArm moves to a
track to retrive
information stored
there

124  /G6C   Theory of Machines
lies on the centre of the pin as shown in Fig. 6.6 ( a). Such a instantaneous centre is of permanent
nature, but if one of the links is fixed, the instantaneous centre will be of fixed type.
2.When the two links have a pure rolling contact ( i.e. link 2 rolls without slipping upon the

fixed link 1 which may be straight or curved), the instantaneous centre lies on their point of contact,as shown in Fig. 6.6 ( b). The velocity of any point A on the link 2 relative to fixed link 1 will be
perpendicular to I
12 A and is proportional to  I12 A . In other words
A1 2
B1 2vI A
vI B=
3.When the two links have a sliding contact, the instantaneous centre lies on the common
normal at the point of contact. We shall consider the following three cases :

(a)When the link 2 (slider) moves on fixed link 1 having straight surface as shown in
Fig. 6.6 (c), the instantaneous centre lies at infinity and each point on the slider have
the same velocity.
(b)When the link 2 (slider) moves on fixed link 1 having curved surface as shown in Fig.
6.6 (d),the instantaneous centre lies on the centre of curvature of the curvilinear path
in the configuration at that instant.
(c)When the link 2 (slider) moves on fixed link 1 having constant radius of curvature as

shown in Fig. 6.6 ( e), the instantaneous centre lies at the centre of curvature i.e. the
centre of the circle, for all configuration of the links.
Fig. 6.6. Location of instantaneous centres.
6.9.  Aronhold Kennedy (or Three Centres in Line) Theorem
The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other ,
they have three instantaneous centres and lie on a straight line.
Consider three kinematic links A, B and C having relative

plane motion. The number of instantaneous centres ( N) is given by
                                      ( – 1) 3(3 – 1)322nnN===
where                                         n = Number of links = 3
The two instantaneous centres at the pin joints of B with A,
and C with A (i.e. Iab and Iac) are the permanent instantaneous centres.
According to Aronhold Kennedy’s theorem, the third instantaneouscentre I
bc must lie on the line joining Iab and Iac. In order to prove this,

Chapter 6 :  Velocity in Mechanisms   /G6C  125
let us consider that the instantaneous centre Ibc lies outside the line joining Iab and Iac as shown in Fig. 6.7.
The point Ibc belongs to both the links B and C. Let us consider the point Ibc on the link B. Its velocity
vBC  must be perpendicular to the line joining Iab and Ibc. Now consider the point Ibc on the link C. Its
velocity vBC must be perpendicular to the line joining Iac and Ibc.

We have already discussed in Art. 6.5, that the velocity of the instantaneous centre is same
whether it is regarded as a point on the first link or as a point on the second link.  Therefore, the velocityof the point I
bc cannot be perpendicular to both lines Iab Ibc and Iac Ibc unless the point Ibc lies on the line
joining the points Iab and Iac. Thus the three instantaneous centres ( Iab, Iac and Ibc) must lie on the same

straight line. The exact location of Ibc on line Iab Iac depends upon the directions and magnitudes of the
angular velocities of B and C relative to A.
The above picture shows ellipsograph which is used to draw ellipses.Central ring
Ellipses drawn by
the ellipsographWinding handle to
operate the deviceDrawing
Pencil
Note : This picture is given as additional information and is not a direct example of the current chapter.
6.10. Method of Locating Instantaneous Centres in a Mechanism

Consider a pin jointed four bar mechanism as shown in Fig. 6.8 ( a). The following procedure
is adopted for locating instantaneous centres.
1.  First of all, determine the number of instantaneous centres ( N) by using the relation
                                     (– 1 ),2nnN=  where n = Number of links.
In the present case,     4(4–1)62N== ...(∵ n = 4)
2.  Make a list of all the instantaneous centres in a mechanism. Since for a four bar mecha-

nism, there are six instantaneous centres, therefore these centres are listed as shown in the followingtable (known as book-keeping table).
Links 1 2 3 4
Instantaneous 12 23 34 –
centres 13 24
(6 in number) 14

126  /G6C   Theory of Machines
3.  Locate the fixed and permanent instantaneous centres by inspection. In Fig. 6.8 ( a), I12
and I14 are fixed instantaneous centres and I23 and I34 are permanent instantaneous centres.
Note. The four bar mechanism has four turning pairs, therefore there are four primary ( i.e. fixed and permanent)
instantaneous centres and are located at the centres of the pin joints.
Fig. 6.8. Method of locating instantaneous centres.

4.  Locate the remaining neither fixed nor permanent instantaneous centres (or secondary
centres) by Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.8 ( b). Mark points
on a circle equal to the number of links in a mechanism. In the present case, mark 1, 2, 3, and 4 on thecircle.
5.  Join the points by solid lines to show that these centres are already found. In the circle
diagram [Fig. 6.8 ( b)] these lines are 12, 23, 34 and 14 to indicate the centres I

12, I23, I34 and I14.
6.  In order to find the other two instantaneous centres, join two such points that the line
joining them forms two adjacent triangles in the circle diagram. The line which is responsible forcompleting two triangles, should be a common side to the two triangles. In Fig. 6.8 ( b), join 1 and 3
to form the triangles 123 and 341 and the instantaneous centre * I
13  will lie on the intersection of  I12

I23  and I14 I34, produced if necessary, on the mechanism. Thus the instantaneous centre I13 is located.
Join 1 and 3 by a dotted line on the circle diagram and mark number 5 on it. Similarly the instanta-neous centre I
24 will lie on the intersection of I12 I14 and I23 I34, produced if necessary, on the mecha-
nism. Thus I24 is located. Join 2 and 4 by a dotted line on the circle diagram and mark 6 on it. Hence

all the six instantaneous centres are located.
Note: Since some of the neither fixed nor permanent instantaneous centres are not required in solving problems,
therefore they may be omitted.
Example 6.1.   In a pin jointed four bar mecha-
nism, as shown in Fig. 6.9, AB = 300 mm, BC =  CD = 360
mm, and AD = 600 mm. The angle BAD  = 60°. The crankAB rotates uniformly at 100 r.p.m. Locate all the instanta-neous centres and find the angular velocity of the link BC.
Solution. Given :  N

AB = 100 r.p.m  or
                  ωAB = 2 π × 100/60 = 10.47 rad/s
Since  the length of crank AB = 300 mm = 0.3 m,
therefore velocity of point B on link AB,
*We may also say as follows: Considering links 1, 2 and 3, the instantaneous centres will be I12, I23 and I13.
The centres I12 and I23 have already been located. Similarly considering links 1, 3 and 4, the instantaneous

centres will be I13, I34 and I14, from which I14  and I34 have already been located. Thus we see that the centre
I13 lies on the intersection of the lines joining the points I12 I23 and I14 I34.Fig. 6.9

Chapter 6 :  Velocity in Mechanisms   /G6C  127
                               vB = ωAB × AB = 10.47 × 0.3 = 3.141 m/s
Location of instantaneous centres
The instantaneous centres are located as discussed below:1.  Since the mechanism consists of four links ( i.e. n = 4 ), therefore number of instantaneous
centres,
                                ( – 1) 4(4 – 1)622nnN===
2.  For a four bar mechanism, the book keeping table may be drawn as discussed in Art. 6.10.

3.  Locate the fixed and permanent instantaneous centres by inspection. These centres are I12,
I23, I34 and I14, as shown in Fig. 6.10.
4.  Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold
Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.11. Mark four points (equal tothe number of links in a mechanism) 1, 2, 3, and 4 on the circle.
Fig. 6.10
5.  Join points 1 to 2, 2 to 3, 3 to 4 and 4 to 1 to indicate the instantaneous centres already

located i.e. I12, I23, I34 and I14.
6.  Join 1 to 3 to form two triangles 1 2 3 and 3 4 1. The side 13, common to both triangles,
is responsible for completing the two triangles. Therefore the instanta-neous centre I
13 lies on the intersection of the lines joining the points I12
I23 and I34 I14 as shown in Fig. 6.10. Thus centre I13 is located. Mark
number 5 (because four instantaneous centres have already been located)on the dotted line 1 3.

7.  Now join 2 to 4 to complete two triangles 2 3 4 and 1 2 4.
The side 2 4, common to both triangles, is responsible for completingthe two triangles. Therefore centre I
24 lies on the intersection of the lines
joining the points I23 I34 and I12 I14 as shown in Fig. 6.10. Thus centre  I24
is located. Mark number 6 on the dotted line 2 4. Thus all the six instan-
taneous centres are located.
Angular velocity of the link BC
Let                           ωBC = Angular velocity of the link BC.

Since B is also a point on link BC, therefore velocity of point B on link BC,
vB=ωBC × I13 BFig. 6.11

128  /G6C   Theory of Machines
By measurement, we find that    I13 B   =   500 mm = 0.5 m
∴                             ωBC=B
133.1416.282 rad/s0.5v
IB==  Ans.
Example 6.2.   Locate all the instantaneous centres of the slider crank mechanism as shown
in Fig. 6.12. The lengths of crank OB and connecting rod AB are 100 mm and 400 mm respectively.If the crank rotates clockwise with an angular velocity of 10 rad/s, find: 1. Velocity of the slider A,
and 2. Angular velocity of the connecting rod AB.

Fig. 6.12
Solution.  Given :     ωOB = 10 rad/s; OB = 100 mm = 0.1 m
We know that linear velocity of the crank OB,
vOB=vB = ωOB × OB = 10 × 0.1 = 1 m/s
The instantaneous centres in a slider crank mechanism are located as discussed below:1.  Since there are four links ( i.e. n = 4), therefore the number of instantaneous centres,
                                     
( – 1) 4(4 – 1)622nnN===

2.  For a four link mechanism, the book keeping table may be drawn as discussed in Art. 6.10.
I23 and I34 as shown in Fig. 6.13. Since the slider (link 4) moves on a straight surface (link 1), there-
fore the instantaneous centre I14 will be at infinity.
Note: Since the slider crank mechanism has three turning pairs and one sliding pair, therefore there will be three

primary ( i.e. fixed and permanent) instantaneous centres.
Slider crank mechanism.Pin
Slider
Connecting
rodCrankBearing block

Chapter 6 :  Velocity in Mechanisms   /G6C  129
4.  Locate the other two remaining neither fixed nor permanent instantaneous centres, by
Aronhold Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.14. Mark four points1, 2, 3 and 4 (equal to the number of links in a mechanism) on the circle to indicate I
Fig. 6.13 Fig. 6.14
5.  Join 1 to 3 to form two triangles 1 2 3 and 3 4 1 in the circle diagram. The side 1 3,

common to both triangles, is responsible for completing the two triangles. Therefore the centre I13
will lie on the intersection of I12 I23 and I14 I34, produced if necessary. Thus centre I13 is located. Join
1 to 3 by a dotted line and mark number 5 on it.
6.  Join 2 to 4 by a dotted line to form two triangles 2 3 4 and 1 2 4. The side 2 4, common
to both triangles, is responsible for completing the two triangles. Therefore the centre I24 lies on the

intersection of I23 I34 and I12 I14. Join 2 to 4 by a dotted line on the circle diagram and mark number 6
on it. Thus all the six instantaneous centres are located.
By measurement, we find that
                            I13 A= 460 mm = 0.46 m; and I13 B = 560 mm = 0.56 m
1.Velocity of the slider A
Let                              vA = Velocity of the slider A.
We know that        AB
13 13vv
IA IB=
or                                            13
AB

AB
130.4610 . 8 2 m / s0.56IAvvIB=× = × =    Ans.
2.Angular velocity of the connecting rod AB
Let                     ωAB= Angular velocity of the connecting rod AB.
AB
13 13vv
IA IB== ω

130  /G6C   Theory of Machines
∴                            B
AB
1311.78 rad/s0.56v
IBω= = =   Ans.
Note:  The velocity of the slider A and angular velocity of the connecting rod AB may also be determined as
follows :
From similar triangles I13 I23 I34 and I12 I23 I24,
                               12 23 23 24
13 23 23 34II II
II II=...(i)
and                                     13 34 12 24
34 23 23 24II II
II II=...(ii)
We know that           BO B
AB
13 13vO B
IB IBω×ω= =...(∵  vB = ωOB × OB)

AB
13 13vO B
                                          12 23 23 24
OB OB
II II=ω × =ω ×  ...[From equation (i)] ...(iii)
Also                            23 24
AA B 1 3 O B 1 3 3 4
23 34.IIvI A I I
II=ω × =ω × ×...[From equation (iii)]
                                           =  ωOB × I12 I24 = ωOB × OD ...[From equation (ii)]
Example 6.3.  A mechanism, as shown in Fig. 6.15, has the following dimensions:

OA = 200 mm; AB = 1.5 m; BC = 600 mm; CD = 500 mm and BE = 400 mm. Locate all the
instantaneous centres.
If crank OA rotates uniformly at 120 r.p.m. cl ockwise, find 1. the velocity of B, C and D,
2.  the angular velocity of the links AB, BC and CD.
The above picture shows a digging machine.Hydraulic
rams
LoadExhaust
waste heat
Engine

Chapter 6 :  Velocity in Mechanisms   /G6C  131
Solution. Given : NOA= 120 r.p.m.   or     ωOA = 2 π × 120/60 = 12.57 rad/s
Since the length of crank OA = 200 mm = 0.2 m, therefore linear velocity of crank OA,
                         vOA=vA = ωOA × OA = 12.57 × 0.2 = 2.514 m/s
Fig. 6.15
The instantaneous centres are located as discussed below:1.  Since the mechanism consists of six links ( i.e. n = 6), therefore the number of instanta-
neous centres,

                                     ( – 1) 6(6 – 1)1522nnN===
2.  Make a list of all the instantaneous centres in a mechanism. Since the mechanism has 15
instantaneous centres, therefore these centres are listed in the following book keeping table.
Links 1 2 3 4 5 6
Instantaneous 12 23 34 45 56
centres 13 24 35 46(15 in number) 14 25 36
15 26
16
Fig. 6.16

132  /G6C   Theory of Machines
3.  Locate the fixed and permanent instantaneous cen-
tres by inspection. These centres are I12 I23, I34, I45, I56, I16 and
I14 as shown in Fig. 6.16.
4.  Locate the remaining neither fixed nor permanent
instantaneous centres by Aronhold Kennedy’s theorem. Draw acircle and mark points equal to the number of links such as 1, 2,3, 4, 5 and 6 as shown in Fig. 6.17. Join the points 12, 23, 34,45, 56, 61 and 14 to indicate the centres I
12, I23, I34, I45, I56, I16

and I14 respectively.
5.  Join point 2 to 4 by a dotted line to form the triangles
1 2 4 and 2 3 4. The side 2 4, common to both triangles, is
responsible for completing the two triangles. Therefore the in-
stantaneous centre I24 lies on the intersection of I12  I14 and I23 I34 produced if necessary. Thus centre
I24 is located. Mark number 8 on the dotted line 24  (because seven centres have already been lo-
cated).

cated).
6.  Now join point 1 to 5 by a dotted line to form the triangles 1 4 5 and 1 5 6. The side 1 5,
common to both triangles, is responsible for completing the two triangles. Therefore the instantaneouscentre I
15 lies on the intersection of I14 I45 and I56 I16 produced if necessary. Thus centre I15 is located.
Mark number 9 on the dotted line 1 5.
7.  Join point 1 to 3 by a dotted line to form the triangles 1 2 3 and 1 3 4. The side 1 3,

common to both triangles, is responsible for completing the two triangles. Therefore the instanta-
neous centre I13 lies on the intersection I12 I23 and I34 I14 produced if necessary. Thus centre I13 is
located. Mark number 10 on the dotted line 1 3.
8.  Join point 4 to 6 by a dotted line to form the triangles 4 5 6 and 1 4 6. The side 4 6,
common to both triangles, is responsible for completing the two triangles. Therefore, centre I46 lies

on the intersection of I45 I56 and I14 I16. Thus centre  I46 is located. Mark number 11 on the dotted line
4 6.
9.  Join point 2 to 6 by a dotted line to form the triangles 1 2 6 and 2 4 6. The side 2 6,
common to both triangles, is responsible for completing the two triangles. Therefore, centre I26 lies
on the intersection of lines joining the points I12 I16 and I24 I46. Thus centre I26 is located. Mark
number 12 on the dotted line 2 6.

10.  In the similar way the thirteenth, fourteenth and fifteenth instantaneous centre ( i.e. I35, I25
and I36) may be located by joining the point 3 to 5, 2 to 5 and 3 to 6 respectively.
I13 A = 840 mm = 0.84 m ; I13 B = 1070 mm = 1.07 m ; I14 B = 400 mm = 0.4 m ;
I14 C = 200 mm = 0.2 m; I15 C = 740 mm = 0.74 m ; I15 D = 500 mm = 0.5 m
1.Velocity of points B , C and D

Let vB, vC and vD = Velocity of the points B, C and D respectively.
13 13vv
IA IB= ...(Considering centre I13)
∴                                 A
B1 3
132.5141.07 3.2 m/s0.84vvI BIA=×= × =     Ans.
Again,                   C B
14 14vv
IB IC= ...(Considering centre I14)Fig. 6.17

Chapter 6 :  Velocity in Mechanisms   /G6C  133
∴                               B
C1 4
143.20.2 1.6m/s0.4vvI CIB=×= × =   Ans.
Similarly,              CD
15 15vv
IC ID=...(Considering centre I15)
∴                               C
D5
151.60.5 1.08 m/s0.74vvI DIC=× =× =   Ans.
2.Angular velocity of the links AB , BC and CD
Let      ωAB, ωBC and ωCD = Angular velocity of the links AB, BC and CD respectively.
We know that         A
AB
132.5142.99 rad/s0.84v
IAω= = =   Ans.

AB
                                  B
BC
143.28r a d / s0.4v
IBω= = =  Ans.
and                                       C
CD
151.62.16 rad/s0.74v
ICω= = =  Ans.
Example 6.4.   The mechanism of a wrapping machine, as shown in Fig. 6.18, has the follow-
ing dimensions :
O1A = 100 mm; AC = 700 mm; BC = 200 mm; O3C = 200 mm; O2E = 400 mm;
O2D = 200 mm and BD = 150 mm.
The crank O1A rotates at a uniform speed of 100 rad/s. Find the velocity of the point E of the

bell crank lever by instantaneous centre method.
Fig. 6.18
Solution.  Given : ωO1A= 100 rad/s; O1 A = 100 mm = 0.1 m
We know that the linear velocity of crank O1 A,
vO1A=vA = ωO1A × O1 A = 100 × 0.1 = 10 m/s
Now let us locate the required instantaneous centres as discussed below :1.  Since the mechanism consists of six links ( i.e. n = 6), therefore number of instantaneous
centres,
                                    ( – 1) 6(6 – 1)1522nnN===

2.  Since the mechanism has 15 instantaneous centres, therefore these centres may be listed in
the book keeping table, as discussed in Example 6.3.

134  /G6C   Theory of Machines
Fig. 6.19 Fig. 6.20
3.  Locate the fixed and the permanent instantaneous centres by inspection. These centres are
I12, I23, I34, I35, I14, I56 and I16 as shown in Fig. 6.19.
4.  Locate the remaining neither fixed nor perm anent instantaneous  centres by Aronhold
Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.20. Mark six points on thecircle (i.e. equal to the number of links in a mechanism), and join 1 to 2, 2 to 3, 3 to 4, 3 to 5, 4 to

1, 5 to 6, and 6 to 1, to indicate the fixed and permanent instantaneous centres i.e. I
12, I23, I34, I35,
I14, I56, and I16 respectively.
5.  Join 1 to 3 by a dotted line to form two triangles 1 2 3 and 1 3 4. The side 1 3, common
to both triangles, is responsible for completing the two triangles. Therefore the instantaneous cen-
tre I13 lies on the intersection of the lines joining the points I12 I23 and I14 I34 produced if necessary.

Thus centre I13 is located. Mark number 8 (because seven centres have already been located) on the
dotted line 1 3.
6.  Join 1 to 5 by a dotted line to form two triangles 1 5 6 and 1 3 5. The side 1 5, common to
both triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I15
lies on the intersection of the lines joining the points I16 I56 and I13 I35 produced if necessary. Thus
centre I15 is located. Mark number 9 on the dotted line 1 5.

Note:  For the given example, we do not require other instantaneous centres.
             I13 A = 910 mm = 0.91 m ; I13 B = 820 mm = 0.82 m ; I15 B = 130 mm = 0.13 m;
             I15 D = 50 mm = 0.05 m ; I16 D = 200 mm = 0.2 m; I16 E = 400 mm = 0.4 m
Velocity of point E on the bell crank lever
Let                    vE= Velocity of point E on the bell crank lever,
                                  vB= Velocity of point B, and

vD= Velocity of point D.
We know that       AB
13 13vv

Chapter 6 :  Velocity in Mechanisms   /G6C  135
∴                               A
B1 3
13100.82 9.01m/s0.91vvI BIA=×= × =   Ans.
and                                    BD
15 15vv
IB ID= ...(Considering centre I15)
∴                               B
D1 5
159.010.05 3.46 m/s0.13vvI DIB=×=× =   Ans.
Similarly,             DE
16 16vv
ID IE= ...(Considering centre I16)
∴                               D
E1 6
163.460.4 6.92 m/s0.2vvI EID=×= × =   Ans.

E1 6
Example 6.5.   Fig. 6.21 shows a sewing needle bar mechanism
O1ABO2CD wherein the different dimensions are as follows :
Crank O1A = 16 mm; ∠β = 45°; Vertical distance between O1 and
O2 = 40 mm; Horizontal distance between O1 and O2 = 13 mm; O2 B = 23
mm; AB = 35 mm; ∠ O2 BC = 90°; BC = 16 mm; CD = 40 mm. D lies
vertically below O1.
Find the velocity of needle at D for the given configuration. The crank
O1A rotates at 400 r.p.m.

Solution. Given : NO1A = 400 r.p.m     or     ωO1A = 2π × 400/60 =
41.9 rad/s; O1 A = 16 mm = 0.016 m
We know that linear velocity of the crank O1A,
                       vO1A = vA = ωO1A × O1A = 41.9 × 0.016 = 0.67 m/s
Now let us locate the required instantaneous centres as discussed
below :
1.  Since the mechanism consists of six links ( i.e. n = 6), therefore number of instantaneous
centres,
                                     (– 1 ) 6 ( 6 – 1 )1522nnN== =

3.  Locate the fixed and permanent instantaneous centres by inspections. These centres are
I12, I23, I34, I45, I56, I16 and I14, as shown in Fig. 6.22.
Fig. 6.22Fig. 6.21

136  /G6C   Theory of Machines
4.  Locate the remaining neither fixed nor permanent in-
stantaneous centres by Aronhold Kennedy’s theorem. This is doneby circle diagram as shown in Fig. 6.23. Mark six points on thecircle (i.e. equal to the number of links in a mechanism) and join 1
to 2, 2 to 3, 3 to 4, 4 to 5, 5 to 6, 6 to 1 and 1 to 4 to indicate thefixed and permanent instantaneous centres i.e. I
12, I23, I34, I45, I56,
I16 and I14 respectively.

5. Join 1 to 3 by a dotted line to form two triangles 1 2
3 and 1 3 4. The side 1 3, common to both the triangles, isresponsible for completing the two triangles. Therefore theinstantaneous centre I
13 lies on the intersection of I12 I23 and
I14 I34  produced if necessary. Thus centre I13 is located. Mark
number 8 (because seven centres have already been located)on the dotted line 1 3.

6.  Join 1 to 5 by a dotted line to form two triangles 1 5 6 and 1 4 5. The side 1 5, common
to both the triangles, is responsible for completing the two triangles. Therefore the instantaneouscentre I
15 lies on the intersection of I16 I56 and I14 I45 produced if necessary. Thus centre I15 is
located. Mark number 9 on the dotted line 1 5.

I13 A= 41 mm = 0.041 m ; I13 B = 50 mm = 0.05 m ; I14 B = 23 mm = 0.023 m ;
I14 C= 28 mm = 0.028 m ; I15 C = 65 mm = 0.065 m ; I15 D = 62 mm = 0.062 m
Let vB= Velocity of point B,
vC= Velocity of point C, and
vD= Velocity of the needle at D.
13 13vv
∴                               A
B1 3
130.670.05 0.817 m/s0.041vvI BIA=×=× =
and                                    C B
14 14vv
IB IC=...(Considering centre I14)

14 14vv
∴                               B
C1 4
140.8170.028 0.995 m/s0.023vvI CIB=×= ×=
Similarly,                CD
15 15vv
∴                               C
D1 5
150.9950.062 0.95 m/s0.065vvI DIC=×=×=  Ans.Fig. 6.23

Chapter 6 :  Velocity in Mechanisms   /G6C  137
Example 6.6.  Fig. 6.24 shows a Whitworth quick
return motion mechanism. The various dimensions inthe mechanism are as follows :
OQ = 100 mm ; OA = 200 mm ; QC = 150 mm ;
and CD = 500 mm.
The crank OA makes an angle of 60° with the
vertical and rotates at 120 r.p.m. in the cl ockwise
direction.
Locate all the instantaneous centres and find the
velocity of ram D .
Solution :  Given. N
OA = 120 r.p.m.   or   ωOA =
2 π × 120 / 60 = 12.57 rad/s

The instantaneous centres are located as discussed below :1.  Since the mechanism consists of six links ( i.e. n = 6), therefore the number of instanta-
2.  Make a list of all the instantaneous centres in a mechanism as discussed in Example 6.3.

I23, I34, I45, I56, I16 and I14 as shown in Fig. 6.25.
Fig. 6.25
Kennedy’s theorem. Draw a circle and mark points equal to the number of links such as 1, 2, 3, 4, 5,Fig. 6.24

138  /G6C   Theory of Machines
and 6 as shown in Fig. 6.26. Join the points 1 2, 2 3, 3 4, 4 5, 5 6,
6 1 and 1 4 to indicate the centres I12, I23, I34, I45, I56, I16 and I14
5.  Join point 1 to 3 by a dotted line to form two triangles
1 2 3 and 1 3 4. The side 1 3, common to both the triangles, isresponsible for completing the two triangles. Therefore the instan-taneous centre I
13 lies on the intersection of I12 I23, and I14 I34  pro-

duced if necessary. Thus centre I13 is located. Mark number 8 on the
dotted line 1 3 (because seven centres have already been located).
6.  Join point 1 to 5 by a dotted line to form two triangles
1 4 5 and 1 5 6. The side 1 5, common to both the triangles, isresponsible for completing the two triangles. Therefore the instan-taneous centre I
15 lies on the intersection of I14 I45 and I56 I16 produced if necessary. Thus centre I15 is

7.  Join point 2 to 4 by a dotted line to form two triangles 1 2 4 and 2 3 4. The side 2 4,
common to both the triangles, is responsible for completing the two triangles. Therefore the instanta-neous centre I
24 lies on the intersection of I12 I14 and I23 I34 produced if necessary. Thus centre I24 is
located. Mark number 10 on the dotted line 2 4.
8.  Join point 2 to 5 by a dotted line to form two triangles 1 2 5 and 2 4 5. The side 2 5,

25 lies on the intersection of I12 I15 and I24 I45 produced  if necessary. Thus centre I25 is
located. Mark number 11 on the dotted line 2 5.
9.  Join point 2 to 6 by a dotted line to form two triangles 1 2 6 and 2 5 6. The side 2 6

26 lies on the intersection of I12 I16 and I25 I56 produced if necessary. Thus centre I26 is
located. Mark number 12 on the dotted line 2 6.
10.  Join point 3 to 5 by a dotted line to form two triangles 2 3 5 and 3 4 5. The side 3 5,
35 lies on the intersection of I23 I25 and I34 I45 produced if necessary. Thus centre I35is
located. Mark number 13 on the dotted line 3 5.

11.  Join point 3 to 6 by a dotted line to form two triangles 1 3 6 and 3 5 6. The side 3 6,
36 lies on the intersection of I13 I16 and I35 I56 produced if necessary. Thus centre I36 is
located. Mark number 14 on the dotted line 3 6.
Note.  The centre I36 may also be obtained by considering the two triangles 2 3 6 and 3 4 6.

12.  Join point 4 to 6 by a dotted line to form two triangles 1 4 6 and 4 5 6. The side 4 6,
46 lies on the intersection of I14 I16 and I45 I56 produced if necessary. Thus centre I46 is
located. Mark number 15 on the dotted line 4 6.
Velocity of ram D
By measurement, we find that I12 I26 = 65 mm = 0.065 m
∴  Velocity of ram,  vD = ωOA × I12 I26 = 12.57 × 0.065 = 0.817 m/s   Ans.

EXERCISES
1.Locate all the instantaneous centres for a four bar mechanism as shown in Fig. 6.27.
The lengths of various links are : AD = 125 mm ; AB = 62.5 mm ; BC = CD = 75 mm.
If the link AB rotates at a uniform speed of 10 r.p.m. in the clockwise direction, find the angular
velocity of the links BC and CD. [Ans. 0.63 rad/s ; 0.65 rad/s]Fig. 6.26

Chapter 6 :  Velocity in Mechanisms   /G6C  139
Fig. 6.27 Fig. 6.28
2.Locate all the instantaneous centres for the crossed four bar mechanism as shown in Fig. 6.28. The dimen-
sions of various links are : CD = 65 mm; CA = 60 mm ; DB = 80 mm ; and
AB = 55 mm.
Find the angular velocities of the links AB and DB, if the crank CA rotates at 100 r.p.m. in the
anticlockwise direction.[Ans. 50 rad/s ; 27 rad/s]
3.Locate all the instantaneous centres

of the mechanism as shown in Fig.6.29. The lengths of various linksare : AB = 150 mm ; BC = 300 mm
; CD = 225 mm ; and CE = 500
mm.
When the crank AB rotates in the
anticlockwise direction at a uni-form speed of 240 r.p.m. ; find 1.Velocity of the slider E, and 2.
Angular velocity of the links BC
and CE.
[Ans. 1.6 m/s ; 2.4 rad/s ; 6.6 rad/s]
4.The crank OA of a mechanism, as shown in Fig. 6.30, rotates clockwise at 120 r.p.m. The lengths of

various links are : OA = 100 mm ; AB = 500 mm ; AC = 100 mm and CD = 750 mm.
Fig. 6.30Fig. 6.29

140  /G6C   Theory of Machines
Find, by instantaneous centre method : 1. Velocity of point C ; 2. Velocity of slider D ; and 3. Angular
velocities of the links AB and CD.  [Ans. 0.115 m/s; 0.065 m/s; 3 rad/s; 1.3 rad/s]
5.A mechanism, as shown in Fig. 6.31, has the following dimensions :
O1 A = 60 mm ; AB = 180 mm ; O2 B = 100 mm ; O2 C = 180 mm and CD = 270 mm.
The crank O1 A rotates clockwise at a uniform speed of 120 r.p.m. The block D moves in vertical

guides. Find, by instantaneous centre method, the velocity of D and the angular velocity of CD.
[Ans. 0.08 m/s ; 1.43 rad/s]
6.The lengths of various links of a mechanism, as shown in Fig. 6.32, are : OA = 0.3 m ;
AB = 1 m ; CD = 0.8 m ; and AC = CB.
Determine, for the given configuration, the velocity of the slider D if the crank OA rotates at 60 r.p.m.
in the clockwise direction. Also find the angular velocity of the link CD. Use instantaneous centre
method. [Ans. 480 mm/s ; 2.5 rad/s]

Fig. 6.31 Fig. 6.32
7.In the mechanism shown in Fig. 6.33, find the instantaneous centres of the links B, C and D.
Fig. 6.33
If  the link  A rotates clockwise at 10 rad/s, find the angular velocity of link E. The lengths of various
links are as follows:
Link A = 25 mm ; Link B = Link C = 100 mm ; Link D = Link E = 50 mm. The link D is hinged to link
B at 25 mm from the left hand end of  link B. [Ans. 1.94 rad/s]

Chapter 6 :  Velocity in Mechanisms   /G6C  141
8.The dimensions of various links in a mechanism, as shown in Fig. 6.34, are as follows :
Fig. 6.34
AB = 25 mm ; BC = 175 mm ; CD = 60 mm ; AD = 150 mm ; BE = EC ; and EF = FG = 100 mm.
The crank AB rotates at 200 r.p.m. When the angle BAD is 135°, determine by instantaneous centre
method : 1. Velocity of G, 2. Angular velocity of EF, and 3. Velocity of sliding of EF in the swivel
block S.
[Ans. 120 mm/s ; 6.5 rad/s ; 400 mm/s]

1.What do you understand by the instantaneous centre of rotation (centro) in kinematic of
machines? Answer briefly.
2.Explain, with the help of a neat sketch, the space centrode and body centrode.
3.Explain with sketch the instantaneous centre method for determination of velocities of links and
mechanisms.
4.Write the relation between the number of instantaneous centres and the number of links in a mechanism.
5.Discuss the three types of instantaneous centres for a mechanism.

6.State and prove the ‘Aronhold Kennedy’s Theorem’ of three instantaneous centres.
1.The total number of instantaneous centres for a mechanism consisting of n links are
(a)2n(b)n
(c)–1
2n
(d)(– 1 )
2nn
2.According to Aronhold Kennedy’s theorem, if three bodies move relatively to each other, their
instantaneous centres will lie on a
(a) straight line ( b) parabolic curve
 (c) ellipse ( d) none of these

142  /G6C   Theory of Machines
3.In a mechanism, the fixed instantaneous centres are those centres which
(a) remain in the same place for all configurations of the mechanism
(b)  vary with the configuration of the mechanism
(c) moves as the mechanism moves, but joints are of permanent nature
(d) none of the above
4.The instantaneous centres which vary with the configuration of the mechanism, are called
(a) permanent instantaneous centres
(b) fixed instantaneous centres

(c) neither fixed nor permanent instantaneous centres
(d) none of these
5.When a slider moves on a fixed link having curved surface, their instantaneous centre lies
(a) on their point of contact ( b) at the centre of curvature
(c) at the centre of circle ( d) at the pin joint
ANSWERS
1.(d) 2.(a) 3.(a) 4.(c) 5.(b)

Chapter 7 : Velocity in Mechanisms    /G6C  143
7.1. Introduction
We have discussed, in the previous chapter, the in-
stantaneous centre method for finding the velocity of variouspoints in the mechanisms. In this chapter, we shall discussthe relative velocity method for determining the velocity ofdifferent points in the mechanism. The study of velocity analy-sis is very important for determining the acceleration of pointsin the mechanisms which is discussed in the next chapter.

7.2. Relative Velocity of Two Bodies
Moving in Straight Lines
Here we shall discuss the application of vectors for
the relative velocity of two bodies moving along parallel linesand inclined lines, as shown in Fig. 7.1 ( a) and 7.2 ( a)
Consider two bodies A and B moving along parallel
lines in the same direction with absolute velocities v
A and
vB  such that vA >  vB , as shown in Fig. 7.1 ( a). The relative
velocity of A with respect to B,

vAB = Vector difference of vA and vB = ABvv−
...(i)
143Velocity in
Mechanisms
(Relative Velocity Method)7
Features
2. Relative Velocity of  Two
Bodies Moving in StraightLines.
3. Motion of a Link.4. Velocity of a Point on a Link
by Relative Velo city Method.
5. Velocities in a Slider Crank
Mechanism.
6. Rubbing V elocity at a Pin
Joint.
7. Forces Acting in a
8. Mechanical Advantage.
TOGGLE
PRESS

144  /G6C   Theory of Machines
From Fig. 7.1 ( b), the relative velocity of A with respect to B (i.e. vAB) may be written in the
vector form as follows :
                                    –ba oa ob=
Fig. 7.1. Relative velocity of two bodies moving along parallel lines.
Similarly, the relative velocity of B with respect to A,
            vBA = Vector difference of vB and ABA–vvv= ...(ii)
or                      –ab ob oa=
Now consider the body B moving in an

inclined direction as shown in Fig. 7.2 ( a). The
relative velocity of A with respect to B may be
obtained by the law of parallelogram of veloci-ties or triangle law of velocities. Take any fixedpoint o and draw vector oa to represent v
A in
magnitude and direction to some suitable scale.Similarly, draw vector ob to represent v
B in mag-
nitude and direction to the same scale. Then vec-tor ba represents the relative velocity of A with

respect to B as shown in Fig. 7.2 ( b). In the simi-
lar way as discussed above, the relative velocityof A with respect to B,
                          v
AB = Vector difference of vA and BAB–vvv=
or                                   –ba oa ob=
Fig. 7.2. Relative velocity of two bodies moving along inclined lines.
 vBA= Vector difference of vB and ABA–vvv=
or                                            ab ob oa=−

Chapter 7 : Velocity in Mechanisms    /G6C  145
From above, we conclude that the relative velocity of point A with respect to B (vAB) and the
relative velocity of point B with respect A (vBA) are equal in magnitude but opposite in direction, i.e.
                                 AB BAor vv b a a b=− =−
Note:  It may be noted that to find vAB, start from point b towards a and for vBA, start from point a towards b.
7.3. Motion of a Link
Consider two points A and B on a rigid link AB, as

shown in Fig. 7.3 ( a). Let one of the extremities ( B) of the link
move relative to A, in a clockwise direction. Since the dis-
tance from A to B remains the same, therefore there can be no
relative motion between A and B, along the line AB. It is thus
obvious, that the relative motion of B with respect to A must
be perpendicular to AB.
Hence velocity of any point on a link with respect to

another point on the same link is always perpendicular tothe line joining these points on the configuration (or space)diagram.
The relative velocity of B with respect to A (i.e. v
BA) is represented by the vector ab and is
perpendicular to the line AB as shown in Fig. 7.3 ( b).
Let                                     ω  = Angular velocity of the link AB about A.
We know that the velocity of the point B with respect to A,
                                      BA . va bA B== ω ...(i)

Similarly, the velocity of any point C on AB with respect to A,
                                      CA . va cA C== ω ...(ii)
                                     CA
BA.
.vac AC AC
vA B A Babω== =ω...(iii)
Thus, we see from equation (iii), that the point c on the vector ab divides it in the same ratio
as C divides the link AB.
Note:  The relative velocity of A with respect to B is represented by ba, although A may be a fixed point. The

motion between A and B is only relative. Moreover, it is immaterial whether the link moves about A in a
clockwise direction or about B in a clockwise direction.
7.4. Velocity of a Point on a Link by Relative Velocity Method
The relative velocity method is based upon the relative velocity of the various points of the
link as discussed in Art. 7.3.
Consider two points A and B on a link as shown in Fig. 7.4 ( a). Let the absolute velocity of the

point A  i.e. vA is known in magnitude and direction and the absolute velocity of the point B i.e. vB is
known in direction only. Then the velocity of B may be determined by drawing the velocity diagram
as shown in Fig. 7.4 ( b). The velocity diagram is drawn as follows :
1.Take some convenient point o, known as the pole.
2.Through o, draw oa parallel and equal to vA, to some suitable scale.
3.Through a, draw a line perpendicular to AB of Fig. 7.4 ( a). This line will represent the

velocity of B with respect to A, i.e. vBA.
4.Through o, draw a line parallel to vB intersecting the line of vBA at b.Fig. 7.3. Motion of a Link.

146  /G6C   Theory of Machines
5.Measure ob, which gives the required velocity of point B ( vB), to the scale.
(a)  Motion of points on a link. ( b)  Velocity diagram.
Fig. 7.4
Notes : 1.   The vector ab which represents the velocity of B with respect to A (vBA) is known as velocity of
image of the link AB.
2.  The absolute velocity of any point C on AB may be determined by dividing vector ab at c in the same
ratio as C divides AB in Fig. 7.4 ( a).
In other words
ac AC
ab AB=

ac AC
ab AB=
Join oc. The *vector oc represents the absolute velocity
of point C (vC) and the vector ac represents the velocity of C
with respect to A i.e. vCA.
3. The absolute velocity of any other point D outside
AB, as shown in Fig. 7.4 ( a), may also be obtained by com-
pleting the velocity triangle abd and similar to triangle ABD,
as shown in Fig. 7.4 ( b).
4. The angular velocity of the link AB may be found
by dividing the relative velocity of B with respect to A (i.e.

vBA) to the length of the link AB. Mathematically, angular
velocity of the link AB,
BA
ABva b
AB ABω= =
7.5. Velocities in Slider Crank Mechanism
In the previous article, we have discused the relative velocity method for the velocity of any
point on a link, whose direction of motion and velocity of some other point on the same link is known.The same method may also be applied for the velocities in a slider crank mechanism.

A slider crank mechanism is shown in Fig. 7.5 ( a). The slider A is attached to the connecting
rod  AB. Let the radius of crank OB be r and let it rotates in a clockwise direction, about the point O
with uniform angular velocity ω rad/s. Therefore, the velocity of B i.e. v
B is known in magnitude and
direction. The slider reciprocates along the line of stroke AO.
The velocity of the slider A (i.e. vA) may be determined by relative velocity method as

1. From any point o, draw vector ob parallel to the direction of vB  (or perpendicular to OB)
such that ob = vB  = ω.r, to some suitable scale, as shown in Fig. 7.5 ( b).
*The absolute velocities of the points are measured from the pole ( i.e. fixed points) of the velocity diagram.

Chapter 7 : Velocity in Mechanisms    /G6C  147
Fig. 7.6. Links connected by pin joints.(a) Slider crank mechanism. ( b) Velocity diagram.
Fig. 7.5
2. Since AB is a rigid link, therefore the velocity of A relative to B is perpendicular to AB.
Now draw vector ba perpendicular to AB to represent the velocity of A with respect to B i.e. vAB.
3. From point o, draw vector oa parallel to the path of motion of the slider A (which is along

AO only). The vectors ba and oa intersect at a. Now oa represents the velocity of the slider A i.e. vA,
to the scale.
The angular velocity of the connecting rod AB (ωAB) may be determined as follows:
BA
ABva b
AB ABω= = (Anticlockwise about A)
The direction of vector ab (or ba) determines the sense of ωAB which shows that it is
anticlockwise.
Note :  The absolute velocity of any other point E on the connecting rod AB may also be found out by dividing

vector ba such that be/ba = BE/BA . This is done by drawing any line bA1 equal in length of BA. Mark bE1 = BE.
Join a A1. From E1 draw a line E1e parallel to a A1. The vector oe now represents the velocity of E and vector ae
represents the velocity of E with respect to A.
7.6. Rubbing Velocity at a Pin Joint
The links in a mechanism are mostly connected by means of pin joints. The rubbing velocity

is defined as the algebraic sum between the angular velocities of the two links which are connected
by pin joints, multiplied by the radius of the pin.
Consider two links OA and OB connected by a pin joint at O as shown in Fig. 7.6.
Let ω1= Angular velocity of the link OA or
the angular velocity of the  point A
 with respect to O.
ω2= Angular velocity of the link OB or
the angular velocity of the point B
with respect to O, and
r= Radius of the pin.

According to the definition,Rubbing velocity at the pin joint O
=(ω
1 – ω2) r, if the links move in the same direction
=(ω1 + ω2) r, if the links move in the opposite direction
Note : When the pin connects one sliding member and the other turning member, the angular velocity of the
sliding member is zero. In such cases,
Rubbing velocity at the pin joint = ω.r
where                             ω  = Angular velocity of the turning member, and

148  /G6C   Theory of Machines
Example 7.1.  In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40
mm long and rotates at 120 r.p.m. cl ockwise, while the link CD  = 80 mm oscillates about D. BC and
AD are of equal length. Find the angular velocity of link CD when angle BAD = 60°.
Solution. Given : NBA  = 120 r.p.m. or ωBA  = 2 π × 120/60 = 12.568 rad/s
Since the length of crank AB  = 40 mm = 0.04 m, therefore velocity of B with respect to A or

velocity of B, (because A is a fixed point),
                      vBA = vB  = ωBA × AB = 12.568 × 0.04 = 0.503 m/s
(a) Space diagram (All dimensions in mm). ( b) Velocity diagram.
Fig. 7.7
First of all, draw the space diagram to some suitable scale, as shown in Fig. 7.7 ( a). Now the
velocity diagram, as shown in Fig. 7.7 ( b), is drawn as discussed below :
1. Since the link AD is fixed, therefore points a and d are taken as one point in the velocity

diagram. Draw vector ab perpendicular to BA, to some suitable scale, to represent the velocity of B
with respect to A or simply velocity of B (i.e. vBA or vB) such that
vector ab = vBA = vB = 0.503 m/s
2. Now from point b, draw vector bc perpendicular to CB to represent the velocity of C with
respect to B (i.e. vCB) and from point d, draw vector dc perpendicular to CD to represent the velocity
of C with respect to D or simply velocity of C (i.e. vCD or vC). The vectors bc and dc intersect at c.

vCD = vC = vector dc = 0.385 m/s
We know that CD = 80 mm = 0.08 m
∴ Angular velocity of link CD,
                               CD
CD0.385
0.08v
CDω= =  = 4.8 rad/s (clockwise about D) Ans.
Example 7.2.  The crank and connecting rod
of a theoretical steam engine are 0.5 m and 2 m longrespectively. The crank makes 180 r.p.m. in theclockwise direction. When it has turned 45° from theinner dead centre position, determine : 1. velocity of

piston, 2. angular velocity of connecting rod,
3. velocity of point E on the connecting rod 1.5 m
from the gudgeon pin, 4. velocities of rubbing at the
pins of the crank shaft, crank and crosshead whenthe diameters of their pins are 50 mm, 60 mm and 30mm respectively,  5. position and linear velocity of any
point G on the connecting rod which has the leastvelocity relative to crank shaft .

Chapter 7 : Velocity in Mechanisms    /G6C  149
Solution. Given : NBO= 180 r.p.m. or ωBO  = 2 π × 180/60 = 18.852 rad/s
Since the crank length OB = 0.5 m, therefore linear velocity of B with respect to O or velocity
of B (because O is a fixed point),
vBO=vB = ωBO × OB = 18.852 × 0.5 = 9.426 m/s
. . . (Perpendicular to BO)
1.  Velocity of piston
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.8 ( a). Now the

velocity diagram, as shown in Fig. 7.8 ( b), is drawn as discussed below :
1.  Draw vector ob perpendicular to BO, to some suitable scale, to represent the velocity of B
with respect to O or velocity of B such that
vector  ob=vBO  = vB = 9.426 m/s
2.  From point b, draw vector bp perpendicular to BP to represent velocity of P with respect
to B (i.e. vPB) and from point o, draw vector op parallel to PO to represent velocity of P with respect

to O (i.e. vPO or simply vP). The vectors bp and op intersect at point p.
By measurement, we find that velocity of piston P,
vP= vector op = 8.15 m/s  Ans.
(a) Space diagram. ( b) Velocity diagram.
Fig. 7.8
2.  Angular velocity of connecting rod
From the velocity diagram, we find that the velocity of P with respect to B,
vPB= vector bp = 6.8 m/s
Since the length of connecting rod PB is 2 m, therefore angular velocity of the connecting rod,
                               PB
PB6.8
2v

PB
PB6.8
2v
PBω= =  = 3.4 rad/s (Anticlockwise)  Ans.
3.  Velocity of point E on the connecting rod
The velocity of point E on the connecting rod 1.5 m from the gudgeon pin ( i.e. PE = 1.5 m)
is determined by dividing the vector bp at e in the same ratio as E divides PB in Fig. 7.8 ( a). This is
done in the similar way as discussed in Art 7.6. Join oe. The vector oe represents the velocity of E. By
measurement, we find that velocity of point E,

vE= vector oe = 8.5 m/s     Ans.
Note :  The point e on the vector bp may also be obtained as follows :
                                 BE be
BP bp=  or   BE bpbeBP×=
4. Velocity of rubbing
We know that diameter of crank-shaft pin at O,
dO= 50 mm = 0.05 m

150  /G6C   Theory of Machines
Diameter of crank-pin at B,
dB= 60 mm = 0.06 m
and diameter of cross-head pin,
dC= 30 mm = 0.03 m
We know that velocity of rubbing at the pin of crank-shaft
       = O
BO0.0518.8522d×ω = ×  = 0.47 m/s   Ans.
Velocity of rubbing at the pin of crank
      =  B
BO PB0.06( ) (18.85 3.4)22dω+ ω= +  = 0.6675 m/s   Ans.
...(/G51 ωBO is clockwise and ωPB is anticlockwise.)
and velocity of rubbing at the pin of cross-head
  = C
PB0.033.422d×ω = ×  = 0.051 m/s   Ans.

= C
...(/G51 At the cross-head, the slider does not rotate and only the connecting rod has angular motion.)
5.Position and linear velocity of point G on the connecting rod which has the least velocity
relative to crank-shaft
The position of point G on the connecting rod which has the least velocity relative to crank-
shaft is determined by drawing perpendicular from o to vector bp. Since the length of og will be the

least, therefore the point g represents the required position of G on the connecting rod.
   vector bg=5   m / s
The position of point G on the connecting rod is obtained as follows:
                                bg BG
bp BP=  or  526.8bgBG BPbp=×=×  = 1.47 m   Ans.
By measurement, we find that the linear velocity of point G,
vG= vector og = 8 m/s   Ans.
Example 7.3.  In Fig. 7.9, the angular velocity of
the crank OA is 600 r.p.m. Det ermine the linear velocity of

the slider D and the angular velocity of the link BD, whenthe crank is inclined at an angle of 75° to the vertical. Thedimensions of various links are : OA = 28 mm ; AB  = 44 mm ;BC  49 mm ; and BD  = 46 mm. The centre distance betweenthe centres of rotation O and C is 65 mm. The path of travelof the slider is 11 mm below the fi xed point C . The slider
moves along a horizontal path and OC is vertical .
Solution. Given: N
AO  = 600 r.p.m.   or
ωAO  = 2 π × 600/60 = 62.84 rad/s

Since OA = 28 mm = 0.028 m, therefore velocity of
A with respect to O or velocity of A (because O is a fixed point),
  vAO=vA = ωAO × OA = 62.84 × 0.028 = 1.76 m/s
. . . (Perpendicular to OA)
Linear velocity of the slider D
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.10 ( a). Now the
velocity diagram, as shown in Fig. 7.10 ( b), is drawn as discussed below :Fig. 7.9

Chapter 7 : Velocity in Mechanisms    /G6C  151
1. Since the points O and C are fixed, therefore these points are marked as one point, in the
velocity diagram. Now from point o, draw vector oa perpendicular to OA, to some suitable scale, to
represent the velocity of A with respect to O or simply velocity of A such that
vector oa=vAO  = vA  = 1.76 m/s
Fig. 7.10
2. From point a, draw vector ab perpendicular to AB to represent the velocity of B with

respect A (i.e. vBA) and from point c, draw vector cb perpendicular to CB to represent the velocity of
B with respect to C or simply velocity of B (i.e. vBC or vB). The vectors ab and cb intersect at b.
3. From point b, draw vector bd perpendicular to BD to represent the velocity of D with
respect to B (i.e. vDB) and from point o, draw vector od parallel to the path of motion of the slider D
which is horizontal, to represent the velocity of D (i.e. vD). The vectors bd and od intersect at d.

By measurement, we find that velocity of the slider D,
vD= vector od = 1.6 m/s  Ans.
Angular velocity of the link BD
By measurement from velocity diagram, we find that velocity of D with respect to B,
vDB= vector bd = 1.7 m/s
Since the length of link BD = 46 mm = 0.046 m, therefore angular velocity of the link BD,
DB
BD1.7
0.046v
BDω= =  = 36.96 rad/s (Clockwise about B)  Ans.
Example 7.4.  The mechanism , as shown in Fig. 7.11, has the dimensions of various links as
follows :

follows :
AB = DE = 150 mm ; BC = CD = 450 mm ; EF = 375 mm.
Fig. 7.11

152  /G6C   Theory of Machines
The crank  AB makes an angle of 45° with the horizontal and rotates about A in the clockwise
direction at a uniform speed of 120 r.p.m. The lever DC  oscillates about the fixed point D, which is
connected to AB by the coupler BC.
The block F moves in the horizontal guides, being driven by the link EF. Determine: 1. velocity of
the block F, 2. angular velocity of DC, and  3. rubbing speed at the pin C which is 50 mm in diameter .

Solution. Given : NBA = 120 r.p.m. or ωBA =  2 π × 120/60 = 4 π rad/s
Since the crank length AB = 150 mm = 0.15 m, therefore velocity of B with respect to A or
simply velocity of B (because A is a fixed point),
vBA=vB = ωBA × AB = 4 π × 0.15 = 1.885 m/s
. . . (Perpendicular to AB)
1.  Velocity of the block F
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.12 ( a). Now the
velocity diagram, as shown in Fig. 7.12 ( b), is drawn as discussed below:

Fig. 7.12
1. Since the points A and D are fixed, therefore these points are marked as one point * as
shown in Fig. 7.12 ( b). Now from point a, draw vector ab perpendicular to AB, to some suitable scale,
to represent the velocity of B with respect to A or simply velocity of B, such that
vector ab=vBA = vB = 1.885 m/s
2. The point C moves relative to B and D, therefore draw vector bc perpendicular to BC to

represent the velocity of C with respect to B (i.e. vCB), and from point d, draw vector dc perpendicular
to DC to represent the velocity of C with respect to D or simply velocity of C (i.e. vCD or vC). The
vectors bc and dc intersect at c.
3. Since the point E lies on DC, therefore divide vector dc in e in the same ratio as E divides
CD in Fig. 7.12 ( a). In other words
ce/cd=CE/CD
The point e on dc may be marked in the same manner as discussed in Example 7.2.

4. From point e, draw vector ef perpendicular to EF to represent the velocity of F with respect
to E (i.e. vFE) and from point d draw vector df parallel to the path of motion of F, which is horizontal,
to represent the velocity of F i.e. vF. The vectors ef and df intersect at f.
By measurement, we find that velocity of the block F,
vF= vector df = 0.7 m/s   Ans.
2.  Angular velocity of  DC
By measurement from velocity diagram, we find that velocity of C with respect to D,

vCD= vector dc = 2.25 m/s
*When the fixed elements of the mechanism appear at more than one place, then all these points lie at one
place in the velocity diagram.

Chapter 7 : Velocity in Mechanisms    /G6C  153
Since the length of link DC = 450 mm = 0.45 m, therefore angular velocity of DC,
CD
DC2.25
0.45v
DCω= =  = 5 rad/s . . . (Anticlockwise about D)
3.  Rubbing speed at the pin C
We know that diameter of pin at C,
dC= 50 mm = 0.05 m   or   Radius , rC   = 0.025 m
From velocity diagram, we find that velocity of C with respect to B,
vCB= vector bc = 2.25 m/s . . . (By measurement)
Length BC= 450 mm = 0.45 m
∴ Angular velocity of BC,
CB
CB2.255rad/s

CB
0.45v
BCω= = = ... (Anticlockwise about B)
We know that rubbing speed at the pin C
= (ωCB – ωCD) rC = (5 – 5) 0.025 = 0 Ans.
Example 7.5. In a mechanism shown in Fig. 7.13, the crank OA is 100 mm long and rotates
clockwise about O at 120 r.p.m. The connceting rod AB is 400 mm long.
Fig. 7.13.
At a point C on AB, 150 mm from A, the rod CE 350 mm long is attached. This rod CE slides

in a slot in a trunnion at D. The end E is connected by a link EF, 300 mm long to the horizontally
moving slider F.
For the mechanism in the position shown, find 1. velocity of F, 2. velocity of sliding of CE in
the trunnion, and 3. angular velocity of CE .
Solution. Given : vAO = 120 r.p.m.  or
ωAO = 2 π × 120/60 = 4 π rad/s
Since the length of crank OA = 100 mm
= 0.1 m, therefore velocity of A with respect to O
or velocity of A (because O is a fixed point),

vAO=vA = ωAO × OA = 4 π × 0.1 = 1.26 m/s
. . . (Perpendicular to AO)
1.  Velocity of F
First of all draw the space diagram, to
some suitable scale, as shown in Fig. 7.14 ( a).
Now the velocity diagram, as shown in Fig. 7.14(b), is drawn as discussed below :
An aircraft uses many mechanisms in engine,
power transmission and steering.
Note : This picture is given as additional information
and is not a direct example of the current chapter.

154  /G6C   Theory of Machines
1. Draw vector oa perpendicular to AO, to some suitable scale, to represent the velocity of A
with respect to O or simply velocity of A (i.e. vAO or vA), such that
vector oa=vAO = vA = 1.26 m/s
2. From point a, draw vector ab perpendicular to AB to represent the velocity of B with respect
to A i.e. vBA, and from point o draw vector ob parallel to the motion of B (which moves along BO

only) to represent the velocity of B i.e. vB . The vectors ab and ob intersect at b.
Fig. 7.14
3. Since the point C lies on AB, therefore divide vector ab at c in the same ratio as C divides AB
in the space diagram. In other words,
ac/ab=AC/AB
4. From point c, draw vector cd perpendicular to CD to represent the velocity of D with respect
to C i.e. vDC, and from point o draw vector od parallel to the motion of CD, which moves along CD

only, to represent the velocity of D, i.e. vD.
5. Since the point E lies on CD produced, therefore divide vector cd at e in the same ratio as E
divides CD in the space diagram. In other words,
cd/ce=CD/CE
6. From point e, draw vector ef perpendicular to EF to represent the velocity of F with respect
to E i.e. vFE, and from point o draw vector of parallel to the motion of F, which is along FD to
represent the velocity of F i.e. vF.
By measurement, we find that velocity of F,

vF= vector of = 0.53 m/s   Ans.
2. Velocity of sliding of CE in the trunnion
Since velocity of sliding of CE in the trunnion is the velocity of D, therefore velocity of sliding
of CE in the trunnion
= vector od  = 1.08 m/s   Ans.
3.  Angular velocity of CE
By measurement, we find that linear velocity of C with respect to E,
vCE= vector ec = 0.44 m/s
Since the length CE = 350 mm = 0.35 m, therefore angular velocity of CE,
CE
CE0.44
0.35v

Chapter 7 : Velocity in Mechanisms    /G6C  155
Example 7.6.  In a mechanism as shown in Fig. 7.15, the various dimensions are : OC = 125
mm ; CP = 500 mm ; PA  = 125 mm ; AQ = 250 mm and QE = 125 mm.
Fig. 7.15. All dimensions in mm.
The slider  P translates along an axis which is 25 mm vertically below point O. The crank OC
rotates uniformly at 120 r.p.m. in the anti-c lockwise direction. The bell crank lever AQE rocks about
fixed centre Q.

Draw the velocity diagram and calculate the absolute velocity of point E of the lever .
Solution.  Given : NCO= 120 r.p.m. or ωCO = 2 π × 120/60 = 12.57 rad/s ;
OC = 125 mm = 0. 125 m
We know that linear velocity of C with respect to O or velocity of C, (because O is
as fixed point)
vCO=vC = ωCO × OC = 12.57 × 0.125 = 1.57 m/s
First of all, draw the space diagram, as shown in Fig. 7.16 ( a), to some suitable scale. Now the

velocity diagram, as shown in Fig. 7.16 ( b) is drawn as discussed below :
1. Since the points O and Q are fixed, therefore these points are taken as one point in the
velocity diagram. From point o, draw vector oc perpendicular to OC, to some suitable scale, to
represent the velocity of C with respect to O or velocity of C, such that
vector oc=vCO = vC = 1.57 m/s
Fig. 7.16

Fig. 7.16
2. From point c, draw vector cp perpendicular to CP to represent the velocity of P with
respect to C (i.e. vPC) and from point o, draw vector op parallel to the path of motion of slider P
(which is horizontal) to represent the velocity of P (i.e. vP). The vectors cp and op intersect at p.

156  /G6C   Theory of Machines
3. From point p, draw vector pa perpendicular to PA to represent the velocity of A with
respect to P (i.e. vAP) and from point q, draw vector qa perpendicular to QA to represent the velocity
of A (i.e. vA ). The vectors pa and qa intersect at a.
4. Now draw vector qe perpendicular to vec-
tor qa in such a way that
QE/QA=qe/qa
By measurement, we find that the velocity of
point E,
               vE  = vector oe = 0.7 m/s Ans.
Example 7.7.  A quick return mechanism of

the crank and slotted lever type shaping machine isshown in Fig. 7.17.
The dimensions of the various links are as
follows :
O
1O2 = 800 mm ; O1B = 300 mm ;
O2D  = 1300 mm ; DR = 400 mm.
The crank O1B makes an angle of 45° with
the vertical and rotates at 40 r.p.m. in the c ounter
clockwise direction. Find : 1. velocity of the ram R, or
the velocity of the cutting tool, and 2. angular velocity
of link O2D.
Solution.  Given: NBO1 = 40 r.p.m. or ωBO1 =

2 π × 40/60 = 4.2 rad/s
Since the length of crank O1B = 300 mm = 0.3m, therefore velocity of B with respect to O1 or
simply velocity of B (because O1 is a fixed point),
vBO1=vB = ωBO1 × O1B = 4.2 × 0.3 = 1.26 m/s . . . (Perpendicular to O1B)
1.  Velocity of the ram R
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.18 ( a). Now the
velocity diagram, as shown in Fig. 7.18 ( b), is drawn as discussed below :

1. Since O1 and O2 are fixed points, therefore these points are marked as one point in the
velocity diagram. Draw vector o1b perpendicular to O1B, to some suitable scale, to represent the
velocity of B with respect to O1 or simply velocity of B, such that
vector o1b=vBO1 = vB = 1.26 m/s
2. From point o2, draw vector o2c perpendicular to O2C to represent the velocity of the
coincident point C with respect to O2 or simply velocity  of C (i.e. vCO2 or vC), and from point b, draw

vector bc parallel to the path of motion of the sliding block (which is along the link O2D) to represent
the velocity of C with respect to B (i.e. vCB). The vectors o2c and bc intersect at c.
3. Since the point D lies on O2C produced, therefore divide the vector o2c at d in the same
ratio as D divides O2C in the space diagram. In other words,
cd / o2d=CD/O2D
4. Now from point d, draw vector dr perpendicular to DR to represent the velocity of R with

respect to D (i.e. vRD), and from point o1 draw vector o1r parallel to the path of motion of R (which is
horizontal) to represent the velocity of R (i.e. vR). The vectors dr and o1r intersect at r.Fig. 7.17.  All dimensions in mm.

Chapter 7 : Velocity in Mechanisms    /G6C  157
By measurement, we find that velocity of the ram R,
vR= vector o1r = 1.44 m/s   Ans.
(a) Space diagram ( b) Velocity diagram.
Fig. 7.18
2.  Angular velocity of link O2D
By measurement from velocity diagram, we find that velocity of D with respect to O2 or
velocity of D,
vDO2=vD  = vector o2d = 1.32 m/s
We know that length of link O2D = 1300 mm = 1.3 m. Therefore angular velocity of the link O2D,
DO2
DO2
21.32
1.3v

DO2
DO2
21.32
1.3v
ODω= = = 1.015 rad/s   (Anticlockwise about O2)     Ans.
The above picture shows prototype of an industrial steam engine. Before to the invention of
electricity, steam engines used to provide the power needed to turn wheels in the factories.

158  /G6C   Theory of Machines
Example 7.8.  In the mechanism , as shown in Fig. 7.19, the crank O1A rotates at a speed of
60 r.p.m. in a clockwise direction imparting vertical reciprocating motion to the rack R, by means oftoothed quadrant Q. O
1 and O2 are fixed centres and the slotted bar BC and quadrant Q are rocking
on O2.
Fig. 7.19. All dimensions are in mm.
Determine : 1. the linear speed of the rack when the crank makes an angle of 30° to the

horizontal,  2. the ratio of the times of lowering and raising the rack, and 3. the length of the stroke of
the rack.
Solution. Given : NAO1 = 60 r.p.m. or ωAO1 = 2 π × 60/60 = 6.28 rad/s
Since crank length O1 A = 85 mm, therefore velocity of A with respect to O1 or velocity of A,
(because O1 is a fixed point),
vAO1=vA = ωAO1 × O1A = 6.28 × 85 = 534 mm/s
. . . (Perpendicular to O1A)
Fig. 7.20

Chapter 7 : Velocity in Mechanisms    /G6C  159
1. Linear speed of the rack
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.20 ( a). Now the
velocity diagram, as shown in Fig. 7.20 ( b), is drawn as discussed below :
1. Since O1 and O2 are fixed points, therefore they are marked as one point in the velocity
diagram. From point o1, draw vector o1a perpendicular to O1A, to some suitable scale, to represent

the velocity of A with respect to O1 or simply velocity of A, such that
vector o1a=vAO1 = vA = 534 mm/s
2. From point a, draw vector ad parallel to the path of motion of D (which is along the slot in
the link BC) to represent the velocity D with respect to A (i.e. vDA), and from point o2 draw vector o2d
perpendicular to the line joining the points O2 and D (because O2 and D lie on the same link) to
represent the velocity of D (i.e. 2DOvor vD). The vectors ad and o2d intersect at d.

Note : The point A represents the point on the crank as well as on the sliding block whereas the point D
represents the coincident point on the lever O2C.
vDO2=vD = vector o2d = 410 mm/s, and O2D = 264 mm
We know that angular velocity of the quadrant Q,
DO2
Q
24101.55 rad/s264v
ODω= = =  (Clockwise about O2)
Radius of the quadrant Q,
rQ= 50 mm
Since the rack and the quadrant have a rolling contact, therefore the linear velocity at the

points of contact will be same as that of quadrant.
∴  Linear speed of the rack,
vR=wQ.rQ = 1.55 × 50    = 77.5 mm/s     Ans.
2.  Ratio of the times of lowering and raising the rack
The two extreme positions of the rack (or AB) are when
the tangent to the circle with centre O1 is also a tangent to the
circle with centre O2, as shown in Fig. 7.21. The rack will be
raising when the crank moves from A1 to A2 through an angle α
and it will be lowering when the crank moves from A2 to A1

through an angle β. Since the times of lowering and raising the
rack is directly proportional to their respective angles, therefore
Timeof lowering 2402Timeof raising 120β°== =α°   Ans.
. . . (By measurement)
3.  Length of stroke of the rack
By measurement, we find that angle  B1O2B2 = 60° =
60 × π/180 = 1.047 rad
We know that length of stroke of the rack= Radius of the quadrant × Angular rotation of the
               quadrant in radians
= r

quadrant in radians
= r
Q × ∠ B1O2B2 in radians = 50 × 1.047 = 52.35 mm    Ans.Fig. 7.21. All dimensions in mm.

160  /G6C   Theory of Machines
Example 7.9.  Fig. 7.22 shows the
structure of Whitworth quick return mecha-nism used in reciprocating machine tools. Thevarious dimensions of the tool are as follows :
OQ = 100 mm ; OP = 200 mm, RQ =
150 mm and RS = 500 mm.
The crank OP makes an angle of 60°
with the vertical. Determine the velocity of theslider S (cutting tool) when the crank rotatesat 120 r.p.m. cl ockwise.
Find also the angular velocity of the

link RS and the velocity of the sliding block Ton the slotted lever QT .
Solution. Given : N
PO = 120 r.p.m. or ωPO = 2 π × 120/60 = 12.57 rad/s
Since the crank OP = 200 mm = 0.2 m, therefore velocity of P with respect to O or velocity
of P (because O is a fixed point),
vPO=vP = ωPO × OP = 12.57 × 0.2 = 2.514 m/s
. . . (Perpendicular to PO)
Velocity of slider S  (cutting tool )

First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.23 ( a). Now the
velocity diagram, as shown in Fig. 7.23 ( b) is drawn as discussed below :
1.  Since O and Q are fixed points, therefore they are taken as one point in the velocity
diagram. From point o, draw vector op perpendicular to OP, to some suitable scale, to represent the
velocity of P with respect to O or simply velocity of P, such that
vector op=vPO = vP = 2.514 m/s

Fig. 7.23
2.  From point q, draw vector qt perpendicular to QT to represent the velocity of T with
respect to Q or simply velocity of T (i.e. vTQ or vT) and from point p draw vector pt parallel to the path
of motion of T (which is parallel to TQ) to represent the velocity of T with respect to P (i.e. vTP). The
vectors qt and pt intersect at t.

Note : The point T is a coincident point with P on the link QT.Fig. 7.22

Chapter 7 : Velocity in Mechanisms    /G6C  161
3.  Since the point R lies on the link TQ produced, therefore divide the vector tq at r in the
same ratio as R divides TQ, in the space diagram. In other words,
                               qr /qt = QR/QT
The vector qr represents the velocity of R with respect to Q or velocity of R (i.e.vRQ or vR).
4.  From point r, draw vector rs perpendicular to RS to represent the velocity of S with respect

to R and from point o draw vector or parallel to the path of motion of S (which is parallel to QS) to
represent the velocity of S (i.e vS). The vectors rs and os intersect at s.
By measurement, we find that velocity of the slider S (cutting tool),
vS= vector os = 0.8 m/s  Ans.
Angular velocity of link RS
From the velocity diagram, we find that the linear velocity of the link RS,
vSR= vector rs = 0.96 m/s
Since the length of link RS = 500 mm = 0.5 m, therefore angular velocity of link RS,
SR

SR
RS0.96
0.5v
RSω= =  = 0.92 rad/s (Clockwise about R) Ans.
Velocity of the sliding block  T on the slotted lever QT
Since the block T moves on the slotted lever with respect to P, therefore velocity of the
sliding block T on the slotted lever QT,
  vTP= vector pt = 0.85 m/s     Ans. . . . (By measurement)
7.7. Forces Acting in a Mechanism
Consider a mechanism of a four bar chain, as shown in
Fig. 7.24. Let force FA newton is acting at the joint  A  in the

direction of the velocity of  A (vA m/s) which is perpendicular to
the link DA. Suppose a force FB newton is transmitted to the
joint B in the direction of the velocity of B
(i.e. vB  m/s) which is perpendicular to the link CB. If we neglect
the effect of friction and the change of kinetic energy of the link(i.e.), assuming the efficiency of transmission as 100%), then
by the principle of conservation of energy,
Input work per unit time
 = Output work per unit time

= Output work per unit time
∴  Work supplied to the joint A
= Work transmitted by the joint B
or F
A.vA =FB.vB   or AA
B
B.FvFv= . . . (i)
If we consider the effect of friction and assuming the efficiency of transmission as η, then
BB
AA. Output
Input .Fv
Fvη= =     or    AA
B
B..FvFvη= . . . (ii)
Notes : 1.   If the turning couples due to the forces FA and FB about D and C are denoted by TA (known as driving

torque) and TB (known as resisting torque) respectively, then the equations  (i) and (ii) may be written as
                              TA.ωA = TB.ωB,  and  BB
AA.
.T
Tωη=ω. . . (iii)
where ωA and ωB  are the angular velocities of the links DA and CB respectively.Fig. 7.24. Four bar mechanism.

162  /G6C   Theory of Machines
2. If the forces FA and FB do not act in the direction of the velocities of the points A and B respectively,
then the component of the force in the direction of the velocity should be used in the above equations.
7.8. Mechanical Advantage
It is defined as the ratio of the load to the effort. In a four bar mechanism, as shown in Fig.
7.24, the link DA is called the driving link and the link CB as the driven link. The force FA acting at A

is the effort and the force FB at B will be the load or the resistance to overcome. We know from the
principle of conservation of energy, neglecting effect of friction,
FA × vA=FB × vB  or BA
ABFv
Fv=
∴ Ideal mechanical advantage,
BA
()
ABM.A.idealFv
Fv==
If we consider the effect of friction, less resistance will be overcome with the given effort.
Therefore the actual mechanical advantage will be less.
Let η= Efficiency of the mechanism.
∴ Actual mechanical advantage,
BA
()
ABM.A.actualFv

BA
()
Fv=η× =η×
Note : The mechanical advantage may also be defined as the ratio of output torque to the input torque.
Let TA= Driving torque,
TB= Resisting torque,
ωA and ωB= Angular velocity of the driving and driven links respectively.
∴  Ideal mechanical advantage,
BA
()
ABM.A.idealT
Tω==
ω. . . (Neglecting effect of friction)
and actual mechanical advantage,
BA
()
ABM.A.actualT
Tω=η× =η×ω. . . (Considering the effect of friction)

Example 7.10.  A four bar mechanism has the following dimensions :
DA = 300 mm ; CB = AB = 360 mm ; DC = 600 mm. The link DC is fixed and the angle ADC
is 60°. The driving link DA rotates uniformly at a speed of 100 r.p.m. clockwise and the constantdriving torque has the magnitude of 50 N-m. Determine the velocity of the point B and angularvelocity of the driven link CB. Also find the actual mechanical advantage and the resisting torque ifthe efficiency of the mechanism is 70 per cent.

AD = 100 r.p.m. or ωAD = 2 π × 100/60 = 10.47 rad/s ; TA = 50 N-m
Since the length of driving link, DA = 300 mm = 0.3 m, therefore velocity of A with respect
to D or velocity of A (because D is a fixed point),
vAD=vA = ωAD × DA = 10.47 × 0.3 = 3.14 m/s
. . . (Perpendicular to DA)
Velocity of point B
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.25 ( a). Now the
velocity diagram, as shown in Fig. 7.25 ( b), is drawn as discussed below :

Chapter 7 : Velocity in Mechanisms    /G6C  163
Fig. 7.261. Since the link DC is fixed, therefore points d and c are taken as one point in the velocity
diagram. Draw vector da perpendicular to DA, to some suitable scale, to represent the velocity of A
with respect to D or simply velocity of A (i.e. vAD or vA) such that
vector da=vAD = vA = 3.14 m/s
2. Now from point a, draw vector ab perpendicular to AB to represent the velocity of B with

respect to A (i.e. vBA), and from point c draw vector cb perpendicular to CB to represent the velocity
of B with respect to C or simply velocity of B (i.e. vBC or vB). The vectors ab and cb intersect at b.
By measurement, we find that velocity of point B,
vB=vBC = vector cb = 2.25 m/s   Ans.
Fig. 7.25
Angular velocity of the driven link CB
Since CB = 360 mm = 0.36 m, therefore angular velocity of the driven link CB,
BC
BC2.25
0.36v

BC
BC2.25
0.36v
BCω= =  = 6.25 rad/s (Clockwise about C)  Ans.
Actual mechanical advantage
We know that the efficiency of the mechanism,
η= 70% = 0.7 . . . (Given)
∴Actual mechanical advantage,
A
()
B10.47M.A. 0.7 1.176.25actualω=η× = × =ω Ans.
...(/G51 ωA = ωAD; and ωB = ωBC)
Resisting torque
Let TB= Resisting torque.
We know that efficiency of the mechanism ( η),
    BB B
B
AA.6 . 2 50.7 0.012. 50 10.47TTTTω×== =ω×
∴ TB = 58.3 N–m  Ans.
Example 7.11.  The dimensions of the various links of a

pneumatic riveter , as shown in Fig. 7.26, are as follows :
OA = 175 mm ; AB = 180 mm ; AD = 500 mm ;
and BC = 325 mm.
Find the velocity ratio between C and ram D when OB
is vertical. What will be the efficiency of the machine if a loadof 2.5 kN on the piston C causes a thrust of 4 kN at the ram D ?

164  /G6C   Theory of Machines
Solution.   Given : WC= 2.5 kN = 2500 N ; WD = 4 kN = 4000 N
Let N= Speed of crank OA.
∴Angular velocity of crank OA,
ωAO=2  π N/60 rad/s
Since the length of crank OA = 175 mm = 0.175 m, therefore velocity of A with respect to O
(or velocity of A) (because O is a fixed point),
                          AO A20.175 0.0183 m/s60Nvv Nπ== × = . . . (Perpendicular to OA)
Fig. 7.27
Velocity ratio between C and the ram D

Fig. 7.27
First of all draw the space diagram, to some
suitable scale, as shown in Fig. 7.27 ( a), Now the
velocity diagram, as shown in Fig. 7.27 ( b), is drawn
as discussed below :
1. Draw vector oa perpendicular to OA to
represent the velocity of A (i.e. vA) such that
        vector oa = vA = 0.0183 N m/s
Since the speed of crank ( N) is not given,
therefore let we take vector oa = 20 mm.
2. From point a, draw a vector ab

perpendicular to AB to represent the velocity of B
with respect to A (i.e. vBA), and from point o draw
vector ob perpendicular to OB to represent the
velocity of B with respect to A or simply velocity of
B (i.e. vBO or vB). The vectors ab and ob intersect at
b.
3. Now from point b, draw vector bc
perpendicular to BC to represent the velocity of C
with respect to B (i.e. vCB) and from point o draw
vector oc parallel to the path of motion of C to

represent the velocity of C (i.e. vC). The vectors bc
and oc intersect at c. We see from Fig. 7.27 ( b) that

Chapter 7 : Velocity in Mechanisms    /G6C  165
the points b and c coincide. Therefore velocity of B with respect to C is zero and velocity of B is equal
to velocity of C, i.e.
vBC=0 . . . (/G51  b and c coincide)
and  vB=vC. . . (/G51 vector ob = vector oc)
4.  From point a, draw vector ad perpendicular to AD to represent velocity of D with respect
to A i.e. vDA, and from point o draw vector ob parallel to the path of motion of D to represent the

velocity of D i.e. vD. The vectors ad and od intersect at d.
By measurement from velocity diagram, we find that velocity of C,
vC= vector oc = 35 mm
and velocity of D, vD= vector od = 21 mm
∴Velocity ratio between C and the ram D
=vC /vD = 35/21 = 1.66  Ans.
Efficiency of the machine
Let η= Efficiency of the machine,
We know that work done on the piston C or input,
=WC × vC = 2500 vC
and work done by the ram D or output,
=WD × vD = 4000 vD
∴D
C4000 Output 4000 1
Input 2500 2500 1.66v
vη= = = ×C

vη= = = ×C
D.... 1.66v
v=
/G51
=0.96  or  96%   Ans.
Example 7.12.   In the toggle mechanism, as
shown in Fig. 7.28, the slider D is constrained to moveon a horizontal path. The crank OA is rotating in thecounter-clockwise direction at a speed of 180 r.p.m.
The dimensions of various links are as follows :OA = 180 mm ; CB = 240 mm ; AB = 360 mm ;
and BD = 540 mm.
For the given configuration, find : 1. Velocity
of slider D, 2. Angular velocity of links AB, CB and BD;

3. Velocities of rubbing on the pins of diameter 30 mm
at A and D, and 4. Torque applied to the crank OA, for
a force of 2 kN at D.
AO = 180 r.p.m. or  ωAO = 2 π × 180/60 = 18.85 rad/s
Since the crank length OA = 180 mm = 0.18 m, therefore velocity of A with respect to O or
velocity of A (because O is a fixed point),
vAO=vA = ωAO × OA = 18.85 × 0.18 = 3.4 m/s
1.  Velocity of slider  D

First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.29 ( a). Now the
velocity diagram, as shown in Fig. 7.29 ( b), is drawn as discussed below :Fig. 7.28

166  /G6C   Theory of Machines
1. Draw vector oa perpendicular to OA, to some suitable scale, to represent the velocity of A
with respect to O or velocity of A (i.e. vAO or vA, ) such that
vector oa=vAO = vA = 3.4 m/s
Fig. 7.29
2. Since point B moves with respect to A and also with respect to C, therefore draw vector ab
perpendicular to AB to represent the velocity of B with respect to A i.e. vBA, and draw vector cb

perpendicular to CB to represent the velocity of B with respect to C, i.e. vBC. The vectors ab and cb
intersect at b.
respect to B i.e. vDB, and from point c draw vector cd parallel to the path of motion of the slider D
(which is along CD) to represent the velocity of D, i.e. vD. The vectors bd and cd intersect at d.
vD= vector cd = 2.05 m/s  Ans.

2.  Angular velocities of links AB , CB and BD
By measurement from velocity diagram, we find that
Velocity of B with respect to A,
vBA= vector ab = 0.9 m/s
Velocity of B with respect to C,
vBC=vB = vector cb = 2.8 m/s
and velocity of D with respect to B,
vDB= vector bd = 2.4 m/s
We know that AB = 360 mm = 0.36 m ; CB = 240 mm = 0.24 m and BD = 540 mm = 0.54 m.
∴ Angular velocity of the link AB,
BA
AB0.92.5 rad/s0.36v
ABω= = =  (Anticlockwise about A)  Ans.

Similarly angular velocity of the link CB,
BC
CB2.811.67 rad/s0.24v
CBω= = =  (Anticlockwise about C)  Ans.
and angular velocity of the link BD,
DB
BD2.44.44 rad/s
0.54v
BDω= = =  (Clockwise about B)   Ans.

Chapter 7 : Velocity in Mechanisms    /G6C  167
3.  Velocities of rubbing on the pins A  and D
Given : Diameter of pins at A and D,
DA=DD = 30 mm = 0.03 m
∴ Radius,  rA=rD = 0.015 m
We know that relative angular velocity at A
=ωBC – ωBA + ωDB = 11.67 – 2.5 + 4.44 = 13.61 rad/s
and relative angular velocity at D
=ωDB = 4.44 rad/s
∴ Velocity of rubbing on the pin A
= 13.61 × 0.015 = 0.204 m/s = 204 mm/s   Ans.
and velocity of rubbing on the pin  D
= 4.44 × 0.015 = 0.067 m/s = 67 mm/s  Ans.

4.  Torque applied to the crank OA
Let TA= Torque applied to the crank OA, in N-m
∴    Power input or work supplied at A
=TA × ωAO = TA × 18.85 = 18.85 TA N-m
We know that force at D,
FD= 2 kN = 2000 N . . . (Given)
∴    Power output or work done by D,
=FD × vD = 2000 × 2.05 = 4100 N-m
Assuming 100 per cent efficiency, power input is equal to power output.
∴ 18.85 TA= 4100   or  TA = 217.5 N-m   Ans.

Example 7.13.  The dimensions of the mechanism, as shown in Fig. 7.30, are as follows :
AB = 0.45 m; BD = 1.5 m : BC = CE = 0.9 m.
Fig. 7.30
The crank  AB turns uniformly at 180 r.p.m. in t he clockwise direction and the blocks at D
and E are working in frictionless guides.
Draw the velocity diagram for the mechanism and find the velocities of the sliders D and E

in their guides. Also determine the turning moment at A if a force of 500 N acts on D in the directionof arrow X and a force of 750 N acts on E in the direction of arrow Y .
BA = 180 r.p.m.  or  ωBA = 2 π × 180/60 = 18.85 rad/s

168  /G6C   Theory of Machines
Since AB = 0.45 m, therefore velocity of B with respect to A or velocity of B (because A is a
fixed point),
vBA=vB = ωBA × AB = 18.85 × 0.45 = 8.5 m/s
Velocities of the sliders D and E
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.31 ( a). Now the
velocity diagram, as shown in Fig. 7.31 ( b), is drawn as discussed below :
Fig. 7.31

Fig. 7.31
1. Draw vector ab perpendicular to AB, to some suitable scale, to represent the velocity of B
with respect to A or simply velocity of B (i.e. vBA or vB), such that
vector ab=vBA = vB = 8.5 m/s
2. From point b, draw vector bd perpendicular to BD to represent the velocity of D with
respect to B (i.e. vDB) and from point a draw vector ad parallel to the motion of D to represent the
velocity of D (vD). The vectors bd and ad intersect at d.

3. Since the point C lies on BD, therefore divide vector bd at c in the same ratio as C divides
BD in the space diagram. In other words,
bc/bd=BC/BD
4. Now from point c, draw vector ce perpendicular to CE to represent the velocity of E with
respect to C (i.e. vEC) and from point a draw vector ae parallel to the path of E to represent the
velocity of E (i.e. vE). The vectors ce and ae intersect at e.
By measurement, we find thatVelocity of slider D,v
D= vector ad = 9.5 m/s   Ans.

Velocity of slider E,vE= vector ae = 1.7 m/s  Ans.
Turning moment at  A
Let TA= Turning moment at A (or at the crank-shaft).
 We know that force at D, FD = 500 N . . . (Given)
and                     Force at E, FE= 750 N . . . (Given)
∴   Power input = FD × vD – FE × vE
. . . (– ve sign indicates that FE  opposes the motion)
= 500 × 9.5 – 750 × 1.7 = 3475 N-m/s
Power output = TA.ωBA = TA × 18.85 TA N-m/s
Neglecting losses, power input is equal to power output.

Chapter 7 : Velocity in Mechanisms    /G6C  169
EXERCISES
1.In a slider crank mechanism, the length of crank OB and connecting rod AB are 125 mm and 500 mm
respectively. The centre of gravity G of the connecting rod is 275 mm from the slider A. The crank
speed is 600 r.p.m. clockwise. When the crank has turned 45° from the inner dead centre position,determine: 1.  velocity of the slider A, 2. velocity of the point G, and 3. angular velocity of the

connecting rod AB. [Ans. 6.45 m/s ; 6.75 m/s  ; 10.8 rad/s]
2.In the mechanism, as shown in Fig. 7.32, OA and OB are two equal cranks at right angles rotating
about O at a speed of 40 r.p.m. anticlockwise. The dimensions of the various links are as follows :
Fig. 7.32
OA = OB = 50 mm ; AC = BD = 175 mm ; DE = CE = 75 mm ; FG = 115 mm and EF = FC.
Draw velocity diagram for the given configuration of the mechanism and find velocity of the slider G.
[Ans. 68 mm/s]

3.The dimensions of various links in a mechanism, as shown in Fig. 7.33, are as follows :
AB = 60 mm ; BC = 400 mm ; CD = 150 mm ; DE = 115 mm ; and EF = 225 mm.
225mm 110
mm
150mm
90°30°
AB
C
DEF
Fig. 7.33
Find the velocity of the slider F when the crank AB rotates uniformly in clockwise direction at a speed
of 60 r.p.m. [Ans. 250 mm/s]
4.In a link work, as shown in Fig. 7.34, the crank AB rotates about A at a uniform speed of 150 r.p.m.

The lever DC oscillates about the fixed point D, being connected to AB by the connecting link BC.
The block F moves, in horizontal guides being driven by the link EF, when the crank AB is at 30°. The
dimensions of the various links are :
AB = 150 mm ; BC = 450 mm ; CE = 300 mm ; DE = 150 mm ; and EF = 350 mm.
Find, for the given configuration, 1. velocity of slider F, 2. angular velocity of DC, and 3. rubbing
speed at pin C which is 50 mm in diameter. [Ans. 500 mm/s ; 3.5 rad/s ; 2.4 m/s]

170  /G6C   Theory of Machines
Fig. 7.34
5.The oscillating link OAB of a mechanism,
as shown in Fig. 7.35, is pivoted at O and
is moving at 90 r.p.m. anticlockwise. If  OA
= 150 mm ;  AB = 75 mm, and
AC = 250 mm, calculate
1. the velocity of the block C;
2. the angular velocity of the link AC;
and
3. the rubbing velocities of the pins at
O, A and C, assuming that these pins
are of equal diameters of 20 mm.[Ans. 1.2 m/s; 1.6 rad/ s
2 clockwise; 21
200 mm/s, 782 mm/s, 160 mm/s]

6.The dimensions of the various links of a
mechanism, as shown in Fig. 7.36, are asfollows :
AB = 30 mm ; BC = 80 mm ; CD = 45 mm ; and CE = 120 mm.
Fig. 7.36
The crank AB rotates uniformly in the clockwise direction at 120 r.p.m. Draw the velocity diagram for
the given configuration of the mechanism and determine the velocity of the slider E and angular
velocities of the links BC, CD and CE.

Also draw a diagram showing the extreme top and bottom positions of the crank DC and the corre-
sponding configurations of the mechanism.
Find the length of each of the strokes.
[Ans. 120 mm/s ; 2.8 rad/s ; 5.8 rad/s ; 2 rad/s ; 10 mm ; 23 mm]Fig. 7.35

Chapter 7 : Velocity in Mechanisms    /G6C  171
7.Fig. 7.37 shows a mechanism in which the crank OA, 100 mm long rotates clockwise about O at 130
r.p.m. The connecting rod AB is 400 mm long. The rod CE, 350 mm long, is attached to AB at C, 150
mm from A. This rod slides in a slot in a trunnion at D. The end E is connected by a link EF, 300 mm
long, to the horizontally moving slider F.
Fig. 7.37

Fig. 7.37
Determine, for the given configuration : 1. velocity of F, 2. velocity of sliding of CE in the trunnion,
and 3. angular velocity of CE. [Ans. 0.54 m/s ; 1.2 m/s ; 1.4 rad/s]
8.Fig. 7.38 shows the mechanism of a quick return motion of the crank and slotted lever type shaping
machine. The dimensions of the various links are as follows :
OA = 200 mm ; AB = 100 mm ; OC = 400 mm ; and CR = 150 mm.

The driving crank AB makes 120° with the vertical and rotates at 60 r.p.m. in the clockwise direction.
Find : 1. velocity of ram R, and 2. angular velocity of the slotted link OC.
[Ans. 0.8 m/s ; 1.83 rad/s]
                      Fig. 7.38 Fig. 7.39
9.In a Whitworth quick return motion mechanism, as shown in Fig. 7.39, the dimensions of various
links are as follows :
OQ = 100 mm ; OA = 200 mm ; BQ = 150 mm and BP = 500 mm.

If the crank OA turns at 120 r.p.m. in clockwise direction and makes an angle of 120° with OQ,
Find : 1. velocity of the block P, and 2. angular velocity of the slotted link BQ.
[Ans. 0.63 m/s ; 6.3 rad/s]
10.A toggle press mechanism, as shown in Fig. 7.40, has the dimensions of various links as follows :
OP = 50 mm ; RQ = RS = 200 mm ; PR = 300 mm.

172  /G6C   Theory of Machines
Fig. 7.41Fig. 7.40
Find the velocity of S when the
crank OP rotates at 60 r.p.m. in
the anticlockwise direction. If thetorque on P is 115 N-m, what
pressure will be exerted at S when
the overall efficiency is 60 per-cent.[Ans. 400 m/s ; 3.9 kN]
11.Fig. 7.41 shows a toggle mecha-
nism in which link D is constained
to move in horizontal direction.For the given configuration, findout : 1. velocities of points bandD; and 2. angular velocities of
links AB, BC, and BD.

The rank OA rotates at 60 r.p.m.
in anticlockwise direction.
[Ans. 0.9  m/s; 0.5 m/s; 0.0016 rad/s (anticlockwise)
0.0075 rad/s (anti-clockwise),0.0044 rad/s (anti-clockwise)]
12.A riveter, as shown in Fig. 7.42, is operated by a piston
F acting through the links EB, AB and BC. The ram D
carries the tool. The piston moves in a line perpen-dicular to the line of motion of D. The length of link
BC is twice the length of link AB. In the position

shown, AB makes an angle of 12° with AC and BE is
at  right angle to AC. Find the velocity ratio of E to D.
If, in the same position, the total load on the piston is
2.2 kN, find the thrust exerted by D when the efficiency
of the mechanism is 72 per cent,
Ans. [3.2 ; 5 kN]
1.Describe the method to find the velocity of a point on a link whose direction (or path) is known and
the velocity of some other point on the same link in magnitude and direction is given.

2.Explain how the velocities of a slider and the connecting rod are obtained in a slider crank
mechanism.Fig. 7.42

Chapter 7 : Velocity in Mechanisms    /G6C  173
Fig. 7.433.Define rubbing velocity at a pin joint. What will be the rubbing velocity at pin joint when the two links
move in the same and opposite directions ?
4.What is the difference between ideal mechanical advantage and actual mechanical advantage ?
1.The direction of linear velocity of any point on a link with respect to another point on the same link is

(a) parallel to the link joining the points ( b) perpendicular to the link joining the points
(c)  at 45° to the link joining the points ( d) none of these
2.The magnitude of linear velocity of a point B on a link AB relative to point A is
(a)ω.AB (b)ω (AB)2
(c)ω2 . AB (d)(ω . AB)2
where ω = Angular velocity of the link AB.
3.The two links OA and OB are connected by a pin joint at O. If the link OA turns with angular velocity

ω1 rad/s in the clockwise direction and the link OB turns with angular velocity
ω2 rad/s in the anti-clockwise direction, then the rubbing velocity at the pin joint O is
(a) ω1.ω2.r (b)(ω1 – ω2) r
(c)(ω1 + ω2) r (d)(ω1 – ω2) 2 r
where r = Radius of the pin at O.
4.In the above question, if both the links OA and OB turn in clockwise direction, then the rubbing
velocity at the pin joint O is

5.In a four bar mechanism, as shown in Fig. 7.43, if a force FA is
acting at point A in the direction of its velocity vA and a force FB is
transmitted to the joint B in the direction of its velocity vB , then the
ideal mechanical advantage is equal to
(a)FB.vA(b)FA.vB
(c)B
BF
v(d)B
AF
F
ANSWERS
1. (b) 2. (a) 3. (c) 4. (b) 5. (d)

174       /G6C               Theory of Machines
174AccelerationAccelerationAccelerationAccelerationAcceleration
in Mechanismsin Mechanismsin Mechanismsin Mechanismsin Mechanisms8
FFFFFeaeaeaeaeaturturturturtureseseseses
2. Acceleration Diagram for a
Link.
3. Acceleration of a Point on a
Link.
4. Acceleration in the Slider
Crank Mechanism.
5. Coriolis Component of
Acceleration.8.1.8.1.8.1.8.1.8.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoduction

We have discussed in the previous chapter the
velocities of various points in the mechanisms. Now we shall
discuss the acceleration of points in the mechanisms. Theacceleration analysis plays a very important role in thedevelopment of machines and mechanisms.
8.2.8.2.8.2.8.2.8.2. Acceleration Diagram for a LinkAcceleration Diagram for a LinkAcceleration Diagram for a LinkAcceleration Diagram for a LinkAcceleration Diagram for a Link
Consider two points A and B on a rigid link as shown

in Fig. 8.1 ( a). Let the point B moves with respect to A, with
an angular velocity of ω rad/s and let α rad/s2 be the angular
acceleration of the link AB.
(a) Link. ( b) Acceleration diagram.
Fig. 8.1.  Acceleration for a link.
Warping Machine

Chapter 8 :  Acceleration in Mechanisms    /G6C  175
We have already discussed that acceleration of a particle whose velocity changes both in
magnitude and direction at any instant has the following two components :
1.  The centripetal or radial component, which is perpendicular to the velocity of the
particle at the given instant.
2.  The tangential component, which is parallel to the velocity of the particle at the given
instant.

instant.
Thus for a link AB, the velocity of point B with respect to A (i.e. vBA) is perpendicular to the
link AB as shown in Fig. 8.1 ( a). Since the point B moves with respect to A with an angular velocity
of ω rad/s, therefore centripetal or radial component of the acceleration of B with respect to A,
               22 2
BA BA Length of link = /raA B A B v A B=ω × ω × =BA...v
ABω=/G51
This radial component of acceleration acts perpendicular to the velocity vBA, In other words,

it acts parallel to the link AB.
We know that tangential component of the acceleration of B with respect to A,
BA Length of the linktaA B A B=α× =α×
This tangential component of acceleration acts parallel to the velocity vBA. In other words,
it acts perpendicular to the link AB.
In order to draw the acceleration diagram for a link AB, as shown in Fig. 8.1 ( b), from any
point b', draw vector b'x parallel to BA  to represent the radial component of acceleration of B with

respect to A i.e. BAraand from point x draw vector xa' perpendicular to BA to represent the tangential
component of acceleration of B with respect to A i.e. BA.taJoin b' a'.  The vector b' a' (known as
acceleration image of the link AB) represents the total acceleration of B with respect to A (i.e. aBA)
and it is the vector sum of radial component BA()raand tangential component BA()taof acceleration.
8.3. Acceleration of a Point on a Link
(a) Points on a Link. ( b) Acceleration diagram.

Fig. 8.2. Acceleration of a point on a link.
Consider two points A and B on the rigid link, as shown in Fig. 8.2 ( a). Let the acceleration
of the point A i.e. aA is known in magnitude and direction and the direction of path of B is given. The
acceleration of the point B is determined in magnitude and direction by drawing the acceleration
diagram as discussed below.

1. From any point o', draw vector o'a' parallel to the direction of absolute acceleration at
point A i.e. aA , to some suitable scale, as shown in Fig. 8.2 ( b).

176  /G6C   Theory of Machines
2. We know that the acceleration of B with
respect to A i.e. aBA has the following two
components:
(i)Radial component of the acceleration
of B with respect to A i.e. BAra, and
(ii)Tangential component of the
acceleration B with respect to A i.e. BA.taThese two
components are mutually perpendicular.
3.  Draw vector a'x parallel to the link AB
(because radial component of the acceleration of B
with respect to  A will pass through  AB), such that
vector 2

vector 2
BA BA /rax a v AB′==
whereBAVelocity of with respect to .vB A=
Note: The value of vBA may be obtained by drawing the
velocity diagram as discussed in the previous chapter.
4. From point x, draw vector xb'
perpendicular to AB or vector a'x (because tangential
component of B with respect to A i.e. BA,tais
perpendicular to radial component BAra) and
through o' draw a line parallel to the path of B to
represent the absolute acceleration of B i.e. aB. The

vectors xb' and o' b' intersect at b'. Now the values
of aB and BAtamay be measured, to the scale.
5. By joining the points a' and b' we may determine the total acceleration of B with respect
to A i.e. aBA. The vector a' b' is known as acceleration image of the link AB.
6. For any other point C on the link, draw triangle a' b' c' similar to triangle ABC. Now
vector b' c' represents the acceleration of C with respect to B i.e. aCB, and vector a' c' represents the

acceleration of C with respect to A i.e. aCA. As discussed above, aCB and aCA will each have two
components as follows :
(i)aCB has two components; CB CBandrtaa as shown by triangle b' zc' in Fig. 8.2 ( b), in
which b' z is parallel to BC and zc' is perpendicular to b' z or BC.
(ii)aCA has two components ; CA CAandrtaa as shown by triangle a' yc' in Fig. 8.2 ( b), in
which a' y is parallel to AC and yc' is perpendicular to a' y or AC.

7. The angular acceleration of the link AB is obtained by dividing the tangential components
of the acceleration of B with respect to A BA()tato the length of the link. Mathematically, angular
acceleration of the link AB,
AB BA /taA Bα=
8.4. Acceleration in the Slider Crank Mechanism
A slider crank mechanism is shown in Fig. 8.3 ( a). Let the crank OB makes an angle θ with
the inner dead centre ( I.D.C) and rotates in a clockwise direction about the fixed point O with

uniform angular velocity ωBO rad/s.
∴ Velocity of B with respect to O or velocity of B (because O is a fixed point),
 BO B BO , acting tangentially at . vv O B B== ω ×
A refracting telescope uses mechanisms to
change directions.
Note : This picture is given as additional
information and is not a direct example of the
current chapter.

Chapter 8 :  Acceleration in Mechanisms    /G6C  177
We know that centripetal or radial acceleration of B with respect to O or acceleration of B
(because O is a fixed point),
2
2 BO
BO B BOr vaa O BOB== ω ×=
Note : A point at the end of a link which moves with constant angular velocity has no tangential component
of acceleration.
(a) Slider crank mechanism. ( b) Acceleration diagram.
Fig. 8.3. Acceleration in the slider crank mechanism.

The acceleration diagram, as shown in Fig. 8.3 ( b), may now be drawn as discussed below:
1. Draw vector o' b' parallel to BO and set off equal in magnitude of BO Braa=, to some
suitable scale.
2. From point b', draw vector b'x parallel to BA. The vector b'x represents the radial component
of the acceleration of A with respect to B whose magnitude is given by :
2
AB AB /rav B A=
Since the point B moves with constant angular velocity, therefore there will be no tangential

component of the acceleration.
3. From point x, draw vector xa' perpendicular to b'x (or AB). The vector xa'  represents the
tangential component of the acceleration of A with respect to B i.e. AB.ta
Note: When a point moves along a straight line, it has  no centripetal or radial  component of the acceleration.
4. Since the point A reciprocates along AO, therefore the acceleration must be parallel to
velocity. Therefore from o', draw o' a' parallel to AO, intersecting the vector xa' at a'.

Now the acceleration of the piston or the slider A (aA) and ABtamay be measured to the scale.
5. The vector b' a', which is the sum of the vectors b' x and x a', represents the total acceleration
of A with respect to B i.e. aAB. The vector b' a' represents the acceleration of the connecting rod AB.
6. The acceleration of any other point on AB such as E may be obtained by dividing the vector
b' a' at e' in the same ratio as E divides AB in Fig. 8.3 ( a). In other words

a' e' / a' b'  =  AE / AB
7. The angular acceleration of the connecting rod AB may be obtained by dividing the
tangential component of the acceleration of A with respect to B()ABtato the length of AB. In other
words, angular acceleration of AB,
                            AB AB / (Clockwise about )taA B Bα=
Example 8.1. The crank of a slider crank mechanism rotates clockwise at a constant speed

of 300 r.p.m. The crank is 150 mm and the connecting rod is 600 mm long. Determine : 1. linear
velocity and acceleration of the midpoint of the connecting rod, and 2. angular velocity and angular
acceleration of the connecting rod, at a crank angle of 45° from inner dead centre position.

178  /G6C   Theory of Machines
Solution. Given : NBO = 300 r.p.m. or ωBO = 2 π × 300/60 = 31.42 rad/s; OB = 150 mm =
0.15 m ; BA = 600 mm = 0.6 m
We know that linear velocity of B with respect to O or velocity of B,
                                vBO = vB = ωBO × OB = 31.42 × 0.15 = 4.713 m/s
...(Perpendicular to BO)
(a) Space diagram. ( b) Velocity diagram. ( c) Acceleration diagram.
Fig. 8.4
1.  Linear velocity of the midpoint of the connecting rod

First of all draw the space diagram, to some suitable scale; as shown in Fig. 8.4 ( a). Now the
velocity diagram, as shown in Fig. 8.4 ( b), is drawn as discussed below:
1. Draw vector ob perpendicular to BO, to some suitable scale, to represent the velocity of
B with respect to O or simply velocity of B i.e. vBO or vB, such that
vector ob = vBO = vB = 4.713 m/s
2. From point b, draw vector ba perpendicular to BA to represent the velocity of A with

respect to B i.e. vAB , and from point o draw vector oa parallel to the motion of A (which is along AO)
to represent the velocity of A i.e. vA. The vectors ba and oa intersect at a.
Note : This picture is given as additional information and is not a direct example of the current chapter.Pushing with fluidsRam moves
outwards
Oil pressure on
lower side of
piston
Load
moves
inwardsOil pressure on
upper side of
pistonRam moves
inwardsLoad moves
outwards

Chapter 8 :  Acceleration in Mechanisms    /G6C  179
By measurement, we find that velocity of A with respect to B,
AB
Avector 3.4 m/s
and Velocity of , vector 4 m/svb a
Av o a==
==
3. In order to find the velocity of the midpoint D of the connecting rod AB, divide the vector
ba at d in the same ratio as D divides AB, in the space diagram. In other words,
bd/ba = BD/BA
Note:  Since D is the midpoint of AB, therefore d is also midpoint of vector ba.

4. Join od. Now the vector od represents the velocity of the midpoint D of the connecting
rod i.e. vD.
                                         vD = vector od = 4.1 m/s Ans.
Acceleration of the midpoint of the connecting rod
We know that the radial component of the acceleration of B with respect to O or the
acceleration of B,
2 2
2 BO
BO B(4.713)148.1m/s0.15r vaaOB== = =
and the radial component of the acceleraiton of A with respect to B,
2 2
2 AB

2 2
2 AB
AB(3.4)19.3 m/s0.6rvaBA== =
Now the acceleration diagram, as shown in Fig. 8.4 ( c) is drawn as discussed below:
1. Draw vector o' b' parallel to BO, to some suitable scale, to represent the radial component
of the acceleration of B with respect to O or simply acceleration of B i.e. BO Bor ,raa such that
                         2
BO B vector 148.1m/srob a a′′== =
Note:  Since the crank OB rotates at a constant speed, therefore there will be no tangential component of the

acceleration of B with respect to O.
2. The acceleration of A with respect to B has the following two components:
(a)The radial component of the acceleration of A with respect to B i.e. AB,raand
(b)The tangential component of the acceleration of A with respect to B i.e. AB.taThese two
Therefore from point b', draw vector b' x parallel to AB to represent 2
AB19.3 m/sra= and

from point x draw vector xa' perpendicular to vector b' x whose magnitude is yet unknown.
3. Now from o', draw vector o' a' parallel to the path of motion of A (which is along AO) to
represent the acceleration of A i.e. aA . The vectors xa' and o' a' intersect at a'. Join a' b'.
4. In order to find the acceleration of the midpoint D of the connecting rod AB, divide the
vector a' b' at d' in the same ratio as D divides AB. In other words

//bd ba B D B A′′ ′′=
Note:  Since D is the midpoint of AB, therefore d' is also midpoint of vector b' a'.
5. Join o' d'. The vector o' d' represents the acceleration of midpoint D of the connecting rod
i.e. aD.
  aD = vector o' d' = 117 m/s2 Ans.

180  /G6C   Theory of Machines
2.  Angular velocity of the connecting rod
We know that angular velocity of the connecting rod AB,
                            2 AB
AB3.45.67 rad/s (Anticlockwise about )
0.6vB
BAω= = =  Ans.
Angular acceleration of the connecting rod
From the acceleration diagram, we find that
2
AB103 m/sta= ...(By measurement)
We know that angular acceleration of the connecting rod AB,
2 AB
AB103171.67 rad/s (Clockwise about )0.6taBBAα= = = Ans.

Example 8.2.  An engine mechanism is shown in Fig. 8.5. The crank CB = 100 mm and the
connecting rod BA = 300 mm with centre of gravity G, 100 mm from B. In the position shown, thecrankshaft has a speed of 75 rad/s and an angular acceleration of 1200 rad/s
2. Find:1. velocity of
G and angular velocity of AB, and 2. acceleration of G and angular acceleration of AB.
Fig. 8.5
Solution. Given :  ωBC = 75 rad/s ; αBC = 1200 rad/s2, CB = 100 mm = 0.1 m; BA = 300 mm
= 0.3 m

= 0.3 m
We know that velocity of B with respect to C or velocity of B,
                             BC B BC 75 0.1 7.5 m/s vv C B== ω × =×= ...(Perpendicular to BC)
Since the angular acceleration of the crankshaft, αBC = 1200 rad/s2, therefore tangential
component of the acceleration of B with respect to C,
                             2
BC BC 1200 0.1 120 m/staC B=α × = × =
Note:  When the angular acceleration is not given, then there will be no tangential component of the acceleration.

1.  Velocity of G and angular velocity of AB
First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.6 ( a). Now the
velocity diagram, as shown in Fig. 8.6 ( b), is drawn as discussed below:
1. Draw vector cb  perpendicular to CB, to some suitable scale, to represent the velocity of
B with respect to C or velocity of B (i.e. vBC or vB), such that
BC B vector 7.5 m/scb v v== =

respect to B i.e. vAB , and from point c, draw vector ca parallel to the path of motion of A (which is
along AC) to represent the velocity of A i.e. vA.The vectors ba and ca intersect at a.
3. Since the point G lies on AB, therefore divide vector ab at g in the same ratio as G divides
AB in the space diagram. In other words,
    //ag ab AG AB=
The vector cg represents the velocity of G.
By measurement, we find that velocity of G,
vG = vector cg = 6.8 m/s  Ans.

Chapter 8 :  Acceleration in Mechanisms    /G6C  181
From velocity diagram, we find that velocity of A with respect to B,
   vAB = vector ba = 4 m/s
We know that angular velocity of AB,
AB
AB413.3 rad/s (Clockwise)
0.3v
Fig. 8.6
2.  Acceleration of G and angular acceleration of AB
We know that radial component of the acceleration of B with
respect to C,
                             *2 2
2 BC
BC(7.5)562.5 m/s0.1rvaCB== =

2 BC
and radial component of the acceleration of A with respect to B,
                               2 2
2 AB
AB453.3 m/s0.3rvaBA== =
Now the acceleration diagram, as shown in Fig. 8.6 ( c), is drawn
as discussed below:
1. Draw vector c' b'' parallel to CB, to some suitable scale, to
represent the radial component  of the acceleration of B with respect to C,
i.e. BC,rasuch that
                   2
BC vector 562.5 m/srcb a′′ ′==

2. From point b'', draw vector b'' b' perpendicular to vector c' b'' or CB to represent the
tangential component of the acceleration of B with respect to C i.e. BCta, such that
2
BC vector 120 m/stbb a′′ ′== ... (Given)
3. Join c' b'. The vector c' b' represents the total acceleration of B with respect to C i.e. aBC.
4. From point b', draw vector b' x parallel to BA to represent radial component of the
acceleration of A with respect to B i.e. ABra such that

2
AB vector 53.3 m/srbx a′==
5. From point x, draw vector xa' perpendicular to vector b'x or BA to represent tangential
component of the acceleration of A with respect to B i.e. AB,tawhose magnitude is not yet known.
6. Now draw vector c' a' parallel to the path of motion of A (which is along AC) to represent
the acceleration of A i.e. aA.The vectors  xa' and c'a' intersect at  a'. Join b' a'. The vector b' a'

represents the acceleration of A with respect to B i.e. aAB.(c) Acceleration diagram.
Fig. 8.6
*When angular acceleration of the crank is not given, then there is no BCta. In that case, BC BC B ,raaa== as
discussed in the previous example.

182  /G6C   Theory of Machines
7. In order to find the acceleratio of G, divide vector a' b' in g' in the same ratio as G divides
BA in Fig. 8.6 ( a). Join c' g'. The vector c' g' represents the acceleration of G.
By measurement, we find that acceleration of G,
 aG = vector c' g' = 414 m/s2 Ans.
From acceleration diagram, we find that tangential component of the acceleration of A with
respect to B,
                                          2
ABvector 546 m/stax a ′ == ...(By measurement)

∴ Angular acceleration of AB,
                                         2 AB
AB5461820 rad/s (Clockwise)0.3ta
BAα= = = Ans.
Example 8.3. In the mechanism shown in Fig. 8.7 , the slider C is
moving to the right with a velocity of 1 m/s and an acceleration of 2.5 m/s2.
The dimensions of various links are AB = 3 m inclined at 45° with the
vertical and BC = 1.5 m inclined at 45° with the horizontal. Determine: 1. the

magnitude of vertical and horizontal component of the acceleration of thepoint B, and 2. the angular acceleration of the links AB and BC.
Solution. Given : v
C = 1 m/s ; aC = 2.5 m/s2; AB = 3 m ; BC = 1.5 m
First of all, draw the space diagram, as shown in Fig. 8.8 ( a), to some
suitable scale. Now the velocity diagram, as shown in Fig. 8.8 ( b), is drawn as
1. Since the points A and D are fixed points, therefore they lie at one place in the velocity

diagram. Draw vector dc parallel to DC, to some suitable scale, which represents the velocity of
slider C with respect to D or simply velocity of C, such that
                                  vector dc = vCD = vC = 1 m/s
2. Since point B has two motions, one with respect to A and the other with respect to C,
therefore from point a, draw vector ab perpendicular to AB to represent the velocity of B with

respect to A, i.e. vBA and from point c draw vector cb perpendicular to CB to represent the velocity
of B with respect to C i.e. vBC .The vectors ab and cb intersect at b.
Fig. 8.8
By measurement, we find that velocity of B with respect to A,
BAvector 0.72 m/sva b==
and velocity of B with respect to C,
BCvector 0.72 m/svc b==Fig. 8.7

Chapter 8 :  Acceleration in Mechanisms    /G6C  183
We know that radial component of acceleration of B with respect to C,
                             2 2
2 BC
BC(0.72)0.346 m/s1.5rvaCB== =
and radial component of acceleration of B with respect to A,
                            2 2
2 BA
BA(0.72)0.173 m/s3rvaAB== =
Now the acceleration diagram, as shown in Fig. 8.8 ( c), is drawn as discussed below:

1. *Since the points A and D are fixed points, therefore they lie at one place in the acceleration
diagram. Draw vector d' c' parallel to DC, to some suitable scale, to represent the acceleration of C
with respect to D or simply acceleration of C i.e. aCD or aC such that
                 2
CD C vector 2.5 m/sdc a a′′== =
2. The acceleration of B with respect to C will have two components, i.e. one radial component

of B with respect to C ()BCraand the other tangential component of B with respect to
()BC.tCaTherefore from point c', draw vector c' x parallel to  CB to represent BCrasuch that
                   2
BC vector 0.346 m/srcx a′==
3. Now from point x, draw vector xb' perpendicular to vector c' x or CB to represent at
BC
whose magnitude is yet unknown.
4. The acceleration of B with respect to A will also have two components, i.e. one radial
component of B with respect to A (ar

BA) and other tangential component of B with respect to A (at BA).
Therefore from point a' draw vector a' y parallel to AB to represent ar
BA, such that
vector a' y = ar
BA = 0.173 m/s2
5. From point y, draw vector yb' perpendicular to vector a'y or AB to represent  BA.taThe
vector yb' intersect the vector xb' at b'. Join a' b' and c' b'. The vector  a' b' represents the acceleration

of point B (aB) and the vector c' b' represents the acceleration of B with respect to C.
1.  Magnitude of vertical and horizontal component of the acceleration of the point B
Draw b' b'' perpendicular to a' c'. The vector b' b'' is the vertical component of the acceleration
of the point B and a' b'' is the horizontal component of the acceleration of the point B. By measurement,
                          vector b' b'' = 1.13 m/s2 and vector a' b'' = 0.9 m/s2  Ans.

2.  Angular acceleration of AB and BC
By measurement from acceleration diagram, we find that tangential component of acceleration
of the point B with respect to A,
2
BAvector 1.41 m/sta yb′ ==
and tangential component of acceleration of the point B with respect to C,
                                    2
BCvector 1.94 m/stax b ′ ==
*If the mechanism consists of more than one fixed point, then all these points lie at the same place in the
velocity and acceleration diagrams.

184  /G6C   Theory of Machines
We know that angular acceleration of AB,
                                   2 BA
AB1.410.47 rad/s3ta
ABα= = = Ans.
and angular acceleration of BC,
                                    2 BA
BC1.941.3 rad/s1.5ta
CBα= = = Ans.
Example 8.4. PQRS is a four bar chain with link PS fixed. The lengths of the links are PQ
= 62.5 mm ; QR = 175 mm ; RS = 112.5 mm ; and PS = 200 mm. The crank PQ rotates at 10 rad/s

clockwise. Draw the velocity and acceleration diagram when angle QPS = 60° and Q and R lie onthe same side of PS. Find the angular velocity and angular acceleration of links QR and RS.
Solution.  Given : ω
QP = 10 rad/s; PQ = 62.5 mm = 0.0625 m ; QR = 175 mm = 0.175 m ;
RS = 112.5 mm = 0.1125 m ; PS = 200 mm = 0.2 m
We know that velocity of Q with respect to P or velocity of Q,
vQP = vQ = ωQP × PQ = 10 × 0.0625 = 0.625 m/s
...(Perpendicular to PQ)
Angular velocity of links QR and RS

First of all, draw the space diagram of a four bar chain, to some suitable scale, as shown in
Fig. 8.9 (a). Now the velocity diagram as shown in Fig. 8.9 ( b), is drawn as discussed below:
Fig. 8.9
1. Since P and S are fixed points, therefore these points lie at one place in velocity diagram.
Draw vector pq perpendicular to PQ, to some suitable scale, to represent the velocity of Q with

respect to P or velocity of Q i.e. vQP or vQ such that
             vector pq = vQP = vQ = 0.625 m/s
2. From point q, draw vector qr perpendicular to QR to represent the velocity of R with
respect to Q (i.e. vRQ) and from point s, draw vector sr perpendicular to SR to represent the velocity
of R with respect to S or velocity of R (i.e. vRS or vR). The vectors qr and sr intersect at r. By
measurement, we find that

                       vRQ = vector qr = 0.333 m/s, and vRS = vR = vector sr = 0.426 m/s
We know that angular velocity of link QR,
                      RQ
QR0.3331.9 rad/s (Anticlockwise)
0.175v
RQω= = = Ans.

Chapter 8 :  Acceleration in Mechanisms    /G6C  185
and angular velocity of link RS,
                       RS
RS0.4263.78 rad/s (Clockwise)
0.1125v
SRω= = = A Ans.
Angular acceleration of links QR and RS
Since the angular acceleration of the crank PQ is not given, therefore there will be no tangential
component of the acceleration of Q with respect to P.
We know that radial component of the acceleration of Q with respect to P (or the acceleration
of Q),
                      2 2QP 2

of Q),
                      2 2QP 2
QP QP Q(0.625)6.25 m/s0.0625rvaaaPQ== == =
Radial component of the acceleration of R with respect to Q,
                       2 2RQ 2
RQ(0.333)0.634 m/s0.175rvaQR== =
and radial component of the acceleration of R with respect to S (or the acceleration of R),
                        2 2
2 RS
RS RS R(0.426)1.613 m/s0.1125r vaaaSR== == =
The acceleration diagram, as shown in Fig. 8.9 ( c) is drawn as follows :

1. Since P and S are fixed points, therefore these points lie at one place in the acceleration
diagram. Draw vector p'q' parallel to PQ, to some suitable scale, to represent the radial component
of acceleration of Q with respect to P or acceleration of Q i.e QP Qorraa such that
            2
QP Q vector 6.25 m/srpqaa′′== =
2. From point q', draw vector q' x parallel to QR to represent the radial component of
acceleration of R with respect to Q i.e. RQrasuch that
             2

2
RQ vector 0.634 m/srqxa′==
3. From point x, draw vector xr' perpendicular to QR to represent the tangential component
of acceleration of R with respect to Q i.e RQtawhose magnitude is not yet known.
4. Now from point s', draw vector s'y parallel to SR to represent the radial component of the
acceleration of R with respect to S i.e. RSrasuch that
              2
RS vector 1.613 m/srsya′==
5. From point y, draw vector yr' perpendicular to SR to represent the tangential component

of acceleration of R with respect to S i.e. RSta.
6. The vectors xr' and yr' intersect at r'. Join p'r and q' r'. By measurement, we find that
                       22
RQ RSvector 4.1m/s and vector 5.3 m/sttax r a yr ′′ == ==
We know that angular acceleration of link QR,
                        RQ 2
QR4.123.43 rad/s (Anticlockwise)QR 0.175taα= = = Ans.
and angular acceleration of link RS,
                        2 RS
RS5.347.1rad/s (Anticlockwise)0.1125ta
SRα= = = Ans.

186  /G6C   Theory of Machines
Example 8.5. The dimensions and
configuration of the four bar mechanism, shown inFig. 8.10, are as follows :
P
1A = 300 mm; P2B = 360 mm; AB = 360
mm, and P1P2 = 600 mm.
The angle AP1P2 = 60°. The crank P1A has
an angular velocity of 10 rad/s and an angularacceleration of  30 rad/s
2, both clockwise.
Determine the angular velocities and angularaccelerations of P
2B, and AB and the velocity and
acceleration of the joint B.

Solution.  Given : ωAP1 = 10 rad/s ; αAP1 = 30 rad/s2; P1A = 300 mm = 0.3 m ; P2B = AB =
360 mm = 0.36 m
We know that the velocity of A with respect to P1 or velocity of A,
                        vAP1 = vA = ωAP1 × P1A = 10 × 0.3 = 3 m/s
Velocity of B and angular velocitites of P2B and AB
First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.11 ( a). Now
the velocity diagram, as shown in Fig. 8.11 ( b), is drawn as discussed below:

1. Since P1 and P2 are fixed points, therefore these points lie at one place in velocity diagram.
Draw vector p1 a perpendicular to P1A, to some suitable scale, to represent the velocity of A with
respect to P1 or velocity of A i.e. vAP1 or vA, such that
               vector p1a = vAP1 = vA = 3 m/s
2. From point a, draw vector ab perpendicular to  AB to represent velocity of  B with respect
to A (i.e. vBA) and from point p2 draw vector p2b perpendicular to P2B to represent the velocity of B

with respect to P2 or velocity of B i.e. vBP2 or vB. The vectors ab and p2b intersect at b.
By measurement, we find that                         v
BP2 = vB = vector p2b = 2.2 m/s  Ans.
and                         vBA = vector ab = 2.05 m/s
We know that angular velocity of P2B,
                       BP2
P2B
22.26.1 rad/s (Clockwise)
0.36v
PBω= = = Ans.
and angular velocity of AB,
                        BA
AB2.055.7 rad/s (Anticlockwise)0.36v
ABω= = = Ans.

Acceleration of B and angular acceleration of P2B and AB
We know that tangential component of the acceleration of A with respect to P1,
                   2
A1 PP1130 0.3 9 m/st
AaP A=α × = × =
Radial component of the acceleration of A with respect to P1,
                   2
AP 22 21
AP AP 111
110 0.3 30 m/srv
aP APA== ω × = × =Fig. 8.10

Chapter 8 :  Acceleration in Mechanisms    /G6C  187
Radial component of the acceleration of B with respect to A.
                    2 2
2 BA
BA(2.05)11.67 m/s0.36rvaAB== =
and radial component of the acceleration of B with respect to P2,
                    2 2BP 2 2
BP2
2(2.2)13.44 m/s0.36rvaPB== =
Fig. 8.11
The acceleration diagram, as shown in Fig. 8.11 ( c), is
drawn as follows:
1. Since P1 and P2 are fixed points, therefore these points

will lie at one place, in the acceleration diagram. Draw vectorp
1' x parallel to P1A, to some suitable scale, to represent the
radial component of the acceleration of A with respect to P1,
such that
          2
1A P 1vector 30 m/srpxa′==
2. From point x, draw vector xa' perpendicular to P1A to
represent the tangential component of the acceleration of A with
respect to P1, such that
       2
AP1vector 9 m/stxa a′==
3. Join p1' a'. The vector p1' a' represents the acceleration

of A. By measurement, we find that the acceleration of A,
                        aA = aAP1 = 31.6 m/s2
4. From point a', draw vector a' y parallel to AB to represent the radial component of the
acceleration of B with respect to A, such that
           2
BA vector 11.67 m/sraya′==
5. From point y, draw vector yb' perpendicular to AB to represent the tangential component
of the acceleration of B with respect to A (i.e. BAta) whose magnitude is yet unknown.

6. Now from point 2,p′draw vector 2pz′parallel to P2B to represent the radial component
of the acceleration B with respect to P2, such that
                          2
2B P 2vector 13.44 m/srpza′==(c) Acceleration diagram
Fig. 8.11

188  /G6C   Theory of Machines
7. From point z, draw vector zb' perpendicular to P2B to represent the tangential component
of the acceleration of B with respect to P2 i.e. BP2.ta
8. The vectors  yb' and zb' intersect at b'. Now the vector p2' b' represents the acceleration of
B with respect to P2 or the acceleration of B i.e. aBP2 or aB. By measurement, we find that
                                      aBP2 = aB = vector p2' b' = 29.6 m/s2 Ans.
Also                     22

Also                     22
BA BP 2vector 13.6 m/s , and vector 26.6 m/sttyb a zb a′′== = =
We know that angular acceleration of P2B,
                                     BP 2 2
P2B226.673.8 rad/s (Anticlockwise)0.36ta
PBα= = = Ans.
and angular acceleration of AB,2 BA
AB13.637.8 rad/s (Anticlockwise)0.36ta
Example 8.6. In the mechanism, as shown in Fig. 8.12, the crank OA rotates at 20 r.p.m.

anticlockwise and gives motion to the sliding blocks B and D. The dimensions of the various linksare OA = 300 mm; AB = 1200 mm; BC = 450 mm and CD = 450 mm.
Fig. 8.12
For the given configuration, determine :  1. velocities of sliding at B and D, 2. angular
velocity of CD , 3. linear acceleration of D, and 4. angular acceleration of CD .
Solution. Given : NAO = 20 r.p.m. or ωAO = 2 π × 20/60 = 2.1 rad/s ; OA = 300 mm = 0.3 m ;
AB = 1200 mm = 1.2 m ; BC = CD = 450 mm = 0.45 m

Bicycle is a common example where simple mechanisms are used.

Chapter 8 :  Acceleration in Mechanisms    /G6C  189
We know that linear velocity of A with respect to O or velocity of A,
                   vAO = vA = ωAO × OA = 2.1 × 0.3 = 0.63 m/s ...(Perpendicular to OA)
1. Velocities of sliding at B and D
First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.13 ( a). Now
the velocity diagram, as shown in Fig. 8.13 ( b), is drawn as discussed below:
(a) Space diagram.
(b) Velocity diagram. ( c) Acceleration diagram.
Fig. 8.13

Fig. 8.13
1. Draw vector oa perpendicular to OA, to some suitable scale, to represent the velocity of
A with respect to O (or simply velocity of A), such that
                              vector oa = vAO = vA = 0.63 m/s
respect to A (i.e. vBA) and from point o draw vector ob parallel to path of motion B (which is along

BO) to represent the velocity of B with respect to O (or simply velocity of B). The vectors ab and ob
3. Divide vector ab at c in the same ratio as C divides AB in the space diagram. In other
words,
BC/CA = bc/ca
4. Now from point c, draw vector cd perpendicular to CD to represent the velocity of D with
respect to C (i.e. vDC) and from point o draw vector od parallel to the path of motion of D (which
along the vertical direction) to represent the velocity of D.

By measurement, we find that velocity of sliding at B,
                                            vB = vector ob = 0.4 m/s  Ans.
and velocity of sliding at D,vD = vector od = 0.24 m/s Ans.
2.  Angular velocity of CD
By measurement from velocity diagram, we find that velocity of D with respect to C,
                                          vDC = vector cd = 0.37 m/s

190  /G6C   Theory of Machines
∴ Angular velocity of CD,
                                 DC
CD0.370.82 rad/s (Anticlockwise).
0.45v
CDω= = = Ans.
3.  Linear acceleration of D
We know that the radial component of the acceleration of A with respect to O or acceleration
of A,
                                  2
22 2 AO
AO A AO (2.1) 0.3 1.323 m/sr vaa O AOA== = ω ×= ×=
Radial component of the acceleration of B with respect to A,
                                  2 2
2 BA

2 2
2 BA
BA(0.54)0.243 m/s1.2rvaAB== =
...(By measurement, vBA = 0.54 m/s)
Radial component of the acceleration of D with respect to C,
                                  2 2
2 DC
DC(0.37)0.304 m/s0.45rvaCD== =
Now the acceleration diagram, as shown in Fig. 8.13 ( c), is drawn as discussed below:
1. Draw vector o' a' parallel to OA, to some suitable scale, to represent the radial component

of the acceleration of A with respect to O or simply the acceleration of A, such that
        2
AO A vector 1.323 m/sroa a a′′== =
2. From point a', draw vector a' x parallel to AB to represent the radial component of the
                        2
BA vector 0.243 m/srax a′==
3. From point x, draw vector xb' perpendicular to AB to represent the tangential component

of the acceleration of B with respect to A (i.e. BAta) whose magnitude is not yet known.
4. From point o', draw vector o' b' parallel to the path of motion of B (which is along BO) to
represent the acceleration of B (aB). The vectors xb' and o' b' intersect at  b'. Join a' b'. The vector
a' b' represents the acceleration of B with respect to A.
5. Divide vector a' b' at c' in the same ratio as C divides AB in the space diagram. In other
words,
                             BC /BA  = b' c'/b' a'

6. From point c', draw vector c'y parallel to CD to represent the radial component of the
acceleration of D with respect to C, such that
                         2
DC vector 0.304 m/srcya′==
7. From point y, draw yd' perpendicular to CD to represent the tangential component of
acceleration of D with respect to C DC(.. )tie awhose magnitude is not yet known.
8. From point o', draw vector o' d' parallel to the path of motion of D (which is along the

vertical direction) to represent the acceleration of D (aD). The vectors yd' and o' d' intersect at  d'.
By measurement, we find that linear acceleration of D,
                                     aD = vector o' d' = 0.16 m/s2         Ans.
4.  Angular acceleration of CD
From the acceleration diagram, we find that the tangential component of the acceleration of
D with respect to C,
                       2
DCvector 1.28 m/sta yd′ == ...(By measurement)

Chapter 8 :  Acceleration in Mechanisms    /G6C  191
∴ Angular acceleration of CD,
          2 DC
CD1.282.84 rad/s (Clockwise)0.45ta
CDα= = = Ans.
Example 8.7.  Find out the acceleration of the slider D and
the angular acceleration of link CD for the engine mechanism shownin Fig. 8.14 .
The crank OA rotates uniformly at 180 r.p.m. in clockwise
direction. The various lengths are: OA = 150 mm ; AB = 450 mm;
PB = 240 mm ; BC = 210 mm ; CD = 660 mm.
Solution.  Given: N

AO = 180 r.p.m., or ωAO = 2π × 180/60 =
18.85 rad/s; OA = 150 mm = 0.15 m ; AB = 450 mm = 0.45 m ;
PB = 240 mm = 0.24 m ; CD = 660 mm = 0.66 m
We know that velocity of A with respect to O or velocity
of A,
vAO= vA = ωAO × OA
= 18.85 × 0.15 = 2.83 m/s
...(Perpendicular to OA)
First of all draw the space diagram, to some suitable scale,
as shown in Fig. 8.15 ( a). Now the velocity diagram, as shown in Fig. 8.15 ( b), is drawn as discussed
below:

below:
Fig. 8.15
1. Since O and P are fixed points, therefore these points lie at one place in the velocity
diagram. Draw vector oa perpendicular to OA, to some suitable scale, to represent the velocity of A
with respect to O or velocity of A (i.e. vAO or vA), such that
vector oa = vAO = vA = 2.83 m/s
2. Since the point B moves with respect to A and also with respect to P, therefore draw

vector ab perpendicular to AB to represent the velocity of B with respect to A i.e. vBA ,and from point
p draw vector pb perpendicular to PB to represent the velocity of B with respect to P or velocity of
B (i.e. vBP or vB). The vectors ab and pb intersect at b.
3. Since the point C lies on PB produced, therefore divide vector pb at c in the same ratio as
C divides PB in the space diagram. In other words, pb/pc = PB/PC.All dimensions in mm.
Fig. 8.14

192  /G6C   Theory of Machines
4. From point c, draw vector cd perpendicular to CD to represent the velocity of D with
respect to C and from point o draw vector od parallel to the path of motion of the slider D (which is
vertical), to represent the velocity of D, i.e. vD.
vD = vector od = 2.36 m/s
Velocity of D with respect to C,
             vDC = vector cd = 1.2 m/s
             vBA = vector ab = 1.8 m/s

vBA = vector ab = 1.8 m/s
and velocity of B with respect to  P, vBP = vector pb = 1.5 m/s
Acceleration of the slider D
We know that radial component of the acceleration of A with respect to O or acceleration
of A,
           222
AO A AO (18.85) 0.15 53.3 m/sraa A O== ω × = × =
           2 2
2 BA
BA(1.8)7.2 m/s0.45rvaAB== =
Radial component of the acceleration of B with respect to P,
           2 2
2 BP

2 2
2 BP
BP(1.5)9.4 m/s0.24rvaPB== =
           2 2
2 DC
DC(1.2)2.2 m/s0.66rvaCD== =
Now the acceleration diagram, as shown in Fig. 8.15 ( c), is drawn as discussed below:
1.  Since O and P are fixed points, therefore these points lie at one place in the acceleration
diagram. Draw vector o' a' parallel to OA, to some suitable scale, to represent the radial component

of the acceleration of A with respect to  O or the acceleration of A (i.e. AOraor aA), such that
2
AO A vector 53.3 m/sroa a a′′== =
acceleration of B with respect to A (i.e. BAra), such that
  2
BA vector 7.2 m/srax a′==
3. From point x, draw vector xb' perpendicular to the vector a'x or AB to represent the
tangential component of the acceleration of B with respect to A i.e.BAtawhose magnitude is yet

unknown.
4. Now from point p', draw vector p' y parallel to PB to represent the radial component of
the acceleration of B with respect to P (i.e. BPra), such that
 2
BP vector 9.4 m/srpya′==
5. From point y, draw vector yb' perpendicular to vector b'y or PB to represent the tangential
component of the acceleration  of B, i.e. BPta.  The vectors xb' and yb' intersect at b'. Join p' b'. The
vector p' b' represents the acceleration of B, i.e. aB.

Chapter 8 :  Acceleration in Mechanisms    /G6C  193
6. Since the point C lies on PB produced, therefore divide vector p'b' at c' in the same ratio
as C divides PB in the space diagram. In other words, p'b'/p'c' = PB/PC
7. From point c', draw vector c'z parallel to CD to represent the radial component of the
acceleration of  D with respect to  C i.e. DC,rasuch that
                       2
DC vector 2.2 m/srcz a′==

8. From point z, draw vector zd' perpendicular to vector c'z or CD to represent the tangential
component of the acceleration of D with respect to C i.e. DC,tawhose magnitude is yet unknown.
9. From point o', draw vector o' d' parallel to the path of motion of D (which is vertical) to
represent the acceleration of D, i.e. aD. The vectors zd' and o' d' intersect at d'. Join c' d'.
By measurement, we find that acceleration of D,

                                   aD = vector o'd' = 69.6 m/s2  Ans.
Angular acceleration of CD
From acceleration diagram, we find that tangential component of the acceleration of D with
2
DCvector 17.4 m/staz d ′ == ...(By measurement)
We know that angular acceleration of CD,
                               2 DC
CD17.426.3 rad/s (Anticlockwise)0.66ta

Example 8.8. In the toggle mechanism shown in Fig. 8.16, the slider D is constrained to
move on a horizontal path. The crank OA is rotating in the counter-clockwise direction at a speed
Fig. 8.16
of 180 r.p.m. increasing at the rate of 50 rad/s2. The dimensions of the various links are as follows:
OA = 180 mm ; CB = 240 mm ; AB = 360 mm ; and BD = 540 mm.
For the given configuration, find 1. Velocity of slider D and angular velocity of BD, and

2. Acceleration of slider D and angular acceleration of BD.
Solution. Given :  NAO = 180 r.p.m. or ωAO = 2 π × 180/60 = 18.85 rad/s ; OA = 180 mm
= 0.18 m ; CB = 240 mm = 0.24 m ; AB = 360 mm = 0.36 m ; BD = 540 mm = 0.54 m
We know that velocity of A with respect to O or velocity of A,
                               vAO = vA = ωAO × OA = 18.85 × 0.18 = 3.4 m/s

194  /G6C   Theory of Machines
1. Velocity of slider D and angular velocity of BD
First of all, draw the space diagram to some suitable scale, as shown in Fig. 8.17 ( a). Now
the velocity diagram, as shown in Fig. 8.17 ( b), is drawn as discussed below:
1. Since O and C are fixed points, therefore these points lie at one place in the velocity

with respect to O or velocity of  A i.e. vAO or vA, such that
                              vector oa = vAO = vA = 3.4 m/s
Fig. 8.17
2. Since B moves with respect to A and also with respect to C, therefore draw vector ab
perpendicular to CB to represent the velocity of B with respect to C ie. vBC. The vectors ab and cb

intersect at  b.
respect  to B i.e. vDB, and from point c draw vector cd parallel to CD (i.e., in the direction of motion
of the slider D) to represent the velocity of D i.e. vD.
vBA = vector ab = 0.9 m/s
vBC = vector cb = 2.8 m/s
Velocity of D with respect to B,
vDB = vector bd = 2.4 m/s

and velocity of slider D,      vD = vector cd = 2.05 m/s Ans.
Angular velocity of BD
We know that the angular velocity of BD,
                           DB
BD2.44.5 rad/s
0.54v
BDω= = = Ans.
2. Acceleration of slider D and angular acceleration of BD
Since the angular acceleration of OA increases at the rate of 50 rad/s2, i.e. αAO = 50 rad/s2,
therefore
Tangential component of the acceleration of A with respect to O,
                            2

Chapter 8 :  Acceleration in Mechanisms    /G6C  195
Radial component of the acceleration of A with respect to O,
2 2
2 AO
AO(3.4)63.9 m/s0.18rvaOA== =
Radial component of the acceleration of B with respect to  A,
     2 2
2 BA
BA(0.9)2.25 m/s0.36rvaAB== =
Radial component of the acceleration of B with respect to C,
                            2 2
2 BC
BC(2.8)32.5 m/s0.24rvaCB== =
and radial component of the acceleration of D with respect to B,
                             2 2
2 DB

2 2
2 DB
DB(2.4)10.8 m/s0.54rvaBD== =
Now the acceleration diagram, as shown in Fig. 8.17 ( c), is drawn as discussed below:
1. Since O and C are fixed points, therefore
these points lie at one place in the accelerationdiagram. Draw vector o'x parallel to OA, to some
suitable scale, to represent the radial componentof the acceleration of A with respect to O i.e.
AO,rasuch that
                     2
AO vector 63.9 m/srox a′==
2. From point x, draw vector xa'

perpendicular to vector o'x or OA to represent the
tangential component of the acceleration of A with
respect to O i.e. AOta,such that
                   2
AO vector 9 m/stxa a′==
3. Join o'a'. The vector o'a' represents the
total acceleration of A with respect to O or
acceleration of A i.e. aAO or aA.
4. Now from point a', draw vector a'y parallel to AB to represent the radial component of the
acceleration of B with respect to A i.e. BA,rasuch that

2
BA vector 2.25 m/sraya′==
 5. From point y, draw vector yb' perpendicular to vector a'y or AB to represent the tangential
component of the acceleration of B with respect to A i.e.BAtawhose magnitude is yet unknown.
6. Now from point c', draw vector c'z parallel to CB to represent the radial component of the
acceleration of B with respect to C i.e. BC,rasuch that
                    2
BC vector 32.5 m/srcz a′==

2
7. From point z, draw vector zb' perpendicular to vector c'z or CB to represent the tangential
component of the acceleration of B with respect to C i.e. BCta. The vectors yb' and zb' intersect at b'.
Join c' b'. The vector c' b' represents the acceleration of B with respect to C i.e. aBC.
8. Now from point b', draw vector b's parallel to BD to represent the radial component of the
acceleration of D with respect to B i.e. DB,rasuch that

2
DB vector 10.8 m/srbs a′==
An experimental IC engine with crank shaft
and cylinders.
Note : This picture is given as additional informa-
tion and is not a direct example of the current
chapter.

196  /G6C   Theory of Machines
9. From point s, draw vector sd' perpendicular to vector b's or BD to represent the tangential
component of the acceleration of D with respect to B i.e. DBtawhose magnitude is yet unknown.
10. From point c', draw vector c'd' parallel to the path of motion of D (which is along CD)
to represent the acceleration of D i.e. aD. The vectors sd' and c'd' intersect at d'.
By measurement, we find that acceleration of slider D,

aD = vector c'd' = 13.3 m/s2 Ans.
Angular acceleration of BD
By measurement, we find that tangential component of the acceleration of D with respect
to B,
                        2
DBvector 38.5 m/stas d ′ ==
We know that angular acceleration of BD,
                        2 DB
BD38.571.3 rad/s (Clockwise)0.54ta
BDα= = = Ans.
Example 8.9. The mechanism of a warping machine, as shown in Fig. 8.18, has the
dimensions as follows:

O1A = 100 mm; AC = 700 mm ; BC = 200 mm ; BD = 150 mm ; O2D = 200 mm ; O2E = 400
mm ; O3C = 200 mm.
Fig. 8.18
The crank O1A rotates at a uniform speed of 100 rad/s. For the given configuration,
determine: 1. linear velocity of the point E on the bell crank lever,  2. acceleration of the points E
and B, and 3. angular acceleration of the bell crank lever.
Solution.  Given : ωAO1 = 100 rad/s ; O1A = 100 mm = 0.1 m

We know that linear velocity of A with respect to O1, or velocity of A,
                       vAO1 = vA = ω AO1 × O1A = 100 × 0.1 = 10 m/s ...(Perpendicular to O1A)
1.  Linear velocity of the point E on bell crank lever
First of all draw the space diagram, as shown in Fig. 8.19 ( a), to some suitable scale. Now
the velocity diagram, as shown in Fig. 8.19 ( b), is drawn as discussed below:
1. Since O1, O2 and O3 are fixed points, therefore these points are marked as one point in the

velocity diagram. From point o1, draw vector o1a perpendicular to O1A to some suitable scale, to
represent the velocity of A with respect to O or velocity of A, such that
               vector o1a = vAO1 = vA = 10 m/s

Chapter 8 :  Acceleration in Mechanisms    /G6C  197
2. From point a, draw vector ac perpendicular to AC to represent the velocity of C with
respect to A (i.e. vCA) and from point o3 draw vector o3c perpendicular to O3C to represent the
velocity of C with respect to O3 or simply velocity of C (i.e. vC). The vectors ac and o3c intersect at
point c.
(b) Velocity diagram.                                       ( c) Acceleration diagram.
Fig. 8.19
3. Since B lies on AC, therefore divide

Fig. 8.19
vector ac at b in the same ratio as B divides AC in
the space diagram. In other words, ab/ac = AB/AC
4. From point b, draw vector bd
perpendicular to BD to represent the velocity of D
with respect to B (i.e. vDB), and from point o2 draw
vector o2d perpendicular to O2D to represent the
velocity of D with respect to O2 or simply velocity
of D (i.e. vD). The vectors bd and o2d intersect at d.
5. From point o2, draw vector o2e

perpendicular to vector o2d in such a way that
         o2e/o2d = O2E/O2D
By measurement, we find that velocity of
point C with respect to A,
vCA = vector ac = 7 m/s
Velocity of point C with respect to O3,
              vCO3 = vC = vector o3c = 10 m/s
Velocity of point D with respect to B,
               vDB = vector bd = 10.2 m/s(a) Space diagram.
Warping machine uses many mechanisms.

198  /G6C   Theory of Machines
Velocity of point D with respect to O2,
vDO2 = vD = vector o2d = 2.8 m/s
and velocity of the point E on the bell crank lever,
vE = vEO2 = vector o2e = 5.8 m/s  Ans.
2.  Acceleration of the points E and B
Radial component of the acceleration of A with respect to O1 (or acceleration of A),
2 2
2 AO1
AO AO1 A2
1101000 m/s0.1r vaaaOA== == =
Radial component of the acceleration of C with respect to A,
2 2
2 CA
CA770 m/s0.7rvaAC== =

2 2
2 CA
Radial component of the acceleration of C with respect to O3,
2 2
CO 2 3
CO3
310500 m/s0.2rvaOC== =
Radial component of the acceleration of D with respect to B,
2 2
2 DB
DB(10.2)693.6 m/s0.15rvaBD== =
Radial component of the acceleration of D with respect to O2,
2 2DO 2 2
DO2
2(2.8)39.2 m/s0.2rv
aOD== =
Radial component of the acceleration of E with respect to O2,
     2 2EO 2 2
EO2
2(5.8)84.1m/s0.4rv
aOE== =

2 2EO 2 2
EO2
aOE== =
Now the acceleration diagram, as shown in Fig. 8.19 ( c), is drawn as discussed below:
acceleration diagram. Draw vector o1' a' parallel to O1A, to some suitable scale, to represent the
radial component of the acceleration of A with respect to O1 (or simply acceleration of A), such that
2
1A O A 1vector 1000 m/sroa a a′′== =

2
2. From point a', draw a'x parallel to AC to represent the radial component of the acceleration
of C with respect to A (i.e. CAra), such that
2
CA m vector 70 /srax a′==
3. From point x, draw vector xc' perpendicular to AC to represent the tangential component
of the acceleration of C with respect to A (i.e. CAta), the magnitude of which is yet unknown.
4. From point o3', draw vector o3' y parallel to O3C to represent the radial component of the

acceleration of C with respect to O3 (i.e. CO3ra), such that
2
CO33m vector 500 /sroya′==
5. From point y, draw vector yc' perpendicular to O3C to represent the tangential component
of the acceleration of C with respect to O3 (i.e. CO3ta). The vectors xc' and yc' intersect at c'.

Chapter 8 :  Acceleration in Mechanisms    /G6C  199
6. Join a' c'. The vector a' c' represents the acceleration of C with respect to A (i.e. aCA).
7. Since B lies on AC, therefore divide vector a'c' at b' in the same ratio as B divides AC in
the space diagram. In other words, a'b'/a'c' = AB/AC. Join b' o2' which represents the acceleration of
point B with respect to O2 or simply acceleration of B. By measurement, we find that
Acceleration of point B = vector o2' b' = 440 m/s2 Ans.

8. Now from point b', draw vector b' z parallel to BD to represent the radial component of
the acceleration of D with respect to B (i.e. DBra), such that
2
DB vector 693.6 m/srbz a′==
9. From point z, draw vector zd' perpendicular to BD to represent the tangential component
of the acceleration of D with respect to B (i.e. DBta), whose magnitude is yet unknown.
10. From point o2' , draw vector o2' z1 parallel to  O2D to represent the radial component of

the acceleration of D with respect to O2 (i.e. DO2ra), such that
2
21 D O 2vector 39.2 m/sroz a′==
11. From point z1, draw vector z1d' perpendicular to O2D to represent the tangential component
of the acceleration of D with respect to O2 (i.e. DO2ta). The vectors zd' and z1d'  intersect at d'.
12. Join o2' d'. The vector o2'd' represents the acceleration of D with respect to O2 or simply
acceleration of D (i.e. aDO2 or aD).

13. From point o2', draw vector o2' e' perpendicular to o2' d' in such a way that
22 22//oe od OEOD′′′′=
Note: The point e' may also be obtained drawing 2EOraand 2EOtaas shown in Fig. 8.19 ( c).
By measurement, we find that acceleration of point E,
aE = aEO2 = vector o' 2 e' = 1200 m/s2 Ans.
3. Angular acceleration of the bell crank lever
By measurement, we find that the tangential component of the acceleration of D with respect
to O2,
2

to O2,
2
D1 1O2vector 610 m/staz d ′ ==
∴ Angular acceleration of the bell crank lever
    2 DO2
26103050 rad/s (Anticlockwise)O0 . 2ta
D== = Ans.
Example 8.10. A pump is driven from an engine
crank-shaft by the mechanism as shown in Fig. 8.20. Thepump piston shown at F is 250 mm in diameter and thecrank speed is 100 r.p.m. The dimensions of various linksare as follows:
OA = 150 mm ; AB = 600 mm ; BC = 350 mm ;
CD = 150 mm; and DE = 500 mm.
Determine for the position shown : 1. The velocity of

the cross-head E,  2. The rubbing velocity of the pins A
and B which are 50 mm diameter. 3. The torque required
at the crank shaft to overcome a presure of 0.35 N/mm
2,
and 4. The acceleration of the cross-head E.
All dimensions in mm.
Fig. 8.20

200  /G6C   Theory of Machines
Solution. Given : NAO = 100 r.p.m. or ωAO = 2 π × 100/60 = 10.47 rad/s; OA = 150 mm = 0.15 m ;
AB = 600 mm = 0.6 m ; BC = 350 mm = 0.35 m ; CD = 150 mm = 0.15 m ; DE = 500 mm = 0.5 m
vAO = vA = ωAO × OA = 10.47 × 0.15 = 1.57 m/s ...(Perpendicular to OA)
1.  Velocity of the cross-head E
First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.21 ( a). Now

the velocity diagram, as shown in Fig. 8.21 ( b), is drawn as discussed below:
                        ( a) Space diagram.           ( b) Velocity diagram. ( c) Acceleration diagram.
Fig. 8.21
1. Since O and C are fixed points, therefore these points are marked as one point in the
velocity diagram. Now draw vector oa perpendicular to OA, to some suitable scale, to represent the
velocity of A with respect ot O or the velocity of A, such that

vector oa = vAO = vA = 1.57 m/s
of B with respect to C (i.e. vBC). The vectors ab and cb intersect at b.
By measurement, we find that                                     v
BA = vector ab = 1.65 m/s
and                                            vBC = vB = vector cb = 0.93 m/s

3. From point c, draw vector cd perpendicular to CD or vector cb to represent the velocity of
D with respect to C or velocity of D, such that
                        vector cd : vector cb = CD: CB      or      vDC : vBC = CD : CB
∴                     DC
DC BC
BC0.15or 0.93 0.4 m/s0.35vCD CDvvvC B C B== × = × =
4. From point d, draw vector de perpendicular to DE to represent the velocity of E with

respect to D (i.e. vED), and from point o draw vector oe parallel to the path of motion of  E (which is
vertical) to represent the velocity of E or F. The vectors oe and de intersect at e.
By measurement, we find that velocity of E with respect to D,
         vED = vector de = 0.18 m/s

Chapter 8 :  Acceleration in Mechanisms    /G6C  201
and velocity of the cross-head E,
                                 vEO = vE = vector oe = 0.36 m/s Ans.
2.  Rubbing velocity of the pins at A and B
We know that angular velocity of A with respect to O,
                                ωAO = 10.47 rad/s ...(Anticlockwise)
Angular velocity of B with respect to A,
                               BA
BA1.652.75 rad/s
0.6v
ABω= = = ...(Anticlockwise)
and angular velocity of B with respect to C,

     BC
BC0.932.66 rad/s
0.35v
CBω= = = ...(clockwise)
We know that diameter of pins at A and B,
                                 dA = dB = 50 mm = 0.05 m ...(Given)
or                               Radius, rA = rB = 0.025 m
∴ Rubbing velocity of pin at A
= (ωAO – ωBA) rA = (10.47 – 2.75) 0.025 = 0.193 m/s Ans.
and rubbing velocity of pin at B
                         = ( ωBA +  ωBC) rB = (2.75 + 2.66) 0.025 = 0.135 m/s Ans.

3.  Torque required at the crankshaft
Given: Pressure to overcome by the crankshaft,
 pF = 0.35 N/mm2
Diameter of the pump piston
  DF = 250 mm
∴ Force at the pump piston at F,
  22
FF FPressure Area ( ) 0.35 (250) 17 183 N44Fp Dππ=× = ×= × =
Let FA = Force required at the crankshaft at A.
Assuming transmission efficiency as 100 per cent,
  Work done at A = Work done at F
               FF
AA FF A
A17183 0.36or 3940 N1.57FvFvFv Fv×××=× = = =
FE...( )vv=/G51
∴ Torque required at the crankshaft,

TA = FA × OA = 3940 × 0.15  =  591  N-m  Ans.
Acceleration of the crosshead E
We know that the radial component of the acceleration of A with respect to O or the
acceleration of A,
                      2 2
2 AO
AO A(1.57)16.43 m/s0.15r vaaOA== = =

202  /G6C   Theory of Machines
        2 2
2 BA
BA(1.65)4.54m/s0.6rvaAB== =
Radial component of the acceleration of B with respect to C.
                      2 2
2 BC
BC(0.93)2.47 m/s0.35rvaCB== =
and radial component of the acceleration of E with respect to D,
                      2 2
2 ED
ED(0.18)0.065 m/s0.5rvaDE== =
Now the acceleration diagram, as shown in Fig. 8.21 ( c), is drawn as discussed below:

acceleration diagram. Draw vector o'a' parallel to OA, to some suitable scale, to represent the radial
component of the acceleration of A with respect to O or the acceleration of A, such that
            2
AO A vector 16.43 m/sroa a a′′== =
2. From point a', draw vector a'x parallel to AB to represent the radial component of the
             2

2
BA vector 4.54 m/srax a′==
of the acceleration of B with respect to A (i.e.BAta) whose magnitude is yet unknown.
4. Now from point c', draw vector c' y parallel to CB to represent the radial component of
the acceleration of B with respect to C (i.e. BCra), such that
2
BC vector 2.47 m/srcya′==
5. From point y, draw vector yb' perpendicular to CB to represent the tangential component

of the acceleration of B with respect to C (i.e. BCta). The vectors yb' and xb' intersect at b'. Join c'b'
and a'b'. The vector c'b' represents the acceleration of B with respect to C (i.e. aBC) or the acceleration
of B (i.e. aB) and vector a'b' represents the acceleration of B with respect to A (i.e. aBA).
By measurement, we find that                       a
BC = aB = vector c'b' = 9.2 m/s2
and                              aBA = vector a'b' = 9 m/s2

6. From point c', draw vector c'd' perpendicular to CD or vector c'b' to represent the
acceleration of D with respect to C or the acceleration of D (i.e. aDC or aD), such that
vector c'd' : vector c'b' = CD : CB   or   aD : aBC = CD : CB
∴                  2 D
DB C
BC0.15or 9.2 3.94 m/s0.35aC D C DaaaC B C B== × = × =
7. Now from point d', draw vector d'z parallel to DE to represent the radial component of E
with respect to  D (i.e. EDra), such that
            2
ED vector 0.065 m/srdz a′==

2
Note:  Since the magnitude of ar
ED is very small, therefore the points d' and z coincide.
8. From point z, draw vector ze' perpendicular to DE to represent the tangential component
of the acceleration of E with respect to D (i.e. EDta) whose magnitude is yet unknown.
9. From point o', draw vector o'e' parallel to the path of motion of E (which is vertical) to
represent the acceleration of E. The vectors ze' and o'e' intersect at e'.

Chapter 8 :  Acceleration in Mechanisms    /G6C  203
By measurement, we find that acceleration of the crosshead E,
                                                        aE = vector o'e' = 3.8 m/s2  Ans.
Example 8.11. Fig. 8.22 shows the mechanism of a radial valve gear. The crank OA turns
uniformly at 150 r.p.m and is pinned at A to rod AB. The point C in the rod is guided in the circularpath with D as centre and DC as radius. The dimensions of various links are:

OA = 150 mm ; AB = 550 mm ; AC = 450 mm ; DC = 500 mm ; BE = 350 mm.
Determine velocity and acceleration of the ram E for the given position of the mechanism.
Fig. 8.22
Solution. Given : NAO = 150 r.p.m. or ωAO = 2 π × 150/60 = 15.71 rad/s; OA = 150 mm = 0.15 m;
AB = 550 mm = 0.55 m ; AC = 450 mm = 0.45 m; DC = 500 mm = 0.5 m ; BE = 350 mm = 0.35 m
vAO = vA = ωAO × OA = 15.71 × 0.15 = 2.36 m/s

Velocity of the ram E
First of all draw the space diagram, as shown in Fig. 8.23 ( a), to some suitable scale. Now
the velocity diagram, as shown in Fig. 8.23 ( b), is drawn as discussed below:
1. Since O and D are fixed points, therefore these points are marked as one point in the
velocity diagram. Draw vector oa perpendicular to OA, to some suitable scale, to represent the

velocity of A with respect to O or simply velocity of A, such that
AO A vector 2.36 m/soa v v== =
respect to A (i.e. vCA), and from point d draw vector dc perpendicular to DC to represent the velocity
of C with respect to D or simply velocity of C (i.e. vCD or vC). The vectors ac and dc intersect at c.
3. Since the point B lies on AC produced, therefore divide vector ac at b in the same ratio as

B divides AC in the space diagram. In other words ac:cb = AC:CB. Join ob. The vector ob represents
the velocity of B (i.e. vB)
4. From point b, draw vector be perpendicular to be to represent the velocity of E with
respect to B (i.e. vEB), and from point o draw vector oe parallel to the path of motion of the ram E
(which is horizontal) to represent the velocity of the ram E. The vectors be and oe intersect at e.
By measurement, we find that velocity of C with respect to A,

vCA = vector ac = 0.53 m/s
Velocity of C with respect to D,
vCD = vC = vector dc = 1.7 m/s

204  /G6C   Theory of Machines
Velocity of E with respect to B,
vEB = vector be = 1.93 m/s
and velocity of the ram E,   vE = vector oe = 1.05 m/s  Ans.
Acceleration of the ram E
 2 2
2 AO
AO A(2.36)37.13 m/s0.15r vaaOA== = =
 2 2
2 CA
CA(0.53)0.624 m/s0.45rvaOA== =
Radial component of the acceleration of C with respect to D,
 2 2
2 CD

2 2
2 CD
CD(1.7)5.78 m/s0.5rvaDC== =
Radial component of the acceleration of E with respect to B,
2 2
2 EB
EB(1.93)10.64 m/s0.35rvaBE== =
The acceleration diagram, as shown in Fig. 8.23 ( c), is drawn as discussed below:
(c) Acceleration diagram.
Fig. 8.23

Chapter 8 :  Acceleration in Mechanisms    /G6C  205
component of the acceleration of A with respect to O or simply the acceleration of  A, such that
2
AO A vector 37.13 m/sroa a a′′== =
2. From point d', draw vector d'x parallel to DC to represent the radial component of the

acceleration of C with respect to D, such that
2
CD vector 5.78 m/srdx a′==
3. From point x, draw vector xc' perpendicular to DC to represent the tangential component
of the acceleration of C with respect to D (i.e. CDta) whose magnitude is yet unknown.
4. Now from point a', draw vector a'y parallel to AC to represent the radial component of
the acceleration of C with respect to A, such that
2
CA vector 0.624 m/sraya′==

2
5. From point y, draw vector yc' perpendicular to AC to represent the tangential component
of acceleration of C with respect to A (i.e. CAta). The vectors xc' and yc' intersect at c'.
6. Join a'c'. The
vector a'c' represents the
acceleration of C with
respect to A (i.e. aCA).
7. Since the point
B lies on AC produced,
therefore divide vector a'c'
at b' in the same ratio as B
divides AC in the space
diagram. In other words, a'
c' : c' b' = AC : CB.
8. From point b',

draw vector b' z parallel to
BE to represent the radial
component of theacceleration of E with
respect to B, such that
2
EB vector 10.64 m/srbz a′==
9. From point z, draw vector ze' perpendicular to BE to represent the tangential component
of the acceleration of E with respect to B (i.e. EBta) whose magnitude is yet unknown.
10. From point o', draw vector o'e' parallel to the path of motion of E (which is horizontal)

to represent the acceleration of the ram E. The vectors ze' and o'e' intersect at e'.
By measurement, we find that the acceleration of the ram E,
2
Evector 3.1m/sao e ′′ == Ans.
Example 8.12. The dimensions of the Andreau diff erential stroke engine mechanism, as
shown in Fig. 8.24, are as follows:
AB = 80 mm ; CD = 40 mm ; BE = DE = 150 mm ; and EP = 200 mm.
The links AB and CD are geared toget her. The speed of the smaller wheel is 1 140 r.p.m.

Determine the velocity and acceleration of the piston P for the given configuration.
A lathe is a machine for shaping a piece of metal, by rotating it rapidly along
its axis while pressing against a fixed cutting or abrading tool.
Note : This picture is given as additional information and is not a direct
example of the current chapter.

206  /G6C   Theory of Machines
Solution. Given: NDC = 1140 r.p.m. or ωDC = 2 π × 1140/60 = 119.4 rad/s ; AB = 80 mm
= 0.08 m ; CD = 40 mm = 0.04 m ; BE = DE = 150 mm = 0.15 m ; EP = 200 mm = 0.2 m
We know that velocity of D with respect to C or velocity of D,
                    vDC = vD = ωDC × CD = 119.4 × 0.04 = 4.77 m/s ...(Perpendicular to CD)
Since the speeds of the gear wheels are inversely proportional to their diameters, therefore
                                BA

BA
DCAngular speed of larger wheel 2
Angular speed of smaller wheel 2CD
ABω==ω
∴ Angular speed of larger wheel,
BA DC0.04119.4 59.7 rad/s0.08CD
ABω= ω× = × =
and velocity of B with respect to A or velocity of B,
BA B BA 59.7 0.08 4.77 m/s vv A B== ω × = × =
...(Perpendicular to AB)
Velocity of the piston P
First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.25 ( a). Now

the velocity diagram, as shown in Fig. 8.25 ( b), is drawn as discussed below:
1. Since A and C are fixed points, therefore these points are marked as one point in the
velocity diagram. Draw vector cd perpendicular to CD, to some suitable scale, to represent the
velocity of D with respect to C or velocity of D (i.e. vDC or vD), such that
vector cd = vDC = vD = 4.77 m/s
2. Draw vector ab perpendicular to AB to represent the velocity of B with respect to A or

velocity of B (i.e. vBA or vB), such that
vector ab = vBA = vB = 4.77 m/s
3. Now from point b, draw vector be perpendicular to BE to represent the velocity of E with
respect to B (i.e. vEB), and from point d draw vector de perpendicular to DE to represent the velocity
of E with respect to D (i.e. vED). The vectors be and de intersect at e.
4. From point e, draw vector ep perpendicular to EP to represent the velocity of P with

respect to E (i.e. vPE), and from point a draw vector ap parallel to the path of motion of P (which is
horizontal) to represent the velocity of P. The vectors ep and ap intersect at p.Fig. 8.24

Chapter 8 :  Acceleration in Mechanisms    /G6C  207
By measurement, we find that velocity of E with respect to B,
 vEB = vector be = 8.1 m/s
Velocity of E with respect to D,
vED = vector de = 0.15 m/s
Velocity of P with respect to E,
vPE = vector ep = 4.7 m/s
and                  velocity of P, vP = vector ap = 0.35 m/s Ans.
               
Fig. 8.25
Acceleration of the piston P
We know that the radial component of the

acceleration of B with respect A  (or the acceleration
of B),
2 2
2 BA
BA B(4.77)284.4 m/s0.08r vaaAB== = =
Radial component of the acceleration of D
with respect to C (or the acceleration of D),
  2 2
2 DC
DC D(4.77)568.8 m/s0.04r vaaCD== = =
Radial component of the acceleration of E with
2 2
2 EB
EB(8.1)437.4 m/s0.15rvaBE== =
Radial component of the acceleration of E with respect  to D,
2 2
2 ED
ED(0.15)0.15 m/s0.15rvaDE== =

2 2
2 ED
and radial component of the acceleration of P with respect to E,
2 2
2 PE
PE(4.7)110.45 m/s0.2rvaEP== =(c) Acceleration diagram.
Fig. 8.25

208  /G6C   Theory of Machines
Now the acceleration diagram, as shown in Fig. 8.25 ( c), is drawn as discussed below:
acceleration diagram. Draw vector a'b' parallel to AB, to some suitable scale, to represent the radial
component of the acceleration of B with respect to A or the acceleration of B, such that
                 2
BA B vector 284.4 m/srab a a′′== =

2. Draw vector c'd' parallel to CD to represent the radial component of the acceleration of D
with respect to C or the acceleration of D, such that
                2
DC D vector 568.8 m/srcd a a′′= = =
3. Now from point b', draw vector b'x parallel to BE to represent the radial component of the
acceleration of E with respect to B, such that
2
EB vector 437.4 m/srbx a′′==

2
4. From point x, draw vector xe' perpendicular to  BE to represent the tangential component
of acceleration of E with respect to B (i.e. EBta) whose magnitude is yet unknown.
5. From point d', draw vector d'y parallel to DE to represent the radial component of the
acceleration of E with respect to D, such that
2
ED vector 0.15 m/srdya′==
Note:  Since the magnitude of EDrais very small ( i.e. 0.15 m/s2), therefore the points d' and y coincide.

6. From point y, draw vector ye' perpendicular to DE to represent the tangential component
of the acceleration of E with respect to D (i.e. EDta). The vectors xe' and ye' intersect at e'.
7. From point e', draw vector e'z parallel to EP to represent the radial component of the
acceleration of P with respect to E, such that
2
PE vector 110.45 m/srez a′==
8. From point z, draw vector zp' perpendicular to EP to represent the tangential component

of the acceleration of P with respect to E (i.e. PEta) whose magnitude is yet unknown.
9. From point a', draw vector a'p' parallel to the path of motion of P (which is horizontal) to
represent the acceleration of P. The vectors zp' and a'p' intersect at p'.
By measurement, we find that acceleration of the piston P,
aP = vector a'p' = 655 m/s2 Ans.
8.5. Coriolis Component of Acceleration
When a point on one link is sliding along another rotating link, such as in quick return

motion mechanism, then the coriolis component of the acceleration must be calculated.
Consider a link OA and a slider B as shown in Fig. 8.26 ( a). The slider B moves along the
link OA. The point C is the coincident point on the link OA.
Let ω= Angular velocity of the link OA at time t seconds.
v= Velocity of the slider B along the link OA at time t seconds.
ω.r= Velocity of the slider B with respect to O (perpendicular to the link OA)
at time t seconds, and

Chapter 8 :  Acceleration in Mechanisms    /G6C  209
(ω + δω), (v + δv) and (ω + δω) (r + δr)
= Corresponding values at time ( t + δt) seconds.
Fig. 8.26.  Coriolis component of acceleration.
Let us now find out the acceleration of the slider B
with respect to O and with respect to its coincident point C
lying on the link OA.
Fig. 8.26 ( b) shows the velocity diagram when their
velocities v and (v + δv) are considered. In this diagram, the

vector bb1 represents the change in velocity in time δt sec ; the
vector bx represents the component of change of velocity bb1
along OA (i.e. along radial direction) and vector xb1 represents
the component of change of velocity bb1 in a direction
perpendicular to OA (i.e. in tangential direction). Therefore
() c o s bx ox ob v v v=−=+ δ δ θ − ↑
Since δθ is very small, therefore substituting
cos δθ = 1, we have
()bx v v v v=+ δ −↑ = δ ↑

                                                         ...(Acting radially outwards)
and1() s i nxb v v=+ δ δ θ
Since δθ is very small, therefore substituting sin δθ =
δθ, we have
1() . .x b v vvv=+ δ δ θ = δ θ + δ δ θ
Neglecting δv.δθ being very small, therefore
1.xb v←
=δ θ ...(Perpendicular to OA and towards left)
Fig. 8.26 ( c) shows the velocity diagram when the velocities ω.r and (ω + δω) (r + δr) are

considered. In this diagram, vector bb1 represents the change in velocity ; vector yb1 represents the
component of change of velocity bb1 along OA (i.e. along radial direction) and vector  by represents
the component of change of velocity bb1 in a direction perpendicular to OA (i.e. in a tangential
direction). Therefore
1() ( ) s i n
(. . . .) s i nyb r r
rrr r=ω + δ ω + δ δ θ ↓
= ω + ωδ + δω + δωδ δθ
A drill press has a pointed tool

which is used for boring holes inhard materials usually by rotatingabrasion or repeated bolows.
information and is not a direct example
of the current chapter.

210  /G6C   Theory of Machines
Since δθ is very small, therefore substituting sin δθ = δθ in the above expression, we have
 1.. . . .. . .y b rrr r= ω δθ + ωδ δθ + δω δθ + δωδ δθ
.. ,r=ω δ θ↓ acting radially inwards ...(Neglecting all other quantities)
and  by=  oy – ob = (ω + δω) (r + δr) cos δθ – ω.r
                       = ( ω.r + ω.δr + δω.r  + δω.δr) cos δθ – ω.r
Since δθ is small, therefore substituting cos δθ = 1, we have

by = ω.r + ω.δr + δω.r + δω.δr – ω.r= ω.δr + r.δω ...(Neglecting δω.δr)
...(Perpendicular to OA and towards left)
Therefore, total component of change of velocity along radial direction
1(. . ) bx yb v r=− = δ − ω δ θ ↑ ...(Acting radially outwards from O to A)
∴ Radial component of the acceleration of the slider B with respect to O on the link OA,
acting radially outwards from O to A,
2
BO..Lt . .r vr d v dd var rtd t d t d tδ− ωδ θ θ= = −ω × = −ω ↑δ

Also, the total component of change of velocity along tangential direction,
1 .( .. ) xb by v r r←←
=+ = δ θ + ω δ + δ ω
∴ Tangential component of acceleration of the slider B with respect to O on the link OA,
acting perpendicular to OA and towards left,
BO.( . . )Lttvr rd d r dav r
td t d t d tδθ + ωδ + δω θ ω== + ω +
δ
.. . ( 2 .. )vv rv r←
=ω + ω+α = ω +α ...(ii)
...( / , and / )dr dt v d dt=ω = α /G51

Now radial component of acceleration of the coincident point C with respect to O, acting in
a direction from C to O,
               2
CO.rar=ω ↑ ...(iii)
and tangential component of acceleraiton of the coincident point C with respect to O, acting in a
direction perpendicular to CO and towards left,
CO.tar←
=α ↑ ...(iv)
Radial component of the slider B with respect to the coincident point C on the link OA,
acting radially outwards,
              ()2 2

              ()2 2
BC BO CO . .rrr dv dvaaa r rdt dt=−= − = ↑ −ω −ω 
and tangential component of the slider B with respect to the coincident point C on the link OA acting
in a direction perpendicular to OA and towards left,
() BC BO CO .2 . 2. .tttaaa r v vr←
=−= − α = ω ω+ α...( / )dd tθ= ω/G51...(i)

Chapter 8 :  Acceleration in Mechanisms    /G6C  211
This tangential component of acceleration of the slider B with respect to the coincident
point C on the link is known as coriolis component of acceleration  and is always perpendicualr to
the link.
∴ Coriolis component of the acceleration of B with respect of C,
BC BC 2.ctaa v== ω
where ω= Angular velocity of the link OA, and
v= Velocity of slider B with respect to coincident point C.

In the above discussion, the anticlockwise direction for  ω and the radially outward direction
for v are taken as  positive. It may be noted that the direction of coriolis component of acceleration
changes sign, if either ω or v is reversed in direction. But the direction of coriolis component of
acceleration will not be changed in sign if both ω and v are reversed in direction. It is concluded that

the direction of coriolis component of acceleration is obtained by rotating v, at 90°, about its origin
in the same direction as that of ω.
Fig. 8.27.  Direction of coriolis component of acceleration.
The direction of coriolis component of acceleration (2 ω.v) for all four possible cases, is
shown in Fig. 8.27. The directions of ω and v are given.
Example 8.13. A mechanism of a crank and slotted lever quick

return motion is shown in Fig. 8.28. If the crank rotates counter clockwiseat 120 r.p.m., determine for the configuration shown, the velocity and
acceleration of the ram D. Also determine the angular acceleration ofthe slotted lever.
Crank, AB = 150 mm ; Slotted arm, OC = 700 mm and link
CD = 200 mm.
Solution.  Given : N
BA = 120 r.p.m or ωBA = 2 π × 120/60
= 12.57 rad/s ; AB = 150 mm = 0.15 m; OC = 700 mm = 0.7 m;
CD = 200 mm = 0.2  m
We know that velocity of B with respect to A,
BA BA

BA BA
12.57 0.15 1.9 m/svA B=ω ×
=× =
Velocity of the ram D
First of all draw the space diagram, to some suitable scale, as
shown in Fig. 8.29 ( a). Now the velocity diagram, as shown in Fig. 8.29Fig. 8.28

212  /G6C   Theory of Machines
(b), is drawn as discussed below:
1. Since O and A are fixed points, therefore these points are marked as one point in velocity
diagram. Now draw vector ab in a direction perpendicular to AB , to some suitable scale, to represent
the velocity of slider B with respect to A i.e.vBA, such that
                    vector ab = vBA = 1.9 m/s
(c) Direction of coriolis component. ( d) Acceleration diagram.
Fig. 8.29

Fig. 8.29
2. From point o, draw vector ob' perpendicular to OB' to represent the velocity of coincident
point B' (on the link OC) with respect to O i.e. vB′O and from point b draw vector bb' parallel to the
path of motion of B' (which is along the link OC) to represent the velocity of coincident point B' with
respect to the slider  B i.e. vB'B. The vectors ob' and bb' intersect at b'.

Note:  Since we have to find the coriolis component of acceleration of the slider B with respect to the coincident
point B', therefore we require the velocity of  B with respect to  B'  i.e. vBB'. The vector  b'b will represent  vBB'
as shown in Fig. 8.29  (b).
3. Since the point C lies on OB' produced, therefore, divide vector ob' at c in the same ratio
as C divides OB' in the space diagram. In other words,
//ob oc OB OC′′=
The vector oc represents the velocity of C with respect to O i.e. vCO.

Chapter 8 :  Acceleration in Mechanisms    /G6C  213
respect to C i.e. vDC ,and from point o draw vector od parallel to the path of motion of D (which is
along the horizontal) to represent the velocity of D i.e. vD.The vectors cd and od intersect at d.
By measurement, we find that velocity of the ram D,
vD = vector od = 2.15 m/s  Ans.

From velocity diagram, we also find thatVelocity of B with respect to B',
v
BB' = vector b'b = 1.05 m/s
vDC = vector cd = 0.45 m/s
Velocity of  B' with respect to O
vB′O = vector ob' = 1.55 m/s
Velocity of C with respect to O,
vCO = vector oc = 2.15 m/s
∴ Angular velocity of the link OC or OB',
CO
CO BO2.153.07 rad/s (Anticlockwise)
0.7v
OC′ ω= ω= = =
Acceleration of the ram D

0.7v
OC′ ω= ω= = =
We know that radial component of the acceleration of B with respect to A,
222
BA BA (12.57) 0.15 23.7 m/sraA B=ω × = × =
Coriolis component of the acceleration of slider B with respect to the coincident point B',
2
BB CO BB2 . 2 . 2 3.07 1.05 6.45 m/scavv ′ ′= ω = ω = × × =
CO BB ...( and ) vv ω=ω = ′/G51
2 2
2 DC
DC(0.45)1.01m/s0.2rvaCD== =

2 2
2 DC
Radial component of the acceleration of the coincident point B' with respect to O,
2 2
2 BO
BO(1.55)4.62 m/s0.52rvaBO′
′== =′...(By measurement B'O = 0.52 m)
Now the acceleration diagram, as shown in Fig. 8.29 ( d), is drawn as discussed below:
1. Since O and A are fixed points, therefore these points are marked as one point in the

component of the acceleration of B with respect to A i.e. BAra or  aB, such that
2
BA B vector 23.7 m/srab a a′′== =
2. The acceleration of the slider B with respect to the coincident point B' has the following
two components :
(i)Coriolis component of the acceleration of B with respect to B' i.e. BB,ca′and
(ii)Radial component of the acceleration of B with respect to B' i.e. BB.ra′
These two components are mutually perpendicular. Therefore from point b' draw vector b'x

perpendicular to B'O i.e. in a direction as shown in Fig. 8.29 ( c) to represent BBca′= 6.45 m/s2. The

214  /G6C   Theory of Machines
direction of BBca′is obtained by rotating vBB′ (represented by vector b'b in velocity diagram) through
90° in the same sense as that of link OC which rotates in the counter clockwise direction. Now from
point x, draw vector xb'' perpendicular to vector b'x (or parallel to B'O) to represent BBra′whose
magnitude is yet unknown.
3. The acceleration of the coincident point B' with respect to O has also the following two

(i)Radial component of the acceleration of coincident point B' with respect to O i.e.
BO’ra′and
(ii)Tangential component of the acceleration of coincident point B' with respect to O,
i.e.BO.ta′
These two components are mutually perpendicular. Therefore from point o', draw vector o'y
parallel to B'O to represent 2
BO4.62 m/sra′= and from point y draw vector yb'' perpendicular to
vector o'y to represent BO.ta′The vectors xb'' and yb'' intersect at b''. Join o'b''. The vector o'b''

represents the acceleration of B' with respect to O, i.e. aB′O.
4. Since the point C lies on OB' produced, therefore divide vector o'b'' at c' in the same ratio
o'b''/o'c' = OB'/OC
5. The acceleration of the ram D with respect to C has also the following two components:
(i)Radial component of the acceleration of D with respect to C i.e. DC,raand
(ii)Tangential component of the acceleration of D with respect to C, i.e. DC.ta

The two components are mutually perpendicular. Therefore draw vector c'z parallel to CD
to represent 2
DC1.01m/sra= and from z draw zd' perpendicular to vector zc' to represent DC,tawhose
6. From point o', draw vector o'd' in the direction of motion of the ram D which is along the
horizontal. The vectors zd' and o'd' intersect at d'. The vector o'd' represents the acceleration of ram
D i.e. aD.
By measurement, we find that acceleration of the ram D,

aD = vector o'd' = 8.4 m/s2 Ans.
Angular acceleration of the slotted lever
By measurement from acceleration diagram, we find that tangential component of the
coincident point B' with respect to O,
2
BOvector 6.4 m/sta yb ′ ′′ ==
We know that angular acceleration of the slotted lever,
2 BO6.412.3 rad/s (Anticlockwise)0.52ta
OB′===′Ans.
Example 8.14. The driving crank AB  of the quick-return mechanism, as shown in Fig. 8.30,

revolves at a uniform speed of 200 r.p.m. Find the velocity and acceleration of the tool-box R, in theposition shown, when the crank makes an angle of 60° with the vertical line of centres PA . What is
the acceleration of sliding of the block at B along the slotted lever PQ ?

Chapter 8 :  Acceleration in Mechanisms    /G6C  215
Solution. Given : NBA = 200 r.p.m. or ωBA = 2 π × 200/60 = 20.95 rad/s; AB = 75 mm = 0.075 m
vBA = ωBA × AB = 20.95 × 0.075 = 1.57 m/s ...(Perpendicular to AB)
Fig. 8.30
Velocity of the tool-box R
First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.31 ( a). Now
the velocity diagram, as shown in Fig. 8.31 ( b), is drawn as discussed below:

1. Since A and P are fixed points, therefore these points are marked as one point in the
velocity diagram. Now draw vector ab in a direction perpendicular to AB, to some suitable scale, to
represent the velocity of B with respect to A or simply velocity of B (i.e. vBA or vB), such that
vector ab = vBA = vB = 1.57 m/s
2. From point p, draw vector pb' perpendicular to PB' to represent the velocity of coincident

point B' with respect to P (i.e. vB'P or vB') and from point b, draw vector bb' parallel to the path of
motion of B' (which is along PQ) to represent the velocity of coincident point B' with respect to the
slider B i.e. vB'B. The vectors  pb' and bb' intersect at b'.
Note. The vector b'b will represent the velocity of the slider B with respect to the coincident point  B' i.e.vBB'.
3. Since the point Q lies on PB' produced, therefore divide vector pb' at q in the same ratio

as Q divides PB'. In other words,
pb'/pq  = PB'/PQ
The vector pq represents the velocity of Q with respect to P i.e. vQP.
4. Now from point q, draw vector qr perpendicular to QR to represent the velocity of R with
respect to Q i.e. vRQ, and from point a draw vector  ar parallel to the path of motion of the tool-box
R (which is along the horizontal), to represent the velocity of R i.e. vR.The vectors qr and ar intersect
at r.
By measurement, we find that velocity of the tool-box R,

vR = vector ar = 1.6 m/s Ans.
We also find that velocity of B' with respect to B,
vB'B = vector bb' = 1.06 m/s
Velocity of B' with respect to P,
 vB'P = vector pb' = 1.13 m/s

216  /G6C   Theory of Machines
Velocity of R with respect to Q,
vRQ = vector qr = 0.4 m/s
Velocity of Q with respect to P,
vQP = vector pq = 1.7 m/s
∴ Angular velocity of the link PQ,
QP
PQ1.74.53 rad/s
0.375v
QPω= = = ...( 0.375 m)PQ=/G51
Fig. 8.31
Acceleration of the tool box R
We know that the radial component of the acceleration of B with respect to A,
22 2

22 2
BA BA (20.95) 0.075 32.9 m/sraA B=ω × = × =
Coriolis component of the acceleration of the slider B with respect to coincident point B′.
          2
BB QP BB2 . 2 2 4.53 1.06 9.6 m/scav v′′=ω=ω × =× × =
() QP BB ... , and vv ω=ω = ′/G51
Radial component of the acceleration of R with respect to Q ,
2 2RQ 2
RQ(0.4)0.32 m/s0.5rvaQR== =

Chapter 8 :  Acceleration in Mechanisms    /G6C  217
Radial component of the acceleration of B' with respect to P,
2 2
2 BP
BP(1.13)5.15 m/s0.248rvaPB′
′== =′
...(By measurement, PB' = 248 mm = 0.248 m)
Now the acceleration diagram, as shown in Fig. 8.31 ( d), is drawn as discussed below:

component of the acceleration of B with respect to A i.e. BA,ra or aB such that
2
BA B vector 32.9 m/srab a a′′== =
two components:
These two components are mutually perpendicular. Therefore from point b', draw vector b'x

perpendicular to BP [i.e. in a direction as shown in Fig. 8.31 ( c)] to represent 2
BB9.6 m/sca′= . The
direction of BBca′is obtained by rotating vBB′ (represented by vector b'b in the velocity diagram)
through 90° in the same sense as that of link PQ which rotates in the clockwise direction. Now from
point x, draw vector xb'' perpendicular to vector b'x (or parallel to B'P) to represent BBra′whose

3. The acceleration of the coincident point B' with respect to P has also the following two
(i)Radial component of the acceleration of B' with respect to P i.e.BP,ra′and
(ii)Tangential component of the acceleration of B' with respect to P i.e. BP.ta′
These two components are mutually perpendicular. Therefore from point p' draw vector p'y
parallel to B'P to represent BPra′= 5.15 m/s2, and from point y draw vector yb'' perpendicular to

vector p'y to represent BP.ta′The vectors xb'' and yb'' intersect at b'', join p'b''. The vector p'b''
represents the acceleration of B' with respect to P i.e. aB'P and the vector b''b' represents the acceleration
of B with respect to B' i.e. aBB' .
4. Since the point Q lies on PB' produced, therefore divide vector p'b'' at q' in the same ratio
as Q divides PB in the space diagram. In other words,
p'b''/p'q' = PB'/PQ

5. The acceleration of the tool-box R with respect to Q has the following two components:
(i)Radial component of the acceleration of R with respect to Q i.e. RQ,raand
(ii)Tangential component of the acceleration of R with respect to Q i.e.RQ.ta
These two components are mutually perpendicular. Therefore from point q', draw vector a'z
parallel to  QR to represent 2
RQ0.32 m/s .ra= Since the magnitude of this component is very small,

therefore the points q' and z coincide as shown in Fig. 8.31 ( d). Now from point z (same as q'), draw
vector zr' perpendicular to vector q'z (or QR) to represent RQtawhose magnitude is yet unknown.
6. From point a' draw vector a'r' parallel to the path of motion of the tool-box R (i.e. along
the horizontal) which  intersects the vector zr' at r'. The vector a'r' represents the acceleration of the
tool-box R i.e. aR.

218  /G6C   Theory of Machines
aR = vector a'r' = 22 m/s2 Ans.
Acceleration of sliding of the block B along the slotted lever PQ
By measurement, we find that the acceleration of sliding of the block B along the slotted
lever PQ
= aBB' = vector b''x = 18 m/s2 Ans.
Example 8.15.  In a Whitworth quick return motion, as shown in Fig. 8.32. OA is a crank

rotating at 30 r.p.m. in a clockwise direction. The dimensions of various links are : OA = 150 mm;OC = 100 mm; CD = 125 mm; and DR = 500 mm.
Determine the acceleraion of the sliding block R and the angular acceleration of the slotted
lever CA.
Fig. 8.32
Solution.  Given : NAO = 30 r.p.m. or ωAO = 2π × 30/60 = 3.142 rad/s ; OA = 150 mm
= 0.15 m; OC = 100 mm = 0.1 m ; CD = 125 mm = 0.125 m; DR = 500 mm = 0.5 m

vAO = vA = ωAO × OA = 3.142 × 0.15 = 0.47 m/s
First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.33 ( a). Now
the velocity diagram, as shown in Fig. 8.33 ( b), is drawn as discussed below:
1. Since O and C are fixed points, therefore these are marked at the same place in velocity
diagram. Now draw vector ca perpendicular to OA, to some suitable scale, to represent the velocity

of A with respect to O  or simply velocity of A i.e. vAO or vA, such that
vector oa = vAO = vA = 0.47 m/s
2. From point c, draw vector cb perpendicular to BC to represent the velocity of the coincident
point B with respect to C i.e. vBC or vB and from point a draw vector ab parallel to the path of motion
of B (which is along BC) to represent the velocity of coincident point B with respect to A i.e.vBA. The
vectors cb and ab intersect at b.

Note: Since we have to find the coriolis component of acceleration of slider A with respect to coincident point
B, therefore we require the velocity of A with respect to B i.e. vAB.The vector ba will represent vAB as shown
in Fig. 8.33 ( b).

Chapter 8 :  Acceleration in Mechanisms    /G6C  219
3. Since D lies on BC produced, therefore divide vector bc at d in the same ratio as D divides
BC in the space diagram. In other words,
bd/bc = BD/BC
Fig. 8.33

respect to D i.e. vRD, and from point c draw vector cr parallel to the path of motion of R (which is
horizontal) to represent the velocity of R i.e.vR.
By measurement, we find that velocity of  B with respect to C,
vBC = vector cb = 0.46 m/s
Velocity of A with respect to B,
vAB = vector ba = 0.15 m/s
and velocity of R with respect to D,
vRD = vector dr = 0.12 m/s
We know that angular velocity of the link BC,
BC
BC0.461.92 rad/s (Clockwise)0.24v
CBω= = =
...(By measurement, CB = 0.24 m)

220  /G6C   Theory of Machines
Acceleration of the sliding block R
We know that the radial component of the acceleration of A with respect to O,
2 2
2 AO
AO(0.47)1.47 m/s0.15rvaOA== =
Coriolis component of the acceleration of slider A with respect to coincident point B,
2
AB BC AB2 2 1.92 0.15 0.576 m/scav=ω × =× × =
2 2
2 BC
BC(0.46)0.88 m/s0.24rvaCB== =
Radial component of the acceleration of R with respect to D,
2 2
2 RD

2 2
2 RD
RD(0.12)0.029 m/s0.5rvaDR== =
Now the acceleration diagram, as shown in Fig. 8.33 ( d), is drawn as discussed below:
1. Since O and C are fixed points, therefore these are marked at the same place in the
component of the acceleration of A with respect to O i.e. AOra,  or aA such that
2
AO A vector 1.47 m/sroa a a′′== =

2
2. The acceleration of the slider A with respect to coincident point B has the following two
(i)Coriolis component of the acceleration of A with respect to B i.e. AB,caand
(ii)Radial component of the acceleration of A with respect to B i.e. AB.ra
These two components are mutually perpendicular. Therefore from point  a' draw vector  a′x
perpendicular to BC to represent ABca= 0.576 m/s2 in a direction as shown in Fig. 8.33 ( c), and draw

vector xb' perpendicular to vector a'x (or parallel to BC) to represent ABrawhose magnitude is yet
unknown.
Note: The direction of ABcais obtained by rotating vAB (represented by vector ba in velocity diagram) through
90° in the same sense as that of ωBC which rotates in clockwise direction.
3. The acceleration of B with respect to C has the following two components:
(i)Radial component of B with respect to C i.e. BC,raand
(ii)Tangential component of B with respect to C i.e. BC.ta

These two components are mutually perpendicular. Therefore, draw vector c'y parallel to
BC to represent 2
BC0.88 m/sra= and from point y draw vector yb' perpendicular to c'y to represent
BC.ta The vectors xb' and yb' intersect at b'. Join b'c'.
4. Since the point D lies on BC produced, therefore divide vector  b'c' at d' in the same ratio
as D divides BC in the space diagram. In other words,
b'd'/b'c' = BD/BC .

5. The acceleration of the sliding block R with respect to D has also the following two
(i)Radial component of R with respect to D i.e. RD,raand
(ii)Tangential component of R with respect to D i.e. RD.ta

Chapter 8 :  Acceleration in Mechanisms    /G6C  221
These two components are mutually perpendicular. Therefore from point d', draw vector
d ′z parallel to DR to represent 2
RD0.029 m/sra= and from z draw zr' perpendicular to d'z to represent
RDtawhose magnitude is yet unknown.
6. From point c', draw vector c'r' parallel to the path of motion of R (which is horizontal).
The vector c'r' intersects the vector zr' at r'. The vector c'r' represents the acceleration of the sliding
block R.

block R.
By measurement, we find that acceleration of the sliding block R,
aR = vector c'r' = 0.18 m/s2 Ans.
Angular acceleration of the slotted lever CA
By measurement from acceleration diagram, we find that tangential component of B with
2
BCvector 0.14 m/sta yb′ ==
We know that angular acceleration of the slotted lever CA,
                            2 CB
CA BC0.140.583 rad/s (Anticlockwise)0.24ta
BCα= α= = = Ans.
Example 8.16. The kinematic diagram of one of the cylinders of a

rotary engine is shown in Fig. 8.34. The crank OA which is vertical andfixed , is 50 mm long. The length of the connecting  rod AB is 125 mm. Theline of the stroke OB is inclined at 50° to the vertical .
The cylinders are rotating at a uniform speed of 300 r.p.m., in a
clockwise direction, about the fixed centre O . Determine: 1. acceleration
of the piston inside the cylinder, and 2. angular acceleration of the
connecting rod.
Solution.  Given: AB = 125 mm = 0.125 m; N
CO = 300 r.p.m.

or ωCO = 2π × 300/60 = 31.4 rad/s
First of all draw the space diagram, as shown in Fig. 8.35 ( a), to some suitable scale. By
measurement from the space diagram, we find that
OC = 85 mm = 0.085 m
∴ Velocity of C with respect to O,
vCO = ωCO × OC = 31.4 × 0.85 = 2.7 m/s
...(Perpendicular to CO)
Now the velocity diagram, as shown in Fig. 8.35 ( b), is drawn as discussed below:
1. Since O and A are fixed points, therefore these are marked at the same place in the

velocity diagram. Draw vector oc perpendicular to OC to represent the velocity of C with respect to
O i.e. vCO, such that
                    vector oc = vCO = vC = 2.7 m/s.
2. From point c, draw vector cb parallel to the path of motion of the piston B (which is along
CO) to represent the velocity of B with respect to C i.e. vBC , and from point  a draw vector ab
perpendicular to AB to represent the velocity of B with respect to A i.e. vBA or vB.

By measurement, we find that velocity of piston B with respect to coincident point C,
                              vBC = vector cb = 0.85 m/sFig. 8.34

222  /G6C   Theory of Machines
and velocity of piston B with respect to A,
                             BA B vector = 2.85 m/s vv a b==
Fig. 8.35
1. Acceleration of the piston inside the cylinder
We know that the radial component of the acceleration of the coincident point C with respect
to O,
                            2 2
2 CO
CO(2.7)85.76 m/s0.085rvaOC== =

2 CO
Coriolis component of acceleration of the piston B with respect to the cylinder or coincident
point C,
2
BC CO BC2 2 31.4 0.85 53.4 m/scav=ω × =× × =
Radial component of acceleration of B with respect to A,
2 2
2 BA
BA(2.85)65 m/s0.125rvaAB== =
The acceleration diagram, as shown in Fig. 8.35 ( d), is drawn as discussed below:

Chapter 8 :  Acceleration in Mechanisms    /G6C  223
1. Since O and A are fixed points, therefore these are marked as one point in the acceleration
diagram. Draw vector o'c' parallel to OC, to some suitable scale, to represent the radial component
of the acceleration of C with respect to O i.e., CO,rasuch that
2
CO vector 85.76 m/sroc a′′==
2. The acceleration of piston B with respect to coincident point C has the following two

(i)Coriolis component of the acceleration of B with respect to C i.e. BCca, and
(ii)Radial component of the acceleration of B with respect to C i.e. BC.ra
These two components are mutually perpendicular. Therefore from point c', draw vector c'x
perpendicular to CO to represent 2
BC53.4 m/sca= in a direction as shown in Fig. 8.35 ( c). The
direction of BCca is obtained by rotating vBC (represented by vector cb in velocity diagram) through

90° in the same sense as that of ωCO which rotates in the clockwise direction. Now from point x,
draw vector xb' perpendicular  to vector c'x (or parallel to OC) to represent BCrawhose magnitude is
yet unknown.
3. The acceleration of B with respect to A has also the following two components:
(i)Radial component of the acceleration of B with respect to  A i.e. BA,raand
(ii)Tangential component of the acceleration of B with respect to A i.e. BA.ta

These two components are mutually perpendicular. Therefore from point a', draw vector a'y
parallel to AB to represent 2
BA65 m/s ,ra= and from point y draw vector yb' perpendicular to vector
a'y to represent BA.taThe vectors xb' and yb' intersect at b'.
4. Join c'b' and a'b'. The vector c'b' represents the acceleration of B with respect to C (i.e.
acceleration of the piston inside the cylinder).
By measurement, we find that acceleration of the piston inside the cylinder,

aBC = vector c'b' = 73.2 m/s2 Ans.
2.  Angular acceleration of the connecting rod
By measurement from acceleration diagram, we find that the tangential component of the
acceleration of B with respect to A,
2
BAvector 37.6 m/sta yb′ ==
∴ Angular acceleration of the connecting rod AB,
              2 BA
AB37.6301rad/s (Clockwise)0.125ta
Example 8.17. In a swivelling joint mechanism, as shown in Fig. 8.36, the driving crank

OA is rotating clockwise at 100  r.p.m. The lengths of various links are : OA = 50 mm ; AB = 350mm; AD = DB ; DE = EF = 250 mm and CB = 125 mm. The horizontal distance between the fixedpoints O and C is 300 mm and the vertical distance between F and C is 250 mm.
For the given configuration, determine: 1. Velocity of the slider block F, 2. Angular velocity
of the link DE, 3. Velocity of sliding of the link DE in the swivel block,and 4. Acceleration of sliding
of the link DE in the trunnion.

224  /G6C   Theory of Machines
Fig. 8.36
Solution. Given: NAO = 100 r.p.m. or ωAO = 2π × 100/60 = 10.47 rad/s ; OA = 50 mm =
0.05 m; AB = 350 mm = 0.35 m ; CB = 125 mm = 0.125 m ; DE = EF = 250 mm = 0.25 m
vAO = vA = ωAO × OA = 10.47 × 0.05 = 0.523 m/s
1. Velocity of slider block F
First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.37 ( a). Now

the velocity diagram, as shown in Fig. 8.37 ( b), is drawn as discussed below:
1. Since O, C and Q are fixed points, therefore these points are marked at one place in the
velocity of A with respect to O or simply velocity of A, i.e. vAO or vA, such that
vector oa = vAO = vA = 0.523 m/sThis machine uses swivelling joint.

Chapter 8 :  Acceleration in Mechanisms    /G6C  225
respect to A i.e. vBA,and from point c draw vector cb perpendicular to CB to represent the velocity of
B with respect to C or simply velocity of B i.e. vBC or vB. The vectors ab and cb intersect at b.
Fig. 8.37

Fig. 8.37
3. Since point D lies on AB, therefore divide vector ab at d in the same ratio as D divides AB
ad/ab = AD/AB
Note:  Since point D is mid-point of AB, therefore d is also mid-point of ab.
4. Now from point d, draw vector ds perpendicular to DS to represent the velocity of S with
respect to D i.e. vSD, and from point q draw vector qs parallel to the path of motion of swivel block

Q (which is along DE) to represent the velocity of S with respect to Q i.e. vSQ. The vectors ds and qs
intersect at s.
Note:  The vector sq will represent the velocity of swivel block Q with respect to S i.e. vQS.
5. Since point E lies on DS produced, therefore divide vector ds at e in the same ratio as E
divides DS in the space diagram. In other words,
   de/ds = DE/DS
6. From point e, draw vector ef perpendicular to EF to represent the velocity of F with

respect to E i.e. vFE , and from point o draw vector of parallel to the path of motion of F (which is
along the horizontal direction) to represent the velocity of F i.e. vF..The vectors ef and of intersect
at f.
vBA = vector ab = 0.4 m/s

226  /G6C   Theory of Machines
vBC = vB = vector cb = 0.485 m/s
Velocity of S with respect to D,
 vSD = vector ds = 0.265 m/s
Velocity of Q with respect to S,
 vQS = vector sq = 0.4 m/s
 vED = vector de = 0.73 m/s
Velocity of F with respect to E,
vFE = vector ef = 0.6 m/s
and velocity of the slider block F, vF = vector of = 0.27 m/s Ans.
2.  Angular velocity of the link DE
We know that angular velocity of the link DE,
ED

ED
DE0.732.92 rad/s (Anticlockwise)0.25v
DEω= = = Ans.
3.  Velocity of sliding of the link DE in the swivel block
The velocity of sliding of the link DE in the swivel block Q will be same as that of velocity
of S i.e. vS.
∴ Velocity of sliding of the link DE in the swivel block,
vS = vSQ = 0.4 m/s Ans.
4.  Acceleration of sliding of the link DE in the trunnion

2 2
2 AO
AO A(0.523)5.47 m/s0.05r vaaOA== = =
2 2
2 BA
BA(0.4)0.457 m/s0.35rvaAB== =
2 2
2 BC
BC(0.485)1.88 m/s0.125rvaCB== =
Radial component of the acceleration of S with respect to D,
2 2
2 SD
SD(0.265)0.826 m/s0.085rvaDS== =
...(By measurement DS = 85 mm = 0.085 m)
Coriolis component of the acceleration of Q with respect to S,
2

Chapter 8 :  Acceleration in Mechanisms    /G6C  227
and radial component of the acceleration of F with respect to E,
2 2
2 FE
FE(0.6)1.44 m/s0.25rvaEF== =
Now the acceleration diagram, as shown in Fig. 8.37 ( d), is drawn as discussed below:
acceleration diagram. Now draw vector o'a' parallel to OA, to some suitable scale, to represent AO,ra
or aA such that
2
AO A vector 5.47 m/sroa a a== =′′

2
Note : Since OA rotates with uniform speed, therefore there will be no tangential component of the
acceleration.
2. The acceleration of B with respect to A has the following two components:
(i)Radial component of the acceleration of B with respect to A i.e. BA,raand
These two components are mutually perpendicular. Therefore from point a', draw vector a'x

BA0.457 m/s ,ra= and from point x draw vector  xb' perpendicular to
vector a'x to represent BAtawhose magnitude is yet unknown.
(i)Radial component of the acceleration of B with respect to C i.e.BC,raand
(ii)Tangential component of the acceleration of B with respect to C i.e.BC.ta
These two components are mutually perpendicular. Therefore from point c', draw vector c'y

parallel to CB to represent BCra = 1.88 m/s2 and from point y draw vector  yb' perpendicular to vector
c'y to represent BCta. The vectors xb' and yb' intersect at b'.
4. Join a'b' and c'b'. The vector a'b' represents the acceleration of B with respect to A i.e.
BAaand the vector c'b' represents the acceleration of B with respect to C or simply the acceleration
of B i.e. aBC or aB, because C is a fixed point.

5. Since the point D lies on AB, therefore divide vector a'b' at d' in the same ratio as D
divides AB in the space diagram. In other words,
a'd'/a'b' = AD/AB
Note:  Since D is the mid-point of AB, therefore d' is also mid-point of vector a'd'.
6. The acceleration of S with respect to D has the following two components:
(i)Radial component of the acceleration of S with respect to D i.e. SDra, and

(ii)Tangential component of the acceleration of S with respect to D i.e.SD.ta
These two components are mutually perpendicular. Therefore from point d′ , draw vector
d′z parallel to DS to represent ar
SD = 0.826 m/s2, and from point z draw vector zs′ perpendicular to
vector d′z to represent at
SD whose magnitude is yet unknown.
7. The acceleration of Q (swivel block) with respect to S (point on link DE i.e. coincident
point) has the following two components:

228  /G6C   Theory of Machines
(i)Coriolis component of acceleration of Q with respect to S i.e. QS,caand
(ii)Radial component of acceleration of Q with respect to S, i.e. QSra.
These two components are mutually perpendicular. Therefore from point q', draw vector
q'z1, perpendicular to DS to represent 2
QS2.336 m/sca= in a direction as shown in Fig. 8.37 ( c). The
direction of QScais obtained by rotating vQS (represented by vector sq in velocity diagram) through

90° in the same sense as that of ωDE which rotates in the anticlockwise direction. Now from z1, draw
vector z1s' perpendicular to vector q'z1 (or parallel to DS) to represent QS.raThe vectors zs' and z1s'
intersect at s'.
8. Join s'q' and d's'. The vector s'q' represents the acceleration of Q with respect to S i.e. aQS
and vector d's' represents the acceleration of S with respect to D i.e. aSD.
By measurement, we find that the acceleration of sliding the link DE in the trunnion,
2

2
QS 1vector 1.55m/sraz s ′ == = Ans.
EXERCISES
1.The engine mechanism shown in Fig. 8.38 has crank OB = 50 mm and length of connecting rod AB
= 225 mm. The centre of gravity of the rod is at G which is 75 mm from B. The engine speed is 200
r.p.m.
For the position shown, in which OB is turned 45° from OA, Find 1. the velocity of G and the
angular velocity of AB, and 2. the acceleration of G and angular acceleration of AB.
[Ans. 6.3 m/s ; 22.6 rad/s ; 750 m/s2 ; 6.5 rad/s2]

2.In a pin jointed four bar mechanism ABCD, the lengths of various links are as follows:
AB = 25 mm ; BC = 87.5 mm ; CD = 50 mm and AD = 80 mm.
The link AD is fixed and the angle BAD = 135°. If the velocity of B is 1.8 m/s in the clockwise
direction, find 1. velocity and acceleration of the mid point of BC, and 2. angular velocity and
angular acceleration of link CB and CD.
[Ans. 1.67 m/s, 110 m/s2 ; 8.9 rad/s, 870 rad/s2 ; 32.4 rad/s, 1040 rad/s2]

3.In a four bar chain ABCD , link AD is fixed and the crank AB rotates at 10 radians per second
clockwise. Lengths of the links are AB = 60 mm ; BC = CD = 70 mm ; DA = 120 mm. When angle
DAB = 60° and both B and C lie on the same side of AD, find 1. angular velocities (magnitude and
direction) of BC and CD ; and 2. angular acceleration of BC and CD.
[Ans.  6.43 rad/s (anticlockwise), 6.43 rad/s (clockwise) ; 10 rad/s2  105 rad/s2]

4.In a mechanism as shown in Fig. 8.39, the link AB rotates with a uniform angular velocity of
30 rad/s. The lengths of various links are :
AB = 100 mm ; BC = 300 mm ; BD = 150 mm ; DE = 250 mm ; EF = 200 mm ; DG = 165 mm.
Determine the velocity and acceleration of G for the given configuration.
[Ans. 0.6 m/s ; 66 m/s2]Fig. 8.38

Chapter 8 :  Acceleration in Mechanisms    /G6C  229
Fig. 8.39 Fig. 8.40
5.In a mechanism as shown in Fig. 8.40, the crank
OA is 100 mm long and rotates in a clockwise
direction at a speed of 100 r.p.m. The straightrod BCD rocks on a fixed point at C. The links
BC and CD are each 200 mm long and the link
AB is 300 mm long. The slider E, which is
driven by the rod DE is 250 mm long. Find the
velocity and acceleration of E.
                            [ Ans. 1.26 m/s; 10.5 m/s
2]

2]
mechanism, as shown in Fig. 8.41, are asfollows:
OA = 80 mm ; AC = CB = CD = 120 mm
If the crank OA rotates at 150 r.p.m. in the anti-
clockwise direction, find, for the given configuration: 1. velocity and acceleration of B and D ; 2.
rubbing velocity on the pin at C, if its diameter is 20 mm ; and 3. angular acceleration of the links AB
and CD.
[Ans.  1.1 m/s ; 0.37 m/s ; 20.2 m/s
2, 16.3 m/s2 ; 0.15 m/s ; 34.6 rad/s2; 172.5 rad/s2]

7.In the toggle mechanism, as shown in Fig. 8.42, D is constrained to move on a horizontal path. The
dimensions of various links are : AB = 200 mm; BC = 300 mm ; OC = 150 mm; and BD = 450 mm.
Fig. 8.42 Fig. 8.43
The crank OC is rotating in a counter clockwise direction at a speed of 180 r.p.m., increasing at the
rate of 50 rad/s2. Find, for the given configuration 1. velocity and acceleration of D, and 2. angular
velocity and angular acceleration of BD.Fig. 8.41

230  /G6C   Theory of Machines
8.In a quick return mechanism, as shown in Fig. 8.43, the driving crank OA is 60 mm long and rotates
at a uniform speed of 200 r.p.m. in a clockwise direction. For the position shown, find 1. velocity ofthe ram R ; 2. acceleration of the ram R, and 3. acceleration of the sliding block A along the slotted
bar CD. [Ans.  1.3 m/s ; 9 m/s
2 ; 15 m/s2]
9.Fig. 8.44 shows a quick return motion mechanism in which the driving crank OA rotates at 120

r.p.m. in a clockwise direction. For the position shown, determine the magnitude and direction of1, the acceleration of the block D ; and 2. the angular acceleration of the slotted bar QB.
[Ans.  7.7 m/s
2 ; 17 rad/s2]
Fig. 8.44
10.In the oscillating cylinder mechanism as shown in Fig. 8.45, the crank OA is 50 mm long while the
piston rod AB is 150 mm long. The crank OA rotates uniformly about O at 300 r.p.m.
Fig. 8.45

Fig. 8.45
Determine, for the position shown : 1. velocity of the piston B relative to the cylinder walls, 2.
angular velocity of the piston rod AB, 3. sliding acceleration of the piston B relative to the cylinder
walls, and 4. angular acceleration of the piston rod AB.
[Ans. 1.5 m/s ; 2.2 rad/s (anticlockwise) ; 16.75 m/s2 ; 234 rad/s2]
11.The mechanism as shown in Fig 8.46 is a marine steering gear, called Rapson’s slide. O2B is the

tiller and AC is the actuating rod. If the velocity of AC is 25 mm/min to the left, find the angular
velocity and angular acceleration of the tiller. Either graphical or analytical technique may be used.
[Ans. 0.125 rad/s; 0.018 rad/s2]
Fig. 8.46

Chapter 8 :  Acceleration in Mechanisms    /G6C  231
1.Explain how the acceleration of a point on a link (whose direction is known) is obtained when the
acceleration of some other point on the same link is given in magnitude and direction.
2.Draw the acceleration diagram of a slider crank mechanism.
3.Explain how the coriolis component of acceleration arises when a point is rotating about some other
fixed point and at the same time its distance from the fixed point varies.

4.Derive an expression for the magnitude and direction of coriolis component of acceleration.
5.Sketch a quick return motion of the crank and slotted lever type and explain the procedure of drawing
the velocity and acceleration diagram, for any given configuration of the mechanism.
1.The component of the acceleration, parallel to the velocity of the particle, at the given instant is
called
(a) radial component ( b) tangential component

(c) coriolis component ( d) none of these
2.A point B on a rigid link AB moves with respect to A with angular velocity ω rad/s. The radial
component of the acceleration of B with respect to A,
(a)vBA  × AB (b)v2
BA × AB (c)BAv
AB(d)2
BAv
AB
where     vBA = Linear velocity of B with respect to A = ω × AB
3.A point B on a rigid link AB moves with respect to A with angular velocity ω rad/s. The angular
acceleration of the link AB is
(a)BAra
AB(b)BAta

(a)BAra
AB(b)BAta
AB(c)vBA × AB (d)2
BAv
AB
4.A point B on a rigid link AB moves with respect to A with angular velocity ω rad/s. The total
acceleration of B with respect to A will be equal to
(a) vector sum of radial component and coriolis component
(b) vector sum of tangential component and coriolis component
(c) vector sum of radial component and tangential component
(d) vector difference of radial component and tangential component

5.The coriolis component of acceleration is taken into account for
(a) slider crank mechanism ( b) four bar chain mechanism
(c) quick return motion mechanism ( d) none of these
ANSWERS
1. (b) 2. (d) 3. (b) 4. (c) 5. (c)

232       /G6C               Theory of Machines
232MMMMMechanismsechanismsechanismsechanismsechanisms
with Lowerwith Lowerwith Lowerwith Lowerwith Lower
PairsPairsPairsPairsPairs9
1. Introduction
2. Pantograph3. Straight Line Mechanism.4. Exact Straight Line Motion
Mechanisms Made up of
Turning Pairs.
5. Exact Straight Line Motion
Consisting of One SlidingPair (Scott Russel’ sMechanism).
6. Approximate Straight Line
Motion Mechanisms.

7. Straight Line Motions for
Engine Indicators.
8. Steering Gear Mechanism.
9. Davis Steering Gear.
10. Ackerman Steering Gear .
11. Universal or Hooke’ s Joint.
12. Ratio of the Shafts Velocities.13. Maximum and Minimum
Speeds of the Driven Shaft.
14. Condition for Equal Speeds
of the Driving and Driven
Shafts.
15. Angular Acceleration of the
Driven Shaft.
16. Maximum Fluctuation of
Speed.

Speed.
17.  Double Hooke’ s Joint.9.1.9.1.9.1.9.1.9.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoduction
We have already discussed, that when the two ele-
ments of a pair have a surface contact and a relative motiontakes place, the surface of one element slides over the sur-face of the other, the pair formed is known as lower pair . In
this chapter we shall discuss such mechanisms with lowerpairs.

9.2.9.2.9.2.9.2.9.2. PantographPantographPantographPantographPantograph
A pantograph is an
instrument used to repro-duce to an enlarged or a re-duced scale and as exactlyas possible the path de-scribed by a given point.
It consists of a
jointed parallelogramABCD  as shown in Fig. 9.1.
It is made up of bars connected by turning pairs. The bars BA
and BC are extended to O and E respectively, such that
OA/OB = AD/BEFig. 9.1.  Pantograph.
Pantograph

Chapter 9 : Mechanisms with Lower Pairs    /G6C  233
Thus, for all relative positions of the bars,
the triangles OAD and OBE are similar and the points
O, D and E are in one straight line. It may be proved
that point E traces out the same path as described by
point D.
From similar triangles OAD and OBE, we
find that
               OD /OE = AD/BE
Let point O be fixed and the points D and E
move to some new positions D′ and E′. Then
                OD/OE = OD′/OE′

OD/OE = OD′/OE′
A little consideration will show that the
straight line DD′ is parallel to the straight line EE′.
Hence, if O is fixed to the frame of a machine by means of a turning pair and D is attached to a point
in the machine which has rectilinear motion relative to the frame, then E will also trace out a straight
line path. Similarly, if E is constrained to move in a straight line, then D will trace out a straight line
parallel to the former.

A pantograph is mostly used for the reproduction of plane areas and figures such as maps,
plans etc., on enlarged or reduced scales. It is, sometimes, used as an indicator rig in order to repro-duce to a small scale the displacement of the crosshead and therefore of the piston of a reciprocatingsteam engine. It is also used to guide cutting tools. A modified form of pantograph is used to collectpower at the top of an electric locomotive.
9.3. Straight Line Mechanisms

One of the most common forms of the constraint mechanisms is that it permits only relative
motion of an oscillatory nature along a straight line. The mechanisms used for this purpose are calledstraight line mechanisms . These mechanisms are of the following two types:
1.in which only turning pairs are used, and
2.in which one sliding pair is used.
These two types of mechanisms may produce exact straight line motion or approximate straight

line motion, as discussed in the following articles.
9.4. Exact Straight Line Motion Mechanisms Made up of Turning Pairs
The principle adopted for a mathematically correct
or exact straight line motion is described in Fig.9.2. Let O
be a point on the circumference of a circle of diameter OP.
Let OA be any chord and B is a point on OA produced, such
that
OA × OB = constant
Then the locus of a point B will be a straight line
perpendicular to the diameter OP. This may be proved as
follows:

follows:
Draw BQ perpendicular to OP produced. Join AP.
The triangles OAP and OBQ are similar.Fig. 9.2. Exact straight line
motion mechanism.
Pantograph.

234  /G6C   Theory of Machines
∴OA OQ
OP OB=
or OP × OQ = OA × OB
orOA OBOQ
OP×=
But OP is constant as it is the diameter of a circle, there-
fore, if OA × OB is constant, then OQ will be constant. Hence
the point B moves along the straight path BQ which is perpen-
dicular to OP.
Following are the two well known types of exact straight
line motion mechanisms made up of turning pairs.
1. Peaucellier mechanism . It consists of a fixed link

OO1 and the other straight links O1A, OC, OD, AD, DB, BC and
CA are connected by turning pairs at their intersections, as shown
in Fig. 9.3. The pin at A is constrained to move along the cir-
cumference of a circle with the fixed diameter OP, by means of
the link O1A. In Fig. 9.3,
 AC = CB  = BD = DA ; OC = OD ; and OO1 = O1A
It may be proved that the product OA × OB remains
constant, when the link O1A rotates. Join CD to bisect AB at R.
Now from right angled triangles ORC and BRC, we have

OC2  = OR2 + RC2...(i)
and                 BC2  = RB2 + RC2...(ii)
Subtracting equation  (ii) from (i), we have
OC2 – BC2 = OR2 – RB2
                 = ( OR + RB) (OR – RB)
                 = OB × OA
Since OC and BC are of constant length, therefore
the product OB × OA remains constant. Hence the point  B
traces a straight path perpendicular to the diameter OP.
2. Hart’s mechanism . This mechanism requires only

six links as compared with the eight links required by thePeaucellier mechanism. It consists of a fixed link OO
1 and other straight links O1A, FC, CD, DE and
EF are connected by turning pairs at their points of intersection, as shown in Fig. 9.4. The links FC
and DE are equal in length and the lengths of the links CD and EF are also equal. The points O, A and
B divide the links FC, CD and EF in the same ratio. A little consideration will show that BOCE is a

trapezium and OA and OB are respectively parallel to * FD and CE.
Hence OAB is a straight line. It may be proved now that the product OA × OB is constant.Fig. 9.3. Peaucellier mechanism.
*In ∆ FCE, O and B divide FC and EF in the same ratio, i.e.
                                       CO /CF = EB/EF
∴ OB is parallel to CE. Similarly, in triangle FCD, OA is parallel to FD.A modified form of pantograph is
used to collect electricity at the
top of electric trains and buses.

Chapter 9 : Mechanisms with Lower Pairs    /G6C  235
From similar triangles CFE and OFB,
                   CE OB
FC OF=        or        CE OFOB
FC×= ...(i)
and from similar triangles FCD and OCA
                   FD OA
FC OC=       or        FD OCOAFC×= ...(ii)
Fig. 9.4. Hart’s mechanism.
Multiplying equations (i) and (ii), we have
2FD OC CE OF OC OFOA OB FD CE
FC FC FC×× ××= × =××
Since the lengths of OC, OF and FC are fixed, therefore
    OA × OB = FD × CE × constant ... (iii)

OA × OB = FD × CE × constant ... (iii)
2... substituting constantOC OF
FC× = 
Now from point E, draw EM parallel to CF and EN perpendicular to FD. Therefore
FD × CE= FD × FM ...(∵ CE = FM )
= (FN + ND) (FN – MN) = FN2 – ND2...(∵ MN = ND)
= (FE2 – NE2) – (ED2 – NE2)
...(From right angled triangles FEN and EDN )
= FE2 – ED2 = constant ... (iv)
...(∵ Length FE and ED are fixed)
From equations (iii) and (iv),
    OA × OB = constant

It therefore follows that if the mechanism is pivoted about O as a fixed point and the point A
is constrained to move on a circle with centre O1, then the point B will trace a straight line perpendicu-
lar to the diameter OP produced.
Note: This mechanism has a great practical disadvantage that even when the path of B is short, a large amount
of space is taken up by the mechanism.
9.5. Exact Straight Line Motion Consisting of One Sliding Pair-Scott
Russell’s Mechanism

It consists of a fixed member and moving member P of a sliding pair as shown in Fig. 9.5.

236  /G6C   Theory of Machines
The straight link PA Q is connected by turning pairs to the link OA and the link P. The link OA rotates
about O. A little consideration will show that the mechanism OAP is same as that of the reciprocating
engine mechanism in which OA is the crank and PA is the
connecting rod. In this mechanism, the straight line mo-tion is not generated but it is merely copied.
In Fig. 9.5, A is the middle point of PQ and
OA = AP = AQ. The instantaneous centre for the link PA Q

lies at I in OA produced and is such that IP is perpendicu-
lar to OP. Join IQ. Then Q moves along the perpendicular
to IQ. Since OPIQ is a rectangle and IQ is perpendicular
to OQ, therefore Q moves along the vertical line OQ for
all positions of QP. Hence Q traces the straight line OQ′. If
OA makes one complete revolution, then P will oscillate
along the line OP through a distance 2 OA on each side of O and Q will oscillate along OQ′ through

the same distance 2 OA above and below O. Thus, the locus of Q is a copy of the locus of P.
Note: Since the friction and wear of a sliding pair is much more than those of turning pair, therefore this
mechanism is not of much practical value.
9.6. Approximate Straight Line Motion Mechanisms
The approximate straight line motion mechanisms are the modifications of the four-bar chain

mechanisms. Following mechanisms to give approximate straight line motion, are important from thesubject point of view :
1. Watt’s mechanism.  It is a crossed four bar chain mechanism and was used by Watt for his
early steam engines to guide the piston rod in a cylinder to have an approximate straight line motion.
Fig. 9.6. Watt’s mechanism.
In Fig. 9.6, OBAO1 is a crossed four bar chain in which O and O1 are fixed. In the mean

position of the mechanism, links OB and O1A are parallel and the coupling rod AB is perpendicular to
O1A and OB. The tracing point P traces out an approximate straight line over certain positions of its
movement, if PB/PA = O1A/OB. This may be proved as follows :
A little consideration will show that in the initial mean position of the mechanism, the instan-
taneous centre of the link BA lies at infinity. Therefore the motion of the point P is along the vertical

line BA . Let OB′ A′O1 be the new position of the mechanism after the links OB and O1A are displaced
through an angle θ and φ respectively. The instantaneous centre now lies at  I. Since the angles θ and
φ are very small, therefore
arc B B′= arc A A′         or      OB × θ = O1 A × φ ...(i)Fig. 9.5. Scott Russell’s mechanism.

Chapter 9 : Mechanisms with Lower Pairs    /G6C  237
∴ OB / O1 A= φ / θ
Also A′P′= IP′ × φ, and B′P′ = IP′ × θ
∴ A′P′ / B′P′= φ / θ ...(ii)
                            
1OB AP AP
OA BP B P′′==′′or1OAPB
OB PA=
Thus, the point  P divides the link AB into two parts whose lengths are inversely proportional
to the lengths of the adjacent links.
2. Modified Scott-Russel mechanism . This mechanism, as shown in Fig. 9.7, is similar to

Scott-Russel mechanism (discussed in Art. 9.5), but in this case AP is not equal to AQ and the points
P and Q are constrained to move in the horizontal and vertical directions. A little consideration will
show that it forms an elliptical trammel, so that any point A
on PQ traces an ellipse with semi-major axis AQ and semi-
minor axis AP.
If the point A moves in a circle, then for point Q to
move along an approximate straight line, the length OA must

be equal ( AP)2 / AQ. This is limited to only small
displacement of P.
3. Grasshopper mechanism . This mechanism is a
modification of modified Scott-Russel’s mechanism withthe difference that the point P does not slide along a straight
line, but moves in a circular arc with centre O.
It is a four bar mechanism and all the pairs are turning pairs as shown in Fig. 9.8. In this
mechanism, the centres O and O
1 are fixed. The link OA oscillates about O through an angle AOA1

which causes the pin P to move along a circular arc with
O1 as centre and O1P as radius. For small angular dis-
placements of OP on each side of the horizontal, the point
Q on the extension of the link PA traces out an approxi-
mately a straight path QQ′, if the lengths are such that OA
= (AP)2 / AQ.
Note: The Grasshopper mechanism was used in early days as
an engine mechanism which gave long stroke with a very short
crank.
4. Tchebicheff’s mechanism . It is a four bar

mechanism in which the crossed links OA and O1B are of
equal length, as shown in Fig. 9.9. The point P, which is
the mid-point of AB traces out an approximately straight
line parallel to OO1. The proportions of the links are, usually, such that point P is exactly above O or
O1 in the extreme positions of the mechanism i.e. when BA lies along OA or when BA lies along BO1.

It may be noted that the point P will lie on a straight line parallel to OO1, in the two extreme positions
and in the mid position, if the lengths of the links are in proportions AB : OO1 : OA = 1 : 2 : 2.5.
5. Roberts mechanism.  It is also a four bar chain mechanism, which, in its mean position, has
the form of a trapezium. The links OA and O1 B are of equal length and OO1 is fixed. A bar PQ is
rigidly attached to the link AB at its middle point P.Fig. 9.7. Modified Scott-Russel
mechanism.

238  /G6C   Theory of Machines
A little consideration will show that if the mechanism is displaced as shown by the dotted
lines in Fig. 9.10, the point Q will trace out an approximately straight line.
Fig. 9.9. Tchebicheff’s mechanism. Fig. 9.10.  Roberts mechanism
9.7. Straight Line Motions for Engine Indicators
The application of straight

line motions is mostly found in theengine indicators. In theseinstruments, the cylinder of theindicator is in directcommunication with the steam orgas inside the cylinder of anengine. The indicator piston rises
and falls in response to pressure

variation within the enginecylinder. The piston is resisted bya spring so that its displacement isa direct measure of the steam orgas pressure acting upon it. Thedisplacement is communicated tothe pencil which traces thevariation of pressure in thecylinder (also known as indicatordiagram) on a sheet of paperwrapped on the indicator drum
which oscillates with angular

motion about its axis, according tothe motion of the engine piston.The variation in pressure isrecorded to an enlarged scale.Following are the various engineindicators which work on thestraight line motion mechanism.
1. Simplex indicator . It closely resembles to the pantograph copying mechanism, as shown
in Fig. 9.11. It consists of a fixed pivot O attached to the body of the indicator. The links AB, BC, CD
Airplane’s Landing Gear.Tyres absorb some
energyLiquid

springHydraulic
cylinder folds
wheels for
storageInternal damper
absorbs shock

Chapter 9 : Mechanisms with Lower Pairs    /G6C  239
and DA form a parallelogram and are pin jointed. The link BC is extended to point P such that O, D
and P lie in one straight line. The point D is attached to the piston rod of the indicator and moves
along the line of stroke of the piston ( i.e. in the vertical direction). A little consideration will show
that the displacement of D is reproduced on an enlarged scale, on the paper wrapped on the indicator

drum, by the pencil fixed at point P which describes the path similar to that of D. In other words,
when the piston moves vertically by a distance DD1, the path traced by P is also a vertical straight line
PP1, as shown in Fig. 9.11.
Fig. 9.11.  Simplex indicator.
The magnification may be obtained by the following relation :
1
1PP OP OB BP
OD OA BC DD===
From the practical point of view, the following are the serious objections to this mechanism:

(a)Since the accuracy of straight line motion of P depends upon the accuracy of motion of
D, therefore any deviation of D from a straight path involves a proportionate deviation of
P from a straight path.
(b)Since the mechanism has five pin joints at O, A, B, C and D, therefore slackness due to
wear in any one of pin joints destroys the accuracy of the motion of P.
2. Cross-by indicator . It is a modified form
of the pantograph copying mechanism, as shown inFig. 9.12.

In order to obtain a vertical straight line
for P, it must satisfy the following two conditions:
1.The point P must lie on the line joining
the points O and A, and
2.The velocity ratio between points P and
A must be a constant.
This can be proved by the instantaneous
centre method as discussed below :
The instantaneous centre I
1 of the link AC
is obtained by drawing a horizontal line from A
to meet the line ED produced at I1. Similarly, the

instantaneous centre I2 of the link BP is obtained by drawing a horizontal line from P to meet the
line BO at I2. We see from Fig. 9.12, that the points I1 and I2 lie on the fixed pivot O. Let vA, vB,
vC and vP be the velocities of the points A, B, C and P respectively.
We know that   C1 2
A1 2vIC IC
vI A I A==...(i)Fig. 9.12.  Cross-by indicator.

240  /G6C   Theory of Machines
andP2
C2vI P
vI C=...(ii)
Multiplying equations  (i) and (ii), we get
                             CP2 2
AC2 2vvI C I P
vvI A I C×= ×     or     P2
A2vI P O P
vI A O A== ...(iii)
...(∵ O and I2 are same points.)
Since AC is parallel to OB, therefore triangles PA C and POB are similar.
∴OP BP
OA BC= ...(iv)
          P
AconstantvO P B P
vO A B C===...(∵ Lengths BP and BC are constant.)

3. Thompson indicator . It consists of the links OB, BD, DE and EO. The tracing point  P lies
on the link  BD produced. A little consideration will show that it constitutes a straight line motion of
the Grasshopper type as discussed in Art.9.6. The link BD gets the motion from the piston rod of the
indicator at C which is connected by the link AC at A to the end of the indicator piston rod. The

condition of velocity ratio to be constant between P and A may be proved by the instantaneous centre
method, as discussed below :
Fig. 9.13.  Thompson indicator.
Draw the instantaneous centres I1 and I2 of the links BD and AC respectively. The line I1P
cuts the links AC at F. Let vA, vC and vP be the velocities of the points A, C and P respectively.
∴  C2
A2vIC
vI A= ...(i)
From similar triangles I1CF and I2CA
                                  21
21IC IC
IA IF=    or     C2 1
A2 1vIC IC

21IC IC
A2 1vIC IC
vI A I F==...(ii)
...[From equation (i)]
Also  P1
C1vI P
vI C= ...(iii)
Multiplying equations (ii) and (iii), we get
                            CP1 1
AC1 1vvI C I P
vvI F I C×= ×   or  P1
A1vI P
vI F= ...(iv)

Chapter 9 : Mechanisms with Lower Pairs    /G6C  241
Now if the links AC and OB are parallel, the triangles PCF and PBI1 are similar.
∴1
1IP B P
IF B C= ...(v)
From equations (iv) and (v),
P1
A1constantvI P B P
vI F B C== =...(∵ Lengths BP and BC are constant)
Note: The links AC and OB can not be exactly parallel, nor the line I1P be exactly perpendicular to the line of
stroke of the piston for all positions of the mechanism. Hence the ratio BP/BC cannot be quite constant. Since

the variations are negligible for all practical purposes, therefore the above relation gives fairly good results.
4. Dobbie Mc Innes indicator . It is similar to Thompson indicator with the difference that the
motion is given to the link DE (instead of BD in Thompson indicator) by the link AC connected to the
indicator piston as shown in Fig. 9.14. Let vA, vC, vD and vP be the velocities of the points A, C, D and

P respectively. The condition of velocity ratio ( i.e. vP  / vA) to be constant between points P and A may
be determined by instantaneous centre method as discussed in Thompson indicator.
Fig. 9.14.  Dobbie McInnes indicator.
cuts the link AC at F. Draw DH perpendicular to I1P. We know that
∴C2
A2vIC
vI A=...(i)
From similar triangles I1CF and I2CA,
                                  21
21IC IC

21
21IC IC
IA IF=   or     C2 1
A2 1vIC IC
vI A I F==...[From equation (i)]   ...(ii)
Again from similar triangles I1CF and I1DH,
                                  11
11IC ID
IF IH=   or    C1
A1vID
vI H= ...[From equation (ii)] ...(iii)
Since the link ED turns about the centre E, therefore
D
CvE D
vE C= ...(iv)

242  /G6C   Theory of Machines
AlsoP1
D1vI P
vI D= ...(v)
Multiplying equations (iii), (iv) and (v), we get
                CDP1 1
ACD1 1vv v ID IP ED
vvvI H E C I D××= × ×     or     P1
A1vI P E D
vI H E C=×                         ... (vi)
From similar triangles I1BP and PDH,
1
1IP P B
IH B D=
∴                                 P
AconstantvP B E D
vB D E C=×=...[From equation (vi)]
...[∵ Lengths PB, BD, ED and EC are constant.]
9.8. Steering Gear Mechanism

The steering gear mechanism is used for
changing the direction of two or more of the wheelaxles with reference to the chassis, so as to move theautomobile in any desired path. Usually the two backwheels have a common axis, which is fixed in direc-
tion with reference to the chassis and the steering is
done by means of the front wheels.
In automobiles, the front wheels are placed

over the front axles, which are pivoted at the pointsA and B, as shown in Fig. 9.15. These points are fixed to the chassis. The back wheels are placed
over the back axle, at the two ends of the differential tube. When the vehicle takes a turn, the
front wheels along with the respective axles turn about the respective pivoted points. The back
wheels remain straight and do not turn. Therefore, the steering is done by means of front wheelsonly.

Chapter 9 : Mechanisms with Lower Pairs    /G6C  243
In order to avoid skidding
(i.e. slipping of the wheels side-
ways), the two front wheels mustturn about the same instant aneous
centre I which lies on the axis of
the back wheels. If the instanta-neous centre of the two frontwheels do not coincide with the in-stantaneous centre of the backwheels, the skidding on the front
or back wheels will definitely take place, which will cause more wear and tear of the tyres.

Thus, the condition for correct steering is that all the four wheels must turn about the same
instantaneous centre. The axis of the inner wheel makes a larger turning angle θ than the angle φ
subtended by the axis of outer wheel.
Let                a  =  Wheel track,
                     b  =  Wheel base, and
                       c  =  Distance between the pivots A and B of the front axle.
Now from triangle IBP,
                
cotBP
IPθ=
and from triangle IAP,

cotBP
IPθ=
                cot cotAP AB BP AB BP c
IP IP IP IP b+φ= = = + = + θ ...(∵ IP = b)
∴ cot φ – cot θ = c / b
This is the fundamental equation for correct steering. If this condition is satisfied, there will
be no skidding of the wheels, when the vehicle takes a turn.
9.9. Davis Steering Gear
The Davis steering gear is shown in Fig. 9.16. It is an exact steering gear mechanism. The

slotted links AM and BH are attached to the front wheel axle, which turn on pivots A and B respec-
tively. The rod CD is constrained to move in the direction of its length, by the sliding members at P
and Q. These constraints are connected to the slotted link AM and BH by a sliding and a turning pair
at each end. The steering is affected by moving CD to the right or left of its normal position. C′D′
shows the position of CD for turning to the left.

Let                 a  = Vertical distance between AB and CD,
                      b  = Wheel base,
                      d  = Horizontal distance between AC and BD,
                       c  = Distance between the pivots A and B of the front axle.
                       x  = Distance moved by AC to AC′ = CC′ = DD′, and
                      α  = Angle of inclination of the links AC and BD, to the vertical.
From triangle AA′ C′,

244  /G6C   Theory of Machines
From triangle AA′C,
tanAC d
AA a′α= =′...(ii)
From triangle BB′D′,
–tan( – )BD d x
BB a′′αθ = =′...(iii)
Fig. 9.16.  Davis steering gear.
We know that        tan tantan( )1 – tan .tanα+ φα+φ =αφ
or               /t a n t a n
1 – / tan – tandx d a da
ad a a d++ φ + φ==×φ φ
...[From equations (i) and (ii)]
             (d + x) (a – d tan φ) = a (d + a tan φ)
a. d – d 2 tan φ + a. x – d.x tan φ = a.d + a2 tan φ
              tan φ (a2 + d2 + d.x) = ax     or    22.tan

.ax
add xφ=
++...(iv)
Similarly, from tan –(–) ,dx
aαθ =  we get
                                    22tan
–.ax
add xθ=
+...(v)
We know that for correct steering,
                       cot – cotc
bφθ =      or    11–tan tanc
b=φθ
             22 22.– .–..add xadd xc
ax ax b++ += ...[From equations (iv) and (v)]

Chapter 9 : Mechanisms with Lower Pairs    /G6C  245
or2.
.dx c
ax b=or2dc
ab=
∴ 2tanc
bα= ortan
2c
bα= ...(∵ d / a = tan α)
Note: Though the gear is theoretically correct, but due to the presence of more sliding members, the wear will
be increased which produces slackness between the sliding surfaces, thus eliminating the original accuracy.
Hence Davis steering gear is not in common use.
Example 9.1.  In a Davis steering gear, the dist ance between the pivots of the front axle is 1.2

metres and the wheel base is  2.7 metres. Find the inclination of the track arm to the longitudinal axis
of the car, when it is moving along a straight path.
Solution. Given :     c = 1.2 m ; b = 2.7 m
Let                  α  = Inclination of the track arm to the longitudinal axis.
We know that   1.2tan 0.22222 2 . 7c
bα= = =×orα = 12.5° Ans.
9.10. Ackerman Steering Gear
The Ackerman steering gear mechanism is much simpler than Davis gear. The difference

between the Ackerman and Davis steering gears are :
1.The whole mechanism of the Ackerman steering gear is on back of the front wheels;
whereas in Davis steering gear, it is in front of the wheels.
2.The Ackerman steering gear consists of turning pairs, whereas Davis steering gear
consists of sliding members.
Fig. 9.17.  Ackerman steering gear.
In Ackerman steering gear, the mechanism ABCD is a four bar crank chain, as shown in Fig.

9.17. The shorter links BC and AD are of equal length and are connected by hinge joints with front
wheel axles. The longer links AB and CD are of unequal length. The following are the only three
positions for correct steering.
1.When the vehicle moves along a straight path, the longer links AB and CD are parallel and
the shorter links BC and AD are equally inclined to the longitudinal axis of the vehicle, as shown by
firm lines in Fig. 9.17.

2.When the vehicle is steering to the left, the position of the gear is shown by dotted lines in
Fig. 9.17. In this position, the lines of the front wheel axle intersect on the back wheel axle at I, for
correct steering.

246  /G6C   Theory of Machines
3.When the vehicle is steering to the right, the similar position may be obtained.
In order to satisfy the fundamental equation for correct steering, as discussed in Art. 9.8, the
links AD and DC are suitably proportioned. The value of  θ and φ  may be obtained either graphically
or by calculations.
9.11. Universal or Hooke’s Joint
A *Hooke’s joint is used to connect two shafts,  which are intersecting at a small angle, as

shown in  Fig. 9.18. The end of each shaft is forked to U-type and each fork provides two bearings
          Fig. 9.18.  Universal or Hooke’s joint.
for the arms of a cross. The arms of the cross are perpendicular to each other. The motion is transmit-

ted from the driving shaft to driven shaft through a cross. The inclination of the two shafts may beconstant, but in actual practice it varies, when the motion is transmitted. The main application of theUniversal or Hooke’s joint is found in the transmission from the **gear box to the differential or back
axle of the automobiles. It is also used for transmission of power to different spindles of multipledrilling machine. It is also used as a knee joint in milling machines.

*This joint was first suggested by  Da Vinci and was named after English physicist and mathematician
Robert Hooke who first applied it to connect two offset misaligned shafts.
**In case of automobiles, we use two Hooke’s joints one at each end of the propeller shaft, connecting the
gear box on one end and the differential on the other end.
Universal Joint.Axis 2
Axis 1Body 1
Body 2

Chapter 9 : Mechanisms with Lower Pairs    /G6C  247
9.12. Ratio of the Shafts Velocities
The top and front views connecting the two shafts by
a universal joint are shown in Fig. 9.19. Let the initial posi-tion of the cross be such that both arms lie in the plane of thepaper in front view, while the arm AB attached to the driving
shaft lies in the plane containing the axes of the two shafts.Let the driving shaft rotates through an angle θ, so that the
arm AB moves in a circle to a new position A

1 B1 as shown in
front view. A little consideration will show that the arm CD
will also move in a circle of the same size. This circle whenprojected in the plane of paper appears to be an ellipse. There-fore the arm CD takes new position C
1D1 on the ellipse, at an
angle θ. But the true angle must be on the circular path. To
find the true angle, project the point C1 horizontally to inter-

sect the circle at C2. Therefore the angle COC2 (equal to φ) is
the true angle turned by the driven shaft. Thus when the driv-ing shaft turns through an angle θ, the driven shaft turns
through an angle φ. It may be noted that it is not necessary
that φ  may be greater than θ or less than θ. At a particular
point, it may be equal to θ.
In triangle OC
1M, ∠ OC1M = θ
∴      
1tanOM
MCθ=                 ... (i)
and in triangle   OC2 N,  ∠ OC2 N = φ
∴    
21tanON ON

∴    
21tanON ON
NC MCφ= =                                                 21...( )NC MC= /G51 ...(ii)
Dividing equation  (i) by (ii),
1
1tan
tanMCOM OM
MC ON ONθ=×=φ
But                  OM  = ON1 cos α = ON cos α
...(where α = Angle of inclination of the driving and driven shafts)
∴               tan coscostanON
ONθα== αφ
or tan θ =  tan φ . cos α ...(iii)
Let ω=  Angular velocity of the driving shaft = dθ / dt
ω1=  Angular velocity of the driven shaft = dφ / dt

Differentiating both sides of equation (iii),
sec2 θ × dθ / dt= cos α . sec2 φ × dφ / dt
sec2 θ × ω= cos α . sec2 φ × ω1
∴2
1
22 2sec 1
cos .sec cos .cos .secω θ==ωαφ θ αφ...(iv)Fig. 9.19. Ratio of shafts
velocities.

248  /G6C   Theory of Machines
We know that       2
22
2tansec 1 tan 1
cosθφ= + φ= +
α...[From equation (iii)]
                                      22 2 2
22 22sin cos .cos sin1
cos .cos cos .cosθθ α + θ=+ =
θα θα
                                      22 2 2 2 2 2
22 22cos (1 – sin ) sin cos – cos .sin sin
cos .cos cos .cosθα + θ θ θ α + θ==
θα θα
22
221–c o s .s i n
cos .cosθα=
θα...(∵ cos2 θ + sin2 θ = 1)
Substituting this value of sec2 φ in equation  (iv), we have veloity ratio,

22
1
22 2 2 21c o s . c o s c o s
cos .cos 1 – cos .sin 1 – cos .sinωθ α α=× =ω θα θ α θ α...(v)
Note:  If N= Speed of the driving shaft in r.p.m., and
N1= Speed of the driven shaft in r.p.m.
Then the equation (v) may also be written as
                  1
22cos.
1–cos .sinN
Nα=
θα
9.13. Maximum and Minimum Speeds of Driven Shaft
We have discussed in the previous article that velocity ratio,
                 1
22cos
1–c o s .s i nω α=
ω θα       or      1 22.cos

1–c o s .s i nωαω=
θα...(i)
The value of ω1 will be maximum for a given value of α, if the denominator of equation  (i) is
minimum. This will happen, when
                          cos2 θ = 1,     i.e.  when θ = 0°, 180°, 360° etc.
∴   Maximum speed of the driven shaft,
                 1( ) 22cos cos
cos 1–s i n c o smaxωα ωα ωω= = =
α αα...(ii)
or                                     1( )cosmaxNN=α...(where N and N1 are in r.p.m.)

Similarly, the value of ω1 is minimum, if the denominator of equation (i) is maximum. This
will happen, when (cos2 θ . sin2 α) is maximum, or
                           cos2 θ = 0, i.e. when θ = 90°, 270° etc.
∴ Minimum speed of the driven shaft,
                           ω1 (min) = ω cos α
or                                      N1 (min) = N cos α ...(where N and N1 are in r.p.m.)
Fig. 9.20, shows the polar diagram depicting the salient features of the driven shaft speed.

Chapter 9 : Mechanisms with Lower Pairs    /G6C  249
From above, we see that
1. For one complete revolution of the driven shaft,
there are two points i.e. at 0° and 180° as shown by points
1 and 2 in Fig. 9.20, where the speed of the driven shaft ismaximum and there are two points i.e. at 90° and 270° as
shown by point 3 and 4 where the speed of the driven shaftis minimum.
2. Since there are two maximum and two mini-

mum speeds of the driven shaft, therefore there are fourpoints when the speeds of the driven and driver shaft aresame. This is shown by points, 5,6,7 and 8 in Fig. 9.20 (SeeArt 9.14).
3. Since the angular velocity of the driving shaft is
usually constant, therefore it is represented by a circle of radius ω. The driven shaft has a variation in
angular velocity, the maximum value being ω/cos α and minimum value is ω cos α. Thus it is repre-

sented by an ellipse of semi-major axis ω/cos α and semi-minor axis ω cos α, as shown in Fig. 9.20.
Note:  Due to the variation in speed of the driven shaft, there will be some vibrations in it, the frequency of
which may be decreased by having a heavy mass (a sort of flywheel) on the driven shaft. This heavy mass of
flywheel does not perform the actual function of flywheel.
9.14. Condition for Equal Speeds of the Driving and Driven Shafts

We have already discussed that the ratio of the speeds of the driven and driving shafts is
1
22cos
ω θα  or22
1(1 – cos .sin )
cosωθ αω=α
For equal speeds,  ω = ω1, therefore
            cos α = 1 – cos2 θ . sin2 α  or cos2 θ . sin2 α = 1 – cos α
and  2
21–c o scos
sinαθ=
α...(i)
We know that 22
221c o s 1 – c o ss in 1–cos 1– 1–
sin 1 – cos−α αθ= θ= =
αα
   1 – cos 1 cos1– 1–(1 cos )(1 – cos ) 1 cos 1 cosαα== =+α α +α+α...(ii)
Dividing equation (ii) by equation  (i),
22

22
2sin cos sin
1c o s 1 – c o scosθα α=×+α αθ
or       22
2
22cos sin cos .sintan cos
1–c o s s i nαα ααθ= = = α
αα
∴         tan cosθ=± α
There are two values of θ corresponding to positive sign and two values corresponding to
negative sign. Hence, there are four values of θ, at which the speeds of the driving and driven shafts
are same. This is shown by point 5, 6, 7 and 8 in Fig. 9.20.
9.15. Angular Acceleration of the Driven Shaft
We know that22 – 1

1 22cos.cos (1 – cos .sin )
1–c o s .s i nωαω= =ω α θ α
θαFig. 9.20. Polar diagram-salient
features of driven shaftspeed.

250  /G6C   Theory of Machines
Differentiating the above expression, we have the angular acceleration of the driven shaft,
22 – 2 2 1cos –1(1 – cos sin ) (2cos sin sin )d d
dt dtω θ =ω α θ α × θ θ α
22
22 2–c o s s i n 2 . s i n
(1 – cos sin )ωα ×θα=
θα...(i)
...( 2 cos θ sin θ = sin 2 θ, and dθ/dt = ω)
The negative sign does not show that there is always retardation. The angular acceleration
may be positive or negative depending upon the value of sin 2 θ. It means that during one complete

revolution of the driven shaft, there is an angular acceleration corresponding to increase in speed ofω
1 and retardation due to decrease in speed of ω1.
For angular acceleration to be maximum, differentiate dω1 / dt with respect to θ and equate to
zero. The result is * approximated as
22
2s i n ( 2–c o s 2 )cos2
2–s i nαθθ=
α
Note:  If the value of α is less than 30°, then cos 2 θ may approximately be written as
2
22sincos2
2–s i nαθ=
α
9.16. Maximum Fluctuation of Speed

2–s i nαθ=
α
We know that the maximum speed of the driven shaft,
 ω1 (max) = ω/cos α
and minimum speed of the driven shaft,
ω1 (min) = ω cos α
∴ Maximum fluctuation of speed of the driven shaft,
                                                    1( ) 1( )–– c o scosmax minqω=ω ω = ω αα
   2211 – c o s s i n–c o scos cos cos  αωα=ω α =ω =    αα α 
= ω tan α . sin α
Since α is a small angle, therefore substituting cos α = 1, and sin α = α radians.

∴ Maximum fluctuation of speed
= ω . α2
Hence, the maximum fluctuation of speed of the driven shaft approximately varies as the
square of the angle between the two shafts.
Note:  If the speed of the driving shaft is given in r.p.m. ( i.e. N r.p.m.), then in the above relations ω may be
replaced by N.
9.17. Double Hooke’s Joint
We have seen in the previous articles, that the velocity of the driven shaft is not constant, but

varies from maximum to minimum values. In order to have a constant velocity ratio of the driving anddriven shafts, an intermediate shaft with a Hooke’s joint at each end as shown in Fig. 9.21, is used.This type of joint is known as double Hooke ’s joint.
*Since the differentiation of dω1/dt is very cumbersome, therefore only the result is given.

Chapter 9 : Mechanisms with Lower Pairs    /G6C  251
Let the driving, intermediate and driven shafts, in the same time, rotate through angles θ, φ
and γ from the position as discussed previously in Art. 9.12.
Now for shafts A and B,tan θ= tan φ . cos α ...(i)
and       for shafts B and C,t a n  γ= tan φ . cos α ...(ii)
From equations (i) and (ii), we see that θ = γ or ωA = ωC.
Fig. 9.21.  Double Hooke’s joint.

This shows that the speed of the driving and driven shaft is constant. In other words, this joint
gives a velocity ratio equal to unity, if
1.The axes of the driving and driven shafts are in the same plane, and
2.The driving and driven shafts make equal angles with the intermediate shaft.
Example. 9.2.   Two shafts with an included angle of 160° are connected by a Hooke’s joint.

The driving shaft runs at a uniform speed of 1500 r. p.m. The driven shaft carries a flywheel of mass
12 kg and 100 mm radius of gyration. Find the maximum angular acceleration of the driven shaftand the maximum torque required.
Solution.  Given :  α = 180° – 160° = 20°; N = 1500 r.p.m.; m = 12 kg ; k = 100 mm = 0.1 m
We know that angular speed of the driving shaft,
ω = 2 π × 1500 / 60 = 157 rad/s
and mass moment of inertia of the driven shaft,
I = m.k
2 = 12 (0.1)2 = 0.12 kg - m2

I = m.k
Maximum angular acceleration of the driven shaft
Let dω1 / dt= Maximum angular acceleration of the driven shaft, and
 θ= Angle through which the driving shaft turns.
We know that, for maximum angular acceleration of the driven shaft,
22
222sin 2sin 20cos2 0.124
2–s i n 2–s i n 2 0α°θ= = =
α°
∴              2 θ = 82.9°     or θ = 41.45°
and                        22
1
22 2cos .sin2 .sin
(1 – cos .sin )d
dtωω α θ α=
θα
22
2

dtωω α θ α=
θα
22
2
22 2(157) cos20 sin82.9 sin 203090 rad/s
(1 – cos 41.45 sin 20 )°× °× °==
°× ° Ans.
Maximum torque required
We know that maximum torque required
= I × d ω1 / dt = 0.12 × 3090 = 371 N-m Ans.

252  /G6C   Theory of Machines
Example. 9.3.  The angle between the axes of two shafts connected by Hooke’s joint is 18° .
Determine the angle turned through by the driving shaft when the velocity ratio is maximum andunity.
Solution.  Given : α = 98°
Let                θ  = Angle turned through by the driving shaft.
When the velocity ratio is maximum
We know that velocity ratio,
1
22cos
ω θα
The velocity ratio will be maximum when cos2 θ is minimum, i.e. when

cos2 θ = 1 or      when θ = 0°    or 180°  Ans.
When the velocity ratio is unity
The velocity ratio ( ω / ω1) will be unity, when
     1 – cos2 θ . sin2 α = cos α or2
21–c o scos
sinαθ=
α
∴                        221–c o s 1–c o s 1cos1c o s sin 1 – cosααθ=± =± =±+α αα
                                 110.71591 cos18 1 0.9510=± =± =±+° +
∴               θ  = 44.3° or     135.7°   Ans.
Example. 9.4.  Two shafts are connected by a H ooke’s joint. The driving shaft revolves

uniformly at 500 r.p.m. If the total permissible variation in speed of the driven shaft is not to exceed
± 6% of the mean speed, find the greatest permissible angle between the centre lines of the shafts.
Solution.  Given : N = 500 r.p.m. or ω = 2 π × 500 / 60 = 52.4 rad/s
Let α = Greatest permissible angle between the centre lines of the shafts.
Since the variation in speed of the driven shaft is ± 6% of the mean speed ( i.e. speed of the

driving speed), therefore total fluctuation of speed of the driven shaft,
q = 12 % of mean speed ( ω) = 0.12 ω
We know that maximum or total fluctuation of speed of the driven shaft ( q),
 21–c o s0.12cos αω=ωα    or   cos2 α + 0.12 cos α – 1 = 0
and                                 2–0.12 (0.12) 4 –0.12 2.0036cos 0.941822±+ ±α= = =
...(Taking + sign)

                                                                                                                                                                                 α = 19.64° Ans.
Example. 9.5.  Two shafts are connected by a universal joint. The driving shaft rotates at a
uniform speed of 1200 r.p.m. Determ ine the greatest permissible angle between the shaft axes so that

the total fluctuation of speed does not exceed 100 r.p. m. Also calculate the maximum and minimum
speeds of the driven shaft.

Chapter 9 : Mechanisms with Lower Pairs    /G6C  253
Solution.  Given : N = 1200 r.p.m.; q = 100 r.p.m.
Greatest permissible angle between the shaft axes
Let                          α  = Greatest permissible angle between the shaft axes.
We know that total fluctuation of speed ( q),
                                   221–c o s 1–c o s100 1200cos cosN  αα==  αα 
∴                     21 – cos 1000.083cos 1200α==α
  cos2 α + 0.083 cos α – 1 = 0

cos2 α + 0.083 cos α – 1 = 0
and                                         2–0.083 (0.083) 4cos 0.95932±+α= = ...(Taking + sign)
∴                          α  = 16.4° Ans.
Maximum and minimum speed of the driven shaft
We know that maximum speed of the driven shaft,
                            N1 (max) = N / cos α = 1200 / 0.9593 = 1251 r.p.m.   Ans.
                              N1 (min) = N  cos α = 1200 × 0.9593 = 1151 r.p.m.   Ans.

Example. 9.6.  The driving shaft of a H ooke’s joint runs at a uniform speed of 240 r.p.m. and
the angle α between the shafts is 20°. The driven shaft with attached masses has a mass of 55 kg at
a radius of gyration of 150 mm.
1. If a steady torque of 200 N-m resists rotation of the driven shaft, find the torque required
at the driving shaft, when θ = 45°.
2. At what value of ‘ α’will the total fluctuation of speed of the driven shaft be limited to 24
r.p.m ?

r.p.m ?
Solution. Given :  N = 240 r.p.m or ω = 2 π × 240/60 = 25.14 rad/s ; α = 20° ; m = 55 kg ;
k = 150 mm = 0.15 m ; T1 = 200 N-m ; θ = 45° ; q = 24 r.p.m.
1.  Torque required at the driving shaft
Let T′= Torque required at the driving shaft.
We know that mass moment inertia of the driven shaft,
I= m.k2 = 55 (0.15)2 = 1.24 kg-m2
and angular acceleration of the driven shaft,
                    22 2 2
1
22 2 2 22– cos .sin2 .sin –(25.14) cos20 sin90 sin 20
(1 – cos sin ) (1– cos 45 sin 20 )d

dtωω α θ α °× °× °==
θα ° °
                              = – 78.4 rad / s2
∴ Torque required to accelerate the driven shaft,
                         1
2 1.24 –78.4 –97.2N mdTI
dtω=× = × = −

254  /G6C   Theory of Machines
and total torque required on the driven shaft,
                          T  = T1 + T2 = 200 – 97.2 = 102.8 N– m
Since the torques on the driving and driven shafts are inversely proportional to their angular
speeds, therefore
                     T ' . ω = T . ω1
or                     1
22.c o s
1–c o s .s i nTTTωα′==ω θα1
2cos...
1–cos .sinωα=ω θα /G51
                        22102.8 cos 20102.6 N-m
1–c o s 4 5 s i n 2 0°==
°° Ans.

°° Ans.
2.  Value of ααααα for the total fluctuation of speed to be 24 r.p.m.
We know that the total fluctuation of speed of the driven shaft ( q),
                        221–c o s 1–c o s24 240cos cosN  αα==  αα 
or         21–c o s 2 40.1cos 240α==α
cos2 α + 0.1 cos α – 1 = 0
2–0.1 (0.1) 4cos 0.952±+α= = ...(Taking + sign)
∴                                     α  = 18.2°  Ans.

Example 9.7.  A double universal joint is used to connect two shafts in the same plane. The
intermediate shaft is inclined at an angle of 20 ° to the driving shaft as well as the driven shaft. Find
the maximum and minimum speed of the intermediate shaft and the driven shaft if the driving shafthas a constant speed of 500 r.p.m.
Solution. Given α = 20° ; N
A = 500 r.p.m.
Maximum and minimum speed of the intermediate shaft

Let  A, B  and C are the driving shaft, intermediate shaft and driven shaft respectively. We
know that for the driving shaft ( A) and intermediate shaft ( B),
Maximum speed of the intermediate shaft,
A
B( )500532.1r.p.mcos cos20maxNN== =α°Ans.
and minimum speed of the intermediate shaft,
NB (min) = NA cos α = 500 × cos 20° = 469.85 r.p.m.   Ans.
We know that for the intermediate shaft ( B) and driven shaft ( C),
Maximum speed of the driven shaft,

                                   B( ) A
C( ) 22500566.25 r.p.m.cos cos cos 20max
maxN NN== ==α α°Ans.

Chapter 9 : Mechanisms with Lower Pairs    /G6C  255
NC (min)= NB (min) × cos α = NA. cos2 α
= 500 × cos2 20° = 441.5 r.p.m. Ans.
EXERCISES
1.Fig. 9.22 shows the link GAB which oscillates on a fixed centre at A and the link FD on a fixed
centre at F. The link AB is equal to AC and DB, BE, EC and CD are equal in length.
Fig. 9.22
(a)Find the length of AF and the position of centre F so that the point E may move in a straight line.

(b)If  the  point  E  is  required  to move in a circle passing through centre A, what will be the path
of point D ? [Ans. AF = FD]
(Hint. The mechanism is similar to Peaucellier’s mechanism)
2.Fig. 9.23 shows a part of the mechanism of a circuit breaker. A and D are fixed centres and the lengths
of the links are : AB = 110 mm, BC = 105 mm, and CD = 150 mm.
            
Fig. 9.23 Fig. 9.24

Find the position of a point P on BC produced that will trace out an approximately straight vertical
path 250 mm long.
3.The mechanism, as shown in Fig. 9.24, is a four bar kinematic chain of which the centres A and B are
fixed. The lengths are : AB = 600 mm, AC = BD = CD = 300 mm. Find the point G on the centre line

of the cross arm of which the locus is an approximately straight line even for considerable displace-ments from the position shown in the figure. [Ans. 400 mm.]
(Hint : It is a Robert’s approximate straight line mechanism. Produce AC and BD to intersect at point
E. Draw a vertical line from E to cut the centre line of cross arm at G. The distance of G from CD is
the required distance).

256  /G6C   Theory of Machines
4.The distance between the fixed centres O and O1 of a Watt’s straight line motion, as shown in Fig. 9.6,
is 250 mm. The lengths of the three moving links OB, BA and AO1 are 150 mm, 75 mm and 100 mm
respectively. Find the position of a point P on BA which gives the best straight line motion.
5.A Watt’s parallel motion has two bars OA and O′B pivoted at O and O′ respectively and joined by the

link AB in the form of a crossed four bar mechanism. When the mechanism is in its mean position, the
bars OA and O′B are perpendicular to the link AB. If OA = 75 mm, O′B = 25 mm and AB = 100 mm,
find the position of the tracing point P and also find how far P is from the straight line given by the
mean position of AB, when
1.  OA and OB are in one straight line, and    2.  O′B and AB are in one straight line.
[Ans. 37.5 mm, 6.5 mm,12 mm]

6.Design a pantograph for an indicator to obtain the indicator diagram of an engine. The distance from
the tracing point of the indicator is 100 mm. The indicator diagram should represent four times the gaspressure inside the cylinder of an engine.
7.In a Davis steering gear, the distance between the pivots of the front axle is 1 metre and the wheel base

is 2.5 metres. Find the inclination of the track arm to the longitudinal axis of the car, when it is movingalong a straight path. [Ans. 11.17°]
8.A Hooke’s joint connects two shafts whose axes intersect at 150°. The driving shaft rotates uni-
formly at 120 r.p.m. T he driven sh aft operates against a steady torque of 150 N-m and carries a
flywheel whose mass is 45 kg and radius of gyration 150 mm. Find the maximum torque which will beexerted by the driving shaft. [Ans. 187 N-m]

(Hint : The maximum torque exerted by the driving shaft is the sum of steady torque and the maxi-
mum accelerating torque of the driven shaft).
9.Two shafts are connected by a Hooke’s joint. The driving shaft revolves uniformly at 500 r.p.m. If the

total permissible variation in speed of a driven shaft is not to exceed  6% of the mean speed, find thegreatest permissible angle between the centre lines of the shafts. Also determine the maximum andminimum speed of the driven shaft. [Ans. 19.6° ; 530 r.p.m. ; 470 r.p.m.]
10.Two inclined shafts are connected by means of a universal joint. The speed of the driving shaft is 1000

r.p.m. If the total fluctuation of speed of the driven shaft is not to exceed 12.5% of this, what is themaximum possible inclination between the two shafts?
With this angle, what will be the maximum acceleration to which the driven shaft is subjected and
when this will occur ? [Ans. 20.4° ; 1570 rad/s
2 ; 41.28°]
1.Sketch a pantograph, explain its working and show that it can be used to reproduce to an enlarged
scale a given figure.

2.A circle has OR as its diameter and a point Q lies on its circumference. Another point P lies on the line
OQ produced. If OQ turns about O as centre and the product OQ × OP remains constant, show that
the point P moves along a straight line perpendicular to the diameter OR.
3.What are straight line mechanisms ? Describe one type of exact straight line motion mechanism
with the help of a sketch.

4.Describe the Watt’s parallel mechanism for straight line motion and derive the condition under which
the straight line is traced.
5.Sketch an intermittent motion mechanism and explain its practical applications.
6.Give a neat sketch of the straight line motion ‘Hart mechanism.’ Prove that it produces an exact
straight line motion.
7.(a) Sketch and describe the Peaucellier straight line mechanism indicating clearly the conditions

under which the point P on the corners of the rhombus of the mechanism, generates a straight
line.
(b) Prove geometrically that the above mechanism is capable of producing straight line.

Chapter 9 : Mechanisms with Lower Pairs    /G6C  257
8.Draw the sketch of a mechanism in which a point traces an exact straight line. The mechanism must be
made of only revolute pairs. Prove that the point traces an exact straight line motion.(Hint. Peaucellier straight line mechanism)
9.Sketch the Dobbie-McInnes indicator mechanism and show that the displacement of the pencil
which traces the indicator diagram is proportional to the displacement of the indicator piston.

10.What is the condition for correct steering ? Sketch and show the two main types of steering gears
and discuss their relative advantages.
11.Explain why two Hooke’s joints are used to transmit motion from the engine to the differential of an
automobile.
12.Derive an expression for the ratio of shafts velocities for Hooke’s joint and draw the polar diagram
depicting the salient features of driven shaft speed.
1.In a pantograph, all the pairs are

(a) turning pairs ( b) sliding pairs
(c) spherical pairs ( d) self-closed pairs
2.Which of the following mechanism is made up of turning pairs ?
(a) Scott Russel’s mechanism ( b) Peaucellier’s mechanism
(c) Hart’s mechanism ( d) none of these
3.Which of the following mechanism is used to enlarge or reduce the size of a drawing ?
(a) Grasshopper mechanism ( b) Watt mechanism
(c) Pantograph ( d) none of these

4.The Ackerman steering gear mechanism is preferred to the Davis steering gear mechanism, because
(a)   whole of the mechanism in the Ackerman steering gear is on the back of the front wheels.
(b) the Ackerman steering gear consists of turning pairs
(c) the Ackerman steering gear is most economical
(d) both (a) and (b)
5.The driving and driven shafts connected by a Hooke’s joint will have equal speeds, if
(a) cos θ = sin α (b)sin tanθ=± α

(c)tan cosθ=± α (d) cot θ = cos α
     where θ=  Angle through which the driving shaft turns, and
α=  Angle of inclination of the driving and driven shafts.
ANSWERS
1. (a) 2. (b), (c) 3. (c) 4. (d) 5. (c)

258  /G6C   Theory of Machines
258Friction10
Features (Main)
2. Types of Friction.4. Friction Between Lubricated
Surfaces.
 5. Limiting Friction.
8. Laws of Solid Friction.9. Laws of Fluid Friction.
10. Coefficient of Friction.
11. Limiting Angle of Friction.
12. Angle of Repose.14. Friction of a Body Lying on
a Rough Inclined Plane.
15.  Efficiency of Inclined Plane.16. Screw Friction.17. Screw Jack.18. Torque Required to Lift the
Load by a Screw Jack.

20. Efficiency of a Screw Jack.21. Maximum Efficiency of a
Screw Jack.
22. Over Hauling and Self
Locking Screws.
23. Efficiency of Self Locking
Screws.
24. Friction of a V-thread.25. Friction in Journal Bearing-
Friction Circle.
26. Friction of Pivot and Collar
Bearing.
27. Flat Pivot Bearing.28. Conical Pivot Bearing.
29. Trapezoidal or Truncated
Conical Pivot Bearing.
30. Flat Collar Bearing.
31. Friction Clutches.

32. Single Disc or Plate Clutch.33. Multiple Disc Clutch.
34. Cone Clutch.
35. Centrifugal Clutches.10.1. Introduction
It has been established since long, that the surfaces

of the bodies are never perfectly smooth. When, even a verysmooth surface is viewed under a microscope, it is found tohave roughness and irregularities, which may not be detectedby an ordinary touch. If a block of one substance is placedover the level surface of the same or of different material, acertain degree of interlocking of the minutely projecting par-ticles takes place. This does not involve any force, so longas the block does not move or tends to move. But wheneverone block moves or

or tends to move. But wheneverone block moves or tends to move tangentially with respectto the surface, on which it rests, the interlocking property ofthe projecting particles opposes the motion. This opposingforce, which acts in the opposite direction of the movementof the upper block, is called the force of friction  or simply

friction. It thus follows, that at every joint in a machine, force
of friction arises due to the relative motion between two partsand hence some energy is wasted in overcoming the friction.Though the friction is considered undesirable, yet it plays animportant role both in nature and in engineering e.g. walk-
ing on a road, motion of locomotive on rails, transmission ofpower by belts, gears etc. The friction between the wheelsand the road is essential for the car to move forward.

10.2. Types of Friction
In general, the friction is of the following two types :

Chapter 10 : Friction    /G6C  259
1.Static friction . It is the friction, experienced by a body, when at rest.
2.Dynamic friction . It is the friction, experienced by a body, when in motion. The dynamic
friction is also called  kinetic friction  and is less than the static friction. It is of the following three
types :
(a)Sliding friction . It is the friction, experienced by a body, when it slides over another
body.

body.
(b)Rolling friction . It is the friction, experienced between the surfaces which has balls or
rollers interposed between them.
(c)Pivot friction.  It is the friction, experienced by a body, due to the motion of rotation  as
in case of foot step bearings.
The friction may further be classified as :
1. Friction between unlubricated surfaces, and
2. Friction between lubricated surfaces.
These are discussed in the following articles.
10.3. Friction Between Unlubricated Surfaces

The friction experienced between two dry and unlubricated surfaces in contact is known as
dry or solid friction . It is due to the surface roughness. The dry or solid friction includes the sliding
friction and rolling friction as discussed above.
10.4. Friction Between Lubricated Surfaces
When lubricant ( i.e. oil or grease) is applied between two surfaces in contact, then the friction

may be classified into the following two types depending upon the thickness of layer of a lubricant.
1. Boundary friction  (or greasy friction or non-viscous friction ). It is the friction,
experienced between the rubbing surfaces, when the surfaces have a very thin l ayer of lubri-
cant. The thickness of this very thin layer is of the molecular dimension. In this type of friction, a

thin layer of lubricant forms a bond between the two rubbing surfaces. The lubricant is absorbed onthe surfaces and forms a thin film. This thin film of the lubricant results in less friction betweenthem. The boundary friction follows the laws of solid friction.
2. Fluid friction  (or film friction or viscous friction ). It is the friction, experienced between

the rubbing surfaces, when the surfaces have a thick layer of the lubrhicant. In this case, the actualsurfaces do not come in contact and thus do not rub against each other. It is thus obvious that fluidfriction is not due to the surfaces in contact but it is due to the viscosity and oiliness of the lubricant.
Note : The viscosity is a measure of the resistance offered to the sliding one layer of the lubricant over an

adjacent layer. The absolute viscosity of a lubricant may be defined as the force required to cause a plate of unitarea to slide with unit velocity relative to a parallel plate, when the two plates are separated by a layer oflubricant of unit thickness.
The oiliness property of a lubricant may be clearly understood by considering two lubricants of equal

viscosities and at equal temperatures. When these lubricants are smeared on two different surfaces, it is foundthat the force of friction with one lubricant is different than that of the other. This difference is due to theproperty of the lubricant known as oiliness. The lubricant which gives lower force of friction is said to havegreater oiliness.
10.5. Limiting Friction
Consider that a body A of weight W is lying on a rough horizontal body B as shown in Fig.

10.1 (a). In this position, the body A is in equilibrium under the action of its own weight W, and the

260  /G6C   Theory of Machines
normal reaction RN (equal to W) of B on A. Now if a small horizontal force P1 is applied to the body
A acting through its centre of gravity as shown in Fig. 10.1 ( b), it does not move because of the
frictional force which prevents the motion. This shows that the applied force P1 is exactly balanced
by the force of friction F1 acting in the opposite direction.
If we now increase the applied force to P2 as shown in Fig. 10.1 ( c), it is still found to be in

equilibrium. This means that the force of friction has also increased to a value F2 = P2. Thus every
time the effort is increased the force of friction also increases, so as to become exactly equal to theapplied force. There is, however, a limit beyond which the force of friction cannot increase as shownin Fig. 10.1 ( d). After this, any increase in the applied effort will not lead to any further increase in the

force of friction, as shown in Fig. 10.1 ( e), thus the body A begins to move in the direction of the
applied force. This maximum value of frictional force, which comes into play, when a body just
begins to slide over the surface of the other body, is known as limiting force of friction or simply
limiting friction . It may be noted that when the applied force is less than the limiting friction, the body

remains at rest, and the friction into play is called static friction  which may have any value between
zero and limiting friction.
Fig. 10.1.  Limiting friction.
10.6. Laws of Static Friction
Following are the laws of static friction :
1.The force of friction always acts in a direction, opposite to that in which the body tends to
move.
2.The magnitude of the force of friction is exactly equal to the force, which tends the body
to move.

to move.
3.The magnitude of the limiting friction ( F ) bears a constant ratio to the normal reaction
(RN) between the two surfaces. Mathematically
        F/RN = constant

Chapter 10 : Friction    /G6C  261
4.The force of friction is independent of the area of contact, between the two surfaces.
5.The force of friction depends upon the roughness of the surfaces.
10.7. Laws of Kinetic or Dynamic Friction
Following are the laws of kinetic or dynamic friction :
1.The force of friction always acts in a direction, opposite to that in which the body is
moving.
2.The magnitude of the kinetic friction bears a constant ratio to the normal reaction between

the two surfaces. But this ratio is slightly less than that in case of limiting friction.
3.For moderate speeds, the force of friction remains constant. But it decreases slightly with
the increase of speed.
10.8. Laws of Solid Friction
Following are the laws of solid friction :
1.The force of friction is directly proportional to the normal load between the surfaces.
2.The force of friction is independent of the area of the contact surface for a given normal
load.

load.
3.The force of friction depends upon the material of which the contact surfaces are made.
4.The force of friction is independent of the velocity of sliding of one body relative to the
other body.
10.9. Laws of Fluid Friction
Following are the laws of fluid friction :
1.The force of friction is almost independent of the load.
2.The force of friction reduces with the increase of the temperature of the lubricant.

3.The force of friction is independent of the substances of the bearing surfaces.
4.The force of friction is different for different lubricants.
10.10. Coefficient of Friction
It is defined as the ratio of the limiting friction ( F) to the normal reaction ( RN) between the
two bodies. It is generally denoted by µ. Mathematically, coefficient of friction,
µ = F/RN
10.11. Limiting Angle of Friction
Consider that a body A of weight ( W) is resting on a horizontal plane B, as shown in Fig. 10.2.

If a horizontal force P is applied to the body, no relative motion will
take place until the applied force P is equal to the force of friction
F, acting opposite to the direction of motion. The magnitude of this
force of friction is F = µ.W = µ.RN, where RN is the normal reaction.
In the limiting case, when the motion just begins, the body will bein equilibrium under the action of the following three forces :
1.Weight of the body ( W),
2.Applied horizontal force ( P), and

3.Reaction ( R) between the body A and the plane B. Fig. 10.2.  Limiting angle of
friction.

262  /G6C   Theory of Machines
The reaction R must, therefore, be equal and opposite to the resultant of W and P and will be
inclined at an angle φ to the normal reaction RN. This angle φ is known as the limiting angle of friction .
It may be defined as the angle which the resultant reaction R makes with the normal reaction RN.
From Fig. 10.2, tan φ = F/RN = µRN/RN = µ
10.12. Angle of Repose
Consider that a body  A of weight ( W) is resting on

an inclined plane B, as shown in Fig. 10.3. If the angle of
inclination  α of the plane to the horizontal is such that the
body begins to move down the plane, then the angle α is
called the angle of repose .
A little consideration will show that the body will
begin to move down the plane when the angle of inclinationof the plane is equal to the angle of friction ( i.e. α = φ). This
may be proved as follows :
The weight of the body ( W) can be re-
solved into the following two components :

1. W  sin α, parallel to the plane B.
This component tends to slide the body downthe plane.
2. W cos α, perpendicular to the plane
B. This component is balanced by the normal
reaction ( R
N) of the body A and the plane B.
The body will only begin to move
down the plane, when
W sin α=F = µ.RN = µ.W cos α ...(∵ RN = W cos α)
∴ tan α=µ = tan φ    or   α = φ ...(∵ µ = tan φ)
10.13. Minimum Force Required to Slide a Body on a Rough Horizontal
  Plane

Plane
Consider that a body A of weight ( W) is resting on a
horizontal plane B as shown in Fig. 10.4. Let an effort P is
applied at an angle θ to the horizontal such that the body A
just moves. The various forces acting on the body are shownin Fig. 10.4. Resolving the force P into two components, i.e.
P sin θ acting upwards and P cos θ acting horizontally. Now
for the equilibrium of the body A,
R
N + P sin θ=W
or RN=W – P sin θ            ...(i)
and P cos θ=F = µ.RN               ... (ii)

...(∵ F = µ.RN)
Substituting the value of RN from equation (i), we have
P cos θ=µ (W – P sin θ) = tan φ (W – P sin θ) ...(∵ µ = tan φ)
()sinsincosWPφ=− θφFig. 10.3.  Angle of repose.
Friction is essential to provide grip between tyres
and road. This is a positive aspect of ‘friction’.
Fig. 10.4.  Minimum force required
to slide a body.

Chapter 10 : Friction    /G6C  263
P cos θ .cos φ=W sin φ – P sin θ.sin φ
P cos θ.cos φ + P sin θ.sin φ=W sin φ
P cos (θ – φ)=W sin φ ...[/G51cos θ. cos φ + sin θ.sin φ = cos (θ – φ)]
sin
cos ( )WPφ=θ−φ      ...(iii)
For P to be minimum, cos ( θ – φ) should be maximum, i.e.
cos (θ – φ) = 1 or θ – φ = 0° or θ = φ
In other words, the effort P will be minimum, if its inclination with the horizontal is equal to
the angle of friction.
∴ Pmin=W sin θ ...[From equation (iii)]

Example 10.1. A body, resting on a rough horizontal plane required a pull of 180  N inclined
at 30º to the plane just to move it. It was found that a push of 220  N inclined at 30º to the plane just
moved the body. Determine the weight of the body and the coefficient of friction.
Solution. Given : θ = 30º
Let W= Weight of the body in newtons,
RN= Normal reaction,
µ= Coefficient of friction, and
F= Force of friction.

First of all, let us consider a pull of 180 N. The force of friction ( F) acts towards left as shown
in Fig. 10.5 ( a).
Resolving the forces horizontally,
F= 180 cos 30º = 180 × 0.866 = 156 N
Fig. 10.5
Now resolving the forces vertically,
RN=W – 180 sin 30º = W – 180 × 0.5 = ( W – 90) N
We know that F=µ.RN     or    156 = µ (W – 90)   ... (i)
Now let us consider a push of  220 N. The force of friction ( F) acts towards right as shown in Fig.
10.5 (b).

10.5 (b).
F= 220 cos 30º = 220 × 0.866 = 190.5 N

264  /G6C   Theory of Machines
RN=W + 220 sin 30º = W + 220 × 0.5 = ( W + 110) N
We know that F=µ.RN     or    190.5 = µ (W + 110) ... (ii)
W= 1000 N,   and   µ = 0.1714   Ans.
10.14. Friction of a Body Lying on a Rough Inclined Plane
Consider that a body of weight ( W) is lying on a plane inclined at an angle α with the horizon-
tal, as shown in Fig. 10.6 ( a) and (b).

(a) Angle of inclination less than ( b) Angle of inclination more than
angle of friction. angle of friction.
Fig. 10.6. Body lying on a rough inclined plane.
A little consideration will show that if the inclination of the plane, with the horizontal, is less
than the angle of friction, the body will be in equilibrium as shown in Fig. 10.6 ( a). If,in this condi-

tion, the body is required to be moved upwards and downwards, a corresponding force is required forthe same. But, if the inclination of the plane is more than the angle of friction, the body will movedown and an upward force ( P) will be required to resist the body from moving down the plane as
shown in Fig. 10.6 ( b).
Let us now analyse the various forces which act on a body when it slides either up or down an
inclined plane.
1.  Considering the motion of the body up the plane

LetW= Weight of the body,
α= Angle of inclination of the plane to the horizontal,
φ= Limiting angle of friction for the contact surfaces,
P= Effort applied in a given direction in order to cause the body to slide with
uniform velocity parallel to the plane, considering friction,
P
0= Effort required to move the body up the plane neglecting friction,
θ= Angle which the line of action of P makes with the weight of the body W,

µ= Coefficient of friction between the surfaces of the plane and the body,
RN= Normal reaction, and
R= Resultant reaction.

Chapter 10 : Friction    /G6C  265
When the friction is neglected, the body is in equilibrium under the action of the three forces,
i.e. P0, W and RN, as shown in Fig. 10.7 ( a). The triangle of forces is shown in Fig. 10.7 ( b). Now
applying sine rule for these three concurrent forces,
                       0or
sin sin( )P W=
αθ − α *0sin
sin ( )WPα=θ−α
...(i)
(a)( b)( c)
Fig. 10.7.  Motion of the body up the plane, neglecting friction.

When friction is taken into account, a frictional force F = µ.RN acts in the direction opposite
to the motion of the body, as shown in Fig. 10.8 ( a). The resultant reaction R between the plane and
the body is inclined at an angle φ with the normal reaction RN. The triangle of forces is shown in Fig.
10.8 (b). Now applying sine rule,
sin( ) sin[ ( )]PW=α+φ θ− α+φ
(a)( b)( c)
Fig. 10.8. Motion of the body up the plane, considering friction.

*1.The effort P0 or (or P) may also be obtained by applying Lami’s theorem to the three forces, as
shown in Fig. 10.7 ( c) and 10.8 ( c). From Fig. 10.7 ( c),
0
sin(180º ) sin[180º ( )]P W=−α − θ−α
or                                                                        0
sin sin( )P W=αθ − α...[same as before]
2.The effort P0 (or P) may also be obtained by resolving the forces along the plane and perpendicular to
the plane and then applying ΣH = 0 and ΣV = 0.

266  /G6C   Theory of Machines
∴           sin( )
sin[ ( )]WPα+φ=θ− α+φ...(ii)
Notes : 1. When the effort applied is horizontal, then  θ = 90º. In that case, the equations (i) and (ii) may be
written as
 0sin sintansin(90º ) cosWWPWαα== = α−α α
and                                                 sin ( ) sin ( )tan ( )
sin[90º ( ) cos( )WWPWα+φ α+φ== = α + φ
−α + φ α + φ
2.  When the effort applied is parallel to the plane, then θ = 90º + α. In that case, the equations (i) and

(ii) may be written as
                                           0sinsin
sin(90º )WPWα== α
+α−α
and                                             sin( ) sin( )
sin[(90º ) ( )] cosWWPα+φ α+φ==
+α − α+φ φ
                                               (sin cos cos sin )(sin cos .tan )
cosWWαφ +αφ== α + α φ
φ
                                                       = W (sin α + µ cos α) ...(  ∵ µ = tan φ)
2.  Considering the motion of the body down the plane

Neglecting friction, the effort required for the motion down the plane will be same as for the
motion up the plane, i.e.
0sin
sin( )WPα=θ−α...(iii)
(a)( b)( c)
Fig. 10.9. Motion of the body down the plane, considering friction.
When the friction is taken into account, the force of friction F = µ.RN will act up the plane and
the resultant reaction R will make an angle φ with RN towards its right as shown in Fig. 10.9 ( a). The
triangle of forces is shown in Fig. 10.9 ( b). Now from sine rule,

sin( ) sin[ ( )]PW=α−φ θ− α−φ
or                sin ( )
sin[ ( )]WPα−φ=θ− α−φ...(iv)

Chapter 10 : Friction    /G6C  267
Notes : 1. The value of P may also be obtained either by applying Lami’s theorem to Fig. 10.9 ( c), or by
resolving the forces along the plane and perpendicular to the plane and then using ΣH = 0 and ΣV = 0 (See Art.
10.18 and 10.19).
2. When P is applied horizontally, then θ = 90º. In that case, equation ( iv) may be written as
                              sin( ) sin( )
sin[90º ( )] cos ( – )WWPα−φ α−φ==−α − φ α φ= W tan (α – φ))

3. When P is applied parallel to the plane, then θ = 90° + α. In that case, equation (iv) may be
written as
sin[90º ) ( )] cosWWPα−φ α−φ==
+α − α−φ φ
                                  (sin cos cos sin )(sin tan cos )
cosWWαφ −αφ== α − φ α
φ
                                   = W (sin α – µ cos α) ...( ∵ tan φ = µ)
10.15. Efficiency of Inclined Plane
The ratio of the effort required neglecting friction ( i.e. P0) to the effort required considering

friction (i.e. P) is known as efficiency of the inclined plane. Mathematically, efficiency of the inclined
plane,
                            0/PPη=
Let us consider the following two cases :
1. For the motion of the body up the plane
 Efficiency,        0sin sin[ ( )]
sin( ) sin( )PW
PWαθ − α + φη= = ×θ−α α+φ
                             sin sin cos( ) cos sin( )
sin cos cos sin sin( )αθ α + φ − θ α + φ=×θα −θα α + φ
Multiplying the numerator and denominator by sin ( α + φ) sin θ, we get

cot( ) cot
cot cotα+φ − θη=α− θ
Notes : 1. When effort is applied horizontally, then θ = 90°.
∴                           tan
tan ( )αη=
α+φ
2. When effort is applied parallel to the plane, then θ = 90º + α.
∴                           cot ( ) cot(90º ) cot( ) tan sin cos
cot cot(90º ) cot tan sin( )α+φ − +α α+φ + α α φη= = =
α− +α α+ α α+φ
2.  For the motion of the body down the plane

Since the value of P will be less than P0, for the motion of the body down the plane, therefore
in this case,
                            
0sin( ) sin( )
sin[ ( )] sinPW
PWα−φ θ−αη= = ×
θ− α−φ α
               sin( ) sin cos cos sin
s i n c o s () c o s s i n () s i nα−φ θ α− θ α=×θα − φ −θα − φ α

268  /G6C   Theory of Machines
Multiplying the numerator and denominator by sin ( α – φ) sin θ, we get
              cot cot
cot( ) cotα− θη=α−φ − θ
Notes : 1. When effort is applied horizontally, then θ = 90º.
∴              cot tan( )
cot( ) tanαα − φη= =α−φ α
∴              cot cot(90º ) cot tan sin( )
cot( ) cot(90º ) cot( ) tan sin cosα− +α α+ α α−φη= = =
α−φ − +α α−φ + α α φ

Example 10.2. An effort of 1500  N is required to just move a certain body up an inclined
plane of angle 12 º, force acting parallel to the plane. If the angle of inclination is increased to 15 º,
then the effort r equired is 1720  N. Find the weight of the body and the coefficient of friction .
Solution. Given : P1 = 1500 N ; α1 = 12º ; α2 = 15º ; P2 = 1720 N
Let        W = Weight of the body in newtons, and
      µ = Coefficient of friction.
(a)( b)
Fig. 10.10

(a)( b)
Fig. 10.10
First of all, let us consider a body lying on a plane inclined at an angle of 12º with the
horizontal and subjected to an effort of 1500 N parallel to the plane as shown in Fig. 10.10 ( a).
Let        RN1 = Normal reaction, and
         F1 = Force of friction.
We know that for the motion of the body up the inclined plane, the effort applied parallel to
the plane ( P1),
                        1500 = W (sin α1 + µ cos α1) = W (sin 12º + µ cos 12º) ... (i)

Now let us consider the body lying on a plane inclined at an angle of 15º with the horizontal
and subjected to an effort of 1720 N parallel to the plane as shown in Fig. 10.10 ( b).
Let                      RN2 = Normal reaction, and
         F2 = Force of friction.
the plane ( P2),
                         1720 = W (sin α2 + µ cos α2) = W (sin 15º + µ cos 15º) ... (ii)
Coefficient of friction

Dividing equation (ii) by equation (i),
                        1720 (sin15º cos15º)
1500 (sin12º cos12º)W
W+µ=+µ

Chapter 10 : Friction    /G6C  269
1720 sin 12º + 1720 µ cos 12º = 1500 sin 15º + 1500 µ cos 15º
µ (1720 cos 12º – 1500 cos 15º) = 1500 sin 15º – 1720 sin 12º
∴1500 sin15º 1720 sin 12º 1500 0.2588 1720 0.2079
1720 cos12º 1500 cos15º 1720 0.9781 1500 0.9659−× − ×µ= =−× − ×
    388.2 357.6 30.60.1311682.3 1448.5 233.8−== =−Ans.
Weight of the body
Substituting the value of µ in equation  (i),
      1500 = W (sin 12º + 0.131 cos 12º)
         = W (0.2079 + 0.131 × 0.9781) = 0.336 W

= W (0.2079 + 0.131 × 0.9781) = 0.336 W
∴       W = 1500/0.336 = 4464 N  Ans.
10.16. Screw Friction
The screws, bolts, studs, nuts etc. are widely used in various machines and structures for

temporary fastenings. These fastenings have screw threads, which are made by cutting a continuoushelical groove on a cylindrical surface. If the threads are cut on the outer surface of a solid rod, theseare known as external threads . But if the threads are cut on the internal surface of a hollow rod, these
are known as internal threads . The screw threads are mainly of two types i.e. V-threads and square

threads. The V-threads are stronger and offer more frictional resistance to motion than square threads.Moreover, the V-threads have an advantage of preventing the nut from slackening. In general, the V-threads are used for the purpose of tightening pieces together e.g. bolts and nuts etc. But the square
threads are used in screw jacks, vice screws etc. The following terms are important for the study ofscrew :

1.  Helix. It is the curve traced by a particle, while describing a circular path at a uniform
speed and advancing in the axial direction at a uniform rate. In other words, it is the curve traced bya particle while moving along a screw thread.Jet engine used in Jet aircraft.

270  /G6C   Theory of Machines
2.  Pitch. It is the distance from a point of a screw to a corresponding point on the next thread,
measured parallel to the axis of the screw.
3.  Lead. It is the distance, a screw thread advances axially in one turn.
4.  Depth of thread . It is the distance between the top and bottom surfaces of a thread (also
known as crest and root of a thread).
5.  Single-threaded screw . If the lead of a screw is equal to its pitch, it is known as single
threaded screw.

6.  Multi-threaded screw. If more than one thread is cut in one lead distance of a screw, it is
known as multi-threaded screw e.g. in a double threaded screw, two threads are cut in one lead length.
In such cases, all the threads run independently along the length of the rod. Mathematically,
                          Lead = Pitch × Number of threads7. Helix angle. It is the slope or inclination of the thread with
the horizontal. Mathematically,

                        Lead of screwtan
Circumference of screwα=
                = p/πd           ...(In single-threaded screw)
                 = n.p/πd          ...(In multi-threaded screw)
where             α = Helix angle,
            p = Pitch of the screw,
                                d  = Mean diameter of the screw, and
                               n   = Number of threads in one lead.
10.17. Screw Jack

The screw jack is a device, for lifting heavy loads, by apply-
ing a comparatively smaller effort at its handle. The principle, onwhich a screw jack works is similar to that of an inclined plane.
(a) Screw jack. ( b) Thrust collar.
Fig. 10.11Screw Jack.

Chapter 10 : Friction    /G6C  271
Fig. 10.11 ( a) shows a common form of a screw jack, which consists of a square threaded rod
(also called screw rod or simply screw) which fits into the inner threads of the nut. The load, to beraised or lowered, is placed on the head of the square threaded rod which is rotated by the applicationof an effort at the end of the lever for lifting or lowering the load.
10.18. Torque Required to Lift the Load by a Screw Jack

If one complete turn of a screw thread by imagined to be unwound, from the body of the
screw and developed, it will form an inclined plane as shown in Fig. 10.12 ( a).
 (a) Development of a screw. ( b) Forces acting on the screw.
Fig. 10.12
Let p= Pitch of the screw,
d= Mean diameter of the screw,
α= Helix angle,
P= Effort applied at the circumference of the screw to lift the
load,
W= Load to be lifted, and
µ= Coefficient of friction, between the screw and nut = tan φ,

where φ is the friction angle.
From the geometry of the Fig. 10.12 ( a), we find that
tan α=p/π d
Since the principle on which a screw jack works is similar to that of an inclined plane, there-
fore the force applied on the lever of a screw jack may be considered to be horizontal as shown in Fig.10.12 (b).
Since the load is being lifted, therefore the force of friction ( F = µ.R
N) will act downwards.
All the forces acting on the screw are shown in Fig. 10.12 ( b).

Resolving the forces along the plane,
P cos α=W sin α + F = W sin α + µ.RN...(i)
and resolving the forces perpendicular to the plane,
RN=P sin α + W cos α ...(ii)
Substituting this value of RN in equation ( i),
P cos α=W sin α + µ (P sin α + W cos α)
=W sin α + µ P sin α + µ W cos α
or P cos α – µ P sin α=W sin α + µ W cos α
or P (cos α – µ sin α)=W (sin α + µ cos α)

272  /G6C   Theory of Machines
∴sin cos
cos sinPWα+µ α=×α−µ α
Substituting the value of µ = tan φ in the above equation, we get
sin tan cos
cos tan sinPWα+ φ α=×α− φ α
Multiplying the numerator and denominator by cos φ,
sin cos sin cos sin( )
cos cos sin sin cos( )PW Wαφ +φα α + φ=× =×αφ −αφ α + φ
tan( )W=α + φ
∴ Torque required to overcome friction between the screw and nut,
1tan( )22ddTP W=×= α + φ

When the axial load is taken up by a thrust collar or a flat surface, as shown in Fig. 10.11 ( b),
so that the load does not rotate with the screw, then the torque required to overcome friction at thecollar,
12
21 1.. .
2RRTW W R+=µ =µ
where R1 and R2= Outside and inside radii of the collar,
R= Mean radius of the collar, and
µ1= Coefficient of friction for the collar.
∴ Total torque required to overcome friction ( i.e. to rotate the screw),

12 1..2dTTT P W R=+=×+ µ
If an effort P1 is applied at the end of a lever of arm length l, then the total torque required to
overcome friction must be equal to the torque applied at the end of the lever, i.e.
1.2dTP P l=×=
Notes : 1. When the  *nominal diameter ( d0) and the **core diameter ( dc) of the screw thread is given, then the
mean diameter of the screw,
                                               0
022 2c
cdd ppdd d+== − = +

022 2c
2. Since the mechanical advantage is the ratio of load lifted ( W) to the effort applied ( P1) at the end of
the lever, therefore mechanical advantage,
                                           
12...WW lMAPp d×==... 1.
2PdPl=/G51
                                                    22
tan ( ) .tan( )Wl l
Wd d×==α+φ α+φ
Example 10.3. An electric motor driven power screw moves a nut in a horizontal plane

against a force of 75 kN at a speed of 300  mm/min. The screw has a single square thread of 6 mm
pitch on a major diameter of 40  mm. The coefficient of friction at the screw threads is 0.1 . Estimate
power of the motor.
*The nominal diameter  of a screw thread is also known as outside diameter  or major diameter.
**The core diameter  of a screw thread is also known as inner diameter or root diameter  or minor diameter.

Chapter 10 : Friction    /G6C  273
Solution. Given : W = 75 kN = 75 × 103 N ; v = 300 mm/min ; p = 6 mm ; d0 = 40 mm ;
µ = tan φ = 0.1
We know that mean diameter of the screw,
d= d0 – p/2 = 40 – 6/2 = 37 mm = 0.037 m
and                                     6tan 0.051637p
dα= = =ππ ×
∴ Force required at the circumference of the screw,
                                            tan tantan( )1 tan .tanPW Wα+ φ=α + φ =−αφ

330.0516 0.175 10 11.43 10 N1 0.0516 0.1+ =× = ×−×
and torque required to overcome friction,
T= P × d/2 = 11.43 × 103 × 0.037/2 = 211.45 N-m
We know that speed of the screw,
Speed of the nut 30050 r.p.m.Pitch of the screw 6N== =
and angular speed,        ω= 2 π × 50/60 = 5.24 rad/s
∴ Power of the motor = T.ω = 211.45 × 5.24 = 1108 W = 1.108 kW Ans.
Example 10.4. A turnbuckle , with right

and left hand single start threads, is used to coupletwo wagons. Its thread pitch is 12  mm and mean
diameter 40  mm. The coefficient of friction between
the nut and screw is  0.16.
1. Determine the work done in drawing the
wagons together a distance of 240 mm, against asteady load of 2500 N.
2. If the load increases from 2500 N to 6000
N over the distance of 240  mm, what is the work to
be done?
Solution. Given : p = 12 mm ; d = 40 mm ;

µ = tan φ = 0.16 ; W = 2500 N
1.  Work done in drawing the wagons together against a steady load of 2500 N
We know that               
12tan 0.095540p
dα= = =ππ ×
∴ Effort required at the circumference of the screw,
  tan tantan ( )1t a n. t a nPW Wα+ φ=α + φ =−αφ
Turnbuckle.

274  /G6C   Theory of Machines
                 0.0955 0.162500 648.7 N1 0.0955 0.16+ ==−×
and torque required to overcome friction between the screw and nut,
                                            /2 648.7 40/2 12 947 N-mm 12.974 N-m TP d=× = × = =
A little consideration will show that for one complete revolution of the screwed rod, the
wagons are drawn together through a distance equal to 2 p, i.e. 2 × 12 = 24 mm. Therefore in order to

draw the wagons together through a distance of 240 mm, the number of turns required are given by
N= 240/24 = 10
∴ Work done = T × 2 π N = 12.974 × 2 π × 10 = 815.36  N-m  Ans.
2.  Work done in drawing the wagons together when load increases from 2500 N to 6000 N
For an increase in load from 2500 N to 6000 N,
                              815.3(6000 2500)Work done = 114.4 N-m
2500−= Ans.
Example 10.5. A 150 mm diameter valve, against which a steam pressure of 2 MN/m2 is

acting, is closed by means of a square threaded screw 50 mm in external diameter with 6 mm pitch.If the coefficient of friction is 0.12 ; find the torque requi red to turn the handle.
Solution. Given : D = 150 mm = 0.15 mm = 0.15 m ;  Ps = 2 MN/m
2 = 2 × 106 N/m2 ;
d0 = 50 mm ; p = 6 mm ; µ = tan φ = 0.12
We know that load on the valve,
W= Pressure × Area = 26 2
S2 10 (0.15) N44pDππ×= × ×
= 35 400 N
Mean diameter of the screw,
d= d0 – p/2 = 50 – 6/2 = 47 mm = 0.047 m

∴                                  6tan 0.040647p
dα= = =ππ ×
We know that force required to turn the handle,
                                            tan tantan( )1t a n. t a nPW Wα+ φ=α + φ =−αφ
                                                0.0406 1235400 5713 N10 . 0 4 0 60 . 1 2+ ==−×
∴  Torque required to turn the handle,
T= P × d/2 = 5713 × 0.047/2 = 134.2 N-m   Ans.

Example 10.6. A square threaded bolt of root diameter 22.5 mm and pitch 5  mm is tightened
by screwing a nut whose mean diameter of bearing surf ace is 50 mm. If coefficient of friction for nut
and bolt is 0.1 and for nut and bearing surface 0.16, find the force required at the end of a spanner500 mm long when the load on the bolt is 10 kN.
Solution. Given : d
c = 22.5 mm ; p = 5 mm ;  D = 50 mm or R = 25 mm ; µ = tan φ = 0.1 ;

µ1 = 0.16 ; l = 500 mm ; W = 10 kN = 10 × 103 N
Let P1= Force required at the end of a spanner in newtons.

Chapter 10 : Friction    /G6C  275
                                            /2 22.5 5/2 25 mmcdd p=+ = + =
∴                                  5tan 0.063625p
dα= = =ππ ×
Force requred at the circumference of the screw,
                                                30.0636 0.110 10 1646 N1 0.06363 0.1+ =× =−×
We know that total torque required,

                                            3
125...1 6 4 6 0 . 1 61 01 0 2 522dTP W R=×+ µ = × + ×× ×
 60575 N - mm= ..(i)
We also know that torque required at the end of a spanner,
T= P1 × l = P1 × 500 = 500 P1 N-mm ... (ii)
P1= 60575/500 = 121.15 N  Ans.
Example 10.7. A vertical screw with single start square threads 50 mm mean diameter and

12.5 mm pitch is raised against a load of 10 kN by means of a hand wheel, the boss of which isthreaded to act as a nut. The axial load is taken up by a thrust collar which supports the wheel bossand has a mean diameter of 60 mm. If the coefficient of friction is 0.15 for the screw and 0.18 for thecollar and the tangential force applied by each hand to the wheel is 100 N ; find suitable diameter ofthe hand wheel.
Solution. Given : d = 50 mm ; p = 12.5 mm ; W = 10 kN = 10 × 10
3 N ; D = 60 mm or

R = 30 mm ; µ = tan φ = 0.15 ; µ1 = 0.18 ; P1 = 100 N
We know that   12.5tan 0.0850p
dα= = =ππ ×
and the tangential force required at the circumference of the screw,
                             tan tantan( )1t a n. t a nPW Wα+ φ=α + φ =−α φ
                                    30.08 0.1510 10 2328 N1 0.08 0.15+=× =−×
Also we know that the total torque required to turn the hand wheel,
                               3

3
150. . 2328 0.18 10 10 3022dTP W R=×+ µ = × + ×× ×
                                    112200 N-mm= ...(i)
Let D1 = Diameter of the hand wheel in mm.
We know that the torque applied to the hand wheel,
                                11
112 2 100 100 N-mm22DDTP D=×= ××= ...(ii)

276  /G6C   Theory of Machines
                                D1 = 112 200/100 = 1222 mm = 1.222 m  Ans.
Example 10.8. The cutter of a broaching machine is pulled by square threaded screw of 55

mm external diameter and 10 mm pitch. The operating nut takes the axial load of 400 N on a flatsurface of 60 mm internal diameter and 90 mm external diameter. If the coefficient of firction is 0.15for all contact surfaces on the nut, determine the power required to rotate the operating nut, whenthe cutting speed is 6 m/min.
0 = 55 mm ; p = 10 mm = 0.01 m ; W = 400 N ; D2 = 60 mm   or
R2 = 30 mm ; D1 = 90 mm or R1 = 45 mm ; µ = tan φ = µ1 = 0.15

                                  d  = d0 – p/2 = 55 – 10/2 = 50 mm
∴                      10tan 0.063750p
dα= = =ππ ×
and force required at the circumference of the screw,
                                tan tantan( )1 tan .tanPW Wα+ φ=α + φ =−αφ
                                    0.0637 0.15400 86.4 N10 . 0 6 3 70 . 1 5+ ==−×
We know that mean radius of the flat surface,
                                12 45 3037.5 mm22RRR+ +===

∴ Total torque required,
                                            150. . 86.4 0.15 400 37.5 N-mm22dTP W R=×+ µ = × + × ×
              4410 N-mm 4.41 N-m== ...(∵ µ1 = µ)
Since the cutting speed is 6 m/min, therefore speed of the screw,
            Cutting speed 6600 r.p.m.
Pitch 0.01N== =
and                angular speed, ω = 2 π × 600/60 = 62.84 rad/s
We know that power required to operate the nut
            . 4.41 62.84 277 W 0.277 kWT=ω = × = = Ans.

10.19. Torque Required to Lower the Load by a Screw Jack
We have discussed in Art. 10.18, that the principle on which the screw jack works is similar
to that of an inclined plane. If one complete turn of a screw thread be imagined to be unwound fromthe body of the screw and developed, it will form an inclined plane as shown in Fig. 10.13 ( a).
P= Effort applied at the circumference of the screw to lower the
load,

Chapter 10 : Friction    /G6C  277
W= Weight to be lowered, and
µ= Coefficient of friction between the screw and nut = tan φ,
(a)( b)
Fig. 10.13
From the geometry of the figure, we find that
tan α=p/πd
Since the load is being lowered, therefore the force of friction ( F = µ.RN) will act upwards.
All the forces acting on the screw are shown in Fig. 10.13  ( b).
P cos α=F – W sin α = µ.RN – W sin α ...(i)

RN=W cos α – P sin α ...(ii)
Substituting this value of RN in equation  (i),
P cos α=µ (W cos α – P sin α) – W sin α
=µ.W cos α – µ.P sin α – W sin α
or P cos α + µ.P sin α=µ.W cos α – W sin α
or P (cos α + µ sin α)=W (µ cos α – sin α)
∴                                        (c o s s i n)
(cos sin )PWµα −α=×α+µ α

(tan cos sin )
(cos tan sin )PWφα −α=×α+ φ α
                                           (sin cos sin cos ) sin( )
(cos cos sin sin ) cos ( )PW Wφα −αφ φ − α=× =×αφ +φα φ − α
                                              tan ( )W=φ − α
tan ( )22ddTP W=×= φ − α
Note : When α > φ, then P = tan (α – φ).

Example 10.9. The  mean diameter of a square threaded screw jack is 50 mm. The pitch of
the thread is 10 mm. The coefficient of friction is 0. 15. What force must be applied at the end of a
0.7 m long lever, which is perpendi cular to the longitudinal axis of the screw to raise a load of 20 kN
and to lower it?

278  /G6C   Theory of Machines
Solution. Given : d = 50 mm = 0.05 m ; p = 10 mm ; µ = tan φ = 0.15 ; l = 0.7 m ; W = 20 kN
= 20 × 103 N
We know that        10tan 0.063750p
dα= = =ππ ×
Let P1= Force required at the end of the lever.
Force required to raise the load
We know that force required at the circumference of the screw,
                                           tan tantan( )1t a n. t a nPW Wα+ φ=α + φ =−αφ

30.0637 0.1520 10 4314 N1 0.0637 0.15+ =× =−×
Now the force required at the end of the lever may be found out by the relation,
P1 × l=P × d/2
∴14314 0.05154 N22 0 . 7PdPl××== =×Ans.
Force required to lower the load
We know that the force required at the circumference of the screw,
tan tantan( )1 tan .tanPW Wφ− α=φ − α =+φα
30.15 0.063720 10 1710 N1 0.15 0.0637−=× =+×

                                     111710 0.05or 61 N22 2 0 . 7dP dPlP Pl×××= × = = =×Ans.
10.20. Efficiency of a Screw Jack
The efficiency of a screw jack may be defined as the ratio between the ideal ef fort (i.e. the
effort required to move the load, neglecting friction) to  the actual effort  (i.e. the effort required to
move the load taking friction into account).

We know that the effort required to lift the load ( W) when friction is taken into account,
P=W tan (α + φ) ... (i)
where α= Helix angle,
φ= Angle of friction, and
µ= Coefficient of friction, between the screw and nut = tan φ.
If there would have been no friction between the screw and the nut, then φ will be equal to
zero. The value of effort P0 necessary to raise the load, will then be given by the equation,

P0=W tan α (i.e. Putting φ = 0 in equation (i)]
                       0 Idealeffort tan tanEfficiency,Actualeffort tan ( ) tan ( )PW
PWαα∴η = = = =α+φ α+φ
which shows that the efficiency of a screw jack, is independent of the load raised.

Chapter 10 : Friction    /G6C  279
In the above expression for efficiency, only the screw friction is considered. However, if the
screw friction and the collar friction is taken into account, then
∴                  Torque required to move the load, neglecting friction
Torque required to move the load, including screw and collar frictionη=
                         00
1/2
/2 . .TP d
TP d W R×==×+ µ

1/2
/2 . .TP d
Note: The efficiency of the screw jack may also be defined as  the ratio of mechanical advantage to the
velocity ratio.
We know that mechanical advantage,
                  
1222..
tan( ) tan( )WW l W l lMA
PP dW d d××== = =
×α + φ α + φ...(Refer Art 10.17)
and velocity ratio,   1Distance moved by the effort ( ), in one revolution..
Distance moved by the load ( ), in one revolutionPVR
W=
                           22 2
tan tanll l
pd dππ== =

tan tanll l
pd dππ== =
α×π α×...(/G8C tan α = p/πd)
∴  Efficiency,  .. 2 t a n t a n
. . tan( ) 2 tan( )MA l d
VR d lα× ×αη= = × =α+φ α+φ
10.21. Maximum Efficiency of a Screw Jack
We have seen in Art. 10.20 that the efficiency of a screw jack,
sin
tan sin cos( ) cos
sin( ) tan( ) cos sin( )
cos( )α
αα × α + φ αη= = =α+φ α+θ α× α+φ
α+φ...(i)
    2s i n c o s( )
2c o s s i n( )α× α+φ=α× α+φ
...(Multiplying the numerator and denominator by 2)
sin(2 ) sin
sin(2 ) sinα+φ − φ=α+φ + φ...(ii)

sin(2 ) sin
2sin cos sin( ) sin ( )...2 cos sin sin ( ) sin ( )A B AB AB
AB A B A B=+ +− 
=+ −− /G51
The efficiency given by equation (ii) is maximum when sin (2 α + φ) is maximum, i.e. when
sin (2α + φ) = 1    or   when 2 α + φ = 90°
∴ 2α= 90º – φ    or    α = 45º – φ/2
Substituting the value of 2 α in equation  (ii), we have maximum efficiency,
                                     sin(90º ) sin sin 90º sin 1 sin

sin(90º ) sin sin 90º sin 1 sinmax− φ + φ −φ −φ −φη= = =− φ + φ +φ +φ+φ
Example 10.10. The pitch of 50 mm mean diameter threaded screw of a screw jack is 12.5
mm.  The coefficient of friction betw een the screw and the nut is 0.13. Determine the torque required
on the screw to raise a load of 25 kN, assuming the l oad to rotate with the screw. Determine the ratio
of the torque required to raise the load to the torque required to lower the load and also the efficiencyof the machine.

280  /G6C   Theory of Machines
Solution. Given : d = 50 mm ; p = 12.5 mm ; µ = tan φ = 0.13 ; W = 25 kN = 25 × 103 N
We know that,         12.5tan 0.0850p
dα= = =ππ ×
and force required on the screw to raise the load,
                                      tan tantan( )1 tan .tanPW Wφ− α=α + φ =+φα
                                          30.08 0.1325 10 5305 N1 0.08 0.13+=× =−×
Torque required on the screw

We know that the torque required on the screw to raise the load,                                   T
1 = P × d/2 = 5305 × 50/2 = 132 625 N-mm  Ans.
Ratio of the torques required to raise and lower the load
We know that the force required on the screw to lower the load,
                                      tan tantan( )1 tan .tanPW Wφ− α=φ − α =+φα
                                          30.13 0.0825 10 1237 N1 0.13 0.08+=× =+×

and torque required to lower the load
                                                T2 = P × d/2 = 1237 × 50/2 = 30 905 N-mm
∴  Ratio of the torques required,
                                        12/ 132625/30925 4.3TT== = Ans.
We know that the efficiency,
                                          tan tan (1 tan .tan ) 0.08(1 0.08 0.13)
tan( ) tan tan 0.08 0.13αα − α φ − ×η= = =α+φ α+ φ +
                                          = 0.377 = 37.7%  Ans.

Example 10.11.  The mean diameter of the screw jack having pitch of 10 mm is 50 mm. A
load of 20 kN is lifted through a distance of 170 mm. Find the work done in lifting the load andefficiency of the screw jack when
1.  the load rotates with the screw, and
2.  the load rests on the loose head which does not rotate with the screw.
The external and internal diameter of the bearing surface of the loose head are 60 mm and

10 mm respectively. The coefficient of friction for the screw as well as the bearing surface may be
taken as 0.08.
Solution.  Given : p = 10 mm ; d = 50 mm ; W = 20 kN = 20 × 10
3 N ; D2 = 60 mm or
R2 = 30 mm ; D1 = 10 mm or R1 = 5 mm ; µ = tan φ = µ1 = 0.08
We know that          10tan 0.063750p
dα= = =ππ ×

Chapter 10 : Friction    /G6C  281
∴  Force required at the circumference of the screw to lift the load,
tan tantan( )1t a n . t a nPW Wα+ φ=α + φ =−αφ
30.0637 0.0820 10 2890 N1 0.0637 0.08+ =× =−×
and torque required to overcome friction at the screw,
                              /2 2890 50/2 72250 N-mm 72.25 N-m TPd=× = × = =
Since the load is lifted through a vertical distance of 170 mm and the distance moved by the

screw in one rotation is 10 mm (equal to pitch), therefore number of rotations made by the screw,
N = 170/10 = 17
1.  When the load rotates with the screw
We know that work done in lifting the load
                                    2 72.25 2 17 7718 N-mTN=× π= × π × = Ans.
and efficiency of the screw jack,
                                tan tan (1 tan .tan )
tan( ) tan tanαα − α φη= =α+φ α+ α
                                    0.0637(1 0.0637 0.08)0.441 or 44.1%0.0637 0.08−×==+Ans.

2.  When the load does not rotate with the screw
We know that mean radius of the bearing surface,
                              12 30 517.5 mm22RRR+ +== =
and torque required to overcome friction at the screw and the collar,
1
3/2 . .
2890 50/2 0.08 20 10 17.5 100 250 N-mm
=100.25 N-mTPd W R=× + µ
=×+ × × × =
∴ Work done by the torque in lifting the load
                                   2 100.25 2 17 10 710 N-mTN=× π= × π × = Ans.

We know that the torque required to lift the load, neglecting friction,
00/2 tan /2 TP d W d=× = α × ...(/G51P0 = W tan α)
= 20 × 103 × 0.0637 × 50/2 = 31 850 N-mm = 31.85 N-m
∴  Efficiency of the screw jack,
0/ 31.85/100.25 0.318 or 31.8%TTη= = = Ans.
10.22. Over Hauling and Self Locking Screws
We have seen in Art. 10.20 that the effort required at the circumference of the screw to lower
the load is
P = W tan (φ – α)

282  /G6C   Theory of Machines
and the torque required to lower the load
In the above expression, if φ < α, then torque required to lower the load will be negative. In
other words, the load will start moving downward without the application of any torque. Such acondition is known as over haulding of screws. If however, φ > α, the torque required to lower the
load will positive, indicating that an effort is applied to lower the load. Such a screw is known as self

locking screw. In other words, a screw will be self locking if the friction angle is greater than helix
angle or coefficient of friction is greater than tangent of helix angle i.e. µ or tan φ > tan α.
10.23. Efficiency of Self Locking Screws
We know that efficiency of the screw,
tan
tan( )αη=α+φ
and for self locking screws,  or .φ≥α α≤φ
∴ Efficiency of self locking screws,
                                            2tan tan tan (1 tan )
tan( ) tan 2 2 tanφφ φ − φη≤ ≤ ≤φ+φ φ φ
 21t a n

 21t a n
22φ≤− 22t a ntan 2...1t a nφ φ=−φ /G51
From this expression we see that efficiency of self locking screws is less than 1
2 or 50%. If
the efficiency is more than 50%, then the screw is said to be overhauling,
Note : It can also be proved as follows :
Let W= Load to be lifted, and
h= Distance through which the load is lifted.
∴ Output = W.h
and                                           Input = Output . Wh=ηη

∴  Work lost in over coming friction.
                                                    1 .1 Input Output . .WhWh Wh− =− = − = η η 
For self locking,,  11..Wh Wh−≤η
∴                                       111 1 or or 50%2−≤ η ≤η
Example 10.12. A load of 10 kN is raised by means of a screw jack, having a square threaded

screw of 12 mm pitch and of mean diameter 50 mm. If a force of 100 N is applied at the end of a leverto raise the load, what should be the length of the lever used? T ake coefficient of friction = 0.15.
What is the mechanical advantage obtained? State whether the screw is self locking.
Solution.  Given : W = 10 kN = 10 × 10
3 N ; p = 12 mm ; d = 50 mm ; P1 = 100 N ;
µ = tan φ = 0.15
Length of the lever
Let l= Length of the lever.

Chapter 10 : Friction    /G6C  283
We know that       12tan 0.076450p
dα= = =ππ ×
∴ Effort required at the circumference of the screw to raise the load,
                                 tan tantan( )1 tan .tanPW Wα+ φ=α + φ =−αφ
                                     30.0764 0.1510 10 22901 0.0764 0.15+ =× =−×N
                                    T = P × d/2 = 2290 × 50/2 = 57 250 N-mm ... (i)

We know that torque applied at the end of the lever,                                   T = P
1 × l = 100 × l N-mm ... (ii)
Equating equations  (i) and (ii)
                                     l = 57 250/100 = 572.5 mm  Ans.
Mechanical advantage
                             3
110 10. . 100100WMAP×== = Ans.
Self locking of the screw
We know that efficiency of the screw jack,
                                   tan tan (1 tan .tan )

tan( ) tan tanαα − α φη= =α+φ α+ φ
                                       0.0764(1 0.0764 0.15) 0.07550.3335 or 33.35%0.0764 0.15 0.2264−×== =+
Since the efficiency of the screw jack is less than 50%, therefore the screw is a self locking
screw.Ans.
10.24. Friction of a V-thread
We have seen Art. 10.18 that the normal reaction in case of a square threaded screw is
                                 RN = W cos α, where α = Helix angle.

But in case of  V-thread (or acme or trapezoidal threads), the normal
reaction between the screw and nut is increased because the axial component ofthis normal reaction must be equal to the axial load W, as shown in Fig. 10.14.
Let                             2 β = Angle of the V-thread, and
                                   β  = Semi-angle of the V-thread.
∴                             
NcosWR=β
and                frictional force, N1..cosWFR W=µ =µ× =µβ

where                                      1,cosµ=µβ known as virtual coefficient of friction.Fig. 10.14.  V-thread.

284  /G6C   Theory of Machines
Notes : 1. When coefficient of friction,1cosµµ=
β is considered, then the V-thread is equivalent to a square
thread.
2. All the equations of square threaded screw also hold good for V-threads. In case of V-threads, µ1
(i.e. tan φ1) may be substituted in place of µ (i.e. tan φ). Thus for V-threads,
1tan ( )PW=α ± φ
where   φ1 =Virtual friction angle, such that tan φ1 = µ1.
Example 10.13. Two co-axial rods are connected by a turn buckle which consists of a box

nut, the one screw being right handed and the other left handed on a pitch diameter of 22 mm, thepitch of thread being 3 mm. The included angle of the thread is 60º. Assuming that the rods do notturn, calculate the torque required on the nut to produce a pull of 40 kN, given that the coeffi cient of
friction is 0.15.
Solution. Given : d = 22 mm ; p = 3 mm ; 2 β = 60º or β = 30º, W = 40 kN = 40 × 10
3 N ; µ = 0.15
We know that         3tan 0.043422p
dα= = =ππ ×

dα= = =ππ ×
and virtual coefficient of friction
                                           110.15tan 0.173cos cos 30ºµµ= φ= = =β
              1
1
1tan tan
tan( )1 tan .tanPW Wα+ φ=α + φ = −αφ
30.0434 0.17340 10 8720 N10 . 0 4 3 40 . 1 7 3+ =× =−×
and torque on one rod,  T = P × d/2 = 8720 × 22/2 = 95 920 N-mm = 95.92 N-m

Since the turn buckle has right and left hand threads and the torque on each rod is T = 95.92
N-m, therefore the torque required on the nut,
T1 = 2T = 2 × 95.92 = 191.84 N-m Ans.
Example 10.14. The mean diameter of a Whitworth bolt having V-threads is 25 mm. The

pitch of the thread is 5 mm and the angle of V is 55º. The bolt is tightened by screwing a nut whosemean radius of the bearing surface is 25 mm. If the coefficient of friction for nut and bolt is 0.1 andfor nut and bearing surfaces 0.16 ; find the force required at the end of a spanner 0.5 m long whenthe load on the bolt is 10 kN.
Solution. Given : d = 25 mm ; p = 5 mm ; 2 β = 55º   or   β = 27.5º ; R = 25 mm ; µ = tan φ
= 0.1; µ
2 = 0.16 ; l = 0.5 m ; W = 10 kN = 10 × 103 N

We know that virtual coefficient of friction,
             110.1 0.1tan 0.113cos cos 27.5º 0.887µµ= φ= = =β
and        5tan 0.06425p
dα= = =ππ ×
∴ Force on the screw,
                                            1
1
1tan tan

Chapter 10 : Friction    /G6C  285
                              30.064 0.11310 10 1783 N1 0.064 0.113+ =× =−×
We know that total torque transmitted,
                          3
225. . 1783 0.16 10 10 25 N-mm22dTP W R=×+ µ = × + ×× ×
 62 300 N-mm 62.3 N-m== ...(i)
Let P1 = Force required at the end of a spanner.
∴  Torque required at the end of a spanner,
 T = P1 × l = P1 × 0.5 = 0.5 P1 N-m ... (ii)
P1 = 62.3/0.5 = 124.6 N  Ans.

10.25. Friction in Journal Bearing-Friction Circle
A journal bearing forms a turning pair as shown in Fig. 10.15 ( a). The fixed outer element of
a turning pair is called a bearing and that portion of the inner element ( i.e. shaft) which fits in the
bearing is called a journal. The journal is slightly less in diameter than the bearing, in order to permit
the free movement of the journal in a bearing.
                    ( a)         ( b)

( a)         ( b)
Fig. 10.15. Friction in journal bearing.
When the bearing is not lubricated (or the journal is stationary), then there is a line contact
between the two elements as shown in Fig. 10.15 ( a). The load W on the journal and normal reaction
RN (equal to W) of the bearing acts through the centre. The reaction RN acts vertically upwards at
point A. This point A is known as seat or point of pressure.

Now consider a shaft rotating inside a bearing in clockwise direction as shown in Fig. 10.15
(b). The lubricant between the journal and bearing forms a thin layer which gives rise to a greasy
friction.Therefore, the reaction  R does not act vertically upward, but acts at another point of pressure
B. This is due to the fact that when shaft rotates, a frictional force F = µ RN acts at  the circumference

of the shaft which has a tendency to rotate the shaft in opposite direction of motion and this shifts the
point A to point B.
In order that the rotation may be maintained, there must be a couple rotating the shaft.
Let φ= Angle between R (resultant of F and RN) and RN,
µ= Coefficient of friction between the journal and bearing,
T= Frictional torque in N-m, and
r= Radius of the shaft in metres.

286  /G6C   Theory of Machines
For uniform motion, the resultant force acting on the shaft must be zero and the resultant
turning moment on the shaft must be zero. In other words,
R= W, and T = W × OC = W × OB sin φ = W.r sin φ
Since φ is very small, therefore substituting sin φ = tan φ
∴ T= W.r tan φ = µ.W.r ...(∵ µ = tan φ)
If the shaft rotates with angular velocity ω rad/s, then power wasted in friction,
P= T.ω = T × 2πN/60 watts
where N= Speed of the shaft in r.p.m.

Notes : 1. If a circle is drawn with centre O and radius OC = r sin φ, then this circle is called the friction circle
of a bearing.
2. The force R exerted by one element of a turning pair on the other element acts along a tangent to the
friction circle.
Example 10.15. A 60 mm diameter shaft running in a bearing carries a load of 2000 N. If

the coefficient of friction between the shaft and bearing is 0.03, find the power transmitted when itruns at 1440 r.p.m.
Solution.  Given : d = 60 mm or r = 30 mm = 0.03 m ; W = 2000 N ; µ = 0.03 ; N = 1440 r.p.m.
or ω = 2π × 1440/60 = 150.8 rad/s
We know that torque transmitted,
T= µ.W.r = 0.03 × 2000 × 0.03 = 1.8 N-m
∴  Power transmitted, P=T.ω = 1.8 × 150.8 = 271.4 W Ans.
10.26. Friction of Pivot and Collar Bearing

The rotating shafts are frequently subjected to axial thrust. The bearing surfaces such as pivot
and collar bearings are used to take this axial thrust of the rotating shaft. The propeller shafts of ships, the
shafts of steam turbines, and vertical machine shafts are examples of shafts which carry an axial thrust.
The bearing surfaces placed at the end of a shaft to take the axial thrust are known as

pivots. The pivot may have a flat surface or conical surface as shown in Fig. 10.16 ( a) and (b)
respectively. When the cone is truncated, it is then known as truncated or trapezoidal pivot asshown in Fig. 10.16 ( c).
The collar may have flat bearing surface or conical bearing surface, but the flat surface is
most commonly used. There may be a single collar, as shown in Fig. 10.16 ( d) or several collars along

the length of a shaft, as shown in Fig. 10.16 ( e) in order to reduce the intensity of pressure.
(a) Flat pivot. ( b) Conical pivot. ( c) Truncated pivot. ( d) Single flat ( e) Multiple flat
collar. collar.
Fig. 10.16. Pivot and collar bearings.
In modern practice, ball and roller thrust bearings are used when power is being transmitted
and when thrusts are large as in case of propeller shafts of ships.

Chapter 10 : Friction    /G6C  287
Fig. 10.17.  Flat pivot or footstep
bearing.A little consideration will show that in a new bear-

ing, the contact between the shaft and bearing may be goodover the whole surface. In other words, we can say that thepressure over the rubbing surfaces is uniformly distributed.But when the bearing becomes old, all parts of the rubbingsurface will not move with the same velocity, because thevelocity of rubbing surface increases with the distance fromthe axis of the bearing. This means that wear may be differentat different radii and this causes to alter the distribution ofpressure. Hence, in

to alter the distribution ofpressure. Hence, in the study of friction of bearings, it is as-sumed that

1.  The pressure is uniformly distributed throughout the bearing surface, and
2.  The wear is uniform throughout the bearing surface.
10.27. Flat Pivot Bearing
When a vertical shaft rotates in a flat pivot bearing
(known as  foot step bearing ), as shown in Fig. 10.17, the
sliding friction will be along the surface of contact betweenthe shaft and the bearing.
Let W=Load transmitted over the bearing surface,
R=Radius of bearing surface,
p=Intensity of pressure per unit area of bear-

ing surface between rubbing surfaces, and
µ=Coefficient of friction.
We will consider the following two cases :1.  When there is a uniform pressure ; and
2.  When there is a uniform wear.
1.  Considering unifrom pressure
When the pressure is uniformly distributed over the bearing area, then
                                              
2Wp
R=
π
Consider a ring of radius r and thickness dr of the bearing area.

∴  Area of bearing surface,   A = 2πr.dr
Load transmitted to the ring,
                                                       δ W  = p × A =  p × 2 π r.dr ...(i)
Frictional resistance to sliding on the ring acting tangentially at radius r,
Fr  = µ.δW = µ p × 2π r.dr = 2π µ.p.r.dr
∴ Frictional torque on the ring,
     Tr = Fr × r = 2π µ p r.dr × r = 2 π µ p r2 dr ...(ii)
Integrating this equation within the limits from 0 to R for the total frictional torque on the
pivot bearing.
Collar bearing.

288  /G6C   Theory of Machines
∴  Total frictional torque,  22
0022RR
T prd r prd r =π µ = π µ∫∫
                                        3 33
0222 . .
33 3RR rpp p R=π µ =π µ× = × π µ
                                        3
222..
33WRW R
R=× π µ × × =× µ
π2...WpR=π/G51
When the shaft rotates at ω rad/s, then power lost in friction,
P= T.ω = T × 2π N/60 ...( 2 /60) Nω= π/G51
where N= Speed of shaft in r.p.m.
2.  Considering uniform wear

We have already discussed that the rate of wear depends upon the intensity of pressure ( p) and
the velocity of rubbing surfaces ( v). It is assumed that the rate of wear is proportional to the product
of intensity of pressure and the velocity of rubbing surfaces ( i.e.  p.v..). Since the velocity of rubbing
surfaces increases with the distance ( i.e. radius r) from the axis of the bearing, therefore for uniform
wear
p.r=C (a constant)     or     p = C/r

wear
and the load transmitted to the ring,
δW=p × 2πr.dr ...[From equation (i)]
    2. 2 .Crdr Cdrr=× π = π
∴ Total load transmitted to the bearing
[]0
02. 2 2 .o r2R
R WWC d r C C R C rR= π =π =π =π∫
We know that frictional torque acting on the ring,
 2222rCTp r d r r d rr=π µ =π µ × × ...Cpr=/G51
=2π µ.C.r  dr ...(iii)
∴  Total frictional torque on the bearing,
2
0 02. . . 2.
2RRrTC r d r C=π µ = π µ∫
2
22. . .2RCC R=π µ × = π µ

2
21..22WRW RR=π µ× × = ×µπ...
2WCR=π /G51
Example 10.16. A vertical shaft 150 mm in diameter rotating at 100 r.p.m. rests on a flat end
footstep bearing. The shaft carries a vertical load of 20 kN. Assuming uniform pressure distributionand coefficient of friction equal to 0.05, estimate power lost in friction.
Solution. Given : D = 150 mm or R = 75 mm = 0.075 m ; N = 100 r.p.m or ω = 2 π × 100/60
= 10.47 rad/s ; W = 20 kN = 20 × 10
3 N ; µ = 0.05

Chapter 10 : Friction    /G6C  289
*The vertical load acting on the ring is also given by
                                   δ W = Vertical component of pn × Area of the ring
                                         = pn sin α × 2πr.dr.cosec α = pn × 2πr.drFig. 10.18.
Conical pivot bearing.We know that for uniform pressure distribution, the total frictional torque,
                                  3 22. . 0.05 20 10 0.075 50 N-m
33TW R=× µ =× × × × =
∴  Power lost in friction,

                                 . 50 10.47 523.5 WPT=ω = × = Ans.
10.28. Conical Pivot Bearing
The conical pivot bearing supporting a shaft carrying a load W is shown in Fig. 10.18.
Let                                Pn= Intensity of pressure normal to
the cone,
α= Semi angle of the cone,
µ= Coefficient of friction
between the shaft and thebearing, and
R= Radius of the shaft.
Consider a small ring of radius r and thickness dr. Let dl is

the length of ring along the cone, such that
                                  dl = dr cosec α
∴  Area of the ring,
                                   A  = 2πr.dl = 2πr.dr cosec α
...(∵ dl = dr cosec α)
1.  Considering uniform pressure
We know that normal load acting on the ring,                              δW
n = Normal pressure × Area
                                      = pn × 2πr.dr cosec α
and vertical load acting on the ring,

                              *δW = Vertical component of δWn = δWn.sin α
                                      = pn × 2πr.dr cosec α. sin α = pn × 2π r.dr
∴  Total vertical load transmitted to the bearing,
                                2 22
0 02. 2 2 .
2 2RR
nn n nR rWpr d rp p R p=× π = π = π × = π∫
or                                         2/npWR=π
We know that frictional force on the ring acting tangentially at radius r,

. . .2 . cosec 2 . .cosec . .rn n nFWp r d r p r d r=µδ =µ π α= πµ α
and frictional torque acting on the ring,
  22. . c o s e c . . 2.c o s e c . .rr n nTFr p rdr r p rd r =× = π µ α × = π µ α

290  /G6C   Theory of Machines
Integrating the expression within the limits from 0 to R for the total frictional torque on the
conical pivot bearing.
∴  Total frictional torque,
    32
0 02. c o s e c . 2. . c o s e c
3RR
nnrTp r d r p=π µ α = π µ α∫
3322 . .cosec . .cosec33nnRRppπ=π µ α × = × µ α ...(i)
Substituting the value of pn in equation ( i),
    3
222cosec . . . cosec33RWTW R
Rπ=× π × ×α = × µ α
π
Note : If slant length ( l ) of the cone is known, then
  2..

2..
3TW l=× µ ...( cosec )lR=α/G51
In Fig. 10.18, let pr be the normal intensity of pressure at a distance r from the central axis.
We know that, in case of uniform wear, the intensity of pressure varies inversely with the distance.
∴           pr.r  = C (a constant)    or pr = C/r
         2. 2. 2 .rCWprdr rdr Cdrrδ= × π =× π = π
∴  Total load transmitted to the bearing,
            []0
02. 2 2.o r2R

[]0
            222 . .cosec . . 2 cosec . .rrCTp r d r r d rr=π µ α =π µ × × α
                 2. . c o s e c . .Cr d r=π µ α
∴  Total frictional torque acting on the bearing,
2
0 02 . .cosec . . 2 . .cosec
2RRrTC r d r C= π µα = π µα∫
2
22 . .cosec . .cosec .2RCC R=π µ α × = π µ α
Substituting the value of C, we have
        211cosec . . . cosec . .
222WTR W R W l

211cosec . . . cosec . .
R= π µ × × α =× µ α =× µ
π
10.29. Trapezoidal or Truncated Conical Pivot Bearing
If the pivot bearing is not conical, but a frustrum of a cone with r1 and r2, the external and
internal radius respectively as shown in Fig. 10.19, then

Chapter 10  : Friction    /G6C  291
Fig.10.19.  Trapezoidal
pivot bearing.Area of the bearing surface,
22
12[( ) ( ) ]Arr=π −
∴  Intensity of uniform pressure,
22
12[( ) ( ) ]nWWpArr==
π−...(i)
The total torque acting on the bearing is obtained by integrating the
value of Tr (as discussed in Art. 10.27) within the limits r1 and r2.
∴  Total torque acting on the bearing,
1
21
2322 . cosec . . 2 . .cosec
3rr
nn
rrrT p rd r p=π µ α = π µ α∫
33

3rr
nn
33
12() ()2.. c o s e c
3nrrp−=π µ α 
Substituting the value of pn from equation (i),
33
12
22
12() ()2. c o s e c
[( ) ( ) ] 3W rrT
rr−=π µ × × α π− 
33
12
22
12() () 2.. c o s e c3 () ()rrW
rr−=× µ α 
−
We have discussed in Art. 10.26 that the load transmitted to the ring,
δW = 2πC.dr
∴  Total load transmitted to the ring,
1
1
2
2122. 2[ ] 2( )r
r
r
rWC d r C r C r r=π = π = π−∫
or
122( )WC
rr=

r
r
or
122( )WC
rr=
π−      ...(ii)
We know that the torque acting on the ring, considering uniform wear, is
Tr = 2π µ.C cosec α.r.dr
∴   Total torque acting on the bearing,
1
21
22
2. s e c . . 2. . c o s e c
2rr
rrrTC c o r d r C=π µ α= π µ α∫
22
12.. c o s e c () () C rr =πµ α −

292  /G6C   Theory of Machines
Substituting the value of C from equation (ii), we get
22
12
12cosec [( ) ( ) ]
2( )WTr r
rr=π µ× × α −
π−
121.( ) c o s e c ..c o s e c2Wr r W R=× µ + α = µ α
where R = Mean radius of the bearing 12
2rr+=
Example 10.17. A conical pivot supports a load of 20 kN, the cone angle is 120º and the
intensity of normal pressure is not to exceed 0.3 N/mm2. The external diameter is twice the internal

diameter. Find the outer and inner radii of the bearing surface. If the shaft rotates at 200 r.p.m. andthe coefficient of friction is 0.1, find the power absorbed in friction. Assume uniform pressure.
Solution.   Given : W = 20 kN = 20 × 10
3 N ; 2 α = 120º  or  α = 60º ; pn = 0.3 N/mm2 ;
N = 200 r.p.m. or ω = 2 π × 200/60 = 20.95 rad/s ; µ = 0.1
Outer and inner radii of the bearing surface
Letr1 and r2= Outer and inner radii of the bearing surface, in mm.

Since the external diameter is twice the internal diameter, therefore
r1=2 r2
We know that intensity of normal pressure ( pn),
33
22 22 2
12 22 220 10 2.12 100.3
[( ) ( ) ] [(2 ) ( ) ] ( )W
rr rr r××== =
π− π −
∴              23 3
22( ) 2.12 10 /0.3 7.07 10 or 84 mmrr=× =× = Ans.
and r1 = 2 r2 = 2 × 84 = 168 mm  Ans.
Power absorbed in friction
We know that total frictional torque (assuming uniform pressure),
                      33
12
22
12() () 2.. c o s e c3 () ()rrTW
−

−
                          33
3
222 (168) (84)0.1 20 10 cosec 60º N-mm3 (168) (84)−=× × × × = −
 301760 N-mm = 301.76 N-m=
∴  Power absorbed in friction,
 P = T.ω = 301.76 × 20.95 = 6322 W = 6.322 kW  Ans.
Example 10.18. A conical pivot bearing supports a vertical shaft of 200 mm diameter. It is

subjected to a load of 30 kN. The angle of the cone is 120º and the coefficient of friction is 0.025.Find the power lost in friction when the speed is 140 r.p.m., assuming 1. uniform pressure ; and
2. uniform wear.
Solution. Given : D = 200 mm or R = 100 mm = 0.1 m ; W = 30 kN = 30 × 10
3 N ; 2 α = 120º
or α = 60º ; µ = 0.025 ; N = 140 r.p.m. or ω = 2 π × 140/160 = 14.66 rad/s
1.  Power lost in friction assuming uniform pressure
We know that total frictional torque,
2... c o s e c3TW R=× µ α

Chapter 10  : Friction    /G6C  293
3 20.025 30 10 0.1 cosec 60º 57.7 N-m3=× × × × × =
P = T.ω = 57.7 × 14.66 = 846 W Ans.
2.  Power lost in friction assuming uniform wear
1... c o s e c2TW R=× µ α
3 10.025 30 10 0.1 cosec 60º 43.3 N-m2=× × × × × =
∴  Power lost in friction,  P = T.ω = 43.3 × 14.66 = 634.8 W Ans.
10.30. Flat Collar Bearing

We have already discussed that collar bearings are used to take the axial thrust of the rotating
shafts. There may be a single collar or multiple collar bearings as shown in Fig. 10.20 ( a) and (b)
respectively. The collar bearings are also known as thrust bearings.  The friction in the collar bear-
ings may be found as discussed below :
(a) Single collar bearing ( b) Multiple collar bearing.
Fig. 10.20. Flat collar bearings.

Consider a single flat collar bearing supporting a shaft as shown in Fig. 10.20 ( a).
Let r1= External radius of the collar, and
r2= Internal radius of the collar.
∴  Area of the bearing surface,
A=π [(r1)2 – (r2)2]

294  /G6C   Theory of Machines
When the pressure is uniformly distributed over the bearing surface, then the intensity of
pressure,
         22
12[) ()]WWpArr==
π−...(i)
We have seen in Art. 10.25, that the frictional torque on the ring of radius r and thickness dr,
         22. . .rT prd r=π µ
Integrating this equation within the limits from r2 to r1 for the total frictional torque on the
collar.
          1
1
2

          1
1
2
2332 312() ()2. . . 2. 2.
3 3r
r
r
rr rrT prd r pp   −=π µ = π µ = π µ    ∫
Substituting the value of p from equation  (i),
         33
12
22
12() ()2
rr−=π µ × π− 
          33
12
22
12() () 2.3 () ()rrW
rr−=× µ 
−
Notes: 1. In order to increase the amount of rubbing surfaces so as to reduce the intensity of pressure, it is better

to use two or more collars, as shown in Fig. 10.20 ( b), rather than one larger collar.
2.  In case of a multi-collared bearings with, say n collars, the intensity of the uniform pressure,
                                                 22
12Load
No. of collars × Bearing area of one collar [( ) ( ) ]Wp
nr r==
π−
3.  The total torque transmitted in a multi collared shaft remains constant i.e.
                                  33
12
22
12() () 2.
3 () ()rrTW
rr−=× µ −

12
22
12() () 2.
3 () ()rrTW
2.  Considering unifrom wear
We have seen in Art. 10.25 that the load transmitted on the ring, considering uniform wear is,
        .2 . 2 . 2 .rCWpr d r r d r C d rrδ= π =× π = π
∴  Total load transmitted to the collar,
       1
21
2122. 2[ ] 2( )r r
rrWC d r C r C r r= π =π =π −∫
or         
122( )WC
rr=
π−...(ii)

Chapter 10  : Friction            /G6C          295
We also know that frictional torque on the ring,
.. 2 . .2 . . .rTW r C d r r C r d r=µδ =µ× π = πµ
11
2222 2
12() ()2. .2 . 2 .
2 2rr
rrrr rTC r d r C C  −=π µ= π µ = π µ   ∫
22
12.[ () () ]Cr r=πµ −
Substituting the value of C from equation (ii),
22
12 1 2
121[( ) ( ) ] . ( )
2( ) 2WTr r W r r
rr=πµ× − = ×µ +
π−
Example 10.19. A thrust shaft of a ship has 6

π−
collars of 600 mm external diameter and 300 mm internal
diameter. The total thrust from the propeller is 100 kN. Ifthe coefficient of friction is 0.12 and speed of the engine90 r.p.m., find the power absorbed in friction at the thrustblock, assuming l. uniform pressure ; and 2. uniform
wear.
Solution. Given : n = 6 ; d
1 = 600 mm or r1 = 300
mm ; d2 = 300 mm or r2 = 150 mm ; W = 100 kN
= 100 × 103 N ; µ = 0.12 ; N = 90 r.p.m. or

ω = 2 π × 90/60 = 9.426 rad/s
1.  Power absorbed in friction, assuming uniform
      pressure
We know that total frictional torque transmitted,
                              33
12
22
12() () 2.3 () ()rrTW
−
33
33
222 (300) (150)0.12 100 10 2800 10 N-mm3 (300) (150)−=× × × = × −
2800 N-m=
∴   Power absorbed in friction,
. 2800 9.426 26 400 W 26.4 kW PT=ω = × = = Ans.
2.  Power absorbed in friction assuming uniform wear

3
1211. ( ) 0.12 100 10 (300 150) N-mm22TW r r=× µ + =× × × +
32700 10 N-mm 2700 N-m=× =
P = T.ω =  2700 × 9.426 = 25 450 W = 25.45 kW  Ans.
Ship propeller.

296  /G6C   Theory of Machines
Example 10.20. A shaft has a number of a collars integral with it. The external diameter of
the collars is 400 mm and the shaft diemater is 250 mm. If the intensity of pressure is 0.35 N/mm2
(uniform) and the coefficient of friction is 0.05, estimate : 1. power absorbed when the shaft runs at
105 r.p.m. ca rrying a load of 150 kN ; and 2. number of collars required.
Solution. Given : d1 = 400 mm  or r1 = 200 mm ; d2 = 250 mm or r2 = 125 mm ; p = 0.35

N/mm2 ; µ = 0.05 ; N = 105 r.p.m or ω = 2 π × 105/60 = 11 rad/s ; W = 150 kN = 150 × 103 N
1.  Power absorbed
We know that for uniform pressure, total frictional torque transmitted,
                   33 33
3 12
22 2 2
12() () 2 2 (200) (125).0 . 0 5 1 5 0 1 0 N - m m33 () () (200) (125)rrTW
rr  − −=× µ =× × ×  − −   
                        35000 248 1240 10 N-mm 1240 N-m=× =× =
∴  Power absorbed,
      . 1240 11 13640 W 13.64 kWPT=ω = × = = Ans.
2.  Number of collars required

Let      n  = Number of collars required.
We know that the intensity of uniform pressure ( p),
              3
22 2 2
12150 10 1.960.35
. [( ) ( ) ] . [(200) (125) ]W
n nr r n×== =
π− π −
∴      1.96/0.35 5.6 say 6n== Ans.
Example 10.21. The thrust of a propeller shaft in a marine engine is taken up by a number

of collars integral with the shaft which is 300 mm in diameter. The thrust on the shaft is 200 kN andthe speed is 75 r.p.m. Taking µ constant and equal to 0.05 and assuming intensity of pressure as
uniform and equal to 0.3 N/mm
2, find the external diameter of the collars and the number of collars
required, if the power lost in friction is not to exceed 16 kW.
Solution. Given : d2 = 300 mm or r2 = 150 mm = 0.15 m ; W = 200 kN = 200 × 103 N ;

N = 75 r.p.m. or ω = 2 π × 75/60 = 7.86 rad/s ; µ = 0.05 ; p = 0.3 N/mm2 ; P = 16 kW = 16 × 103 W
Let      T = Total frictional torque transmitted in N-m.
We know that power lost in friction ( P),
       16 × 103 = T.ω = T × 7.86  or  T = 16 × 103/7.86 = 2036 N-m
External diameter of the collar
Let        d1 = External diameter of the collar in metres = 2 r1.
We know that for uniform pressure, total frictional torque transmitted ( T ),
      33 22
12 12 1 2
22

33 22
12 12 1 2
22
12 12() () () () . 222036 .33 () ()rr rrr rWW
rr rr  − ++=× µ =× µ ×  + −  *
      22
311
1( ) (0.15) 0.15 20.05 200 103 0.15rr
r++ ×=× × × + 
    32
11 12036 3( 0.15) 20 10 [( ) 0.15 0.0225] rr r×+ = × + +
*33 22 22
1 2 1 2 1 2 12 1 2 12
23
12 12 12 12() () ( ) [ () () .] () () .
() () () ()rr r r rrr rrrr r
rr rr rr rr−− + + + +==+− + −

Chapter 10  : Friction    /G6C  297
Dividing throughout by 20 × 103,
           0.305 ( r1 + 0.15) = ( r1)2 + 0.15 r1 + 0.0225
(r1)2 – 0.155 r1 – 0.0233 = 0
Solving this as a quadratic equation,
     2
10.155 (0.155) 4 0.0233 0.155 0.342
22r±+ × ±==
    = 0.2485 m = 248.5 mm ...(Taking + ve sign)
∴     d1 = 2 r1 = 2 × 248.5 = 497 mm  Ans.
Number of collars
Let      n = Number of collars.
We know that intensity of pressure ( p),
     3
22 2 2
12200 10 1.620.3
[ ) ( ) ] [(248.5) (150) ]W

n nr r n×== =
π− π −
∴      1.62/0.3 5.4 or 6n== Ans.
10.31. Friction Clutches
A friction clutch has its principal application in the transmission of power of shafts and

machines which must be started and stopped frequently. Its application is also found in cases in whichpower is to be delivered to machines partially or fully loaded. The force of friction is used to start thedriven shaft from rest and gradually brings it up to the proper speed without excessive slipping of thefriction surfaces. In automobiles, friction clutch is used to connect the engine to the driven shaft. Inoperating such a clutch, care should be taken so that the friction surfaces engage

be taken so that the friction surfaces engage easily and graduallybrings the driven shaft up to proper speed. The proper alignment of the bearing must be maintainedand it should be located as close to the clutch as possible. It may be noted that

1. The contact surfaces should develop a frictional force that may pick up and hold the load
with reasonably low pressure between the contact surfaces.
2. The heat of friction should be rapidly dissipated and tendency to grab should be at a
minimum.
3. The surfaces should be backed by a material stiff enough to ensure a reasonably uniform
distribution of pressure.

The friction clutches of the following types are important from the subject point of view :1. Disc or plate clutches (single disc or multiple disc clutch),
2. Cone clutches, and
3. Centrifugal clutches.
We shall now discuss, these clutches, in detail, in the following pages. It may be noted that
the disc and cone clutches are based on the same theory as the pivot and collar bearings.
10.32. Single Disc or Plate Clutch

A single disc or plate clutch, as shown in Fig. 10.21, consists of a clutch plate whose both
sides are faced with a friction material (usually of Ferrodo). It is mounted on the hub which is free tomove axially along the splines of the driven shaft. The pressure plate is mounted inside the clutchbody which is bolted to the flywheel. Both the pressure plate and the flywheel rotate with the engine

298  /G6C   Theory of Machines
crankshaft or the driving shaft. The pressure plate

pushes the clutch plate towards the flywheel by a setof strong springs which are arranged radially insidethe body. The three levers (also known as releaselevers or fingers) are carried on pivots suspendedfrom the case of the body. These are arranged in sucha manner so that the pressure plate moves away fromthe flywheel by the inward movement of a thrustbearing. The bearing is mounted upon a forked shaftand moves forward when the clutch pedal is pressed.

linkage forces the thrust release bearing to move intowards the flywheel and pressing the longer ends of the levers inward. The levers are forced to turnon their suspended pivot and the pressure plate moves away from the flywheel by the knife edges,thereby compressing the clutch springs. This action removes the pressure from the clutch plate andthus moves back from the flywheel and the driven shaft becomes stationary. On the other hand, whenthe foot is taken off from the clutch pedal, the

foot is taken off from the clutch pedal, the thrust bearing moves back by the levers. This allows thesprings to extend and thus the pressure plate pushes the clutch plate back towards the flywheel.

Fig. 10.21. Single disc or plate clutch.
The axial pressure exerted by the spring provides a frictional force in the circumferential
direction when the relative motion between the driving and driven members tends to take place. If thetorque due to this frictional force exceeds the torque to be transmitted, then no slipping takes placeand the power is transmitted from the driving shaft to the driven shaft.

Now consider two friction surfaces, maintained in contact by an axial thrust W, as shown in
Fig. 10.22 ( a).
Single disc clutch

Chapter 10  : Friction    /G6C  299
Let T= Torque transmitted by the clutch,
p= Intensity of axial pressure with which the contact surfaces are held
together,
r1 and r2= External and internal radii of friction faces, and
µ= Coefficient of friction.
Consider an elementary ring of radius r and thickness dr as shown in Fig. 10.22 ( b).
We know that area of contact surface or friction surface,
=2  π r.dr
∴  Normal or axial force on the ring,
δW= Pressure × Area = p × 2 π r.dr

and the frictional force on the ring acting tangentially at radius r,
Fr=µ.δW = µ.p × 2 π r.dr
∴  Frictional torque acting on the ring,
                          Tr = Fr × r= µ.p × 2 π r.dr × r = 2 π × µ .p.r2 dr
                                                          (a)                                            (b)
Fig. 10.22.  Forces on a single disc or plate clutch.
We shall now consider the following two cases :

1.  When there is a uniform pressure, and
When the pressure is uniformly distributed over the entire area of the friction face, then the
intensity of pressure,
22
12[( ) ( ) ]Wp
rr=
π−...(i)
where W= Axial thrust with which the contact or friction surfaces are held together.
We have discussed above that the frictional torque on the elementary ring of radius r and
thickness dr is

Tr=2  π µ.p.r2 dr
Integrating this equation within the limits from r2 to r1 for the total frictional torque.

300  /G6C   Theory of Machines
∴  Total frictional torque acting on the friction surface or on the clutch,
                          12
12332 3 12() ()2. . . 2 2
3 3rr
rrr rrT prd r pp  −=π µ = π µ = π µ   ∫
Substituting the value of p from equation (i),
                          33
12
22
12() ()23 [( ) ( ) ]rr WT
rr−=π µ × ×
π−
33
12
22
12() () 2.. .3 () ()rrWW R
rr−=× µ = µ 
−
where                              R = Mean radius of friction surface
33
12
22
12() ()2
3() ()rr

33
12
22
12() ()2
3() ()rr
rr−=
−
In Fig. 10.22, let p be the normal intensity of pressure at a distance r from the axis of the
clutch. Since the intensity of pressure varies inversely with the distance, therefore
p.r. = C (a constant)   or   p = C/r ...(i)
and the normal force on the ring,
.2 . 2 . 2 .CWpr d r C d r C d rrδ= π =× π = π
∴  Total force acting on the friction surface,
[]1
1
2
21222 2 ( )r
r
r
rWC d r C rC r r=π= π = π−∫
or
122( )WC
rr=
π−

r
or
122( )WC
rr=
π−
We know that the frictional torque acting on the ring,
222 . .2 .2 . . .rCT pr dr r dr Crdrr=π µ =π µ × × =π µ
...(∵  p = C/r)
∴  Total frictional torque on the friction surface,
1 1
2 222 2
12() ()2. . . 2. 2.
2 2r r
r rrr rTC r d r C C  −=π µ = π µ = π µ   ∫
22 22
12 12
12.[ () ()] () ()2( )WCr r rrrr =π µ − =π µ× −π−
121.( ) ..2Wr r WR=× µ + = µ
where R = Mean radius of the friction surface 12
2rr+=

2rr+=
Notes : 1. In general, total frictional torque acting on the friction surface (or on the clutch) is given by
T=n.µ.W.R

Chapter 10  : Friction    /G6C  301
where n= Number of pairs of friction or contact surfaces, and
R= Mean radius of friction surface
33
12
22
12() ()2
3() ()rr
rr−=−...(For uniform pressure)
12
2rr+= ...(For uniform wear)
2.  For a single disc or plate clutch, normally both sides of the disc are effective. Therefore, a single disc
clutch has two pairs of surfaces in contact, i.e. n = 2.

3.  Since the intensity of pressure is maximum at the inner radius ( r2) of the friction or contact surface,
therefore equation (i) may be written as
pmax × r2 = Corpmax = C/r2
4.  Since the intensity of pressure is minimum at the outer radius ( r1) of the friction or contact surface,
pmin × r1 = Corpmin = C/r1
5.  The average pressure ( pav) on the friction or contact surface is given by
22
12Total force on friction surface

22
Cross-sectional area of friction surface [( ) ( ) ]avWp
rr==
π−
6.  In case of a new clutch, the intensity of pressure is approximately uniform but in an old clutch the
uniform wear theory is more approximate.
7.  The uniform pressure theory gives a higher frictional torque than the uniform wear theory. Therefore
in case of friction clutches, uniform wear should be considered, unless otherwise stated.
10.33. Multiple Disc Clutch

A multiple disc clutch, as shown in Fig. 10.23, may be used when a large torque is to be
transmitted. The inside discs (usually of steel) are fastened to the driven shaft to permit axial motion
(except for the last disc). The outside discs (usually of bronze) are held by bolts and are fastened to
the housing which is keyed to the driving shaft. The multiple disc clutches are extensively used inmotor cars, machine tools etc.
Let n

Let n
1= Number of discs on the driving shaft, and
n2= Number of discs on the driven shaft.
Dual Disc Clutches.

302  /G6C   Theory of Machines
∴  Number of pairs of contact surfaces,
n=n1 + n2 – 1
and total frictional torque acting on the friction surfaces or on the clutch,
T=n.µ.W.R
where R= Mean radius of the friction surfaces
   33
12
22
12() ()2
3() ()rr
rr−=
−...(For uniform pressure)
12
Fig. 10.23. Multiple disc clutch.
Example 10.22. Determine the maximum, minimum and average pressure in plate clutch

when the axial force is 4 kN. The inside radius of the contact surface is 50 mm and the outside radiusis 100 mm. Assume uniform wear.
Solution. Given : W = 4 kN = 4 × 10
3 N ; r2 = 50 mm ; r1 = 100 mm
Maximum pressure
Let pmax= Maximum pressure.
Since the intensity of pressure is maximum at the inner radius ( r2), therefore
pmax × r2=C   or  C = 50 pmax
We know that the total force on the contact surface ( W),
4 × 103=2  π C (r1 – r2) = 2 π × 50 pmax (100 – 50) = 15 710 pmax

∴ pmax= 4 × 103/15 710 = 0.2546 N/mm2  Ans.
Minimum pressure
Let pmin= Minimum pressure.
Since the intensity of pressure is minimum at the outer radius ( r1), therefore
pmin × r1=C    or    C = 100 pmin

Chapter 10  : Friction    /G6C  303
4 × 103=2  π C (r1 – r2) = 2π × 100 pmin (100 – 50) = 31 420 pmin
∴ pmin= 4 × 103/31 420 = 0.1273 N/mm2  Ans.
Average pressure
We know that average pressure,
Total normal force on contact surface
Cross-sectional area of contact surfacesavp=
3
2
22 2 2
1241 00.17 N/mm
[( ) ( ) ] [(100) (50) ]W
rr×== =
π− π −Ans.
Example 10.23. A single plate clutch, with both sides effective, has outer and inner

diameters 300 mm and 200 mm respecti vely. The maximum inten sity of pressure at any point in the
contact surface is not to exceed 0.1 N/mm2. If the coefficient of friction is 0.3, determine the power
transmitted by a clutch at a speed 2500 r.p.m .
Solution. Given : d1 = 300 mm or r1 = 150 mm ; d2 = 200 mm or r2 = 100 mm ; p = 0.1 N/mm2 ;
µ = 0.3 ; N = 2500 r.p.m. or ω = 2π × 2500/60 = 261.8 rad/s
Since the intensity of pressure ( p) is maximum at the inner radius ( r2), therefore for uniform

wear,
p.r2=C    or   C = 0.1 × 100 = 10 N/mm
We know that the axial thrust,
W=2  π C (r1 – r2) = 2 π × 10 (150 – 100) = 3142 N
and mean radius of the friction surfaces for uniform wear,
12150 100125 mm 0.125m22rrR+ +== = =
T = n.µ.W.R = 2 × 0.3 × 3142 × 0.125 = 235.65 N-m
...( 2,for bothsidesof plateeffective)n=/G51
∴ Power transmitted by a clutch,
P = T.ω = 235.65 × 261.8 = 61 693 W = 61.693 kW Ans.

Example 10.24. A single plate clutch, effective on both sides, is required to transmit 25 kW
at 3000 r.p.m. D etermine the outer and inner radii of frictional surface if the coefficient of friction is
0.255, the ratio of radii is 1.25 and the maximum pressure is not to exceed 0.1 N/mm2. Also deter-
mine the axial thrust to be provided by springs. Ass ume the theory of uniform wear.
Solution. Given: n = 2 ; P = 25 kW = 25 × 103 W ; N = 3000 r.p.m. or ω = 2π × 3000/60

= 314.2 rad/s ; µ = 0.255 ; r1/r2 = 1.25 ; p = 0.1 N/mm2
Outer and inner radii of frictional surface
Let r1 and r2= Outer and inner radii of frictional surfaces, and
T= Torque transmitted.
Since the ratio of radii ( r1/r2) is 1.25, therefore
r1= 1.25 r2
We know that the power transmitted ( P),
25 × 103=T.ω = T × 314.2
∴ T= 25 × 103/314.2 = 79.6 N-m = 79.6 × 103 N-mm

304  /G6C   Theory of Machines
p.r2=C         or        C = 0.1 r2 N/mm
and the axial thrust transmitted to the frictional surface,
W=2  π C (r1 – r2) = 2 π × 0.1 r2 (1.25 r2 – r2) = 0.157 ( r2)2...(i)
We know that mean radius of the frictional surface for uniform wear,
12 22
21.251.12522rr rrRr++== =
 We know that torque transmitted ( T),
79.6 × 103=n.µ.W.R = 2 × 0.255 × 0.157 ( r2)2 × 1.125 r2 = 0.09 (r2)3

∴ (r2)3= 79.6 × 103/0.09 = 884 × 103   or  r2 = 96 mm  Ans.
and r1= 1.25 r2 = 1.25 × 96 = 120 mm  Ans.
Axial thrust to be provided by springs
We know that axial thrust to be provided by springs,
W= 2 π C (r1 – r2) = 0.157 ( r2)2...[From equation (i)]
= 0.157 (96)2 = 1447 N Ans.
Example 10.25. A single dry plate clutch transmits 7.5 kW at 900 r.p.m. The axial pressure
is limited to 0.07 N/mm2. If the coefficient of friction is 0.25, find 1. Mean radius and face width of

the friction lining assuming the ratio of the mean radius to the face width as 4, and 2. Outer and
inner radii of the clutch plate.
Solution.  Given : P = 7.5 kW = 7.5 × 103 W ; N = 900 r.p.m or ω = 2 π × 900/60 = 94.26 rad/s ;
p = 0.07 N/mm2 ; µ = 0.25
1.  Mean radius and face width of the friction lining
Let R= Mean radius of the friction lining in mm, and
w= Face width of the friction lining in mm,
Ratio of mean radius to the face width,
R/w= 4 ...(Given)

We know that the area of friction faces,
A=2  π R.w
∴ Normal or the axial force acting on the friction faces,
W=A × p = 2 π R.w.p
We know that torque transmitted (considering uniform wear),
.. . . ( 2 ..)Tn W Rn R w p R=µ =µ π
3.. . .22 4RnR n p RRpπ =µ = ×µπ× ×...(∵ w = R/4)
332 0.25 0.07 0.055 N-mm2RRπ=×× × = ...(i)
...(∵  n = 2, for single plate clutch)

Chapter 10  : Friction    /G6C  305
We also know that power transmitted ( P),
7.5 × 103 = T.ω = T × 94.26
∴ T = 7.5 × 103/94.26 = 79.56 N-m = 79.56 × 103 N-mm ... (ii)
R3 = 79.56 × 103/0.055 = 1446.5 × 103  or  R = 113 mm Ans.
and w = R/4 = 113/4 = 28.25mm Ans.
2.  Outer and inner radii of the clutch plate
Let r1 and r2 = Outer and inner radii of the clutch plate respectively.

Since the width of the clutch plate is equal to the difference of the outer and inner radii,
therefore
w = r1 – r2 = 28.25 mm ... (iii)
Also for uniform wear, the mean radius of the clutch plate,
                              12
12or 2 2 113 226 mm2rrRr r R+=+ = = × = ...(iv)
r1 = 127.125 mm ; and r2 = 98.875  Ans.
Example 10.26. A dry single plate clutch is to be designed for an automotive vehicle whose

engine is rated to give 100 kW at 2400 r.p.m. and maximum torque 500 N-m. The outer radius of
friction plate is 25% more than the inner radius. The intensity of pressure between the plate  is not toexceed 0.07 N/mm
2. The coefficient of friction may be assumed equal to 0.3. The helical springs

required by this clutch to provide axial force necessary to engage the clutch are eight. If each springhas stiffness equal to 40 N /mm, determine the initial compression in the springs and dimensions ofthe friction plate.
Solution. Given : P = 100 kW = 100 × 10
3 W ; T = 500 N-m = 500 × 103 N-mm ;
p = 0.07 N/mm2 ; µ = 0.3 ; Number of springs = 8 ; Stiffness = 40 N/mm
Dimensions of the friction plate
Let r1 and r2= Outer and inner radii of the friction plate respectively.

Since the outer radius of the friction plate is 25% more than the inner radius, therefore
r1=1 . 2 5  r2
We know that, for uniform wear,
p.r2=C     or    C = 0.07 r2 N/mm
and load transmitted to the friction plate,
W=2  π C (r1 – r2) = 2 π × 0.07 r2 (1.125 r2 – r2) = 0.11 (r2)2 N
         ...(i)
We know that mean radius of the plate for uniform wear,
12 22
∴  Torque transmitted ( T ),
500 × 103=n.µ.W.R = 2 × 0.3 × 0.11 ( r2)2 × 1.125 r2 = 0.074 ( r2)3
...(∵  n = 2)

...(∵  n = 2)
             ∴ (r2)3= 500 × 103/0.074 = 6757 × 103  or  r2 = 190 mm   Ans.

306  /G6C   Theory of Machines
and r1= 1.25 r2 = 1.25 × 190 = 273.5 mm Ans.
Initial compression of the springs
We know that total stiffness of the springs,
s= Stiffness per spring × No. of springs = 40 × 8 = 320 N/mm
Axial force required to engage the clutch,
W=0 . 1 1  (r2)2 = 0.11 (190)2 = 3970 N ...[From equation (i)]
∴  Initial compression in the springs
=W/s= 3970/320 = 12.5 mm  Ans.
Example 10.27. A rotor is driven by a co-axial motor  through a single plate clutch, both

sides of the plate being effective. The external  and internal diameters of the plate are respectively
220 mm and 160 mm and the total spring load pressing the plates together is 570 N. The motorarmature and shaft has a mass of 800 kg with an ef fective radius of gyration of 200 mm. The rotor
has a mass of 1300 kg with an effective radius of gyration of 180 mm. The coefficient of friction forthe clutch is 0.35.

The driving motor is brought up to a speed of 1250 r.p.m. when the current is switched off
and the clutch suddenly engaged. Determine
1. The final speed of motor and rotor, 2. The time to reach this speed, and 3. The kinetic
energy lost during the period of slipping.
How long would slipping continue if it is assumed that a constant resisting torque of 60 N-m

were present? If instead of a resisting torque, it is assumed that a constant driving torque of 60 N-mis maintained on the armature shaft, what would then be slipping time?
1 = 220 mm or r1 = 110 mm ; d2 = 160 mm or r2 = 80 mm ; W = 570 N ;
m1 = 800 kg ; k1 = 200 mm = 0.2 m ; m2 = 1300 kg ; k2 = 180 mm = 0.18 m ; µ = 0.35 ; N1 = 1250 r.p.m.
or ω1 = π × 1250/60 = 131 rad/s
1. Final speed of the motor and rotor
Let ω3= Final speed of the motor and rotor in rad/s.

We know that moment of inertia for the motor armature and shaft,
I1=m1 (k1)2 = 800 (0.2)2 = 32 kg-m2
and moment of inertia for the rotor,
I2=m2 (k2)2 = 1300 (0.18)2 = 42.12 kg-m2
Since the angular momentum before slipping is equal to the angular momentum after slip-
ping, therefore
I1.ω1 + I2.ω2= (I1 + I2) ω3
32 × 131 + I2 × 0 = (32 + 42.12) ω3 = 74.12 ω3...(∵ ω2 = 0)
∴ω3= 32 × 131 /74.12 = 56.56 rad/s Ans.
2. Time to reach this speed
Let t= Time to reach this speed i.e. 56.56 rad/s.

We know that mean radius of the friction plate,
12110 8095 mm 0.095 m22rrR+ +== ==

Chapter 10  : Friction    /G6C  307
and total frictional torque,
T=n.µ.W.R = 2 × 0.35 × 570 × 0.095 = 37.9 N-m ...(∵  n = 2)
Considering the rotor, let α2, ωI and ωF be the angular acceleration, initial angular speed and
the final angular speed of the rotor respectively.
We know that the torque ( T ),
37.9 =I2.α2 = 42.12 α2    or    α2 = 37.9/42.12 = 0.9  rad/s2
Since the angular acceleration is the rate of change of angular speed, therefore
FI FI
2
256.56 0or 62.8s0.9ttω− ω ω− ω −α= = = =αAns.

2
...(∵  ωF = ω3 = 56.56 rad/s, and ω1 = 0)
3.  Kinetic energy lost during the period of slipping
We know that angular kinetic energy before impact,
22
11 2 2 1 111(( ) ( )22EI I I2
11=ω ) +ω =ω2...(∵  ω2 = 0)
2 132(131) 274 576 N-m2=× =
and angular kinetic energy after impact,
                            22
21 2 311( )( ) (32 42.12)(56.56) 118 556 N-m22EI I=+ ω =+ =
∴  Kinetic energy lost during the period of slipping,

= E1 – E2 = 274 576 – 118 556 = 156 020 N-m  Ans.
Time of slipping assuming constant resisting torque
Let t1= Time of slipping, and
ω2= Common angular speed of armature and rotor shaft = 56.56 rad/s
When slipping has ceased and there is exerted a constant torque of 60 N-m on the armature
shaft, then
Torque on armature shaft,
T1= – 60 – 37.9 = – 97.9 N-m
Torque on rotor shaft,
T2=T = 37.9 N-m
Considering armature shaft,
                            1
31 1 11 1 1 1

1
31 1 11 1 1 1
197.9. 131 131 3.0632Ttt t tIω= ω+ α = ω+ × = − × = − ...(i)
Considering rotor shaft,
2
32 1 1 1 1
237.9.0 . 942.12Ttt t tIω= α = × = × =       ...(ii)
11 1131 3.06 0.9 or 3.96 131 tt t−= =
∴1131/3.96 33.1st== Ans.

308  /G6C   Theory of Machines
Time of slipping assuming constant driving torque of 60 N-m
In this case, T1= 60 – 37.9 = 22.1 N-m
Since    12
111
12, thereforeTTttIIω+ × = ×
        11 1 122.1 37.9131 or 131 0.69 0.9
32 42.12tt t t +× = × + =
∴0.9 t1 – 0.69 t1= 131      or      t1 = 624 s  Ans.
Example 10.28. A multiple disc clutch has five plates having four pairs of active friction
surfaces. If the intensity of pressure is not to exceed 0.127 N/mm2, find the power transmitted at 500

r.p.m. The outer and inner radii of friction surfaces are 125 mm and 75 mm respectively. Assumeuniform wear and take coefficient of friction = 0.3.
Solution. Given : n
1 + n2 = 5 ; n = 4 ; p = 0.127 N/mm2 ; N = 500 r.p.m. or ω = 2π × 500/60
= 52.4 rad/s ; r1 = 125 mm ; r2 = 75 mm ; µ = 0.3
Since the intensity of pressure is maximum at the inner radius r2, therefore
p.r2=C      or     C = 0.127 × 75 = 9.525 N/mm
We know that axial force required to engage the clutch,

W=2  π C (r1 – r2) = 2 π × 9.525 (125 – 75) = 2990 N
and mean radius of the friction surfaces,
12125 75100 mm 0.1m22rrR+ +== = =
T=n.µ.W.R = 4 × 0.3 × 2990 × 0.1 = 358.8 N-m
∴ Power transmitted,
P=T.ω = 358.8 × 52.4 = 18 800 W = 18.8 kW  Ans.
Example 10.29. A multi-disc clutch has three discs on the driving shaft and two on the

driven shaft. The outside diameter of the contact surfaces is 240 mm and inside diameter 120 mm.Assuming uniform wear and coefficient of friction as 0.3, find the maximum axial intensity of pres-sure between the discs for transmitting 25 kW at 1575 r.p.m.
1 = 3 ; n2 = 2 ; d1 = 240 mm or  r1 = 120 mm ; d2 = 120 mm or r2 = 60 mm ;
µ = 0.3 ; P = 25 kW = 25 × 103 W ; N = 1575 r.p.m. or ω = 2 π × 1575/60 = 165 rad/s
Let T= Torque transmitted in N-m, and

W= Axial force on each friction surface.
25 × 103= T.ω = T × 165    or    T = 25 × 103/165 = 151.5 N-m
Number of pairs of friction surfaces,
n=n1 + n2 – 1 = 3 + 2 – 1 = 4
and mean radius of friction surfaces for uniform wear,
12120 6090 mm 0.09 m22rrR+ +== ==

Chapter 10  : Friction    /G6C  309
We know that torque transmitted ( T ),
151.5 =n.µ.W.R = 4 × 0.3 × W × 0.09 = 0.108 W
∴ W= 151.5/0.108 = 1403 N
Let p= Maximum axial intensity of pressure.
wear
p.r2=C   or    C = p × 60 = 60 p N/mm
We know that the axial force on each friction surface ( W),
1403 = 2 π.C (r1 – r2) = 2 π × 60 p (120 – 60) = 22 622 p
∴ p= 1403/22 622 = 0.062 N/mm2  Ans.

Example 10.30. A plate clutch has three discs on the driving shaft and two discs on the
driven shaft, providing four pairs of contact surfaces. The outside diameter of the contact surfaces is240 mm and inside diameter 120 mm. Assuming uniform pressure and µ = 0.3; find the total spring
load pressing the plates together to transmit 25 kW at 1575 r.p.m.
If there are 6 springs each of stiffness 13 kN/m and each of the contact surfaces has worn

away by 1.25 mm, find the maximum power that can be transmitted, assuming uniform wear.
1 = 3 ; n2 = 2 ; n = 4 ; d1 = 240 mm or r1 = 120 mm ; d2 = 120 mm or
r2 = 60 mm ; µ = 0.3 ; P = 25 kW = 25 × 103 W ; N = 1575 r.p.m. or ω = 2 π × 1575/60 = 165 rad/s
Total spring load
Let W= Total spring load, and
We know that power transmitted ( P),
25 × 103=T.ω = T × 165  or   T = 25 × 103/165 = 151.5 N-m
Mean radius of the contact surface, for uniform pressure,

33 33
12
22 2 2
12() ()2 2 (120) (60)93.3 mm 0.0933 m33() () (120) (60)rrR
rr  − −== = = − −   
and torque transmitted ( T ),
151.5 =n.µ.W.R = 4 × 0.3 W × 0.0933 = 0.112 W
∴ W= 151.5/0.112 = 1353 N  Ans.
Maximum power transmitted
Given : No of springs = 6∴  Contact surfaces of the spring
= 8
Wear on each contact surface
= 1.25 mm
∴            Total wear = 8 × 1.25 = 10 mm = 0.01 m
Stiffness of each spring = 13 kN/m = 13 × 10
3 N/m

3 N/m
∴  Reduction in spring force
= Total wear × Stiffness per spring × No. of springs= 0.01 × 13 × 10
3 × 6 = 780 N

310  /G6C   Theory of Machines
∴ New axial load,  W= 1353 – 780 = 573 N
We know that mean radius of the contact surfaces for uniform wear,
                              12120 6090 mm 0.09 m22rrR+ +== ==
∴ Torque transmitted,
T= n.µ.W.R. = 4 × 0.3 × 573 × 0.09 = 62 N-m
and maximum power transmitted,
 P=T. ω = 62 × 155 = 10 230 W = 10.23 kW  Ans.
10.34. Cone Clutch
A cone clutch, as shown in Fig. 10.24, was extensively used in automobiles but now-a-days it

has been replaced completely by the disc clutch.
Fig. 10.24. Cone clutch.
It consists of one pair of friction surface only. In a cone clutch, the driver is keyed to the
driving shaft by a sunk key and has an inside conical surface or face which exactly fits into the outsideconical surface of the driven. The driven member resting on the feather key in the driven shaft, maybe shifted along the shaft by a forked lever provided at B, in order to engage the clutch by bringing the

two conical surfaces in contact. Due to the frictional resistance set up at this contact surface, thetorque is transmitted from one shaft to another. In some cases, a spring is placed around the drivenshaft in contact with the hub of the driven. This spring holds the clutch faces in contact and maintainsthe pressure between them, and the forked lever is used only for disengagement of the clutch. Thecontact surfaces of the clutch may be metal to metal contact, but more often the driven member

metal contact, but more often the driven member islined with some material like wood, leather, cork or asbestos etc. The material of the clutch faces ( i.e.

contact surfaces) depends upon the allowable normal pressure and the coefficient of friction.
Consider a pair of friction surface as shown in Fig. 10.25 ( a). Since the area of contact of a
pair of friction surface is a frustrum of a cone, therefore the torque transmitted by the cone clutch maybe determined in the similar manner as discussed for conical pivot bearings in Art. 10.28.
Let p
n= Intensity of pressure with which the conical friction surfaces are held

together ( i.e. normal pressure between contact surfaces),
r1 and r2= Outer and inner radius of friction surfaces respectively.

Chapter 10  : Friction    /G6C  311
R= Mean radius of the friction surface 12,2rr+=
α= Semi angle of the cone (also called face angle of the cone) or the
angle of the friction surface with the axis of the clutch,
µ= Coefficient of friction between contact surfaces, and
b= Width of the contact surfaces (also known as face width or clutch
face).
Fig. 10.25. Friction surfaces as a frustrum of a cone.
Consider a small ring of radius r and thickness dr, as shown in Fig. 10.25 ( b). Let dl is length

of ring of the friction surface, such that
dl=dr.cosec α
A=2π r.dl = 2πr.dr cosec α
We shall consider the following two cases :1.When there is a uniform pressure, and
2.When there is a uniform wear.
We know that normal load acting on the ring,
δW
n= Normal pressure × Area of ring = pn × 2 π r.dr.cosec α
and the axial load acting on the ring,
δW= Horizontal component of δWn (i.e. in the direction of W)

=δWn × sin α = pn × 2π r.dr. cosec α × sin α = 2π × pn.r.dr
∴  Total axial load transmitted to the clutch or the axial spring force required,
                            11
2222 2
12() ()2. .2 2
2 2rr
nn n
rrrr rW prd r pp  −= π =π =π   ∫
22
12() ()nprr=π −
∴ 22
12[( ) ( ) ]nWp
rr=
π−...(i)

312  /G6C   Theory of Machines
Fr=µ.δWn = µ.pn × 2 π r.dr.cosec α
                                 Tr = Fr × r = µ.pn × 2 π r.dr. cosec α.r = 2 π µ.pn.cosec α.r2 dr
Integrating this expression within the limits from r2 to r1 for the total frictional torque on the
clutch.
211
2322 . .cosec . . 2 .cosec
3rr
nn
rrrT p rd r p=π µ α= π µ α∫
33

3rr
nn
33
12() ()2. c o s e c3nrrp−=π µ α 
Substituting the value of pn from equation  (i), we get
33
12
22
12() ()2c o s e c3 [( ) ( ) ]rr WT
rr−=π µ × × α  π−
 33
12
22
−..(ii)
In Fig. 10.25, let pr be the normal intensity of pressure at a distance r from the axis of the

clutch. We know that, in case of uniform wear, the intensity of pressure varies inversely with thedistance.
∴ p
r.r=C (a constant)    or    pr = C/r
We know that the normal load acting on the ring,
δWn= Normal pressure × Area of ring = pr × 2πr.dr cosec α
and the axial load acting on the ring ,
                            δ W = δWn × sin α = pr.2 π r.dr.cosec α .sin α = pr × 2 π r.dr
2. 2 .Crdr Cd rr=× π = π ...(∵  pr = C/r)
∴   Total axial load transmitted to the clutch,
[]1
1
2
2122. 2 2( )r

[]1
1
2
r
r
rWC d r C C r r r =π= π = π−∫
or                          
122( )WC
rr=
π−...(iii)
We know that frictional force acting on the ring,
Fr=µ.δWn = µ.pr × 2 π r × dr cosec α
Tr=Fr × r = µ.pr  × 2 π r.dr.cosec α × r
22 . .cosec 2 . cosecCrd r C r d rr=µ× × π α= πµ α×

Chapter 10  : Friction    /G6C  313
∴  Total frictional torque acting on the clutch,
 11
222
2. . c o s e c . 2. . c o s e c
2rr
rrrTC r d r C=π µ α = π µ α∫
22
12() ()2. . c o s e c
2rrC−=π µ α 
Substituting the value of C from equation (i), we have
22
12
12() ()2c o s e c2( ) 2W rrTrr−=π µ × × α π− 
12 . cosec . . cosec
2rrWW R+=µ α =µ α ...(iv)
where12
2rrR+==  Mean radius of friction surface

where12
Since the normal force acting on the friction surface, Wn = W/sin α, therefore the equation
(iv) may be written as
T=µ.Wn.R ...(v)
The forces on a friction surface, for steady operation of the clutch and after the clutch is
engaged, is shown in Fig. 10.26.
Fig. 10.26. Forces on a friction surface.
From Fig. 10.26 ( a), we find that
12
12 12sin ; and or 22rrrrb R rr R+−= α = +=

314  /G6C   Theory of Machines
∴  From equation, (i), normal pressure acting on the friction surface,
22
12 12 12() () 2 . . s i n [( ) ( ) ]nWW Wprr rr R b rr== =π+ − π α π−
or W=pn × 2 π R.b sin α = Wn sin α
where Wn= Normal load acting on the friction surface = pn × 2 π R.b
Now the equation  (iv) may be written as,
2(2 . s i n ) c o s e c2 . .nnTpRb R pRb =µ × π α α= πµ
The following points may be noted for a cone clutch :

1. The above equations are valid for steady operation of the clutch and after the clutch is
engaged.
2. If the clutch is engaged when one member is stationary and the other rotating ( i.e. during
engagement of the clutch) as shown in Fig. 10.26 ( b), then the cone faces will tend to slide on each
other due to the presence of relative motion. Thus an additional force (of magnitude equal to µ.Wn.cos α)

acts on the clutch which resists the engagement and the axial force required for engaging the clutchincreases.
∴   Axial force required for engaging the clutch,
W
e=W + µ.Wn cos α = Wn sin α + µ.Wn cos α
=Wn (sin α + µ cos α)
3. Under steady operation of the clutch, a decrease in the semi-cone angle ( α) increases the
torque produced by the clutch ( T ) and reduces the axial force ( W ). During engaging period, the axial

force required for engaging the clutch ( We) increases under the influence of friction as the angle α
decreases. The value of α can not be decreased much because smaller semi-cone angle ( α) requires
larger axial force for its disengagement.
For free disengagement of the clutch, the value of tan α must be greater than µ. In case the
value of tan α is less than µ, the clutch will not disengage itself and the axial force required to
disengage the clutch is given by
Wd=Wn (µ cos α – sin α)

Example 10.31. A conical friction clutch is used to transmit 90 kW at 1500 r.p.m. The semi-
cone angle is 20º and the coefficient of friction is 0.2. If the mean diameter of the bearing surface is375 mm and the intensity of normal pressure is not to exceed 0.25 N/mm
2, find the dimensions of the
conical bearing surface and the axial load required.
Solution. Given : P = 90 kW = 90 × 103 W ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 156

rad/s ; α = 20º ; µ = 0.2 ; D = 375 mm or R = 187.5 mm ; pn = 0.25 N/mm2
Dimensions of the conical bearing surface
Let r1 and r2= External and internal radii of the bearing surface respectively,
b= Width of the bearing surface in mm, and
90 × 103=T.ω = T × 156
∴ T= 90 × 103/156 = 577 N-m = 577 × 103 N-mm

Chapter 10  : Friction    /G6C  315
and the torque transmitted ( T),
577 × 103=2  π µ pn.R2.b = 2π × 0.2 × 0.25 (187.5)2 b = 11 046 b
∴ b= 577 × 103/11 046 = 52.2 mm  Ans.
We know that r1 + r2=2R = 2 × 187.5 = 375 mm ... (i)
and r1 – r2=b sin α = 52.2 sin 20º = 18 mm ... (ii)
r1= 196.5 mm, and r2 = 178.5 mm  Ans.
Axial load required
Since in case of friction clutch, uniform wear is considered and the intensity of pressure is

maximum at the minimum contact surface radius ( r2), therefore
pn.r2=C (a constant) or C = 0.25 × 178.5 = 44.6 N/mm
We know that the axial load required,
W=2πC (r1 – r2) = 2π × 44.6 (196.5 – 178.5) = 5045 N  Ans.
Example 10.32. An engine developing 45 kW at 1000 r.p.m.  is fitted with a cone clutch built

inside the flywheel. The cone has a face angle of 12.5º and a maximum mean diameter of 500 mm.The coefficient of friction is 0.2. The normal pressure on the clutch face is not to exceed 0.1 N/mm
2.
Determine : 1. the axial spring force necessary to engage to clutch, and 2. the face width required.
Solution. Given : P = 45 kW = 45 × 103 W ; N = 1000 r.p.m. or ω = 2π × 1000/60 = 104.7
rad/s ; α = 12.5º ; D = 500 mm or R = 250 mm = 0.25 m ; µ = 0.2 ; pn = 0.1 N/mm2

1.  Axial spring force necessary to engage the clutch
First of all, let us find the torque ( T ) developed by the clutch and the normal load ( Wn) acting
on the friction surface.
We know that power developed by the clutch ( P),
45 × 103=T.ω = T × 104.7   or   T = 45 × 103/104.7 = 430 N-m
We also know that the torque developed by the clutch ( T),
430 =µ.Wn.R = 0.2 × Wn × 0.25 = 0.05 Wn
∴ Wn= 430/0.05 = 8600 N
and axial spring force necessary to engage the clutch,
We=Wn (sin α + µ cos α)

= 8600 (sin 12.5º + 0.2 cos 12.5º) = 3540 N   Ans.
2.  Face width required
Let b= Face width required.
We know that normal load acting on the friction surface ( Wn),
8600 =pn × 2 π R.b = 0.1 × 2 π × 250 × b = 157 b
∴ b= 8600/157 = 54.7 mm    Ans.
Example 10.33. A leather faced conical clutch has a cone angle of 30º. If the intensity of
pressure between the contact surfaces is limited to 0.35 N/mm2 and the breadth of the conical surface

is not to exceed one-third of the mean radius, find the dimensions of the contact surfaces to transmit22.5 kW at 2000 r.p.m. Assume uniform rate of wear and take coefficient of friction as 0.15.
Solution. Given : 2 α = 30º  or  α = 15º  ; p
n = 0.35 N/mm2; b = R/3 ; P = 22.5 kW =
22.5 × 103 W ; N = 2000 r.p.m. or ω = 2 π × 2000/60 = 209.5 rad/s ; µ = 0.15
Let r1= Outer radius of the contact surface in mm,

316  /G6C   Theory of Machines
r2= Inner radius of the contact surface in mm,
R= Mean radius of the the contact surface in mm,
b= Face width of the contact surface in mm = R/3, and
T= Torque transmitted by the clutch in N-m.
22.5 × 103=T.ω = T × 209.5
∴ T= 22.5 × 103/209.5 = 107.4 N-m = 107.4 × 103 N-mm
We also know that torque transmitted ( T),
107.4 × 103=2π µ pn.R2. b = 2π × 0.15 × 0.35 × R2 × R/3 = 0.11 R3

∴ R3= 107.4 × 103/0.11 = 976.4 × 103    or    R = 99 mm   Ans.
The dimensions of the contact surface are shown in Fig. 10.27.
Fig. 10.27
From Fig. 10.27, we find that
                     1299sin sin sin 15º 8.54 mm33Rrrb−= α =× α = × = ...(i)
and122 2 99 198 mm rr R+= = ×= ...(ii)
r1= 103.27 mm,  and r2 = 94.73 mm  Ans.
Example 10.34. The contact surfaces in a cone clutch have an effective diameter of 75 mm.

The semi-angle of the cone is 15º. The coefficient of friction is 0.3. Find the torque required to
produce slipping of the clutch if an axial force applied is 180 N.
This clutch is employed to connect an electric motor running uniformly at 1000 r.p.m. with a
flywheel which is initially stationary. The flywheel has a mass of 13.5 kg and its radius of gyration is
150 mm. Calculate the time required for the flywheel to attain full speed and also the energy lost inthe slipping of the clutch.

Solution. Given : D = 75 mm or R = 37.5 mm = 0.0375 m ; α = 15º ; µ = 0.3 ; W = 180 N ;
N
F = 1000 r.p.m. or ωF = 2π × 1000/60 = 104.7 rad/s ; m = 13.5 kg ; k = 150 mm = 0.15 m
Torque required to produce slipping
We know that torque required to produce slipping,
T=µ.W.R.cosec α = 0.3 × 180 × 0.0375 × cosec 15º = 7.8 N-m  Ans.
Time required for the flywheel to attain full speed
Let tF= Time required for the flywheel to attain full speed in seconds, and

αF= Angular acceleration of the flywheel in rad/s2.
We know that the mass moment of inertia of the flywheel,
IF=m.k2 = 13.5 × (0.15)2 = 0.304 kg-m2

Chapter 10  : Friction    /G6C  317
∴  Torque required ( T ),
         7.8 = IF.αF = 0.304 αF   or   αF = 7.8/0.304 = 25.6 rad/s2
and angular speed of the flywheel ( ωF),
      104.7 = αF.tF = 25.6 tF   or    tF = 104.7/25.6 = 4.1 s   Ans.
Energy lost in slipping of the clutch
We know that the angle turned through by the motor and flywheel ( i.e. clutch) in time 4.1 s
from rest,
              FF11Average angular velocity × time = 104.7 4.1 214.6 rad22Wt θ= × × = × × =

∴  Energy lost in slipping of the clutch,
               = T.θ = 7.8 × 214.6 = 1674 N-m  Ans.
10.35. Centrifugal Clutch
The centrifugal clutches are usually incorporated into the motor pulleys. It consists of a
number of shoes on the inside of a rim of the pulley, as shown in Fig. 10.28. The outer surface of theshoes are covered with a friction material. These shoes, which can move radially in guides, are held
Fig. 10.28. Centrifugal clutch.

against the boss (or spider) on the driving shaft by means of
springs. The springs exert a radially inward force which isassumed constant. The mass of the shoe, when revolving, causesit to exert a radially outward force ( i.e. centrifugal force). The

magnitude of this centrifugal force depends upon the speed atwhich the shoe is revolving. A little consideration will showthat when the centrifugal force is less than the spring force, theshoe remains in the same position as when the driving shaftwas stationary, but when the centrifugal force is equal to thespring force, the shoe is just floating. When the centrifugalforce exceeds the spring force, the shoe moves outward andcomes into contact with the driven member and presses againstit. The

with the driven member and presses againstit. The force with which the shoe presses against the drivenmember is the difference of the centrifugal force and the springforce. The increase of speed causes the shoe to press harder

318  /G6C   Theory of Machines
and enables more torque to be transmitted.
In order to determine the mass and size of the shoes, the following procedure is adopted :
1.  Mass of the shoes
Consider one shoe of a centrifugal clutch as shown in Fig. 10.29.
Let m= Mass of each shoe,
n= Number of shoes,
r= Distance of centre of gravity of
the shoe from the centre of thespider,
R= Inside radius of the pulley rim,
N= Running speed of the pulley in
r.p.m.,
ω= Angular running speed of the

r.p.m.,
pulley in rad/s = 2 πN/60 rad/s,
ω
1= Angular speed at which the
engagement begins to take place,and
µ= Coefficient of friction between
the shoe and rim.
We know that the centrifugal force acting on each shoe at the running speed,
*P
c=m.ω2.r
and the inward force on each shoe exerted by the spring at the speed at which engagement begins to
take place,
Ps=m (ω1)2 r

∴  The net outward radial force ( i.e. centrifugal  force) with which the shoe presses against
the rim at the running speed
=Pc – Ps
and the frictional force acting tangentially on each shoe,
F=µ (Pc – Ps)
∴  Frictional torque acting on each shoe,
=F × R = µ (Pc – Ps) R
and total frictional torque transmitted,
T=µ (Pc – Ps) R × n = n.F.R
From this expression, the mass of the shoes ( m) may be evaluated.
2.  Size of the shoes
Let l= Contact length of the shoes,

b= Width of the shoes,
*The radial clearance between the shoe and the rim being very small as compared to r, therefore it is neglected.
If, however, the radial clearance is given, then the operating radius of the mass centre of the shoe from the axisof the clutch,
                             r
1 = r + c, where c = Radial clearance.
Then                              Pc = m.ω2.r1, and Ps = m (ω1)2 r1Fig. 10.29. Forces on a shoe of
centrifugal clutch.

Chapter 10  : Friction    /G6C  319
R= Contact radius of the shoes. It is same as the inside radius of the rim
of the pulley.
θ= Angle subtended by the shoes at the centre of the spider in radians.
p= Intensity of pressure exerted on the shoe. In order to ensure reason-
able life, the intensity of  pressure may be taken as 0.1 N/mm2.
We know that         θ = l/R rad     or     l = θ.R
∴  Area of contact of the shoe,
A=l.b
and the force with which the shoe presses against the rim
=A × p = l.b.p

Since the force with which the shoe presses against the rim at the running speed is ( Pc – Ps),
therefore
l.b.p=Pc – Ps
From this expression, the width of shoe ( b) may be obtained.
Example 10.35. A centrifugal clutch is to transmit 15 kW at 900 r.p.m. The shoes are four in

number. The speed at which the engagement begins is 3/4th of the running speed. The inside radiusof the pulley rim is 150 mm and the centre of gravity of the shoe lies at 120 mm from the centre of thespider. The shoes are lined with Ferrodo for which the coefficient of friction may be taken as 0.25.
Determine : 1. Mass of the shoes, and 2. Size of the shoes, if angle subtended by the shoes at the
centre of the spider is 60º and the pressure exerted on the shoes is 0.1 N/mm
2.

2.
Solution. Given : P = 15 kW = 15 × 103 W ; N = 900 r.p.m. or ω = 25 × 900/60 = 94.26 rad/s ;
n = 4 ; R = 150 mm = 0.15 m ; r = 120 mm = 0.12 m ; µ = 0.25
Since the speed at which the engagement begins ( i.e. ω1) is 3/4th of the running speed ( i.e.
ω), therefore
                          13394.26 70.7 rad/s44ω= ω = × =
Let T= Torque transmitted at the running speed.
15 × 103=T.ω = T × 94.26    or    T = 15 × 103/94.26 = 159 N-m

Let m= Mass of the shoes in kg.
We know that the centrifugal force acting on each shoe,
Pc=m.ω2.r = m (94.26)2 × 0.12 = 1066 m N
and the inward force on each shoe exerted by the spring i.e. the centrifugal force at the engagement
speed ω1,
Ps=m (ω1)2 r = m (70.7)2 × 0.12 = 600 m N
∴  Frictional force acting tangentially on each shoe,
F=µ (Pc – Ps) = 0.25 (1066 m – 600 m) = 116.5 m N
We know that the torque transmitted ( T ),

159 =n.F.R = 4 × 116.5 m × 0.15 = 70 m    or   m = 2.27 kg  Ans.
Let l= Contact length of shoes in mm,
b= Width of the shoes in mm,

320  /G6C   Theory of Machines
θ= Angle subtended by the shoes at the centre of the spider in radians
= 60º = π/3 rad, and ...(Given)
p= Pressure exerted on the shoes in N/mm2 = 0.1 N/mm2...(Given)
We know that         . 150 157.1mm3lRπ=θ = × =
and l.b.p=Pc – Ps = 1066 m – 600 m = 466 m
∴    157.1 × b × 0.1 = 466 × 2.27 = 1058
or b= 1058/157.1 × 0.1 = 67.3  mm  Ans.
Example 10.36. A centrifugal clutch has four shoes which slide radially in a spider keyed to

the driving shaft and make contact with the internal cylindrical surface of a rim keyed to the drivenshaft. When the clutch is at rest, each shoe is pulled against a stop by a spring so as to leave a radialclearance of 5 mm between the shoe and the rim. The pull exerted by the spring is then 500 N. Themass centre of the shoe is 160 mm from the axis of the clutch.
If the internal diameter of the rim is 400 mm, the mass of each shoe is 8 kg, the stiffness of

each spring is 50 N/mm and the coefficient of friction between the shoe and the rim is 0.3 ; find thepower transmitted by the clutch at 500 r.p.m.
Solution. Given : n = 4 ; c = 5 mm ; S = 500 N ; r = 160 mm ; D = 400 mm or R = 200 mm
= 0.2 m ; m = 8 kg ; s = 50 N/mm ; µ = 0.3 ; N = 500 r.p.m. or ω = 2 π × 500/60 = 52.37 rad/s
We know that the operating radius,
r
1=r + c = 160 + 5 = 165 mm = 0.165 m
Centrifugal force on each shoe,
Pc=m.ω2.r1 = 8 (52.37)2 × 0.165 = 3620 N

and the inward force exerted by the spring,
P4=S + c.s = 500 + 5 × 50 = 750 N
F=µ (Pc – Ps) = 0.3 (3620 – 750) = 861 N
We know that total frictional torque transmitted by the clutch,
T=n.F.R = 4 × 861 × 0.2 = 688.8 N-m
∴  Power transmitted,
P=T.ω = 688.8 × 52.37 = 36 100 W = 36.1 kW  Ans.
EXERCISES
1.Find the force required to move a load of 300 N up a rough plane, the force being applied parallel to

the plane. The inclination of the plane is such that a force of 60 N inclined at 30º to a similar smooth
plane would keep the same load in equilibrium. The coefficient of friction is 0.3. [Ans. 146 N]
2.A square threaded screw of mean diameter 25 mm and pitch of thread 6 mm is utilised to lift a weight
of 10 kN by a horizontal force applied at the circumference of the screw. Find the magnitude of theforce if the coefficient of friction between the nut and screw is 0.02. [Ans. 966 N]

3.A bolt with a square threaded screw has mean diameter of 25 mm and a pitch of 3 mm. It carries an
axial thrust of 10 kN on the bolt head of 25 mm mean radius. If µ = 0.12, find the force required at the
end of a spanner 450 mm long, in tightening up the bolt.  [Ans. 110.8 N]
4.A turn buckle, with right and left hand threads is used to couple two railway coaches. The threads

which are square have a pitch of 10 mm and a mean diameter of 30 mm and are of single start type.Taking the coefficient of friction as 0.1, find the work to be done in drawing the coaches together adistance of 200 mm against a steady load of 20 kN.  [Ans. 3927 N-m]

Chapter 10  : Friction    /G6C  321
5.A vertical two start square threaded screw of a 100 mm mean diameter and 20 mm pitch supports a

vertical load of 18 kN. The axial thrust on the screwis taken by a collar bearing of 250 mm outside diam-eter and 100 mm inside diameter. Find the force re-quired at the end of a lever which is 400 mm long inorder to lift and lower the load. The coefficient offriction for the vertical screw and nut is 0.15 andthat for collar bearing is 0.20.                                              [Ans. 1423 N ; 838 N]
6.A sluice gate weighing 18 kN is raised and lowered

by means of square threaded screws, as shown inFig.10.30. The frictional resistance induced by waterpressure against the gate when it is in its lowestposition is 4000 N.
The outside diameter of the screw is 60 mm and pitch
is 10 mm. The outside and inside diameter of washeris 150 mm and 50 mm respectively. The coefficientof friction between the screw and nut is 0.1 and forthe washer and seat is 0.12. Find :

1. The maximum force to be exerted at the ends of the lever for raising and lowering the gate, and
2. Efficiency of the arrangement. [Ans. 114 N ; 50 N ; 15.4%]
7.The spindle of a screw jack has single start square threads with an outside diameter of 45 mm and a

pitch of 10 mm. The spindle moves in a fixed nut. The load is carried on a swivel head but is not freeto rotate. The bearing surface of the swivel head has a mean diameter of 60 mm. The coefficient offriction between the nut and screw is 0.12 and that between the swivel head and the spindle is 0.10.Calculate the load which can be raised by efforts of 100 N each applied at the end of two levers eachof effective length of 350 mm. Also determine the velocity ratio and the efficiency of the

the velocity ratio and the efficiency of the liftingarrangement.  [Ans. 9943 N ; 218.7 N ; 39.6%]

8.The lead screw of a lathe has acme threads of 50 mm outside diameter and 10 mm pitch. The included

angle of the thread is 29°. It drives a tool carriage and exerts an axial pressure of 2500 N. A collarbearing with outside diameter 100 mm and inside diameter 50 mm is provided to take up the thrust. Ifthe lead screw rotates at 30 r.p.m., find the efficiency and the power required to drive the screw. Thecoefficient of friction for screw threads is 0.15 and for the collar is 0.12. [Ans. 16.3% ; 75.56 W]

9.A flat foot step bearing 225 mm in diameter supports a load of 7.5 kN. If the coefficient of friction is
0.09 and r.p.m is 60, find the power lost in friction, assuming 1. Uniform pressure, and 2. Uniformwear.  [Ans. 318 W ; 239 W]
10.A conical pivot bearing 150 mm in diameter has a cone angle of 120º. If the shaft supports an axial

load of 20 kN and the coefficient of friction is 0.03, find the power lost in friction when the shaftrotates at 200 r.p.m., assuming 1. Uniform pressure, and 2. uniform wear.
[Ans. 727.5 W ; 545.6 W]
11.A vertical shaft supports a load of 20 kN in a conical pivot bearing. The external radius of the cone is
3 times the internal radius and the cone angle is 120º. Assuming uniform intensity of pressure as 0.35MN/m
2, determine the dimensions of the bearing.

If the coefficient of friction between the shaft and bearing is 0.05 and the shaft rotates at 120 r.p.m.,
find the power absorbed in friction.  [Ans. 47.7 mm ; 143 mm ; 1.50 kW]
12.A plain collar type thrust bearing having inner and outer diameters of 200 mm and 450 mm is sub-

jected to an axial thrust of 40 kN. Assuming coefficient of friction between the thrust surfaces as0.025, find the power absorbed in overcoming friction at a speed of 120 r.p.m. The rate of wear isconsidered to be proportional to the pressure and rubbing speed.  [Ans. 4.1 kW]
13.The thrust on the propeller shaft of a marine engine is taken up by 8 collars whose external and
internal diameters are 660 mm and 420 mm respectively. The thrust pressure is 0.4 MN/m
2 and mayFig. 10.30

322  /G6C   Theory of Machines
be assumed uniform. The coefficient of friction between the shaft and collars is 0.04. If the shaft
rotates at 90 r.p.m. ; find 1. total thrust on the collars ; and 2. power absorbed by friction at the bearing.
[Ans. 651 kN ; 68 kW]
14.A shaft has a number of collars integral with it. The external diameter of the collars is 400 mm and the
shaft diameter is 250 mm. If the uniform intensity of pressure is 0.35 N/mm2 and its coefficient of

friction is 0.05, estimate : 1. power absorbed in overcoming friction when the shaft runs at 105 r.p.m.and carries a load of 150 kN, and 2. number of collars required. [Ans. 13.4 kW ; 6]
15.A car engine has its rated output of 12 kW. The maximum torque developed is 100 N-m. The clutch
used is of single plate type having two active surfaces. The axial pressure is not to exceed 85 kN/m
2.

2.
The external diameter of the friction plate is 1.25 times the internal diameter. Determine the dimen-sions of the friction plate and the axial force exerted by the springs. Coefficient of friction = 0.3.
[Ans. 129.5 mm ; 103.6 mm ; 1433 N ]
16.A single plate clutch (both sides effective) is required to transmit 26.5 kW at 1600 r.p.m. The outer
diameter of the plate is limited to 300 mm and intensity of pressure between the plates is not to exceed68.5 kN/m

2. Assuming uniform wear and a coefficient of friction 0.3, show that the inner diameter of
the plates is approximately 90 mm.
17.A multiplate clutch has three pairs of contact surfaces. The outer and inner radii of the contact sur-
faces are 100 mm and 50 mm respectively. The maximum axial spring force is limited to 1 kN. If thecoefficient of friction is 0.35 and assuming uniform wear, find the power transmitted by the clutch at1500 r.p.m.  [Ans. 12.37 kW]

18.A cone clutch is to transmit 7.5 kW at 900 r.p.m. The cone has a face angle of 12º. The width of the
face is half of the mean radius and the normal pressure between the contact faces is not to exceed 0.09N/mm
2. Assuming uniform wear and the coefficient of friction between contact faces as 0.2, find the
main dimensions of the clutch and the axial force required to engage the clutch.
[Ans. R = 112 mm, b = 56 mm, r1 = 117.8 mm, r2 = 106.2 mm ; 1433 N]

19.A cone clutch with cone angle 20º is to transmit 7.5 kW at 750 r.p.m. The normal intensity of pressure
between the contact faces is not to exceed 0.12 N/mm2. The coefficient of friction is 0.2. If face width
is 15th of mean diameter, find : 1. the main dimensions of the clutch, and 2. axial force required
while running.  [Ans. R = 117 mm ; b = 46.8 mm ; r1 = 125 mm ; r2 = 109 mm ; 1395 N]

20.A centrifugal friction clutch has a driving member consisting of a spider carrying four shoes which are
kept from contact with the clutch case by means of flat springs until increase of centrifugal forceovercomes the resistance of the springs and the power is transmitted by friction between the shoes andthe case.
Determine the necessary mass of each shoe if 22.5 kW is to be transmitted at 750 r.p.m. with

engagement beginning at 75% of the running speed. The inside diameter of the drum is 300 mm andthe radial distance of the centre of gravity of each shoe from the shaft axis is 125 mm. Assumeµ = 0.25. [Ans. 5.66 kg]
1.Discuss briefly the various types of friction experienced by a body.
2.State the laws of
(i) Static friction ; ( ii)  Dynamic friction ;
(iii) Solid friction ; and ( iv) Fluid friction.
3.Explain the following :
(i) Limiting friction, ( ii) Angle of friction, and

(iii) Coefficient of friction.
4.Derive from first principles an expression for the effort required to raise a load with a screw jack
taking friction into consideration.
5.Neglecting collar friction, derive an expression for mechanical advantage of a square threaded screw
moving in a nut, in terms of helix angle of the screw and friction angle.

Chapter 10  : Friction    /G6C  323
6.In a screw jack, the helix angle of thread is α and the angle of friction is φ. Show that its efficiency is
maximum, when 2 α = (90º – φ).
7.For a screw jack having the nut fixed, derive the equation ( with usual notations),
                                                 tan.
tan ( ) . .mrrαη=
α+φ +µ
8.Neglecting collar friction, from first principles, prove that the maximum efficiency of a square threaded
screw moving in a nut is 1s i n,1s i n−φ

+φ where φ is the friction angle.
9.Write a short note on journal bearing.
10.What is meant by the expression ‘friction circle’? Deduce an expression for the radius of friction
circle in terms of the radius of the journal and the angle of friction.
11.From first principles, deduce an expression for the friction moment of a collar thrust bearing, stating
clearly the assumptions made.

12.Derive an expression for the friction moment for a flat collar bearing in terms of the inner radius r1,
outer radius r2, axial thrust W and coefficient of friction µ. Assume uniform intensity of pressure.
13.Derive from first principles an expression for the friction moment of a conical pivot assuming
(i) Uniform pressure, and ( ii) Uniform wear.
14.A truncated conical pivot of cone angle φ rotating at speed N supports a load W. The smallest and

largest diameter of the pivot over the contact area are ‘ d’ and ‘D’ respectively. Assuming uniform
wear, derive the expression for the frictional torque.
15.Describe with a neat sketch the working of a single plate friction clutch.
16.Establish a formula for the maximum torque transmitted by a single plate clutch of external and
internal  radii r1 and r2, if the limiting coefficient of friction is µ and the axial spring load is W. Assume

that the pressure intensity on the contact faces is uniform.
17.Which of the two assumptions-uniform intensity of pressure or uniform rate of wear, would you make
use of in designing friction clutch and why ?
18.Describe with a neat sketch a centrifugal clutch and deduce an equation for the total torque transmitted.
1.The angle of inclination of the plane, at which the body begins to move down the plane, is called

(a) angle of friction ( b) angle of repose ( c) angle of projection
2.In a screw jack, the effort required to lift the load W is given by
(a)P = W tan (α – φ)( b)P = W tan (α + φ)
(c)P = W cos (α – φ)( d)P = W cos (α + φ)
whereα = Helix angle, and
φ = Angle of friction.
3.The efficiency of a screw jack is given by
(a)tan( )
tanα+φ
α(b)tan
tan ( )α
α+φ
(c)tan( )
tanα−φ
α(d)tan
tan( )α
α−φ
4.The radius of a friction circle for a shaft of radius r rotating inside a bearing is

324  /G6C   Theory of Machines
5.The efficiency of a screw jack is maximum, when
(a)45º2φα= + (b)45º2φα= − (c)90ºα= +φ (d)90ºα= −φ
6.The maximum efficiency of a screw jack is
(a)1s i n
1s i n−φ
+φ(b)1s i n
1s i n+φ
−φ(c)1t a n
1t a n−φ
+φ(d)1t a n
1t a n+φ
−φ
7.The frictional torque transmitted in a  flat pivot bearing, considering uniform pressure, is
(a)1..2WR×µ (b)2..3WR×µ (c)3..4WR×µ (d)..WRµ
whereµ = Coefficient of friction,
W = Load over the bearing, and
R = Radius of the bearing surface.

8.The frictional torque transmitted in a conical pivot bearing, considering uniform wear, is
(a)1..c o s e c2WR×µ α (b)2..c o s e c3WR×µ α
(c)3..c o s e c4WR×µ α (d)µ . W.R cosec α
whereR = Radius of the shaft, and
α = Semi-angle of the cone.
9.The frictional torque transmitted by a disc or plate clutch is same as that of
(a) flat pivot bearing ( b) flat collar bearing
(c) conical pivot bearing ( d) trapezoidal pivot bearing

10.The frictional torque transmitted by a cone clutch is same as that of
ANSWERS
1. (a) 2.(b) 3. (b) 4. (a) 5. (b)
6. (a) 7. (b) 8. (a) 9. (b) 10. (d)

Chapter 11 : Belt, Rope and Chain Drives    /G6C  325
11.1. Introduction
The belts or ropes are used to transmit power from
one shaft to another by means of pulleys which rotate at thesame speed or at different speeds. The amount of power trans-mitted depends upon the following factors :
1.The velocity of the belt.
2.The tension under which the belt is placed on the
pulleys.
3.The arc of contact between the belt and the smaller
pulley.
4.The conditions under which the belt is used.

(a)The shafts should be properly in line to insure
uniform tension across the belt section.
(b)The pulleys should not be too close together, in
order that the arc of contact on the smaller pul-ley may be as large as possible.
(c)The pulleys should not be so far apart as to cause
the belt to weigh heavily on the shafts, thus in-creasing the friction load on the bearings.
325Belt, Rope
and Chain
Drives11

and Chain
Drives11
4. Types of Belts.
5. Material used for Belts.6. Types of Flat Belt Drives.
7. Velocity Ratio of Belt Drive.
11. Length of an Open Belt Drive.
13. Power Transmitted by a Belt.
14. Ratio of Driving Tensions for
Flat Belt Drive.
16. Centrifugal Tension.
17. Maximum T ension in the Belt.
19. Initial Tension in the Belt.20. V-belt Drive.
22. Ratio of Driving Tensions for
V-belt.
23. Rope Drive.
24. Fibre Ropes.

V-belt.
26. Sheave for Fibre Ropes.27. Wire Ropes.
28. Ratio of Driving Tensions for
Rope Drive.
29. Chain Drives.
30. Advantages and
Disadvantages of ChainDrive Over Belt or Rope
Drive.
31. Terms Used in Chain Drive.32. Relation Between Pitch and
Pitch Circle Diameter.
33. Relation Between Chain
Speed and Angular Velocity of
Sprocket.
34. Kinematic of Chain Drive.35. Classification of Chains.
36. Hoisting and Hauling Chains.

37. Conveyor Chains.38. Power Transmitting Chains.
39. Length of Chains.

326  /G6C   Theory of Machines
(d)A long belt tends to swing from side to side, causing the belt to run out of the pulleys,
which in turn develops crooked spots in the belt.
(e)The tight side of the belt should be at the bottom, so that whatever sag is present on the
loose side will increase the arc of contact at the pulleys.
(f) In order to obtain good results with flat belts, the maximum distance between the shafts

should not exceed 10 metres and the minimum should not be less than 3.5 times thediameter of the larger pulley.
11.2. Selection of a Belt Drive
Following are the various important factors upon which the selection of a belt drive depends:
1.Speed of the driving and driven shafts, 2.Speed reduction ratio,
3.Power to be transmitted, 4.Centre distance between the shafts,
5.Positive drive requirements, 6. Shafts layout,
7.Space available, and 8.Service conditions.
11.3. Types of Belt Drives

The belt drives are usually classified into the following three groups :
1.  Light drives . These are used to transmit small powers at belt speeds upto about 10 m/s, as
in agricultural machines and small machine tools.
2.  Medium drives . These are used to transmit medium power at belt speeds over 10 m/s but
up to 22 m/s, as in machine tools.
3.  Heavy drives . These are used to transmit large powers at belt speeds above 22 m/s, as in
compressors and generators.

11.4. Types of Belts
(a)  Flat belt. ( b)  V-belt. ( c)  Circular belt.
Fig. 11.1. Types of belts.
Though there are many types of belts used these days, yet the following are important from
the subject point of view :
1.  Flat belt. The flat belt, as shown in Fig. 11.1 ( a), is mostly used in the factories and
workshops, where a moderate amount of power is to be transmitted, from one pulley to another whenthe two pulleys are not more than 8 metres apart.

2.  V-belt. The V-belt, as shown in Fig. 11.1 ( b), is mostly used in the factories and work-
shops, where a moderate amount of power is to be transmitted, from one pulley to another, when thetwo pulleys are very near to each other.
3.  Circular belt or rope . The circular belt or rope, as shown in Fig. 11.1 ( c), is mostly used
in the factories and workshops, where a great amount of power is to be transmitted, from one pulleyto another, when the two pulleys are more than 8 meters apart.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  327
If a huge amount of power is to be transmitted, then a single belt may not be sufficient. In
such a case, wide pulleys (for V-belts or circular belts) with a number of grooves are used. Then a beltin each groove is provided to transmit the required amount of power from one pulley to another.
11.5. Material used for Belts
The material used for belts and ropes must be strong, flexible, and durable. It must have a

high coefficient of friction. The belts, according to the material used, are classified as follows :
1.  Leather belts . The most important material for the belt is leather. The best leather belts are

made from 1.2 metres to 1.5 metres long strips cut from either side of the back bone of the top gradesteer hides. The hair side of the leather is smoother and harder than the flesh side, but the flesh side isstronger. The fibres on the hair side are perpendicular to the surface, while those on the flesh side areinterwoven and parallel to the surface. Therefore for these reasons, the hair side of a belt should be incontact with the pulley surface, as shown in Fig. 11.2. This gives a more

surface, as shown in Fig. 11.2. This gives a more intimate contact between thebelt and the pulley and places the greatest tensile strength of the belt section on the outside, where thetension is maximum as the belt passes over the pulley.

(a)  Single layer belt. ( b)  Double layer belt.
Fig. 11.2.  Leather belts.
The leather may be either oak-tanned or mineral salt tanned e.g. chrome tanned. In order to
increase the thickness of belt, the strips are cemented together. The belts are specified according tothe number of layers e.g. single, double or triple ply and according to the thickness of hides used e.g.
light, medium or heavy.
The leather belts must be periodically cleaned and dressed or treated with a compound or

dressing containing neats foot or other suitable oils so that the belt will remain soft and flexible.
2.  Cotton or fabric belts . Most of the fabric belts are made by folding canvass or cotton duck

to three or more layers (depending upon the thickness desired) and stitching together. These belts arewoven also into a strip of the desired width and thickness. They are impregnated with some filler likelinseed oil in order to make the belts water proof and to prevent injury to the fibres. The cotton beltsare cheaper and suitable in warm climates, in damp atmospheres and in exposed positions. Since the

cotton belts require little attention, therefore these belts are mostly used in farm machinery, belt
conveyor etc.
3.  Rubber belt . The rubber belts are made of layers of fabric impregnated with rubber com-

position and have a thin layer of rubber on the faces. These belts are very flexible but are quicklydestroyed if allowed to come into contact with heat, oil or grease. One of the principal advantage ofthese belts is that they may be easily made endless. These belts are found suitable for saw mills, papermills where they are exposed to moisture.
4.  Balata belts . These belts are similar to rubber belts except that balata gum is used in place

of rubber. These belts are acid proof and water proof and it is not effected by animal oils or alkalies.The balata belts should not be at temperatures above 40° C because at this temperature the balatabegins to soften and becomes sticky. The strength of balata belts is 25 per cent higher than rubberbelts.

328  /G6C   Theory of Machines
11.6. Types of Flat Belt Drives
The power from one pulley to another may be transmitted by any of the following types of
belt drives:
1.  Open belt drive . The open belt drive, as shown in Fig. 11.3, is used with shafts arranged
parallel and rotating in the same direction. In this case, the driver A pulls the belt from one side ( i.e.
lower side RQ) and delivers it to the other side ( i.e. upper side LM). Thus the tension in the lower side

belt will be more than that in the upper side belt. The lower side belt (because of more tension) isknown as  tight side whereas the upper side belt (because of less tension) is known as slack side , as
shown in Fig. 11.3.
Fig. 11.3.   Open belt drive.
2.  Crossed or twist belt drive . The crossed or twist belt drive, as shown in Fig. 11.4, is used
with shafts arranged parallel and rotating in the opposite directions.
Fig. 11.4.   Crossed or twist belt drive.

In this case, the driver pulls the belt from one side ( i.e. RQ) and delivers it to the other side
(i.e. LM). Thus the tension in the belt RQ will be more than that in the belt LM. The belt RQ (because
of more tension) is known as tight side, whereas the belt LM (because of less tension) is known as
slack side , as shown in Fig. 11.4.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  329
A little consideration will show that at a point where the belt crosses, it rubs against each
other and there will be excessive wear and tear. In order to avoid this, the shafts should be placed ata maximum distance of 20 b, where b is the width of belt and the speed of the belt should be less than
15 m/s.
3.  Quarter turn belt drive . The quarter turn belt drive also known as right angle belt drive, as

shown in Fig. 11.5 ( a), is used with shafts arranged at right angles and rotating in one definite direc-
tion. In order to prevent the belt from leaving the pulley, the width of the face of the pulley should begreater or equal to 1.4 b, where b is the width of belt.
In case the pulleys cannot be arranged, as shown in Fig. 11.5 ( a), or when the reversible
motion is desired, then a quarter turn belt drive with guide pulley , as shown in Fig. 11.5 ( b), may be
used.

used.
(a)  Quarter turn belt drive. ( b)  Quarter turn belt drive with guide pulley.
Fig. 11.5
4.  Belt drive with idler pulleys . A belt drive with an idler pulley, as shown in Fig. 11.6 ( a), is
used with shafts arranged parallel and when an open belt drive cannot be used due to small angle ofcontact on the smaller pulley. This type of drive is provided to obtain high velocity ratio and when therequired belt tension cannot be obtained by other means.

(a)  Belt drive with single idler pulley. ( b)  Belt drive with many idler pulleys.
Fig. 11.6
When it is desired to transmit motion from one shaft to several shafts, all arranged in parallel,
a belt drive with many idler pulleys, as shown in Fig. 11.6 ( b), may be employed.

330  /G6C   Theory of Machines
Fig. 11.9.   Fast and loose pulley drive.5.  Compound belt drive . A compound belt drive, as shown in Fig. 11.7, is used when power
is transmitted from one shaft to another through a number of pulleys.
Fig.  11.7.   Compound belt brive.
6.  Stepped or cone pulley drive . A stepped or cone pulley drive, as shown in Fig. 11.8, is

used for changing the speed of the driven shaft while the main or driving shaft runs at constant speed.This is accomplished by shifting the belt from one part of the steps to the other.
7.  Fast and loose pulley drive . A fast and loose pulley drive, as shown in Fig. 11.9, is used
when the driven or machine shaft is to be started or stopped when ever desired without interferingwith the driving shaft. A pulley which is keyed to the machine shaft is called fast pulley  and runs at

the same speed as that of machine shaft. A loose pulley runs freely over the machine shaft and isincapable of transmitting any power. When the driven shaft is required to be stopped, the belt ispushed on to the loose pulley by means of sliding bar having belt forks.
Fig. 11.8.   Stepped or cone pulley drive.
11.7. Velocity Ratio of Belt Drive
It is the ratio between the velocities of the driver and the follower or driven.  It may be
expressed, mathematically, as discussed below :

Let d1= Diameter of the driver,
d2= Diameter of the follower,

Chapter 11 : Belt, Rope and Chain Drives            /G6C          331
N1= Speed of the driver in r.p.m., and
N2= Speed of the follower in r.p.m.
∴  Length of the belt that passes over the driver, in one minute
=π d1.N1
Similarly, length of the belt that passes over the
follower, in one minute
=π d2 . N2
Since the length of belt that passes over the
driver in one minute is equal to the length of belt that
passes over the follower in one minute, therefore
π d1 . N1=π d2 . N2
∴ Velocity ratio, 21

12Nd
Nd=
When the thickness of the belt ( t) is considered,
then velocity ratio,
 21
12Nd t
Nd t+=+
Note: The velocity ratio of a belt drive may also be obtained as discussed below :
We know that peripheral velocity of the belt on the driving pulley,
11
1.m/s60π=dNv
and peripheral velocity of the belt on the driven or follower pulley,
 22
2.m/s60dNvπ=
When there is no slip, then v1 = v2.
∴  11 2 2..
60 60dN d Nππ= or21
12Nd
Nd=

∴  11 2 2..
12Nd
Nd=
11.8.11.8.11.8.11.8.11.8. VVVVVelocity Raelocity Raelocity Raelocity Raelocity Ra tio of a Compound Belt Drtio of a Compound Belt Drtio of a Compound Belt Drtio of a Compound Belt Drtio of a Compound Belt Dr iviviviviveeeee
Sometimes the power is transmitted from one shaft to another, through a number of pulleys as
shown in Fig. 11.7. Consider a pulley 1 driving the pulley 2. Since the pulleys 2 and 3 are keyed to the

same shaft, therefore the pulley 1 also drives the pulley 3 which, in turn, drives the pulley 4.
Let d1= Diameter of the pulley 1,
N1= Speed of the pulley 1 in r.p.m.,
d2, d3, d4, and N2, N3, N4= Corresponding values for pulleys 2, 3 and 4.
We know that velocity ratio of pulleys 1 and 2,
                                21
12Nd
Nd=...(i)
Similarly, velocity ratio of pulleys 3 and 4,
                                3 4
34d N
Nd=...(ii)
Multiplying equations (i) and (ii),

                      3 24 1
13 2 4d NNd
NNdd×= ×

332  /G6C   Theory of Machines
or                         134
12 4ddN
Nd d×=×...(∵ N2 = N3, being keyed to the same shaft)
A little consideration will show, that if there are six pulleys, then
                                61 3 5
12 4 6Nd d d
Nd d d××=××
or               Speed of last driven Product of diameters of drivers
Speed of first driver Product of diameters of drivens=
11.9. Slip of Belt
In the previous articles, we have discussed the motion

of belts and shafts assuming a firm frictional grip between thebelts and the shafts. But sometimes, the frictional grip becomesinsufficient. This may cause some forward motion of the driverwithout carrying the belt with it. This may also cause someforward motion of the belt without carrying the driven pulleywith it. This is called slip of the belt  and is generally expressed
as a percentage.
The result of the belt slipping is to reduce the velocity

ratio of the system. As the slipping of the belt is a commonphenomenon, thus the belt should never be used where adefinite velocity ratio is of importance (as in the case of hour,minute and second arms in a watch).
Let s
1 % = Slip between the
driver and the belt, and
s2 % = Slip between the belt and the follower.
∴   Velocity of the belt passing over the driver per second
                                     11 11 1 11 1.. .–1 –60 60 100 60 100dN dN s dN svππ π =× = ...(i)

and velocity of the belt passing over the follower per second,
                          22 2 2.–1 –60 100 100dN s svv vπ =× = 
Substituting the value of v from equation ( i),
                          22 11 1 21– 1–60 60 100 100dN dN s sππ  =   
                                 21 1 2
121– –100 100Nd s s
Nd=12... Neglecting 100 100ss  ×
× 
                                        11 2 1
221– 1–100 100ds s d s
dd+  ==   

... (where s = s1 + s2, i.e. total percentage of slip)
If thickness of the belt ( t) is considered, then
                                 21
121–100Nd t s
Nd t+=+

Chapter 11 : Belt, Rope and Chain Drives    /G6C  333
Example 11.1.  An engine, running at 150 r.p.m., dri ves a line shaft by means of a belt. The
engine pulley is 750 mm diameter and the pulley on the line shaft being 450 mm. A 900 mm diameterpulley on the line shaft drives a 150 mm diameter pulley keyed to a dynamo shaft. Find the speed ofthe dynamo shaft, when 1. there is no slip, and 2. there is a slip of 2% at each drive.

1 = 150 r.p.m. ; d1 = 750 mm ; d2 = 450 mm ; d3 = 900 mm ; d4 = 150 mm
The arrangement of belt drive is shown in Fig. 11.10.Let N
4= Speed of the dynamo shaft .
Fig. 11.10
1. When there is no slip
We know that134
12 4ddN
Nd d×=×         or   4750 90010150 450 150N ×==×
∴ N4= 150 × 10 = 1500 r.p.m.  Ans.
2. When there is a slip of  2% at each drive
We know that        1341 2
12 41– 1–100 100ddNs s
Nd d× =  × 

                            4750 900 2 21– 1– 9 . 6150 450 150 100 100N × ==  × 
∴ N4= 150 × 9.6 = 1440 r.p.m.  Ans.
11.10. Creep of Belt
When the belt passes from the slack side to the tight side, a certain portion of the belt extends

and it contracts again when the belt passes from the tight side to slack side. Due to these changes oflength, there is a relative motion between the belt and the pulley surfaces. This relative motion istermed as creep. The total effect of creep is to reduce slightly the speed of the driven pulley or
follower. Considering creep, the velocity ratio is given by
                             
2 21
12 1E Nd
Nd E+σ=×
+σ

2 21
12 1E Nd
Nd E+σ=×
+σ
where                   σ1 and σ2= Stress in the belt on the tight and slack side respectively, and
E= Young’s modulus for the material of the belt.

334  /G6C   Theory of Machines
Example 11.2.  The power is transmitted from a pulley 1 m diameter running at 200 r.p.m. to
a pulley 2.25 m diameter by means of a belt. Find the speed lost by the driven pulley as a result ofcreep, if the stress on the tight and slack side of the belt is 1.4 MPa and 0.5 MPa respectively. TheYoung’s modulus for the material of the belt is 100 MPa.
Solution.  Given : d
1 = 1 m ; N1 = 200 r.p.m. ; d2 = 2.25 m ; σ1 = 1.4 MPa = 1.4 × 106 N/m2;

σ2 = 0.5 MPa = 0.5 × 106 N/m2 ; E = 100 MPa = 100 × 106 N/m2
Let N2= Speed of the driven pulley.
Neglecting creep, we know that
                              21
12Nd
Nd=  or  1
21
21200 88.9 r.p.m.2.25dNNd=× = × =
Considering creep, we know that
                             2 21
12 1E Nd
Nd E+σ=×
+σ
or                                         66
266100 10 0.5 101200 88.7r.p.m.2.25100 10 1.4 10N×+ ×=× × =
×+ ×
∴   Speed lost by driven pulley due to creep
= 88.9 – 88.7 = 0.2 r.p.m.   Ans.

11.11. Length of an Open Belt Drive
Fig. 11.11.   Length of an open belt drive.
We have already discussed in Art. 11.6 that in an open belt drive, both the pulleys rotate in the
same direction as shown in Fig. 11.11.
Let r1 and r2= Radii of the larger and smaller pulleys,
x= Distance between the centres of two pulleys ( i.e. O1 O2), and
L= Total length of the belt.
Let the belt leaves the larger pulley at E and G and the smaller pulley at F and H as shown in

Fig. 11.11. Through O2, draw O2 M parallel to FE.
From the geometry of the figure, we find that O2 M will be perpendicular to O1 E.
Let the angle MO2 O1=α radians.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  335
We know that the length of the belt,
L=A r c  GJE + EF + Arc FKH + HG
= 2 (Arc JE + EF + Arc FK) ... (i)
11 1 2
12 12––sinOM OE E M r r
OO OO xα= = =
Since α is very small, therefore putting
sin α = α (in radians)12–rr
x= ...(ii)
∴ Arc 12JE rπ=+ α...(iii)
Similarly Arc 2–2FK rπ=α...(iv)
and22 2 2
21 21 1 2() – () – ( – ) EF MO O O O M x r r== =
2
12–1–rrxx= 

2
Expanding this equation by binomial theorem,
2 2
12 12–( – )11– . . . . –22rr rrEF x xxx=+ =...(v)
Substituting the values of arc JE from equation (iii), arc FK from equation  (iv) and EF from
equation (v) in equation  (i), we get
2
12
12(– )2– –22 2rrLr x rxππ =+ α + +α  
2
12
11 22(– )2. – – .22 2rrrr x rrxππ=× + α + + × α

2
12
12 12(– )2( ) ( – ) –22rrrr rr xxπ=+ + α+

2
12
12 12(– )( ) 2(– ) 2–rrrr rr xx=π + + α +

2
12
Substituting the value of 12–rr
xα=  from equation (ii),
2
12 12
12 12(– ) (– )() 2 ( – ) 2 –rr rrLr r r rxxx=π + + × × +
22
12 12
122( – ) ( – )() 2 –rr rrrr xxx=π + + +
2
12
12(– )() 2rrrr xx=π + + + ...(In terms of pulley radii)
2
12
12(– )() 224dddd xxπ=+ + + ...(In terms of pulley diameters)

336  /G6C   Theory of Machines
11.12. Length of a Cross Belt Drive
We have already discussed in Art. 11.6 that in a cross belt drive, both the pulleys rotate in
opposite directions as shown in Fig. 11.12.
Fig. 11.12.  Length of a cross belt drive.
Let the belt leaves the larger pulley at  E and G and the smaller pulley at F and H, as shown in

Fig. 11.12. Through O2, draw O2M parallel to FE.
From the geometry of the figure, we find that O2M will be perpendicular to O1E.
Let the angle MO2 O1 = α radians.
11 1 2
12 12sinOM OE E M r r
OO OO x++α= = =
 sin α = α (in radians) 12rr
x+= ...(ii)

Similarly Arc 22FK rπ=+ α...(iv)
and22 2 2
21 2 1 1 2() – () – ( ) EF MO OO O M x r r== = +
2
121–rrxx+= 

Chapter 11 : Belt, Rope and Chain Drives    /G6C  337
2 2
12 12 () 11– . . . –22rr rrEF x xxx ++=+ =...(v)
equation (v) in equation (i), we get
2
12
12()2–22 2rrLr x rx + ππ =+ α + + + α  
2
12
11 2 2()2. – .22 2rrrr x r rx + ππ=× + α + + × + α

2
12
12 1 2()2( ) ( ) –22rrrr rr xx + π=+ + α + +

2
12


2
12
12 12()() 2 () 2 –rrrr rr xx+=π + + α + +
Substituting the value of 12rr
x+α=  from equation  (ii),
2
12 12
12 122( ) ( )() () 2 –rr rrLr r r r xxx++=π + + × + +
22
12 12
122( ) ( )() 2 –rr rrrr xxx++=π + + +
2
12
12()() 2rrrr xx+=π + + + ...(In terms of pulley radii)
2
12
12()() 224dddd xx+ π=+ + + ...(In terms of pulley diameters)
It may be noted that the above expression is a function of
(r1 + r2). It is thus obvious that if sum of the radii of the two pulleys

be constant, then length of the belt required will also remain con-stant, provided the distance between centres of the pulleys remainunchanged.
Example 11.3.  A shaft which rotates at a constant speed of
160 r.p.m. is connected by belting to a parallel shaft 720 mm apart,
which has to run at 60, 80 and 100 r.p.m. The smallest pulley on thedriving shaft is 40 mm in radius. Determine the remaining radii ofthe two stepped pulleys for 1. a crossed belt, and 2. an open belt.

Neglect belt thickness and slip.
1 = N3 = N5 = 160 r.p.m. ; x = 720 mm ;
N2 = 60 r.p.m.; N4 = 80 r.p.m.; N6  = 100 r.p.m. ; r1 = 40 mmFig. 11.13.

338  /G6C   Theory of Machines
Let r2, r3, r4, r5 and r6 be the radii of the pulleys 2, 3, 4, 5, and 6 respectively, as shown in Fig.
11.13.
1.  For a crossed belt
We know that for pulleys 1 and 2,
             21
12Nr
Nr=
or              1
21
216040 106.7 mm60NrrN=× = × =  Ans.
and for pulleys 3 and 4,
             34
34rN
Nr=  or  3
43 3 3
4160280Nrr r rN=× =× =
We know that for a crossed belt drive,

                                           r1 + r2 = r3 + r4 = r5 + r6 = 40 + 106.7 = 146.7 mm ... (i)
∴ r3 + 2 r3= 146.7   or    r3 = 146.7/3 = 48.9 mm   Ans.
and      r4 = 2 r3 = 2 × 48.9 = 97.8 mm  Ans.
Now for pulleys 5 and 6,
             65
56Nr
Nr=  or 5
65 5 5
61601.6100Nrr r rN=× =× =
From equation (i),
r5 + 1.6 r5= 146.7  or  r5 = 146.7/2.6 = 56.4 mm  Ans.
and r6=1 . 6  r5 = 1.6 × 56.4 = 90.2 mm  Ans.
2.  For an open belt

 21
12Nr
Nr=  or  1
21
34
34rN
Nr=  or  3
43 3 3
We know that length of belt for an open belt drive,
2
21
12(–)() 2rrLr r xx=π + + +
2(106.7 – 40)(40 106.7) 2 720 1907 mm720=π + + + × =
Since the length of the belt in an open belt drive is constant, therefore for pulleys 3 and 4,
length of the belt ( L),
2
43
34(–)1907 ( ) 2rrrr xx=π + + +

Chapter 11 : Belt, Rope and Chain Drives    /G6C  339
              2
33
33(2 – )(2 ) 2 7 2 0720rrrr=π + + + ×
            = 9.426 r3 + 0.0014 ( r3)2 + 1440
or      0.0014 ( r3)2 + 9.426 r3 – 467 = 0
∴      2
3–9.426 (9.426) 4 0.0014 467
2 0.0014r±+ × ×=×
             –9.426 9.564= 49.3 mm0.0028±=  Ans. ...(Taking +ve sign)
and               r4 =  2 r3 = 2 × 49.3 = 98.6 mm  Ans.
      65
56Nr
Nr=  or
          5
655 5
61601.6100Nrrr rN=× =× =
and length of the belt ( L),

                 2
65
56(–)1907 ( ) 2rrrr xx=π + + +
                          2
55
55(1.6 – )(1 . 6 ) 2 7 2 0720rrrr=π + + + ×
                        = 8.17 r5 + 0.0005 ( r5)2 + 1440
or       0.0005 ( r5)2 + 8.17 r5 – 467 = 0
∴             2
5–8.17 (8.17) 4 0.0005 467
2 0.0005r±+ × ×=×
                         –8.17 8.23=6 0m m0.001±=  Ans.
...(Taking +ve sign)
and r6=1 . 6  r5 = 1.6 × 60 = 96 mm Ans.
11.13. Power Transmitted by a Belt

Fig. 11.14 shows the driving pulley (or driver) A and the driven pulley (or follower) B. We
have already discussed that the driving pulley pulls the belt from one side and delivers the same to theother side. It is thus obvious that the tension on the former side ( i.e. tight side) will be greater than the
latter side ( i.e. slack side) as shown in Fig. 11.14.
Let T
1 and T2= Tensions in the tight and slack side of the belt respectively in
newtons,

newtons,
Milling machine is used for dressing
surfaces by rotary cutters.

340  /G6C   Theory of Machines
r1 and r2= Radii of the driver and follower respectively, and
v= Velocity of the belt in m/s.
Fig. 11.14.   Power transmitted by a belt.
The effective turning (driving) force at the  circumference of the follower is the difference
between the two  tensions ( i.e. T1 – T2).
∴  Work done per second = ( T1 – T2) v N-m/s
and power transmitted,        P = (T1 – T2) v W ...(∵ 1 N-m/s = 1 W)

A little consideration will show that the torque exerted on the driving pulley is ( T1 – T2) r1.
Similarly, the torque exerted on the driven pulley i.e. follower is ( T1 – T2) r2.
11.14. Ratio of Driving Tensions For Flat Belt Drive
Consider a driven pulley rotating in the clockwise direction as shown in Fig. 11.15.
Fig. 11.15.  Ratio of driving tensions for flat belt.
Let T1= Tension in the belt on the tight side,
T2= Tension in the belt on the slack side, and

θ= Angle of contact in radians ( i.e. angle subtended by the arc AB, along
which the belt touches the pulley at the centre).
Now consider a small portion of the belt PQ, subtending an angle δθ at the centre of the
pulley as shown in Fig. 11.15. The belt PQ is in equilibrium under the following forces :
1.Tension T in the belt at P,
2.Tension (T + δ T) in the belt at Q,
3.Normal reaction RN, and

4.Frictional force,  F = µ × RN , where µ is the coefficient of friction between the belt and
pulley.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  341
Resolving all the forces horizontally and equating the same,
N() s i n s i n22RT T Tδθ δθ=+ δ + ...(i)
Since the angle δθ is very small, therefore putting sin δ θ / 2 = δθ / 2 in equation (i),
N.. .() .22 2 2 2TTTRT T T Tδθ δθ δθ δ δθ δθ=+ δ + ×= + + = δ θ ...(ii)
.... Neglecting
2Tδδ θ 

Now resolving the forces vertically, we have
N( ) cos – cos22RT T Tδθ δθµ× = +δ ...(iii)

Since the angle δ θ is very small, therefore putting cos δ θ / 2 = 1 in equation (iii),
µ × RN  = T + δT – T = δT  or  NTRδ=µ...(iv)
Equating the values of RN from equations (ii) and (iv),
.TTδδθ=µ  or  .T
Tδ=µδθ
Integrating both sides between the limits T2  and T1 and from 0 to θ respectively,
i.e.1
20T
TT
Tθδ=µ δθ∫∫                  or       1
2log .eT
T=µθ
  or  . 1
2TeTµθ=...(v)

2log .eT
T=µθ
  or  . 1
Equation (v) can be expressed in terms of corresponding logarithm to the base 10, i.e.
1
22.3log .T
T=µθ

The above expression gives the relation between the tight side and slack side tensions, in terms
of coefficient of friction and the angle of contact.
11.15. Determination of Angle of Contact
When the two pulleys of different diameters are connected by means of an open belt as

shown in Fig. 11.16 ( a), then the angle of contact or lap ( θ) at the smaller pulley must be taken into
consideration.
Let r1= Radius of larger pulley,
r2= Radius of smaller pulley, and
x= Distance between centres of two pulleys ( i.e. O1 O2).
From Fig. 11.16 ( a),
11 1 2
12 12––sinOM OE M E r r
OO OO xα= = = ...(∵ ME = O2 F = r2)
∴ Angle of contact or lap,
(180 – 2 ) rad
180πθ= ° α

342  /G6C   Theory of Machines
A little consideration will show that when the two pulleys are connected by means of a crossed
belt as shown in Fig. 11.16 ( b), then the angle of contact or lap ( θ) on both the pulleys is same. From
Fig. 11.16 ( b),
11 1 2
12 12sinOM OE M E r r
∴ Angle of contact or lap, (180 2 ) rad180πθ= °+ α
(a) Open belt drive.
(b)  Crossed belt drive.
Fig. 11.16
Example 11.4. Find the power transmitted by a belt running over a pulley of 600 mm

diameter at 200 r.p.m. The coefficient of friction betw een  the belt and the pulley is 0.25, angle of lap
160° and maximum tension in the belt is 2500 N .
Solution. Given : d = 600 mm = 0.6 m ; N = 200 r.p.m. ; µ = 0.25 ; θ = 160° = 160 × π / 180
= 2.793 rad ; T1 = 2500 N
We know that velocity of the belt,
. 0.6 2006.284 m/s
60 60dNvππ × ×== =
Let T2= Tension in the slack side of the belt.
We know that1
22.3log . 0.25 2.793 0.6982T
T=µθ= × =


Chapter 11 : Belt, Rope and Chain Drives    /G6C  343
1
20.6982log 0.30362.3T
T==

∴1
22.01T
T=...(Taking antilog of 0.3036)
and1
225001244N
2.01 2.01TT== =
We know that power transmitted by the belt,
P=(T1 – T2) v = (2500 – 1244) 6.284 = 7890 W
= 7.89 kW  Ans.
Example 11.5.  A casting weighing 9 kN hangs freely from a rope which makes 2.5 turns
round a drum of 300 mm diameter revolving at 20 r.p.m.  The other end of the rope is pulled by a man.

The coefficient of friction is 0.25. Determine 1. The force required by the man, and  2. The power to
raise the casting .
Solution. Given : W = T1 = 9 kN = 9000 N ; d = 300 mm = 0.3 m ; N = 20 r.p.m. ; µ = 0.25
1.  Force required by the man
Let T2= Force required by the man.
Since the rope makes 2.5 turns round the drum, therefore angle of contact,
θ=2 . 5  ×  2  π = 5 π radAnother model of milling machine.

344  /G6C   Theory of Machines
22.3log . 0.25 5 3.9275T
T=µθ= × π=

1
23.9275log 1.712.3T
T==
  or  1
251T
T=
...(Taking antilog of 1.71)
∴1
29000=176.47 N51 51TT==  Ans.
2.  Power to raise the casting
We know that velocity of the rope,
.0 . 3 2 00.3142 m/s60 60dNvππ × ×== =
∴  Power to raise the casting,
P=(T1 – T2) v = (9000 – 176.47) 0.3142 = 2772 W
= 2.772 kW   Ans.
Example 11.6.  Two pulleys, one 450 mm diameter  and the other 200 mm diameter are on

parallel shafts 1.95 m apart and are connected by a crossed belt. Find the length of the belt requiredand the angle of contact between the belt and each pulley.
What power can be transmitted by the belt when the larger pulley rotates at 200 rev/min, if
the maximum permissible tension in the belt is 1 kN, and the coefficient of friction between the beltand pulley is 0.25 ?
1 = 450 mm = 0.45 m or r1 = 0.225 m ; d2 = 200 mm = 0.2 m or

r2 = 0.1 m ; x = 1.95 m ; N1 = 200 r.p.m. ; T1 = 1 kN = 1000 N ; µ = 0.25
We know that speed of the belt,
11. 0.45 2004.714 m/s
60 60dNvπ π× ×== =
Length of the belt
We know that length of the crossed belt,
2
12
12()() 2rrLr r xx+=π + + +
2(0.225 0.1)(0.225 0.1) 2 1.95 4.975m1.95+=π + + × + =  Ans.
Angle of contact between the belt and each pulley
Let θ= Angle of contact between the belt and each pulley.
120.225 0.1sin 0.1667
1.95rr

1.95rr
x+ +α= = =   or   α = 9.6°
∴θ = 180° + 2 α = 180° + 2 × 9.6° = 199.2°
199.2 3.477 rad180π=× = Ans.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  345
Power transmitted
1
22.3 log . 0.25 3.477 0.8692T

1
20.8692log 0.3782.3T
T==
  or  1
22.387T
T= ...(Taking antilog of 0.378)
∴1
21000419 N
2.387 2.387TT===
We know that power transmitted,
P=(T1 – T2) v = (1000 – 419) 4.714 = 2740 W = 2.74 kW   Ans.
11.16. Centrifugal Tension
Since the belt continuously runs over the pulleys, there-

fore, some centrifugal force is caused, whose effect is to increasethe tension on both, tight as well as the slack sides. The tensioncaused by centrifugal force is called centrifugal tension . At lower
belt  speeds (less than 10 m/s), the centrifugal tension is verysmall, but at higher belt speeds (more than 10 m/s), its effect isconsiderable and thus should be taken intoaccount.
Consider a small portion PQ of the belt
subtending an angle dθ the centre of the pulley as shown in Fig.
11.17.

11.17.
Let m= Mass of the belt per unit length in kg,
v= Linear velocity of the belt in m/s,
r= Radius of the pulley over which the belt runs in metres, and
T
C= Centrifugal tension acting tangentially at P and Q in newtons.
We know that length of the belt PQ
=r. dθ
and mass of the belt PQ                       = m. r. dθ
∴ Centrifugal force acting on the belt PQ,
2
2
C(.. ) . .vFm r d m d vr=θ = θ
The centrifugal tension TC acting tangentially at P and Q keeps the belt in equilibrium.

Now resolving the forces ( i.e. centrifugal force and centrifugal tension) horizontally and
equating the same, we have
2
CCCsin sin . .
22ddTTF m d vθθ += = θ  
Since the angle dθ is very small, therefore, putting sin ,22ddθθ= in the above expression,Fig. 11.17.   Centrifugal tension.

346  /G6C   Theory of Machines
2
C2. .2dTm d vθ=θ  or   TC = m . v2
Notes : 1.  When the centrifugal tension is taken into account, then total tension in the tight side,
Tt1=T1 + TC
and total tension in the slack side,
Tt2=T2 + TC
             2.    Power transmitted, P=(Tt1 – Tt2) v ...(in watts)
=[ (T1 + TC) – (T2 + TC)] v = (T1 – T2) v ...(same as before)
Thus we see that centrifugal tension has no effect on the power transmitted.

3.    The ratio of driving tensions may also be written as
1C
2C–2.3log .–t
tTT
TT=µθ 

where Tt1= Maximum or total tension in the belt.
11.17. Maximum Tension in the Belt
A little consideration will show that the maximum tension in the belt ( T) is equal to the total
tension in the tight side of the belt ( Tt1).
Let σ= Maximum safe stress in N/mm2,
b= Width of the belt in mm, and
t= Thickness of the belt in mm.
We know that maximum tension in the belt,

T= Maximum stress × cross-sectional area of belt = σ. b. t
When centrifugal tension is neglected, then
T (or Tt1)=T1, i.e. Tension in the tight side of the belt
and when centrifugal tension is considered, then
T (or Tt1)=T1 + TC
11.18. Condition For the Transmission of Maximum Power
We know that power transmitted by a belt,
P=(T1 – T2) v ...(i)
where T1= Tension in the tight side of the belt in newtons,

T2= Tension in the slack side of the belt in newtons, and
From Art. 11.14, we have also seen that the ratio of driving tensions is
. 1
2TeTµθ=  or  1
2.TT
eµθ= ...(ii)
Substituting the value of T2 in equation (i),
1
11 1..1–1 – . .TPT v T v T v C
eeµθ µθ  == =   ...(iii)

Chapter 11 : Belt, Rope and Chain Drives    /G6C  347
where
.11–C
eµθ=
We know that T1=T – TC
where T= Maximum tension to which the belt can be subjected in
newtons, and
TC= Centrifugal tension in newtons.
Substituting the value of T1 in equation (iii),
P=(T – TC) v.C
=(T – m.v2) v.C = (T.v – m v3) C... (Substituting TC = m. v2)
For maximum power, differentiate the above expression with respect to v and equate to zero,
i.e.
0dP
dv=         or          3(.– ) 0dTv m v C
dv=
∴ T – 3 m . v2=0

dv=
or T – 3 TC= 0  or  T = 3 TC...(iv)
It shows that when the power transmitted is maximum, 1/3rd of the maximum tension is
absorbed as centrifugal tension.
Notes :  1.  We know that T1 = T– TC and for maximum power, C3TT=.
∴12–
33TTTT==
              2.   From equation  (iv), the velocity of the belt for the maximum power,
3Tvm=
Example. 11.7.  A shaft rotating at 200 r.p.m. drives another shaft at 300 r.p.m. and transmits

6 kW through a belt. The belt is 100 mm wide and 10 mm thick. The distance between the shafts is 4m.The smaller pulley is 0.5 m in diameter . Calculate the stress in the belt, if it is 1. an open belt drive,
and 2. a cross belt drive. Take µ = 0.3.
1 = 200 r.p.m. ; N2 = 300 r.p.m. ; P = 6 kW = 6 × 103 W ; b = 100 mm ;
t = 10 mm ; x = 4 m ; d2 = 0.5 m ; µ = 0.3
Let σ= Stress in the belt.
1.  Stress in the belt for an open belt drive

First of all, let us find out the diameter of larger pulley ( d1). We know that
21
12Nd
Nd=  or  22
1
1. 300 0.50.75m200NddN×== =
and velocity of the belt,22. 0.5 3007.855 m/s60 60dNvπ π× ×== =
Now let us find the angle of contact on the smaller pulley. We know that, for an open belt
drive,
12 1 2– – 0.75 – 0.5sin 0.0312522 4rr dd
xxα= = = =×  or  α = 1.8°

348  /G6C   Theory of Machines
∴  Angle of contact, θ= 180° – 2 α = 180 – 2 × 1.8 = 176.4°
= 176.4 × π / 180 = 3.08 rad
Let T1= Tension in the tight side of the belt, and
T2= Tension in the slack side of the belt.
1
22.3log . 0.3 3.08 0.924T

∴1
20.924log 0.40172.3T
T==
  or  1
22.52T
T= ...(i)
...(Taking antilog of 0.4017)
6 × 103=(T1 – T2) v = (T1 – T2) 7.855
∴ T1 – T2= 6 × 103 / 7.855 = 764 N ... (ii)

T1= 1267 N, and T2 = 503 N
We know that maximum tension in the belt ( T1),
1267 =σ . b. t = σ × 100 × 10 = 1000 σ
∴σ = 1267 / 1000 = 1.267 N/mm2 = 1.267 MPa  Ans.
...[∵ 1 MPa = 1 MN/m2 = 1 N/mm2]
Stress in the belt for a cross belt drive
We know that for a cross belt drive,
12 1 2 0.75 0.5sin 0.156222 4rr dd
xx++ +α= = = =×  or  α = 9°
∴  Angle of contact, θ= 180° + 2 α = 180 + 2 × 9 = 198°
= 198 × π / 180 = 3.456 rad

1
22.3log . 0.3 3.456 1.0368T

1
21.0368log 0.45082.3T
T==
  or  1
22.82T
T= ...(iii)
...(Taking antilog of 0.4508)
From equations (ii) and (iii),
T1= 1184 N and T2 = 420 N
1184 =σ. b. t = σ × 100 × 10 = 1000 σ
∴σ = 1184 / 1000 = 1.184 N/mm2 = 1.184 MPa  Ans.
Example 11.8.  A leather belt is required to transmit 7.5 kW from a pulley 1.2 m in diameter,

running at 250 r.p.m. The angle embraced is 165° and the coeffici ent of friction between the belt and
the pulley is 0.3. If the safe working stress for the leather belt is 1.5 MPa, density of leather 1 Mg/m3
and thickness of belt 10 mm, determine the width of the belt taking centrifugal tension into account.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  349
Solution. Given : P = 7.5 kW = 7500 W ; d = 1.2 m ; N = 250 r.p.m. ; θ = 165° = 165 × π / 180
= 2.88 rad ; µ = 0.3 ; σ = 1.5 MPa = 1.5 × 106 * N/m2 ; ρ = 1 Mg/m3 = 1 × 106 g/m3 = 1000 kg/m3;
t = 10 mm = 0.01 m
Let b= Width of belt in metres,
T1= Tension in the tight side of the belt in N, and
T2= Tension in the slack side of the belt in N.
v=π d . N / 60 = π × 1.2 × 250/60 = 15.71 m/s

and power transmitted ( P),
7500 = ( T1 – T2) v = (T1 – T2) 15.71
∴ T1 – T2= 7500 / 15.71 = 477.4 N ... (i)
1
22.3log . 0.3 2.88 0.864T

1
20.864log 0.37562.3T
T==
  or  1
22.375T
T= ...(ii)
...(Taking antilog of 0.3756)
T1= 824.6 N,   and   T2 = 347.2 N
We know that mass of the belt per metre length,
m= Area × length × density = b.t.l.ρ
=b × 0.01 × 1 × 1000 = 10 b kg

∴  Centrifugal tension,
TC=m. v2 = 10 b (15.71)2 = 2468 b N
and maximum tension in the belt,
T=σ. b. t = 1.5 × 106 × b × 0.01 = 15 000 b N
We know that T=T1 + TC  or  15000 b = 824.6 + 2468 b
15 000 b – 2468 b= 824.6  or  12 532 b = 824.6
∴ b= 824.6 / 12532 = 0.0658 m = 65.8 mm  Ans.
Example. 11.9.  Determine the width of a 9.75 mm thick leather belt required to transmit

15 kW from a motor running at 900 r.p.m. The diam eter of the driving pulley of the motor is 300 mm.
The driven pulley runs at 300 r.p.m. and the d istance between the centre of two pulleys is 3 metres.
The density of the leather is 1000 kg/m3. The maximum allowable stress in the leather is 2.5 MPa.
The coefficient of friction between the leather and pulley is 0.3. Assume open belt drive and neglectthe sag and slip of the belt.
Solution. Given : t = 9.75 mm = 9.75 × 10

–3 m ; P = 15 kW = 15 × 103 W ; N1 = 900 r.p.m. ;
d1 = 300 mm = 0.3 m ; N2 = 300 r.p.m. ; x = 3m ;  ρ = 1000 kg/m3 ; σ = 2.5 MPa = 2.5 × 106 N/m2 ;
µ = 0.3
*1 MPa = 1 × 106 N/m2

350  /G6C   Theory of Machines
First of all, let us find out the diameter of the driven pulley ( d2). We know that
21
12Nd
Nd=  or  11
2
2900 0.30.9 m300NddN××== =
and velocity of the belt,11 0.3 90014.14 m/s60 60dNvπ π× ×== =
For an open belt drive,
21 2 1–– 0 . 9 – 0 . 3sin 0.122 3rrdd
xxα= = = =×...(∵ d2 > d1)
or α= 5.74°
∴   Angle of lap, θ= 180° – 2 α = 180 – 2 × 5.74 = 168.52°
= 168.52 × π / 180 = 2.94 rad

1
22.3 log . 0.3 2.94 0.882T

1
20.882log 0.38352.3T
T==
  or  1
22.42T
T= ...(i)
... (Taking antilog of 0.3835)
15 × 103=(T1 – T2) v = (T1 – T2) 14.14
∴ T1 – T2= 15 × 103 / 14.14 = 1060 N .... (ii)
T1= 1806 N
Let b= Width of the belt in metres.

=b × 9.75 × 10–3 × 1 × 1000 = 9.75 b kg
TC=m.v2 = 9.75 b (14.14)2 = 1950  b N
Maximum tension in the belt,
T=σ. b. t = 2.5 × 106 × b × 9.75 × 10–3 = 24 400 b N
We know that T=T1 + TC  or  T – TC = T1
24400 b – 1950 b= 1806  or  22 450 b = 1806
∴ b= 1806 / 22 450 = 0.080 m = 80 mm  Ans.
Example. 11.10. A pulley is driven by a flat belt, the angle of lap being 120°. The belt is 100

mm wide by 6 mm thick and density1000 kg/m3. If the coefficient of friction is 0.3 and the maximum
stress in the belt is not to exceed 2 MPa, find the greatest power which the belt can transmit and thecorresponding speed of the belt.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  351
Solution. Given : θ = 120° = 120 × π / 180 = 2.1 rad ; b = 100 mm = 0.1 m ; t = 6 mm
= 0.006 m ; ρ = 1000 kg / m3 ; µ = 0.3 ; σ = 2 MPa = 2 × 106 N/m2
Speed of the belt for greatest power
We know that maximum tension in the belt,                                   T  = σ. b. t = 2 × 10
6 × 0.1 × 0.006 = 1200 N
and mass of the belt per metre length,
                                  m  = Area × length × density = b. t. l. ρ

= 0.1 × 0.006 × 1 × 1000 = 0.6 kg/m
∴  Speed of the belt for greatest power,
                                  1200= 25.82 m/s33 0 . 6Tvm==×Ans.
Greatest power which the belt can transmit
We know that for maximum power to be transmitted, centrifugal tension,
                               TC = T/3 = 1200/3 = 400 N
and tension in the tight side of the belt,                               T
1 = T – TC  = 1200 – 400 = 800 N

Let                            T2 =Tension in the slack side of the belt.
1
22.3log . 0.3 2.1 0.63T
T= µ θ =×=

1
20.63log 0.27392.3T
T==
  or  1
21.88T
T= ...(Taking antilog of 0.2739)
and1
2800425.5 N1.88 1.88TT===
∴ Greatest power which the belt can transmit,
P=(T1 – T2) v = (800 – 425.5) 25.82 = 9670 W = 9.67 kW  Ans.
Example 11.11.  An open belt drive connects two pulleys 1.2 m and 0.5 m diameter, on

parallel shafts 4 metres apart. The mass of the belt is 0.9 kg per metre length and the maximumtension is not to exceed 2000 N.The coefficient of friction is 0.3. The 1.2 m pulley, which is the driver,
runs at 200 r.p.m. Due to belt slip on one of the pulleys, the velocity of the driven shaft is only 450
r.p.m. Calculate the torque on each of the two shafts, the power transmitted, and power lost infriction. What is the efficiency of the drive ?

1 = 1.2 m or r1 = 0.6 m ; d2 = 0.5 m or r2 = 0.25 m ; x = 4 m ; m = 0.9 kg/m;
T = 2000 N ; µ = 0.3 ; N1 = 200 r.p.m. ; N2 = 450 r.p.m.
11. 1.2 20012.57 m/s
and centrifugal tension, TC=m.v2 = 0.9 (12.57)2 = 142 N
∴  Tension in the tight side of the belt,
T1=T – TC = 2000 – 142 = 1858 N

352  /G6C   Theory of Machines
We know that for an open belt drive,
12–0.6 – 0.25sin 0.08754rr
xα= = =   or  α = 5.02°
∴  Angle of lap on the smaller pulley,
θ= 180° – 2 α = 180° – 2 × 5.02° = 169.96°
= 169.96 × π / 180 = 2.967 rad
1
22.3log . 0.3 2.967 0.8901T

1
20.8901log 0.3872.3T
T==
  or  1
22.438T
T=
...(Taking antilog of 0.387)
∴1
21858762N2.438 2.438TT===
Torque on the shaft of larger pulley

We know that torque on the shaft of larger pulley,
TL=(T1 – T2) r1 = (1858 – 762) 0.6 = 657.6 N-m Ans.
Torque on the shaft of smaller pulley
We know that torque on the shaft of smaller pulley,
TS=(T1 – T2) r2 = (1858 – 762) 0.25 = 274 N-m Ans.
We know that the power transmitted,
P=(T1 – T2) v = (1858 – 762) 12.57 = 13780 W
= 13.78 kW Ans.
Power lost in friction
We know that input power,
L1
12 657.6 2 20013780W 13.78kW60 60TNP×π ×π ×== = =

and output power,S2
22 274 2 45012 910W 12.91kW60 60TNP×π ×π ×== = =
∴ Power lost in friction = P1 – P2 = 13.78 – 12.91 = 0.87 kW Ans.
Efficiency of the drive
We know that efficiency of the drive,
Output power 12.910.937 or 93.7%Input power 13.78η= = =   Ans.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  353
11.19. Initial Tension in the Belt
When a belt is wound round the two pulleys ( i.e. driver and follower), its two ends are joined

together ; so that the belt may continuously move over the pulleys, since the motion of the belt fromthe driver and the follower is governed by a firm grip, due to friction between the belt and the pulleys.In order to increase this grip, the belt is tightened up. At this stage, even when the pulleys are station-ary, the belt is subjected to some tension, called initial tension .
When the driver starts rotating, it pulls the belt from one side (increasing tension in the belt

on this side) and delivers it to the other side (decreasing the tension in the belt on that side). Theincreased tension in one side of the belt is called tension in tight side and the decreased tension in theother side of the belt is called tension in the slack side.
Let T
0= Initial tension in the belt,
T1= Tension in the tight side of the belt,
T2= Tension in the slack side of the belt, and
α= Coefficient of increase of the belt length per unit force.

A little consideration will show that the increase of tension in the tight side
=T1 – T0
and increase in the length of the belt on the tight side
=α (T1 – T0) ... (i)
Similarly, decrease in tension in the slack side
=T0 – T2
and decrease in the length of the belt on the slack side
=α (T0 – T2) ... (ii)
Assuming that the belt material is perfectly elastic such that the length of the belt remains

constant, when it is at rest or in motion, therefore increase in length on the tight side is equal todecrease in the length on the slack side. Thus, equating equations (i) and (ii),
α (T
1 – T0)=α (T0 – T2)   or   T1 – T0 = T0 – T2
∴12
02TTT+= ...(Neglecting centrifugal tension)
12 C2
2TT T++= ...(Considering centrifugal tension)
Example. 11.12.  In a flat belt drive the initial tension is 2000 N. The coefficient of friction

between the belt and the pulley is 0.3 and the angle of lap on the smaller pulley is 150°. The smallerpulley has a radius of 200 mm and rotates at 500 r.p.m.  Find the power in kW transmitted by the belt.
0 = 2000 N ; µ0 = 0.3 ; θ = 150° = 150° × π / 180 = 2.618 rad ; r2 = 200 mm
or d2 = 400 mm = 0.4 m ; N2 = 500 r.p.m.
22. 0.4 50010.47m/s

354  /G6C   Theory of Machines
We know that initial tension ( T0),
                    1220002TT+=          or      T1 + T2 = 4000 N ... (i)
We also know that
         1
22.3log . 0.3 2.618 0.7854T

                                 1
20.7854log 0.34152.3T
T==

or                                 1
22.2T
T=  ...(ii)
...(Taking antilog of 0.3415)
T1= 2750 N ;
and                           T2 = 1250 N
∴ Power transmitted,   P = (T1 – T2) v

                                     = (2750 – 1250) 10.47
= 15 700 W =15.7 kW  Ans.
Example 11.13. Two parallel shafts whose centre lines are 4.8 m apart, are connected by
open belt drive. The diameter of the larger pulley is  1.5 m and that of smaller pulley 1 m. The initial
tension in the belt when stationary is 3 kN. The mass of the belt is 1.5 kg/m  length. The coefficient

of friction between the belt and the pulley is 0. 3. Taking centrifugal tension into account, calculate
the power transmitted, when the smaller pulley rotates at 400 r.p.m.
Solution. Given : x = 4.8 m ; d1 = 1.5 m ; d2 = 1 m ; T0 = 3 kN = 3000 N ; m = 1.5 kg / m ;
µ = 0.3 ; N2 = 400 r.p.m.
22. 1 40021m/s
and centrifugal tension, TC=m.v2 = 1.5 (21)2 = 661.5 N
Let T1= Tension in the tight side, and
T2= Tension in the slack side.

12 C 122 26 6 1 . 5300022TT T TT ++ ++ ×==
∴ T1 + T2= 3000 × 2 – 2 × 661.5 = 4677 N ... (i)
12 1 2–– 1 . 5 – 1sin 0.052122 4 . 8rr dd
xxα= = = =×  or   α = 3°
∴ Angle of lap on the smaller pulley,
θ= 180° – 2 α = 180° – 2 × 3° = 174°
= 174° × π / 180 = 3.04 rad
A military tank uses chain, belt and gear drives
for its movement and operation.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  355
1
22.3log . 0.3 3.04 0.912T

1
20.912log 0.39652.3T
T==
  or   1
22.5T
T= ...(ii)
...(Taking antilog of 0.3965)
                          T1 =3341 N ;  and  T2 = 1336 N
∴Power transmitted,
  P = (T1 – T2) v = (3341 – 1336) 21 = 42 100 W = 42.1 kW  Ans.
Example 11.14.  An open flat belt drive connects two parallel shafts 1.2 metres apart. The

driving and the driven shafts rotate at 350 r.p.m. and 140 r.p.m. respectively and the driven pulley is
400 mm in diameter. The belt is 5 mm thick and 80 mm wide. The coef ficient of friction between the
belt and pulley is 0.3 and the maximum permissible tension in the belting is 1.4 MN/m2. Determine :
1. diameter of the driving pulley, 2. maximum power that may be transmitted by the belting,
and 3. required initial belt tension.

Solution. Given : x = 1.2 m ; N1 = 350 r.p.m. ; N2 = 140 r.p.m. ; d2 = 400 mm = 0.4 m ;
t = 5 mm = 0.005 m ; b = 80 mm = 0.08 m ; µ = 0.3 ; σ = 1.4 MN/m2 = 1.4 × 106 N/m2
1.  Diameter of the driving pulley
Let d1= Diameter of the driving pulley.
We know that              21
12Nd
Nd=  or  22
1
1.140 0.40.16m350NddN×== =  Ans.
2.  Maximum power transmitted by the belting

First of all, let us find the angle of contact of the belt on the smaller pulley (or driving
pulley).
Let                                  θ  = Angle of contact of the belt on the driving pulley.
Fig. 11.18

356  /G6C   Theory of Machines
From Fig. 11.18, we find that
22 1 2 1
12– – 0.4 – 0.16sin 0.122 1 . 2OM r r d d
OO x xα= = = = =×
or α= 5.74°
∴θ = 180° – 2 α = 180° – 2 × 5.74° = 168.52°
1
22.3log . 0.3 2.94 0.882T

1
T==
 or  1
22.42T
T= ...(i)
...(Taking antilog of 0.3835)

22.42T
T= ...(i)
We know that maximum tension to which the belt can be subjected,
T1=σ × b × t = 1.4 × 106 × 0.08 × 0.005 = 560 N
∴1
2560231.4N
2.42 2.42TT=== ...[From equation (i)]
Velocity of the belt,11. 0.16 3502.93 m/s60 60dNvπ π× ×== =
∴ Power transmitted, P=(T1 – T2) v = (560 – 231.4) 2.93 = 963 W = 0.963 kW   Ans.
3.  Required initial belt tension
We know that the initial belt tension,
12
0560 231.4395.7N22TTT++== =  Ans.

12
Example 11.15.  An open belt running over two pulleys 240 mm and 600 mm diameter connects
two parallel shafts 3 metres apart and transmits 4 kW from the smaller pulley that rotates at 300r.p.m. Coefficient of friction between the belt and the pulley is 0.3 and the safe working tension is10N per mm width. Determine : 1. minimum width of the belt, 2. initial belt tension, and 3. length of
the belt required.

2 = 240 mm = 0.24 m ; d1 = 600 mm = 0.6 m ;  x = 3 m ; P = 4 kW = 4000 W;
N2 = 300 r.p.m. ; µ = 0.3; T1 = 10 N/mm width
1.  Minimum width of belt
22. 0.24 3003.77 m/s
∴ Power transmitted ( P),
4000 = (T1 – T2) v = (T1 – T2) 3.77
or T1 – T2= 4000 / 3.77 = 1061 N ... (i)

Chapter 11 : Belt, Rope and Chain Drives    /G6C  357
12 1 2– – 0.6 – 0.24sin 0.0622 3rr dd
xxα= = = =×  or  α = 3.44°
and angle of lap on the smaller pulley,
θ= 180° – 2 α = 180° – 2 × 3.44° = 173.12°
= 173.12 × π / 180 = 3.022 rad
1
22.3log . 0.3 3.022 0.9066T

1
20.9066log 0.39422.3T
T==
  or  1
22.478T
T= ...(ii)
...(Taking antilog of 0.3942)
T1= 1779 N,  and  T2 = 718 N

Since the safe working tension is 10 N per mm width, therefore minimum width of the belt,
11779177.9mm
10 10Tb== =   Ans.
2.  Initial belt tension
We know that initial belt tension,
12
01779 718=1248.5N22TTT+ +==  Ans.
3.  Length of the belt required
We know that length of the belt required,
2
12
12(– )(– )224ddLd dxxπ=+ +
2(0.6 – 0.24)(0.6 0.24) 2 324 3π=+ + × +×
= 1.32 + 6 + 0.01 = 7.33 m  Ans.
Example 11.16.  The following data refer to an open belt drive  :

Diameter of larger pulley = 400 mm ; Diameter of smaller pulley = 250 mm ; Distance
between two pulleys = 2 m ; Coefficient of friction between smaller pulley surface and belt = 0.4 ;
Maximum tension when the belt is on the point of slipping = 1200 N.
Find the power transmitted at speed of 10 m/s. It is desired to increase the power. Which of
the following two methods you will select ?
1.  Increasing the initial tension in the belt by 10 per cent.

2.  Increasing the coefficient of friction betw een the smaller pulley surface and belt by 10 per
cent by the application of suitable dressing on the belt.
Find, also, the percentage increase in power possible in each case.

358  /G6C   Theory of Machines
Solution. Given : d1 = 400 mm = 0.4 m ; d2 = 250 mm = 0.25 m ; x = 2 m ; µ = 0.4 ;
T = 1200 N ; v = 10 m/s
12 1 2– – 0.4 – 0.25sin 0.037522 2rr dd
xxα= = = =×  or  α = 2.15°
∴   Angle of contact,
θ= 180° – 2 α = 180° – 2 × 2.15° = 175.7°
= 175.7 × π / 180 = 3.067 rad
Neglecting centrifugal tension,

T1=T = 1200 N ...(Given)
1
22.3log . 0.4 3.067 1.2268T

1
21.2268log 0.53342.3T
T==
  or  1
23.41T
T=
...(Taking antilog of 0.5334)
and1
21200352N
3.41 3.41TT===
P=(T1 – T2) v = (1200 – 352) 10 = 8480 W = 8.48 kW  Ans.
Power transmitted when initial tension is increased by 10%
We know that initial tension,
12
01200 352776N22TTT++== =
∴  Increased initial tension,
0776 10776 853.6 N100T×′=+ =

Let T1 and T2 be the corresponding tensions in the tight side and slack side of  the belt
∴12
02TTT+′=
or T1 + T2=2  T'0 = 2 × 853.6 = 1707.2 N ... (i)
Since the ratio of tensions is constant, therefore
1
23.41T
T=...(ii)

Chapter 11 : Belt, Rope and Chain Drives    /G6C  359
T1= 1320.2 N ; and T2 = 387 N
∴ Power transmitted, P = (T1 – T2) v = (1320.2 – 387) 10 = 9332 W = 9.332 kW
Power transmitted when coefficient  of friction is increased by 10%
We know that coefficient of friction,
µ=0 . 4
∴ Increased coefficient of friction,
100.4 0.4 0.44
100′µ= + × =
Let T1 and T2 be the corresponding tensions in the tight side and slack side respectively.
1

1
22.3log . 0.44 3.067 1.3495T
T′=µ θ= × =

1
21.3495log 0.58672.3T
T==
  or  1
23.86T
... (Taking antilog of 0.5867)
Here the initial tension is constant, i.e.
12
02TTT+=    or   T1 + T2 = 2 T0 = 2 × 776 = 1552 N ... (iv)
T= 1232.7 N and T2 = 319.3 N
P=(T1 – T2) v = (1232.7 – 319.3) 10 = 9134 W = 9.134 kW
Since the power transmitted by increasing the initial tension is more, therefore in order to

increase the power transmitted we shall adopt the method of increasing the initial tension. Ans.
Percentage increase in power
We know that percentage increase in power when the initial tension is increased
9.332 – 8.48100 10.05%8.48=× =   Ans.
and percentage increase in power when coefficient of friction is increased
9.134 – 8.48100 7.7%8.48=× =   Ans.
11.20. V-belt drive
We have already discussed that a V-belt is mostly used in factories and workshops where a

great amount of power is to be transmitted from one pulley to another when the two pulleys are verynear to each other.

360  /G6C   Theory of Machines
*The wedging action of the V-belt in the groove of the pulley results in higher forces of friction. A little

consideration will show that the wedging action and the transmitted torque will be more if the groove angleof the pulley is small. But a smaller groove angle will require more force to pull the belt out of the groovewhich will result in loss of power and excessive belt wear due to friction and heat. Hence a selective grooveangle is a compromise between the two. Usually the groove angles of 32° to 38° are used.The V-belts are made of fabric and cords moulded in rubber and covered with fabric and

rubber, as shown in Fig. 11.19 ( a). These belts are moulded to a trapezoidal shape and are made
endless. These are particularly suitable for short drives i.e. when the shafts are at a short distance

apart. The included angle for the V-belt is usually from 30° – 40°. In case of flat belt drive, the beltruns over the pulleys whereas in case of V-belt drive, the rim of the pulley is grooved in which theV-belt runs. The effect of the groove is to increase the frictional grip of the V-belt on the pulley andthus to reduce the tendency of slipping. In order to have a good grip on the pulley, the V-belt is incontact with the side faces of the groove and not at the bottom. The power is transmitted

and not at the bottom. The power is transmitted by the*wedging action between the belt and the V-groove in the pulley.

(a)  Cross-section of a V-belt. ( b)  Cross-section of a V-grooved pulley.
Fig. 11.19. V-belt and V-grooved pulley.
A clearance must be provided at the bottom of the groove, as shown in Fig. 11.19 ( b), in order

to prevent touching to the bottom as it becomes narrower from wear. The V-belt drive, may beinclined at any angle with tight side either at top or bottom. In order to increase the power output,several V- belts may be operated side by side. It may be noted that in multiple V-belt drive, all thebelts should stretch at the same rate so that the load is equally divided between them. When one of theset of belts break, the entire set should be replaced at the same time. If only one belt is replaced,

at the same time. If only one belt is replaced, thenew unworn and unstressed belt will be more tightly stretched and will move with different velocity.

11.21. Advantages and Disadvantages of V-belt Drive Over Flat Belt Drive
Following are the advantages and disadvantages of the V-belt drive over flat belt drive.
Advantages
1.The V-belt drive gives compactness due to the small distance between the centres of pulleys.
2.The drive is positive, because the slip between the belt and the pulley groove is negligible.
3.Since the V-belts are made endless and there is no joint trouble, therefore the drive is
smooth.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  361
5.It can be easily installed and removed.
6.The operation of the belt and pulley is quiet.
7.The belts have the ability to cushion the shock when machines are started.
8.The high velocity ratio (maximum 10) may be obtained.
9.The wedging action of the belt in the groove gives high value of limiting ratio of tensions.

Therefore the power transmitted by V-belts is more than flat belts for the same coefficient of friction,arc of contact and allowable tension in the belts.
10.The V-belt may be operated in either direction with tight side of the belt at the top or
bottom. The centre line may be horizontal, vertical or inclined.
Disadvantages
1.The V-belt drive cannot be used with large centre distances.
2.The V-belts are not so durable as flat belts.

3.The construction of pulleys for V-belts is more complicated than pulleys for flat belts.
4.Since the V-belts are subjected to certain amount of creep, therefore these are not suitable
for constant speed application such as synchronous machines, and timing devices.
5.The belt life is greatly influenced with temperature changes, improper belt tension and
mismatching of belt lengths.

6.The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above 50m/s.
11.22. Ratio of Driving Tensions for V-belt
A V-belt with a grooved pulley is shown in Fig. 11.20.
Let R1= Normal reaction between the belt and
sides of the groove.
R= Total reaction in the plane of the groove.
2 β= Angle of the groove.
µ = Coefficient of friction between the belt
and sides of the groove.
Resolving the reactions vertically to the groove,

R=R1 sin β + R1 sin β = 2 R1 sin β
or12sinRR=β
We know that the frictional force
1.2. 2 . c o s e c2sin sinRRRRµ=µ =µ × = = µ βββ
Consider a small portion of the belt, as in Art. 11.14, subtending an angle δθ at the centre.
The tension on one side will be  T and on the other side T + δT. Now proceeding as in Art. 11.14, we
get the frictional resistance equal to µ. R cosec β instead of µ . R. Thus the relation between T1 and T2

for the V-belt drive will be
1
22.3 log . cosecT
T=µθ β
Fig. 11.20.

362  /G6C   Theory of Machines
Example 11.17.  A belt drive consists of two V-belts in parallel, on grooved pulleys of the
same size. The angle of the groove is 30°. The cross-sectional area of each belt is 750 mm2 and
µ. = 0.12. The density of the belt material is 1.2 Mg/m3 and the maximum safe stress in the material

is 7 MPa. Calculate the power that can be transmitted between pulleys 300 mm diameter rotating at1500 r.p.m. Find also the shaft speed in r.p.m. at which the power transm itted would be maximum.
Solution. Given : 2 β = 30° or β = 15° ; α = 750 mm
2 = 750 × 10–6 m2 ; µ = 0.12 ; ρ = 1.2 Mg/m3
= 1200 kg/m3 ; σ = 7 MPa = 7 × 106 N/m2 ; d = 300 mm = 0.3 m ; N = 1500 r.p.m.
                . 0.3 150023.56 m/s

       m = Area × length × density = 750 × 10–6 × 1 × 1200 = 0.9 kg/m
∴ Centrifugal tension,
      TC = m.v2 = 0.9 (23.56)2 = 500 N
         T = Maximum stress × cross-sectional area of belt = σ × a
              = 7 × 106 × 750 × 10–6 = 5250 N
   T1 = T – TC = 5250 – 500 = 4750 N
Let         T2 = Tension in the slack side of the belt.

Since the pulleys are of the same size, therefore angle of contact, θ = 180° = π rad.
                1
22.3log . cosec 0.12 cosec15 1.457T
T=µθ β= ×π× °=

                1
21.457log 0.63342.3T
T==
  or  1
24.3T
T=
...(Taking antilog of 0.6334)
and                                        1
247501105N4.3 4.3TT== =
                                    P =(T1 – T2) v × 2 ...(∵ No. of belts = 2)

= (4750 – 1105) 23.56 × 2 = 171  752 W = 171.752 kW  Ans.
Shaft speed
Let                             N1= Shaft speed in r.p.m., and
                                    v1= Belt speed in m/s.
We know that for maximum power, centrifugal tension,                                T
C =  T / 3  or  m (v1)2 = T / 3   or   0.9 ( v1)2 = 5250 / 3 = 1750
∴                           ( v1)2 = 1750 / 0.9 = 1944.4  or  v1 = 44.1 m/s

Chapter 11 : Belt, Rope and Chain Drives    /G6C  363
We know that belt speed ( v1),
11
1.0 . 344.1 0.0157N
60 60dN Nππ × ×== =
∴ N1= 44.1 / 0.0157 = 2809 r.p.m.  Ans.
Example 11.18.  Power is transmitted using a V-belt drive. The included angle of V-groove
is 30°. The belt is 20 mm deep and maximum width is 20 mm. If the mass of the belt is 0.35 kg permetre length and maximum allowable stress is 1.4 MPa, determine the maximum power transmittedwhen the angle of lap is 140°. µ = 0.15.

Solution.  Given : 2 β = 30° or β = 15° ; t = 20 mm = 0.02 m ; b = 20 mm = 0.02 m ;
m = 0.35 kg/m ; σ = 1.4 MPa = 1.4 × 10
6 N/m2 ; θ = 140° = 140° × π / 180 = 2.444 rad ; µ = 0.15
T=σ. b. t = 1.4 × 106 × 0.02 × 0.02 = 560 N
and for maximum power to be transmitted, velocity of the belt,
56023.1m/s3 3 0.35Tvm== =×
1

1
22.3log . cosec 0.15 2.444 cosec15 1.416T
T=µθ β= × × °=

1
21.416log 0.6162.3T
T==
  or  1
24.13T
T= ...(i)
...(Taking antilog of 0.616)
Centrifugal tension,C560187N
33TT== =
and T1=T – TC  = 560 – 187 = 373 N
1
237390.3 N
4.13 4.13TT=== ...[From equation  (i)]
We know that maximum power transmitted,
P=(T1 – T2) v = (373 – 90.3) 23.1 = 6530 W = 6.53 kW  Ans.
Example 11.19.  A compressor, requiring 90 kW is to run at about 250 r.p.m. The drive is by

V-belts from an electric motor running at 750 r.p.m.  The diameter of the pulley on the compressor
shaft must not be greater than 1 metre while the centre distance between the pulleys is limited to 1.75metre. The belt speed should not exceed 1600 m/min.
Determine the number of V-belts required to transmit the power if each belt has a cross-
sectional area of 375 mm
2, density 1000 kg/m3 and an allowable tensile stress of 2.5 MPa. The

groove angle of the pulley is 35°. The coefficient of friction between the belt and the pulley is 0.25.Calculate also the length required of each belt.
Solution. Given : P = 90 kW ; N
2 = 250 r.p.m. ; N1 = 750 r.p.m. ; d2 = 1 m ; x = 1.75 m ;
v = 1600 m/min = 26.67 m/s ; a = 375 mm2 = 375 × 10–6 m2 ; ρ = 1000 kg/m3 ; σ = 2.5 MPa
= 2.5 × 106 N/m2 ; 2 β = 35° or β = 17.5° ; µ = 0.25

364  /G6C   Theory of Machines
First of all, let us find the diameter of pulley on the motor shaft ( d1). We know that
21
12Nd
Nd=  or  22
1
1. 250 10.33m750NddN×== =
We know that the mass of the belt per metre length,
m= Area × length × density
= 375 × 10–6 × 1 × 1000 = 0.375 kg
∴   Centrifugal tension, TC=m.v2 = 0.375 (26.67)2 = 267 N
T=σ. a = 2.5 × 106 × 375 × 10–6 = 937.5 N
∴ Tension in the tight side of the belt,
T1=T – TC = 937.5 – 267 = 670.5 N

For an open belt drive, as shown in Fig. 11.21,
22 1 2 1
12– – 1 – 0.33sin 0.191422 1 . 7 5OM r r d d
∴α =1 1 °
and angle of lap on smaller pulley ( i.e. pulley on motor shaft),
θ= 180° – 2 α = 180° – 2 × 11° = 158°
= 158 × π / 180 = 2.76 rad
Fig. 11.21
1
22.3log . cosec 0.25 2.76 cosec17.5 2.295T

1
22.295log 0.9982.3T
T==
  or  1
29.954T

1
T==
  or  1
29.954T
T=      ...(Taking antilog of 0.998)
and1
2670.567.36N
9.954 9.954TT===
Number of V-belts
We know that power transmitted per belt
=(T1 – T2) v = (670.5 – 67.36) 26.67 = 16  086 W
= 16.086 kW
∴Total power transmitted 90Number of V-belts = 5.6 or 6Power transmitted per belt 16.086== Ans.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  365
Length of each belt
2
21
21(–)() 224ddLd dxxπ=+ + +
2(1 – 0.33)(1 0.33) 2 1.752 4 1.75π=+ + ×+×
= 2.1 + 3.5 + 0.064 = 5.664 m  Ans.
11.23. Rope Drive
The rope drives are widely used where a large amount of power is to be transmitted, from one

pulley to another, over a considerable distance. It may be noted that the use of flat belts is limited forthe transmission of moderate power from one pulley to another when the two pulleys are not morethan 8 metres apart. If large amounts of power are to be transmitted by the flat belt, then it wouldresult in excessive belt cross-section. It may be noted that frictional grip in case of rope drives is morethan that in V-drive. One of the main advantage of rope drives is that a number of separate

of rope drives is that a number of separate drives maybe taken from the one driving pulley. For example, in many spinning mills, the line shaft on each flooris driven by ropes passing directly from the main engine pulley on the ground floor.

The rope drives use the following two types of ropes :1.  Fibre ropes, and 2. Wire ropes.
The fibre ropes operate successfully when the pulleys are about 60 metres apart, while the
wire ropes are used when the pulleys are upto 150 metres apart.
11.24. Fibre Ropes
The ropes for transmitting power are usually made from fibrous materials such as hemp,

manila and cotton. Since the hemp and manila fibres are rough, therefore the ropes made from thesefibres are not very flexible and possesses poor mechanical properties. The hemp ropes have lessstrength as compared to manila ropes. When the hemp and manila ropes are bent over the sheave (orpulley), there is some sliding of fibres, causing the rope to wear and chafe internally. In order tominimise this defect, the rope fibres are lubricated with a tar, tallow or graphite. The lubrication

with a tar, tallow or graphite. The lubrication alsomakes the rope moisture proof. The hemp ropes are suitable only for hand operated hoisting machin-ery and as tie ropes for lifting tackle, hooks etc.

The cotton ropes are very soft and smooth. The lubrication of cotton ropes is not necessary.
But if it is done, it reduces the external wear between the rope and the grooves of its sheaves. It maybe noted that manila ropes are more durable and stronger than cotton ropes. The cotton ropes arecostlier than manila ropes.
Note :The diameter of manila and cotton ropes usually ranges from 38 mm to 50 mm. The size of the rope is
usually designated by its circumference or  ‘girth’.

11.25. Advantages of Fibre Rope Drives
The fibre rope drives have the following advantages :
1.  They give smooth, steady and quiet service.
2.  They are little affected by out door conditions.
3.  The shafts may be out of strict alignment.
4.  The power may be taken off in any direction and in fractional parts of the whole amount.
5.  They give high mechanical efficiency.

366  /G6C   Theory of Machines
11.26. Sheave for Fibre Ropes
The fibre ropes are usually circular in cross-section as shown in Fig. 11.22 ( a). The sheave
for the fibre ropes is shown in Fig. 11.22 ( b).  The groove angle of the pulley for rope drives is usually
45°. The grooves in the pulleys are made narrow at the bottom and the rope is pinched between theedges of the V-groove to increase the holding power of the rope on the pulley.

(a)  Cross-section of a rope. ( b)  Sheave (Grooved pulley) for ropes.
Fig.  11.22.  Rope and sheave.
11.27. Wire Ropes
When a large amount of power is to be transmitted over long distances from one pulley to
another (i.e. when the pulleys are upto 150 metres apart), then wire ropes are used. The wire ropes are
widely used in elevators, mine hoists, cranes, conveyors, hauling devices and suspension bridges.

The wire ropes run on grooved pulleys but they rest on the bottom of the *grooves and are not wedged
between the sides of the grooves. The wire ropes have the following advantage over cotton ropes.
*The fibre ropes do not rest at the bottom of the groove.This electric hoist uses wire ropes.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  367
1. These are lighter in weight, 2. These offer silent operation,  3. These can withstand shock
loads, 4. These are more reliable, 5. They do not fail suddenly,  6. These are more durable,  7. The
efficiency is high, and 8. The cost is low.
11.28. Ratio of Driving Tensions for Rope Drive
The ratio of driving tensions for the rope drive may be obtained in the similar way as V-belts.

We have discussed in Art. 11.22, that the ratio of driving tensions is
1
22.3log . cosecT
T=µθ β

where, µ, θ and β have usual meanings.
Example 11.20.  A rope drive transmits 600 kW from a pulley of effective diameter 4 m,
which runs at a speed of 90 r.p.m. The  angle of lap is 160° ; the angle of groove 45° ; the coefficient
of friction 0.28 ; the mass of rope 1.5 kg /m and the allowable tension in each rope 2400 N. Find the
number of ropes required.

Solution. Given : P = 600 kW ; d = 4 m ; N = 90 r.p.m. ; θ = 160° = 160 × π / 180 = 2.8 rad;
2 β = 45° or β = 22.5° ; µ = 0.28 ; m = 1.5 kg / m ; T = 2400 N
.4 9 018.85 m/s60 60dNvππ × ×== =
∴  Centrifugal tension, TC=m.v2 = 1.5 (18.85)2 = 533 N
and tension in the tight side of the rope,
T1=T – TC = 2400 – 533 = 1867 N
Let T2= Tension in the slack side of the rope.
1
22.3log . cosec 0.28 2.8 cosec22.5 2.05T

1

1
22.05log 0.89132.3T
T==
  or  1
27.786T
T=
...(Taking antilog of 0.8913)
and1
21867240N
7.786 7.786TT===
We know that power transmitted per rope
=(T1 – T2) v = (1867 – 240) 18.85 = 30  670 W = 30.67 kW
∴Total power transmitted 600Number of ropes = 19.56 or 20Power transmitted per rope 30.67==   Ans.
Example 11.21.  A pulley used to transmit power by means of ropes has a diameter of 3.6

metres and has 15 grooves of 45° angle. The angle of contact is 170° and the coefficient of frictionbetween the ropes and the groove sides is 0.28. The maximum possible tension in the ropes is 960 Nand the mass of the rope is 1.5 kg per metre length. What is the speed of pulley in r.p.m. and the powertransmitted if the condition of maximum power prevail ?
Solution. Given : d = 3.6 m ; No. of grooves = 15 ; 2 β = 45° or β = 22.5° ; θ = 170°

368  /G6C   Theory of Machines
Speed of the pulley
Let                          N  = Speed of the pulley in r.p.m.
We know that for maximum power, velocity of the rope or pulley,
96014.6 m/s33 1 . 5Tvm== =×
∴60 14.6 60= 77.5r.p.m.3.6vNd××==ππ ×  Ans. ...
60dNvπ=/G51
We know that for maximum power, centrifugal tension,
TC=T / 3 = 960 / 3 = 320 N
∴ Tension in the tight side of the rope,
T1=T – TC  = 960 – 320 = 640 N

22.3log . cosec 0.28 2.967 cosec22.5 2.17T

1
22.17log 0.94382.3T
T==
  or  1
28.78T
T=
...(Taking antilog of 0.9438)
and1
264073 N
8.78 8.78TT===
∴ Power transmitted per rope = ( T1 – T2) v = (640 – 73) 14.6 = 8278 W = 8.278 kW
Since the number of grooves are 15, therefore total power transmitted
= 8.278 × 15 = 124.17 kW  Ans.
Example 11.22.  Following data is given for a rope pulley transmitting 24 kW  :

Diameter of pulley = 400 mm ; Speed = 110 r.p.m.; angle of groove = 45° ; Angle of lap on
smaller pulley = 160° ; Coefficient  of friction = 0.28 ; Number of ropes = 10 ; Mass in kg/m length
of ropes = 53 C2 ; and working tension is limited to 122 C2 kN, where C is girth of rope in metres.
Find initial tension and diameter of each rope .
Solution. Given : PT = 24 kW ; d = 400 mm = 0.4 m ; N = 110 r.p.m. ; 2 β = 45° or β = 22.5°;

θ = 160° = 160 × π / 180 = 2.8 rad ; n = 0.28 ; n = 10 ; m = 53 C2 kg/m ; T = 122 C2 kN
= 122 × 103 C2  N
Initial tension
We know that power transmitted per rope,
T Total power transmitted 242.4 kW = 2400WNo. of ropes 10PPn== = =
and velocity of the rope,.0 . 4 1 1 02.3m/s60 60dNvππ × ×== =
Let T= Tension in the tight side of the rope, and
T2= Tension in the slack side of the rope.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  369
We know that power transmitted per rope ( P)
2400 = ( T1 – T2) v = (T1 – T2) 2.3
∴ T1 – T2= 2400 / 2.3 = 1043.5 N ... (i)
1

1
T==
  or  1
27.786T
T= ...(ii)
T1= 1197.3 N, and T2 = 153.8 N
We know that initial tension in each rope,
12
01197.3 153.8675.55N22TTT+ +== = Ans.

12
Diameter of each rope
Let d1= Diameter of each rope,
We know that centrifugal tension,
TC=m.v2 = 53 C2 (2.3)2 = 280.4 C2 N
and working tension ( T),
122 × 103 C2=T1 + TC = 1197.3 + 280.4 C2
122 × 103 C2 – 280.4 C2= 1197.3
∴ C2= 9.836 × 10–3  or  C = 0.0992 m = 99.2 mm
We know that girth ( i.e. circumference) of rope ( C),
99.2 =π d1  or  d1 = 99.2 / π = 31.57 mm  Ans.
11.29. Chain Drives
We have seen in belt and rope

drives that slipping may occur. In order
to avoid slipping, steel chains are used.The chains are made up of rigid linkswhich are hinged together in order to
provide the necessary flexibility for
warping around the driving and drivenwheels. The wheels have projecting teethand fit into the corresponding recesses,
in the links of the chain as shown in Fig.

11.23. The wheels and the chain are thusconstrained to move together withoutslipping and  ensures perfect velocity
ratio. The toothed wheels are known as
sprocket wheels  or simply sprockets .
These wheels resemble to spur gears.

370  /G6C   Theory of Machines
The chains are mostly used to transmit mo-
tion and power from one shaft to another, when thedistance between the centres of the shafts is short suchas in bicycles, motor cycles, agricultural   machinery,road rollers, etc.
11.30. Advantages and Disadvantages
of Chain Drive Over Belt or RopeDrive
Following are the advantages and disadvan-
tages of chain drive over belt or rope drive :
 Advantages

1.  As no slip takes place during chain drive, hence perfect velocity ratio is obtained.
2.  Since the chains are made of metal, therefore they occupy less space in width than a belt
   or rope drive.
3.  The chain drives may be used when the distance between the shafts is less.
4.  The chain drive gives a high transmission efficiency (upto 98 per cent).
5.  The chain drive gives less load on the shafts.

6.  The chain drive has the ability of transmitting motion to several shafts by one chain only.
 Disadvantages
1.  The production cost of chains is relatively high.
2.  The chain drive needs accurate mounting and careful maintenance.
3.  The chain drive has velocity fluctuations especially when unduly stretched.
11.31. Terms Used in Chain Drive
The following terms are frequently used in chain drive.
1.  Pitch of the chain : It is the distance between the hinge centre of a link and the corre-

sponding hinge centre of the adjacent link as shown in Fig. 11.24. It is usually denoted by p.
Fig. 11.24. Pitch of the chain. Fig. 11.25.  Pitch circle diameter of the chain sprocket.
2.  Pitch circle diameter of the chain sprocket . It is the diameter of the circle on which the
hinge centres of the chain lie, when the chain is wrapped round a sprocket as shown in Fig. 11.25. Thepoints A, B, C, and D are the hinge centres of the chain and the circle drawn through these centres is

called pitch circle and its diameter ( d) is known as pitch circle diameter.Fig. 11.23.  Sprocket and chain.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  371
11.32. Relation Between Pitch and Pitch Circle Diameter
A chain wrapped round the sprocket is shown in Fig. 11.25. Since the links of the chain are
rigid, therefore pitch of the chain does not lie on the arc of the pitch circle. The pitch length becomesa chord. Consider one pitch length AB of the chain subtending an angle θ at the centre of sprocket (or
pitch circle).
Let d= Diameter of the pitch circle, and

T= Number of teeth on the sprocket.
From Fig. 11.25, we find that pitch of the chain,
2s i n 2 s i n s i n22 2 2dpA B A O dθθ θ  == = × =     
We know that360
T°θ=
∴360 180sin sin2pd dTT°° ==  
or180cosecdpT°= 
11.33. Relation Between
 Chain Speed and Angular Velocity of  Sprocket
Since the links of the chain

are rigid, therefore they will havedifferent positions on the sprocketat different instants. The relationbetween the chain speed ( v) and
angular velocity of the sprocket ( ω)
also varies with the angular posi-tion of the sprocket. The extremepositions are shown in Fig. 11.26(a) and (b).
Fig. 11.26.  Relation between chain speed and angular velocity of sprocket.

372  /G6C   Theory of Machines
For the angular position of the sprocket as shown in Fig. 11.26 ( a),
v=ω × OA
and for the angular position of the sprocket as shown in Fig. 11.26 ( b),
cos cos22vO XO C O Aθθ =ω× =ω× =ω×   ...(∵ OC = OA)
11.34. Kinematic of Chain Drive
Fig. 11.27 shows an arrangement of a chain drive in which the smaller or driving sprocket has

6 teeth and the larger or driven sprocket has 9 teeth. Though this is an impracticable case, but this isconsidered to bring out clearly the kinematic conditions of a chain drive. Let both the sprockets rotateanticlockwise and the angle subtended by the chain pitch at the centre of the driving and drivensprockets be α and φ respectively. The lines AB and A
1B1 show the positions of chain having mini-

mum and maximum inclination respectively with the line of centres O1O2 of the sprockets. The points
A, B2 and B are in one straight line and the points A1,C and B1 are in one straight line. It may be noted
that the straight length of the chain between the two sprockets must be equal to exact number ofpitches.
Fig. 11.27. Kinematic of chain drive.
Let us now consider the pin centre on the driving sprocket in position A. The length of the

chain AB will remain straight as the sprockets rotate, until A reaches A1 and B reaches B1. As the
driving sprocket continues to turn, the link A1C of the chain turns about the pin centre C and the
straight length of the chain between the two sprockets reduces to CB1. When the pin centre C moves
to the position A1, the pin centre A1 moves to the position A2. During this time, each of the sprockets

rotate from its original position by an angle corresponding to one chain pitch. During the first part ofthe angular displacement, the radius O
1 A moves to O1 A1 and the radius O2 B moves to O2 B1. This
arrangement is kinematically equivalent to the four bar chain O1ABO2.
During the second part of the angular displacement, the radius O1 A1 moves to O1A2 and the
radius O2 B1 moves to O2 B2. This arrangement is kinematically equivalent to the four bar chain

O1CB1O2. The ratio of the angular velocities, under these circumstances, cannot be constant. This
may be easily shown as discussed below :
First of all, let us find the instantaneous centre for the two links O1 A and O2 B. This lies at
point I which is the intersection of BA and O2 O1 produced as shown in Fig. 11.28. If ω1 is the angular
velocity of the driving sprocket and ω2 is the angular velocity of the driven sprocket, then
ω1 × O1 I=ω2 × O2 I
or12 2 1 1 2 1
21 1 11OI OO OI OO

Chapter 11 : Belt, Rope and Chain Drives    /G6C  373
The distance between the centres of two sprockets O1 O2 is constant for a given chain drive,
but the distance O1 I varies periodically as the two sprockets rotate. This period corresponds to a
rotation of the driving sprocket by an angle α. It is clear from the figure that the line AB has minimum
inclination with line O1 O2. Therefore the distance O1 I is maximum and thus velocity ratio ( ω1 / ω2)

is minimum. When the chain occupies the position A1 B1, the inclination of line A1 B1 is maximum
with the line O1 O2. Therefore the distance O1 I1 is minimum and thus the velocity ratio ( ω1 / ω2) is
maximum.
Fig. 11.28.   Angular velocities of the two sprockets.
In actual practice, the smaller sprocket have a minimum of 18 teeth and hence the actual
variation of velocity ratio ( ω1/ω2) from the mean value is very small.
11.35. Classification of Chains

The chains, on the basis of their use, are classified into the following three groups :
1.  Hoisting and hauling (or crane) chains,
2.  Conveyor (or tractive) chains, and
3.  Power transmitting (or driving) chains.
These chains are discussed, in detail, in the following pages.
11.36. Hoisting and Hauling Chains
These chains are used for hoisting
and hauling purposes. The hoisting andhauling chains are of the following twotypes :

1.  Chain with oval links . The links
of this type of chain are of oval shape, asshown in Fig. 11.29 ( a). The joint of each
link is welded. The sprockets which are usedfor this type of chain have receptacles to re-ceive the links. Such type of chains are usedonly at low speeds such as in chain hoists and in anchors for marine works.
(a) Chain with oval links. ( b) Chain with square links.
Fig. 11.29.   Hoisting and hauling chains.

2.  Chain with square links . The links of this type of chain are of square shape, as shown in
Fig. 11.29 ( b). Such type of chains are used in hoists, cranes, dredges. The manufacturing cost of this
type of chain is less than that of chain with oval links, but in these chains, the kinking occurs easily onoverloading.

374  /G6C   Theory of Machines
11.37. Conveyor Chains
These chains are used for elevating and conveying the materials continuously. The conveyor
chains are of the following two types :
1.  Detachable or hook joint type chain, as shown in Fig. 11.30 ( a), and
2.  Closed joint type chain, as shown in Fig. 11.30 ( b).
(a) Detachable or hook joint type chain. ( b) Closed joint type chain.
Fig. 11.30.   Conveyor chains.

The conveyor chains are usually made of malleable cast iron. These chains do not have
smooth running qualities. The conveyor chains run at slow speeds of about 3 to 12 km.p.h.
11.38. Power Transmitting Chains
These chains are used for transmission of power, when the distance between the centres of
shafts is short. These chains have provision for efficient lubrication. The power transmitting chainsare of the following three types.

1.  Block chain . A block chain, as shown in Fig. 11.31, is also known as bush chain.  This
type of chain was used in the early stages of development in the power transmission.
Fig. 11.31.   Block chain.
It produces noise when approaching or leaving the teeth of the sprocket because of rubbing
between the teeth and the links. Such type of chains are used to some extent as conveyor chain atsmall speed.

2.  Bush roller chain . A bush roller chain, as shown in Fig. 11.32, consists of outer plates or
pin link plates, inner plates or roller link plates, pins, bushes and rollers. A pin passes through thebush which is secured in the holes of the roller between the two sides of the chain. The rollers are freeto rotate on the bush which protect the sprocket wheel teeth against wear.
A bush roller chain is extremely strong and simple in construction. It gives good service

under severe conditions. There is a little noise with this chain which is due to impact of the rollers onthe sprocket wheel teeth. This chain may be used where there is a little lubrication. When one of thesechains elongates slightly due to wear and stretching of the parts, then the extended chain is of greaterpitch than the pitch of the sprocket wheel teeth. The rollers then fit unequally into the cavities of the

Chapter 11 : Belt, Rope and Chain Drives    /G6C  375
wheel. The result is that the total load falls on one teeth or on a few teeth. The stretching of the parts
increase wear of the surfaces of the roller and of the sprocket wheel teeth.
Fig. 11.32.   Bush roller chain.
3.  Inverted tooth or silent chain . An inverted

tooth or silent chain is shown in Fig. 11.33. It is designedto eliminate the evil effects caused by stretching and toproduce noiseless running. When the chain stretches andthe pitch of the chain increases, the links ride on the teethof the sprocket wheel at a slightly increased radius. Thisautomatically corrects the small change in the pitch. Thereis no relative sliding between the teeth of the inverted toothchain and the sprocket wheel teeth. When properlylubricated, this chain gives durable

When properlylubricated, this chain gives durable service and runs verysmoothly and quietly.

11.39. Length of Chain
An open chain drive system connecting the two sprockets is shown in Fig. 11.34. We have
already discussed in Art. 11.11 that the length of belt for an open belt drive connecting the two pulleysof radii r
1 and r2 and a centre distance x, is
2
12
12(– )() 2rrLr r xx=π + + + (i)
Fig. 11.34.   Length of chainFig. 11.33.   Inverted tooth or silent chain.

376  /G6C   Theory of Machines
If this expression is used for determining the length of chain, the result will be slightly greater
than the required length. This is due to the fact that the pitch lines A B C D E F G  and P Q R S of the
sprockets are the parts of a polygon and not that of a circle. The exact length of the chain may bedetermined as discussed below :
Let T
1= Number of teeth on the larger sprocket,
T2= Number of teeth on the smaller sprocket, and
p= Pitch of the chain.

We have discussed in Art. 11.32, that diameter of the pitch circle,
180cosecdpT°=   or  180cosec2prT°= 
∴  For larger sprocket,
1
1180cosec2prT°= 

and for smaller sprocket, 2
2180cosec2prT°= 

Since the term π (r1 + r2) is equal to half the sum of the circumferences of the pitch circles,
therefore the length of chain corresponding to
12 12() ( )2prr TTπ+ = +

Substituting the values of r1, r2 and π (r1 + r2) in equation  (i), the length of chain is given by
2
12
12180 180cosec – cosec22
() 2
2pp
TT pLT Tx
x °°
 
 =+ + +
If  x = m.p, then
2
12 12180 180cosec – cosec
()2.24TT TTLp m p Km °° +  =+ + =


where K= Multiplying factor
2
()224TT TTmm °°
 +  =+ +

The value of multiplying factor ( K) may not be a complete integer. But the length of the chain
must be equal to an integer number of times the pitch of the chain. Thus, the value of K should be
rounded off to the next higher integral number.
Example 11.23.  A chain drive is used for reduction of speed from 240 r.p.m. to 120 r.p.m.

The number of teeth on the driving sprocket is 20. Find the number of teeth on the driven sprocket. Ifthe pitch circle diameter of the driven sprocket is 600 mm and centre to centre distance between thetwo sprockets is 800 mm, determine the pitch and length of the chain.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  377
Solution. Given : N1 = 240 r.p.m ; N2 = 120 r.p.m ; T1 = 20 ; d2 = 600 mm or r2 = 300 mm
= 0.3 m ; x = 800 mm = 0.8 m
Number of teeth on the driven sprocket
Let T2= Number of teeth on the driven sprocket.
N1. T1=N2. T2           or          11
2
2.2 4 0 2 040120NTTN×== =   Ans.
Pitch of the chain
Let p= Pitch of the chain.
We know that pitch circle radius of the driven sprocket ( r2),

2180 1800.3 cosec cosec 6.3722 4 0pppT°° ===   
∴ p= 0.3 / 6.37 = 0.0471 m = 47.1 mm  Ans.
Length of the chain
We know that pitch circle radius of the driving sprocket,
1
1180 47.1 180cosec cosec 150.5mm22 2 0prT°° == =   
and x=m.p       or    m = x / p = 800 / 47.1 = 16.985
We know that multiplying factor,
2
()2
24TT TTKm
m °°
2180 180cosec – cosec20 40 (20 40)2 16.9852 4 16.985° °   +  =+ × +×

2(6.392 – 12.745)30 33.97 64.56 say6567.94=+ + =
∴  Length of the chain,
L=p.K = 47.1 × 65 = 3061.5 mm = 3.0615 m  Ans.
EXERCISES
1.An engine shaft running at 120 r.p.m. is required to drive a machine shaft by means of a belt. The
pulley on the engine shaft is of 2 m diameter and that of the machine shaft is 1 m diameter. If the belt
thickness is 5 mm ; determine the speed of the machine shaft, when 1. there is no slip ; and 2. there isa slip of 3%. [Ans. 239.4 r.p.m. ; 232.3 r.p.m.]

2.Two parallel shafts 6 metres apart are provided with 300 mm and 400 mm diameter pulleys and are
connected by means of a cross belt. The direction of rotation of the follower pulley is to be reversed bychanging over to an open belt drive. How much length of the belt has to be reduced ?
[Ans. 203.6 mm]
3.A pulley is driven by a flat belt running at a speed of 600 m/min. The coefficient of friction

between the pulley and the belt is 0.3 and the angle of lap is 160°. If the maximum tension in the beltis 700 N ; find the power transmitted by a belt. [Ans. 3.983 kW]

378  /G6C   Theory of Machines
4.Find the width of the belt, necessary to transmit 7.5 kW to a pulley 300 mm diameter, if the pulley
makes 1600 r.p.m and the coefficient of friction between the belt and the pulley is 0.22. Assume theangle of contact as 210° and the maximum tension in the belt is not to exceed 8 N/mm width.
[Ans. 67.4 mm]
5.An open belt 100 mm wide connects two pulleys mounted on parallel shafts with their centres 2.4 m

apart. The diameter of the larger pulley is 450 mm and that of the smaller pulley 300 mm. The coeffi-cient of friction between the belt and the pulley is 0.3 and the maximum stress in the belt is limited to14 N/mm width. If the larger pulley rotates at 120 r.p.m., find the maximum power that can be trans-mitted. [Ans. 2.39 kW]
6.A leather belt 125 mm wide and 6 mm thick, transmits power from a pulley 750 mm diameter which

runs at 500 r.p.m. The angle of lap is 150° and µ  = 0.3. If the mass of 1 m
3 of leather is 1 Mg and the
stress in the belt is not to exceed 2.75 MPa, find the maximum power that can be transmitted.
[Ans. 19 kW]
7.A flat belt is required to transmit 35 kW from a pulley of 1.5 m effective diameter running at 300

r.p.m. The angle of contact is spread over 11/24 of the circumference and the coefficient of frictionbetween belt and pulley surface is 0.3. Determine, taking centrifugal tension into account, width ofthe belt required. It is given that the belt thickness is 9.5 mm, density of its material is 1.1 Mg/m
3 and
the related permissible working stress is 2.5 MPa. [Ans. 143 mm]
8.A blower is driven by an electric motor though a belt drive. The motor runs at 750 r.p.m. For this

power transmission, a flat belt of 8 mm thickness and 250 mm width is used. The diameter of themotor pulley is 350 mm and that of the blower pulley 1350 mm. The centre distance between thesepulleys is 1350 mm and an open belt configuration is adopted. The pulleys are made out of cast iron.The frictional coefficient between the belt and pulley is 0.35 and the permissible stress for the beltmaterial can be taken as 2.5 N/mm
2 with sufficient factor of safety. The mass of a belt is 2 kg per metre

length. Find the maximum power transmitted without belt slipping in any one of the pulleys.
[Ans. 35.9 kW]
9.An open belt drive connects two pulleys 1.2 m and 0.5 m diameter on parallel shafts 3.6 m apart. The

belt has a mass of 1 kg/m length and the maximum tension in it is not to exceed 2 kN. The 1.2 mpulley, which is the driver, runs at 200 r.p.m. Due to the belt slip on one of the pulleys, the velocity ofthe driven shaft is only 450 r.p.m. If the coefficient of friction between the belt and the pulley is 0.3,find : 1. Torque on each of the two shafts, 2. Power transmitted, 3. Power lost in friction, and 4.Efficiency of the drive. [Ans. 648.6 N-m, 270.25 N-m ; 13.588 kW ; 0.849 kW ; 93.75%]

10.The power transmitted between two shafts 3.5 metres apart by a cross belt drive round the two pulleys
600 mm and 300 mm in diameters, is 6 kW. The speed of the larger pulley (driver) is 220 r.p.m. Thepermissible load on the belt is 25 N/mm width of the belt which is 5 mm thick. The coefficient offriction between the smaller pulley surface and the belt is 0.35. Determine : 1. necessary length of thebelt ; 2. width of the belt, and 3. necessary initial tension in the belt.

[Ans. 8.472 m ; 53 mm ; 888 N]
11.A flat belt, 8 mm thick and 100 mm wide transmits power between two pulleys, running at
1600 m/min. The mass of the belt is 0.9 kg/m length. The angle of lap in the smaller pulley is 165° andthe coefficient of friction between the belt and pulley is 0.3. If the maximum permissible stress in thebelt is 2 MN/m
2, find : 1. maximum power transmitted ; and 2. initial tension in the belt
[Ans. 14.83 kW ; 1002 N]

12.An open belt connects two flat pulleys. The smaller pulley is 400 mm diameter and runs at 200 r.p.m.
The angle of lap on this pulley is 160° and the coefficient of friction between the belt and pulley faceis 0.25. The belt is on the point of slipping when 3 kW is being transmitted. Which of the followingtwo alternatives would be more effective in order to increase the power :

1.  Increasing the initial tension in the belt by 10 per cent, and2.  Increasing the coefficient of friction by 10 per cent by the application of a suitable dressing to the
belt? [Ans. First method is more ef fective]

Chapter 11 : Belt, Rope and Chain Drives    /G6C  379
13.A V-belt drive consists of three V-belts in parallel on grooved pulleys of the same size. The angle of
groove is 30° and the coefficient of friction 0.12. The cross-sectional area of each belt is 800 mm2 and
the permissible safe stress in the material is 3 MPa. Calculate the power that can be transmittedbetween two pulleys 400 mm in diameter rotating at 960 r.p.m. [Ans. 111.12 kW]

14.Power is transmitted between two shafts by a V-belt whose mass is 0.9 kg/m length. The maximum
permissible tension in the belt is limited to 2.2 kN. The angle of lap is 170° and the groove angle 45°.If the coefficient of friction between the belt and pulleys is 0.17, find : 1. velocity of the belt formaximum power ; and 2. power transmitted at this velocity. [Ans. 28.54 m/s ; 30.7 kW]
15.Two shafts whose centres are 1 m apart are connected by a V-belt drive. The driving pulley is supplied

with 100 kW and has an effective diameter of 300 mm. It runs at 1000 r.p.m. while the driven pulleyruns at 375 r.p.m. The angle of groove on the pulleys is 40°. The permissible tension in 400 mm
2
cross-sectional area belt is 2.1 MPa. The density of the belt is 1100 kg/m3. The coefficient of friction
between the belt and pulley is 0.28. Estimate the number of belts required. [Ans. 10]
16.A rope drive is required to transmit 230 kW from a pulley of 1 metre diameter running at 450 r.p.m.

The safe pull in each rope is 800 N and the mass of the rope is 0.46 kg per metre length. The angle oflap and the groove angle is 160° and 45° respectively. If the coefficient of friction between the ropeand the pulley is 0.3, find the number of ropes required. [Ans. 21]
17.Power is transmitted between two shafts, 3 metres apart by an open wire rope passing round two

pulleys of 3 metres and 2 metres diameters respectively, the groove angle being 40°. If the rope has amass of 3.7 kg per metre length and the maximum working tension in rope is 20 kN, determine themaximum power that the rope can transmit and the corresponding speed of the smaller pulley. Thecoefficient of friction being 0.15. [Ans. 400 kW ; 403.5 r.p.m.]
18.A rope drive transmits 75 kW through a 1.5 m diameter, 45° grooved pulley rotating at 200 r.p.m. The

coefficient of friction between the ropes and the pulley grooves is 0.3 and the angle of lap is 160°.Each rope has a mass of 0.6 kg/m and can safely take a pull of 800 N. Taking centrifugal tension intoaccount determine : 1. the number of ropes required for the drive, and 2. initial rope tension.
[Ans. 9 ; 510.2 N]
19.The reduction of speed from 360 r.p.m. to 120 r.p.m. is desired by the use of chain drive. The driving

sprocket has 10 teeth. Find the number of teeth on the driven sprocket. If the pitch radius of the drivensprocket is 250 mm and the centre to centre distance between the two sprocket is 400 mm, find thepitch and length of the chain. [Ans. 30 ; 52.25 mm ; 1.93 m]
1.Discuss briefly the various types of belts used for the transmission of power.
 2.How does the velocity ratio of a belt drive effect, when some slip is taking place between the belt and
the two pulleys ?

3.Obtain an expression for the length of a belt in 1. an open belt drive ; and 2. a cross belt drive.
4.Explain the phenomena of ‘slip’ and ‘creep’ in a belt drive.
5.For a flat belt, prove that 1
2,TeTµθ=where
T2= Tension in the slack side of the belt,
µ= Coefficient of friction between the belt and the pulley, and
θ= Angle of contact between the belt and the pulley (in radians.)

6.What is centrifugal tension in a belt ? How does it affect the power transmitted.
7.Derive the condition for transmitting the maximum power in a flat belt drive.

380  /G6C   Theory of Machines
8.It is stated that the speed at which a belt or rope should be run to transmit maximum power is that at
which the maximum allowable tension is three times the centrifugal tension in the belt or rope at thatspeed. Prove the statement.
9.Explain what do you understand by ‘initial tension in a belt’.
10.Derive an expression for the ratio of the driving tensions in a rope drive assuming the angle of the
groove of the pulley to be as 2 β.

11.Discuss relative merits and demerits of belt, rope and chain drive for transmission of power.
12.What are different types of chains ? Explain, with neat sketches, the power transmission chains.
13.Obtain an expression for the length of a chain.
1.The velocity ratio of two pulleys connected by an open belt or crossed belt is
(a)  directly proportional to their diameters
(b)  inversely proportional to their diameters

(c)  directly proportional to the square of their diameters
(d)  inversely proportional to the square of their diameters
2.Two pulleys of diameters d1 and d2 and at distance x apart are connected by means of an open belt
drive. The length of the belt is
2
12
12()() ( ) 224ddad d xxπ+++ +2
12
12(–)() ( – ) 224ddbd d xxπ++
2
12
12(–)() ( ) 224ddcd d xxπ++ +2
12
12()() ( – ) 224dddd d xxπ+++

12
3.In a cone pulley, if the sum of radii of the pulleys on the driving and driven shafts is constant, then
(a)  open belt drive is recommended
(b)  cross belt drive is recommended
(c)  both open belt drive and cross belt drive are recommended
(d)  the drive is recommended depending upon the torque transmitted
4.Due to slip of the belt, the velocity ratio of the belt drive
(a) decreases ( b) increases ( c) does not change

5.When two pulleys of different diameters are connected by means of an open belt drive, then the angle
of contact taken into consideration should be of the
(a) larger pulley ( b) smaller pulley ( c) average of two pulleys
6.The power transmitted by a belt is maximum when the maximum tension in the belt ( T) is equal
to
(a) TC(b) 2TC(c) 3TC(d) 4TC
where TC = Centrifugal tension.
7.The velocity of the belt for maximum power is
(a)3T
m(b) 4T

(a)3T
m(b) 4T
m(c)  5T
m(d) 6T
m
where      m = Mass of the belt in kg per metre length.

Chapter 11 : Belt, Rope and Chain Drives    /G6C  381
8.The centrifugal tension in belts
(a) increases power transmitted
(b) decreases power transmitted
(c) have no effect on the power transmitted
(d) increases power transmitted upto a certain speed and then decreases
9.When the belt is stationary, it is subjected to some tension, known as initial tension. The value of this
tension is equal to the
(a)  tension in the tight side of the belt
(b)  tension in the slack side of the belt

(c)  sum of the tensions in the tight side and slack side of the belt
(d)  average tension of the tight side and slack side of the belt
10.The relation between the pitch of the chain ( p) and pitch circle diameter of the sprocket ( d) is given by
(a)  60sinpdT°=(b)  90sinpdT°=
(c)  120sinpdT°=(d)  180sinpdT°=
where T  = Number of teeth on the sprocket.
ANSWERS
1.(b) 2.(c) 3.(b) 4.(a) 5.(b)
6.(c) 7.(a) 8.(c) 9.(d) 10.(d)

382       /G6C               Theory of Machines
382TTTTToothedoothedoothedoothedoothed
GearingGearingGearingGearingGearing12
2. Friction Wheels.3. Advantages and
Disadvantages of Gear Drive.
4. Classification of Toothed
Wheels.
5. Terms Used in Gears.6. Gear Materials.7. Law of Gearing.8. Velocity of Sliding of Teeth.9. Forms of Teeth.
10. Cycloidal Teeth.
11. Involute Teeth.
12. Effect of Altering the Centre
Distance.

Distance.
13. Comparison Between Involute
and Cycloidal Gears.
14. Systems of Gear Teeth.15. Standard Proportions of Gear
Systems.
16. Length of Path of Contact.17. Length of Arc of Contact.18. Contact Ratio19. Interference in Involute
Gears.
20. Minimum Number of Teeth on
the Pinion.
21. Minimum Number of Teeth on
the Wheel.
22. Minimum Number of Teeth on
a Pinion for Involute Rack inOrder to Avoid Interference.

23. Helical Gears.24. Spiral Gears.25. Centre Distance For a Pair of
Spiral Gears.
26. Efficiency of Spiral Gears.12.1.12.1.12.1.12.1.12.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoduction
We have discussed in the previous chapter, that the

slipping of a belt or rope is a common phenomenon, in thetransmission of motion or power between two shafts. Theeffect of slipping is to reduce the velocity ratio of the system.In precision machines, in which a definite velocity ratio is ofimportance (as in watch mechanism), the only positive driveis by means of gears  or toothed wheels . A gear drive is also
provided, when the distance between the driver and the fol-lower is very small.

12.2.12.2.12.2.12.2.12.2. Friction WheelsFriction WheelsFriction WheelsFriction WheelsFriction Wheels
The motion and power transmitted by gears is kine-
matically equivalent to that transmitted by friction wheels ordiscs. In order tounderstand howthe motion can betransmitted bytwo toothedwheels, considertwo plain circularwheels A and B
mounted onshafts, having sufficient rough surfaces and pressing againsteach other as shown in Fig. 12.1 ( a).

Chapter 12 : Toothed Gearing    /G6C  383
Let the wheel A be keyed to the rotating shaft and the wheel B to the shaft, to be rotated. A
little consideration will show, that when the wheel A is rotated by a rotating shaft, it will rotate the
wheel B in the opposite direction as shown in Fig. 12.1 ( a).
The wheel B will be rotated (by the wheel A) so long as the tangential force exerted by the
wheel A does not exceed the maximum frictional resistance between the two wheels. But when the

tangential force ( P) exceeds the *frictional resistance ( F), slipping will take place between the two
wheels. Thus the friction drive is not a positive drive.
        
(a) Friction wheels. ( b) Toothed wheels.
Fig. 12.1
In order to avoid the slipping, a number of projections (called teeth) as shown in
Fig. 12.1 ( b), are provided on the periphery of the wheel A, which will fit into the corresponding

recesses on the periphery of the wheel B. A friction wheel with the teeth cut on it is known as toothed
wheel or gear. The usual connection to show the toothed wheels is by their **pitch circles.
Note :  Kinematically, the friction wheels running without slip and toothed gearing are identical. But due to the
possibility of slipping of wheels, the friction wheels can only be used for transmission of small powers.
12.3. Advantages and Disadvantages of Gear Drive

The following are the advantages and disadvantages of the gear drive as compared to belt,
rope and chain drives :
1.It transmits exact velocity ratio.
2.It may be used to transmit large power.
3.It has high efficiency.
4.It has reliable service.
5.It has compact layout.
1.The manufacture of gears require special tools and equipment.
2.The error in cutting teeth may cause vibrations and noise during operation.

*The frictional force F is equal to µ. RN, where µ = Coefficient of friction between the rubbing surface of
two wheels, and RN = Normal reaction between the two rubbing surfaces.
**For details, please refer to Art. 12.4.

384  /G6C   Theory of Machines
12.4. Classification of Toothed Wheels
The gears or toothed wheels may be classified as follows :
1.  According to the position of axes of the shafts . The axes of the two shafts between which
the motion is to be transmitted, may be
(a)  Parallel,   (b)  Intersecting, and   (c)  Non-intersecting and non-parallel.
The two parallel and co-planar shafts connected by the gears is shown in Fig. 12.1. These

gears are called spur gears  and the arrangement is known as spur gearing . These gears have teeth
parallel to the axis of the wheel as shown in Fig. 12.1. Another name given to the spur gearing ishelical gearing , in which the teeth are inclined to the axis. The single and double helical gears con-
necting parallel shafts are shown in Fig. 12.2 ( a) and (b) respectively. The double helical gears are

known as herringbone gears . A pair of spur gears are kinematically equivalent to a pair of cylindrical
discs, keyed to parallel shafts and having a line contact.
The two non-parallel or intersecting, but coplanar shafts connected by gears is shown in Fig.
12.2 (c). These gears are called bevel gears  and the arrangement is known as bevel gearing . The
bevel gears, like spur gears, may also have their teeth inclined to the face of the bevel, in which casethey are known as helical bevel  gears.

The two non-intersecting and non-parallel  i.e. non-coplanar shaft connected by gears is shown
in Fig. 12.2 ( d). These gears are called skew bevel gears  or spiral gears and the arrangement is
known as skew bevel gearing  or spiral gearing . This type of gearing also have a line contact, the
rotation of which about the axes generates the two pitch surfaces known as hyperboloids .
Notes : ( a)When equal bevel gears (having equal teeth) connect two shafts whose axes are mutually perpen-

dicular, then the bevel gears are known as mitres.
(b)A hyperboloid is the solid formed by revolving a straight line about an axis (not in the same
plane), such that every point on the line remains at a constant distance from the axis.
(c)The worm gearing is essentially a form of spiral gearing in which the shafts are usually at  right
angles.
(a) Single helical gear. ( b) Double helical gear. ( c) Bevel gear. ( d) Spiral gear.
Fig. 12.2

Fig. 12.2
2.  According to the peripheral velocity of the gears . The gears, according to the peripheral
velocity of the gears may be classified as :
(a)  Low velocity,  (b)  Medium velocity, and  (c)  High velocity.
  The gears having velocity less than 3 m/s are termed as low velocity  gears and gears having
velocity between 3 and 15 m/s are known as medium velocity gears . If the velocity of gears is more
than 15 m/s, then these are called high speed gears .

Chapter 12 : Toothed Gearing    /G6C  385
3.  According to the type of gearing . The gears, according to the type of gearing may be
classified as :
(a) External gearing,  (b) Internal gearing, and  (c) Rack and pinion.
In external gearing , the gears of the two shafts mesh externally with each other as shown in Fig.
12.3 (a). The larger of these two wheels is called spur wheel and the smaller wheel is called pinion. In

an external gearing, the motion of the two wheels is always unlike, as shown in Fig. 12.3 ( a).
(a) External gearing. ( b) Internal gearing.
Fig. 12.3 Fig. 12.4. Rack and pinion.
In internal gearing , the gears of the two shafts mesh internally  with each other as shown in
Fig. 12.3 ( b). The larger of these two wheels is called annular wheel  and the smaller wheel is called
pinion. In an internal gearing, the motion of the two wheels is always like, as shown in Fig. 12.3 ( b).

386  /G6C   Theory of Machines
Sometimes, the gear of a shaft meshes externally and internally with the gears in a *straight
line, as shown in Fig. 12.4. Such type of gear is called rack and pinion . The straight line gear is called
rack and the circular wheel is called pinion. A little consideration will show that with the help of arack and pinion, we can convert linear motion into rotary motion and vice-versa  as shown in Fig.
12.4.

12.4.
4.  According to position of teeth on the gear surface . The teeth on the gear surface may be
(a) straight,  (b) inclined, and  (c) curved.
We have discussed earlier that the spur gears have straight teeth where as helical gears have
their teeth inclined to the wheel rim. In case of spiral gears, the teeth are curved over the rim surface.
12.5. Terms Used in Gears
The following terms, which will be mostly used in this chapter, should be clearly understood

at this stage. These terms are illustrated in Fig. 12.5.
Fig. 12.5.  Terms used in gears.
1. Pitch circle .  It is an imaginary circle which by pure rolling action, would give the same
motion as the actual gear.
Internal gears
*A straight line may also be defined as a wheel of infinite radius.Rack and pinion

Chapter 12 : Toothed Gearing    /G6C  387
2. Pitch circle diameter .  It is the diameter of the pitch circle. The size of the gear is usually
specified by the pitch circle diameter. It is also known as pitch diameter .
3. Pitch point .  It is a common point of contact between two pitch circles.
4. Pitch surface .  It is the surface of the rolling discs which the meshing gears have replaced
at the pitch circle.

5. Pressure angle or angle of obliquity .  It is the angle between the common normal to two
gear teeth at the point of contact and the common tangent at the pitch point. It is usually denoted by φ.
The standard pressure angles are 1
214°and 20°.
6. Addendum .  It is the radial distance of a tooth from the pitch circle to the top of the tooth.
7. Dedendum .  It is the radial distance of a tooth from the pitch circle to the bottom of the tooth.

8. Addendum circle .  It is the circle drawn through the top of the teeth and is concentric with
the pitch circle.
9. Dedendum circle .  It is the circle drawn through the bottom of the teeth. It is also called
root circle.
Note :  Root circle diameter = Pitch circle diameter × cos φ, where φ is the pressure angle.
10. Circular pitch .  It is the distance measured on the circumference of the pitch circle from

a point of one tooth to the corresponding point on the next tooth. It is usually denoted by pc.
Mathematically,
Circular pitch, pc=π D/T
where D= Diameter of the pitch circle, and
T= Number of teeth on the wheel.
A little consideration will show that the two gears will mesh together correctly, if the two
wheels have the same circular pitch.
Note :  If D1 and D2 are the diameters of the two meshing gears having the teeth T1 and T2 respectively, then for
them to mesh correctly,
12 1 1

12 1 1
12 2 2orcDD D TpTT D Tππ== =
11. Diametral pitch.  It is the ratio of number of teeth to the pitch circle diameter in millimetres.
It is denoted by pd . Mathematically,
Diametral pitch, d
cTpDpπ== ...cDpTπ=/G51
where T= Number of teeth, and
D= Pitch circle diameter.
12. Module.  It is the ratio of the pitch circle diameter in millimeters to the number of teeth.
It is usually denoted by m. Mathematically,
Module,  m=D /T

Note : The recommended series of modules in Indian Standard are 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16,
and 20. The modules 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14 and 18 are of second choice.
13. Clearance .  It is the radial distance from the top of the tooth to the bottom of the tooth, in
a meshing gear. A circle passing through the top of the meshing gear is known as clearance circle .

14. Total depth .  It is the radial distance between the addendum and the dedendum circles of
a gear. It is equal to the sum of the addendum and dedendum.

388  /G6C   Theory of Machines
*For details, see Art. 12.16.
**For details, see Art. 12.17.15. Working depth .  It is the radial distance from the addendum circle to the clearance circle.
It is equal to the sum of the addendum of the two meshing gears.
16. Tooth thickness .  It is the width of the tooth measured along the pitch circle.
17. Tooth space .  It is the width of  space betw een the two adjacent teeth measured along the pitch
circle.

circle.
18. Backlash.  It is the difference between the tooth space and the tooth thickness, as mea-
sured along the pitch circle. Theoretically, the backlash should be zero, but in actual practice somebacklash must be allowed to prevent jamming of the teeth due to tooth errors and thermal expansion.
19. Face of tooth .  It is the surface of the gear tooth above the pitch surface.
20. Flank of tooth .  It is the surface of the gear tooth below the pitch surface.

21. Top land.  It is the surface of the top of the tooth.
22. Face width .  It is the width of the gear tooth measured parallel to its axis.
23. Profile.  It is the curve formed by the face and flank of the tooth.
24. Fillet radius .  It is the radius that connects the root circle to the profile of the tooth.
25. Path of contact .  It is the path traced by the point of contact of two teeth from the
beginning to the end of engagement.

26. *Length of the path of contact .  It is the length of the common normal cut-off by the
addendum circles of the wheel and pinion.
27. ** Arc of contact .  It is the path traced by a point on the pitch circle from the beginning
to the end of engagement of a given pair of teeth. The arc of contact consists of two parts, i.e.
(a) Arc of approach .  It is the portion of the path of contact from the beginning of the
engagement to the pitch point.

(b) Arc of recess .  It is the portion of the path of contact from the pitch point to the end of the
engagement of a pair of teeth.
Note :   The ratio of the length of arc of contact to the circular pitch is known as contact ratio  i.e. number of pairs
of teeth in contact.
12.6. Gear Materials
The material used for the manufacture of gears depends upon the strength and service condi-

tions like wear, noise etc. The gears may be manufactured from metallic or non-metallic materials.The metallic gears with cut teeth are commercially obtainable in cast iron, steel and bronze. The non-metallic materials like wood, raw hide, compressed paper and synthetic resins like nylon are used forgears, especially for reducing noise.
The cast iron is widely used for the manufacture of gears due to its good wearing properties,

excellent machinability and ease of producing complicated shapes by casting method. The cast irongears with cut teeth may be employed, where smooth action is not important.
The steel is used for high strength gears and steel may be plain carbon steel or alloy steel. The
steel gears are usually heat treated in order to combine properly the toughness and tooth hardness.
The phosphor bronze is widely used for worm gears in order to reduce wear of the worms

which will be excessive with cast iron or steel.
12.7. Condition for Constant Velocity Ratio of Toothed Wheels–Law of
Gearing
Consider the portions of the two teeth, one on the wheel 1 (or pinion) and the other on the

Chapter 12 : Toothed Gearing    /G6C  389
wheel 2, as shown by thick line curves in Fig. 12.6. Let the two teeth
come in contact at point Q, and the wheels rotate in the directions as
shown in the figure.
Let T T be the common tangent and MN be the
common normal to the curves at the point of contact Q. From the
centres O1 and O2 , draw O1M and O2N perpendicular to MN. A
little consideration will show that the point Q moves in the direction

QC, when considered as a point on wheel 1, and in the direction
QD when considered as a point on wheel 2.
Let v1 and v2 be the velocities of the point Q on the wheels
1 and 2 respectively. If the teeth are to remain in contact, then thecomponents of these velocities along the common normal MN must
be equal.
∴
12cos cosvvα= β
      11 22
12
11 22 11 22
12() c o s ( ) c o s
()( ) o rOQ OQ
OM ONOQ OQ OM ON
OQ OQω× α=ω× β
ω× =ω× ω× = ω×
∴12
21ON
OMω=ω…(i)
Also from similar triangles O1MP and O2NP,

22
11ON OP
OM OP= ...(ii)
Combining equations  (i) and (ii), we have
12 2
21 1ON OP
OM OPω==ω...(iii)
From above, we see that the angular velocity ratio is inversely proportional to the ratio of the
distances of the point  P from the centres O1 and O2, or the common normal to the two surfaces at the
point of contact Q intersects the line of centres at point P which divides the centre distance inversely
as the ratio of angular velocities.

Therefore in order to have a constant angular velocity ratio for all positions of the wheels, the
point P must be the fixed point (called pitch point) for the two wheels. In other words, the common
normal at the point of contact between a pair of teeth must always pass through the pitch point .
This is the fundamental condition which must be satisfied while designing the profiles for the teeth ofgear wheels. It is also known as  law of gearing.

Notes : 1. The above condition is fulfilled by teeth of involute form, provided that the root circles from which
the profiles are generated are tangential to the common normal.
2.If the shape of one tooth profile is arbitrarily chosen and another tooth is designed to satisfy the
above condition, then the second tooth is said to be conjugate to the first. The conjugate teeth are not in commonuse because of difficulty in manufacture, and cost of production.
3.If D

3.If D
1 and D2 are pitch circle diameters of wheels 1 and 2 having teeth T1 and T2 respectively, then
velocity ratio,
            12 2 2
21 1 1OP D T
OP D Tω== =ωFig. 12.6. Law of gearing.
or

390  /G6C   Theory of Machines
12.8. Velocity of Sliding of Teeth
The sliding between a pair of teeth in contact at Q occurs along the common tangent T T to
the tooth curves as shown in Fig. 12.6. The velocity of sliding is the velocity of one tooth relative to
its mating tooth along the common tangent at the point of contact .
The velocity of point Q, considered as a point on wheel 1, along the common tangent T T is
represented by EC. From similar triangles QEC and O1MQ,

11
1or .EC vEC MQMQ OQ== ω = ω
Similarly, the velocity of point Q, considered as a point on wheel 2, along the common tan-
gent T T is represented by ED. From similar triangles QCD and O2 NQ,
2
22
2or .vEDED QNQN O Q== ω = ω
LetSVelocity of sliding at .vQ=
∴S2 1 .. vE D E C Q N M Q=−= ω − ω
21() ()QP PN MP QP=ω + −ω −
12 2 1() . . QP PN MP = ω +ω +ω −ω ...(i)
Since 12
12
21or . . ,OP P NMP PNOP M Pω== ω= ωωtherefore equation (i) becomes
S1 2()vQ P=ω+ ω ...(ii)

Notes : 1. We see from equation  (ii), that the velocity of sliding is proportional to the distance of the point
of contact from the pitch point.
2.Since the angular velocity of wheel 2 relative to wheel 1 is ( ω1 + ω2 ) and P is the instantaneous
centre for this relative motion, therefore the value of vs may directly be written as vs  (ω1 + ω2 ) QP, without the
above analysis.
12.9. Forms of Teeth
We have discussed in Art. 12.7 (Note 2)

that conjugate teeth are not in common use.Therefore, in actual practice following are the twotypes of teeth commonly used :
1.  Cycloidal  teeth ; and 2. Involute teeth.
We shall discuss both the above mentioned
types of teeth in the following articles. Both theseforms of teeth satisfy the conditions as discussedin Art. 12.7.
12.10. Cycloidal Teeth
A cycloid is the curve traced by a point on the circumference of a circle which rolls without

slipping on a fixed straight line. When a circle rolls without slipping on the outside of a fixed circle,the curve traced by a point on the circumference of a circle is known as epi-cycloid . On the other
hand, if a circle rolls without slipping on the inside of a fixed circle, then the curve traced by a pointon the circumference of a circle is called hypo-cycloid .

Chapter 12 : Toothed Gearing    /G6C  391
In Fig. 12.7 ( a), the fixed line or pitch line of a rack is shown. When the circle C rolls without
slipping above the pitch line in the direction as indicated in Fig. 12.7 ( a), then the point  P on the circle
traces epi-cycloid PA. This represents the face of the cycloidal tooth profile. When the circle  D rolls
without slipping below the pitch line, then the point P on the circle D traces hypo-cycloid PB, which

represents the flank of the cycloidal tooth. The profile BPA is one side of the cycloidal rack tooth.
Similarly, the two curves P' A'  and  P'B' forming the opposite side of the tooth profile are traced by
the point P' when the circles C and D roll in the opposite directions.
In the similar way, the cycloidal teeth of a gear may be constructed as shown in Fig. 12.7 ( b).
The circle C is rolled without slipping on the outside of the pitch circle and the point P on the circle

C traces epi-cycloid PA, which represents the face of the cycloidal tooth. The circle D is rolled on the
inside of pitch circle and the point P on the circle D traces hypo-cycloid PB, which represents the
flank of the tooth profile. The profile BPA is one side of the cycloidal tooth. The opposite side of the
tooth is traced as explained above.
The construction of the two mating cycloidal teeth is shown in Fig. 12.8. A point on the circle

D will trace the flank of the tooth T1 when circle  D rolls without slipping on the inside of pitch circle
of wheel 1 and face of tooth T2 when the circle D rolls without slipping on the outside of pitch circle
of wheel 2. Similarly, a point on the circle C will trace the face of tooth T1 and flank of tooth T2. The
rolling circles C and D may have unequal diameters, but if several wheels are to be interchangeable,
they must have rolling circles of equal diameters.

Fig. 12.8.  Construction of two mating cycloidal teeth.
A little consideration will show, that the common normal XX at the point of contact between
two cycloidal teeth always passes through the pitch point, which is the fundamental condition for aconstant velocity ratio.  (a)( b)
   Fig. 12.7.  Construction of cycloidal teeth of a gear.

392  /G6C   Theory of Machines
12.11. Involute Teeth
An involute of a circle is a plane curve generated by a
point on a tangent, which rolls on the circle without slipping orby a point on a taut string which in unwrapped from a reel asshown in Fig. 12.9. In connection with toothed wheels, the circleis known as base circle. The involute is traced as follows :
Let A be the starting point of the involute. The base
circle is divided into equal number of parts e.g. AP
1, P1P2,

1, P1P2,
P2P3 etc. The tangents at P1, P2, P3 etc. are drawn and the
length P1A1, P2A2, P3A3 equal to the arcs AP1, AP2 and AP3 are
set off. Joining the points A, A1, A2, A3 etc. we obtain the involute
curve AR. A little consideration will show that at any instant
A3, the tangent A3T to the involute is perpendicular to P3A3 and P3A3 is the normal to the involute. In
other words, normal at any point of an involute is a tangent to the circle .

Now, let O1 and O2 be the fixed centres of the two base circles as shown in Fig. 12.10 ( a). Let
the corresponding involutes AB and A1B1 be in contact at point Q. MQ and NQ are normals to the
involutes at Q and are tangents to base circles. Since the normal of an involute at a given point is the
tangent drawn from that point to the base circle, therefore the common normal MN at Q is also the
common tangent to the two base circles. We see that the common normal MN intersects the line of

centres O1O2 at the fixed point P (called pitch point). Therefore the involute teeth satisfy the
fundamental condition of constant velocity ratio.
(a)( b)
Fig. 12.10.  Involute teeth.
From similar triangles O2NP and O1MP,
11 2
221OM OP
ON OPω==ω... (i)
which determines the ratio of the radii of the two base circles. The radii of the base circles is given by
11 22 cos , and cos OM OP ON OP=φ = φ
Also the centre distance between the base circles,
12 12

12 12
12 1 2cos cos cosOM ON OM O NOO OP OP+=+= + =φφ φFig. 12.9. Construction of involute.

Chapter 12 : Toothed Gearing    /G6C  393
where φ is the pressure angle or the angle of obliquity. It is the angle which the common normal to the
base circles  ( i.e. MN) makes with the common tangent to the pitch circles.
When the power is being transmitted, the maximum tooth pressure (neglecting friction at the

teeth) is exerted along the common normal through the pitch point. This force may be resolved intotangential and radial or normal components. These components act along and at right angles to thecommon tangent to the pitch circles.
If F is the maximum tooth pressure as shown in Fig. 12.10 ( b), then
Tangential force, F
T = F cos φ
and radial or normal force, FR = F sin φ.
∴   Torque exerted on the gear shaft
= FT × r, where r is the pitch circle radius of the gear.

Note : The tangential force provides the driving torque and the radial or normal force produces radial deflection
of the rim and bending of the shafts.
12.12.Effect of Altering the Centre Distance on the Velocity Ratio for
Involute Teeth  Gears
In the previous article, we have seen that the velocity ratio for the involute teeth gears is given by
11 2
221OM OP
ON OPω==ω...(i)
Let, in Fig. 12.10 ( a), the centre of rotation of one of the gears (say wheel 1) is shifted from

O1 to O1' . Consequently the contact point shifts from Q to Q '. The common normal to the teeth at the
point of contact Q ' is the tangent to the base circle, because it has a contact between two involute
curves and they are generated from the base circle. Let the tangent M' N'  to the base circles intersects
1O′O2 at the pitch point P' . As a result of this, the wheel continues to work * correctly.
Now from similar triangles O2NP and O1MP,
11
22OM OP
ON OP=...(ii)

11
22OM OP
and from similar triangles O2N'P' and O1'M'P',
11
22OM OP
ON OP′′′′=′′...(iii)
But O2N = O2N', and O1M = O1' M'. Therefore from equations (ii) and (iii),
11
22OP OP
OP OP′′=′...[Same as equation  (i)]
Thus we see that if the centre distance is changed within limits, the velocity ratio remains
unchanged. However, the pressure angle increases (from φ to φ′) with the increase in the centre
distance.

distance.
Example 12.1.   A single reduction gear of 120 kW with a pinion 250 mm pitch circle diameter
and speed 650 r.p.m. is supported in bearings on eit her side. Calculate the total load due to the
power transmitted, the pressure angle being 20 °.
Solution.  Given : P = 120 kW = 120 × 103 W ; d = 250 mm or r = 125 mm = 0.125 m ;
N = 650 r.p.m. or ω = 2π × 650/60 = 68 rad/s ; φ = 20°
*It is not the case with cycloidal teeth.

394  /G6C   Theory of Machines
Let T = Torque transmitted in N-m.
120 × 103 = T.ω = T × 68        or     T = 120 × 103/68 = 1765 N-m
and tangential load on the pinion,
FT = T /r = 1765 / 0.125 = 14 120 N
∴  Total load due to power transmitted,
F = FT / cos φ = 14 120 / cos 20° = 15 026 N = 15.026 kN Ans.
12.13. Comparison Between Involute and Cycloidal Gears
In actual practice, the involute gears are more commonly used as compared to cycloidal

gears, due to the following advantages :
 Advantages of involute gears
Following are the advantages of involute gears :
1.  The most important advantage of the involute gears is that the centre distance for a pair of
involute gears can be varied within limits without changing the velocity ratio. This is not true forcycloidal gears which requires exact centre distance to be maintained.
2.  In involute gears, the pressure angle, from the start of the engagement of teeth to the end

of the engagement, remains constant. It is necessary for smooth running and less wear of gears. But incycloidal gears, the pressure angle is maximum at the beginning of engagement, reduces to zero atpitch point, starts decreasing and again becomes maximum at the end of engagement. This results inless smooth running of gears.
3.  The face and flank of involute teeth are generated by a single curve where as in cycloidal

gears, double curves ( i.e. epi-cycloid and hypo-cycloid) are required for the face and flank respec-
tively. Thus the involute teeth are easy to manufacture than cycloidal teeth. In involute system, thebasic rack has straight teeth and the same can be cut with simple tools.
Note :  The only disadvantage of the involute teeth is that the interference occurs (Refer Art. 12.19) with pinions
having smaller number of teeth. This may be avoided by altering the heights of addendum and dedendum of the

mating teeth or the angle of obliquity of the teeth.
Advantages of cycloidal gears
Following are the advantages of cycloidal gears :1.  Since the cycloidal teeth have wider flanks, therefore the cycloidal gears are stronger than
the involute gears, for the same pitch. Due to this reason, the cycloidal teeth are preferred speciallyfor cast teeth.
2.  In cycloidal gears, the contact takes place between a convex flank and concave surface,

whereas in involute gears, the convex surfaces are in contact. This condition results in less wear incycloidal gears as compared to involute gears. However the difference in wear is negligible.
3.  In cycloidal gears, the interference does not occur at all. Though there are advantages of
cycloidal gears but they are outweighed by the greater simplicity and flexibility of the involutegears.
12.14. Systems of Gear Teeth
The following four systems of gear teeth are commonly used in practice :
1. 1

1. 1
214° Composite system,  2. 1
214° Full depth involute system,  3. 20° Full depth involute
system, and  4. 20° Stub involute system.
The 1
214°composite system  is used for general purpose gears. It is stronger but has no inter-

Chapter 12 : Toothed Gearing    /G6C  395
changeability. The tooth profile of this system has cycloidal curves at the top and bottom and involute
curve at the middle portion. The teeth are produced by formed milling cutters or hobs. The tooth
profile of the 1
214°full depth involute system  was developed for use with gear hobs for spur and
helical gears.
The tooth profile of the  20° full depth involute system  may be cut by hobs. The increase of
the pressure angle from 1

214°to 20° results in a stronger tooth, because the tooth acting as a beam is
wider at the base. The 20° stub involute system  has a strong tooth to take heavy loads.
12.15. Standard Proportions of Gear Systems
The following table shows the standard proportions in module ( m) for the four gear systems
as discussed in the previous article.
Table 12.1. Standard proportions of gear systems.
S. No. Particulars1
2°14composite or full 20° full depth 20° stub involute

depth involute system in volute system system
1. Addenddm 1 m 1 m 0.8 m
2. Dedendum 1.25 m 1.25 m 1 m
3. Working depth 2 m 2 m 1.60 m
4. Minimum total depth 2.25 m 2.25 m 1.80 m
5. Tooth thickness 1.5708 m 1.5708 m 1.5708 m
6. Minimum clearance 0.25 m 0.25 m 0.2 m
7. Fillet radius at root 0.4 m 0.4 m 0.4 m
12.16. Length of Path of Contact
Consider a pinion driving the wheel as shown in Fig. 12.11. When the pinion rotates in

clockwise direction, the contact between a pair of involute teeth begins at K (on the flank near the
base circle of pinion or the outer end of the tooth face on the wheel) and * ends at L (outer end of the
tooth face on the pinion or on the flank near the base circle of wheel). MN is the common normal at
the point of contacts and the common tangent to the base circles. The point K is the intersection of the

addendum circle of wheel and the common tangent. The point  L is the intersection of the addendum
circle of pinion and common tangent.
Fig. 12.11.  Length of path of contact.
*If the wheel is made to act as a driver and the directions of motion are reversed, then the contact between
a pair of teeth begins at L and ends at K.

396  /G6C   Theory of Machines
We have discussed in Art. 12.4 that the length
of path of contact is the length of common normal cut-off by the addendum circles of the wheel and the pinion.Thus the length of path of contact is KL which is the sum
of the parts of the path of contacts KP and PL. The part
of the path of contact KP is known as path of approach
and the part of the path of contact PL is known as path
of recess.
Let r
A=O1L= Radius of addendum
circle of pinion,

Let r
RA=O2K= Radius of addendum
circle of wheel,
r=O1P= Radius of pitch circle of
pinion, and
R=O2P= Radius of pitch circle of
                                      wheel.
From Fig. 12.11, we find that radius of the base circle of pinion,
           O1M = O1P cos φ = r cos φ
and radius of the base circle of wheel,
           O2N = O2P cos φ = R cos φ
Now from right angled triangle O2KN,
()2 22 2 2
22 A ()() c o sKN O K O N R R =− = − φ

()2 22 2 2
and2sin sin PN O P R=φ = φ
∴  Length of the part of the path of contact, or the path of approach,
()222
A cos sin KP KN PN R R R =−= − φ − φ
Similarly from right angled triangle O1ML,
and22 2 2 2
11 A()( ) ( ) c o sML O L O M r r=−= − φ
1sin sin MP O P r=φ = φ
 ∴   Length of the part of the path of contact, or path of recess,
222
A() c o s s i n PL ML MP r r r=−= − φ −φ
 ∴  Length of the path of contact,
22 2 2 2 2

∴  Length of the path of contact,
22 2 2 2 2
AA() c o s ( ) c o s ( ) s i n KL KP PL R R r r R r=+= − φ + − φ − + φ
12.17. Length of Arc of Contact
We have already defined that the arc of contact is the path traced by a point on the pitch circle
from the beginning to the end of engagement of a given pair of teeth. In Fig. 12.11, the arc of contactis EPF or GPH. Considering the arc of contact GPH, it is divided into two parts i.e. arc GP and arc

PH. The arc GP is known as arc of approach  and the arc PH is called arc of recess . The angles
subtended by these arcs at O
1 are called angle of approach  and angle of recess  respectively.
Bevel  gear

Chapter 12 : Toothed Gearing    /G6C  397
We know that the length of the arc of approach (arc GP)
Length of path of approach
cos cosKP==φφ
and the length of the arc of recess (arc PH)
L e n g t ho fp a t ho fr e c e s s
cos cosPL==φφ
Since the length of the arc of contact GPH is equal to the sum of the length of arc of approach
and arc of recess, therefore,
Length of the arc of contact
arc arccos cos cos
Length of path of contact
cosKP PL KLGP PH= + =+=φφφ
=φ

=φ
12.18. Contact Ratio (or Number of Pairs of Teeth in Contact)
The contact ratio or the number of pairs of teeth in contact is defined as the ratio of the
length of the arc of contact to the circular pitch. Mathematically,
Contact ratio or number of pairs of teeth in contact
  Length of the arc of contact
cp=
where Circular pitch , andcp m == π
Module.m=

cp=
Module.m=
Notes :  1.  The contact ratio, usually, is not a whole number. For example, if the contact ratio is 1.6, it does not
mean that there are 1.6 pairs of teeth in contact. It means that there are alternately one pair and two pairs of teeth
in contact and on a time basis the average is 1.6.
2.The theoretical minimum value for the contact ratio is one, that is there must always be at least one
pair of teeth in contact for continuous action.

3.Larger the contact ratio, more quietly the gears will operate.
Example 12.2.   The number of teeth on each of the two equal spur gears in mesh are 40. The
teeth have 20° involute profile and the module is 6 mm. If the arc of contact is 1.75 times the circularpitch, find the addendum.
Solution.  Given : T = t = 40 ; φ = 20° ; m = 6 mm
We know that the circular pitch,
p
c = π m = π × 6 = 18.85 mm
∴ Length of arc of contact

= 1.75 pc = 1.75 × 18.85 = 33 mm
and length of path of contact
= Length of arc of contact × cos φ = 33 cos 20° = 31 mm
Let RA = rA = Radius of the addendum circle of each wheel.
We know that pitch circle radii of each wheel,
R = r = m.T / 2 = 6 × 40/2 = 120 mm

398  /G6C   Theory of Machines
2 2 2 222
AA 31 ( ) cos ( ) cos ( ) sinRR rr R r=−φ + −φ − + φ
22 2
A2() c o s s i nRR R= −φ − φ...(∵ R = r, and RA = rA)
22 2
A31( ) (120) cos 20 120 sin 202R=− ° − °
215.5 ( ) 12 715 41AR=− −
22
A (15.5 41) ( ) 12 715 R += −
2
AA 3192 12 715 ( ) or 126.12 mm RR += =
We know that the addendum of the wheel,
               =A126.12 120 6.12 mm RR−= − =  Ans.
Example 12.3.   A pinion having 30 teeth drives a

gear having 80 teeth. The profile of the gears is involutewith 20° pressure angle, 12 mm module and 10 mmaddendum. Find the length of path of contact, arc of contactand the contact ratio.
Solution. Given : t = 30 ; T  = 80 ; φ  = 20° ;
m = 12 mm ; Addendum = 10 mm
We know that pitch circle radius of pinion,
r = m.t / 2 = 12 × 30 / 2 = 180 mm
and pitch circle radius of gear,
R = m.T / 2 = 12 × 80 / 2 = 480 mm

∴  Radius of addendum circle of pinion,
r
A = r +  Addendum = 180 + 10 = 190 mm
and radius of addendum circle of gear,
RA = R + Addendum = 480 + 10 = 490 mm
We know that length of the path of approach,
22 2
A() c o s s i nKP R R R=−φ − φ ...(Refer Fig.12.11)
22 2(490) (480) cos 20 480 sin 20=− ° − ° 191.5 164.2 27.3mm=−=
and length of the path of recess,
222
A() c o s s i nPL r r r=−φ − φ
22 2(190) (180) cos 20 180 sin20=− ° − ° 86.6 61.6 25 mm=−=

We know that length of path of contact,
KL = KP + PL = 27.3 + 25 = 52.3 mm  Ans.
Worm.

Chapter 12 : Toothed Gearing    /G6C  399
Length of arc of contact
We know that length of arc of contact
Length of path of contact 52.355.66 mmcos cos20== =φ° Ans.
Contact ratio
We know that circular pitch,
pc = π.m = π × 12 = 37.7 mm
∴    Length of arc of contact 55.66Contact ratio = 1.5 say 237.7cp==  Ans.
Example 12.4.   Two involute gears of 20° pressure angle are in mesh. The number of teeth

on pinion is 20 and the gear ratio is 2. If the pitch expressed in module is 5 mm and the pitch linespeed is 1.2 m/s, assuming addendum as standard and equal to one module, find  :
1. The angle turned through by pinion when one pair of teeth is in mesh ; and
2. The maximum velocity of sliding.
Solution. Given : φ = 20° ; t = 20; G = T/t = 2; m = 5 mm ; v = 1.2 m/s ; addendum = 1 module
= 5 mm
1. Angle turned through by pinion when one pair of teeth is in mesh

r = m.t / 2 = 5 × 20 / 2 = 50 mm
and pitch circle radius of wheel,
R = m.T/2 = m.G.t / 2 = 2 × 20 × 5 / 2 = 100 mm
...(. )TG t=/G51
∴ Radius of addendum circle of pinion,
rA = r + Addendum = 50 + 5 = 55 mm
and radius of addendum circle of wheel,
RA = R + Addendum = 100 + 5 = 105 mm
We know that length of the path of approach ( i.e. the path of contact when engagement
occurs),
22 2

22 2= (105) (100) cos 20 100 sin20 −° − °
46.85 34.2 12.65 mm=− =
and the length of path of recess ( i.e. the path of contact when disengagement occurs),
222
22 2(55) (50) cos 20 50 sin 20 28.6 17.1 11.5 mm=− ° − ° = − =
∴ Length of the path of contact,
KL = KP + PL = 12.65 + 11.5 = 24.15  mm

400  /G6C   Theory of Machines
and length of the arc of contact
Length of path of contact 24.1525.7 mmcos cos20== =φ°
We know that angle turned through by pinion
Length of arc of contact ×360° 25.7 36029.45Circumference of pinion 2 50×°== = °π× Ans.
2.  Maximum velocity of sliding
Let ω1 = Angular speed of pinion, and
ω2 = Angular speed of wheel.
We know that pitch line speed,
v = ω1.r = ω2.R
∴ω1 = v/r  =  120/5 = 24 rad/s
and ω2 = v/R = 120/10 = 12 rad/s
 ∴Maximum velocity of sliding,

∴Maximum velocity of sliding,
vS = (ω1 + ω2) KP ...( )KP PL>/G51
= (24 + 12) 12.65 = 455.4 mm/s  Ans.
Example 12.5.  A pair of gears, having 40 and 20 teeth respectively, are rotating in mesh,
the speed of the smaller being 2000 r.p.m. Determin e the velocity of sliding between the gear teeth
faces at the point of engagement, at the pitch point, and at the point of disengagement if the smallergear is the driver. Assume that the gear teeth are 20 ° involute form, addendum length is 5 mm and the

module is 5 mm.
Also find the angle through which the pinion turns while any pairs of teeth are in contact.Solution.  Given : T  = 40 ; t  = 20 ; N
1  = 2000 r.p.m. ; φ = 20° ; addendum = 5 mm ; m = 5 mm
We know that angular velocity of the smaller gear,
1
12 22 0 0 0209.5 rad/s60 60Nπ π×ω= = =
and angular velocity of the larger gear,
2120209.5 104.75 rad/s
40t
Tω= ω× = × =2
1...t
Tω=ω/G51
Pitch circle radius of the smaller gear,
r = m.t / 2 = 5 × 20/2 = 50 mm

and pitch circle radius of the larger gear,
R = m.t / 2 = 5 × 40/2 = 100 mm
∴  Radius of addendum circle of smaller gear,
and radius of addendum circle of larger gear,
RA = R + Addendum = 100 + 5 = 105  mm
The engagement and disengagement of the gear teeth is shown in Fig. 12.11. The point K is
the point of engagement, P is the pitch point and L is the point of disengagement. MN is the common
tangent at the points of contact.

Chapter 12 : Toothed Gearing    /G6C  401
We know that the distance of point of engagement K from the pitch point P or the length of
the path of approach,
22 2
A() c o s s i nKP R R R=−φ − φ
22 2(105) (100) cos 20 100 sin 20=− ° − °
and the distance of the pitch point P from the point of disengagement L or the length of the path of
recess,
222

Velocity of sliding at the point of engagement
We know that velocity of sliding at the point of engagement K,
SK 1 2( ) (209.5 104.75)12.65 3975 mm/svK P=ω+ ω = + =  Ans.
Velocity of sliding at the pitch point
Since the velocity of sliding is proportional to the distance of the contact point from the pitch
point, therefore the velocity of sliding at the pitch point is zero. Ans.
Velocity of sliding at the point of disengagement
We know that velocity of sliding at the point of disengagement L,

SL 1 2( ) (209.5 104.75)11.5 3614 mm/svP L=ω+ ω = + =  Ans.
Angle through which the pinion turns
We know that length of the path of contact,
            KL   = KP + PL = 12.65 + 11.5 = 24.15 mm
and length of arc of contact 24.1525.7 mmcos cos20KL== =φ°
Circumference of the smaller gear or pinion
= 2 π r = 2π × 50 = 314.2 mm
∴  Angle through which the pinion turns
360Length of arc of contactCircumference of pinion°=×
36025.7 29.45
314.2°=× =°  Ans.

Example 12.6.  The following data relate to a pair of 20° involute gears in mesh :
Module = 6 mm, Number of teeth on pinion = 17, Number of teeth on gear = 49 ; Addenda
on pinion and gear wheel = 1 module.
Find : 1. The number of pairs of teeth in contact ;  2. The angle turned through by the pinion
and the gear wheel when one pair of teeth is in contact, and 3. The ratio of sliding to rolling motion

when the tip of a tooth on the larger wheel  (i) is just making contact,  (ii) is just leaving contact with
its mating tooth, and (iii) is at the pitch point.

402  /G6C   Theory of Machines
Solution. Given : φ  = 20° ; m  = 6 mm ; t = 17 ; T = 49 ; Addenda on pinion and gear wheel
= 1 module = 6 mm
1. Number of pairs of teeth in contact
r = m.t / 2 = 6 × 17 / 2 = 51 mm
r = m.T / 2 = 6 × 49 / 2 = 147 mm
rA = r + Addendum = 51 + 6 = 57 mm
RA = R + Addendum = 147 + 6 = 153 mm

We know that the length of path of approach ( i.e. the path of contact when engagement
occurs),
22 2
A() c o s s i nKP R R R=−φ − φ ...(Refer Fig. 12.11)
22 2(153) (147) cos 20 147 sin 20=− ° − °
65.8 50.3 15.5 mm=−=
and length of path of recess ( i.e. the path of contact when disengagement occurs),
222
22 2(57) (51) cos 20 51sin 20=− ° − °
30.85 17.44 13.41mm=−=
∴  Length of path of contact,

15.5 13.41 28.91mm KL KP PL=+= + =Racks

Chapter 12 : Toothed Gearing    /G6C  403
and length of arc of contact Length of path of contact 28.9130.8 mmcos cos 20== =φ°
. 6 18.852 mmcpm=π =π× =
∴ Number of pairs of teeth in contact (or contact ratio)
Length of arc of contact 30.81.6 say 2Circular pitch 18.852== =  Ans.
2.  Angle turned through by the pinion and gear wheel when one pair of teeth is in contact
We know that angle turned through by the pinion

Length of arc of contact 360° 30.8 36034.6Circumference of pinion 2 51××== = °π× Ans.
and angle turned through by the gear wheel
Length of arc of contact 360° 30.8 36012Circumference of gear 2 147××== = °π× Ans.
3.  Ratio of sliding to rolling motion
Let ω1 = Angular velocity of pinion, and
ω2 = Angular velocity of gear wheel.
We know that  12 2 1 1 1/ / or / 17/49 0.347Tt tT ωω= ω= ω × = ω × = ω

and rolling velocity,R1 2 1 1. . 51 51 mm/svr R=ω =ω =ω × = ω
(i) At the instant when the tip of a tooth on the larger wheel is just making contact with its
mating teeth ( i.e. when the engagement commences), the sliding velocity
S1 2 1 1 1() ( 0.347)15.5 20.88 mm/s vK P=ω+ ω =ω+ ω = ω
∴ Ratio of sliding velocity to rolling velocity,
S1
R120.880.4151v
vω==ω Ans.
(ii)  At the instant when the tip of a tooth on the larger wheel is just leaving contact with its

mating teeth ( i.e. when engagement terminates), the sliding velocity,
S1 2 1 1 1() ( 0.347)13.41 18.1 mm/s vP L=ω+ ω =ω+ ω = ω
∴ Ratio of sliding velocity to rolling velocity
S1
R118.10.35551v
vω==ω Ans.
(iii) Since at the pitch point, the sliding velocity is zero, therefore the ratio of sliding velocity
to rolling velocity is zero. Ans.
Example 12.7.   A pinion having 18 teeth engages with an internal gear having 72 teeth. If

the gears have involute profiled teeth with 20° pressure angle, module of 4 mm and the addenda onpinion and gear are 8.5 mm and 3.5 mm respectivel y, find the length of path of contact.
Solution. Given : t = 18 ; T = 72 ; φ = 20° ; m = 4 mm ; Addendum on pinion = 8.5 mm ;
Addendum on gear = 3.5 mm

404  /G6C   Theory of Machines
Fig. 12.12 shows a pinion with centre O1, in mesh with internal gear of centre O2. It may be
noted that the internal gears have the addendum circle and the tooth faces inside the pitch circle.
We know that the length of path of contact is the length of the common tangent to the two

base circles cut by the addendum circles. From Fig. 12.12, we see that the addendum circles cut thecommon tangents at points K and L. Therefore the length of path of contact is KL which is equal to the
sum of KP (i.e. path of approach) and PL (i.e. path of recess).
Fig. 12.12
We know that pitch circle radius of the pinion,
1. /2 4 18/2 36 mm rO Pm t== = × =
and pitch circle radius of the gear,
2. /2 4 72/2 144 mm RO Pm T== = × =
∴ Radius of addendum circle of the pinion,

  A1 1rO L O P==+ Addendum on pinion = 36 + 8.5 = 44.5 mm
and radius of addendum circle of the gear,
A2 2 Addendum on wheel =144 – 3.5 =140.5 mm RO K O P==−
From Fig. 12.12, radius of the base circle of the pinion,
11 cos cos 36 cos 20 33.83 mm OM OP r=φ = φ = ° =
and radius of the base circle of the gear,
22 cos cos 144 cos 20 135.32 mm ON OP R=φ = φ = ° =
22
22 2sin 20 ( ) ( ) KP PN KN O P O K O N=−= ° − −

= 144 × 0.342 – 22(140.5) (135.32)−  = 49.25 – 37.8 = 11.45 mm

Chapter 12 : Toothed Gearing    /G6C  405
                   22
111()( ) s i n 2 0 P L M L M P OL OM OP=−= − − °
 22(44.5) (33.83) 36 0.342 28.9 12.3 16.6 mm=−− × = − =
                     11.45 16.6 28.05 mm KL KP PL=+= + =  Ans.
12.19. Interference in Involute Gears
Fig. 12.13 shows a pinion with centre O1, in mesh with wheel or gear with centre O2. MN is

the common tangent to the base circles and KL is the path of contact between the two mating teeth.
Fig. 12.13.  Interference in involute gears.
A little consideration will show, that if the radius of the addendum circle of pinion is
increased to O1N, the point of contact L will move from L to N. When this radius is further increased,
the point of contact L will be on the inside of base circle of wheel and not on the involute profile of

tooth on wheel. The tip of tooth on the pinion will then undercut the tooth on the wheel at the root andremove part of the involute profile of tooth on the wheel. This effect is known as interference , and
occurs when the teeth are being cut. In brief, the phenomenon when the tip of tooth undercuts the
root on its mating gear is known as interference .
Similarly, if the radius of the addendum circle of the wheel increases beyond O
2M, then the

tip of tooth on wheel will cause interference with the tooth on pinion. The points M and N are called
interference points . Obviously, interference may be avoided if the path of contact does not extend
beyond interference points. The limiting value of the radius of the addendum circle of the pinion is*O
1N and of the wheel is O2M.
From the above discussion, we conclude that the interference may only be avoided, if the

point of contact between the two teeth is always on the involute profiles of both the teeth. In other
*From Fig. 12.13, we see that
22 2 2
11() ( )( ) [) s i n ]b ON OM M N r r R=+ = + + φ
where rb = Radius of base circle of pinion = O1P cos φ = r cos φ
and222 2
22() () ( ) [ ) s i n ]b OM ON M N R r R=+ = + + φ
where Rb = Radius of base circle of wheel = O2P cos φ = R cos φ

406  /G6C   Theory of Machines
words, interference may only be prevented, if the addendum circles of the two mating gears cut the
common tangent to the base circles between the points of tangency .
When interference is just avoided, the maximum length of path of contact is MN when the
maximum addendum circles for pinion and wheel pass through the points of tangency N and M re-
spectively as shown in Fig. 12.13. In such a case,
Maximum length of path of approach,
        MP = r sin φ

MP = r sin φ
and maximum length of path of recess,
       PN = R sin φ
∴  Maximum length of path of contact,
       MN = MP + PN = r sin φ + R sin φ = (r + R) sin φ
and maximum length of arc of contact
        () s i n() t a ncosrRrR+φ== + φφ
Note : In case the addenda on pinion and wheel is such that the path of approach and path of recess are half of
their maximum possible values, then
   Path of approach,         1
2KP MP=
or         22 2
Asin() c o s s i n2rRR Rφ−φ − φ =

or         22 2
and path of recess,                         1
2PL PN=
or             222
Asin() c o s s i n2Rrr rφ−φ − φ =
∴ Length of the path of contact
           11
22() s i n
2rRKP PL MP PN+φ=+= + =
Example 12.8.   Two mating gears have 20 and 40 involute teeth of module 10 mm and 20°

pressure angle. The addendum on each wheel is to be made of such a length that the line of contacton each side of the pitch point has half the maximum possible length. Determine the addendumheight for each gear wheel, length of the path of contact, arc of contact and contact ratio.
Solution. Given : t  = 20 ; T  = 40 ; m  = 10 mm ; φ = 20°
Addendum height for each gear wheel
We know that the pitch circle radius of the smaller gear wheel,
r = m.t / 2 = 10 × 20 / 2 = 100 mm

and pitch circle radius of the larger gear wheel,
R = m.T / 2 = 10 × 40 / 2 = 200 mm
Let R
A = Radius of addendum circle for the larger gear wheel, and
rA = Radius of addendum circle for the smaller gear wheel.
Since the addendum on each wheel is to be made of such a length that the line of contact on
each side of the pitch point ( i.e. the path of approach and the path of recess) has half the maximum
possible length, therefore

Chapter 12 : Toothed Gearing    /G6C  407
Path of approach,           1
2KP MP= ...(Refer Fig. 12.13)
or     22 2
A.sin() c o s s i n2rRR Rφ−φ − φ =
or        22 2
A100 sin20( ) (200) cos 20 200 sin20 50 sin202R×°−° − ° == °
   2
A( ) 35 320 50 sin20 200 sin20 250 0.342 85.5R− = °+ °= × =
    22
A( ) 35 320 (85.5) 7310R−== ...(Squaring both sides)
   (RA)2 = 7310 + 35 320 = 42 630     or    RA = 206.5 mm
∴ Addendum height for larger gear wheel
   A 206.5 200 6.5 mm RR=− = −=  Ans.

A 206.5 200 6.5 mm RR=− = −=  Ans.
Now path of recess,    1
2PL PN=
or222
A.sin() c o s s i n2Rrr rφ−φ − φ =
or22 2
A200 sin 20( ) (100) cos 20 100 sin 20 100 sin 202r°−° − ° = = °
    22 2
A( ) (100) cos 20 100 sin20 100 sin20 200 0.342 68.4r− ° =° +° = × =
   22
A( ) 8830 (68.4) 4680r−= = ...(Squaring both sides)
    2
AA( ) 4680 8830 13510 or 116.2 mmrr=+= =
∴  Addendum height for smaller gear wheel
    A116.2 100 6.2 mmrr=− = − = Ans.
Length of the path of contact

We know that length of the path of contact
   11( ) s i n
22 2rRKP PL MP PN+φ=+= + =
   (100 200) sin 2051.3 mm2+°==  Ans.
We know that length of the arc of contact
   Length of the path of contact 51.354.6 mmcos cos20== =φ°Ans.
We know that circular pitch,                                       P
c = π m  =  π  × 10 = 31.42 mm
∴                   Length of the path ofcontact 54.6Contact ratio = 1.74 say 231.42cp== Ans.

408  /G6C   Theory of Machines
12.20. Minimum Number of Teeth on the Pinion in Order to Avoid
Interference
We have already discussed in the previous article that in order to avoid interference, the
addendum circles for the two mating gears must cut the common tangent to the base circles betweenthe points of tangency. The limiting condition reaches, when the addendum circles of pinion andwheel pass through points N and M (see Fig. 12.13) respectively.
Let t= Number of teeth on the pinion,,

T= Number of teeth on the wheel,
m= Module of the teeth,
r= Pitch circle radius of pinion = m.t / 2
G= Gear ratio = T / t = R / r
φ= Pressure angle or angle of obliquity.
From triangle O
1NP,
22 2
11 1 1
22 2()()( )2 c o s
sin 2 . sin cos (90 )ON OP P N OP P N OP N
rR r R=+− × ×
=+ φ − φ ° + φ
2 ...( sin sin )PN O P R=φ = φ/G51
22 2 2
22 2
22 2
2sin 2 . sin
sin 2 sin11 2 s i nrR r R
RR R Rrrrr r r=+ φ + φ
 φφ   =+ + = + + φ  

∴ Limiting radius of the pinion addendum circle,
22
1.12 s i n12 s i n2RR m t TTON rrr tt =+ + φ = + + φ   
   Let AP.m= Addendum of the pinion, where  AP  is a fraction by which the standard
addendum  of  one  module  for the pinion should be multiplied in orderto avoid interference.
We know that the addendum of the pinion
= O
1N – O1P
∴2
P...1 2 s i n22mt T T mtAmtt=+ +φ − 1...( . /2)OP r mt==/G51
2 .1s i n 1 22mt TT

tt= +φ −+ 
or2
P1s i n 1 22tTTA
tt=+φ −+ 
∴PP
2222
1+ ( +2)sin 11s i n 1 2AAt
TT GG
tt==
 φ−+φ −+

Chapter 12 : Toothed Gearing    /G6C  409
This equation gives the minimum number of teeth required on the pinion in order to avoid
interference.
Notes :  1.  If the pinion and wheel have equal teeth, then G  = 1. Therefore the above equation reduces to
22
13 s i n 1pAt=
+φ −
2. The minimum number of teeth on the pinion which will mesh with any gear (also rack) without
interference are given in the following table :
Table 12.2.  Minimum number of teeth on the pinion

S. No. System of gear teeth M inimum number of teeth on the pinion
1.1
214°Composite 12
2.1
214°Full depth involute 32
3. 20° Full depth involute 18
4. 20° Stub involute 14
12.21. Minimum Number of Teeth on the Wheel in Order to Avoid
Let T= Minimum number of teeth required on the wheel in order to avoid
interference,
and AW.m= Addendum of the wheel, where AW  is a fraction by which the standard
addendum for the wheel should be multiplied.

Using the same notations as in Art. 12.20, we have from triangle O2MP
222
22 2 2
222( )()()2 c o s
sin 2 . sin cos (90 )OM OP P M OP P M OP M
Rr R r=+− × ×
=+ φ − φ ° + φ
1 ...( sin )PM OP r=φ =/G51
= R2 + r2 sin2 φ + 2R.r sin2 φ
22
22 2
2sin 2 sin11 2 s i nrr r rRR
RR R R2 φφ   =+ + =++ φ    
∴ Limiting radius of wheel addendum circle,
22
2.1 sin 1 sin 222rm T trtOM RRTRT =+ φ = + φ ++  

We know that the addendum of the wheel
= O2M – O2P
∴2
W..1s i n 222mT t mT tAmTT=+ φ − +2...( . /2)O PRm T==/G51
2 .1s i n 1 22mT tt
TT= +φ −+ 
or2
W 1s i n 1 22T ttA
TT=+φ −+ 

410  /G6C   Theory of Machines
∴WW
2 222
111s i n 1 1s i n 1 22AAT
tt
TGTG==
 +φ − +φ −++  
Notes : 1.  From the above equation, we may also obtain the minimum number of teeth on pinion.
Multiplying both sides by ,t
T
W
22
111s i n 1 2tAt TTT
GG×
×=
+φ −+
W
22
111s i n 1 2At
G
GG=+φ −+ 
2. If wheel and pinion have equal teeth, then G  = 1, and
W
22
13 s i n 1AT=
+φ −
Example 12.9.  Determine the minimum number of teeth required on a pinion, in order to

avoid interference which is to gear with,
1. a wheel to give a gear ratio of 3 to 1 ; and 2. an equal wheel.
The pressure angle is 20° and a standard addendum of 1 module for the wheel may be
assumed.
Solution. Given : G = T / t = 3 ; φ = 20° ; AW  = 1 module
1. Minimum number of teeth for a gear ratio of 3 : 1
We know that minimum number of teeth required on a pinion,
W
22
G
GG×=
+φ −+ 
221 215.04 or16
0.133 1131s i n 2 0 1 233×== =
+° −+   Ans.

2. Minimum number of teeth for equal wheel
We know that minimum number of teeth for equal wheel,
                                  W
222 21 2
0.1621 3 sin 1 1 3 sin 20 1At× ×== =
+φ − + ° −
                                    12.34 or13=  Ans.
Example 12.10.   A pair of spur gears with involute teeth is to give a gear ratio of 4 : 1. The
arc of approach is not to be less than the circular pitch and smaller wheel is the driver. The angle of

pressure is 14.5°. Find : 1. the least number of teeth that can be used on each wheel, and 2. the
addendum of the wheel in terms of the circular pitch ?

Chapter 12 : Toothed Gearing    /G6C  411
Solution. Given : G  = T/t = R/r  =  4 ; φ = 14.5°
1. Least number of teeth on each wheel
Let t = Least number of teeth on the smaller wheel i.e. pinion,
T = Least number of teeth on the larger wheel i.e. gear, and
r = Pitch circle radius of the smaller wheel i.e. pinion.
We know that the maximum length of the arc of approach
Maximum length of the path of approach sintancos cosrrφ== = φφφ
and circular pitch,2
crpm
tπ=π =2...rmt=/G51

crpm
Since the arc of approach is not to be less than the circular pitch, therefore
22 2tan or 24.3say 25tan tan14.5rrttππ πφ= = = =φ°Ans.
and . 4 25 100TG t== × = Ans. ...( / )GT t=/G51
2. Addendum of the wheel
We know that addendum of the wheel
2 100 25251s i n 1 4 . 5 1 22 100100m× = +° − +   
50 0.017 0.85 0.85 / 0.27cc mm p p=× = =× π = Ans.
...( / ) cmp=π/G51

Example 12.11.   A pair of involute spur gears with 16° pressure angle and pitch of module
6 mm is in mesh. The number of teeth on pinion is 16 and its rotational speed is 240 r.p.m. W hen the
gear ratio is 1.75, find in order that the interference is just avoided ;  1. the addenda on pinion and
gear wheel ;  2. the length of path of contact ; and 3. the maximum velocity of sliding of teeth on either
side of the pitch point.

Solution. Given : φ = 16° ; m = 6 mm ; t = 16; N1 = 240 r.p.m. or ω1 = 2π × 240/60
= 25.136 rad/s; G = T / t = 1.75 or T = G.t = 1.75 × 16 = 28
1. Addenda on pinion and gear wheel
We know that addendum on pinion
       2
2.1s i n 1 22
61 6 28281s i n 1 6 1 22 1616mt TT
× = +° − +   
                     48 (1.224 1) 10.76 mm=− =  Ans.
2 .and addendum on wheel 1s i n 1 22mT tt

412  /G6C   Theory of Machines
       2 62 8 16161s i n 1 6 1 22 2828× = +° − +   
       84 (1.054 1) 4.56 mm=− =  Ans.
2. Length of path of contact
We know that the pitch circle radius of wheel,
. /2 6 28/2 84 mmRm T== × =
and pitch circle radius of pinion,
. /2 6 16/2 48 mmrm t== × =
∴  Addendum circle radius of wheel,
A Addendum of wheel 84 10.76 94.76 mm RR=+ = + =
and addendum circle radius of pinion,
rA = r + Addendum of pinion = 48 + 4.56 = 52.56 mm

We know that the length of path of approach,
22 2
22 2(94.76) (84) cos 16 84 sin16=− ° − °
49.6 23.15 26.45 mm=− =
and the length of the path of recess,
22 2
22 2(52.56) (48) cos 16 48sin16=− ° − °
25.17 13.23 11.94 mm=−=
26.45 11.94 38.39 mm KL KP PL=+= + = Ans.
3. Maximum velocity of sliding of teeth on either side of pitch point
Let ω2 = Angular speed of gear wheel.

We know that           1
21.75T
tω==ω or1
225.13614.28rad/s1.75 1.75ωω= = =
∴ Maximum velocity of sliding of teeth on the left side of pitch point i.e. at point K
12( ) (25.136 14.28) 26.45 1043 mm/s KP =ω+ ω = + = Ans.
and maximum velocity of sliding of teeth on the right side of pitch point i.e. at point L
12( ) (25.136 14.28)11.94 471mm/s PL =ω+ ω = + = Ans.
Example 12.12.   A pair of 20° full depth involute spur gears having 30 and 50 teeth respec-

tively of module 4 mm are in mesh. The smaller gear rotates at 1000 r.p.m. Determine : 1. sliding
velocities at engagement and at disengagement of pair of a teeth, and 2. contact ratio.
Solution. Given: φ = 20° ; t = 30 ; T = 50 ; m = 4 ; N1 = 1000 r.p.m. or ω1 = 2π × 1000/60
= 104.7 rad/s

Chapter 12 : Toothed Gearing    /G6C  413
1. Sliding velocities at engagement and at disengagement of pair of a teeth
First of all, let us find the radius of addendum circles of the smaller gear and the larger gear.
Addendum of the smaller gear,
tt = +φ −+   
2 43 0 50501s i n 2 0 1 22 3030× = +° − +   
60(1.31 1) 18.6 mm=− =
and addendum of the larger gear,
2
2.1s i n 1 22
45 0 30301s i n 2 0 1 22 5050
100(1.09 1) 9 mmmT tt

TT= +φ −+   
=− =
. /2 4 30/2 60 mmrm t== × =
∴ Radius of addendum circle of the smaller gear,
rA = r + Addendum of the smaller gear = 60 + 18.6 = 78.6 mm
Pitch circle radius of the larger gear,
R = m.T / 2 = 4 × 50 / 2 = 100 mm
∴  Radius of addendum circle of the larger gear,
RA = R + Addendum of the larger gear = 100 + 9 = 109 mm

We know that the path of approach ( i.e. path of contact when engagement occurs),
22 2
22 2(109) (100) cos 20 100 sin 20 55.2 34.2 21mm=− ° − ° = − =
and the path of recess ( i.e. path of contact when disengagement occurs),
222
22 2(78.6) (60) cos 20=− ° – 60 sin 20° = 54.76 – 20.52 = 34.24 mm
Let ω2 = Angular speed of the larger gear in rad/s.
We know that11
2
210.47 30or 62.82 rad/s50t T
tTωω × ×=ω = = =ω

2
∴ Sliding velocity at engagement of a pair of teeth
12( ) (104.7 62.82)21 3518 mm/s KP =ω+ ω = + =
= 3.518 m/s Ans.

414  /G6C   Theory of Machines
and sliding velocity at disengagement of a pair of teeth
12( ) (104.7 62.82)34.24 5736 mm/s PL =ω+ ω = + =
= 5.736m/s Ans.
2. Contact ratio
We know that the length of the arc of contact
Length of thepath of contact 21 34.24
cos cos cos 20
58.78 mmKP PL++== =
φφ °
=
and Circular pitch = π × m = 3.142 × 4 = 12.568 mm
∴Length of arc of contact 58.78Contact ratio = 4.67 say5Circular pitch 12.568== Ans.

Example 12.13.   Two gear wheels mesh externally and are to give a velocity ratio of 3 to 1.
The teeth are of involute form ; module = 6 mm, addendum = one module, pressure angle = 20°. Thepinion rotates at 90 r.p.m. Determine :  1. The number of teeth on the pinion to avoid interference on
it and the corresponding number of teeth on the wheel, 2. The length of path and arc of contact,
3.The number of pairs of teeth in contact, and  4. The maximum velocity of sliding .

Solution.  Given : G = T / t = 3 ; m  = 6 mm ; A
P = AW = 1 module = 6 mm ;  φ = 20° ;
N1 = 90 r.p.m.  or  ω1 = 2π × 90 / 60 = 9.43 rad/s
1. Number of teeth on the pinion to avoid interference on it and the corresponding number of teeth
    on the wheel
We know that number of teeth on the pinion to avoid interference,
P
222 26
1 ( 2)sin 1 1 3(3 2)sin 20 1At
GG×==
++ φ − + + ° −
= 18.2 say 19 Ans.
and corresponding number of teeth on the wheel,
              T = G.t = 3 × 19 = 57 Ans.

T = G.t = 3 × 19 = 57 Ans.
2. Length of path and arc of contact
r  =  m.t / 2  =  6  × 19/2 = 57 mm
rA = r + Addendum on pinion ( AP) = 57 + 6 = 63 mm
R = m.T / 2 = 6 × 57 / 2 = 171 mm
∴  Radius of addendum circle of wheel,
AWAddendum on wheel ( ) 171 6 177 mm RR A=+ = +=
22 2

22 2
22 2(177) (171) cos 20=− ° – 171 sin 20° = 74.2 – 58.5 = 15.7 mm

Chapter 12 : Toothed Gearing    /G6C  415
222
A
22 2() c o s s i n
(63) (57) cos 20 57sin 20 33.17 19.5 13.67 mmPL r r r=−φ − φ
=− ° − ° = − =
∴ Length of path of contact,
15.7 13.67 29.37 mm KL KP PL=+= + = Ans.
Length of path of contact 29.3731.25 mmcos cos20== =φ°Ans.
3. Number of pairs of teeth in contact
6 18.852 mmcpm=π× =π× =

∴ Number of pairs of teeth in contact
Length of arc of contact 31.251.66 say218.852cp== = Ans.
4. Maximum velocity of sliding
2
1
21
2Let Angular speed of wheel in rad/s.
19We knowthat or 9.43 3.14 rad/s
57Tt
tTω=
ω=ω = ω × = × =
ω
∴ Maximum velocity of sliding,
S1 2()vK P=ω+ ω ...( )KP PL>/G51
(9.43 3.14)15.7 197.35 mm/s=+ = Ans.
12.22. Minimum Number of Teeth on a Pinion for Involute Rack in Order to
Avoid Interference
A rack and pinion in mesh is shown in Fig. 12.14.

Fig. 12.14.  Rack and pinion in mesh.
Let t = Minimum number of teeth on the pinion,

416  /G6C   Theory of Machines
r = Pitch circle radius of the pinion = m.t / 2, and
φ = Pressure angle or angle of obliquity, and
AR.m = Addendum for rack, where AR is the fraction by which the standard
            addendum of one module for the rack is to be multiplied.
We know that a rack is a part of toothed wheel of infinite diameter. Therefore its base circle

diameter and the profiles of the involute teeth are straight lines. Since these straight profiles aretangential to the pinion profiles at the point of contact, therefore they are perpendicular to the tangentPM. The point M is the interference point.
Addendum for rack,
R.s i nAm L H P L== φ
2(s i n ) s i n s i nOP OP=φ φ =φ ...(∵ PL = OP sin φ)
22sin.sin2rmt=φ= × φ
∴R
22
sintA=
φ
Example 12.14.  A pinion of 20 involute teeth and 125 mm pitch circle diameter drives a

rack. The addendum of both pinion and rack is 6.25 mm. What is the least pressure angle which canbe used to avoid interference ? With this pressure angl e, find the length of the arc of contact and the
minimum number of teeth in contact at a time.
Solution. Given : T = 20 ; d = 125 mm or r = OP = 62.5 mm ; LH = 6.25 mm
Least pressure angle to avoid interference
Let φ = Least pressure angle to avoid interference.
We know that for no interference, rack addendum,

22 6.25sin or sin 0.16.25LHLH rr=φ φ = = =
∴ sin 0.3162φ=           or      18.435φ= °  Ans.
22()()KLOK OL= − ...(Refer Fig. 12.14)
22
22( 6 . 2 5 )(c o s )
(62.5 6.25) (62.5 cos18.435 )
4726.56 3515.62 34.8 mmOP OP=
=+− φ
=+− °
−=
∴ Length of the arc of contact
Length of the path of contact 34.836.68 mmcos cos18.435= ==φ°Ans.
Minimum number of teeth

/ 125/20 19.64 mmcpdT=π =π× =

Chapter 12 : Toothed Gearing    /G6C  417
Fig. 12.15.  Helical gear.
Crossed helical gears.and the number of pairs of teeth in contact
Length of the arc of contact 36.681.87Circular pitch ( ) 19.64cp= ==
 ∴ Minimum number of teeth in contact
= 2 or one pair Ans.
12.23. Helical Gears
A helical gear has teeth in the form

of helix around the gear. Two such gearsmay be used to connect two parallel shaftsin place of spur gear. The helixes may beright handed on one wheel and left handedon the other. The pitch surfaces are cylin-drical as in spur gearing, but the teeth in-stead of being parallel to the axis, windaround the cylinders helically like screwthreads. The teeth of helical gears withparallel axis have line contact, as in spurgearing. This provides gradual engage-ment and continuous contact of the engaging

and continuous contact of the engaging teeth. Hence helical gears give smooth drive with ahigh efficiency of transmission.

We have already discussed that the helical gears may be
of single helical type or double helical type. In case of singlehelical gears, there is some axial thrust between the teeth, whichis a disadvantage. In order to eliminate this axial thrust, doublehelical gears are used. It is equivalent to two single helical gears,in which equal and opposite thrusts are produced on each gearand the resulting axial thrust is zero.
The following definitions may be clearly understood in

connection with a helical gear as shown in Fig. 12.15.
1. Normal pitch . It is the distance between similar faces of adjacent teeth, along a helix on
the pitch cylinder normal to the teeth. It is denoted by p
N.
2. Axial pitch . It is the distance measured parallel to the axis, between similar faces of adja-
cent teeth. It is the same as circular pitch and is therefore denoted by pc. If α is the helix angle, then
circular pitch,
N
coscpp=α

N
coscpp=α
Note : The helix angle  is also known as  spiral angle of the teeth.
12.24. Spiral Gears
 We have already discussed that spiral gears (also known as skew gears  or screw gears ) are

used to connect and transmit motion between two non-parallel and non-intersecting shafts. The pitchsurfaces of the spiral gears are cylindrical and the teeth have point contact. These gears are onlysuitable for transmitting small power. We have seen that helical gears, connected on parallel shafts,are of opposite hand. But spiral gears may be of the same hand or of opposite hand.

418  /G6C   Theory of Machines
12.25. Centre Distance for a Pair of  Spiral Gears
The centre distance, for a pair of spiral gears,
is the shortest distance between the two shafts makingany angle between them. A pair of spiral gears 1 and 2,both having left hand helixes ( i.e. the gears are of the
same hand) is shown in Fig. 12.16. The shaft angle θ is

the angle through which one of the shafts must be rotatedso that it is parallel to the other shaft, also the two shaftsbe rotating in opposite directions.
Let    α
1 and α2 = Spiral angles of gear
  teeth for gears 1 and 2   respectively,
           p
c1 and pc2 = Circular pitches of gears 1
 and 2,
             T1 and T2 = Number of teeth on gears
1 and 2,
         d1 and d2= Pitch circle diameters of
  gears 1 and 2,
            N1 and N2 = Speed of gears 1 and 2,

N1 and N2 = Speed of gears 1 and 2,
                        G  = Gear ratio = 21
12,TN
TN=
                       pN = Normal pitch, and
             L  = Least centre distance
                                                  between the axes of shafts.
Since the normal pitch is same for both the spiral gears, therefore
NN
12
12,a n dcos cosccpppp==ααFig. 12.16. Centre distance for
a pair of spiral gears.
Helical gears

Chapter 12 : Toothed Gearing    /G6C  419
We know that11 1
11
1,o rc
cpT dp dT× π==π
and22 2
22
2,o rc
∴12 11 221
22cc dd pTpTL+ ××== + ππ
NN 2 11
12
11 2
N1
12cos cos 22
1
cos cos2ccpp T TTpp GT
G PT +× + × ==  αα ππ 
×+ = αα π
Notes : 1.  If the pair of spiral gears have teeth of the same hand, then
θ = α1 + α2
and for a pair of spiral gears of opposite hand,
θ = α1 – α2
2.  When θ = 90°, then both the spiral gears must have teeth of the same hand.

12.26. Efficiency of Spiral Gears
A pair of spiral gears 1 and 2 in mesh is shown in Fig. 12.17. Let the gear 1 be the driver and
the gear 2 the driven. The forces acting on each of a pair of teeth in contact are shown in Fig. 12.17.The forces are assumed to act at the centre of the width of each teeth and in the plane tangential to thepitch cylinders.
Fig. 12.17.  Efficiency of spiral gears.
Let F1 = Force applied tangentially on the driver,

F2 = Resisting force acting tangentially on the driven,
Fa1 = Axial or end thrust on the driver,

420  /G6C   Theory of Machines
Fa2 = Axial or end thrust on the driven,
RN = Normal reaction at the point of contact,
φ = Angle of friction,
R = Resultant reaction at the point of contact, and
θ = Shaft angle = α1+ α2
...( Both gears are of the same hand)/G51
From triangle OPQ, F1 = R cos (α1 – φ)
∴ Work input to the driver
= F1× π d1.N1 = R cos (α1 – φ) π d1.N1
From triangle OST,F2 = R cos (α2 + φ)
∴ Work output of the driven
= F2 × π d2.N2 = R cos (α2 + φ) π d2.N2
∴ Efficiency of spiral gears,

22 2
11 1cos( ) . Work output
Work input cos( ) .Rd N
Rd Nα+ φ πη= =α− φπ
22 2
11 1cos ( ) .
cos ( ) .dN
dNα+ φ=α− φ...(i)
We have discussed in Art. 12.25, that pitch circle diameter of gear 1,
11 N 1
1
1coscpTP Td×== ×πα π
and pitch circle diameter of gear 2,
22 N 2
2
2coscpTP Td×== ×πα π
∴22 1
11 2cos
cosdT
dTα=α...(ii)
We know that21
12NT
NT= ...(iii)
Multiplying equations  (ii) and (iii), we get,
22 1
11 2.c o s
.c o sdN
dNα=α

22 1
11 2.c o s
.c o sdN
dNα=α
Substituting this value in equation (i), we have
                                   21
12cos ( ) cos
cos ( ) cosα+ φ αη=α− φ α...(iv)
                                      12 12
21 21cos ( ) cos ( )
cos ( ) cos ( )α+ α+ φ+ α− α− φ=α+ α− φ+ α− α+ φ
1...cos cos [cos( ) cos ( )]2AB A B A B=+ + − /G51

Chapter 12 : Toothed Gearing    /G6C  421
12
21cos ( ) cos ( )
cos ( ) cos ( )θ+φ + α −α −φ=θ−φ + α −α +φ...(v)
12 ...( )θ=α +α/G51
Since the angles θ and φ are constants, therefore the efficiency will be maximum, when
cos (α1 – α2 – φ) is maximum, i.e.
cos (α1 – α2  – φ) = 1 or α1 – α2  – φ  = 0
∴α1 = α2  + φand  α2 = α1 – φ
Since  α1 + α2  = θ, therefore
121 1 or2θ+φα= θ − α= θ − α+ φ α=
Similarly, 22θ−φα=
Substituting α1 = α2 + φ  and  α2 = α1 – φ, in equation (v), we get
cos ( ) 1

cos ( ) 1
cos ( ) 1maxθ+φ +η=θ−φ +...(vi)
Note: From Fig. 12.17, we find that12
N
12cos cosFFR==
αα
∴  Axial thrust on the driver, Fa1 = RN.sin α1 = F1.tan α1
and axial thrust on the driven, Fa2 = RN.sin α2 = F2.tan α2
Example 12.15.  A pair of spiral gears is required to connect two shafts 175 mm apart, the

shaft angle being 70°. The velocity ratio is to be 1.5 to 1, the faster wheel having 80 teeth and a pitchcircle diameter of 100 mm. Find the spiral angles for each wheel. If the torque on the faster wheel is75 N-m ; find the axial thrust on each shaft, neglecting friction.
Solution. Given : L = 175 mm = 0.175 m ; θ = 70° ; G = 1.5 ; T
2 = 80 ; d2 = 100 mm = 0.1 m
or r2 = 0.05 m ; Torque on faster wheel = 75 N-m
Spiral angles for each wheel
Let α1 = Spiral angle for slower wheel, and

α2 = Spiral angle for faster wheel.
We know that velocity ratio,21
121.5NTGNT== =
∴  No. of teeth on slower wheel,
T1 = T2 × 1.5 = 80 × 1.5 = 120
We also know that the centre distance between shafts ( L),
12 1 0.10.17522dd d++==
∴ d1 = 2 × 0.175 – 0.1 = 0.25 m
and22 1 1 1
11 2 2 2cos 80 cos 2 cos 0.1orcos 0.25 120 cos 3 cosdT
dTαα α== =αα α

422  /G6C   Theory of Machines
∴1
12
2cos 0.1 30.6 or cos 0.6 coscos 0.25 2α×== α = αα×...(i)
We know that,        α1 + α2 = θ = 70°   or    α2 = 70° – α1
Substituting the value of α2 in equation  (i),
cos α1 = 0.6 cos (70° –  α1) = 0.6 (cos 70° cos α1 + sin 70° sin α1)
[]...cos( ) cos cos sin sinAB A B A B−= + /G51
11 0.2052 cos 0.5638 sin=α +α
cos α1 – 0.2052 cos α1 = 0.5638 sin α1
11 0.7948 cos 0.5638 sin α= α
∴1
11
1sin 0.7948tan 1.4097 or 54.65cos 0.5638αα= = = α= °α

and270 54.65 15.35α = °− °= ° Ans.
Axial thrust on each shaft
We know that Torque = Tangential force × Pitch circle radius∴ Tangential force at faster wheel,
2
2Torqueon thefaster wheel 751500 NPitchcircleradius( ) 0.05Fr== =
and normal reaction at the point of contact,
RN = F2 / cos α2 = 1500/cos 15.35° = 1556 N
We know that axial thrust on the shaft of slower wheel,
Fa1 = RN. sin α1 = 1556 × sin 54.65° = 1269 N Ans.
and axial thrust on the shaft of faster wheel,

Fa2 = RN. sin α2 = 1556 × sin 15.35° = 412 N Ans.
Example 12.16.   In a spiral gear drive connecting two shafts, the approximate centre
 distance is 400 mm and the speed ratio = 3. The angle between the two shafts is 50° and the normalpitch is 18 mm. The spiral angle for the driving and driven wheels are equal. Find : 1. Number of
teeth on each wheel, 2. Exact centre distance, and  3. Efficiency of the drive, if friction angle = 6°.

Solution. Given : L  = 400 mm  = 0.4 m ; G  = T
2 / T1 = 3 ; θ = 50° ; pN = 18 mm ; φ = 6°
1. Number of teeth on each wheel
Let T1 = Number of teeth on wheel 1 ( i.e. driver), and
T2 = Number of teeth on wheel 2 ( i.e. driven).
Since the spiral angle α1 for the driving wheel is equal to the spiral angle α2 for the driven
wheel, therefore
          α1 = α2 = θ/2 = 25°12...( 50 )α+ α= θ = °/G51
We know that centre distance between two shafts ( L),
           N. 1 N 1

N. 1 N 1
12 111 .400cos cos cos22GG pT pT +   + ==  αα α ππ  12...( )α= α/G51

Chapter 12 : Toothed Gearing    /G6C  423
1
113 1812.64cos252TT+×== ° π
∴ T1 = 400/12.64  =  31.64  or  32  Ans.
and T2 = G.T1 = 3 × 32 = 96 Ans.
2. Exact centre distance
We know that exact centre distance,
N1 N1
1
12 11.. 1
cos cos22 c o sG pT pT GL+  + == ααππ α 12...( )α= α/G51
13 18 32404.5 mmcos252+ ×== ° πAAns.
3. Efficiency of the drive
21 1
12 1cos ( ) cos cos ( )
cos ( ) cos cos ( )α+ φ α α+ φη= =α− φ α α− φ12...( )α= α/G51

cos (25 6 ) cos 31 0.85720.907 90.7%cos (25 6 ) cos19 0.9455°+ ° °== = = =°− ° °Ans.
Example 12.17.  A drive on a machine tool is to be made by two spiral gear wheels, the
spirals of which are of the same hand  and has normal pitch of 12.5 mm. The wheels are of equaldiameter and the centre distance between the axes of the shafts is approximately 134 mm. The anglebetween the shafts is 80° and the speed ratio 1.25. Determine : 1. the spiral angle of each wheel,

2. the number of teeth on each wheel, 3. the efficiency of the drive, if the friction angle is 6°, and
4. the maximum efficiency.
Solution.  Given : p
N  = 12.5 mm ; L = 134 mm ; θ =  80° ; G = N2 / N1 = T1 / T2 = 1.25
1. Spiral angle of each wheel
Let α1 and α2 = Spiral angles of wheels 1 and 2 respectively, and
d1 and d2 = Pitch circle diameter of wheels 1 and 2 respectively.
We know that22 1
12 21
11 2cosor cos coscosdTTTdTα=α = αα12...( )dd=/G51
∴11
12
22cos1.25 or cos 1.25 coscosT

∴11
12
Tα== α = αα...(i)
12 2 1 80 or 80 α+ α= θ = ° α= ° − α
Substituting the value of α2 in equation (i),
cos α1 = 1.25 cos (80° – α1) = 1.25 (cos 80° cos α1 + sin 80° sin α1)
= 1.25 (0.1736 cos α1 + 0.9848 sin α1)
= 0.217 cos α1 + 1.231 sin α1
cos α1 – 0.217 cos α1 = 1.231 sin α1     or     0.783 cos α1 = 1.231 sin α1
∴ tan α1 = sin α1 / cos α1 = 0.783 / 1.231 = 0.636    or   α1 = 32.46° Ans.
and α2 = 80° – 32.46° = 47.54°  Ans.

424  /G6C   Theory of Machines
2. Number of teeth on each wheel
Let T1 = Number of teeth on wheel 1, and
T2 = Number of teeth on wheel 2.
 We know that centre distance between the two shafts ( L),
12
12 134 or 134 mm2dddd+== =12...( )dd=/G51
1
1..
coscNpTpTd==ππ α
∴11
1
N.cos 134 cos 32.4628.4 or 3012.5dTpπα π × × °== = Ans.
and1
230241.25 1.25TT=== Ans.
 21
12cos ( ) cos cos (47.54 6 ) cos 32.46

21
cos ( ) cos cos (32.46 6 ) cos 47.54α+ φ α ° +° °η= =α− φ α ° −° °
     0.5943 0.84370.83 or 83%0.8952 0.6751×==×Ans.
4. Maximum efficiency
We know that maximum efficiency,
          cos ( ) 1 cos (80 6 ) 1 1.0698
cos ( ) 1 cos (80 6 ) 1 1.2756maxθ+φ + °+ ° +η= = =θ−φ + °− ° +
     0.838 or 83.8%= Ans.
EXERCISES
1.The pitch circle diameter of the smaller of the two spur wheels which mesh externally and have

involute teeth is 100 mm. The number of teeth are 16 and 32. The pressure angle is 20° and the
addendum is 0.32 of the circular pitch. Find the length of the path of contact of the pair of teeth.
[Ans. 29.36 mm]
2.A pair of gears, having 40 and 30 teeth respectively are of 25° involute form. The addendum length is

5 mm and the module pitch is 2.5 mm. If the smaller wheel is the driver and rotates at 1500 r.p.m., findthe velocity of sliding at the point of engagement and at the point of disengagement.
[Ans. 2.8 m/s ; 2.66 m/s]
3.Two gears of module 4mm have 24 and 33 teeth. The pressure angle is 20° and each gear has a
standard addendum of one module. Find the length of arc of contact and the maximum velocity ofsliding if the pinion rotates at 120 r.p.m. [Ans. 20.58 mm ; 0.2147 m/s]

4.The number of teeth in gears 1 and 2 are 60 and 40 ; module = 3 mm ; pressure angle = 20° and
addendum = 0.318 of the circular pitch. Determine the velocity of sliding when the contact is at the tipof the teeth of gear 2 and the gear 2 rotates at 800 r.p.m. [Ans. 1.06 m/s]
5.Two spur gears of 24 teeth and 36 teeth of 8 mm module and 20° pressure angle are in mesh. Adden-

dum of each gear is 7.5 mm. The teeth are of involute form. Determine : 1. the angle through which thepinion turns while any pair of teeth are in contact, and 2. the velocity of sliding between the teethwhen the contact on the pinion is at a radius of 102 mm. The speed of the pinion is 450 r.p.m.
[Ans. 20.36°, 1.16 m/s]

Chapter 12 : Toothed Gearing    /G6C  425
6.A pinion having 20 involute teeth of module pitch 6 mm rotates at 200 r.p.m. and transmits 1.5 kW to
a gear wheel having 50 teeth. The addendum on both the wheels is 1/4 of the circular pitch. The angle
of obliquity is 20°. Find ( a) the length of the path of approach ; ( b) the length of the arc of approach;
(c) the normal force between the teeth at an instant where there is only pair of teeth in contact.
[Ans. 13.27 mm ; 14.12 mm ; 1193 N]

7.Two mating involute spur gear of 20° pressure angle have a gear ratio of 2. The number of teeth on the
pinion is 20 and its speed is 250 r.p.m. The module pitch of the teeth is 12 mm.
If the addendum on each wheel is such that the path of approach and the path of recess on each side are

half the maximum possible length, find : 1. the addendum for pinion and gear wheel ; 2. the length ofthe arc of contact ; and 3. the maximum velocity of sliding during approach and recess.Assume pinion to be the driver. [Ans. 19.5 mm, 7.8 mm ; 65.5 mm ; 807.5 mm/s, 1615 mm/s]
8.Two mating gears have 20 and 40 involute teeth of module 10 mm and 20° pressure angle. If the
addendum on each wheel is such that the path of contact is maximum and interference is just avoided,

find the addendum for each gear wheel, path of contact, arc of contact and contact ratio.
[Ans. 14 mm ; 39 mm ; 102.6 mm ; 109.3 mm ; 4]
9.A 20° involute pinion with 20 teeth drives a gear having 60 teeth. Module is 8 mm and addendum of
each gear is 10 mm.1.  State whether interference occurs or not. Give reasons.2.  Find the length of path of approach and arc of approach if pinion is the driver.
[Ans. Interference does not occur ; 25.8 mm, 27.45 mm]

10.A pair of spur wheels with involute teeth is to give a gear ratio of 3 to 1. The arc of approach is not to
be less than the circular pitch and the smaller wheel is the driver. The pressure angle is 20°. What isthe least number of teeth that can be used on each wheel ? What is the addendum of the wheel in termsof the circular pitch ? [Ans. 18, 54 ; 0.382 P
c]
11.Two gear wheels mesh externally and are to give a velocity ratio of 3. The teeth are of involute form

of module 6. The standard addendum is 1 module. If the pressure angle is 18° and pinion rotates at 90
r.p.m., find : 1. the number of teeth on each wheel, so that the interference is just avoided, 2. the lengthof the path of contact, and 3. the maximum velocity of sliding between the teeth.
[Ans. 19, 57 ; 31.5 mm ; 213.7 mm/s]
12.A pinion with 24 involute teeth of 150 mm of pitch circle diameter drives a rack. The addendum of the

pinion and rack is 6 mm. Find the least pressure angle which can be used if under cutting of the teeth
is to be avoided. Using this pressure angle, find the length of the arc of contact and the minimum
number of teeth in contact at one time. [Ans. 16.8° ; 40 mm ; 2 pairs of teeth]
13.Two shafts, inclined at an angle of 65° and with a least distance between them of 175 mm are to be

connected by spiral gears of normal pitch 15 mm to give a reduction ratio 3 : 1. Find suitable diam-eters and numbers of teeth. Determine, also, the efficiency if the spiral angles are determined by thecondition of maximum efficiency. The friction angle is 7°.
 [Ans. 88.5 mm ; 245.7 mm ; 15, 45 ; 85.5 %]
14.A spiral wheel reduction gear, of ratio 3 to 2, is to be used on a machine, with the angle between the

shafts 80°. The approximate centre distance between the shafts is 125 mm. The normal pitch of theteeth is 10 mm and the wheel diameters are equal. Find the number of teeth on each wheel, pitch circlediameters and spiral angles. Find the efficiency of the drive if the friction angle is 5°.
[Ans. 24, 36 ; 128 mm ; 53.4°, 26.6° ; 85.5 %]
15.A right angled drive on a machine is to be made by two spiral wheels. The wheels are of equal

diameter with a normal pitch of 10 mm and the centre distance is approximately 150 mm. If the speedratio is 2.5 to 1, find : 1. the spiral angles of the teeth, 2. the number of teeth on each wheel, 3.the exact
centre distance, and 4. transmission efficiency, if the friction angle is 6°.
[Ans. 21.8°, 68.2° ; 18 , 45 ; 154 mm ; 75.8 %]
1.Explain the terms :  (i) Module, (ii) Pressure angle, and (iii) Addendum.

2.State and prove the law of gearing. Show that involute profile satisfies the conditions for correct
gearing.
3.Derive an expression for the velocity of sliding between a pair of involute teeth. State the advantages
of involute profile as a gear tooth profile.

426  /G6C   Theory of Machines
4.Prove that the velocity of sliding is proportional to the distance of the point of contact from the pitch
point.
5.Prove that for two involute gear wheels in mesh, the angular velocity ratio does not change if the
centre distance is increased within limits, but the pressure angle increases.
6.Derive an expression for the length of the arc of contact in a pair of meshed spur gears.
7.What do you understand by the term ‘interference’ as applied to gears?

8.Derive an expression for the minimum number of teeth required on the pinion in order to avoid
interference in involute gear teeth when it meshes with wheel.
9.Derive an expression for minimum number of teeth required on a pinion to avoid interference when it
gears with a rack.
10.Define (i) normal pitch, and ( ii) axial pitch relating to helical gears.
11.Derive an expression for the centre distance of a pair of spiral gears.

12.Show that, in a pair of spiral gears connecting inclined shafts, the efficiency is maximum when the
spiral angle of the driving wheel is half the sum of the shaft and friction angles.
1.The two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the
shaft. This arrangement is called
(a) spur gearing ( b) helical gearing ( c) bevel gearing ( d) spiral gearing

2.The type of gears used to connect two non-parallel non-intersecting shafts are
(a) spur gears ( b) helical gears ( c) spiral gears ( d) none of these
3.An imaginary circle which by pure rolling action, gives the same motion as the actual gear, is called
(a) addendum circle ( b) dedendum circle ( c) pitch circle ( d) clearance circle
4.The size of a gear is usually specified by
(a) pressure angle ( b) circular pitch ( c) diametral pitch ( d) pitch circle diameter

5.The radial distance of a tooth from the pitch circle to the bottom of the tooth, is called
(a) dedendum ( b) addendum ( c) clearance ( d) working depth
6.The product of the diametral pitch and circular pitch is equal to
(a)1 ( b)1 /π (c)π (d)2π
7.The module is the reciprocal of
(a) diametral pitch ( b) circular pitch ( c) pitch diameter ( d) none of these
8.Which is the incorrect relationship of gears?
(a) Circular pitch × Diametral pitch = π (b) Module = P.C.D/No.of teeth

(c) Dedendum = 1.157 module ( d) Addendum = 2.157 module
9.If the module of a gear be m, the number of teeth T and pitch circle diameter D, then
(a)m = D/T (b)D = T/m (c)m = D/2T(d) none of these
10.Mitre gears are used for
(a) great speed reduction ( b) equal speed
(c) minimum axial thrust ( d) minimum backlash
11.The condition of correct gearing is
(a) pitch line velocities of teeth be same
(b) radius of curvature of two profiles be same

(c) common normal to the pitch surface cuts the line of centres at a fixed point
12.Law of gearing is satisfied if
(a) two surfaces slide smoothly
(b) common normal at the point of contact passes through the pitch point on the line joining the
centres of rotation
(c) number of teeth = P.C.D./ module
(d) addendum is greater than dedendum

Chapter 12 : Toothed Gearing    /G6C  427
13.Involute profile is preferred to cyloidal because
(a) the profile is easy to cut
(b) only one curve is required to cut
(c) the rack has straight line profile and hence can be cut accurately
14.The contact ratio for gears is
(a) zero ( b) less than one ( c) greater than one
15.The maximum length of arc of contact for two mating gears, in order to avoid interference, is

(a)(r + R) sin φ(b)(r + R) cos φ(c)(r + R) tan  φ(d) none of these
where    r = Pitch circle radius of pinion,
    R = Pitch circle radius of driver, and
φ = Pressure angle.
16.When the addenda on pinion and wheel is such that the path of approach and path of recess are half of
their maximum possible values, then the length of the path of contact is given by
(a)() s i n
2rR+φ(b)() c o s
2rR+φ(c)() t a n
2rR+φ(d) none of these

17.Interference can be avoided in involute gears with 20° pressure angle by
(a) cutting involute correctly
(b) using as small number of teeth as possible
(c) using more than 20 teeth
(d) using more than 8 teeth
18.The ratio of face width to transverse pitch of a helical gear with α as the helix angle is normally
(a) more than 1.15/tan α (b) more than 1.05/tan α
(c) more than 1/tan α (d) none of these
19.The maximum efficiency for spiral gears is

(a)sin ( ) 1
cos ( ) 1θ+φ +
θ−φ +(b)cos ( ) 1
sin ( ) 1θ−φ +
θ+φ +
 (c)cos ( ) 1
θ−φ +(d)cos ( ) 1
cos ( ) 1θ−φ +
θ+φ +
where       θ = Shaft angle, and φ = Friction angle.
20.For a speed ratio of 100, smallest gear box is obtained by using
(a) a pair of spur gears
(b) a pair of helical and a pair of spur gear compounded
(c) a pair of bevel and a pair of spur gear compounded
(d) a pair of helical and a pair of worm gear compounded

ANSWERS
1.(a)  2.(c) 3.(c) 4.(d) 5.(a)
6.(c) 7.(a) 8.(d) 9.(a) 10.(b)
11.(c) 12.(b) 13.(b) 14.(c) 15.(c)
16.(a) 17.(c) 18.(a) 19.(c) 20.(d)

428       /G6C               Theory of Machines
Gear Gear Gear Gear Gear TTTTTrainsrainsrainsrainsrains13
2. Types of Gear Trains.3. Simple Gear Train.4. Compound Gear Train.5. Design of Spur Gears.6. Reverted Gear Train.7. Epicyclic Gear Train.8. Velocity Ratio of Epicyclic
Gear Train.
9. Compound Epicyclic Gear
Train (Sun and Planet
Wheel).
10. Epicyclic Gear Train With
Bevel Gears.
11. Torques in Epicyclic Gear

Trains.13.1.13.1.13.1.13.1.13.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoduction
Sometimes, two or more gears are made to mesh with
each other to transmit power from one shaft to another. Sucha combination is called gear train or train of toothed wheels .
The nature of the train used depends upon the velocity ratiorequired and the relative position of the axes of shafts. Agear train may consist of spur, bevel or spiral gears.

13.2.13.2.13.2.13.2.13.2. TTTTTypes of Gear ypes of Gear ypes of Gear ypes of Gear ypes of Gear TTTTTrainsrainsrainsrainsrains
Following are the different types of gear trains, de-
pending upon the arrangement of wheels :
1. Simple gear train, 2. Compound gear train, 3. Re-
verted gear train, and 4. Epicyclic gear train.
In the first three types of gear trains, the axes of the

shafts over which the gears are mounted are fixed relative toeach other. But in case of epicyclic gear trains, the axes ofthe shafts on which the gears are mounted may move relativeto a fixed axis.
13.3.13.3.13.3.13.3.13.3. Simple Gear Simple Gear Simple Gear Simple Gear Simple Gear TTTTTrainrainrainrainrain
When there is only one gear on each shaft, as shown
in Fig. 13.1, it is known as  simple gear train . The gears are
represented by their pitch circles.

When the distance between the two shafts is small,
the two gears 1 and 2 are made to mesh with each other to
428

Chapter 13 : Gear Trains    /G6C  429
transmit motion from one shaft to the other, as shown in Fig. 13.1 ( a). Since the gear 1 drives the gear
2, therefore gear 1 is called the driver and the gear 2 is called the  driven or follower. It may be noted
that the motion of the driven gear is opposite to the motion of driving gear.
 
(a)( b)( c)
  Fig. 13.1. Simple gear train.
Let N1 = Speed of gear 1(or driver) in r.p.m.,
N2 = Speed of gear 2 (or driven or follower) in r.p.m.,

T1 = Number of teeth on gear 1, and
T2 = Number of teeth on gear 2.
Since the speed ratio (or velocity ratio) of gear train is the ratio of the speed of the driver to
the speed of the driven or follower and ratio of speeds of any pair of gears in mesh is the inverse oftheir number of teeth, therefore
                                   Speed ratio 
12
21NT
NT==
It may be noted that ratio of the speed of the driven or follower to the speed of the driver is

known as  train value of the gear train. Mathematically,
                                   Train value 21
12NT
NT==
From above, we see that the train value is the reciprocal of speed ratio.
Sometimes, the distance between the two gears is large. The motion from one gear to another,
in such a case, may be transmitted by either of the following two methods :
1. By providing the large sized gear, or 2. By providing one or more intermediate gears.

A little consideration will show that the former method ( i.e. providing large sized gears) is
very inconvenient and uneconomical method ; whereas the latter method ( i.e. providing one or more
intermediate gear) is very convenient and economical.
It may be noted that when the number of intermediate gears are odd, the motion of both the
gears (i.e. driver and driven or follower) is like as shown in Fig. 13.1 ( b).

But if the number of intermediate gears are even, the motion of the driven or follower will be
in the opposite direction of the driver as shown in Fig. 13.1 ( c).
Now consider a simple train of gears with one intermediate gear as shown in Fig. 13.1 ( b).
Let N1 = Speed of driver in r.p.m.,
N2 = Speed of intermediate gear in r.p.m.,

430  /G6C   Theory of Machines
N3 = Speed of driven or follower in r.p.m.,
T1 = Number of teeth on driver,
T2 = Number of teeth on intermediate gear, and
T3 = Number of teeth on driven or follower.
Since the driving gear 1 is in mesh with the intermediate gear 2, therefore speed ratio for
these two gears is
12
21=NT
NT...(i)
Similarly, as the intermediate gear 2 is in mesh with the driven gear 3, therefore speed ratio
for these two gears is
3 2
32=TN
NT...(ii)

3 2
32=TN
NT...(ii)
The speed ratio of the gear train as shown in Fig. 13.1 ( b) is obtained by multiplying the
equations (i) and (ii).
∴3 12 2
23 1 2×= ×T NNT
NNT Tor3 1
31=TN
NT
i.e.Speed of driver No. of teeth on drivenSpeed ratio = =Speed of driven No. of teeth on driver
andSpeed of driven No. of teeth on driverTrain value = =Speed of driver No. of teeth on driven
Similarly, it can be proved that the

above equation holds good even if there areany number of intermediate gears. Fromabove, we see that the speed ratio and thetrain value, in a simple train of gears, is in-dependent of the size and number of inter-mediate gears. These intermediate gears arecalled idle gears , as they do not effect the
speed ratio or train value of the system. Theidle gears are used for the following two pur-poses :
1. To connect gears where a large

centre distance is required, and
2. To obtain the desired direction of
motion of the driven gear ( i.e. clockwise or
anticlockwise).
13.4. Compound Gear Train
When there are more than one gear on a shaft, as shown in Fig. 13.2, it is called a  compound
train of gear .
We have seen in Art. 13.3 that the idle gears, in a simple train of gears do not effect the speed

ratio of the system. But these gears are useful in bridging over the space between the driver and thedriven.
Gear trains inside a mechanical watch

Chapter 13 : Gear Trains    /G6C  431
But whenever the distance between the driver and the driven or follower has to be bridged

over by intermediate gears and at the same time a great ( or much less ) speed ratio is required, thenthe advantage of intermediate gears is intensified by providing compound gears on intermediate shafts.In this case, each intermediate shaft has two gears rigidly fixed to it so that they may have the samespeed. One of these two gears meshes with the driver and the other with the driven or followerattached to the next shaft as shown in Fig.13.2.
Fig. 13.2. Compound gear train.

In a compound train of gears, as shown in Fig. 13.2, the gear 1 is the driving gear mounted on
shaft A, gears 2 and 3 are compound gears which are mounted on shaft B. The gears 4 and 5 are also
compound gears which are mounted on shaft C and the gear 6 is the driven gear mounted on shaft D.
Let N1 = Speed of driving gear 1,
T1 = Number of teeth on driving gear 1,
N2 ,N3 ..., N6 = Speed of respective gears in r.p.m., and

T2 ,T3..., T6 = Number of teeth on respective gears.
Since gear 1 is in mesh with gear 2, therefore its speed ratio is
12
21NT
NT= ...(i)
Similarly, for gears 3 and 4, speed ratio is
34
43NT
NT= ...(ii)
and for gears 5 and 6, speed ratio is
56
65NT
The speed ratio of compound gear train is obtained by multiplying the equations (i), (ii) and (iii),
∴35 612 4
2 4 6135NN TNT T
NNNT TT××= × × or*2461
61 3 5TTTN
NT T T××=××

61 3 5TTTN
NT T T××=××
*Since gears 2 and 3 are mounted on one shaft B, therefore N2 = N3. Similarly gears 4 and 5 are mounted on
shaft C, therefore N4 = N5.

432  /G6C   Theory of Machines
i.e.Speed of the first driverSpeed ratio = Speed of the last driven or follower
Product of the number of teeth on the drivens= 
Product of the number of teeth on the drivers 
andSpeed of the last driven or followerTrain value = Speed of the first driver 
Product of the number of teeth on the drivers= 
Product of the number of teeth on the drivens 
The advantage of a compound train over a simple gear train is that a much larger speed

reduction from the first shaft to the last shaft can be obtained with small gears. If a simple gear trainis used to give a large speed reduction, the last gear has to be very large. Usually for a speed reductionin excess of 7 to 1, a simple train is not used and a compound train or worm gearing is employed.
Note: The gears which mesh must have the same circular pitch or module. Thus gears 1 and 2 must have the
same module as they mesh together. Similarly gears 3 and 4, and gears

5 and 6 must have the same module.
Example 13.1.  The gearing of a machine tool is shown
in Fig. 13.3. The motor shaft is connected to gear A and rotates
at 975 r.p.m. The gear wheels B , C, D and E are fixed to parallel
shafts rotating together . The final gear F is fixed on the output
shaft. What is the speed of gear F ? The number of teeth on
each gear are as given below :
Gear A B C D E FNo. of teeth 20 50 25 75 26 65Solution.  Given : N
A = 975 r.p.m. ;

TA = 20 ; TB = 50 ; TC = 25 ; TD = 75 ; TE = 26 ;
TF = 65
From Fig. 13.3, we see that gears A, C
and E are drivers while the gears B, D and F are
driven or followers. Let the gear A rotates in
clockwise direction. Since the gears B and C are
mounted on the same shaft, therefore it is acompound gear and the direction or rotation ofboth these gears is same ( i.e. anticlockwise).
Similarly, the gears D and E are mounted on the

same shaft, therefore it is also a compound gearand the direction of rotation of both these gearsis same (i.e. clockwise). The gear F will rotate in
anticlockwise direction.
Let N
F  = Speed of gear F, i.e. last driven or follower.
Speed of the first driver Product of no. of teeth on drivens=Speed of the last driven Product of no. of teeth on driversFig. 13.3
Battery Car:  Even though it is run by batteries,

the power transmission, gears, clutches,brakes, etc. remain mechanical in nature.

Chapter 13 : Gear Trains    /G6C  433
orAB D F
FA C E50 75 6518.7520 25 26NT T T
NT T T×× ××===×× ××
∴A
F97552 r. p. m.
18.75 18.75NN===  Ans.
13.5. Design of Spur Gears
Sometimes, the spur gears ( i.e. driver and driven) are to be designed for the given velocity
ratio and distance between the centres of their shafts.
Let x = Distance between the centres  of two shafts,
N1 = Speed of the driver,
T1 = Number of teeth on the driver,
d1 = Pitch circle diameter of the driver,

N2 , T2 and d2 = Corresponding values for the driven or follower, and
pc = Circular pitch.
We know that the distance between the centres of two shafts,
12
2ddx+= ...(i)
and speed ratio or velocity ratio,
122
211NdT
NdT== ...(ii)
From the above equations, we can conveniently find out the values of d1 and d2 (or T1 and T2)
and the circular pitch ( pc ). The values of T1 and T2, as obtained above, may or may not be whole

numbers. But in a gear since the number of its teeth is always a whole number, therefore a slightalterations must be made in the values of x, d
1 and d2, so that the number of teeth in the two gears may
be a complete number.
Example 13.2.  Two parallel shafts , about 600 mm apart are to be connected by spur gears.
One shaft is to run at 360 r.p.m. and the other at  120 r.p.m. Design the gears, if the circular pitch is
to be 25 mm.

to be 25 mm.
Solution. Given : x = 600 mm ; N1 = 360 r.p.m. ; N2 = 120 r.p.m. ; pc = 25 mm
Let d1 = Pitch circle diameter of the first gear, and
d2 = Pitch circle diameter of the second gear.
We know that speed ratio,
12
213603120Nd
Nd== = ord2 = 3d1...(i)
and centre distance between the shafts ( x),
121600 ( )2dd=+ ord1 + d2 = 1200 ... (ii)
From equations  (i) and (ii), we find that
d1 = 300 mm, and d2 = 900 mm
∴ Number of teeth on the first gear,
2
130037.725cdTpππ ×== =

434  /G6C   Theory of Machines
and number of teeth on the second gear,
2
2
c900113.125dTpππ ×== =
Since the number of teeth on both the gears are to be in complete numbers, therefore let us
make the number of teeth on the first gear as 38. Therefore for a speed ratio of 3, the number of teethon the second gear should be 38 × 3 = 114.
Now the exact pitch circle diameter of the first gear,
1
138 25302.36 mmcTpd× ×′===
ππ
and the exact pitch circle diameter of the second gear,
2

2
2114 25907.1mmcTpd× ×′== =
ππ
∴ Exact distance between the two shafts,
12 302.36 907.1604.73 mm22ddx′′++′== =
Hence the number of teeth on the first and second gear must be 38 and 114 and their pitch
circle diameters must be 302.36 mm and 907.1 mmrespectively. The exact distance between the two shaftsmust be 604.73 mm. Ans.
13.6. Reverted Gear Train
When the axes of the first gear ( i.e. first driver)
and the last gear ( i.e. last driven or follower) are co-axial,

then the gear train is known as reverted gear  train as
shown in Fig. 13.4.
We see that gear 1 ( i.e. first driver) drives the
gear 2 (i.e. first driven or follower) in the opposite direc-
tion. Since the gears 2 and 3 are mounted on the sameshaft, therefore they form a compound gear and the gear3 will rotate in the same direction as that of gear 2. Thegear 3 (which is now the second driver) drives the gear 4(i.e. the last driven or follower) in the same direction as

that of gear 1. Thus we see that in a reverted gear train,the motion of the first gear and the last gear is like.
Let  T
1 = Number of teeth on gear 1,
 r1 = Pitch circle radius of gear 1, and
N1 = Speed of gear 1 in r.p.m.
Similarly,
     T2, T3, T4 = Number of teeth on respective gears,
      r2, r3, r4 = Pitch circle radii of  respective gears, and
   N2, N3, N4 = Speed of respective gears in r.p.m.Fig. 13.4.  Reverted gear train.

Chapter 13 : Gear Trains    /G6C  435
Since the distance between the centres of the shafts of gears 1 and 2 as well as gears 3 and 4
is same, therefore
r1 + r2 = r3 + r4...(i)
Also, the circular pitch or module of all the gears is assumed to be same, therefore number of
teeth on each gear is directly proportional to its circumference or radius.
∴ *T1 + T2 = T3 + T4...(ii)
andProductof number of teeth on drivensSpeed ratio =
Product of number of teeth on drivers
or12 4
41 3×=×NT T
NT T... (iii)

or12 4
41 3×=×NT T
From equations (i), (ii) and (iii), we can determine the number of teeth on each gear for the
given centre distance, speed ratio and module only whenthe number of teeth on one gear is chosen arbitrarily.
The reverted gear trains are used in automotive trans-
missions, lathe back gears, industrial speed reducers, and inclocks (where the minute and hour hand shafts are co-axial).
Example 13.3.  The speed ratio of the reverted gear

train, as shown in Fig. 13.5, is to be 12. The module pitch ofgears A and B is 3.125 mm and of gears C and D is 2.5 mm.Calculate the suitable numbers of teeth for the gears. Nogear is to have less than 24 teeth.
Solution.  Given : Speed ratio, N
A/ND = 12 ;
mA = mB = 3.125 mm ; mC = mD = 2.5 mm
Let NA = Speed of gear A,
TA = Number of teeth on gear A,
rA = Pitch circle radius of gear A,
NB, NC , ND = Speed of respective gears,
TB, TC , TD = Number of teeth on respective gears, and

rB, rC , rD = Pitch circle radii of respective gears.Fig. 13.5
*We know that circular pitch,
                            2
crpmTπ== π         or      .
2mTr=, where m is the module.
∴                         1
1.
2mTr= ; 2
2.
2mTr= ; 3
3.
2mTr= ; 4
4.
2mTr=
Now from equation (i),
                    3 12 4 . .. .
2222mT mT mT mT+=+
                            T1 + T2 = T3 + T4

436  /G6C   Theory of Machines
Since the speed ratio between the gears A and B and between the gears C and D are to be
same, therefore
*C A
BD12 3.464NN
NN===
Also the speed ratio of any pair of gears in mesh is the inverse of their number of teeth,
therefore
BD
AC3.464TT
TT== ...(i)
We know that the distance between the shafts
x = rA + rB = rC + rD = 200 mm
orCC AA BB DD . .. .2002222mT mT mT mT+=+=....
2mTr=/G51
3.125 (TA + TB) = 2.5 (TC + TD) = 400 ...(∵ mA = mB, and mC = mD)

∴ TA + TB = 400 / 3.125 = 128 ... (ii)
and TC  + TD = 400 / 2.5 = 160 ... (iii)
From equation (i), TB = 3.464 TA. Substituting this value of TB in equation (ii),
TA + 3.464 TA = 128 or TA = 128 / 4.464 = 28.67 say 28  Ans.
and TB = 128 – 28 = 100 Ans.
Again from equation (i), TD = 3.464 TC. Substituting this value of TD in equation (iii),
TC + 3.464 TC = 160 or TC = 160 / 4.464 =  35.84 say 36  Ans.
and TD = 160 – 36 = 124 Ans.

Note : The speed ratio of the reverted gear train with the calculated values of number of teeth on each gear is
AB D
DA C100 12412.328 36NT T
NT T××== =××
13.7. Epicyclic Gear Train
We have already discussed that in an epicyclic gear train, the axes of the shafts, over which
the gears are mounted, may move relative to a fixed axis. A simple epicyclic gear train is shown inFig. 13.6, where a gear A and the arm C have a common axis at O

1 about which they can rotate. The
gear B meshes with gear A and has its axis on the arm at O2, about which the gear B can rotate. If the
* We know that speed ratioA
DvSpeed of first driver12Speed of last dri enN
N== =
AlsoC AA
DBDN NN
NNN=× ...(NB = NC, being on the same shaft)
For A
BN
N and C
DN
N to be same, each speed ratio should be 12 so that
C AA
DBD12 12 12N NN
NNN=×=×=

Chapter 13 : Gear Trains    /G6C  437
arm is fixed, the gear train is simple and gear A can drive gear B
or vice- versa , but if gear A is fixed and the arm is rotated about
the axis of gear A (i.e. O1), then the gear B is forced to rotate
upon and around gear A. Such a motion is called epicyclic and
the gear trains arranged in such a manner that one or more oftheir members move upon and around another member areknown as  epicyclic gear trains  (epi. means upon and cyclic

means around). The epicyclic gear trains may be simple or com-
pound.
The epicyclic gear trains are useful for transmitting
high velocity ratios with gears of moderate size in a compara-tively lesser s pace. The epi cyclic gear trains are used in the
back gear of lathe, differential gears of the automobiles, hoists,pulley blocks, wrist watches etc.
13.8. Velocity Ratioz of Epicyclic Gear Train
The following two methods may be used for finding out the velocity ratio of an epicyclic
gear train.

1. Tabular method, and 2. Algebraic method.
These methods are discussed, in detail, as follows :1. Tabular method . Consider an epicyclic gear train as shown in Fig. 13.6.
Let T
A = Number of teeth on  gear A, and
TB = Number of teeth on gear B.
First of all, let us suppose that
the arm is fixed. Therefore the axes ofboth the gears are also fixed relative toeach other. When the gear A makes one
revolution anticlockwise, the gear B will
make *T
A / TB revolutions, clockwise.

make *T
Assuming the anticlockwise rotation aspositive and clockwise as negative, wemay say that when gear A makes + 1
revolution, then the gear B will make
(– T
A / TB) revolutions. This statement
of relative motion is entered in the firstrow of the table (see Table 13.1).
Secondly, if the gear A makes
+ x revolutions, then the gear B will
make – x × T
A / TB  revolutions. This

make – x × T
statement is entered in the second rowof the table. In other words, multiplythe each motion (entered in the first row) by x.
Thirdly, each element of an epicyclic train is given + y revolutions and entered in the third
row. Finally, the motion of each element of the gear train is added up and entered in the fourth row.
*We know that NB / NA = TA / TB. Since NA = 1 revolution, therefore NB = TA / TB.
Fig. 13.6.  Epicyclic gear train.

Inside view of a car engine.
Note : This picture is given as additional information and is not
a direct example of the current chapter.

438  /G6C   Theory of Machines
Arm fixed-gear A rotates through + 1
revolution i.e. 1 rev. anticlockwise
Arm fixed-gear A rotates through + x
revolutions
Add + y revolutions to all elements
Total motionTable 13.1. Table of motions
Revolutions of elements
Step No. Conditions of motion Arm C Gear A Gear B
1. 0 + 1A
B–T
T
2. 0 + xA
B–TxT×
3. + y + y + y
4. + yx  + yA
B–TyxT×
A little consideration will show that when two conditions about the motion of rotation of any

two elements are known, then the unknown speed of the third element may be obtained by substitut-ing the given data in the third column of the fourth row.
2. Algebraic method.  In this method, the motion of each element of the epicyclic train relative
to the arm is set down in the form of equations. The number of equations depends upon the number ofelements in the gear train. But the two conditions are, usually, supplied in any epicyclic train viz. some

element is fixed and the other has specified motion. These two conditions are sufficient to solve all theequations ; and hence to determine the motion of any element in the epicyclic gear train.
Let the arm C be fixed in an epicyclic gear train as shown in Fig. 13.6. Therefore speed of the
gear A relative to the arm C
= N
A – NC
and speed of the gear B relative to the arm C,
= NB – NC
Since the gears A and B are meshing directly, therefore they will revolve in  opposite directions.
∴BC A

∴BC A
AC B–––NN T
NN T=
Since the arm C is fixed, therefore its speed, NC = 0.
∴BA
AB–NT
NT=
If the gear A is fixed, then NA = 0.
BC A
CB––0–NN T
NT= orBA
CB1NT
NT=+
Note : The tabular method is easier and hence mostly used in solving problems on epicyclic gear train.
Example 13.4.  In an epicyclic gear train, an arm carries
two gears A and B having 36 and 45 teeth respectively. If the armrotates at 150 r.p.m. in the anticl ockwise direction about the centre

of the gear A which is fixed, determine the speed of gear B. If thegear A instead of being fixed, makes 300 r.p.m. in the clockwisedirection, what will be the speed of gear B ?
A = 36 ; TB = 45 ; NC = 150 r.p.m.
(anticlockwise)
The gear train is shown in Fig. 13.7. Fig. 13.7

Chapter 13 : Gear Trains    /G6C  439
revolution ( i.e. 1 rev. anticlockwise)
Total motionWe shall solve this example, first by tabular method and then by algebraic method.
1. Tabular method
First of all prepare the table of motions as given below :
Table 13.2. Table of motions.
1. 0 + 1A
B–T
T

1. 0 + 1A
B–T
T
2. 0 + xA
B–TxT×
3. + y + y + y
4. + yx  + yA
B–TyxT×
Speed of gear B when gear A is fixed
Since the speed of arm is 150 r.p.m. anticlockwise, therefore from the fourth row of the table,
y = + 150 r.p.m.
Also the gear A is fixed, therefore
x + y = 0 or x = – y = – 150 r.p.m.
∴ Speed of gear B,   A
B
B36– 150 150 270 r.p.m.45TNy xT=× = + × = +
= 270 r.p.m. (anticlockwise)  Ans.
Speed of gear  B when gear A makes 300 r .p.m. clockwise

Since the gear A makes 300 r.p.m.clockwise, therefore from the fourth row of the table,
x + y = – 300 or x = – 300 – y = – 300 – 150 = – 450 r.p.m.
∴ Speed of gear B,
       A
B
B36– 150 450 510 r.p.m.45TNy xT=× = + × = +
      = 510 r.p.m. (anticlockwise)   Ans.
2.  Algebraic method
Let NA = Speed of gear A.
NB = Speed of gear B, and
NC = Speed of arm C.
Assuming the arm C to be fixed, speed of gear A relative to arm C
= NA – NC
and speed of gear B relative to arm  C = NB – NC

440  /G6C   Theory of Machines
Since the gears A and B revolve in opposite directions, therefore
BC A
AC B–––NN T
NN T= ...(i)
Speed of gear  B when gear  A is fixed
When gear A is fixed, the arm rotates at 150 r.p.m. in the anticlockwise direction, i.e.
NA = 0, and NC = + 150 r.p.m.
∴B–150 36–– 0 . 80–1 5 0 4 5N== ...[From equation (i)]
or NB = – 150 × – 0.8 + 150 = 120 + 150 =  270 r.p.m.  Ans.
Speed of gear  B when gear  A makes 300 r.p.m. clockwise

Since the gear A makes 300 r.p.m. clockwise, therefore
NA = – 300 r.p.m.
∴B–150 36–– 0 . 8–300 –150 45N==
or NB = – 450 × – 0.8 + 150 = 360 + 150 = 510 r.p.m.  Ans.
Example 13.5. In a reverted epicyclic gear
train, the arm A carries two gears B and C and a
compound gear D - E. The gear B meshes with gear E

and the gear C meshes with gear D. The number of teethon gears B, C and D are 75, 30 and 90 respectively.Find the speed and direction of gear C when gear B isfixed and the arm A ma kes 100 r.p.m.  clockwise.
B = 75 ; TC = 30 ; TD  = 90 ;
NA = 100 r.p.m. (clockwise)
The reverted epicyclic gear train is
shown in Fig. 13.8. First of all, let us find thenumber of teeth on gear E (T
E). Let dB , dC , dD
and dE be the pitch circle diameters of gears B,

C, D and E respectively. From the geometry of
the figure,
 dB + dE = dC + dD
Since the number of teeth on each gear,
for the same module, are proportional to theirpitch circle diameters, therefore
 T
B + TE = TC + TD
      ∴        TE = TC + TD – TB = 30 + 90 – 75 = 45
The table of motions is drawn as
follows :Fig. 13.8
A gear-cutting machine is used to cut gears.

Chapter 13 : Gear Trains    /G6C  441
Arm fixed-compound gear D-E
rotated through + 1 revolution ( i.e.
1 rev. anticlockwise)
rotated through + x revolutions
Total motion Table 13.3. Table of motions.
Step Conditions of motion Arm A Com pound Gear B Gear C
No. gear D-E
1. 0 + 1E
B–T
TD
C–T
T
2. 0 + xE
B–TxT×D
C–TxT×
3. + y + y + y + y
4. + yx  + yE
B–TyxT×D
C–TyxT×

4. + yx  + yE
B–TyxT×D
C–TyxT×
Since the gear B is fixed, therefore from the fourth row of the table,
E
B–0TyxT×= or45–0
75yx×=
∴ y – 0.6 = 0 ... (i)
Also the arm A makes 100 r.p.m. clockwise, therefore
y = – 100 ... (ii)
Substituting y = – 100 in equation (i), we get
– 100 – 0.6 x = 0 or x = – 100 / 0.6 = – 166.67
 Model of sun and planet gears.
INPUT
Spline to AcceptMotor ShaftHousing OD Designed to meet
RAM Bore Dia, and Share MotorCoolant SupplyOUTPUT- External Spline to

SpindleRatio Detection Switches Hydraulic or Pneumatic Speed
Change Actuator
Round Housing With O-ring
Seated Cooling Jacket
Motor Flange
Hollow Through Bore for
Drawbar Integration

442  /G6C   Theory of Machines
From the fourth row of the table, speed of gear C,
D
C
C90– –100 166.67 400 r.p.m.30TNy xT=× =+ × = +
= 400 r.p.m. (anticlockwise)  Ans.
13.9. Compound Epicyclic Gear Train—Sun and Planet Gear
A compound epicyclic
gear train is shown in Fig. 13.9.It consists of two co-axial shaftsS
1 and S2, an annulus gear A which
is fixed, the compound gear (orplanet gear) B-C, the sun gear D
and the arm H. The annulus gear

has internal teeth and the com-pound gear is carried by the armand revolves freely on a pin of thearm H. The sun gear is co-axial
with the annulus gear and the armbut independent of them.
The annulus gear A
meshes with the gear B and the
sun gear D meshes with the gear
C. It may be noted that when the
annulus gear is fixed, the sun gearprovides the drive and when thesun gear is fixed, the annulus gearprovides the drive. In both cases, the arm acts as a follower.

Note : The gear at the centre is called the sun gear and the gears whose axes move are called planet gears .
Fig. 13.9. Compound epicyclic gear train.Sun and Planet gears.Speed Change
Shift Axis
Bearing Housing
Output Belt Pulley
Slide Dog
Clutch
Output Sun
GearMotor
FlangeInput Sun
GearPlanet
GearsOil
Collector

Chapter 13 : Gear Trains    /G6C  443
Arm fixed-gear C rotates thr ough
+ 1 revolution ( i.e. 1 rev.
anticlockwise)
Arm fixed-gear C rotates through
+ x revolutions
Add + y revolutions to all
elements
Total motionArm fixed-gear D rotates
through + 1 revolution
Arm fixed-gear D rotates
through + x revolutions
elements
Total motionLet TA , TB , TC , and TD be the teeth and NA, NB, NC  and ND  be the speeds for the gears A, B,

C and D respectively. A little consideration will show that when the arm is fixed and the sun gear D is
turned anticlockwise, then the compound gear B-C and the annulus gear A will rotate in the clockwise
The motion of rotations of the various elements are shown in the table below.
Table 13.4. Table of motions.
Step Conditions of motion Arm G ear D Com pound gear Gear A
No. B-C
1. 0 + 1D
C–T
TDB
CA–TT
TT×
2. 0 + xD
C–Tx
T× DB
CA–TTxTT××
3. + y + y + y  + y

C–Tx
T× DB
4. + yx  + yD
C–Tyx
T×DB
CA–TTyx
TT××
Note : If the annulus gear A is rotated through one revolution anticlockwise with the arm fixed, then the
compound gear rotates through TA / TB revolutions in the same sense and the sun gear D rotates through
 TA / TB × TC / TD revolutions in clockwise direction.
Example 13.6. An epicyclic gear consists of three gears A, B and C as shown in Fig. 13.10.

The gear A has 72 internal teeth and gear C has 32 external teeth. The gear B meshes with both Aand C and is carried on an arm EF which rotates about the centre of A at 18 r.p.m.. If the gear A is
fixed, determine the speed of gears B and C.
A = 72 ; TC = 32 ; Speed of arm EF = 18 r.p.m.
Considering the relative motion of rotation as shown in Table 13.5.
Table 13.5. Table of motions.
Step No. Conditions of motion Arm EF Gear C Gear B G ear A
1. 0 + 1C

1. 0 + 1C
B–T
TCB C
BA A––TT T
TT T×=
2. 0 + xC
B–Tx
T×C
A–Tx
T×
3. + y + y+ y + y
4. + yx  + yC
B–Tyx
T× C
A–Tyx
T×

444  /G6C   Theory of Machines
Speed of gear C
We know that the speed of the arm is 18 r.p.m. therefore,
y = 18 r.p.m.
and the gear A is fixed, therefore
C
A–0TyxT×= or3218 – 072x×=
∴ x = 18 × 72 / 32 = 40.5
∴  Speed of gear C         = x + y = 40.5 + 18
= + 58.5 r.p.m.= 58.5 r.p.m. in the direction
of arm.  Ans.
Speed of gear B
Let d
A, dB and dC be the pitch circle diameters of gears
A, B and C respectively. Therefore, from the geometry of Fig. 13.10,
                CA

CA
B22ddd+= or 2 dB + dC = dA
Since the number of teeth are proportional to their pitch circle diameters, therefore
               2 TB  + TC = TA         or            2 TB + 32 = 72 or TB = 20
∴  Speed of gear B          C
B32– 1 8–4 0 . 5 –4 6 . 8r . p . m .20TyxT=× = × =
               =  46.8 r.p.m. in the opposite direction of arm.   Ans.
Example 13.7.  An epicyclic train of gears is arranged as shown in

Fig.13.11 . How many revolutions does the arm , to which the pinions B and
C are attached , make :
1. when A makes one revolution clockwise and D makes half a
revolution anticlockwise , and
2. when A makes one revolution clockwise and D is stationary  ?
The number of teeth on the gears A and D are 40 and 90
respectively .
Solution. Given : TA = 40 ; TD = 90
First of all, let us find the number of teeth on gears B and C (i.e. TB and TC). Let  dA, dB, dC

and dD be the pitch circle diameters of gears A, B, C and D respectively. Therefore from the geometry
of the figure,
dA + dB + dC = dDordA + 2 dB = dD...(/G51 dB = dC)
Since the number of teeth are proportional to their pitch circle diameters, therefore,
TA + 2 TB = TDor 40 + 2 TB = 90
∴ TB = 25, and TC = 25 ...(/G51 TB = TC)Fig. 13.10
Fig. 13.11

Chapter 13 : Gear Trains    /G6C  445
The table of motions is given below :
Table 13.6. Table of motions.
Step No. Conditions of motion Arm Gear A Com pound G ear D
gear B-C
1. 0 –  1A
BT
T+AB A
BD DTT T
TT T+×= +
2. 0 – xA
BTx
T+×A
DTx
T+×
3. – y – y – y – y
4. – y– x – yA
B–Txy
T×A
D–Txy
T×
1. Speed of arm when A makes  1 revolution clockwise and D makes half revolution anticlockwise

Since the gear A makes 1 revolution clockwise, therefore from the fourth row of the table,
– x – y = –1 or x + y = 1 ... (i)
Also, the gear D makes half revolution anticlockwise, therefore
A
D1–2TxyT×= or40 1–
90 2xy×=
∴ 40 x – 90 y = 45 or x – 2.25 y = 1.125 ... (ii)
From equations  (i) and (ii),  x = 1.04 and y = – 0.04
∴            Speed of arm = – y = – (– 0.04) = + 0.04
= 0.04 revolution anticlockwise  Ans.
2. Speed of arm when A makes  1 revolution clockwise and D is stationary

Since the gear A makes 1 revolution clockwise, therefore from the fourth row of the
table,
– x – y = – 1 or x + y = 1 ... (iii)
Also the gear D is stationary, therefore
A
D–0TxyT×= or40–0
90xy×=
∴ 40 x – 90 y = 0 or x – 2.25 y = 0 ... (iv)
x = 0.692 and y = 0.308
∴     Speed of arm = – y = – 0.308 = 0.308 revolution clockwise Ans.Arm fixed , gear A rotates
through – 1 revolution ( i.e. 1
rev. clockwise)
Arm fixed, gear A rotates
through – x revolutions

Add – y revolutions to all
elements
Total motion

446  /G6C   Theory of Machines
Example 13.8.  In an epicyclic gear train , the internal wheels A and B and compound wheels
C and D rotate independently about axis O . The wheels E and F rotate on pins fixed to the arm G . E
gears with A and C and F gears with B and D . All the wheels have
the same module and the number of teeth are :  TC = 28;  TD = 26;
TE = TF = 18.
1. Sketch the arrangement ; 2. Find the number of teeth on
A and B ; 3. If the arm G makes 100 r.p.m. cl ockwise and A is fixed,

find the speed of B ; and 4. If the arm G makes 100 r.p.m. clockwise
and wheel A makes 10 r.p.m. counter clockwise ; find the speed ofwheel B.
C = 28 ; TD = 26 ; TE = TF = 18
1. Sketch the arrangement
The arrangement is shown in Fig. 13.12.
2. Number of teeth on wheels A and B
Let              TA = Number of teeth on wheel A, and
            TB = Number of teeth on wheel B.
If dA , dB , dC , dD , dE and dF are the pitch circle diameters of wheels A, B, C, D, E and F

respectively, then from the geometry of Fig. 13.12,
            dA = dC + 2 dE
and             dB  = dD + 2 dF
Since the number of teeth are proportional to their pitch circle diameters, for the same
module, therefore
           TA = TC + 2 TE = 28 + 2 × 18 = 64 Ans.
and            TB = TD + 2 TF = 26 + 2 × 18 = 62 Ans.
3. Speed of wheel B when arm G makes 100 r .p.m. clockwise and wheel A is fixed
First of all, the table of motions is drawn as given below :
Table 13.7. Table of motions.

Step Conditions of Arm Wheel Wheel Com pound W heel F W heel B
No. motion G A E wheel C-D
1. 0 + 1A
ET
T+AE
EC–TT
TT×AD
CFTT
TT+×ADF
CFBTTT
TTT+××
2. 0 + xA
ETx
T+×A
C–Tx
T×AD
CFTTx
TT+× ×AD
CBTTx
TT+× ×
3. + y+ y+ y + y + y + y
4. + yx + yA
ETyx
T+×A
C–Tyx
T×AD
CFTTyx
TT+× ×AD
CBTTyx
TT+× ×Fig. 13.12
Arm fixed- wheel A
rotates through + 1revolution ( i.e. 1 rev.
Arm fixed-wheel A
rotates through + x

Add + y revolutions
to all elements
Total motionA
C–T
T=AD
CBTT
TT=+ ×

Chapter 13 : Gear Trains    /G6C  447
Since the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the table,
y = – 100 ... (i)
Also, the wheel A is fixed, therefore from the fourth row of the table,
x + y = 0 or x = – y = 100 ... (ii)
∴   Speed of wheel AD
CB64 26–100 100 –100 95.8 r.p.m.28 62TTByxTT= + × × =+ × × =+
= – 4.2 r.p.m. = 4.2 r.p.m. clockwise  Ans.
4.Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A makes  10 r.p.m. counter
clockwise

clockwise
Since the arm  G makes 100 r.p.m. clockwise, therefore from the fourth row of the
table
y = – 100 ... (iii)
Also the wheel A makes 10 r.p.m. counter clockwise, therefore from the fourth row of the
table,
x + y = 10 or x = 10 – y = 10 + 100 =  110 ... (iv)
∴Speed of wheel AD
CB64 26–100 110 –100 105.4 r.p.m.28 62TTByxTT=+ × × = + × × = +
= + 5.4 r.p.m. = 5.4 r.p.m. counter clockwise  Ans.
Example 13.9.  In an epicyclic gear of the ‘sun and planet’ type shown

in Fig. 13.13, the pitch circle diameter of the internally toothed ring is to be224 mm and the module 4 mm. When the ring D is stationary, the spider A,which carries three planet wheels C of equal size, is to make one revolution inthe same sense as the sunwheel B for every five revolutions of the drivingspindle carrying the sunwheel B. Determine suitable numbers of teeth for allthe wheels.
D = 224 mm ; m = 4 mm ; NA = NB / 5

Let TB , TC  and TD be the number of teeth on the sun wheel B,
planet wheels C and the internally toothed ring D. The table of motions is given below :
Table 13.8. Table of motions.
Step No. Conditions of motion Spider A Sun wheel B Planet wheel C Internal gear D
1. 0 + 1B
C–T
TBC B
CD D––TT T
TT T×=
2. 0 + xB
C–Tx
T×B
D–Tx
T×
4. + yx  + yB
C–Tyx
T×B
D–Tyx
T×Fig. 13.13
Spider A fixed, sun wheel
B rotates through + 1

revolution ( i.e. 1 rev.
B rotates through + x
elements

448  /G6C   Theory of Machines
We know that when the sun
wheel B makes + 5 revolutions, the spi-
der A makes + 1 revolution. Therefore
from the fourth row of the table,
          y = + 1 ;  and  x + y = + 5
∴       x = 5 – y = 5 – 1 = 4
Since the internally toothed ring
D is stationary, therefore from the fourth
row of the table,
B
D–0TyxT×=
orB
D1–4 0T
T×=
∴B
D1
4T
T=orTD = 4 TB...(i)
We know that TD = dD / m = 224 / 4 = 56 Ans.
∴ TB = TD / 4 = 56 / 4 = 14 Ans. ...[From equation (i)]

Let dB, dC and dD be the pitch circle diameters of sun wheel B, planet wheels C and internally
toothed ring D respectively. Assuming the pitch of all the gears to be same, therefore from the geom-
etry of Fig. 13.13,
dB + 2 dC = dD
TB + 2 TC = TDor 14 + 2 TC = 56
∴ TC = 21 Ans.
Example 13.10.  Two shafts A  and B are co-axial . A gear C (50 teeth) is rigidly mounted

on shaft A. A compound gear D-E gears with C and an internal gear G. D has 20 teeth and gearswith C and E has 35 teeth and gears with an internal gear G. The gear G is fixed and is concen-tric with the shaft axis. The compound gear D-E is mounted on a pin which projects from an armkeyed to the shaft B. Sketch the arrangement and find the number of teeth on internal gear Gassuming that all gears have the same module. If  the shaft A rotates at  110 r.p.m., find the speed
of shaft B.

of shaft B.
C = 50 ; TD = 20 ; TE = 35 ; NA = 110 r.p.m.
The arrangement is shown in Fig. 13.14.
Number of teeth on internal gear G
Let dC , dD , dE and dG  be the pitch circle diameters of gears C, D, E and G respectively. From
the geometry of the figure,
GCDE
2222dddd=++
or dG = dC + dD + dE
Power transmission in a helicopter is essentially through
gear trains.
Note : This picture is given as additional information and is not a

direct example of the current chapter.Main rotor Tail rotor
Tail boom
Landing skids Engine, transmis-
sion fuel, etc.CockpitDrive shaft

Chapter 13 : Gear Trains    /G6C  449
Let TC , TD , TE and TG be the number of teeth on gears C, D, E and G respectively. Since all
the gears have the same module, therefore number of teeth are proportional to their pitch circlediameters.
   ∴ T
G = TC + TD + TE = 50 + 20 + 35 = 105 Ans.
Fig. 13.14
Speed of shaft B
Table 13.9. Table of motions.
Step Conditions of motion Arm Gear C (or Com pound G ear G
No. shaft A) gear D-E
1. 0 + 1C

1. 0 + 1C
D–T
TCE
DG–TT
TT×
2. 0 + xC
D–TxT×CE
DG–TTxTT××
4. + yx  + yC
D–TyxT×CE
DG–TTyxTT××
Since the gear G is fixed, therefore from the fourth row of the table,
CE
DG–0TTyxTT××= or50 35–020 105yx×× =
∴5–06yx= ...(i)Arm fixed - gear C rotates through + 1
revolution
Arm fixed - gear C rotates through + x

450  /G6C   Theory of Machines
Since the gear C is rigidly mounted on shaft A, therefore speed of gear C and shaft A is same.
We know that speed of shaft A is 110 r.p.m., therefore from the fourth row of the table,
x + y = 100 ... (ii)
From equations (i) and (ii),x = 60, and    y = 50
∴               Speed of shaft B = Speed of arm = + y = 50 r.p.m. anticlockwise  Ans.
Example 13.11.   Fig. 13.15 shows diagrammatically a compound

epicyclic gear train. Wheels A , D and E are free to rotate independentlyon spindle O, while B and C are compound and rotate together on spindleP, on the end of arm OP. All the teeth on different wheels have the samemodule. A has 12 teeth, B has 30 teeth and C has 14 teeth cut externally.Find the number of teeth on wheels D and E which are cut internally.
If the wheel A is driven clockwise at 1 r.p.s. while D is driven

counter clockwise at 5 r.p.s., determine the magnitude and direction ofthe angular velocities of arm OP and wheel E.
A = 12 ;TB = 30 ;TC = 14 ; NA = 1 r.p.s. ; ND = 5 r.p.s.
Number of teeth on wheels D and E
Let TD and TE  be the number of teeth on wheels D and E respectively. Let dA , dB , dC , dD and dE
be the pitch circle diameters of wheels A, B, C, D and E respectively. From the geometry of the figure,
dE = dA + 2dBanddD = dE – (dB – dC)

Since the number of teeth are proportional to their pitch circle diameters for the same module,
therefore
TE = TA + 2TB = 12 + 2 × 30 = 72  Ans.
and TD = TE – (TB – TC) = 72 – (30 – 14) = 56  Ans.
Magnitude and direction of angular velocities of arm OP and wheel E
The table of motions is drawn as follows :
Table 13.10. Table of motions.
Step Conditions of motion Arm W heel A Compound Wheel D Wh eel E
No. wheel B-C
1. 0 – 1A
BT
T+AC

1. 0 – 1A
BT
T+AC
BDTT
TT+×AB
BETT
TT+×
A
ET
T=+
 2. 0 – xA
BTxT+×AC
BDTTxTT+× ×A
ETxT+×
3. – y– y – y – y – y
4. – y– x – yA
B–TxyT×AC
BD–TTxyTT××A
E–TxyT×Fig. 13.15
Arm fixed A rotated through
– 1 revolution ( i.e. 1 revolu-
tion clockwise)
Arm fixed-wheel A rotated
Add – y revolutions to all ele-
ments

Chapter 13 : Gear Trains    /G6C  451
Since the wheel A makes 1 r.p.s. clockwise, therefore from the fourth row of the table,
– x – y = – 1 or x + y = 1 ... (i)
Also, the wheel D makes 5 r.p.s. counter clockwise, therefore
CA
BD–5TTxyTT×× = or12 14–530 56xy×× =
∴ 0.1 x – y = 5 ... (ii)
x = 5.45 and y = – 4.45
∴ Angular velocity of arm OP
          = – y = –(– 4.45) = 4.45 r.p.s
= 4.45 × 2 π = 27.964 rad/s (counter clockwise) Ans.
and angular velocity of wheel A

E12– 5.45 – (– 4.45) 5.36 r.p.s.72TEx yT=× = × =
= 5.36 × 2 π = 33.68 rad/s (counter clockwise)   Ans.
Example 13.12.  An internal wheel B with 80 teeth is keyed to a shaft F. A fixed internal
wheel C with 82 teeth is concentricwith B. A compound wheel D-Egears with the two internal wheels;D has 28 teeth and gears with Cwhile E gears with B. The compoundwheels revol ve freely on a pin which

projects from a disc keyed to a shaftA co-axial with F. If the wheels havethe same pitch and the shaft A makes800 r.p.m., what is the speed of theshaft F ? Sketch the arrangement.
B = 80 ; TC
= 82 ; TD = 28 ; NA = 500 r.p.m.
The arrangement is shown in Fig. 13.16.
Fig. 13.16
First of all, let us find out the number of teeth on wheel E (TE). Let dB , dC , dD and dE be the
pitch circle diameter of wheels B, C, D and E respectively. From the geometry of the figure,

dB = dC – (dD – dE )
Helicopter
direct example of the current chapter.

452  /G6C   Theory of Machines
or dE = dB + dD – dC
Since the number of teeth are proportional to their pitch circle diameters for the same pitch,
therefore
TE = TB + TD – TC = 80 + 28 – 82 = 26
Table 13.11. Table of motions.
Step Conditions of motion Arm (or Wheel B (or Com pound W heel C
No. shaft A) shaft F)  gear D-E
1. 0 + 1B
ET
T+BD
ECTT
TT+×
2. 0 + xB
ETxT+×BD
ECTTxTT+× ×
4. + yx  + yB
ETyxT+×BD
ECTTyxTT+× ×

4. + yx  + yB
ETyxT+×BD
Since the wheel C is fixed, therefore from the fourth row of the table,
BD
EC0TTyxTT+× × = or80 280
26 82yx+× × =
∴ y + 1.05 x = 0 ... (i)
Also, the shaft A (or the arm) makes 800 r.p.m., therefore from the fourth row of the table,
y = 800 ... (ii)
x = – 762
∴ Speed of shaft F = Speed of wheel B = x + y = – 762 + 800 = + 38 r.p.m.
= 38 r.p.m. (anticlockwise) Ans.
Example 13.13.  Fig. 13.17 shows an epicyclic gear

train known as Ferguson’s paradox. Gear A is fixed to theframe and is, therefore, stationary. The arm B and gears Cand D are free to rotate on the shaft S. Gears A, C and D have100, 101 and 99 teeth respectively. The planet gear has 20teeth. The pitch circle diameters of all are the same so that theplanet gear P meshes with all of them. Determine therevolutions of gears C and D for one revolution of the arm B .
A = 100 ; TC = 101 ; TD = 99 ;

TP = 20 Fig. 13.17Arm fixed - wheel B rotated
through + 1 revolution ( i.e. 1
revolution anticlockwise)
Arm fixed - wheel B rotated
elements

Chapter 13 : Gear Trains    /G6C  453
Table 13.12. Table of motions.
Step No. Conditions of motion Arm B Gear A Gear C Gear D
1. 0 + 1A
CT
T+AC A
CD DTT T
TT T+×= +
2. 0 + xA
CTxT+×A
DTxT+×
4. + yx  + yA
CTyx
T+×A
DTyx
T+×
The arm B makes one revolution, therefore
y = 1
Since the gear A is fixed, therefore from the fourth row of the table,
x + y = 0 or x = – y = – 1

Let NC and ND  = Revolutions of gears C and D respectively.
From the fourth row of the table, the revolutions of gear C,
A
C
C100 11–1101 101TNy xT=+× = × = +Ans.
and the revolutions of gear D,
A
D
D100 11–99 99TNy xT=+× = = – Ans.
From above we see that for one revolution of the arm B, the gear C rotates through 1/101
revolutions in the same direction and the gear D rotates through 1/99 revolutions in the opposite

Example 13.14.  In the gear drive as shown in Fig.
13.18, the driving shaft A rotates at 300 r.p.m. in the clock-wise direction, when seen from left hand. The shaft B is thedriven shaft. The casing C is held stationary. The wheels E
and H are keyed to the central vertical spindle and wheel Fcan rotate freely on this spindle. The wheels K and L arerigidly fixed to each other and rotate together freely on apin fitted on the underside of F. The wheel L meshes with

internal teeth on the casing C. The numbers of teeth on thedifferent wheels are indicated within brackets in Fig. 13.18.
Find the number of teeth on wheel C and the speed
and direction of rotation of shaft B .
A = 300 r.p.m. (clockwise) ;
TD = 40 ; TB = 30 ; TF = 50 ; TG = 80 ; TH = 40 ; TK = 20 ; TL = 30
In the arrangement shown in Fig. 13.18, the wheels D and G are auxillary gears and do not
form a part of the epicyclic gear train.Fig. 13.18Arm B fixed, gear A rotated

Arm B fixed, gear A rotated
elements

454  /G6C   Theory of Machines
Speed of wheel E, D
EA
E40300 400 r.p.m.(clockwise)30TNNT=×=×=
Number of teeth on wheel C
Let TC = Number of teeth on wheel C.
Assuming the same module for all teeth and since the pitch circle diameter is proportional to
the number of teeth ; therefore from the geometry of Fig.13.18,
TC = TH + TK + TL = 40 + 20 + 30 = 90 Ans.
Speed and direction of rotation of shaft B
The table of motions is given below. The wheel F acts as an arm.
Table 13.13. Table of motions.

Step Conditions of motion Arm or Wheel E Wheel H Com pound W heel C
No. wheel F wheel K-L
1. 0 – 1H
KT
T+HL
KCTT
TT+×
2. 0 – x – xH
KTx
T+×HL
KCTTx
TT+× ×
3. – y – y – y – y – y
4. – y– x – y– x – yH
K–TxyT×HL
KC–TTxyTT××
Since the speed of wheel E is 400 r.p.m. (clockwise), therefore from the fourth row of the table,
– x – y = – 400 or x + y = 400 ... (i)
Also the wheel C is fixed, therefore
                HL
KC–0TTxyTT×× =

HL
or                    40 30–020 90xy×× =
∴                        2–03xy= ...(ii)
x = 240 and y = 160
∴ Speed of wheel F,NF = – y = – 160 r.p.m.
Since the wheel F is in mesh with wheel G, therefore speed of wheel G or speed of shaft B
F
F
G50– – – 160 100 r.p.m.80TNT=× = × = 
...(/G51 Wheel G will rotate in opposite direction  to that of wheel F.)
= 100 r.p.m. anticlockwise i.e. in opposite direction of

shaft A. Ans.Arm  fixed-wheel  E
rotated through – 1revolution ( i.e. 1
revolution clockwise)
Arm fixed-wheel E
rotated through – x
Add – y revolutions to
all elements
Total motion– 1(
/G51E and
H are on the
same shaft)

Chapter 13 : Gear Trains    /G6C  455
Example 13.15.  Fig. 13.19 shows a compound epicyclic gear in which the casing C contains
an epicyclic train and this casing is inside the larger casing D.
Determine the velocity ratio of the output shaft B to the input shaft A when the casing D is
held stationary. The number of teet h on various wheels are as follows :
Wheel on A = 80 ; Annular wheel on B = 160 ; Annular wheel on C = 100 ; Annular wheel

on D = 120 ; Small pinion on F = 20 ; Large pinion on F = 66.
Fig. 13.19
Solution. Given : T1 = 80 ;  T8 = 160 ;  T4 = 100;  T3 = 120 ;  T6 = 20;  T7 = 66
First of all, let us consider the train of wheel 1 (on A), wheel 2 (on E), annular wheel 3 (on D)
and the arm i.e. casing C. Since the pitch circle diameters of wheels are proportional to the number of
teeth, therefore from the geometry of Fig. 13.19,
T1 + 2 T2 = T3or 80 + 2 T2 = 120
∴ T2 = 20

∴ T2 = 20
The table of motions for the train considered is given below :
Table 13.14. Table of motions.
Step No. Conditons of motion Arm W heel 1 Wheel 2 Wheel 3
1. 0 + 11
2–T
T12 1
23 3––TT T
TT T×=
2. 0 + x1
2–TxT×1
3–TxT×
4. yx  + y1
2–TyxT×1
3–TyxT×Arm fixed - wheel 1 rotated
through + 1 revolution(anticlockwise)
Arm fixed - wheel 1 rotated
Add + y  revolutions to all
elements

456  /G6C   Theory of Machines
Let us assume that wheel 1 makes 1 r.p.s. anticlockwise.
∴ x + y = 1 ... (i)
Also the wheel 3 is stationary, therefore from the fourth row of the table,
1
3–0TyxT×= or80–0120yx×=
∴2–0
3yx= ...(ii)
From equations  (i) and (ii),x = 0.6,   and y = 0.4
∴  Speed of arm or casing  C = y = 0.4  r.p.s.
and speed of wheel 2 or arm E1
280– 0 . 4–0 . 6 –2r . p . s .20TyxT=× = × =
= 2 r.p.s. (clockwise)

Let us now consider the train of annular wheel 4 (on C), wheel 5 (on E), wheel 6 (on F) and
arm E. We know that
T6 + 2 T5  = T4or 20 + 2 T5 = 100
∴ T5 = 40
Table 13.15. Table of motions.
Step Conditions of motion Arm E or Wheel 6 Wheel 5 Wheel 4
No. wheel 2
1. 0 + 16
5–T
T65 6
54 4––TT T
TT T×=
2. 0 x16
1
5–TxT×6
1
4–TxT×
3. + y1+ y1+ y1+ y1
4. + y1x1 + y16
11
5–TyxT×6
11
4–TyxT×

11
5–TyxT×6
11
4–TyxT×
We know that speed of arm E = Speed of wheel 2 in the first train
∴ y1 = – 2 ... (iii)
             Also speed of wheel 4 = Speed of arm or casing C in the first train
∴6
11
4–0 . 4TyxT×= or 120–2 – 0.4100x×= ...(iv)
or 1100(–2 – 0.4) –12
20x==Arm  fixed, wheel  6  rotated
Arm  fixed, wheel  6  rotated
through + x1 revolutions
Add + y1 revolutions to all
elements

Chapter 13 : Gear Trains    /G6C  457
∴ Speed of wheel 6 (or F)
= x1 + y1 = – 12 – 2 = – 14 r.p.s. = 14 r.p.s. (clockwise)
Now consider the train of wheels 6 and 7 (both on F), annular wheel 8 (on B) and the arm i.e.
casing C. The table of motions is given below :
Table 13.16. Table of motions.
Step No. Conditions of motion Arm Wheel 8 Wheel 7
1. 0 + 18
7T
T+
2. 0 + x28
2
7Tx
T+×
3. + y2+ y2+ y2
4. y2x2 + y28
22
7Tyx
T+×

7Tx
T+×
22
7Tyx
T+×
We know that the speed of C in the first train is 0.4 r.p.s., therefore
y2 = 0.4 ... (v)
Also the speed of wheel 7 is equal to the speed of F or wheel 6 in the second train, therefore
8
22
7–14TyxT+× = or21600.4 –14
66x+× = ...(vi)
∴ 266( 14 0.4) 5.94
160x=− − = −
∴ Speed of wheel 8 or of the shaft B
x2 + y2 = – 5.94 + 0.4 = – 5.54 r.p.s. = 5.54 r.p.s. (clockwise)

We have already assumed that the speed of wheel 1 or the shaft A is 1 r.p.s. anticlockwise
∴ Velocity ratio of the output shaft B to the input shaft A
= – 5.54 Ans.
Note : The – ve sign shows that the two shafts A and B rotate in opposite directions.
13.10. Epicyclic Gear Train with Bevel Gears
The bevel gears are used to make a more compact epicyclic system and they permit a very

high speed reduction with few gears. The useful application of the epicyclic gear train with bevelgears is found in Humpage’s speed reduction gear and differential gear of an automobile as discussedbelow :
1. Humpage’s speed reduction gear . The Humpage’s speed reduction gear was originally

designed as a substitute for back gearing of a lathe, but its use is now considerably extended to allkinds of workshop machines and also in electrical machinery. In Humpage’s speed reduction gear, asshown in Fig. 13.20, the driving shaft X and the driven shaft Y are co-axial. The driving shaft carries
a bevel gear A and driven shaft carries a bevel gear E. The bevel gear B meshes with gear A (also

known as pinion) and a fixed gear C. The gear E meshes with gear D which is compound with gear B.Arm  fixed, wheel  8  rotated through
+ 1 revolution
Arm  fixed, wheel  8  rotated through
+ x2 revolutions
Add  + y2  revolutions  to  all
elements

458  /G6C   Theory of Machines
This compound gear B-D is mounted on the arm or spindle F which is rigidly connected with a hollow
sleeve G. The sleeve revolves freely loose on the axes of the driving and driven shafts.
Fig. 13.20. Humpage’s speed reduction gear.
2. Differential gear of an automobile . The differential gear used in the rear drive of an
automobile is shown in Fig. 13.21. Its function is
(a) to transmit motion from the engine shaft to the rear driving wheels, and

(b) to rotate the rear wheels at different speeds while the automobile is taking a turn.
As long as the automobile is running on a straight path, the rear wheels are driven directly by
the engine and speed of both the wheels is same. But when the automobile is taking a turn, the outerwheel will run faster than the * inner wheel because at that time the outer rear wheel has to cover more

distance than the inner rear wheel. This is achieved by epicyclic gear train with bevel gears as shownin Fig. 13.21.
The bevel gear A (known as pinion) is keyed to
the propeller shaft driven from the engine shaft throughuniversal coupling. This gear A drives the gear B (known
as crown gear) which rotates freely on the axle P. Two
equal gears C and D are mounted on two separate parts P
and Q of the rear axles respectively. These gears, in turn,

mesh with equal pinions E and F which can rotate freely
on the spindle provided on the arm attached to gear B.
When the automobile runs on a straight path, the
gears C and D must rotate together. These gears are rotated
through the spindle on the gear B. The gears E and F do
not rotate on the spindle. But when the automobile is takinga turn, the inner rear wheel should have lesser speed thanthe outer rear wheel and due to relative speed of the inner and outer gears D and C, the gears E and F

start rotating about the spindle axis and at the same time revolve about the axle axis.
Due to this epicyclic effect, the speed of the inner rear wheel decreases by a certain amount
and the speed of the outer rear wheel increases, by the same amount. This may be well understood bydrawing the table of motions as follows :Fig. 13.21. Differential gear of an automobile.
*This difficulty does not arise with the front wheels as they are greatly used for steering purposes and are

mounted on separate axles and can run freely at different speeds.

Chapter 13 : Gear Trains    /G6C  459
Table 13.17. Table of motions.
Step No. Conditions of motion Gear B Gear C Gear E Gear D
1. 0 + 1C
ET
T+CE
ED–– 1TT
TT×=
()CDTT=/G51
2. 0 + xC
ETxT+×–x
4. + yx  + yC
ETyxT+× –yx
From the table, we see that when the gear B, which derives motion from the engine shaft,
rotates at y revolutions, then the speed of inner gear D (or the rear axle Q) is less than y by x  revolu-

tions and the speed of the outer gear C (or the rear axle P) is greater than y by x revolutions. In other
words, the two parts of the rear axle and thus the two wheels rotate at two different speeds. We alsosee from the table that the speed of gear B is the mean of speeds of the gears C and D.
Example 13.16.  Two bevel gears A and B (having 40 teeth and 30 teeth) are rigidly mounted

on two co-axial shafts X and Y. A bevel gear C (having50 teeth) meshes with A and B and rotates freely on oneend of an arm. At the other end of the arm is welded asleeve and the sleeve is riding freely loose on the axes ofthe shafts X and Y. Sketch the arrangement.
If the shaft X rotates at 100  r.p.m. clockwise  and
arm rotates at 100 r.p.m.anitcl ockwise, find the
speed of shaft Y.
A = 40 ; TB = 30 ; TC = 50 ; NX
= NA = 100 r.p.m. (clockwise) ; Speed of arm = 100 r.p.m.

The arangement is shown in Fig. 13.22.The table of motions is drawn as below :
Table 13.18. Table of motions.
Step No. Conditions of motion Arm Gear A Gear C Gear B
1. 0 + 1A
CT
T±AC A
CB B––TT T
TT T×=
2. 0 + xA
CTxT±×A
B–TxT×
4. + yx  + yA
CTyxT±×A
B–TyxT×Fig. 13.22
*The ± sign is given to the motion of the wheel C because it is in a different plane. So we cannot indicate the

direction of its motion specifically, i.e. either clockwise or anticlockwise.Gear B fixed-Gear C rotated
through + 1 revolution ( i.e.
1 revolution anticlockwise )
Gear B fixed-Gear C rotated
elements
elements
Total motion*

460  /G6C   Theory of Machines
Since the speed of the arm is 100 r.p.m. anticlockwise, therefore from the fourth row of the
table,
y = + 100
Also, the speed of the driving shaft X or gear A is 100 r.p.m. clockwise.
∴ x + y = – 100 or x = – y – 100 = – 100 – 100 = – 200
∴ Speed of the driven shaft i.e. shaft Y,
NY = Speed of gear A
B40– 100 – – 20030TByxT=× = × 
= + 366.7 r.p.m. = 366.7 r.p.m. (anticlockwise) Ans.
Example 13.17.  In a gear train , as

shown in Fig . 13.23, gear B is connected to the
input shaft and gear F is connected to the outputshaft. The arm A carrying the compound wheels
D and E, turns freely on the output shaft . If the
input speed is 1000 r.p. m. counter- clockwise
when seen from the right, determine the speed ofthe output shaft under the following conditions :
1. When gear C is fixed, and  2. when
gear C is rotated at 10 r .p.m. counter clockwise .

B = 20 ; TC = 80 ;
TD = 60 ; TE = 30 ; TF = 32 ; NB = 1000 r.p.m.
(counter-clockwise)
Table 13.19. Table of motions.
Step Conditions of motion Arm A Gear B Com pound Gear C Gear F (or
No. (or input wheel D-E output shaft)
shaft)
1. 0 + 1B
DT
T+BD
DC–TT
TT×BE
DF–TT
TT×
B
C–T
T=
2. 0 + xB
DTx
T+×B
C–Tx
T×BE
DF–TTx
TT××
3. + y+ y + y + y + y
4. + yx + yB
DTyxT+×B
C–TyxT×BE
DF–TTyxTT××Fig. 13.23

DTyxT+×B
C–TyxT×BE
Arm fixed, gear B rotated
1 revolution anticlockwise)
elements

Chapter 13 : Gear Trains    /G6C  461
1. Speed of the output shaft when gear C is fixed
Since the gear C is fixed, therefore from the fourth row of the table,
B
C–0TyxT×= or20–080yx×=
∴ y – 0.25 x = 0 ... (i)
We know that the input speed (or the speed of gear B) is 1000 r.p.m. counter clockwise,
therefore from the fourth row of the table,
x + y = + 1000 ... (ii)
From equations (i) and (ii),x = + 800, and y = + 200
∴  Speed of output shaft = Speed of gear BE
DF–TTFyxTT=× ×

20 30200 – 800 200 – 187.5 12.5 r.p.m.
80 32=× × = =
= 12.5 r.p.m. (counter clockwise) Ans.
2. Speed of the output shaft when gear C is rotated at  10 r .p.m. counter clockwise
Since the gear C is rotated at 10 r.p.m. counter clockwise, therefore from the fourth row of the table,
B
C–1 0TyxT×= + or20–1 0
80yx×=
∴ y – 0.25 x = 10 ... (iii)
x = 792, and y = 208
∴   Speed of output shaft
= Speed of gear BE
DF20 30– 208 – 79280 32TTFyxTT=× × = × ×

= 208 – 185.6 = 22.4 r.p.m. = 22.4 r.p.m. (counter clockwise)  Ans.
Example 13.18.  Fig. 13.24 shows a diff erential
gear used in a motor car. The pinion A on the propellershaft has 12 teeth and gears with the crown gear B whichhas 60 teeth. The shafts P and Q form the rear axles to
which the road wheels are attached. If the propellershaft rotates at 1000 r.p.m.  and the road wheel attached
to axle Q has a speed of 210 r.p.m. while taking a turn,

find the speed of road wheel attached to axle P.
A = 12 ; TB = 60 ; NA = 1000
r.p.m. ; NQ = ND = 210 r.p.m.
Since the propeller shaft or the pinion A rotates at
1000 r.p.m., therefore speed of crown gear B,
                                  A
BA
B12100060TNNT=×= ×
                                 = 200 r.p.m.
The table of motions is given below :Fig. 13.24

462  /G6C   Theory of Machines
Table 13.20. Table of motions.
1. 0 + 1C
ET
T+CE
ED–– 1TT
TT×=
CD()TT=/G51
2. 0 + xC
ETx
T+×– x
4. + yx  + yC
ETyx
T+×y – x
Since the speed of gear B is 200 r.p.m., therefore from the fourth row of the table,
y = 200 ... (i)
Also, the speed of road wheel attached to axle Q or the speed of gear D is 210 r.p.m., there-
fore from the fourth row of the table,

y – x = 210 or x = y – 210 = 200 – 210 = – 10
∴  Speed of road wheel attached to axle P
= Speed of gear C = x + y
= – 10 + 200 = 190 r.p.m. Ans.
13.11. Torques in Epicyclic Gear Trains
Fig. 13.25. Torques in epicyclic gear trains.
When the rotating parts of an epicyclic gear train, as shown in Fig. 13.25, have no angular
acceleration, the gear train is kept in equilibrium by the three externally applied torques, viz.

1. Input torque on the driving member ( T1),
2. Output torque or resisting or load torque on the driven member ( T2),
3. Holding or braking or fixing torque on the fixed member ( T3).Gear B fixed-Gear C rotated
elements

Chapter 13 : Gear Trains    /G6C  463
The net torque applied to the gear train must be zero. In other words,
T1 + T2 + T3  = 0 ... (i)
∴F1.r1 + F2.r2 + F3.r3 = 0 ... (ii)
where F1, F2 and F3 are the corresponding externally applied forces at radii r1, r2 and r3.
Further, if ω1, ω2 and ω3 are the angular speeds of the driving, driven and fixed members
respectively, and the friction be neglected, then the net kinetic energy dissipated by the gear trainmust be zero, i.e.
T

T
1.ω1 + T2.ω2 + T3.ω3 = 0 ... (iii)
But, for a fixed member, ω3 = 0
∴ T1.ω1 + T2.ω2 = 0 ... (iv)
Notes : 1. From equations (i) and (iv), the holding or braking torque T3 may be obtained as follows :
1
21
2–TTω=×ω...[From equation (iv)]
and T3 = – (T1+ T2 ) ...[From equation (i)]
11
11
22–1 –1NTTN ω== ω 
2. When input shaft (or driving shaft) and output shaft (or driven shaft) rotate in the same direction,

then the input and output torques will be in opposite directions. Similarly, when the input and output shafts
rotate in opposite directions, then the input and output torques will be in the same direction.
Example 13.19.  Fig. 13.26 shows an epicyclic gear train. Pinion

A has 15 teeth and is rigidly fixed to the motor shaft. The wheel B has 20teeth and gears with A and also with the annular fixed wheel E. PinionC has 15 teeth and is integral with B (B, C being a compound gearwheel). Gear C meshes with annular wheel D, which is keyed to themachine shaft. The arm rotates about the same shaft on which A is fixedand carries the compound wheel B, C. If the motor runs at 1000 r.p.m.,find the speed of the machine shaft. Find the torque exerted on themachine shaft, if

Find the torque exerted on themachine shaft, if the motor develops a torque of 100 N-m.

A = 15 ; TB = 20 ; TC = 15 ; NA = 1000 r.p.m.; Torque developed by motor (or
pinion A) = 100 N-m
First of all, let us find the number of teeth on wheels D and E. Let TD and TE be the number of
teeth on wheels D and E respectively. Let dA, dB, dC, dD and dE be the pitch circle diameters of wheels
A, B, C, D and E respectively. From the geometry of the figure,
dE = dA + 2 dBanddD = dE – (dB – dC)

TE = TA + 2 TB = 15 + 2 × 20 = 55
and TD = TE – (TB – TC) = 55 – (20 – 15) = 50
Speed of the machine shaft
The table of motions is given below :Fig. 13.26

464  /G6C   Theory of Machines
Table 13.21. Table of motions.
Step Conditions of motion Arm Pinion Compound W heel D Wh eel E
No. A wheel B-C
1. 0 + 1A
B–T
TAC
BD–TT
TT×AB A
BE ETT T
TT T−×= −
2. 0 + xA
B–TxT×AC
BD–TTxTT××A
ETxT−×
4. + yx + yA
B–TyxT×AC
BD–TTyxTT××A
ETyxT−×
We know that the speed of the motor or the speed of the pinion A is 1000 r.p.m.
Therefore
x + y = 1000 ... (i)
Also, the annular wheel E is fixed, therefore
A
E–0TyxT×= orA

A
E150.27355Tyx x xT=× =× = ...(ii)
x = 786 and y = 214
∴  Speed of machine shaft = Speed of wheel D,
CA
D
BD15 15– 214 – 786 37.15 r.p.m.20 50TTNy xTT=× × = × × = +
= 37.15 r.p.m. (anticlockwise)  Ans.
Torque exerted on the machine shaft
We know thatTorque developed by motor × Angular speed of motor
= Torque exerted on machine shaft
       × Angular speed of machine shaft
or                      100 × ω
A = Torque exerted on machine shaft × ωD

∴  Torque exerted on machine shaft
AA
DD1000100 100 100 2692 N-m37.15N
Nω=×=×=× =ωAns.Arm fixed-pinion A
rotated through + 1revolution(anticlockwise)
Arm fixed-pinion A
rotated through + x
Add + y revolutions to

Chapter 13 : Gear Trains    /G6C  465
Fig. 13.27Example 13.20.  An epicyclic gear train consists of a sun wheel
S, a stationary internal gear E and three identical planet wheels Pcarried on a star- shaped planet carrier C. The size of different toothed
wheels are such that the planet carrier C rotates at 1/5th of the speedof the sunwheel S. The minimum number of teeth on any wheel is 16.The driving torque on the sun wheel is 100 N-m. Determine : 1. num-

ber of teeth on different wheels of the train, and 2. torque necessary to
keep the internal gear stationary.
Solution.  Given :     
S
C5NN=
1. Number of teeth on different wheels
The arrangement of the epicyclic gear train is shown in Fig. 13.27. Let TS and TE be the
number of teeth on the sun wheel S and the internal gear E respectively. The table of motions is
given below :
Table 13.22. Table of motions.
Step Conditions of motion P lanet Sun Planet Internal gear E

No. carrier C w heel S wheel P
1. 0 + 1S
P–T
TSP S
PE E––TT T
TT T×=
2. 0 + xS
P–TxT×S
E–TxT×
4. + yx  + yS
P–TyxT×S
E–TyxT×
We know that when the sunwheel S makes 5 revolutions, the planet carrier C makes 1
revolution. Therefore from the fourth row of the table,
y = 1, and x + y = 5 or x = 5 – y = 5 – 1 = 4
Since the gear E is stationary, therefore from the fourth row of the table,
S
E–0TyxT×= orS
E1–4 0T
T×= orS
E1
4T
T=
∴ TE = 4TS

E1–4 0T
T×= orS
E1
4T
T=
∴ TE = 4TS
Since the minimum number of teeth on any wheel is 16, therefore let us take the number of
teeth on sunwheel, TS = 16
∴ TE = 4 TS = 64 Ans.
Let dS, dP and dE be the pitch circle diameters of wheels S, P and E respectively. Now from the
geometry of Fig. 13.27,
dS + 2 dP = dEPlanet carrier C fixed, sunwheel S
rotates through + 1 revolution ( i.e.
Planet carrier C fixed, sunwheel S
rotates through + x revolutions

Total  motion

466  /G6C   Theory of Machines
Assuming the module of all the gears to be same, the number of teeth are proportional to their
pitch circle diameters.
TS + 2 TP = TEor 16 + 2 TP = 64 or TP = 24 Ans.
2. Torque necessary to keep the internal gear stationary
We know thatTorque on S × Angular speed of S
= Torque on C × Angular speed of C
100 × ω
S = Torque on C × ωC
∴ Torque on SS
CC100 100 100 5 500 N-mNCNω=×=×=× =ω
∴ Torque necessary to keep the internal gear stationary
= 500 – 100 = 400 N-m  Ans.

Example 13.21.  In the epicyclic gear train, as
shown in Fig. 13.28, the driving gear A rotating in clock-wise direction has 14 teeth and the fixed annular gear Chas 100 teeth. The ratio of teeth in gears E and D is 98 :41. If 1.85 kW is supplied to the gear A rotating at 1200r.p.m., find : 1. the speed and direction of rotation of gear
E, and 2. the fixing torque required at C, assuming 100
per cent efficiency throughout and that all teeth have thesame pitch.

Solution.  Given : T
A = 14 ; TC = 100 ; TE / TD
= 98 / 41 ; PA = 1.85 kW = 1850 W ; NA = 1200 r.p.m.
Let dA, dB and dC be the pitch circle diameters of gears A, B and C respectively. From Fig.
13.28,
dA + 2 dB = dCFig. 13.28
Gears are extensively used in trains for power transmission.

Chapter 13 : Gear Trains    /G6C  467
Arm fixed-Gear A rotated
through – 1 revolution ( i.e.
1 revolution clockwise)
elements
Total motionSince teeth of all gears have the same pitch and the number of teeth are proportional to their
pitch circle diameters, therefore
TA + 2TB = TCorCA
B–100 –144322TTT===
The table of motions is now drawn as below :
Table 13.23. Table of motions.

Step Conditions of motion Arm Gear Com pound Gear C Gear E
No. A gear B-D
1. 0 – 1A
BT
T+AB
BCTT
TT+×AD
BETT
TT+×
A
CT
T=+
2. 0 – xA
BTx
T+×A
CTx
T+×AD
BETTx
TT+× ×
4. –  y– y – xA
B–Tyx
T+×A
C–Tyx
T+×AD
BE–TTyx
TT+× ×
Since the annular gear C is fixed, therefore from the fourth row of the table,
A
C–0TyxT+× = or14–0
100yx+× =
∴ – y + 0.14 x = 0 ... (i)
Also, the gear A is rotating at 1200 r.p.m., therefore
– x – y = 1200 ... (ii)

From equations (i) and (ii), x = – 1052.6, and y = – 147.4
1. Speed and direction of rotation of gear E
From the fourth row of the table, speed of gear E,
AD
E
BE14 41– 147.4 –1052.643 98TTNy xTT=+ ××= × ×
= 147.4 – 143.4 = 4 r.p.m.
= 4 r.p.m. (anticlockwise) Ans.
2. Fixing torque required at C
We know that torque on A A
A60 1850 6014.7 N-m2 2 1200P
N××== =ππ ×
Since the efficiency is 100 per cent throughout, therefore the power available at E (PE) will

468  /G6C   Theory of Machines
∴ Torque on EA
E60 1850 604416 N-m22 4P
N××== =π× π×
∴ Fixing torque required at C
= 4416 – 14.7 = 4401.3 N-m  Ans.
Example 13.22.  An over drive for a vehicle consists of an

epicyclic gear train, as shown in Fig. 13.29, with compound planetsB-C. B has 15 teeth and meshes with an annulus A which has 60teeth. C has 20 teeth and meshes with the sunwheel D which is fixed.The annulus is keyed to the propeller shaft Y which rotates at 740rad /s. The spider which carries the pins upon which the planetsrevolve, is driven directly from main gear box by shaft X, this shaftbeing relatively free to rotate with respect to wheel D. Find the

speed of shaft X, when all the teeth have the same module.
When the engine develops 130 kW, what is the holding
torque on the wheel D ? Assume 100 per cent effici ency
throughout.
B = 15 ; TA = 60 ; TC = 20 ; ωY = ωA = 740 rad /s ; P = 130 kW = 130 × 103 W
First of all, let us find the number of teeth on the sunwheel D (TD). Let dA , dB , dC and dD be
the pitch circle diameters of wheels A, B, C and D respectively. From Fig. 13.29,
CDB A
2222ddd d++= ordD + dC + dB = dA

CDB A
Since the module is same for all teeth and the number of teeth are proportional to their pitch
circle diameters, therefore
TD + TC + TB = TAorTD = TA – (TC + TB) = 60 – (20 + 15) = 25
Table 13.24. Table of motions.
Step Conditions of motion Arm (or W heel D Compound Wheel A
No. shaft X) wheel C-B (or shaft Y)
1. 0 + 1D
C–T
TDB
CA–TT
TT×
2. 0 + xD
C–TxT×DB

2. 0 + xD
C–TxT×DB
4. + yx  + yD
C–Tyx
T×DB
CA–TTyx
TT××
Since the shaft Y or wheel A rotates at 740 rad/s, therefore
DB
CA–7 4 0TTyxTT××= or25 15–7 4 0
20 60yx××=
y – 0.3125 x = 740 ... (i)Arm  fixed-wheel D rotated
Arm  fixed-wheel D rotated
Add + y revolutions to all ele-
ments
Total motionFig. 13.29

Chapter 13 : Gear Trains    /G6C  469
Also the wheel D is  fixed, therefore
x + y = 0 or y = – x ...(ii)
x = – 563.8 and y = 563.8
Speed of shaft X
Since the shaft X will make the same number of revolutions as the arm, therefore
Speed of shaft X, ωX = Speed of arm = y = 563.8 rad/s Ans.
Holding torque on wheel D
We know that torque on A = P/ωA = 130 × 103 / 740 = 175.7 N-m
and          Torque on   X = P/ωX = 130 × 103/563.8 = 230.6 N-m
∴ Holding torque on wheel D

= 230.6 – 175.7 = 54.9 N-m Ans.
Example 13.23.  Fig. 13.30 shows some details of a compound epicyclic gear drive where I
is the driving or input shaft and O is the driven or output shaft which carries two arms A and Brigidly fixed to it. The arms carry planet wheels which mesh with annular wheels P and Q and thesunwheels X and Y. The sun wheel X is a part of Q. Wheels Y and Z are fixed to the shaft I. Z engages

with a planet wheel carried on Q and this planet wheel engages the fixed annular wheel R. Thenumbers of teeth on the wheels are :
P = 114, Q = 120, R = 120, X = 36, Y = 24 and Z = 30.
Fig. 13.30.
The driving shaft I makes 1500 r.p.m.clockwise looking from our right and the input at I is
7.5 kW.
1. Find the speed and direction of rotation of the driven shaft O and the wheel P .
2. If the mechanical efficiency of the drive is 80%, find the torque tending to rotate the fixed
wheel R.

wheel R.
Solution. Given :TP =144 ; TQ = 120 ; TR = 120 ; TX = 36 ; TY = 24 ; TZ = 30 ; NI = 1500
r.p.m. (clockwise) ; P = 7.5 kW = 7500 W ; η = 80% = 0.8
First of all, consider the train of wheels Z,R and Q (arm). The revolutions of various wheels
are shown in the following table.

470  /G6C   Theory of Machines
Arm fixed-wheel Z rotates through + 1
revolution (anticlockwise)
Arm fixed-wheel Z rotates through + x revo-
lutions
Total motionTable 13.25. Table of motions.
Step No. Conditions of motion Q (Arm) Z (also I) R (Fixed)
1. 0 + 1Z
R–T
T
2. 0 + xZ
R–TxT×
3. + y + y + y
4. + yx  + yZ
R–TyxT×
Since the driving shaft I as well as wheel Z rotates at 1500 r.p.m. clockwise, therefore
x + y = – 1500 ... (i)

Also, the wheel R is fixed. Therefore
Z
R–0TyxT×= orZ
R300.25120Tyx x xT=× =× = ...(ii)
x = – 1200, and y = – 300
Now consider the train of wheels Y, Q, arm A, wheels P and X. The revolutions of various
elements are shown in the following table.
Table 13.26. Table of motions.
Step Conditions of motion Arm A, B W heel Y Compound W heel P
No. and Shaft O wheel Q-X
1. 0 + 1Y
Q–T
TYX
QPTT
TT+×
2. 0 + x1Y1
Q–Tx
T×YX1
QPTTx

Q–T
TYX
QPTT
TT+×
2. 0 + x1Y1
Q–Tx
T×YX1
QPTTx
TT+× ×
4. + y1x1 + y1Y11
Q–Tyx
T×YX11
QPTTyx
TT+× ×
Since the speed of compound wheel Q-X is same as that of Q, therefore
Y
11
Q– –300Tyx y
T×= =
or1124–– 3 0 0120yx×=Arm A fixed-wheel Y
rotates through + 1revolution (anticlockwise)
Arm A fixed-wheel Y rotates
through + x
1 revolutions
elements

Chapter 13 : Gear Trains    /G6C  471
∴ y1 = 0.2 x1 – 300 ... (iii)
Also Speed of wheel Y =  Speed of wheel Z or shaft I
∴ x1 + y1 = x + y = – 1500 ... (iv)
x1 + 0.2 x1 – 300 = – 1500 ...[From equation (iii)]
1.2 x1= – 1500 + 300 = – 1200
or x1 = – 1200/1.2 = – 1000
and y1 = – 1500 – x1 = – 1500 + 1000 = – 500
1.  Speed and direction of the driven shaft O  and the wheel P
Speed of the driven shaft O,
NO = y1 = – 500 = 500 r.p.m. clockwise  Ans.
and Speed of the wheel P, YX
P11
QP24 36–500 –1000

P11
120 144TTNy x
TT=+× × = × ×
= – 550 = 550 r.p.m. clockwise  Ans.
2.  Torque tending to rotate the fixed wheel R
We know that the torque on shaft I or input torque
1
160 7500 6047.74 N-m2 2 1500PTN××== =π× π×
and torque on shaft O or output torque,
2
O60 0.8 7500 60114.58 N-m2 2 500PTNη× × × ×== =π× π×
Since the input and output shafts rotate in the same direction ( i.e. clockwise), therefore input
and output torques will be in opposite direction.

∴ Torque tending to rotate the fixed wheel R
= T2 – T1 = 114.58 – 47.74 =  66.84 N-m  Ans.
Example 13.24.  An epicyclic bevel gear train (known as  Humpage’s reduction gear) is shown
in Fig. 13.31. It consists of a fixed wheel C, thedriving shaft X and the driven shaft Y. The compoundwheel B-D can revolve on a spindle F which canturn freely about the axis X and Y.
Show that  (i) if the ratio of tooth numbers
T

T
B / TD is greater than TC / TE , the wheel E will ro-
tate in the same direction as wheel A , and (ii) if the
ratio TB / TD is less than TC / TE, the direction of E is
reversed.
If the numbers of teeth on wheels A, B, C, D

and E are 34, 120, 150, 38 and 50 respectively and7.5 kW is put into the shaft X at 500 r.p.m., what isthe output torque of the shaft Y, and what are theforces (tangential to the pitch cones) at the contactpoints between wheels D and E and between wheels B and C, if the module of all wheels is 3.5 mm ?
A = 34 ; TB = 120 ; TC = 150 ; TD = 38 ; TE = 50 ; PX = 7.5 kW = 7500 W ;
NX = 500 r.p.m. ; m = 3.5 mmFig. 13.31

472  /G6C   Theory of Machines
Spindle fixed, wheel A
is rotated through + 1revolution
is rotated through + x
Total motionThe table of motions is given below :
Table 13.27. Table of motions.
Step Conditions of motion Spindle Wheel A Compound W heel C Wh eel E (or
No. F (or shaft X) wheel B-D shaft Y)
1. 0 + 1A
BT
T+AB
BCTT
TT−×AD
BETT
TT−×
A
C–T
T=
2. 0 + xA
BTxT+×A
C–TxT×AD
BE–TTxTT××

A
C–T
T=
2. 0 + xA
BTxT+×A
C–TxT×AD
3. + y + y + y + y + y
4. + yx + yA
BTyxT+×A
C–TyxT×AD
BE–TTyxTT××
Let us assume that the driving shaft X rotates through 1 revolution anticlockwise, therefore
the wheel A will also rotate through 1 revolution anticlockwise.
∴ x + y = + 1 or y = 1 – x ...(i)
We also know that the wheel C is fixed, therefore
A
C–0TyxT×= orA
C(1 – ) – 0TxxT×= ...[From equation  (i)]
A
C1– 1 0TxT+=
orCA
C1TTxT+= 

andC
CATxTT=+...(ii)
CA


andC
CA
CACA1– 1–T TyxTTTT== =++...(iii)
We know that speed of wheel E,
C AD A AD
E
BECACABE––T TT T TTNy xTTTTTTTT=× × = × ×++
C AD
CA BE1–T TT
TT TT=× +...(iv)
and the speed of wheel A,
NA = x + y = + 1 revolution
(i)  If CB
DETT
TT> or TB × TE > TC × TD , then the equation  (iv) will be positive. Therefore the
wheel E will rotate in the same direction as wheel A.  Ans.

Chapter 13 : Gear Trains    /G6C  473
(ii)  If CB
DETT
TT< or TB × TE < TC × TD , then the equation  (iv) will be negative. Therefore the
wheel E will rotate in the opposite direction as wheel A.  Ans.
Output torque of shaft Y
We know that the speed of the driving shaft X (or wheel A) or input speed is 500 r.p.m.,
x + y = 500 or y = 500 – x ...(v)
Since the wheel C is fixed, therefore
 A
C–0TyxT×= or34(500 – ) – 0150xx×= ...[From equation (v)]

∴ 500 – x – 0.227 x = 0 or x = 500/1.227 = 407.5 r.p.m.
and                                           y = 500 – x = 500 – 407.5 = 92.5 r.p.m.
Since the speed of the driven or output shaft Y (i.e. NY) is equal to the speed of wheel E
(i.e. NE), therefore
                  AD
YE
BE34 38– 92.5 – 407.5120 50TTNNy xTT== ××= ××
               = 92.5 – 87.75 = 4.75 r.p.m.
Assuming 100 per cent efficiency of the gear train, input power PX is equal to output power
(PY), i.e.

(PY), i.e.
                PY = PX = 7.5 kW = 7500 W
∴  Output torque of shaft Y,
                  Y
Y60 7500 6015 076 N-m 15.076 kN-m2 2 4.75P
N××== = =ππ × Ans.
Tangential force between wheels D and E
We know that the pitch circle radius of wheel E,
                 E
E3.5 5087.5 mm 0.0875 m22mTr××== = =
∴  Tangential force between wheels D and E,
                  Torque on wheel 15.076172.3 kN
Pitch circle radius of wheel 0.0875E
E== =  Ans.
...(∴ Torque on wheel E = Torque on shaft Y)

Tangential force between wheels B and C
We know that the input torque on shaft X or on wheel A
                 X
X60 7500 60143 N-m22 5 0 0P
N××== =ππ ×
∴  Fixing torque on the fixed wheel C
                  = Torque on wheel E – Torque on wheel A
              = 15 076 – 143 = 14 933 N-m = 14.933  kN-m

474  /G6C   Theory of Machines
Pitch circle radius of wheel C,
                  C
C3.5 150262.5 mm 0.2625 m22mTr××== = =
                 
CFixing torque on wheel 14.93357 kN0.2625C
r== = Ans.
EXERCISES
1.A compound train consists of six gears. The number of teeth on the gears are as follows :
Gear : ABCDEF
No. of teeth: 60 40 50 25 30 24The gears B and C are on one shaft while the gears D and E are on another shaft. The gear A drives gear

B, gear C drives gear D and gear E drives gear F. If the gear A transmits 1.5 kW at 100 r.p.m. and the gear
train has an efficiency of 80 per cent, find the torque on gear F. [Ans. 30.55 N-m]
2.Two parallel shafts are to be connected by spur gearing. The approximate distance between the shafts
is 600 mm. If one shaft runs at 120 r.p.m. and the other at 360 r.p.m., find the number of teeth on eachwheel, if the module is 8 mm. Also determine the exact distance apart of the shafts.

[Ans. 114, 38 ; 608 mm]
3.In a reverted gear train, as shown in Fig. 13.32, two shafts A and B are
in the same straight line and are geared together through an interme-diate parallel shaft C. The gears connecting the shafts A and C have a
module of 2 mm and those connecting the shafts C and B have a
module of 4.5 mm. The speed of shaft A is to be about but greater than
12 times the speed of shaft B, and the ratio at each reduction is same.

Find suitable number of teeth for gears. The number of teeth of eachgear is to be a minimum but not less than 16. Also find the exactvelocity ratio and the distance of shaft C from A and B.
                                          [Ans. 36, 126, 16, 56 ; 12.25 ; 162 mm]
4.In an epicyclic gear train, as shown in Fig.13.33, the number of teeth
on wheels A, B and C are 48, 24 and 50 respectively. If the arm rotates at 400 r.p.m., clockwise,

find : 1. Speed of wheel C when A is fixed, and 2. Speed of wheel A when C is fixed.
[Ans. 16 r.p.m. (clockwise) ; 16.67 (anticlockwise)]
Fig. 13.33 Fig. 13.34Fig. 13.32

Chapter 13 : Gear Trains    /G6C  475
5.In an epicyclic gear train, as shown in Fig. 13.34, the wheel C is keyed to the shaft B and wheel F is
keyed to shaft A. The wheels D and E rotate together on a pin fixed to the arm G. The number of teeth
on wheels C, D, E and F are 35, 65, 32 and 68 respectively.
If the shaft A rotates at 60 r.p.m. and the shaft B rotates at 28 r.p.m. in the opposite direction, find

the speed and direction of rotation of arm G.[Ans. 90 r.p.m., in the same direction as shaft A]
6.An epicyclic gear train, as shown in Fig. 13.35, is composed of a fixed annular wheel A having 150
teeth. The wheel A is meshing with wheel B which drives wheel D through an idle wheel C, D being
concentric with A. The wheels B and C are carried on an arm which revolves clockwise at 100 r.p.m.
about the axis of A and D. If the wheels B and D have 25 teeth and 40 teeth respectively, find the

number of teeth on C and the speed and sense of rotation of C.[Ans. 30 ; 600 r.p.m. clockwise]
                            
Fig. 13.35 Fig. 13.36
7.Fig. 13.36, shows an epicyclic gear train with the following details :
A has 40 teeth external (fixed gear) ; B has 80 teeth internal ; C - D is a compound wheel having 20 and
50 teeth (external) respectively, E-F is a compound wheel having 20 and 40 teeth (external) respec-
tively, and G has 90 teeth (external).

The arm runs at 100 r.p.m. in clockwise direction. Determine the speeds for gears C, E, and B.
[Ans. 300 r.p.m. clockwise ; 400 r .p.m. anticlockwise ; 150 r. p.m. clockwise]
8.An epicyclic gear train, as shown in Fig. 13.37, has a sun wheel S of 30 teeth and two planet wheels
P-P of 50 teeth. The planet wheels mesh with the internal teeth of a fixed annulus A. The driving shaft

carrying the sunwheel, transmits 4 kW at 300 r.p.m. The driven shaft is connected to an arm whichcarries the planet wheels. Determine the speed of the driven shaft and the torque transmitted, if the
overall efficiency is 95%. [Ans. 56.3 r.p.m. ; 644.5 N-m]
Fig. 13.37 Fig. 13.38
9.An epicyclic reduction gear, as shown in Fig. 13.38, has a shaft A fixed to arm B. The arm B has a pin
fixed to its outer end and two gears C and E which are rigidly fixed, revolve on this pin. Gear C

meshes with annular wheel D and gear E with pinion F. G is the driver pulley and D is kept stationary.
The number of teeth are : D = 80 ; C = 10 ; E = 24 and F = 18.
If the pulley G runs at 200 r.p.m. ; find the speed of shaft A.
[Ans. 17.14 r.p.m. in the same direction as that of G]

476  /G6C   Theory of Machines
10.A reverted epicyclic gear train for a hoist block is shown in
Fig. 13.39. The arm E is keyed to the same shaft as the load
drum and the wheel A is keyed to a second shaft which car-
ries a chain wheel, the chain being operated by hand. Thetwo shafts have common axis but can rotate independently.The wheels B and C are compound and rotate together on a
pin carried at the end of arm E. The wheel D has internal

teeth and is fixed to the outer casing of the block so that itdoes not rotate.
The wheels A and B have 16 and 36 teeth respectively with a
module of 3 mm. The wheels C and D have a module of 4
mm. Find : 1. the number of teeth on wheels C and D when
the speed of A is ten times the speed of arm E, both rotating
in the same sense, and 2. the speed of wheel D when the
wheel A is fixed and the arm E rotates at 450 r.p.m.
anticlockwise.[Ans. T
C = 13 ; TD = 52 ; 500 r.p.m. anticlockwise]

11.A compound epicyclic gear is shown diagrammatically in Fig. 13.40. The gears A, D and E are free to
rotate on the axis P. The compound gear B and C rotate together on the axis Q at the end of arm F. All
the gears have equal pitch. The number of external teeth on the gears A, B and C are 18, 45 and 21
respectively. The gears D and E are annular gears. The gear A rotates at 100 r.p.m. in the anticlockwise

direction and the gear D rotates at 450 r.p.m. clockwise. Find the speed and direction of the arm and
the gear E. [Ans. 400 r.p.m. clockwise ; 483.3 r.p.m. clockwise]
12.In an epicyclic gear train of the ‘sun and planet type’ as shown in Fig. 13.41, the pitch circle diameter
of the internally toothed ring D is to be 216  mm and the module 4 mm. When the ring D is stationary,
the spider A, which carries three planet wheels C of equal size, is to make one revolution in the same

sense as the sun wheel B for every five revolutions of the driving spindle carrying the sunwheel B.
Determine suitable number of teeth for all the wheels and the exact diameter of pitch circle of the ring.
[Ans. TB = 14 , TC = 21 , TD = 56 ; 224 mm]
Fig. 13.40 Fig. 13.41
13.An epicyclic train is shown in Fig. 13.42. Internal gear A is keyed to the driving shaft and has 30 teeth.
Compound wheel C and D of 20 and 22 teeth respectively are free to rotate on the pin fixed to the arm

P which is rigidly connected to the driven shaft. Internal gear B which has 32 teeth is fixed. If the
driving shaft runs at 60 r.p.m. clockwise, determine the speed of the driven shaft. What is the directionof rotation of driven shaft with reference to driving shaft? [Ans. 1980 r.p.m. clockwise]Fig. 13.39

Chapter 13 : Gear Trains    /G6C  477
Fig. 13.42 Fig. 13.43
14.A shaft Y is driven by a co-axial shaft X by means of an epicyclic gear train, as shown in Fig. 13.43.
The wheel A is keyed to X and E to Y. The wheels B and D are compound and carried on an arm F
which can turn freely on the common axes of X and Y. The wheel C is fixed. If the numbers of teeth
on A, B, C, D and E are respectively 20, 64, 80, 30 and 50 and the shaft X makes 600 r.p.m.,

determine the speed in r.p.m. and sense of rotation of the shaft Y.
[Ans. 30 r.p.m. in the same sense as shaft X]
15.An epicyclic bevel gear train, as shown in Fig. 13.44, has fixed gear B meshing with pinion C. The
gear E on the driven shaft meshes with the pinion D. The pinions C and D are keyed to a shaft,
which revolves in bearings on the arm A. The arm A is keyed to the driving shaft. The number of

teeth are : TB = 75, TC = 20, TD = 18, and TE = 70. Find the speed of the driven shaft, if 1. the driving
shaft makes 1000 r.p.m., and 2. the gear B turns in the same sense as the driving shaft at 400
r.p.m., the driving shaft still making 1000 r.p.m.
[Ans. 421.4 r.p.m. in the same di rection as driving shaft]
16.The epicyclic gear train is shown in Fig. 13.45. The wheel D is held stationary by the shaft A and the

arm B is rotated at 200 r.p.m. The wheels E (20 teeth) and F (40 teeth) are fixed together and rotate
freely on the pin carried by the arm. The wheel G (30 teeth) is rigidly attached to the shaft C. Find the
speed of shaft C stating the direction of rotation to that of B.
If the gearing transmits 7.5 kW, what will be the torque required to hold the shaft A stationary, neglect-
ing all friction losses?
[Ans. 466.7 r.p.m. in opposite direction of B; 511.5 N-m in opposite direction of B]

Fig. 13.44 Fig. 13.45
17.An epicyclic gear train, as shown in Fig. 13.46, consists of two sunwheels A and D with 28 and 24
teeth respectively, engaged with a compound planet wheels B and C with 22 and 26 teeth. The sunwheel

478  /G6C   Theory of Machines
D is keyed to the driven shaft and the sunwheel A is a fixed wheel co-axial with the driven shaft. The
planet wheels are carried on an arm E from the driving shaft which is co-axial with the driven shaft.
Find the velocity ratio of gear train. If 0.75 kW is transmitted and input speed being 100 r.p.m.,
determine the torque required to hold the sunwheel A. [Ans. 2.64 ; 260.6 N-m]
Fig. 13.46 Fig. 13.47

18.In the epicyclic reduction gear, as shown in Fig. 13.47, the sunwheel D has 20 teeth and is keyed
to the input shaft. Two planet wheels B, each having 50 teeth, gear with wheel D and are carried
by an arm A fixed to the output shaft. The wheels B also mesh with an internal gear C which is
fixed. The input shaft rotates at 2100 r.p.m. Det ermine the speed of the output shaft and the torque
required to fix C when the gears are transmitting 30 kW.

[Ans. 300 r.p.m. in the same sense as the input shaft ; 818.8 N-m]
19.An epicyclic gear train for an electric motor is shown in Fig. 13.48. The wheel S has 15 teeth and is
fixed to the motor shaft rotating at 1450 r.p.m. The planet P has 45 teeth, gears with fixed annulus A
and rotates on a spindle carried by an arm which is fixed to the output shaft. The planet P also gears
with the sun wheel S. Find the speed of the output shaft. If the motor is transmitting 1.5 kW,  find the

torque required to fix the annulus A. [Ans. 181.3 r.p.m. ; 69.14 N-m]
Fig. 13.48 Fig. 13.49
20.An epicyclic gear consists of bevel wheels as shown in Fig. 13.49. The driving pinion A has 20 teeth
and meshes with the wheel B which has 25 teeth. The wheels B and C are fixed together and turn freely
on the shaft F. The shaft F can rotate freely about the main axis XX. The wheel C has 50 teeth and
meshes with wheels D and E, each of which has 60 teeth. Find the speed and direction of E when A

rotates at 200 r.p.m., if
1. D is fixed, and 2. D rotates at 100 r.p.m., in the same direction as A.
In both the cases, find the ratio of the torques transmitted by the shafts of the wheels A and E, the
friction being neglected.
[Ans. 800 r.p.m. in the opposite direction of A ; 300 r.p.m. in the opposite
direction of A ; 4 ; 1.5]

Chapter 13 : Gear Trains    /G6C  479
1.What do you understand by ‘gear train’? Discuss the various types of gear trains.
2.Explain briefly the differences between simple, compound, and epicyclic gear trains. What are the
special advantages of epicyclic gear trains ?
3.Explain the procedure adopted for designing the spur wheels.
4.How the velocity ratio of epicyclic gear train is obtained by tabular method?
5.Explain with a neat sketch the ‘sun and planet wheel’.

6.What are the various types of the torques in an epicyclic gear train ?
1.In a simple gear train, if the number of idle gears is odd, then the motion of driven gear will
(a) be same as that of driving gear
(b) be opposite as that of driving gear
(c) depend upon the number of teeth on the driving gear
2.The train value of a gear train is
(a) equal to velocity ratio of a gear train ( b) reciprocal of velocity ratio of a gear train

(c) always greater than unity ( d) always less than unity
3.When the axes of first and last gear are co-axial, then gear train is known as
(a) simple gear train ( b) compound gear train
(c) reverted gear train ( d) epicyclic gear train
4.In a clock mechanism, the gear train used to connect minute hand to hour hand, is
(a) epicyclic gear train ( b) reverted gear train
(c) compound gear train ( d) simple gear train

5.In a gear train, when the axes of the shafts, over which the gears are mounted, move relative to a fixed
axis, is called
6.A differential gear in an automobile is a
(a) simple gear train ( b) epicyclic gear train
(c) compound gear train ( d) none of these
7.A differential gear in automobilies is used to
(a) reduce speed ( b) assist in changing speed

(c) provide jerk-free movement of vehicle ( d) help in turning
ANSWERS
1.(a) 2.(b) 3.(c) 4.(b) 5.(d)
6.(b) 7.(d)

480       /G6C               Theory of Machines
480GyrGyrGyrGyrGyroscopicoscopicoscopicoscopicoscopic
Couple andCouple andCouple andCouple andCouple and
PrPrPrPrPrecessionalecessionalecessionalecessionalecessional
MotionMotionMotionMotionMotion14
2. Precessional Angular
Motion.
3. Gyroscopic Couple.4. Effect of Gyroscopic Couple
on an Aeroplane.
5. Terms Used in a Naval Ship.6. Effect of Gyroscopic Couple
on a Naval Ship during
Steering.

Steering.
7. Effect of Gyroscopic Couple
on a Naval  Ship duringPitching.
8. Effect of Gyroscopic Couple
on a Navalship duringRolling.
9. Stability of  a Four Wheel
drive Moving in a CurvedPath.
10. Stability of a Two Wheel
Vehicle Taking a Turn.
11. Effect of Gyroscopic Couple
on a Disc Fixed Rigidly at aCertain Angle to a RotatingShaft.14.1.14.1.14.1.14.1.14.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoduction
We have already discussed that,

1.  When a body moves along a curved path with a
uniform linear velocity, a force in the direction of centripetal
acceleration (known as centripetal force)   has to be applied
externally over the body, so that it moves along the requiredcurved path. This external force applied is known as active
force.
2.  When a body, itself, is moving with uniform lin-

ear velocity along a circular path, it is subjected to the cen-trifugal force * radially outwards. This centrifugal force is
called reactive force.  The action of the reactive or centrifu-
gal force is to tilt or move the body along radially outwarddirection.
Note:  Whenever the effect of any force or couple over a moving or
rotating body is to be considered, it should be with respect to thereactive force or couple and not with respect  to active force orcouple.

*Centrifugal force is equal in magnitude to centripetal force but
opposite in direction.

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  481
14.2. Precessional Angular Motion
We have already discussed that the angular acceleration is the rate of change of angular
velocity with respect to time. It is a vector quantity and may be represented by drawing a vectordiagram with the help of right hand screw rule (see chapter 2, Art. 2.13).
Fig. 14.1.  Precessional angular motion.
Consider a disc, as shown in Fig. 14.1 ( a), revolving or spinning about the axis OX (known as

axis of spin ) in anticlockwise when seen from the front, with an angular velocity ω in a plane at right
angles to the paper.
After a short interval of time δt, let the disc be spinning about the new axis of spin OX′ (at an
angle δθ) with an angular velocity ( ω + δω). Using the right hand screw rule, initial angular velocity
of the disc ( ω) is represented by vector ox; and the final angular velocity of the disc ( ω + δω) is

represented by vector ox′ as shown in Fig. 14.1 ( b). The vector xx′ represents the change of angular
velocity in time δt i.e. the angular acceleration of the disc. This may be resolved into two components,
one parallel to ox and the other perpendicular to ox.
Component of angular acceleration in the direction of ox,
–c o s –
txr or ox ox ox
tt t′δθα= = =δδ δ
( ) cos – cos cos –
ttω + δω δθ ω ω δθ + δω δθ ω==δδ
Since δθ is very small, therefore substituting cos  δθ = 1, we have
–

–
tttω+δ ω ω δ ωα= =
δδGyroscopic inertia prevents a spinning top from falling sideways.
Spin axis Input axis
Gyorscope
will resistmovementin thesedirections
Output axis
GimbalsWheel
Axle

482  /G6C   Theory of Machines
In the limit, when 0tδ→,
0Ltttd
td t δ→δω ωα= = δ
Component of angular acceleration in the direction perpendicular to ox,
sin ( )sin sin .sin
crx ox
tt t t′′ δθ ω + δω δθ ω δθ + δω δθα= = = =
δδ δ δ
Since δθ in very small, therefore substituting sin δθ = δθ, we have
.. .
cttωδ θ+δ ωδ θ ωδ θα= =δδ
...(Neglecting δω.δθ, being very small)
In the limit when δt → 0,
P0.Lt .ctd
td t δ→ωδ θ θα= = ω × = ω ωδ... P Substitutin gd
dtθ =ω 

∴  Total angular acceleration of the disc
= vector xx′ = vector sum of  αt and αc
P.dd d
dt dt dtωθ ω=+ ω ×=+ ω ω
where dθ/dt is the angular velocity of the axis of spin about a certain axis, which is perpendicular to
the plane in which the axis of spin is going to rotate. This angular velocity of the axis of spin ( i.e.
dθ/dt) is known as angular velocity of precession   and is denoted by ωP. The axis, about which the

axis of spin is to turn, is known as axis of precession.  The angular motion of the axis of spin about the
axis of precession is known as precessional angular motion.
Notes:1. The axis of precession is perpendicular to the plane in which the axis of spin is going to rotate.
2.  If the angular velocity of the disc remains constant at all positions of the axis of spin, then dθ/dt is
zero; and thus αc is zero.

3. If the angular velocity of the disc changes the direction, but remains constant in magnitude, then
angular acceleration of the disc is given by
αc=ω.dθ/dt = ω.ωP
     The angular acceleration αc is known as gyroscopic acceleration.
This experimental car burns hydrogen fuel in an ordinary piston engine. Its exhaust gases cause no pollution,
because they contain only water vapour.EngineEvaporators change liquid
hydrogen to gasFuel tank

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  483
14.3. Gyroscopic Couple
Consider a disc spinning with an angular velocity ω rad/s about the axis of spin OX, in
anticlockwise direction when seen from the front, as shown in Fig. 14.2 ( a). Since the plane in which
the disc is rotating is parallel to the plane YOZ, therefore it is called plane of spinning.  The plane
XOZ  is a horizontal plane and the axis of spin rotates in a plane parallel to the horizontal plane about

an axis OY . In other words, the axis of spin is said to be rotating or processing about an axis OY . In
other words, the axis of spin is said to be rotating or processing about an axis  OY (which is perpendicular
to both the axes OX and OZ) at an angular velocity ωP rap/s. This horizontal plane XOZ is called
plane of precession  and OY is the axis of precession.
Let I= Mass moment of inertia of the disc about OX, and
ω= Angular velocity of the disc.
∴  Angular momentum of the disc
=I.ω

=I.ω
Since the angular momentum is a vector quantity, therefore it may be represented by the
vector ox→, as shown in Fig. 14.2 ( b). The axis of spin OX is also rotating anticlockwise when seen
from the top about the axis OY . Let the axis OX is turned in the plane XOZ through a small angle δθ
radians to the position OX′, in time δt seconds. Assuming the angular velocity ω to be constant, the
angular momentum will now be represented by vector ox′.

Fig. 14.2.   Gyroscopic couple.
∴  Change in angular momentum
–.ox ox xx ox→→ → →′′== = δ θ ...(in the direction of xx→′)
=I. ω.δθ
and rate of change of angular momentum
.Idtδθ=ω ×
Since the rate of change of angular momentum will result by the application of a couple to the
disc, therefore the couple applied to the disc causing precession,
P0Lt . . . .
tdCI I I
td t δ→δθ θ=ω × = ω × = ω ω
δ... Pd
dtθ=ω/G51

484  /G6C   Theory of Machines
where ωP = Angular velocity of precession of the
axis of spin or the speed of rotation of the axis ofspin about the axis of precession OY .
In S.I. units, the units of C is N-m when I is
in kg-m
2.
It may be noted that1.  The couple I.ω.ω
p, in the direction of
the vector xx′ (representing the change in angular
momentum) is the active gyroscopic couple,  which
has to be applied over the disc when the axis ofspin is made to rotate with angular velocity ω
P about

P about
the axis of precession. The vector xx′ lies in the
plane XOZ or the horizontal plane. In case of a very
small displacement δθ, the vector xx′ will be
perpendicular to the vertical plane XOY . Therefore
the couple causing this change in the angularmomentum will lie in the plane XOY .  The vector
xx
′, as shown in Fig. 14.2 ( b), represents an
anticlockwise couple in the plane XOY . Therefore, the plane XOY  is called the  plane of active

gyroscopic couple  and the axis OZ perpendicular to the plane XOY , about which the couple acts, is
called the axis of active gyroscopic couple.
2.  When the axis of spin itself moves with angular velocity ωP, the disc is subjected to
reactive couple whose magnitude is same ( i.e. I. ω.ωP) but opposite in direction to that of active

couple. This reactive couple to which the disc is subjected when the axis of spin rotates about the axisof precession is known as  reactive gyroscopic couple.  The axis of the reactive gyroscopic couple is
represented by OZ′ in Fig. 14.2 ( a).
3.  The gyroscopic couple is usually applied through the bearings which support the shaft.
The bearings will resist equal and opposite couple.
4.  The gyroscopic principle is used in an instrument or toy known as gyroscope.  The

gyroscopes are installed in ships in order to minimize the rolling and pitching effects of waves. Theyare also used in aeroplanes, monorail cars, gyrocompasses etc.
Example 14.1.  A uniform disc of diameter 300 mm and of mass 5 kg is mounted on one end
of an arm of length 600 mm. The other end of the arm is free to rotate in a universal bearing. If the
disc rotates about the arm with a speed of 300 r.p.m. cl ockwise,  looking from the front, with what

speed will it precess about the vertical axis ?
Solution. Given: d = 300 mm or  r = 150 mm = 0.15 m; m = 5 kg; l =  600 mm = 0.6 m ;
N = 300 r.p.m. or ω = 2π × 300/60 = 31.42 rad/s
We know that the mass moment of inertia of the disc, about an axis through its centre of
gravity and perpendicular to the plane of disc,
I=m.r
2/2 = 5(0.15)2/2 = 0.056 kg-m2
and couple due to mass of disc,
C=m.g.l = 5 × 9.81 × 0.6 = 29.43 N-m
Let ωP= Speed of precession.
We know that couple ( C),

29.43 = I.ω.ωP = 0.056 × 31.42 × ωP = 1.76 ωP
∴ωP= 29.43/1.76 = 16.7 rad/s Ans.
Above picture shows an aircraft propeller.
These rotors play role in gyroscopic couple.

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  485
Example 14.2.  A uniform disc of 150 mm diameter has a
mass of 5 kg. It is mounted centrally in bearings which maintainits axle in a horizontal plane. The disc spins about it axle with aconstant speed of 1000 r.p.m. while the axle precesses uniformly
about the vertical at 60 r.p.m. The  directions of rotation are as

shown in Fig. 14.3. If the distance between the bearings is 100mm, find the resultant reaction at each bearing  due to the massand gyroscopic effects.
Solution. Given:d = 150 mm or r = 75 mm = 0.075 m; m = 5 kg ; N = 1000 r.p.m. or
ω = 2π × 1000/60 = 104.7 rad/s (anticlockwise); N
P = 60 r.p.m. or ωP = 2π × 60/60 = 6.284 rad/s
(anticlockwise); x = 100 mm = 0.1 m
We know that mass moment of inertia of the disc, about an axis through its centre of gravity
and perpendicular to the plane of disc,

I=m.r2/2 = 5 (0.075)2/2 = 0.014 kg m2
∴  Gyroscopic couple acting on the disc,
C=I. ω. ωP = 0.014 × 104.7 × 6.284 = 9.2 N-m
The direction of the reactive gyroscopic couple is shown in Fig.14.4 ( b). Let F be the force at
each bearing due to the gyroscopic couple.
∴ F=C/x = 9.2/0.1 = 92 N
The force F will act in opposite directions at the bearings as shown in Fig. 14.4 ( a). Now let

RA and RB be the reaction at the bearing A and B respectively due to the weight of the disc. Since the
disc is mounted centrally in bearings, therefore,
RA=RB = 5/2 = 2.5 kg = 2.5 × 9.81 = 24.5 N
(a)( b)
Fig. 14.4
Resultant reaction at each bearing
Let RA1 and RB1= Resultant reaction at the bearings A and B respectively.
Since the reactive gyroscopic couple acts in clockwise direction when seen from the front,

therefore its effect is to increase the reaction on the left hand side bearing ( i.e. A) and to decrease the
reaction on the right hand side bearing ( i.e. B).
∴ RA1=F + RA = 92 + 24.5 = 116.5 N (upwards)  Ans.
and RB1= F – RB = 92 – 24.5 = 67.5 N (downwards) Ans.Fig.  14.3

486  /G6C   Theory of Machines
14.4. Effect of the Gyroscopic Couple on an Aeroplane
The top and front view of an aeroplane are shown in Fig 14.5 ( a). Let engine or propeller
rotates in the clockwise direction when seen from the rear or tail end and the aeroplane takes a turn tothe left.
Let ω= Angular velocity of the engine in rad/s,
m= Mass of the engine and the propeller in kg,
k= Its radius of gyration in metres,
I= Mass moment of inertia of the engine and the propeller in kg-m
2
= m.k2,

2
= m.k2,
v= Linear velocity of the aeroplane in m/s,
R= Radius of curvature in metres, and
ωP= Angular velocity of precession v
R= rad/s
∴ Gyroscopic couple acting on the aeroplane,
C=I.ω.ωP
Fig. 14.5.   Aeroplane taking a left turn.

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  487
Before taking the left turn, the angular momentum vector is represented by ox. When it takes
left turn, the active gyroscopic couple will change the direction of the angular momentum vector fromox to ox′ as shown in Fig. 14.6 ( a). The vector xx′, in the limit, represents the change of angular
momentum or the active gyroscopic couple and is perpendicular to ox. Thus the plane of active

gyroscopic couple XOY  will be perpendicular to xx′ , i.e. vertical in this case, as shown in Fig 14.5
(b). By applying right hand screw rule to vector xx′, we find that the direction of active gyroscopic
couple is clockwise as shown in the front view of Fig. 14.5 ( a). In other words, for left hand turning,
the active gyroscopic couple on the aeroplane in the axis OZ will be clockwise as shown in Fig. 14.5

(b).The reactive gyroscopic couple (equal in magnitude of active gyroscopic couple) will act in the
opposite direction ( i.e. in the anticlockwise direction) and the effect of this couple is, therefore, to
raise the nose  and dip the tail  of the aeroplane.
(a)  Aeroplane taking left turn. ( b)  Aeroplane taking right turn.
Fig. 14.6.  Effect of gyroscopic couple on an aeroplane.
Notes :  1. When the aeroplane takes a right turn under similar conditions as discussed above, the effect of the

reactive gyroscopic couple will be to  dip the nose and raise the tail of the aeroplane.
2.  When the engine or propeller rotates in anticlockwise direction when viewed from the rear or tail
end and the aeroplane takes a  left turn, then the effect of reactive gyroscopic couple will be to dip the nose  and
raise the tail  of the aeroplane.
3.  When the aeroplane takes a right turn  under similar conditions as mentioned in note 2 above, the

effect of reactive gyroscopic couple will be to raise the nose  and dip the tail of the aeroplane.
4.  When the engine or propeller rotates in clockwise direction when viewed from the front and the
aeroplane takes a left turn, then the effect of reactive gyroscopic couple will be to  raise the tail  and dip the nose
of the aeroplane.
5.  When the aeroplane takes a  right turn under similar conditions as mentioned in note 4-above, the

effect of reactive gyroscopic couple will be to  raise the nose and dip the tail of the aeroplane.
Example 14.3.  An aeroplane makes a complete half circle of 50 metres radius, towards left,
when flying at 200 km per hr. The rotary engine and the propeller of the plane has a mass of 400 kg
and a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m. clockwise w hen viewed from the
rear. Find the gyroscopic couple on the aircraft and state its effect on it.

Solution.  Given : R = 50 m ; v = 200 km/hr = 55.6 m/s; m = 400 kg;  k = 0.3 m;
N = 2400 r.p.m. or ω = 2π × 2400/60 = 251 rad/s
We know that mass moment of inertia of the engine and the propeller,
I=m.k2 = 400(0.3)2 = 36 kg-m2
and angular velocity of precession,
ωP=v/R = 55.6/50 = 1.11 rad/s
We know that gyroscopic couple acting on the aircraft,
C=I. ω. ωP = 36 × 251.4 × 1.11 = 100 46 N-m
= 10.046 kN-m Ans.

We have discussed in Art. 14.4 that when the aeroplane turns towards left, the effect of the
gyroscopic couple is to lift the nose upwards and tail downwards. Ans.

488  /G6C   Theory of Machines
14.5. Terms Used in a Naval Ship
The top and front views of a naval ship are shown in Fig 14.7. The fore end of the ship is
called bow  and the rear end is known as  stern or aft. The left hand and right hand sides of the ship,
when viewed from the stern are called port and star-board respectively. We shall now discuss the
effect of gyroscopic couple on the naval ship in the following three cases:
1.  Steering,  2.  Pitching, and  3.  Rolling.

Fig. 14.7.  Terms used in a naval ship.
14.6. Effect of Gyroscopic Couple on a Naval Ship during Steering
Steering is the turning of a complete ship in a curve towards left or right, while it moves

forward. Consider the ship taking a left turn, and rotor rotates in the clockwise direction when viewedfrom the stern, as shown in Fig. 14.8. The effect of gyroscopic couple on a naval ship during steeringtaking left or right turn may be obtained in the similar way as for an aeroplane as discussed in Art.14.4.
Fig. 14.8.   Naval ship taking a left turn.
When the rotor of the ship rotates in the clockwise direction when viewed from the stern, it will have

its angular momentum vector in the direction ox as shown in Fig. 14.9 ( a). As the ship steers to the
left, the active gyroscopic couple will change the angular momentum vector from ox to ox′. The
vector xx′ now represents the active gyroscopic couple and is perpendicular to ox. Thus the plane of
active gyroscopic couple is perpendicular to xx′ and its direction in the axis OZ for left hand turn is

clockwise as shown in Fig. 14.8. The reactive gyroscopic couple of the same magnitude will act in the

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  489
opposite direction ( i.e. in anticlockwise direction). The  effect of this reactive gyroscopic couple is to
raise the bow and lower the stern.
Notes: 1.  When the ship
steers to the right under simi-lar conditions as discussedabove, the effect of the reac-tive gyroscopic couple, asshown in Fig. 14.9 ( b), will
be to raise the stern and
lower the bow.
2.  When the rotor rates in

the anticlockwise direction,when viewed from the stern and the ship is steering to theleft, then the effect of reactive gyroscopic couple will beto lower the bow  and raise the stern.
3.  When the ship is steering to the right under similar
conditions as discussed in note 2 above, then the effect ofreactive gyroscopic couple will be to raise the bow  and
lower the stern.
4.  When the rotor rotates in the clockwise direction when

viewed from the bow or fore end and the ship is steeringto the left, then the effect of reactive gyroscopic couple will be to raise the stern  and lower the bow.
5.  When the ship is steering to the right under similar conditions as discussed in note 4 above, then the effect of
reactive gyroscopic couple will be to  raise the bow  and lower the stern.
6.  The effect of the reactive gyroscopic couple on a boat propelled by a turbine taking left or right turn is similar
as discussed above.

14.7. Effect of Gyroscopic Couple on a Naval Ship during Pitching
Pitching is the movement of a complete ship up and down in a vertical plane about transverse
axis, as shown in Fig. 14.10 ( a). In this case, the transverse axis is the axis of precession. The pitching
of the ship is assumed to take place with simple harmonic motion i.e. the motion of the axis of spin
about transverse axis is simple harmonic.

Fig. 14.10.  Effect of gyroscopic couple on a naval ship during pitching.
Fig. 14.9.  Effect of gyroscopic couple on a naval ship during steering.

490  /G6C   Theory of Machines
∴∴∴∴∴  Angular displacement of the axis of spin from mean position after time t seconds,
θ=φ  sin ω1. t
where φ=Amplitude of swing i.e. maximum angle turned from the mean
position in radians, and
ω1= Angular velocity of S.H.M.
22rad/s
TimeperiodofS.H.M. in secondsptππ==
Angular velocity of precession,
P1 1 1 (s i n .) c o sddtt
dt dtθω= = φ ω = φ ω ω
The angular velocity of precession will be maximum, if cos ω1.t = 1.
∴  Maximum angular velocity of precession,

ωPmax=φ.ω1 = φ × 2π/tp...(Substituting cos ω1.t = 1)
Let I= Moment of inertia of the rotor in kg-m2, and
ω= Angular velocity of the rotor in rad/s.
∴ Mamimum gyroscopic couple,
Cmax=I. ω. ωPmax
When the pitching is upward, the effect of the reactive gyroscopic couple, as shown in Fig. 14.10
(b),  will try to move the ship toward star-board. On the other hand, if the pitching is downward, the effect

of the reactive gyroscopic couple, as shown in Fig. 14.10 ( c), is to turn the ship towards port side.
Notes : 1.  The effect of the gyroscopic couple is always given on specific position of the axis of spin i.e.
whether it is pitching downwards or upwards.
2.  The pitching of  a ship produces forces on the bearings which act horizontally and perpendicular to
the motion of the ship.
3.  The maximum gyroscopic couple tends to shear the holding-down bolts.

4.  The angular acceleration during pitching,
2
2
11 2–( )s i ndt
dtθα= = φω ω ...Differentiating with respect todtdtθ 
    The angular acceleration is maximum, if sin ω1t = 1.
     ∴  Maximum angular acceleration during pitching,
       αmax = (ω1)2Gryroscopic couple plays its role during ship’s turning and pitching.

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  491
14.8. Effect of Gyroscopic Couple on a Naval Ship during Rolling
We know that, for the effect of gyroscopic couple to occur, the axis of precession should
always be perpendicular to the axis of spin. If, however, the axis of precession becomes parallel to theaxis  of spin, there will be no effect of the gyroscopic couple acting on the body of the ship.

In case of rolling of a ship, the axis of precession ( i.e. longitudinal axis) is always parallel to
the axis of spin for all positions. Hence, there is no effect of the gyroscopic couple acting on the bodyof a ship.
Example 14.4.  The turbine rotor of a ship has a mass of 8 tonnes and a radius of gyration
0.6 m. It rotates at 1800 r.p.m. clockwise, when looking from the st ern. Determine the gyroscopic
couple, if the ship travels at 100 km/hr and steer to the left in a curve of 75 m radius.

Solution. Given:  m = 8 t = 8000 kg ; k = 0.6 m; N = 1800 r.p.m. or ω = 2π × 1800/60
= 188.5 rad/s ; v = 100 km/h = 27.8 m/s; R = 75 m
We know that mass moment of inertia of the rotor,
I=m.k
2 = 8000 (0.6)2 = 2880 kg-m2
ωP=v/R = 27.8/75 = 0.37 rad/s
We know that gyroscopic couple,
C=I.ω.ωP= 2880 × 188.5 × 0.37 = 200 866 N-m
= 200.866 kN-m  Ans.
We have discussed in Art. 14.6, that when the rotor rotates in clockwise direction when

looking from the stern and the ship steers to the left, the effect of the reactive gyroscopic couple is toraise the bow and lower the stern.
Example 14.5.  The heavy turbine

rotor of a sea vessel rotates at 1500 r.p.m.clockwise looking from the stern, its massbeing 750 kg. The vessel pitches with anangular velocity of 1 rad/s. Determine thegyroscopic couple transmitted to the hull whenbow is rising, if the radius of gyration for therotor is 250 mm. Also show in what directionthe couple acts on the hull?
Solution. Given: N = 1500 r.p.m. or
ω = 2π × 1500/60 = 157.1 rad/s; m = 750 kg;
ω
P = 1 rad/s; k = 250 mm = 0.25 m

ω
We know that mass moment of inertia
of the rotor,
I =m.k2 = 750 (0.25)2 = 46.875 kg-m2
∴  Gyroscopic couple transmitted to
the hull (i.e. body of the sea vessel),
C = I.ω.ωP = 46.875 × 157.1 × 1= 7364 N-m = 7.364 kN-mShip’s propeller shown as a separate part. A ship’s
propeller is located at backside (stern) of the ship
below the water surface.

492  /G6C   Theory of Machines
We have discussed in Art. 14.7, that when the bow is rising i.e. when the pitching is upward,
the reactive gyroscopic couple acts in the clockwise direction which moves the sea vessel towardsstar-board.
Example 14.6.   The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration
of 0.45 m and a speed of 3000 r.p. m. clockwise when looking from stern. Determine the gyroscopic
couple and its effect upon the ship:

1.  when the ship is steering to the left on a curve of 100 m radius at a speed of 36 km/h.
2.  when the ship is pitching in a simple harmonic motion, the bow falling with its maximum
velocity. The period of pitching is 40 seconds and the total angular displacement between the twoextreme positions of pitching is 12 degrees.
Solution. Given : m = 3500 kg ; k = 0.45 m; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.2 rad/s
1.  When the ship is steering to the left

Given:       R =100 m; v = km/h = 10 m/s
I=m.k
2 = 3500 (0.45)2 = 708.75 kg-m2
ωP=v/R = 10/100 = 0.1 rad/s
∴  Gyroscopic couple,
C=I.ω.ωP = 708.75 × 314.2 × 0.1 = 22 270 N-m
= 22.27 kN-m Ans.
We have discussed in Art. 14.6, that when the rotor rotates clockwise when looking from the

stern and the ship takes a left turn, the effect of the reactive gyroscopic couple is to raise the bow andlower the stern. Ans.
2.  When the ship is pitching with the bow falling
Given:  t
p = 40 s
Since the total angular displacement between the two extreme positions of pitching is 12°
(i.e. 2φ = 12°), therefore amplitude of swing,
φ= 12/2 = 6° = 6 × π/180 = 0.105 rad
and angular velocity of the simple harmonic motion,
ω1=2 π/tp = 2π/40 = 0.157 rad/s

We know that maximum angular velocity of precession,
ωP=φ.ω1 = 0.105 × 0.157 = 0.0165 rad/s
∴Gyroscopic couple,
C=I.ω.ωP = 708.75 × 314.2 × 0.0165 = 3675 N-m
= 3.675 kN-m Ans.
We have discussed in Art. 14.7, that when the bow is falling ( i.e. when the pitching is down-
ward), the effect of the reactive gyroscopic couple is to move the ship towards port side. Ans.
Example 14.7.  The mass of the turbine rotor of a ship is 20 tonnes and has a radius of

gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6°  below the horizontalposition. A complete oscillation takes 30 seconds and the motion is simple harmonic. Determine thefollowing:

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  493
1. Maximum gyroscopic couple, 2. Maximum angular acceleration of the ship during pitch-
ing, and 3. The direction in which the bow will tend to turn when rising, if the rotation of the rotor is
clockwise when looking from the left.
Solution. Given:  m = 20 t = 20 000 kg ; k = 0.6 m;  N = 2000 r.p.m. or ω = 2π × 2000/60 =
209.5 rad/s; φ = 6° = 6 × π/180 = 0.105 rad; tp = 30 s
1.  Maximum gyroscopic couple

I=m.k2 = 20 000 (0.6)2 = 7200 kg-m2
ω1=2π/tp = 2π/30 = 0.21 rad/s
ωPmax=φ.ω1 = 0.105 × 0.21 = 0.022 rad/s
We know that maximum  gyroscopic couple,
Cmax=I.ω.ωPmax = 7200 × 209.5 × 0.022 = 33 185 N-m
= 33.185 kN-m  Ans.
2.  Maximum angular acceleration during pitching
We know that maximum angular acceleration during pitching

=φ(ω1)2 = 0.105 (0.21)2 = 0.0046 rad/s2
3.  Direction in which the bow will tend to turn when rising
We have discussed in Art. 14.7, that when the rotation of the rotor is clockwise when looking
from the left ( i.e. rear end or stern) and when the bow is rising ( i.e. pitching is upward), then the
reactive gyroscopic couple acts in the clockwise direction which tends to turn the bow towards right(i.e. towards star-board). Ans.

Example 14.8.  A ship propelled by a turbine rotor which has a mass of 5 tonnes and a speed
of 2100 r.p.m. The rotor has a radius of gyration of 0.5 m and rotates in a clockwise direction whenviewed from the stern. Find the gyroscopic effects in the following conditions:
1.  The ship sails at a speed of 30 km/h and steers to the left in a curve having 60 m radius.
2.  The ship pitches 6 degree above and 6 degree below the horizontal position. The bow is

descending with its maximum velocity . The motion due to pitching is simple harmonic and the periodic
time is 20 seconds.
3.  The ship rolls and at a certain instant it has an angular velocity of 0.03 rad/s clockwise
when viewed from stern.
Determine also the maximum angular acceleration during pitching. Explain how the direction
of motion due to gyroscopic effect is determined in each case.
Solution. Given: m = 5 t = 5000 kg; N = 2100 r.p.m. or ω = 2π × 2100/60 = 220 rad/s;
k = 0.5 m

k = 0.5 m
1.  When the ship steers to the left
Given:     v = 30 km / h = 8.33 m/ s ;  R = 60 m
We know that angular velocity of precession,
ω
P=v/R = 8.33/60 = 0.14 rad/s
and mass moment of inertia of the rotor,
I=m.k2 = 5000(0.5)2 = 1250 kg-m2

494  /G6C   Theory of Machines
C=I.ω.ωP = 1250 × 220 × 0.14 = 38500 N-m = 38.5 kN-m
We have discussed in Art. 14.6, that when the rotor in a clockwise direction when viewed
from the stern and the ship steers to the left, the effect of reactive gyroscopic couple is to raise thebow and lower the stern. Ans.
2.  When the ship pitches with the bow descending
Given:  φ = 6° = 6 × π/180 = 0.105 rad/s; t
p = 20 s
We know that angular velocity of simple harmonic motion,

ω1=2π/tp = 2π/20 = 0.3142 rad/s
and maximum angular velocity of precession,
ωPmax=φ.ω1 = 0.105 × 0.3142 = 0.033 rad/s
∴  Maximum gyroscopic couple,
Cmax=I.ω.ωPmax = 1250 × 220 × 0.033 = 9075 N-m
Since the ship is pitching with the bow descending, therefore the effect of this maximum
gyroscopic couple is to turn the ship towards port side. Ans.
3.  When the ship rolls
Since the ship rolls at an angular velocity of 0.03 rad /s, therefore angular velocity of precession
when the ship rolls,

ωP= 0.03 rad /s
C=I.ω.ωP = 1250 × 220 × 0.03 = 8250 N-m
In case of rolling of a ship, the axis of precession is always parallel to the axis of spin for all
positions, therefore there is no effect of gyroscopic couple.  Ans.
Maximum angular acceleration during pitching
We know that maximum angular acceleration during pitching.
αmax=φ (ω1)2 = 0.105 (0.3142)2 = 0.01 rad/s2 Ans.

Example 14.9. The turbine rotor of a ship has a mass of 2000 kg and rotates at a speed of
3000 r.p.m. cl ockwise when looking from a stern. The radius of gyration of the rotor is 0.5 m .
Determine the gyroscopic couple and its effects upon the ship when the ship is steering to the
right in a curve of 100 m radius at a speed of 16.1 knots (1 knot = 1855 m/hr) .

Calculate also the torque and its effects when the ship is pitching in simple harmonic motion,
the bow falling with its maximum velocity. The period of pitching is 50 seconds and the total angulardisplacement between the two extreme positions of pitching is 12°. Find the maximum accelerationduring pitching motion .
Solution. Given : m = 2000 kg ; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.2 rad/s;
k = 0.5 m; R = 100 m ; v = 16.1 knots = 16.1 × 1855 / 3600 = 8.3 m/s
Gyroscopic couple

I=m.k
2 = 2000 (0.5)2 = 500 kg-m2
ωP=v/R = 8.3/100 = 0.083 rad /s

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  495
C=I.ω.ωP = 500 × 314.2 × 0.083 = 13 040 N-m = 13.04 kN-m
We have discussed in Art. 14.6, that when the rotor rotates clockwise when looking from a
stern and the ship steers to the right, the effect of the reactive gyroscopic couple is to raise the sternand lower the bow. Ans.
Torque during pitching
Given :  t
p =50 s ; 2 φ = 12° or φ = 6° × π/180 = 0.105 rad

ω1=2π /tp = 2π /50 = 0.1257 rad/s
ωPmax=φ.ω1 = 0.105 × 0.1257 = 0.0132 rad/s
∴ Torque or maximum gyroscopic couple during pitching,
Cmax=I.ω.ωP max = 500 × 314.2 × 0.0132 = 2074 N-m Ans.
We have discussed in Art. 14.7, that when the pitching is downwards, the effect of the reac-
tive gyroscopic couple is to turn the ship towards port side.

Maximum acceleration during pitching
We know that maximum acceleration during pitching
αmax=φ (ω1)2 = 0.105 (0.1257)2 = 0.00166 rad/s2 Ans.
14.9. Stability of a Four Wheel Drive Moving in a Curved Path
Consider the four wheels A, B, C and D of an
automobile locomotive taking a turn towards left as shownin Fig. 14.11. The wheels A and C are inner wheels, whereas
B and D are outer wheels. The centre of gravity ( C.G.) of
the vehicle lies vertically above the road surface.

Letm= Mass of the vehicle in kg,
W= Weight of the vehicle in newtons = m.g,
r
W= Radius of the wheels in metres,
R= Radius of curvature in metres
(R > rW),
h= Distance of centre of gravity, vertically
above the road surface in metres,
x= Width of track in metres,
IW= Mass moment of inertia of one of the
wheels in kg-m2,
ωW= Angular velocity of the wheels or ve-
locity of spin in rad/s,
IE= Mass moment of inertia of the rotating
parts of the engine in kg-m2,

ωE= Angular velocity of the rotating parts of
the engine in rad/s,
G= Gear ratio = ωE /ωW,
v= Linear velocity of the vehicle in m/s = ωW.rWFig. 14.11.   Four wheel drive
moving in  a curved path.

496  /G6C   Theory of Machines
A little considereation will show,
that the weight of the vehicle ( W) will be
equally distributed over the four wheelswhich will act downwards. The reactionbetween each wheel and the road surfaceof the same magnitude will act upwards.Therefore
Road reaction over each wheel
       = W/4 = m.g /4 newtons
Let us now consider the effect of
the gyroscopic couple and centrifugal couple on the vehicle.
1.  Effect of the gyroscopic couple

Since the vehicle takes a turn towards left due to the precession and other rotating parts,
therefore a gyroscopic couple will act.
We know that velocity of precession,
ω
P=v/R
∴  Gyroscopic couple due to 4 wheels,
CW=4  IW.ωW.ωP
and gyroscopic couple due to the rotating parts of the engine,
CE=IE.ωE.ωP = IE.G.ωW.ωP ... (/G8C G = ωE/ωW)
∴  Net gyroscopic couple,
C=CW ± CE = 4 IW.ωW.ωP ± IE.G.ωW.ωP
=ωW.ωP (4 IW ± G.IE)

The positive sign is used when the wheels and rotating parts of the engine rotate in the same
direction. If the rotating parts of the engine revolves in opposite direction, then  negative sign is used.
Due to the gyroscopic couple, vertical reaction on the road surface will be produced. The

reaction will be vertically upwards on the outer wheels and vertically downwards on the inner wheels.Let the magnitude of this reaction at the two outer or inner wheels be P newtons. Then
P × x=C   or   P = C/x
∴  Vertical reaction at each of the outer or inner wheels,
P/2=C/ 2x
Note:     We have discussed above that when rotating parts of the engine rotate in opposite directions, then –ve
sign is used, i.e. net gyroscopic couple,
C=CW – CE

C=CW – CE
When CE > CW, then C will be –ve. Thus the reaction will be vertically downwards on the outer wheels
and vertically upwards on the inner wheels.
2.  Effect of the centrifugal couple
Since the vehicle moves along a curved path, therefore centrifugal force will act outwardly at
the centre of gravity of the vehicle. The effect of this centrifugal force is also to overturn the vehicle.We know that centrifugal force,
2
CmvFR×=

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  497
∴  The couple tending to overturn the vehicle or overturning couple,
2
OC.mvCF h hR=× = ×
This overturning couple is balanced by vertical reactions, which are vertically upwards on
the outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction atthe two outer or inner wheels be Q. Then
Q × x=C
O   or   2
O..
.Cmv hQxR x==
2..
22 .Qm v h

2..
22 .Qm v h
Rx=
∴  Total vertical reaction at each of the outer wheel,
O422WPQP=++
and total vertical reaction at each of the inner wheel,
I422WPQP=−−
A little consideration will show that when the vehicle is running at high speeds, PI may be
zero or even negative. This will cause the inner wheels to leave the ground thus tending to overturnthe automobile. In order to have the contact between the inner wheels and the ground, the sum ofP/2 and Q/2 must be less than W/4.

Example 14.10.  A four-wheeled trolley car of mass
2500 kg runs on rails, which are 1.5 m apart and travelsaround a curve of 30 m radius at  24 km / hr. The rails are at
the same level. Each wheel of the trolley is 0.75 m in diameterand each of the two axles is driven by a motor running in adirection opposite to that of the wheels at a speed of fivetimes the speed of rotation of the wheels. The moment ofinertia of each axle with gear and wheels is 18 kg-m
2. Each

2. Each
motor with shaft and gear pinion has a moment of inertia of12 kg-m
2. The centre of gravity of the car is 0.9 m above the
rail level. Determine the vertical force exerted by each wheelon the rails taking into consideration the centrifugal andgyroscopic effects. State the centrifugal and gyroscopic ef fects
on the trolley .
Solution. Given : m = 2500 kg ; x = 1.5 m ; R = 30 m ;
v = 24 km/h = 6.67 m/s ; d
W = 0.75 m or rW = 0.375 m ; G = ωE/ωW = 5 ; IW = 18 kg-m2 ;

IE = 12 kg-m2 ; h = 0.9 m
The weight of the trolley ( W = m.g) will be equally distributed over the four wheels, which
will act downwards. The reaction between the wheels and the road surface of the same magnitude willact upwards.
∴  Road reaction over each wheel = W/4 = m.g/4 = 2500 × 9.81/4 = 6131.25 N
We know that angular velocity of the wheels,
ω
W=v/rW = 6.67/0.375 = 17.8 rad/s

498  /G6C   Theory of Machines
and angular velocity of precession, ωP=v/R = 6.67/30 = 0.22 rad/s
∴Gyroscopic couple due to one pair of wheels and axle,
CW=2  IW.ωW.ωP = 2 × 18 × 17.8 × 0.22 = 141 N-m
and gyroscopic couple due  to the rotating parts of the motor and gears,
CE=2  IE.ωE.ωP = 2 IE.G.ωW.ωP... (/G51 ωE = G. ωW)
= 2 × 12 × 5 × 17.8 × 0.22 = 470 N-m
∴  Net gyroscopic couple, C=CW – CE = 141 – 470 = – 329 N-m
... (–ve sign is used due to opposite direction of motor)

Due to this net gyroscopic couple, the vertical reaction on the rails will be produced. Since
CE is greater than CW, therefore the reaction will be vertically downwards on the outer wheels and
vertically upwards on the inner wheels. Let the magnitude of this reaction at each of the outer or innerwheel be P/2 newton.
∴ P/ 2 =C/2x = 329 / 2 × 1.5 = 109.7 N
We know that centrifugal force, F
C=m.v2/R = 2500 (6.67)2/30 = 3707 N
∴  Overturning couple, CO=FC × h = 3707 × 0.9 = 3336.3 N-m

This overturning couple is balanced by the vertical reactions which are vertically upwards on
the outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction ateach of the outer or inner wheels be Q/2 newton.
∴ Q/ 2 =C
O / 2x = 3336.3 / 2 × 1.5 = 1112.1 N
We know that vertical force exerted on each outer wheel,
O422WPQP=−+ = 6131.25 – 109.7 + 1112.1 = 7142.65 N  Ans.
and vertical force exerted on each inner wheel,

I422WPQP=+− = 6131.25 + 109.7 – 1112.1 = 5128.85 N  Ans.
Example 14.11.  A rear engine automobile is travelling along a track of 100 metres mean
radius. Each of the four road wheels has a moment of inertia of 2.5 kg-m2 and an effective diameter
of 0.6 m. The rotating parts of the engine have a moment of inertia of 1.2 kg-m2. The engine axis is

parallel to the rear axle and the crankshaft rotates in the same sense as the road wheels. The ratio ofengine speed to back axle speed is 3 : 1. The automobile has a mass of 1600 kg and has its centre ofgravity 0.5 m above road level. The width of the track of the vehicle is 1.5 m.
Determine the limiting speed of the vehicle around the curve for all four wheels to maintain

contact with the road surface. Assume that the road surface is not cambered and centre of gravity ofthe automobile lies centrally with respect to the four wheels .
Solution. Given : R = 100 m ; I
W = 2.5 kg-m2 ; dW = 0.6 m or rW = 0.3 m ; IE = 1.2 kg-m2;
G = ωE/ωW = 3 ; m = 1600 kg ; h = 0.5 m ; x = 1.5 m
The weight of the vehicle ( m.g) will be equally distributed over the four wheels which will act

downwards. The reaction between the wheel and the road surface of the same magnitude will actupwards.
∴  Road reaction over each wheel
=W/4 = m.g / 4 = 1600 × 9.81/4 = 3924 N

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  499
Let v= Limiting speed of the vehicle in m/s.
W
W3.33 rad/s0.3vvvrω= = =
P 0.01 rad/s100vvvRω= = =
CW=4 IW.ωW.ωP = 242 . 5 0 . 3 3N - m0.3 100vvv ×× × =
and gyroscopic couple due to rotating parts of the engine,
CE=IE.ωE.ωP = IE.G.ωW.ωP
= 1.2 × 3 × 3.33 v × 0.01v = 0.12 v2 N-m
∴  Total gyroscopic couple,

C=CW + CE = 0.33 v2 + 0.12 v2 = 0.45 v2 N-m
Due to this gyroscopic couple, the vertical reaction on the rails will be produced. The reac-
tion will be vertically upwards on the outer wheels and vertically downwards on the inner wheels. Letthe magnitude of this reaction at each of the outer or inner wheel be P/2 newtons.
∴ P/2 =C/2x = 0.45v
2/2 × 1.5 = 0.15 v2 N
We know that centrifugal force,
FC=m.v2/R = 1600 × v2/100 = 16 v2 N

∴Overturning couple acting in the outward direction,
CO=FC × h = 16 v2 × 0.5 = 8 v2 N-m
This overturning couple is balanced by vertical reactions which are vertically upwards on the
outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction at eachof the outer or inner wheels be Q/2 newtons.
∴ Q / 2 =C
O / 2x = 8 v2/ 2 × 1.5 = 2.67 v2 N
We know that total vertical reaction at each of the outer wheels,
O422WPQP=++ ...(i)

and total vertical reaction at each of the inner wheels,
I–– –4224 22WPQ W PQP== + ...(ii)
From equation (i), we see that there will always be contact between the outer wheels and
the road surface because W/4, P/2 and Q/2 are vertically upwards. In order to have contact between
the inner wheels and road surface, the reactions should also be vertically upwards, which is onlypossible if
224PQW+≤

500  /G6C   Theory of Machines
i.e. 0.15 v2 + 2.67 v2 ≤ 3924       or          2.82 v2 ≤ 3924
∴ v2 ≤ 3924/2.82 = 1391.5
or v ≤ 37.3 m/s = 37.3 × 3600 / 1000 = 134.28 km/h Ans.
Example 14.12.  A four wheeled motor car of mass 2000 kg has a wheel base 2.5 m, track
width 1.5 m and height of centre of gravity 500 mm above the ground level and lies at 1 metre fromthe front axle. Each wheel has an ef fective diameter of 0.8 m and a moment of inertia of 0.8 kg-m
2.

2.
The drive shaft, engine flywheel and transmission are rotating at 4 times the speed of road wheel, ina clockwise direction when viewed from the front, and is equivalent to a mass of 75 kg having aradius of gyration of 100 mm. If the car is taking a right turn of 60 m radius at 60 km/h, find the loadon each wheel .
Solution. Given : m = 2000 kg : b = 2.5 m ; x = 1.5 m ; h = 500 mm = 0.5 m ; L = 1 m ; d
W =

W =
0.8 m or rW = 0.4 m ; IW = 0.8 kg-m2 ; G = ωE / ωW = 4 ; mE = 75 kg ; kE = 100 mm = 0.1 m ;
R = 60 m ; v = 60 km/h = 16.67 m/s
Since the centre of gravity of the car lies at 1 m from the front axle and the weight of the car
(W = m.g) lies at the centre of gravity, therefore weight on the front wheels and rear wheels will be
different.
Let W1= Weight on the front wheels, and
W2= Weight on the rear wheels.
Taking moment about the front wheels,

W2 × 2.5 =W × 1 = m.g × 1 = 2000 × 9.81 × 1 = 19 620
∴ W2= 19 620 / 2.5 = 7848 N
We know that weight of the car or on the four wheels,
W=W1 + W2 = m.g = 2000 × 9.81 = 19 620 N
or W1=W – W2 = 19 620 – 7848 = 11 772 N
∴  Weight on each of the front wheels
=W1 / 2 = 11 772 / 2 = 5886 N
and weight on each of the rear wheels
=W2 /2 = 7874 / 2 = 3924 N
Since the weight of the car over the four wheels will act

downwards, therefore the reaction between each wheel and theroad surface of the same magnitude will act upwards as shownin Fig. 14.12.
Let us now consider the effect of gyroscopic couple
due to four wheels and rotating parts of the engine.
We know angular velocity of wheels,
ω
W=v/rW = 16.67 / 0.4 = 41.675 rad /s
ωP=v/R = 16.67 / 60 = 0.278 rad /sFig. 14.12

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  501
∴  Gyroscopic couple due to four wheels,
= 4 × 0.8 × 41.675 × 0.278 = 37.1 N-m
This gyroscopic couple tends to lift the inner wheels and to press the outer wheels. In other
words, the reaction will be vertically downward on the inner wheels ( i.e. wheels 1 and 3) and verti-
cally upward on the outer wheels ( i.e. wheels 2 and 4) as shown in Fig. 14.12. Let P/2 newtons be the

magnitude of this reaction at each of the inner or outer wheel.
∴ P / 2 =CW / 2x = 37.1 / 2 × 1.5 = 12.37 N
We know that mass moment of inertia of rotating parts of the engine,
IE=mE (kE)2 = 75 (0.1)2 = 0.75 kg-m2...(∵ I = m.k2)
∴  Gyroscopic couple due to rotating parts of the engine,
CE=IE.ωE.ωP = mE (kE)2 G. ωW.ωP
= 75 (0.1)2 4 × 41.675 × 0.278 = 34.7 N-m
This gyroscopic couple tends to lift the front wheels and to press the outer wheels. In other

words, the reaction will be vertically downwards on the front wheels and vertically upwards on therear wheels as shown in Fig. 14.12. Let F/2 newtons be the magnitude of this reaction on each of the
front and rear wheels.
∴ F / 2 =C
E / 2b = 34.7/2 × 2.5 = 6.94 N
Now let us consider the effect of centrifugal couple acting on the car. We know that centrifugal
force,
FC=m.v2 / R = 2000 (16.67)2/ 60 = 9263 N
∴  Centrifugal couple tending to overturn the car or over turning couple,

CO=FC × h = 9263 × 0.5 = 4631.5 N-m
This overturning couple tends to reduce the pressure on the inner wheels and to increase on
the outer wheels. In other words, the reactions are vertically downward on the inner wheels andvertically upwards on the outer wheels. Let Q/2 be the magnitude of this reaction on each of the inner
and outer wheels.
∴ Q / 2 =C
O / 2x = 4631.5 / 2 × 1.5 = 1543.83 N
From Fig. 14.12, we see thatLoad on the front wheel 1

1–––2222W PFQ=  = 5886 – 12.37 – 6.94 – 1543.83 = 4322.86 N  Ans.
Load on the front wheel 2
1–2222W PFQ=+ +  = 5886 + 12.37 – 6.94 + 1543.83 = 7435.26 N Ans.
Load on the rear wheel 3
2
2222WP F Q=− + −  = 3924 – 12.37 + 6.94 – 1543.83 = 2374.74 N Ans.
Load on the rear wheel 4
2
2222W PFQ=+ + +  = 3924 + 12.37 + 6.94 + 1543.83 = 5487.14 N Ans.

502  /G6C   Theory of Machines
Example 14.13.  A four-wheeled trolley car of total mass 2000 kg running on rails of 1.6 m

gauge, rounds a curve of 30 m radius at 54 km/h. The track is banked at 8°. The wheels have anexternal diameter of 0.7 m and each pair with axle has a mass of 200 kg. The radius of gyration foreach pair is 0.3 m. The height of centre of gravity of the car above the wheel base is 1 m. Determine,allowing for centrifugal force and gyroscopic couple actions, the pressure on each rail .
Solution. Given : m = 2000 kg ; x = 1.6 m ; R = 30 m ; v = 54 km/ h = 15 m / s ; θ = 8° ;
d

d
W = 0.7 m or rW = 0.35 m ; m1 = 200 kg ; k = 0.3 m ; h = 1 m
First of all, let us find the reactions RA and RB at the wheels A and B respectively. The various
forces acting on the trolley car are shown in Fig. 14.13.
Resolving the forces perpendicular to the track,
RA + RB=W cos θ + FC sin θ = m.g cos θ + 2.sinmv
Rθ
22000 (15)2000 9.81cos 8 sin 830=× ° + × °
= 19 620 × 0.9903 + 15 000 × 0.1392 = 21 518 N
Fig. 14.13
Now taking moments about B,

Fig. 14.13
RA × x=(W cos θ + FC sin θ) 2x+ W sin θ × h – FC cos θ × h
∴22
A.1 .. cos sin . sin – cos2mv mv hRm g m gRR x =θ +θ +θ θ   
22000 (15) 12000 9.81cos 8 sin 830 2=× ° + °
22000 (15) 12000 9.81sin 8 cos 830 1.6+× ° − °
=  (19 620 × 0.9903 + 15 000 × 0.1392) 1
2
+ (19 620 × 0.1392 – 15 000 × 0.9903) 1
1.6

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  503
11(19 430 2088) (2731 – 14 855)21 . 6=+ +
= 10 759 – 7577 = 3182 N
∴ RB=(RA + RB) – RA = 21 518 – 3182 = 18 336 N
We know that angular velocity of wheels,
W
W1542.86 rad/s0.35v
rω= = =
P150.5 rad/s
30v
Rω= = =
C=* I ωW cos θ × ωP = mI.k2.ωW cos θ.ωP...(∵ I = mI.k2)
= 200 (0.3)2 42.86 cos 8° × 0.5 = 382 N-m

Due to this gyroscopic couple, the car will tend to overturn about the outer wheels. Let P be
the force at each pair of wheels or each rail due to the gyroscopic couple,
∴ P=C / x = 382 / 1.6 = 238.75 N
We know that pressure (or total reaction) on the inner rail,
PI=RA – P = 3182 – 238.75 = 2943.25 N Ans.
and pressure on the outer rail,
PO=RB + P = 18 336 + 238.75 = 18 574.75  N Ans.

Example 14.14.  A pair of locomotive driving wheels with the axle, have a moment of inertia
of 180 kg-m2. The diameter of the wheel treads is 1.8 m and the distance between wheel centres is

1.5 m. When the locomotive is travelling on a level track at 95 km/h, defective ballasting causes onewheel to fall 6 mm and to rise again in a total time of 0.1 s. If the displacement of the wheel takesplace with simple harmonic motion, find : 1. The gyroscopic couple set up, and 2. The reaction
between the wheel and rail due to this couple .
Solution. Given : I = 180 kg-m
2 ; D = 1.8 m or R = 0.9 m ; x = 1.5 m ; v = 95 km/ h = 26.4 m/s
1.  Gyroscopic couple set up

We know that angular velocity of the locomotive,
ω=v/R = 26.4/ 0.9 = 29.3 rad /s
Since the defective ballasting causes one wheel to fall 6 mm and to rise again in a total time
(t) of 0.1 s, therefore
         Amplitude,111Fall Rise 6 3 mm222A=== × =
and maximum velocity while falling,
223 118.5 mm/s 0.1885 m/s
0.1maxvA
tππ=× =× = =
∴  Maximum angular velocity of tilt of the axle or angular velocity of precession,
           P0.18850.126 rad/s
1.5max
maxv
xω= = =

P0.18850.126 rad/s
1.5max
maxv
xω= = =
*Angular momentum about axle = I.ωW
∴  Angular momentum about horizontal = I.ωW cos θ

504  /G6C   Theory of Machines
We know that gyroscopic couple set up,
C = I.ω.ωPmax = 180 × 29.3 × 0.126 = 664.5  N-m Ans.
The gyroscopic couple will act in a horizontal plane and this couple will tend to produce
swerve i.e. it tends to turn the locomotive aside.
2.  Reaction between the wheel and rail due to the gyroscopic couple
We know that the reaction between the wheel and rail due to the gyroscopic couple is
P=C / x = 664.5 / 1.5 =  443 N Ans.

14.10. Stability of a Two Wheel Vehicle Taking a Turn
Consider a two wheel vehicle (say a scooter or motor cycle) taking a right turn as shown in
Fig. 14.14 ( a).
(a)( b)( c)
Fig. 14.14.   Stability of a two wheel vehicle taking a turn.
Let m= Mass of the vehicle and its
rider in kg,
W= Weight of the vehicle and
its rider in newtons = m.g,
h= Height of the centre of
gravity of the vehicle andrider,
r
W= Radius of the wheels,
R= Radius of track or
curvature,

IW= Mass moment of inertia of each wheel,
IE= Mass moment of inertia of the rotating parts of the engine,
ωW= Angular velocity of the wheels,
ωE= Angular velocity of the engine,
G= Gear ratio = ωE / ωW,
Motorcycle taking a turn.

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  505
v= Linear velocity of the vehicle = ωW × rW,
θ= Angle of heel. It is inclination of the vehicle to the vertical for equilibrium.
Let us now consider the effect of the gyroscopic couple and centrifugal couple on the vehicle,
as discussed below.
1.  Effect of gyroscopic couple
We know that v=ωW × rW           or       ωW = v / rW
and EW
W.vGGrω= ω = ×
∴  Total ( I × ω)=2  IW × ωW ± IE × ωE
WE W E

WE W E
WW W2( 2 . )vv vII G I G Irr r=× ± × × = ±
and velocity of precession, ωP = v /R
A little consideration will show that when the wheels move over the curved path, the vehicle
is always inclined at an angle θ with the vertical plane as shown in Fig. 14.14 ( b). This angle is known
as angle of heel . In other words, the axis of spin is inclined to the horizontal at an angle θ, as shown

in Fig. 14.14 ( c). Thus the angular momentum vector Iω due to spin is represented by OA inclined to
OX at an angle θ. But the precession axis is vertical. Therefore the spin vector is resolved along OX.
1P W E
W.c o s ( 2 .) c o svvCI I G IrR=ω θ × ω= ± θ ×
2
WE
W(2 . ) cos.vIG IRr=± θ
Notes : (a)  When the engine is rotating in the same direction as that of wheels, then the positive sign is used in

the above expression and if the engine rotates in opposite direction, then  negative sign is used.
(b)  The gyroscopic couple will act over the vehicle outwards i.e. in the anticlockwise direction
when seen from the front of the vehicle. The tendency of this couple is to overturn the vehicle in outwarddirection.
An aircraft of 1920’s model.

506  /G6C   Theory of Machines
2.  Effect of centrifugal couple
2
C.mvFR=
This force acts horizontally through the centre of gravity ( C.G.) along the outward direction.
∴  Centrifugal couple,
2
2C.cos cosmvCFh hR=× θ = θ 
Since the centrifugal couple has a tendency to overturn the vehicle, therefore
Total overturning couple,
CO= Gyroscopic couple + Centrifugal couple
()22
WE
W.2. c o s c o s.vm vIG I hRr R=+ θ + × θ
2
WE
W2..c o sIG IvmhRr+=+ θ

2
WE
We know that balancing couple = m.g.h sin θ
The balancing couple acts in clockwise direction when seen from the front of the vehicle.
Therefore for stability, the overturning couple must be equal to the balancing couple, i.e.
     2
WE
W2..c o s . . s i nIG Ivmh mghRr++θ =θ 

From this expression, the value of the angle of heel ( θ) may be determined, so that the vehicle
does not skid.

Example 14.15.  Find the angle of inclination with respect to the vertical of a two wheeler
negotiating a turn. Given : combined mass of the vehicle with its rider 250 kg ; moment of inertia ofthe engine flywheel 0.3 kg-m
2 ; moment of inertia of each road wheel 1 kg-m2 ; speed of engine
flywheel 5 times that of road wheels and in the same direction ; height of centre of gravity of riderwith vehicle 0.6 m ; two wheeler speed 90 km/h ; wheel radius 300 mm ; radius of turn 50 m.

Solution. Given : m = 250 kg ; I
E = 0.3 kg-m2 ; IW = 1 kg-m2 ; ωE = 5 ωW or E
W5; Gω==ωh = 0.6 m ; v = 90 km/h = 25 m/s ; rW = 300 mm = 0.3 m ; R = 50 m
Let θ= Angle of inclination with respect to the vertical of a two wheeler.
22
1W E
W(25)(2 . ) cos (2 1 5 0.3) cos50 0.3vCI G IRr=+ θ = × + × θ××
= 146 cos θ N-m
and centrifugal couple,22
2. 250 (25)cos 0.6 cos50mvChR=×θ = × θ = 1875 cos θ N-m
∴  Total overturning couple,

=C1 + C2 = 146 cos θ + 1875 cos θ = 2021 cos θ N-m
We know that balancing couple
=m.g.h sin θ = 250 × 9.81 × 0.6 sin θ = 1471.5 sin θ N-m

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  507
Since the overturning couple must be equal to the balancing couple for equilibrium condi-
tion, therefore
2021 cos θ= 1471.5 sin θ
∴ tan θ=s i n  θ / cos θ = 2021 / 1471.5 = 1.3734 or θ = 53.94° Ans.
Example 14.16.  A gyrowheel D of mass 0.5 kg, with a radius

of gyration of 20 mm, is mounted in a pivoted frame C as shown inFig. 14.15. The axis AB of the pivots passes through the centre ofrotation O of the wheel, but the centre of gravity G of the frame C is10 mm below O. The frame has a mass of 0.30 kg and the speed ofrotation of the wheel is 3000 r.p.m. in the anticlockwise direction asshown.
The entire unit is mounted on a vehicle so that the axis AB is

parallel to the direction of motion of the vehicle. If the vehicle travelsat 15 m/s in a curve of 50 metres radius, find the inclination of thegyrowheel from the vertical, when
1. The vehicle moves in the direction of the arrow ‘X’ taking
a left hand turn along the curve, and
2. The vehicle reverse at the same speed in the direction of arrow ‘Y’ along the same path.
1 = 0.5 kg ; k = 20 mm = 0.02 m ; OG = h = 10 mm = 0.01 m ;

m2 = 0.3 kg ; N = 3000 r.p.m. or ω = 2 π × 3000 / 60 = 314.2 rad/s ; v = 15 m/s ; R = 50 m
We know that mass moment of inertia of the gyrowheel,
I=m1.k2 = 0.5 (0.02)2 = 0.0002 kg-m2
ωP=v/R = 15 / 50 = 0.3 rad /s
Let θ= Angle of inclination of gyrowheel from the vertical.
1.  When the vehicle moves in the direction of arrow X taking a left turn along the curve
We know that gyroscopic couple about O,
C1=I ω.ωP cos θ = 0.0002 × 314.2 × 0.3 cos θ N-m

= 0.019 cos θ N-m (anticlockwise)
and centrifugal couple about O,
2 2
2
2. 0.3 (15)cos 0.01cos N-m50mvChR=× θ = × θ
= 0.0135 cos θ N-m (anticlockwise)
∴  Total overturning couple
=C1 – C2 = 0.019 cos θ – 0.0135 cos θ
... (– ve sign due to opposite direction)
= 0.0055 cos θ N-m (anticlockwise)
We know that balancing couple due to weight ( W2 = m2.g) of the frame about O,
=m2.g.h sin θ = 0.3 × 9.81 × 0.01 sin θ N-m

= 0.029 sin θ N-m (clockwise)Fig. 14.15

508  /G6C   Theory of Machines
0.0055 cos θ= 0.029 sin θ
or tan θ=s i n  θ / cos θ = 0.0055 / 0.029 = 0.1896
∴θ = 10.74° Ans.
Fig. 14.16
2.  When the vehicle reverses at the same speed in the direction of arrow Y along the same path
When the vehicle reverses at the same speed in the direction of arrow Y, then the gyroscopic

and centrifugal couples ( C1 and C2) will be in clockwise direction about O and the balancing couple
due to weight ( W2 = m2.g) of the frame about O will be in anticlockwise direction.
=C1 + C2 = 0.019 cos θ + 0.0135 cos θ = 0.0325 cos θ N-m
Equating the total overturning couple to the balancing couple, we have
0.0325 cos θ= 0.029 sin θ
or tan θ=s i n  θ / cos θ = 0.0325 / 0.029 = 1.1207
∴θ = 48.26° Ans.

14.11. Effect of Gyroscopic Couple on a Disc Fixed Rigidly at a Certain
Angle to a Rotating Shaft
Consider a disc fixed rigidly to a rotating shaft such that the polar axis of the disc makes an
angle θ with the shaft axis, as shown in Fig. 14.17. Let the shaft rotates with an angular velocity ω

rad/s in the clockwise direction when viewed from the front. A little consideration will show that thedisc will also rotate about OX with the same angular velocity ω rad/s. Let OP be the polar axis and
OD the diametral axis of the disc.
Fig. 14.17.   Effect of gyroscopic couple on a disc fixed rigidly at a certain angle to a rotating shaft.

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  509
∴  Angular velocity of the disc about the polar axis OP or the angular velocity of spin
=ω cos θ ... (Component of ω in the direction of OP)
Since the shaft rotates, therefore the point P will move in a plane perpendicular to the plane
of paper. In other words, precession is produced about OD.
∴  Angular velocity of the disc about the diametral axis OD or the angular velocity of precession
=ω sin θ

=ω sin θ
If IP is the mass moment of inertia of the disc about the polar axis OP, then gyroscopic couple
acting on the disc,
CP=IP.ω cos θ.ω sin θ = 1
2 × IP.ω2 sin 2θ
... (∵ 2 sin θ cos θ = sin 2θ)
The effect of this gyroscopic couple is to turn the disc in the anticlockwise when viewed from
the top, about an axis through O in the plane of paper.
Now consider the movement of point D about the polar axis OP. In this case, OD is axis of
spin and OP is the axis of precession.

∴  Angular velocity of disc about OD or angular velocity of spin
=ω sin θ
and angular velocity of D about OP or angular velocity of precession
=ω cos θ
If ID is the mass moment of inertia of the disc about the diametral axis OD, then gyroscopic
couple acting on the disc,
CD=ID.ω sin θ.ω cos θ = 1
2× ID.ω2 sin 2θ
The effect of this couple will be opposite to that of CP.
∴  Resultant gyroscopic couple acting on the disc,
C=CP – CD = 1
2× ω2 sin 2θ (IP – ID)

C=CP – CD = 1
This resultant gyroscopic couple will act in the anticlockwise direction as seen from the top.
In other words, the shaft tends to turn in the plane of paper in anticlockwise direction as seen from thetop, as a result the horizontal force is exerted on the shaft bearings.
Notes: 1.  The mass moment of inertia of the disc about polar axis OP,
IP=m.r2/2
and mass moment of inertia of the disc about diametral axis OD,
22
D12 4lrIm=+
where m= Mass of disc,

22
r= Radius of disc, and
l= Width of disc.
2.  If the disc is thin, l may be neglected. In such a case
ID=m.r2/4
∴22
22 21. .sin 2 – . sin 2
22 4 8mr mr mCr=× ω θ =× ω θ 

510  /G6C   Theory of Machines
Example 14.17.  A shaft carries a uniform thin disc of 0.6 m diameter and mass 30 kg. The
disc is out of truth and makes an angle of 1° with a plane at right angles to the axis of the shaft. Findthe gyroscopic couple acting on the bearing when the shaft rotates at 1200 r.p.m.
Solution. Given : d = 0.6 m or r = 0.3 m , m = 30 kg ; θ = 1° ; N = 1200 r.p.m. or
ω = 2 π × 1200/60 = 125.7 rad /s
We know that gyroscopic couple acting on the bearings,

22 2 2 30. sin 2 (125.7) (0.3) sin 2
88mCr=× ω θ = °  = 186 N-m Ans.
EXERCISES
1.A flywheel of mass 10 kg and radius of gyration 200 mm is spinning about its axis, which is horizontal
and is suspended at a point distant 150 mm from the plane of rotation of the flywheel. Determine theangular velocity of precession of the flywheel. The spin speed of flywheel is 900 r.p.m.
[Ans. 0.39 rad/s]
2.A horizontal axle AB, 1 m long, is pivoted at the mid point C. It carries a weight of 20 N at A and a

wheel weighing 50 N at B. The wheel is made to spin at a speed of 600 r.p.m in a clockwise direction
looking from its front. Assuming that the weight of the flywheel is uniformly distributed around therim whose mean diameter is 0.6 m, calculate the angular velocity of precession of the system aroundthe vertical axis through C. [Ans. 0.52 rad/s]
3.An aeroplane runs at 600 km/ h. The rotor of the engine weighs 4000 N with radius of gyration of

1 metre. The speed of rotor is 3000 r.p.m. in anticlockwise direction when seen from rear side of theaeroplane.
If the plane takes a loop upwards in a curve of 100 metres radius, find : 1. gyroscopic couple devel-
oped; and 2. effect of reaction gyroscopic couple developed on the body of aeroplane.
[Ans. 213.5 kN-m]
4.An aeroplane makes a complete half circle of 50 metres radius, towards left, when flying at 200 km

per hour. The rotary engine and the propeller of the plane has a mass of 400 kg with a radius ofgyration of 300 mm. The engine runs at 2400 r.p.m. clockwise, when viewed from the rear. Find thegyroscopic couple on the aircraft and state its effect on it. What will be the effect, if the aeroplaneturns to its right instead of to the left ? [Ans. 10 kN-m]
5.Each paddle wheel of a steamer have a mass of 1600 kg and a radius of gyration of 1.2 m. The steamer

turns to port in a circle of 160 m radius at 24 km / h, the speed of the paddles being 90 r.p.m. Find themagnitude and effect of the gyroscopic couple acting on the steamer. [Ans. 905.6 N-m]
6.The rotor of the turbine of a yacht makes 1200 r.p.m. clockwise when viewed from stern. The rotor has
a mass of 750 kg and its radius of gyration is 250 mm. Find the maximum gyroscopic couple transmit-

ted to the hull (body of the yacht) when yacht pitches with maximum angular velocity of 1 rad /s. What isthe effect of this couple ? [Ans. 5892 N-m]
7.The rotor of a turbine installed in a boat with its axis along the longitudinal axis of the boat makes

1500 r.p.m. clockwise when viewed from the stern. The rotor has a mass of 750 kg and a radius ofgyration of 300 mm. If at an instant, the boat pitches in the longitudinal vertical plane so that the bowrises from the horizontal plane with an angular velocity of 1 rad /s, determine the torque acting on theboat and the direction in which it tends to turn the boat at the instant. [Ans. 10.6 kN-m]

8.The mass of a turbine rotor of a ship is 8 tonnes and has a radius of gyration 0.6 m. It rotates at 1800
r.p.m. clockwise when looking from the stern. Determine the gyroscopic effects in the followingcases:
1. If the ship travelling at 100 km / h strees to the left in a curve of 75 m radius, 2. If the ship is pitching

and the bow is descending with maximum velocity. The pitching is simple harmonic, the periodic timebeing 20 seconds and the total angular movement between the extreme positions is 10°, and 3. If theship is rolling and at a certain instant has an angular velocity of 0.03 rad/s clockwise when lookingfrom stern.

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  511
In each case, explain clearly how you determine the direction in which the ship tends to move as a
result of the gyroscopic action. [Ans. 201 kN-m ; 14.87 kN-m ; 16.3 kN-m]
9.The turbine rotor of a ship has a mass of 20 tonnes and a radius of gyration of 0.75 m. Its speed is 2000

r.p.m. The ship pitches 6° above and below the horizontal position. One complete oscillation takes 18seconds and the motion is simple harmonic. Calculate :
1. the maximum couple tending to shear the holding down bolts of the turbine, 2. the maximum
angular acceleration of the ship during pitching, and 3. the direction in which the bow will tend to turnwhile rising, if the rotation of the rotor is clockwise when looking from rear.
[Ans. 86.26 kN-m ; 0.0128 rad /s
2, towards star-board]

10.A motor car takes a bend of 30 m radius at a speed of 60 km / hr. Determine the magnitudes of
gyroscopic and centrifugal couples acting on the vehicle and state the effect that each of these has onthe road reactions to the road wheels. Assume that :
Each road wheel has a moment of inertia of 3 kg-m
2 and an effective road radius of 0.4 m.
The rotating parts of the engine and transmission are equivalent to a flywheel of mass 75 kg with a radius

of gyration of 100 mm. The engine turns in a clockwise direction when viewed from the front.
The back-axle ratio is 4 : 1, the drive through the gear box being direct. The gyroscopic effects of the
half shafts at the back axle are to be ignored.
The car has a mass of 1200 kg and its centre of gravity is 0.6 m above the road wheel.The turn is in a right hand direction.If the turn has been in a left hand direction, all other details being unaltered, which answers, if any,

need modification. [Ans. 347.5 N-m : 6670 N-m]
11.A rail car has a total mass of 4 tonnes. There are two axles, each of which together with its wheels and
gearing has a total moment of inertia of 30 kg-m
2. The centre distance between the two wheels on an
axle is 1.5 metres and each wheel is of 375 mm radius. E ach axle is driven by a motor , the speed ratio
between the two being 1 : 3. Each motor with its gear has a moment of inertia of 15 kg-m2 and runs

in a direction opposite to that of its axle. The centre of gravity of the car is 1.05 m above the rails.
Determine the limiting speed for this car, when it rounding a curve of 240 metres radius such that no
wheel leaves the rail. Consider the centrifugal and gyroscopic effects completely. Assume that no cantis provided for outer rail. [Ans. 144 km / h]
12.A racing car weighs 20 kN. It has a wheel base of 2 m, track width 1 m and height of C.G. 300 mm

above the ground level and lies midway between the front and rear axle. The engine flywheel rotatesat 3000 r.p.m. clockwise when viewed from the front. The moment of inertia of the flywheel is4 kg-m
2 and moment of inertia of each wheel is 3 kg-m2. Find the reactions between the wheels and
the ground when the car takes a curve of 15 m radius towards right at 30 km / h, taking into consider-ation the gyroscopic and the centrifugal effects. Each wheel radius is 400 mm.

[Ans. Front inner wheel = 3341.7 N ; Front outer wheel = 6309.5 N ;
Rear inner wheel = 3690.5 N ; Rear outer wheel = 6658.3 N]
13.A four wheel trolley car of total mass 2000 kg running on rails of 1 m gauge, rounds a curve of 25 m

radius at 40 km / h. The track is banked at 10°. The wheels have an external diameter of 0.6 m andeach pair of an axle has a mass of 200 kg. The radius of gyration for each pair is 250 mm. The heightof C.G. of the car above the wheel base is 0.95 m. Allowing for centrifugal force and gyroscopiccouple action, determine the pressure on each rail. [Ans. 4328 N ; 16 704 N]
14.A 2.2 tonne racing car has a wheel base of 2.4 m and a track of 1.4 m from the rear axle. The equiva-

lent mass of engine parts is 140 kg with radius of gyration of 150 mm. The back axle ratio is 5. Theengine shaft and flywheel rotate clockwise when viewed from the front. Each wheel has a diameter of0.8 m and a moment of inertia of 0.7 kg-m
2. Determine the load distribution on the wheels when the
car is rounding a curve of 100 m radius at a speed of 72 km/ h to the left.
15.A disc has a mass of 30 kg and a radius of gyration about its axis of symmetry 125 mm while its radius

of gyration about a diameter of the disc at right angles to the axis of symmetry is 75 mm. The disc ispressed on to the shaft but due to incorrect boring, the angle between the axis of symmetry and theactual axis of rotation is 0.25°, though both these axes pass through the centre of gravity of the disc.Assuming that the shaft is rigid and is carried between bearings 200 mm apart, determine the bearingforces due to the misalignment at a speed of 5000 r.p.m. [Ans. 1810 N]

512  /G6C   Theory of Machines
16.A wheel of a locomotive, travelling on a level track at 90 km / h, falls in a spot hole 10 mm deep and
rises again in a total time of 0.8 seconds. The displacement of the wheel takes place with simpleharmonic motion. The wheel has a diameter of 3 m and the distance between the wheel centres is 1.75m. The wheel pair with axle has a moment of inertia of 500 kg-m
2. Determine the magnitude and the
effect of gyrocouple produced in this case. [Ans. 186.6 N-m]

17.Each road wheel of a motor cycle has a mass moment of inertia of 1.5 kg-m2. The rotating parts of the
engine of the motor cycle have a mass moment of inertia of 0.25 kg-m2. The speed of the engine is 5
times the speed of the wheels and is in the same sense. The mass of the motor cycle with its rider is 250kg and its centre of gravity is 0.6 m above the ground level.
Find the angle of heel if the cycle is travelling at 50 km /h and is taking a turn of 30 m radius. The

wheel diameter is 0.6 m. [Ans. 35.7°]
18.A racing motor cyclist travels at 140 km/h round a curve of 120 m radius measured horizontally. The
cycle and rider have mass of 150 kg and their centre of gravity lies at 0.7 m above the ground levelwhen the motor cycle is vertical. Each wheel is 0.6 m in diameter and has moment of inertia about itsaxis of rotation 1.5 kg-m
2. The engine has rotating parts whose moment of inertia about their axis of

rotation is 0.25 kg-m2 and it rotates at five times the wheel speed in the same direction. Find : 1. the
correct angle of banking of the track so that there is no tendency to side slip, and 2. the correct angleof inclination of the cycle and rider to the vertical. [Ans. 52.12°; 55.57°]
[Hint. In calculating the angle of banking of the track, neglect the effect of gyroscopic couple]
1.Write a short note on gyroscope.

2.What do you understand by gyroscopic couple ? Derive a formula for its magnitude.
3.Explain the application of gyroscopic principles to aircrafts.
4.Describe the gyroscopic effect on sea going vessels.
5.Explain the effect of the gyroscopic couple on the reaction of the four wheels of a vehicle negotiating
a curve.
6.Discuss the effect of the gyroscopic couple on a two wheeled vehicle when taking a turn.

7.What will be the effect of the gyroscopic couple on a disc fixed at a certain angle to a rotating shaft ?
1.A disc is spinning with an angular velocity ω rad/s about the axis of spin. The couple applied to the
disc causing precession will be
(a)21.2Iω (b)I.ω2(c) P1..2Iωω (d)I.ω.ωP
where I= Mass moment of inertia of the disc, and
ωP= Angular velocity of precession of the axis of spin.

2.A disc spinning on its axis at 20 rad/s will undergo precession when a torque 100 N-m is applied about
an axis normal to it at an angular speed, if mass moment of inertia of the disc is the 1 kg-m2
(a) 2 rad/s ( b) 5 rad/s ( c) 10 rad/s ( d) 20 rad/s
3.The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane
takes a turn to the left. The effect of the gyroscopic couple on the aeroplane will be

(a) to raise the nose and dip the tail ( b) to dip the nose and raise the tail
(c) to raise the nose and tail ( d) to dip the nose and tail

Chapter 14 : Gyroscopic Couple and Precessional Motion    /G6C  513
4.The air screw of an aeroplane is rotating clockwise when looking from the front. If it makes a left turn,
the gyroscopic effect will
(a) tend to depress the nose and raise the tail
(b) tend to raise the nose and depress the tail
(c) tilt the aeroplane
5.The rotor of a ship rotates in clockwise direction when viewed from the stern and the ship takes a left

turn. The effect of the gyroscopic couple acting on it will be
(a) to raise the bow and stern ( b) to lower the bow and stern
(c) to raise the bow and lower the stern ( d) to lower the bow and raise the stern
6.When the pitching of a ship is upward, the effect of gyroscopic couple acting on it will be
(a) to move the ship towards port side ( b) to move the ship towards star-board
(c) to raise the bow and lower the stern ( d) to raise the stern and lower the bow

7.In an automobile, if the vehicle makes a left turn, the gyroscopic torque
(a) increases the forces on the outer wheels ( b) decreases the forces on the outer wheels
(c) does not affect the forces on the outer wheels
8.A motor car moving at a certain speed takes a left turn in a curved path. If the engine rotates in the
same direction as that of wheels, then due to the centrifugal forces
(a) the reaction on the inner wheels increases and on the outer wheels decreases

(b) the reaction on the outer wheels increases and on the inner wheels decreases
(c) the reaction on the front wheels increases and on the rear wheels decreases
(d) the reaction on the rear wheels increases and on the front wheels decreases
ANSWERS
1.(d) 2.(b) 3.(a) 4.(b)
5.(c) 6.(b) 7.(a) 8.(b)

514  /G6C   Theory of Machines
514Inertia
Forces in
Reciprocating
Parts15
Features
2. Resultant Effect of a System of
Forces Acting on a Rigid Body.
3. D-Alembert’s P rinciple.
4. Velocity and Acceleration of
the Reciprocating Parts inEngines.
5. Klien’s Construction.
6. Ritterhaus’s Construction.
7. Bennett’s Construction.
8. Approximate Analytical
Method for V elocity and
Acceleration of the Piston.
9. Angular Velocity and
Acceleration of the ConnectingRod.

10. Forces on the Reciprocating
Parts of an Engine NeglectingWeight of the Connecting Rod.
11. Equivalent Dynamical System.
12. Determination of Equivalent
Dynamical System of TwoMasses by Graphical Method.
13. Correction Couple to be
Applied to Make the Two MassSystems DynamicallyEquivalent.
14. Inertia Forces in a
Reciprocating EngineConsidering t he Weight of
Connecting Rod.
15. Analytical Method for Inertia
Torque.15.1. Introduction

The inertia force is an imaginary force, which when
acts upon a rigid body, brings it in an equilibrium position. Itis numerically equal to the accelerating force in magnitude,but opposite in direction. Mathematically,
Inertia force   =  – Accelerating force = – m.a
where                       m =  Mass of the body, and
                                 a =   Linear acceleration of the centre
                                          of gravity of the body.

Similarly, the inertia torque is an imaginary torque,
which when applied upon the rigid body, brings it in equilib-rium position. It is equal to the accelerating couple in magni-tude but opposite in direction.
15.2. Resultant Effect of a System of Forces
Acting on a Rigid Body
Consider a rigid body acted upon by a system of
forces. These forces may be reduced to a single resultant force

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  515
F whose line of action is at a distance h from the centre of
gravity G. Now let us assume two equal and opposite forces
(of magnitude F ) acting through G, and parallel to the
resultant force, without influencing the effect of theresultant force F, as shown in Fig. 15.1.
A little consideration will show that the body is
now subjected to a couple (equal to  F × h) and a force,

equal and parallel to the resultant force F passing through
G. The force F through G causes linear acceleration of the
c.g. and the moment of the couple ( F × h) causes angular
acceleration of the body about an axis passing through G
and perpendicular to the point in which the couple acts.
Let α= Angular acceleration of the rigid body due to couple,
h= Perpendicular distance between the force and centre of gravity of the
body,
m= Mass of the body,

body,
k= Least radius of gyration about an axis through G, and
I= Moment of inertia of the body about an axis passing through its centre
of gravity and perpendicular to the point in which the couple acts= m.k
2
           Force,           F= Mass × Acceleration = m.a ...(i)
and                                      F.h= m.k2.α = I.α2...( . )Im k=/G51 ...(ii)
From equations (i) and (ii), we can
find the values of a and α, if the values of F,

m, k, and h are known.
15.3. D-Alembert’s Principle
Consider a rigid body acted upon by
a system of forces. The system may bereduced to a single resultant force acting onthe body whose magnitude is given by theproduct of the mass of the body and the linearacceleration of the centre of mass of the body.According to Newton’s second law ofmotion,
F=m.a ...(i)
where F= Resultant force acting on the body,
m= Mass of the body, and

a= Linear acceleration of the centre of mass of the body.
The equation  (i) may also be written as:
F – m.a= 0 ... (ii)
A little consideration will show, that if the quantity– m.a be treated as a force, equal, oppositeFig. 15.1.  Resultant effect of a system of
forces acting on a rigid body.
The above picture  shows the reciprocating parts
of a 19th century oil engine.

516  /G6C   Theory of Machines
and with the same line of action as the resultant force F, and include this force with the system of
forces of which F is the resultant, then the complete system of forces will be in equilibrium. This
principle is known as D-Alembert’s principle. The equal and opposite force – m.a is known as reversed
effective force  or the inertia force (briefly written as FI). The equation  (ii) may be written as
F + FI= 0 ... (iii)

Thus, D-Alembert’s principle states that the resultant force acting on a body together with
the reversed effective force (or inertia force ), are in equilibrium.
This principle is used to reduce a dynamic problem into an equivalent static problem.
15.4. Velocity and Acceleration of the Reciprocating Parts in Engines
The velocity and acceleration of the reciprocating parts of the steam engine or internal

combustion engine (briefly called as I.C. engine) may be determined by graphical method or analyticalmethod. The velocity and acceleration, by graphical method, may be determined by one of the followingconstructions:
1. Klien’s construction,  2. Ritterhaus’s construction, and 3. Bennett’s construction.
We shall now discuss these constructions, in detail, in the following pages.
15.5. Klien’s Construction
Let OC be the crank and PC the connecting rod of a reciprocating steam engine, as shown in

Fig. 15.2 ( a). Let the crank makes an angle θ with the line of stroke PO and rotates with uniform
angular velocity ω rad/s  in a clockwise direction. The Klien’s velocity and acceleration diagrams are
drawn as discussed below:
             (a) Klien’s acceleration diagram.      ( b) Velocity diagram. ( c) Acceleration diagram.
Fig. 15.2.  Klien’s construction.
Klien’s velocity diagram
First of all, draw OM perpendicular to OP; such that it intersects the line PC produced at M.

The triangle OCM is known as Klien’s velocity diagram . In this triangle OCM,
OM may be regarded as a line perpendicular to PO,
CM may be regarded as a line parallel to PC, and ...( It is thesame line.)/G51
CO may be regarded as a line parallel to CO.
We have already discussed that the velocity diagram for given configuration is a triangle ocp

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  517
as shown in Fig. 15.2 ( b). If this triangle is revolved through 90°, it will be a triangle oc1 p1, in which
oc1 represents vCO (i.e. velocity of C with respect to O or velocity of crank pin C) and is paralel to OC,
op1 represents vPO (i.e. velocity of P with respect to O or velocity of cross-head or piston P)
and is perpendicular to OP, and
c1p1 represents vPC (i.e. velocity of P with respect to C) and is parallel to CP.

A little consideration will show, that the triangles oc1p1 and OCM are similar. Therefore,
11 1 1(a constant)oc op c p
OC OM CM=== ω
or                                    CO PO PCvvv
∴                      vCO  =  ω × OC; vPO = ω × OM, and vPC = ω × CM
Thus, we see that by drawing the Klien’s velocity diagram, the velocities of various points
may be obtained without drawing a separate velocity diagram.
Klien’s acceleration diagram
The Klien’s acceleration dia-

gram is drawn as discussed below:
1. First of all, draw a circle
with C as centre and CM as radius.
2. Draw another circle with
PC as diameter. Let this circle inter-
sect the previous circle at K    and L.
3. Join KL and produce it to
intersect PO at N. Let KL intersect
PC at Q. This forms the quadrilateral
CQNO, which is known as Klien’s
acceleration diagram .
We have already discussed

that the acceleration diagram for the given configuration is as shown in Fig. 15. 2 ( c). We know that
(i)o'c' represents  COra(i.e. radial component of the acceleration of crank pin C with respect
to O ) and is parallel to CO;
(ii)c'x represents PCra(i.e. radial component of the acceleration of crosshead or piston P
with respect to crank pin C) and is parallel to CP or CQ;

(iii)xp' represents PCta(i.e. tangential component of the acceleration of P with respect to C )
and is parallel to QN (because QN is perpendicular to CQ); and
(iv)o'p' represents aPO (i.e. acceleration of P with respect to O or the acceleration of piston
P) and is parallel to PO or NO.
A little consideration will show that the quadrilateral o'c'x p' [Fig. 15.2 ( c)] is similar to
quadrilateral CQNO [Fig. 15.2 ( a)]. Therefore,
2(a constant)oc cx x p op
OC CQ QN NO′′ ′ ′ ′′==== ω

518  /G6C   Theory of Machines
or                              2 CO PC PC POrrtaaaa
OC CQ QN NO==== ω
∴                22
CO PC ;rraO C aC Q=ω × =ω ×
                                22
PC PO ;a n dtaQ N aN O=ω × =ω ×
Thus we see that by drawing the Klien’s acceleration diagram, the acceleration of various
points may be obtained without drawing the separate acceleration diagram.
Notes: 1. The acceleration of piston P with respect to crank pin C (i.e. aPC) may be obtained from:

22 PCorcp a
CN CN′′=ω =ω
∴    aPC  =  ω2 × CN
2. To find the velocity of any point D on the connecting rod PC, divide CM at D1 in the same ratio as D
divides CP. In other words,
                                  1CD CD
CM CP=
∴ Velocity of D, vD = ω × OD1
3. To find the acceleration of any point D on the connecting rod PC, draw a line from a point D parallel
to PO which intersects CN at D2 .
∴  Acceleration of D,   aD  =  ω2 × OD2

4.  If the crank position is such that the point N  lies on the right of O instead of to the left as shown in
Fig. 15.2 ( a), then the acceleration of the piston is negative. In other words, the piston is under going retardation.
5.  The acceleration of the piston P is zero and its velocity is maximum, when N coincides with O.
There is no simple graphical method of finding the corresponding crank position, but it can be shown that for N

and O to coincide, the angle between the crank and the connecting rod must be slightly less than 90°. For most
practical purposes, it is assumed that the acceleration of piston P is zero, when the crank OC and  connecting rod
PC are at right angles to each other.
15.6. Ritterhaus’s Construction
Let OC be the crank and PC the connecting rod of a rciprocating steam engine, as shown in
Fig. 15.3. Let the crank makes an angle θ with the line of stroke PO and rotates with uniform angular

velocity ω rad/s in a clockwise direction. The Ritterhaus’s velocity and acceleration diagrams are
Fig. 15.3.  Ritterhaus’s construction.
Ritterhaus’s velocity diagram
Draw OM perpendicular to the line of stroke PO, such that it intersects the line PC produced
at M. The triangle OCM is known as Ritterhaus’s velocity diagram . It is similar to Klien’s velocity
diagram.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  519
∴  Velocity of C with respect to O or the velocity of crank pin C,
vCO  =  vC = ω × OC
Velocity of P with respect to O or the velocity of crosshead or piston P,
      vPO  =  vP = ω × OM
and velocity of P with respect to C,      vPC  =  ω × CM
Ritterhaus’s acceleration diagram
The Ritterhaus’s acceleration diagram is drawn as discussed below:1.  From point M, draw MK parallel to the line of stroke PO, to interect OC produced at K.

2.  Draw KQ parallel to MO. From Q draw QN perpendicular to PC.
3.  The quadrilateral CQNO is known as Ritterhaus’s acceleration diagram . This is similar
to Klien’s acceleration diagram.
∴ Radial component of the acceleration of C with respect to O or the acceleration of crank
pin C,
                                                  
2
CO Craa O C== ω ×
Radial component of the acceleration of the crosshead or piston P with respect to crank
pin C,

pin C,
                                                  2
PCraC Q=ω ×
Tangential component of the acceleration of P with respect to C,
     2
PCtaQ N=ω ×
and acceleration of P with respect to O or the acceleration of piston P,
2
PO Paa N O== ω ×
Notes : 1. The acceleration of piston P with respect to crank pin C is given by
                                                      aPC  =  ω2 × CN

                                                    1CD CD
CM CP=
∴  Velocity of D                       vD  =  ω × OD1
3. To find the acceleration of any point D on the connecting rod PC, draw DD2 parallel to the line of
stroke PO, which intersects CN at D2. The acceleration of D is given by
      aD  =  ω2 × OD2
15.7. Bennett’s Construction

Let OC be the crank and PC the connecting rod of reciprocating steam engine, as shown in
Fig. 15.4. Let the crank makes an angle θ with the line of stroke PO and rotates with uniform angular
velocity ω rad/s in the clockwise direction. The Bennett’s velocity and acceleration diagrams are
Bennett’s velocity diagram
When the crank OC is at right angle to the line of stroke, it occupies the postition OC1 and the

crosshead P moves to the position P1, as shown in Fig. 15.4. Now, produce PC to intersect OC1 at M.
The triangle OCM is known as Bennett’s velocity diagram . It is similar to Klien’s velocity diagram.

520  /G6C   Theory of Machines
Fig. 15.4.  Bennett’s construction.
                                                vCO  =  vC = ω × OC
    vPO  =  vP = ω × OM
and velocity of  P with respect to C,     vPC = ω × CM
Bennett’s acceleration diagram
The Bennett’s acceleration diagram is drawn as discussed below:1. From O, draw OL

1 perpendicular to P1C1 (i.e. position of connecting rod PC when crank is
at right angle).  Mark the position of point L on the connecting rod PC such that CL = C1L1.
2. From L, draw LK perpendicular to PC and from point K draw KQ perpendicular to the line
of stroke PO. From point C, draw CN perpendicular to the line of stroke PO. Join NQ. A little
consideration will show that NQ is perpendicular to PC.
3. The quadrilateral CQNO is known as Bennett’s acceleration diagram. It is similar to

Klien’s acceleration diagram.
∴  Radial component of the acceleration of C with respect to O or the acceleration of the
crank pin C,
2
pin C,
    2
PCraC Q=ω ×
    2
PCtaQ N=ω ×
    2

2
Notes : 1.  The acceleration of piston  P with respect to crank pin  C is given by
     2
PCaC N=ω ×
2. The velocity and acceleration of any point D on the connecting rod PC may be obtained in the similar
way, as discussed in the previous articles, i.e.
Velocity of D, vD = ω × OD1
and           Acceleration of D, aD = ω2 × OD2

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  521
Example 15.1. The crank and connecting rod of a reciprocating engine are 200 mm and 700
mm respectively. The crank is rotating in clockwise direction at 120 rad/s. Find with the help ofKlein’s construction:  1. Velocity and acceleration of the piston, 2. Velocity and acceleration of the
mid point of the connecting rod, and 3. Angular velocity and angular acceleration of the connecting

rod, at the instant when the crank is at 30° to I.D.C. (inner dead centre ).
Solution. Given: OC = 200 mm = 0.2 m ; PC = 700 mm = 0.7 m ; ω = 120 rad/s
Fig. 15.5
The Klein’s velocity diagram OCM and Klein’s acceleration diagram CQNO as shown in Fig.
15.5 is drawn to some suitable scale, in the similar way as discussed in Art. 15.5. By measurement, wefind that
OM = 127 mm = 0.127 m  ; CM = 173 mm = 0.173 m ; QN = 93 mm = 0.093 m ; NO = 200 mm
= 0.2 m
1. Velocity and acceleration of the piston

We know that the velocity of the piston P,
v
P = ω × OM = 120 × 0.127 = 15.24 m/s   Ans.
and acceleration of the piston P,
                                   aP = ω2 × NO = (120)2 × 0.2 = 2880 m/s2  Ans.
2. Velocity and acceleration of the mid-point of the connecting rod
In order to find the velocity of the mid-point D of the connecting rod, divide CM at D1 in the

same ratio as D divides CP. Since D is the mid-point of CP, therefore D1 is the mid-point of CM, i.e.
CD1 = D1M. Join OD1. By measurement,
                                OD1 = 140 mm = 0.14 m
∴        Velocity of D, vD = ω × OD1 = 120 × 0.14 = 16.8 m/s  Ans.
In order to find the acceleration of the mid-point of the connecting rod, draw a line DD2
parallel to the line of stroke PO which intersects CN at D2. By measurement,
                               OD2 = 193 mm = 0.193 m

∴ Acceleration of D,
                                  aD = ω2 × OD2 = (120)2 × 0.193 = 2779.2 m/s2 Ans.
3. Angular velocity and angular acceleration of the connecting rod
We know that the velocity of the connecting rod PC (i.e. velocity of P with respect to C),
                                 vPC = ω × CM = 120 × 0.173 = 20.76 m/s

522  /G6C   Theory of Machines
∴  Angular acceleration of the connecting rod PC,
                                 PC
PC20.7629.66 rad/s
0.7v
PCω= = = Ans.
We know that the tangential  component of the acceleration of P with respect to C,
                                  22 2
PC (120) 0.093 1339.2 m/staQ N=ω × = × =
∴ Angular acceleration of the connecting rod PC,
   2 PC
PC1339.21913.14 rad/s0.7ta
PCα= = =  Ans.

2 PC
Example 15.2.  In a slider crank mechanism, the length of the crank and connecting rod are
150 mm and 600 mm respectivel y. The crank position is 60° from inner dead centre. The crank shaft
speed is 450 r.p.m. clockw ise. Using Ritterhaus’s c onstruction, determine 1. Velocity and accelera-
tion of the slider, 2. Velocity and acceleration of point D on the connecting rod which is 150 mm from

crank pin C , and 3. angular velocity and angular acceleration of the connecting rod.
Solution. Given: OC = 150 mm = 0.15m; PC = 600 mm = 0.6 m ; CD = 150 mm = 0.15 m;
N = 450 r.p.m. or ω = 2π × 450/60 = 47.13 rad/s
The Ritterhaus’s velocity diagram OCM and acceleration diagram CQNO, as shown in
Fig. 15.6, is drawn to some suitable scale in the similar way as discussed in Art. 15.6. By measure-ment, we find that
OM = 145 mm = 0.145 m ; CM = 78 mm = 0.078 m ; QN = 130 mm = 0.13 m ; and

NO = 56 mm = 0.056 m
Fig. 15.6
1. Velocity and acceleration of the slider
We know that the velocity of the slider P,
P 47.13 0.145 6.834 m/s vO M=ω× = × = Ans.
and acceleration of the slider P,
             22 2
P (47.13) 0.056 124.4 m/s aN O=ω × = × = Ans.
2. Velocity and acceleration of point D  on the connecting rod
In order to find the velocity of point D on the connecting rod, divide CM at D1 in the same
ratio as D divides CP. In other words,
         1
1150or 78 19.5 mm
600CDCD CDCD CM

1
CM CP CP== × = × =
Join OD1. By measurement, OD1 = 145 mm = 0.145 m
∴ Velocity of point D,
           D1 47.13 0.145 6.834 m/s vO D=ω× = × = Ans.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  523
In order to find the acceleration of point D on the connecting rod, draw DD2 parallel to the
line of stroke PO. Join OD2. By measurement, we find that OD2 = 120 mm = 0.12 m.
∴ Acceleration of point D,
            22 2
D2 (47.13) 0.12 266.55 m/s aO D=ω × = × = Ans.
3.  Angular velocity and angular acceleration of the connecting rod
We know that the velocity of the connecting rod PC (or the velocity of point P with respect
to C),

to C),
          PC 47.13 0.078 3.676 m/s vC M=ω× = × =
∴ Angular velocity of the connecting rod,
          PC
PC3.6766.127 rad/s
0.6v
We know that the tangential component of the acceleration of P with respect to C,
           22 2
PC (47.13) 0.13 288.76 m/staQ N=ω × = × =
          2 PC
PC288.76481.27 rad/s0.6ta
PCα= = = Ans.
15.8. Approximate Analytical Method for Velocity and Acceleration of the
Piston

Piston
Consider the motion of a crank and connecting rod of a reciprocating steam engine as shown
in Fig. 15.7. Let OC be the crank and PC the connecting rod. Let the crank rotates with angular
velocity of ω rad/s and the crank turns through an angle θ from the inner dead centre (briefly written
as I.D.C). Let x be the displacement of a reciprocating body P from I.D.C. after time t seconds, during
which the crank has turned through an angle θ.

Fig. 15.7.  Motion of a crank and connecting rod of a reciprocating steam engine.
Let l= Length of connecting rod between the centres,
r= Radius of crank or crank pin circle,
φ= Inclination of connecting rod to the line of stroke PO, and
n= Ratio of length of connecting rod to the radius of crank = l/r.
From the geometry of Fig. 15.7,
      () ( ) xP PO PO P P C C O P QQ O′′ ′ ′ ′==−= + −+

() ( c o s c o s )lr l r=+− φ + θcos ,...and cosPQ l
QO r=φ 
=θ /G51

524  /G6C   Theory of Machines
                       (1 cos ) (1 cos ) (1 cos ) (1 cos )lrlr
r=−θ +−φ = −θ + −φ
                                       [(1 cos ) (1 cos )]rn=− θ +− φ ...(i)
From triangles CPQ and CQO,
                CQ   =  l sin φ = r sin θ  or  l/r  = sin θ/sin φ
∴                                n  =  sin θ/sin φ   or  sin φ = sin θ/n ...(ii)
We know that,       ()1
1 2222
2sincos 1 1s i n
n θφ= = − −φ 
Expanding the above expression by binomial theorem, we get

2
21s i ncos 1 .....2nθφ= − × + ...(Neglecting higher terms)
or                                 2
2sin1c o s
2nθ−φ = ...(iii)
Substituting the value of (1 – cos φ) in equation (i), we have
2 2
2sin sin(1 cos ) (1 cos )2 2xr r n
n n θ θ==−θ + × −θ +  
 ..(iv)
Differentiating equation  (iv) with respect to θ,
1 sin2sin 2 sin .cos sin2 2dxrrn dnθ  θ+ × θ θ== θ +  θ ... (v)
(2 s i n . c o s s i n 2 )θθ =θ /G51

∴  Velocity of P with respect to O or velocity of the piston P,
     PO Pdx dx d dxvv
dt d dt dθ===×=× ω
θθ
...( Ratio of change of angular velocity / ) dd t=θ = ω /G51
Substituting the value of dx/dθ from equation (v), we have
                                    PO Psin2.s i n2vv rnθ == ω θ +...(vi)
Note:We know that by Klien’s construction,
        PvO M=ω×
Comparing this equation with equation (vi), we find that
     sin 2sin
2OM rnθ θ+=

sin 2sin
Since the acceleration is the rate of change of velocity, therefore acceleration of the piston P,
       PP P
Pdv dv dv da
dt d dt dθ==× =× ω
θθ

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  525
Differentiating equation (vi) with respect to θ,
     P cos 2 2 cos2cos.. cos2dvrrn d nθ×  θ θ+ =ω =ω θ+  θ  
Substituting the value of Pdv
dθin the above equation, we have
                                        2
Pcos 2 cos 2..cos cos ar r
nnθθ  =ω ×ω=ω θ+ θ+  ...(vii)
Notes: 1.  When crank is at the inner dead centre (I.D.C.), then θ = 0°.
∴         22
Pcos 0 1..cos 0 1 ar r

∴         22
nn°   =ω =ω °+ +    
2. When the crank is at the outer dead centre (O.D.C.), then θ = 180°.
∴         22
Pcos 2 180 1..cos180 1 ar r
nn×°   =ω =ω °+ − +    
As the direction of motion is reversed at the outer dead centre therefore changing the sign of the above
expression,
        2
P1.1 ar
n=ω −
15.9. Angular Velocity and Acceleration of the Connecting Rod

Consider the motion of a connecting rod and a crank as shown in Fig. 15.7.From the geometry
of the figure, we find that
                             CQ   =  l sin φ = r sin θ
Above picture shows a diesel engine. Steam engine, petrol engine and diesel engine, all
have reciprocating parts such as piston, piston rod, etc.

526  /G6C   Theory of Machines
∴                        sinsin sinr
lnθφ= × θ= ...lnr=/G51
Differentiating both sides with respect to time t,
                 cos coscosdd
dt n dt nφθ θθφ× = × = ×ω ...d
Since the angular velocity of the connecting rod PC is same as the angular velocity of point P
with respect to C and is equal to dφ/dt, therefore angular velocity of the connecting rod
                             PCcos cos
cos cosd
dt n nφθ ω ωθω= = × =×φφ
We know that, ()1

1 2222
n θφ= = − −φ sin...sinnθ φ=/G51
∴             PC 122 1 / 222
2cos cos
1(s i n )sin1nnnn
nωθ ω θω=× =×
−θ  θ−
                                    22 1 / 2cos
(s i n )nωθ=
−θ...(i)
Angular acceleration of the connecting rod PC,
                               αPC = Angular acceleration of P with respect to PC()dC
dtω=
                      PC PC PC() () ()dd d d
dt d dt dωω ω θ=× =× ωθθ...(ii)

...( / )dd tθ= ω/G51
Now differentiating equation (i), we get
PC
22 1 / 2cos ()
(s i n )d d
n ddωθ ω=−θ θθ 
                       2 2 1/2 2 2 –1/2 1
2
22
22 1 / 2 22 – 1 / 2 2
22
22( s i n)(s i n) ][ ( c o s ) ( s i n) 2 s i nc o s
sin
(s i n ) ( s i n ) (s i n ) s i n c o s
sin
(s i n )(s i n ) c o ssin
sinnn
n
nn
n
nn
n−θ − θ −θ × −θ× − θθ=ω
−θ 
−θ − θ +−θ θ θ=ω−θ 
−θ −−θ θ=−ω θ −θ 
22 2
22 3 / 2(s i n ) c o ssin

22 2
(s i n )n
n−θ − θ=−ω θ −θ22 1 / 2...[Dividing and multiplying by ( sin ) ] n−θ

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  527
             2
22 2
22 3 / 2 22 3 / 2sin sin ( 1)(sin cos )(s i n ) (s i n )nnnn−ω θ − ω θ − == −θ +θ−θ −θ
22...( sin cos 1) θ+ θ= /G51
22
PC
PC 22 3 / 2() sin ( 1)
(s i n )d n
d nω −ω θ −∴α = ×ω=θ −θ           ...[From equation (ii)] ... (iii)
The negative sign shows that the sense of the acceleration of the connecting rod is such that it tends to
reduce the angle φ.

Notes: 1. Since sin2 θ is small as compared to n2, therefore it may be neglected. Thus, equations  (i) and (iii) are
reduced to
22
PC PC 3cos sin ( 1),a n dn
n nωθ − ωθ−ω= α=
2. Also in equation (iii), unity is small as compared to n2, hence the term unity may be neglected.
∴2
PCsin
n−ω θα=
Example 15.3. If the crank and the connecting rod are 300 mm and 1 m long respectively
and the crank rotates at a constant speed of 200 r.p.m. , determine :1. The crank angle at which the

maximum velocity occurs, and 2. Maximum velocity of the piston.
Solution. Given : r = 300 mm = 0.3 m; l = 1 m; N = 200 r.p.m. or ω = 2 π × 200/60 = 20.95 rad/s
1. Crank angle at which the maximum velocity occurs
Let θ= Crank angle from the inner dead centre at which the maximum
velocity occurs.
We know that ratio of length of connecting rod to crank radius,
n=l/r = 1/0.3 = 3.33
and velocity of the piston,
                                Psin2sin.2vrnθ θ+ =ω...(i)

For maximum velocity of the piston,
               P 2c o s20. . . 0 cos2dvie rd nθ =ω = θ+θ 
or n cos θ + 2 cos2 θ – 1 = 0  2...( cos 2 2 cos 1) θ= θ− /G51
          2 cos2 θ + 3.33 cos θ – 1 = 0
∴                        23.33 (3.33) 4 2 1cos 0.2622−± + × ×θ= =×...(Taking + ve sign)
or                                θ   =  75º Ans.
2. Maximum velocity of the piston
Substituting the value of θ = 75° in equation (i), maximum velocity of the piston,

P( )sin150 0.5. 20.95 0.3 m/ssin75 0.96623 . 3 3maxvr
n°   =ω = × °+ +    
                                        =  6.54  m/s   Ans.

528  /G6C   Theory of Machines
Example 15.4.  The crank and connecting rod of a steam engine are 0.3 m and 1.5 m in
length. The crank rotates at 180 r. p.m. clockwise. Determine the velocity and acceleration of the
piston when the crank is at 40 degrees from the inner dead centre position. Also determine theposition of the crank for zero acceleration of the piston.
Solution. Given : r = 0.3; l = 1.5 m ; N = 180 r.p.m. or ω = π × 180/60 = 18.85 rad/s; θ = 40°

We know that ratio of lengths of the connecting rod and crank,                                n  =  l/r = 1.5/0.3 = 5
∴ Velocity of the piston,
Psin80 sin2sin40 . 18.85 0.3 m/ssin25 2vr
n°  θ °+ =ω = × θ+ ×  
      =4 . 1 9m / s  Ans.
We know that acceleration of piston,
                                   22 2
Pcos 2 cos80. (18.85) 0.3 m/scos cos 405ar
nθ°   =ω = × θ+ °+    
                    285.35 m/s= Ans.

285.35 m/s= Ans.
Position of the crank for zero acceleration of the piston
Let                           θ1 =  Position of the crank from the inner dead centre for zero acceleration
     of the piston.
                              2 1
P 1cos 2.cos ar
nθ =ω θ+
or                                         2
11.0( c o s c o s 2 )rnnω=θ + θ P...( 0)a=/G51
∴   n cos θ1 + cos 2θ1 = 0

           5 cos θ1 + 2 cos2 θ1 – 1 = 0           or 2 cos2 θ1 + 5 cos θ1 – 1 = 0
∴                     2
1554 1 2cos 0.186222−± +××θ= =×...(Taking + ve sign )
or                            θ1 = 79.27° or 280.73°   Ans.
Example 15.5. In a slider crank mechanism, the length of the crank and connecting rod are

speed is 450 r.p.m. (clockwise ). Using analytical method, determine:  1. Velocity and acceleration of
the slider, and 2. Angular velocity and angular acceleration of the connecting rod.
Solution. Given: r = 150 mm = 0.15 m ; l = 600 mm = 0.6 m ; θ = 60°; N = 400 r.p.m or
ω = π × 450/60 = 47.13 rad/s
We know that ratio of the length of connecting rod and crank,                        n   =  l/r = 0.6/ 0.15 = 4

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  529
∴ Velocity of the slider,
           Psin120 sin2sin 60 . 47.13 0.15 m/ssin24 2vr
  =6 . 9m / s  Ans.
and acceleration of the slider,
                          22 2
P
2cos 2 cos120. (47.13) 0.15 m/scos cos 604
124.94 m/sar
nθ°  =ω = × θ+ °+   
=
2.  Angular velocity and angular acceleration of the connecting rod
We know that angular velocity of the connecting rod,

PCcos 47.13 cos 605.9 rad/s
4 nωθ × °ω= = = Ans.
and angular acceleration of the connecting rod,
                        22
2
PCsin (47.13) sin 60481rad/s4 nωθ × °α= = = Ans.
15.10. Forces on the Reciprocating Parts of an Engine, Neglecting the
Weight of the Connecting Rod
The various forces acting on the reciprocating parts of a horizontal engine are shown in Fig.
15.8. The expressions for these forces, neglecting the weight of the connecting rod, may be derived as

1. Piston effort . It is the net force acting on the piston or crosshead pin, along the line of
stroke. It is denoted by FP in Fig. 15.8.
Fig. 15.8.  Forces on the reciprocating parts of an engine.
Let                     mR =  Mass of the reciprocating parts, e.g. piston, crosshead pin or
   gudgeon pin etc., in kg, and
                           WR =   Weight of the reciprocating parts in newtons = mR.g
We know that acceleration of the reciprocating parts,

530  /G6C   Theory of Machines
∴ *Accelerating force or inertia force of the reciprocating parts,
                                   2
IR RRcos 2.. . cos Fm a m r
nθ == ω θ+
It may be noted that in a horizontal
engine, the reciprocating parts areaccelerated from rest, during the latter halfof the stroke ( i.e. when the piston moves
from inner dead centre to outer deadcentre). It is, then, retarded during the latterhalf of the stroke ( i.e. when the piston

moves from outer dead centre to inner deadcentre). The inertia force due to theacceleration of the reciprocating parts,opposes  the force on the piston due to thedifference of pressures in the cylinder onthe two sides of the piston. On the otherhand, the inertia force due to retardation of the reciprocating parts, helps the force on the piston.Therefore,
Piston effort,     
PNet load on the piston Inertia forceF= /G6D

                                                     LIFF=/G6D ...(Neglecting frictional resistance)
                                                         LIFFFR=−/G6D ...(Considering frictional resistance)
where      RF = Frictional resistance.
The –ve sign is used when the piston is accelerated, and +ve sign is used when the piston is
retarded.
In a double acting reciprocating steam engine, net load on the piston,

FL=p1A1 – p2 A2 = p1 A1 – p2 (A1 – a)
where                                p1, A1= Pressure and cross-sectional area on the back end side of the
piston,
p2, A2= Pressure and cross-sectional area on the crank end side of the
piston,
a= Cross-sectional area of the piston rod.
Notes : 1. If ‘p’ is the net pressure of steam or gas on the piston and D is diameter of the piston, then
Net load on the piston,  FL  =  Pressure × Area2
4pDπ=××

4pDπ=××
2. In case of a vertical engine, the weight of the reciprocating parts assists the piston effort during the
downward stroke ( i.e. when the piston moves from top dead centre to bottom dead centre) and opposes during
the upward stroke of the piston ( i.e. when the piston moves from bottom dead centre to top dead centre).
∴ Piston effort,     PLIRFFFF WR=± −/G6D
2. Force acting along the connecting rod. It is denoted by FQ in Fig. 15.8. From the geom-
etry of the figure, we find that

                              P
QcosFF=φ
Connecting rod of a petrol engine.
*The acceleration of the reciprocating parts by Klien’s construction is,
        aP = ω2 × NO
          ∴                                             FI  =  mR.. ω2  ×  NO

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  531
We know that    2
2sincos 1
nθφ= −
∴                           P
Q2
2sin1FF
n=
θ−
3. Thrust on the sides of the cylinder walls or normal reaction on the guide bars. It is
denoted by FN in Fig. 15.8. From the figure, we find that
                              P
NQ Psin sin tancosFFF F=φ = × φ =φφP
Q...cosFF=φ /G51
4. Crank-pin effort and thrust on crank shaft bearings . The force acting on the connecting

rod FQ may be resolved into two components, one perpendicular to the crank and the other along the
crank. The component of FQ perpendicular to the crank is known as crank-pin effort and it is denoted
by FT in Fig. 15.8. The component of FQ along the crank produces a thrust on the crank shaft bearings
and it is denoted by FB in Fig. 15.8.
Resolving FQ perpendicular to the crank,
                               P
TQsin ( ) sin ( )cosFFF=θ + φ = × θ + φφ
and resolving FQ along the crank,

                               P
BQcos ( ) cos ( )cosFFF=θ + φ = × θ + φφ
5. Crank effort  or turning moment or torque on the crank shaft.  The product of the crank-
pin effort ( FT) and the crank pin radius ( r) is known as crank effort  or turning moment or torque on
the crank shaft . Mathematically,
 P
T
P
Psin ( )Crank effort,cos
(sin cos cos sin )
cos
sinsin coscosFTFr r
Fr
Frθ+φ=× = ×φ
θφ +θφ=×φ
φ θ+ θ× =×φ 

Fr
θφ +θφ=×φ
                                     P(sin cos tan )Fr=θ + θ φ × ...(i)
We know that l sin φ  = r sin θ
                           sinsin sinr
lnθφ= θ= ...lnr=/G51
and                          2
22 2
2sin 1cos 1 sin 1 sin nnnθφ= − φ= − = − θ
∴                       
22 22sin sin sintancossin sinn
nnnφθ θφ= = × =φ−θ −θ

532  /G6C   Theory of Machines
Substituting the value of tan φ in equation (i), we have crank effort,
                                   P 22cos sinsin
sinTF r
nθθ θ+ =×−θ
                                       P 22sin 2sin
2s i nFr
nθ θ+ =×−θ...(ii)
...( 2 cos sin sin 2 ) θθ = θ /G51
Note:Since sin2 θ is very small as compared to n2 therefore neglecting sin2 θ, we have,
Crank effort,                   PPsin 2sin2TFr FO M
nθ =× =× θ+
We have seen in Art. 15.8, that

                                   sin 2sin2OM r
nθ =θ+
Therefore, it is convenient to find OM instead of solving the large expression.
Example 15.6. Find the inertia force for the following data of an I.C. engine.
Bore = 175 mm, stroke = 200 mm, engine speed = 500 r.p.m., length of connecting rod =
400 mm, crank angle = 60° fro m T.D.C and mass of reciprocating parts = 180 kg.

Solution. Given: *D =175 mm; L = 200 mm = 0.2 m or r = L/ 2 = 0.1 m ; N = 500 r.p.m. or
ω = 2π × 500/60 =52.4 rad/s; l = 400 mm = 0.4 m; mR = 180 kg
The inertia force may be calculated by graphical method or analytical method as discussed
below:
1. Graphical method
First of all, draw the Klien’s acceleration diagram OCQN to some suitable scale as shown in
Fig. 15.9. By measurement,
              ON  =  38 mm = 0.038 m
∴ Acceleration of the reciprocating parts,
     aR = ω2 × ON

aR = ω2 × ON
          = (52.4)2 × 0.038 = 104.34 m/s
We know that inertia force,
    FI = mR × aR = 180 × 104.34 N
        = 18 780  N = 18.78 kN  Ans.
2. Analytical method
We know that ratio of lengths of connecting rod and crank,
     n  = l/r = 0.4/0.1 = 4
∴ Inertia force,         2
IRcos 2..cos Fm r
                                       2 cos120180 (52.4) 0.1 18 530 N cos 604° =× × = °+
         =18.53 kN Ans.Fig. 15.9
*Superfluous data.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  533
Example 15.7. The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the
reciprocating parts is 250 kg. When the crank has travelled 60° from I.D.C. , the difference between
the driving and the back pressures is 0.35 N/mm2. The connecting rod length between centres is 1.2 m
and the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter

is neglected, calculate : 1. pressure on slide bars,  2. thrust in the connecting rod, 3. tangential force
on the crank-pin, and 4. turning moment on the crank shaft.
Solution. Given: r = 300 mm = 0.3 m; mR = 250 kg; θ = 60°; p1 – p2 = 0.35 N/mm2;
l = 1.2 m ; D = 0.5 m = 500 mm ; N = 250 r.p.m. or ω = 2 π × 250/60 = 26.2 rad/s
First of all, let us find out the piston effort ( FP).
We know that net load on the piston,
                               22

22
L1 2( ) 0.35 (500) 68730 N44Fp p Dππ=− ×= × =
...( Force = Pressure×Area)/G51
Ratio of length of connecting rod and crank,
                                  /1 . 2 / 0 . 3 4nl r== =
and accelerating or inertia force on reciprocating parts,
2
IRcos 2.c o sFm rnθ =ω θ + 
       2 cos120250 (26.2) 0.3 19306 N cos 604° == °+
∴ Piston effort,      FP = FL – FI = 68 730 – 19 306 = 49 424 N = 49.424 kN
1. Pressure on slide bars

Let                              φ  =  Angle of inclination of the connecting rod to the line of stroke.
We know that,     sin sin 60 0.866sin 0.216544 nθ°φ= = = =
∴                              φ  = 12.5°
We know that pressure on the slide bars,                             F
N  =  FP tan φ = 49.424 × tan 12.5° = 10.96 kN   Ans.
2. Thrust  in the connecting rod
We know that thrust in the connecting rod,
                               P

P
Q49.42450.62 kNcos cos12.5FF== =φ°Ans.
3.  Tangential force on the crank-pin
We know that tangential force on the crank pin,
                               TQsin ( ) 50.62 sin (60 12.5 ) 48.28 kN FF=θ + φ = ° + ° = Ans.
4. Turning moment on the crank shaft
We know that turning moment on the crank shaft,
   T48.28 0.3 14.484 kN-m TFr=× = ×= Ans.

534  /G6C   Theory of Machines
Example 15.8. A vertical double acting steam engine has a cylinder 300 mm diameter and
450 mm stroke and runs at 200 r.p.m. The reciprocating parts has a mass of 225 kg and the piston rodis 50 mm diameter. The connecti ng rod is 1.2 m long. When the crank has turned through 125°
 from the top dead centre, the steam pressure above the piston is 30 kN/m
2 and below the piston is 1.5
kN/m2. Calculate the effective turning moment on the crank shaft.

Solution. Given : D = 300 mm = 0.3 m ; L = 450 mm or r = L/2 = 225 mm = 0.225 m;
N = 200 r.p.m. or ω = 2 π × 200/60 = 20.95 rad/s ; mR= 225 kg; d = 50 mm = 0.05 m ; l = 1.2 m ;
θ = 125°; p1 = 30 kN/m2 = 30 × 103 N/m2; p2 = 1.5 kN/m2 = 1.5 × 103 N/m2
We know that area of the piston,
                                 222
1 (0.3) 0.0707 m44ADππ=× =× =
and area of the piston rod,      22 2(0.05) 0.00196 m44adππ=× =× =
∴ Force on the piston due to steam pressure,

                                L1 12 1
33.( )
30 10 0.0707 1.5 10 (0.0707 0.001 96)N
= 2121 – 103 = 2018 NFp Ap A a=− −
=× × −× −
Ratio of lengths of connecting rod and crank,
   n  =  l/r  =  1.2/0.225 = 5.33
and inertia force on the reciprocating parts,
                                 2
1R
2cos 2..cos
cos 250225 (20.95) 0.225 14 172 N cos1255.33Fm r
° =× = − °+

We know that for a vertical engine, net force on the piston or piston effort,
  FP  =  FL – FI + mR.g
      =  2018 – (– 14 172) + 225 × 9.81 = 18 397 N
Let       φ =  Angle of inclination of the connecting rod to the line of stroke.
We know that,      sin sin125 0.8191sin 0.1537
5.33 5.33nθ°φ= = = =
∴                               8.84φ= °
We know that effective turning moment on the crank shaft,

Psin ( ) 18397 sin(125 8.84 )0.225 N-mcos cos 8.84FTr×θ + φ ° +°=× = ×φ°
                                     = 3021.6 N-m Ans.
Example 15.9. The crank and connecting rod of a petrol engine, running at 1800 r.p.m.are
50 mm and 200 mm respectively.  The diameter of the piston is 80 mm and the mass of the reciprocat-
ing parts is 1 kg. At a point during the power stroke, the pressure on the piston is 0.7 N/mm2, when it

has  moved 10 mm from the inner dead centre. Determine : 1. Net load on the gudgeon pin, 2. Thrust
in the connecting rod, 3. Reaction between the piston and cylinder, and  4. The engine speed at which
the above values become zero.
Solution. Given : N = 1800 r.p.m. or ω = 2π × 1800/60 = 188.52 rad/s ; r = 50 mm = 0.05 m;
l = 200 mm; D = 80 mm; mR = 1 kg; p = 0.7 N/mm2; x = 10 mm

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  535
1. Net load on the gudgeon pin
We know that load on the piston,
   22
L (80) 0.7 3520 N44FD pππ=× = ×× =
Fig. 15.10
When the piston has moved 10 mm from the inner dead centre, i.e. when P1P = 10 mm, the
crank rotates from OC1 to OC through an angle θ as shown in Fig. 15.10.
By measurement, we find that *θ = 33°.
     n  =  l/r = 200 /50 = 4

n  =  l/r = 200 /50 = 4
    2
IR RR
2cos 2.. . cos
cos 661 (188.52) 0.05 1671 N cos 33
4Fm a m r
° =× × = °+
We know that net load on the gudgeon pin,
   PLI 3520 1671 1849 N FFF=−= − = Ans.
2. Thrust in the connecting rod
Let φ= Angle of inclination of the connecting rod to the line of
stroke.
We know that,sin sin 33 0.5446sin 0.136144 nθ°φ= = = =
∴      φ   =  7.82°
*The angle θ may also be obtained as follows:

We know that   22sin 1 cos(1 cos ) (1 cos )22xr r
nn   θ− θ==−θ + −θ +  
  
   2
2 50 1c o s10 50 (1 cos ) (8 8 cos 1 cos8 24 −θ ==−θ + −θ + − θ  × 
  = 50 – 50 cos θ + 6.25 – 6.25 cos2 θ
or 6.25 cos2 θ + 50 cos θ – 56.25 = 0
Solving this quadratic equation, we get θ = 33.14°

536  /G6C   Theory of Machines
P
Q18491866.3Ncos cos 7.82FF== =φ°Ans.
3. Reaction between the piston and cylinder
We know that reaction between the piston and cylinder,
NPtan 1849 tan 7.82 254 N FF=φ = ° = AAns.
4. Engine speed at which the above values will become zero
A little consideration will show that the above values will become zero, if the inertia force on

the reciprocating parts ( FI) is equal to the load on the piston ( FL). Let ω1 be the speed in rad/s, at
which FI = FL .
∴     22
R1cos 2()cos4mr D p
nπθ ω= × θ+
             22 2
1 1cos 661( ) 0.05 (80) 0.7 or 0.0 47 ( ) 3520 cos 3344π° ω× =× × ω= °+
∴  (ω1)2 = 3520 / 0.047 = 74 894 or ω1 = 273.6 rad/s
∴  Corresponding speed in r.p.m.,
N1 = 273.6 × 60/2 π = 2612 r.p.m.  Ans.
Example 15.10. During a trial on steam engine, it is found that the acceleration of the piston

is 36 m/s2 when the crank has moved 30° from the inner dead centre position. The net effective steam
pressure on the piston is 0.5 N/mm2 and the frictional resistance is equivalent to a force of 600 N. The

diameter of the piston is 300 mm and the mass of the reciprocating parts is 180 kg. If the length ofthe crank is 300 mm and the ratio of the connecting rod length to the crank length is 4.5, find:1. Reaction on the guide bars, 2. Thrust on the crank shaft bearings, and 3. Turning moment on the
crank shaft.
Solution. Given: a
P = 36 m/s2 ; θ = 30°;  p = 0.5 N/mm2 ; RF = 600 N; D = 300 mm;
mR = 180 kg; r = 300 mm = 0.3 m ; n = l/r = 4.5
1. Reaction on the guide bars

First of all, let us find the piston effort ( FP). We know that load on the piston,
22
L 0.5 (300) 35350 N44Fp Dππ=×× = ×× =
Twin-cylinder aeroplane engine.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  537
and inertia force due to reciprocating parts,
 FI = mR × aP = 180 × 36 = 6480 N
∴ Piston effort,    FP = FL – FI – RF = 35 350 – 6480 – 600 = 28 270 N = 28.27 kN
Let φ = Angle of inclination of the connecting rod to the line of stroke.
We know that sin φ = sin θ/n = sin 30°/4.5 = 0.1111
∴  φ  = 6.38°
We know that reaction on the guide bars,
 FN = FP tan φ = 28.27 tan 6.38° = 3.16 kN Ans.
2. Thrust on the crank shaft bearing

We know that thrust on the crank shaft bearings,
P
Bcos( ) 28.27 cos (30 6.38 )22.9 kNcos cos 6.38FFθ+φ °+ °== =φ°Ans.
3.  Turning moment on the crank shaft
Psin ( ) 28.27 sin (30 6.38 )0.3 kN-mcos cos 6.38FTrθ+φ °+ °=× = ×φ°
                                          5.06 kN-m=
Example 15.11.   A vertical petrol engine 100 mm diameter and 120 mm stroke has a

connecting rod 250 mm long. The mass of the piston is 1.1 kg. The speed is 2000 r.p.m. On theexpansion stroke with a crank 20° from top dead centre, the gas pressure is 700 kN/m
2. Determine:
1. Net force on the piston,  2. Resultant load on the gudgeon pin,
3. Thrust on the cylinder walls, and 4. Speed above which, other things re-
maining same, the gudgeon pin load would be reversed in direction.
Solution. Given: D = 100 mm = 0.1 m; L = 120 mm = 0.12 m or

r = L/2 = 0.06 m ; l = 250 mm = 0.25 m ; mR = 1.1 kg; N = 2000 r.p.m. or
ω = 2 π × 2000/60 = 209.5 rad/s ; θ = 20°; p = 700 kN/m2
1. Net force on the piston
The configuration diagram of a vertical engine is shown in Fig. 15.11.
We know that force due to gas pressure,
22
L 700 (0.1) 5.5 kN44
= 5500 NFp Dππ=×× = ×× =
and ratio of lengths of the connecting rod and crank,
n = l/r = 0.25 /0.06 = 4.17
∴ Inertia  force on the piston,
2
IR
cos 401.1 (209.5) 0.06 cos 204.17
3254 NFm r

3254 NFm r
° =× × × °+
=Fig. 15.11

538  /G6C   Theory of Machines
We know that for a vertical engine, net force on the piston,
PLIRLIR .
5500 3254 1.1 9.81 2256.8 NFFF WFFm g=−+=−+
=−+ × =
2. Resultant load on the gudgeon pin
Let             φ  =  Angle of inclination of the connecting rod to the line of stroke.
We know that,
sin φ = sin θ / n = sin 20°/4.17 = 0.082
∴φ   =  4.7°
We know that resultant load on the gudgeon pin,
P
Q2256.82265 Ncos cos 4.7FF== =φ°Ans.
3. Thrust on the cylinder walls

We know that thrust on the cylinder walls,
       NPtan 2256.8 tan 4.7 185.5 N FF=φ =×° = Ans.
4. Speed, above which, the gudgeon pin load would be reversed in direction
Let N1 =  Required speed, in r.p.m.
The gudgeon pin load i.e. FQ will be reversed in direction, if FQ becomes negative. This is only
possible when FP is negative. Therefore, for FP to be negative, FI must be greater than ( FL + WR),
i.e.2
R1cos 2( ) 5500 1.1 9.81cos mr
nθ ω> + × θ+
2

2
1cos 401.1 ( ) 0.06 5510.8 cos 204.17° ×ω × > °+
22
110.074 ( ) 5510.8 or ( ) 5510.8/0.074 or 74 470ω> ω>
or1273 rad/sω>
1273 60/2 or 2606 r.p.m.N>×π Ans.
Example 15. 12. A horizontal steam engine running at 120 r.p.m. has a bore of 250 mm and

a stroke of 400 mm. The connecting rod is 0.6 m and mass of the reciprocating parts is 60 kg. Whenthe crank has turned through an angle of 45° from the inner dead centre, the steam pressure on thecover end side is 550 kN/m
2 and that on the crank end side is 70 kN/m2. Considering the diameter of
the piston rod equal to 50 mm, determine:
1. turning moment on the crank shaft, 2. thrust on the bearings, and 3. acceleration of the

flywheel, if the power of the engine is 20 kW, mass of the flywheel 60 kg and radius of gyration 0.6 m.
Solution. Given : N = 120 r.p.m. or ω = 2π × 120/60  = 12.57 rad/s; D = 250 mm = 0.25 m;
L = 400 mm = 0.4 m or r = L/2 = 0.2 m ; l = 0.6 m; mR = 60 kg; θ = 45° ; d = 50 mm = 0.05 m;
p1 = 550 kN/m2 = 550 × 103 N/m2; p2 = 70 kN/m2 = 70 × 103 N/m2Ans.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  539
1. Turning moment on the crankshaft
First of all, let us find the net load on the piston ( FP).
We know that area of the piston on the cover end side,
                         22 2
1 = (0.25) = 0.049 m44ADππ=× ×
and area of piston rod,        222(0.05) 0.00196 m44adππ=× =× =
∴  Net load on the piston,
       L1 12 21 12 1
33.. .( )
550 10 0.049 70 10 (0.049 0.00196) 23657 NFp Ap Ap A p A a=− =− −
=×× − × − =

=×× − × − =
We know that ratio of lengths of the connecting rod and crank,
                                               n = l/r = 0.6/0.2 = 3
                                              2
IR
cos 9060 (12.57) 0.2 1340 N cos 453Fm r
∴  Net force on the piston or piston effort,
                                            PLI 23657 1340 22 317 N 22.317 kN FFF=−= − = =

Let                             φ  =  Angle of inclination of the connecting rod to the line of stroke.
We know that,     sin  φ =  sin θ/n = sin 45°/3 = 0.2357
∴                               φ = 13.6°
We know that turning moment on the crankshaft,
         Psin ( ) 22.317 sin (45 13.6 )0.2 kN-m
cos cos13.6
= 3.92 kN-m = 3920 N-mFTrθ+φ ×° + °=× = ×
φ°
Ans.High-pressure steam in
Valve rod Slide valveExhaust steam
Cross-head guideCross head
Piston Rod Piston
Cylinder

540  /G6C   Theory of Machines
2. Thrust on the bearings
We know that thrust on the bearings,
P
Bcos ( ) 22.317 cos (45 13.6 )11.96 kNcos cos13.6FFθ+φ × °+ °== =φ°Ans.
3. Acceleration of the flywheel
Given: P = 20 kW = 20 × 103 W; m = 60 kg; k = 0.6 m
Let α = Acceleration of the flywheel in rad/s2.
I  =  m.k2 = 60 × (0.6) 2 = 21.6 kg-m2
∴  Accelerating torque, TA = I.α = 21.6 α N-m ... (i)
and resisting torque,                   3

R60 20 10 601591 N-m22 1 2 0PTN×× ×== =ππ ×2
60NTPπ ∴=
Since the accelerating torque is equal to the difference of torques on the crankshaft or turning
moment ( T) and the resisting torque ( TR), therefore, accelerating torque,
TA = T – TR = 3920 – 1591 = 2329 N-m ... (ii)
From equation (i) and (ii),
α = 2329/21.6 = 107.8 rad/s2   Ans.
Example 15.13. A vertical, single cylinder,

single acting diesel engine has a cylinder diameter300 mm, stroke length 500 mm, and connecting rodlength 4.5 times the crank length. The engine runsat 180 r.p.m. The m ass of the reciprocating parts is
280 kg. The compression ratio is 14 and the pressureremains constant during the injection of the oil for1/10th of the stroke. If the compression andexpansion follows the law p.V
1.35 = constant, find :

1. Crank-pin effort, 2. Thrust on the bearings, and
3. Turning moment on the crank shaft, when the
crank displacement is 45° from the inner dead centreposition during expansion stroke.
The suction pressure may be taken as
0.1 N/mm
2.
Solution. Given : D  = 300 mm = 0.3 m ;
L = 500 mm = 0.5 m or  r = 0.25 m ; l =  4.5 r or n = l/r  = 4.5;  N = 180 r.p.m.   or
ω = 2π × 180/60 = 18.85 rad/s;  mR = 280 kg; 1
214;V
V=θ = 45°; p1 = 0.1 N/mm2

214;V
The pressure-volume ( i.e. p-V) diagram for a *diesel engine is shown in Fig 15.12, in which
1-2 represents the compression, 2-3 represents the injection of fuel, 3-4 represents the expan-
sion, and 4-1 represents the exhaust.
Let p1, p2, p3, and  p4 = Pressures corresponding to points 1, 2, 3 and 4 respectively, and
V1, V2, V3, and      V4 = Volumes corresponding to points 1, 2, 3 and 4 respectively.Fig. 15.12

*In a diesel engine, the compression and expanssion are isentropic i.e. according to the law p.Vγ = constant.
The injection of fuel takes place at constant pressure and the exhaust is at constant volume.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  541
Since the compression follows the law p.V1.35 = constant, therefore
1.35 1.35
11 22() ()pV pV =
or1.35
11 . 3 5 2
21
20.1 (14) 3.526 N/mmVppV== × =
We know that swept volume,
22 3
S (0.3) 0.5 0.035 m44VD Lππ=× ×=× × =
and        compression ratio,CS S1
2C C1VV VV
VV V+== = + 2C...( )VV=/G51
∴3
C
C0.035 0.03514 1 or 0.0027 m14 1VV=+ = =−

∴3
C
Since the injection of fuel takes place at constant pressure ( i.e.  p2 = p3) and continues up to
1/10th of the stroke, therefore volume at the end of the injection of fuel,
3
3C S10 . 0 3 50.0027 0.0062 m10 10VV V=+×= + =
When the crank displacement is 45° ( i.e. when θ = 45°) from the inner dead centre during
expansion stroke, the corresponding displacement of the piston (marked by point 4' on the p-V diagram)
is given by

is given by
22sin sin 45(1 cos ) (1 cos 45 )22 4 . 5xr r
n   θ°==−θ + − ° +  ×   
   0.50.25 0.087 m(1 0.707)9== −+
∴                 22 2
4C 0.0027 (0.3) 0.087 0.0088 m44VV Dxππ′=+ ×× = + × × =
Since the expansion follows the law p.V1.35 = constant, therefore,
                 1.35 1.35
33 44() ()pV pV′′=
∴                1.35 1.35
32
34
40.00623.526 2.2 N/mm
0.0088VppV′
′ == =   
Difference of pressures on two sides of the piston,
              26 2

26 2
14 2.2 0.1 2.1 N/mm 2.1 10 N/m pp p′=− =−= =×
∴ Net load on the piston,
              26 2
L 2.1 10 (0.3) 148 460N44Fp Dππ=×× = × ×× =
Inertia force on the reciprocating parts,
               2
IR
cos 90280 (18.85) 0.25 17585 N cos454.5Fm r

542  /G6C   Theory of Machines
We know that net force on the piston or piston effort,
        PLIRLIR .
148460 17858 280 9.81 133622 NFFF WFFm g =−+ =−+
=− + × =
1. Crank-pin effort
Let             φ  = Angle of inclination of the connecting rod to the line of stroke.
We know that,  sin φ = sin θ/n = sin 45°/4.5 = 0.1571
∴           φ   =  9.04°
We know that crank-pin effort,
        P
Tsin ( ) 133622 sin (45 9.04 )109522 N
cos cos 9.04
=109.522 kNFFθ+φ ×° + °== =
φ°
2.  Thrust on the bearings

φ°
        P
B.cos ( ) 133622 sin(45 9.04 )79456 N
= 79.956 kNFFθ+φ ×° + °== =
φ°
3. Turning moment on the crankshaft
We know that the turning moment on the crankshaft,
        T109.522 0.25 27.38 kN-m TFr=× = × = Ans.
Example 15.14. A vertical double acting steam engine

has cylinder diameter 240 mm, length of stroke 360 mm andlength of connecting rod 0.6 m. The crank rotates at 300 r.p.m.and the mass of  the reciprocating parts is 160 kg. The steam isadmitted at a pressure of 8 bar gauge and cut-off takes place at1/3rd of the stroke. The expansion of steam is hyperbolic. Theexhaust of steam takes place at a pressure of –0.75 bar gauge.The frictional resistance is equivalent to a force of 500 N. Deter-mine the turning moment on the crankshaft, when the piston

turning moment on the crankshaft, when the piston is75° from the top dead centre. Neglect the effect of clearance andassume the atmospheric presssure as 1.03 bar.

Solution. Given D = 240 mm = 0.24 m; L = 360 mm
= 0.36 m or r = L/2 = 0.18 m ; l = 0.6 m; l= 0.6 m;
N = 300 r.p.m. or ω = 2π × 300/60= 31.42 rad/s; m
R = 160 kg;
pA = 8 + 1.03 = 9.03 bar = 903 × 103 N/m2 ; pE = – 0.75 + 1.03
= 0.28 bar = 28 × 103 N/m2 ; FR = 500 N ; θ = 75°
First of all, let us find the piston effort ( FP).
The pressure-volume ( p-V) diagram for a steam engine,
neglecting clearance, is shown in FIg. 15.13, in which AB

represents the admission of steam, BC the expansion and DE the exhaust  of steam. The steam is cut-
off at point B.
We know that the stroke volume,
22 3
S (0.24) 0.36 0.0163 m44VD Lππ=× ×=× × =Fig. 15.13Ans.
Ans.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  543
Since the admission of steam is cut-off at 1/3rd of the stroke, therefore volume of steam at
cut-off,
            VB = VS / 3 = 0.0163/3 = 0.00543 m3
We know that ratio of the lengths of the connecting rod and crank,
              / 0.6/0.18 3.33nl r== =
When the crank position is 75° from the top dead centre ( i.e. when θ = 75°), the displacement
of the piston (marked by point C' on the expansion curve BC) is given by

22sin sin 750.18 (1 cos ) 1 cos 752 2 3.33
0.1586 mxr
=
∴              3
CS0.15860.0163 0.0072 m
0.36xVV
L′=×= × =
Since the expansion is hyperbolic ( i.e. according to the law pV = constant), therefore
             BB C C..pV pV ′′=
or               3
32 BB
C
C903 10 0.00543681 10 N/m0.0072pVp
V×× ×′== = ×
′
∴  Difference of pressures on the two sides of the piston,
              33 3 2
CE 681 10 28 10 653 10 N/m pp p′=−=×− ×=×

              22 3
L (0.24) 653 10 29545 N44FD pππ=× ×=× × × =
2
IR
cos150160 (31.42) 0.18 36 N cos753.33Fm r
° =× × = − °+
∴  Piston effort,  PLIRRFFF WF=−+−
29545 ( 36) 160 9.81 500 30651 N=− − + × − =
Turning moment on the crankshaft
Let φ  =  Angle of inclination of the connecting rod to the line of stroke.

We know that sin φ  = sin θ/n = sin 75°/3.33 = 0.29
∴φ   = 16.86°
We know that turning moment on the crankshaft
Psin( ) 30651sin (75 16.86 )0.18 N-m
cos cos16.86
= 5762 N-mFTrθ+φ °+ °=× = ×
φ°
15.11.  Equivalent Dynamical System
In order to determine the motion of a rigid body, under the action of external forces, it is
usually convenient to replace the rigid  body by two masses placed at a fixed distance apart, in sucha way that,Ans.

544  /G6C   Theory of Machines
1.the sum of their masses is equal to the total mass of the body;
2.the centre of gravity of the two masses coincides with that of the body; and
3.the sum of mass moment of inertia of the masses about their centre of gravity is equal to
the mass moment of inertia of the body.
When these three conditions are satisfied, then it is said to be an equivalent dynamical system.
Consider a rigid body, having its centre of gravity at G, as shown in Fig. 15.14.

Let  m = Mass of the body,
kG = Radius of gyration about
   its centre of gravity G,
m1 and m2 = Two masses which form a
                  dynamical equivalent system,
l1 = Distance of mass m1 from G,
l2 = Distance of mass m2 from G,
   and
L = Total distance between the
   masses m1 and m2.
Thus, for the two masses to be dynamically equivalent,
                                          12mmm+= ...(i)
                                                11 22..ml ml= ...(ii)

and                    222
11 22 G() () ( )ml ml mk+= ...(iii)
2
1
12.lmmll=+...(iv)
and1
2
12.lmmll=+...(v)
Substituting the value of m1 and m2 in equation (iii), we have
22 2 2 21 1 2 1 2
12 G G
12 12 12.. . ( )( ) ( ) () o r ()lm lm ll l lll m k kll ll ll++= =++ +
∴2
12 G.( )ll k= ...(vi)
This equation gives the essential condition of placing the two masses, so that the system

becomes dynamical equivalent. The distance of one of the masses ( i.e. either l1 or l2) is arbitrary
chosen and the other distance is obtained from equation  (vi).
Note : When the radius of gyration kG is not known, then the position of the second mass may be obtained by
considering the body as a compound pendulum. We have already discussed, that the length of the simple pendu-
lum which gives the same frequency as the rigid body ( i.e. compound pendulum) is
22 2 2
GG 1

22 2 2
GG 1
1() () ( )kh k lLhl++== ..(Replacing h by l1)
We also know that2
12 G.( )ll k=
∴2
12 1
21
1.( )ll lLl l
l+== +
This means that the second mass is situated at the centre of oscillation  or percussion of the
body, which is at a distance of l2 = (kG)2/l1.Fig. 15.14.  Equivalent
dynamical system.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  545
Fig. 15.16
Fig. 15.17Fig. 15.15.  Determination of equivalent
dynamical system by graphical method.15.12. Determination of Equivalent Dynamical System of Two Masses by
Graphical Method
Consider a body of mass m, acting at G as
shown in Fig. 15.15. This mass m, may be replaced
by two masses m1 and m2 so that the system becomes
dynamical equivalent. The position of mass m1 may
be fixed arbitrarily at A. Now  draw perpendicular

CG at G, equal in length of the radius of gyration of
the body, kG . Then join AC and draw CB
perpendicular to AC intersecting AG produced in
B. The point B now fixes the position of the second
mass m2.
triangles ACG and BCG are similar. Therefore,
2 G2
G1 2
1Gor ( ) .klkl llk==
                    ...(Same as before)
Example 15.15. The connecting rod of a gasoline engine

is 300 mm long between its centres. It has a mass of 15 kg andmass moment of inertia of 7000 kg-mm
2. Its centre of gravity is at
200 mm from its small end centre. Determine the dynamicalequivalent two-mass system of the connecting rod if one of themasses is located at the small end centre.
Solution. Given: l = 300 mm ; m = 15 kg; I = 7000 kg-mm
2;
l1 = 200 mm
The connecting rod is shown in Fig. 15.16.Let   k
G = Radius of gyration of the connecting rod

           about an axis passing through its centre of          gravity G.
We know that mass moment of inertia ( I),
7000 = m (k
G)2 = 15 (kG)2
∴(kG)2 =  7000/15  =  466.7 mm2 or kG = 21.6 mm
It is given that one of the masses is located at the small end
centre. Let the other mass is placed at a distance l2 from the centre
of gravity G, as shown in Fig. 15.17.
We know that for a dynamical equivalent system,
                2
∴    2
G
2

2
∴    2
G
2
1()466.72.33 mm200kll===
Let   m1 =  Mass placed at the small end
             centre, and
 m2 =     Mass placed at a distance l2 from
            the centre of gravity G.

546  /G6C   Theory of Machines
         2
1
12.2 . 3 3 1 50.17 kg200 2.33lmmll×== =++Ans.
and         1
2
12. 200 1514.83 kg200 2.33lmmll×== =++Ans.
Example 15.16.  A connecting rod is suspended from a point 25 mm above the centre of

small end, and 650 mm above its centre of gravity, its mass being 37.5 kg. When permitted to oscil-late, the time period is found to be 1.87 seconds. Find the dynamical equivalent system constituted oftwo masses, one of which is located at the small end centre.
Solution. Given : h = 650 mm = 0.65 m ; l
1 = 650 – 25 = 625 mm
= 0.625 m ; m = 37.5 kg ; tp = 1.87 s
First of all, let us find the radius of gyration ( kG) of the connect-

ing rod (considering it is a compound pendulum), about an axis passingthrough its centre of gravity, G.
We know that for a compound pendulum, time period of
oscillation ( t
p),
          22 2 2
GG() () ( 0 . 6 5 ) 1.871.87 2 or
. 2 9.81 0.65kh k
gh++=π =
π×
Squaring both sides, we have
                2
G( ) 0.42250.08856.38k+=
                  22
G( ) 0.0885 6.38 0.4225 0.1425 mk=× −=
∴       kG = 0.377 m
It is given that one of the masses is located at the small end centre.

Let the other mass is located at a distance l2 from the centre of gravity G,
as shown in Fig. 15.19. We know that, for a dynamically equivalent system,
                       l1.l2 = (kG)2
∴        2
G
2
1()0.14250.228 m0.625kll== =
Let       m1 = Mass placed at the small end
                   centre A, and
                        m2 =  Mass placed at a distance l2 from
  G, i.e. at B.
We know that, for a dynamically equivalent system,
                      2
1

2
1
12. 0.228 37.510 kg0.625 0.228lmmll×== =++ Ans.
and        1
2
12. 0.625 37.527.5 kg0.625 0.228lmmll×== =++Ans.Fig. 15.18
Fig. 15.19

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  547
Example 15.17. The following data relate to a connecting rod of a reciprocating engine :
Mass = 55 kg; Distance between bearing centres = 850 mm; Diameter of small end bearing
= 75 mm; Diameter of big end bearing = 100 mm; Ti me of oscillation when the connecting rod is
suspended from small end = 1.83 s ; Time of  oscillation when the connecting rod is suspended from
big end = 1.68 s.

Determine: 1. the radius of gyration of the rod about an axis passing through the centre of
gravity and perpendicular to the plane of oscillation; 2. the moment of inertia of the rod about the
same axis; and 3. the dynamically equivalent system for the connecting rod, constituted of two masses,
one of which is situated at the small end centre.
 Solution. Given : m = 55 kg; l = 850 mm = 0.85 m ; d1 = 75 mm = 0.075 m ;
d2 = 100 mm = 0.1 m ; tp1 = 1.83 s; tp2 = 1.68 s

First of all, let us find the lengths of the equivalent simple
pendulum when suspended
(a)  from the top of small end bearing; and
(b)  from the top of big end bearing.
Let L1= Length of equivalent simple pendulum
when  suspended from the top of smallend bearing,
L
2= Length of equivalent simple pendulum
when suspended from the top of big endbearing,
h
1= Distance of centre of gravity, G, from the
top of small end bearing, and

h2= Distance of centre of gravity, G, from the
top of big end bearing.
We know that for a simple pendulum
2
1 11
12o r
2p
pt LLtgg=π = π...(Squaring both sides)
∴  2 2
1
11.839.81 0.832 m
2 2ptLg == =  π π
Similarly,2 2
2
21.689.81 0.7 m
1. Radius of gyration of the rod about an axis passing through the centre of gravity and perpen-
dicular to the plane of oscillation
Let kG = Required radius of gyration of the rod.

We know that the length of equivalent simple pendulum,
22
22 G
G()or ( ) . ( )khLk L h h h L hh+== − = −
∴ When the rod is suspended from the top of small end bearing,
()2
11 1 G ()hL hk=− ...(i)Fig. 15.20

548  /G6C   Theory of Machines
and when the rod is suspended from the top of big end bearing,
                ()2
22 2 G ()hL hk=− ...(ii)
Also, from the geometry of the Fig. 15.20,
                12
120.075 0.10.85 0.9375 m22 2 2ddhh l+=+ += + + =
∴                210.9375hh=− ...(iii)
                11 1 22 2() ( )hL h hL h−= −
Substituting the value of h2 from equation  (iii),
                 [] 11 1 1
22
11 11
11(0.832 ) (0.9375 ) 0.7 (0.9375 )

22
11 11
0.832 ( ) 0.223 1.175 ( )
0.343 0.233 or 0.223 /0.343 0.65 mhh h h
hh hh
hh−= − −−
−= −+ −
== =
                2
GG( ) 0.65 (0.832 0.65) 0.1183 or 0.343 mkk=− = = Ans.
2. Moment of inertia of the rod
We know that moment of inertia of the rod,
               2 2
G( ) 55 0.1183 6.51kg-mIm k== × = Ans.
3. Dynamically equivalent system for the rod

Since one of the masses ( m1) is situated at the centre of small end bearing, therefore its
distance from the centre of gravity, G, is
             l1 =  h1 – 0.075/2 = 0.65 – 0.0375 = 0.6125 m
Let                  m2 =  Magnitude of the second mass, and
              l2 = Distance of the second mass from the centre of gravity, G,
                     towards big  end bearing.
For a dynamically equivalent system,
                2
2 G
12 G 2

2
2 G
12 G 2
1()0.1183.( ) o r 0 . 1 9 3 m0.6125kll k ll== = =
We know that2
1
12. 0.193 5513.18 kg0.6125 0.193lmmll×== =++Ans.
and1
2
12. 0.6125 5541.82 kg0.6125 0.193lmmll×== =++Ans.
15.13. Correction Couple to be Applied to Make Two Mass System
Dynamically Equivalent
In Art. 15.11, we have discussed the conditions for equivalent dynamical system of two
bodies. A little consideration will show that when two masses are placed arbitrarily *, then the condi-

*When considering the inertia forces on the connecting rod in a mechanism, we replace the rod by two
masses arbitrarily. This is discussed in Art. 15.14.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  549
tions (i) and (ii) as given in Art. 15.11 will only be satisfied. But the condition ( iii) is not possible to
satisfy. This means that the mass moment of inertia of these two masses placed arbitrarily, will differthan that of mass moment of inertia of the rigid body.
Fig. 15.21.  Correction couple to be applied to make the two-mass system dynamically equivalent.

Consider two masses, one at A and the other at D be placed arbitrarily, as shown in Fig. 15.21.
Let l3= Distance of mass placed at D from G,
I1= New mass moment of inertia of the two masses;
k1= New radius of gyration;
α= Angular acceleration of the body;
I= Mass moment of inertia of a dynamically equivalent system;
kG= Radius of gyration of a dynamically equivalent system.
We know that the torque required to accelerate the body,
T=I.α = m (kG)2 α ...(i)

Similarly, the torque required to accelerate the two-mass system placed arbitrarily,
T1=I1.α = m (k1)2 α ...(ii)
∴ Difference between the torques required to accelerate the two-mass system and the torque
required to accelerate the rigid body,
T'=T1–T = m (k1)2 α – m (kG)2 α = m [(k1)2 – (kG)2] α ...(iv)
The difference of the torques T' is known as correction couple . This couple must be applied,

when the masses are placed arbitrarily to make the system dynamical equivalent. This, of course, willsatisfy the condition ( iii) of Art. 15.11.
Note:We know that ( kG)2=l1.l2 ,    and   ( k1)2 = l1.l3
∴ Correction couple, T'=m (l1.l3 – l1.l2) α = m.ll (l3 – l2) α
But l3 – l2=l – L
∴ T'=m.l1 (l – L) α
where l= Distance between the two arbitrarily masses, and
L= Distance between the two masses for a true dynamically equivalent

system. It is the equivalent length of a simple pendulum when a bodyis suspended from an axis which passes through the position of massm, and perpendicular to the plane of rotation of the two mass system.
22
G1
1() ( )kl
l+=

550  /G6C   Theory of Machines
Example 15.18. A connecting rod of an I.C. engine has a mass of 2 kg and the distance
between the centre of gudgeon pin and centre of crank pin is 250 mm. The C.G. falls at a point 100  mm
from the gudgeon pin along the line of centres. The radius of gyration about an axis through the C.G.perpendicular to the plane of rotation is 110 mm. Find the equivalent dynamical system if only one ofthe masses is located at gudgeon pin.

If the connecting rod is replaced by two masses, one at the gudgeon pin and the other at the
crank pin and the angular acceleration of the rod is 23 000 rad/s
2 clockwise, determine  the correc-
tion couple applied to the system to reduce it to a dynamically equivalent system.
Solution. Given : m = 2 kg ; l = 250 mm = 0.25 m ; l1 = 100 mm = 0.1m ; kG = 110 mm = 0.11 m;
α = 23 000 rad/s2
Equivalent dynamical system

It is given that one of the masses is located at the gudgeon pin. Let the other mass be located
at a distance l2 from the centre of gravity. We know that for an equivalent dynamical system.
                                          2 2
2 G
12 G 2
1()(0.11). ( ) or 0.121m0.1kll k ll== = =
Let                          m1 = Mass placed at the gudgeon pin, and
                               m2 = Mass placed at a distance l2 from C.G.

We know that        2
1
12. 0.121 21.1kg0.1 0.121lmmll×== =++ Ans.
and                                     1
2
12.0 . 1 20.9 kg0.1 0.121lmmll×== =++Ans.
Correction couple
Since the connecting rod is replaced by two masses located at the two centres ( i.e. one at the
gudgeon pin and the other at the crank pin), therefore,
                                  l = 0.1 m,   and   l3 = l – l1 = 0.25 – 0.1 = 0.15 m
Let                           k1 = New radius of gyration.

We know that        ( k1)2= l1.l3 = 0.1 × 0.15 = 0.015 m2
∴  Correction couple,
                    22 21G( ) 2 23 000 133.4 N-m 0.015 (0.11) Tm kk′  =− α = = − Ans.
Note : Since T' is positive, therefore, the direction of correction couple is same as that of angular acceleration
i.e. clockwise.
15.14. Inertia Forces in a Reciprocating Engine, Considering the Weight of
 Connecting Rod
In a reciprocating engine, let OC be the crank and PC, the connecting rod whose centre of

gravity lies at G. The inertia forces in a reciprocating engine may be obtained graphically as discussed
below:
1. First of all, draw the acceleration diagram OCQN by Klien’s construction. We know that
the acceleration of the piston P with respect to O,
                             aPO = aP = ω2 × NO,

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  551
acting in the direction from N to O. Therefore, the inertia force FI of the reciprocating parts will act in
the opposite direction as shown in Fig. 15.22.
Fig. 15.22.  Inertia forces is reciprocating engine, considering the weight of connecting rod.
2. Replace the connecting rod by dynamically equivalent system of two masses as discussed

in Art. 15.12. Let one of the masses be arbitrarily placed at P. To obtain the position of the other mass,
draw GZ perpendicular to CP such that GZ = k, the radius of gyration of the connecting rod. Join PZ
and from Z draw perpendicular to DZ which intersects CP at D. Now, D is the position of the second
mass.
Note: The position of the second mass may also be obtained from the equation,
GP × GD=k2
3. Locate the points G and D on NC which is the acceleration image of the connecting rod.

This is done by drawing parallel lines from G and D to the line of stroke PO. Let these parallel lines
intersect NC at g and d respectively. Join gO and dO. Therefore, acceleration of G with respect  to O,
in the direction from g to O,
aGO=aG = ω2 × gO
and acceleration of D with respect to O, in the direction from d to O,
aDO=aD = ω2 × dO
4. From D, draw DE parallel to dO which intersects the line of stroke PO at E. Since the

accelerating forces on the masses at P and D intersect at E, therefore their resultant must also pass
through E. But their resultant is equal to the accelerang force on the rod, so that the line of action of
the accelerating force on the rod, is given by a line drawn through E and parallel to gO, in the direc-
tion from g to O. The inertia force of the connecting rod FC therefore acts through E and in the

opposite direction as shown in Fig. 15.22. The inertia force of the connecting rod is given by
FC=mC × ω2 × gO ...(i)
where mC= Mass of the connecting rod.
A little consideration will show that the forces acting on the connecting rod are :
(a)Inertia force of the reciprocating parts ( FI) acting along the line of stroke PO,

552  /G6C   Theory of Machines
(b)The side thrust between the
crosshead and the guide bars(F
N) acting at P and right angles
to line of stroke PO,
(c)The weight of the connecting rod
(WC = mC.g),
(d)Inertia force of the connecting
rod (FC),
(e)The radial force ( FR) acting
through O and parallel to the
crank OC,
(f)The force ( FT) acting perpen-
dicular to the crank OC.
Now, produce the lines of action of
FR and FN to intersect at a point I, known as
instantaneous centre. From I draw I X and I

Y, perpendicular to the lines of action of FC
and WC. Taking moments about I, we have
FT × IC=FI × IP + FC × I X + WC × I Y ...(ii)
The value of FT may be obtained from this equation and from the force polygon as shown in
Fig. 15.22, the forces FN and FR may be calculated. We know that, torque exerted on the crankshaft to
overcome the inertia of the moving parts = FT × OC
Note : When the mass of the reciprocating parts is neglected, then FI is zero.

15.15.  Analytical Method for Inertia Torque
The effect of the inertia of the connecting rod on the crankshaft torque may be obtained as
discussed in the following steps:
Fig. 15.23.  Analytical method for inertia torque.
1. The mass of the connecting rod ( mC) is divided into two masses. One of the mass is placed
at the crosshead pin P and the other at the crankpin C as shown in Fig. 15.23, so that the centre of
gravity of these two masses coincides with the centre of gravity of the rod G.

2. Since the inertia force due to the mass at C acts radially outwards along the crank OC,
therefore the mass at C has no effect on the crankshaft torque.
3. The inertia force of the mass at P may be obtained as follows:
Let mC= Mass of the connecting rod,
l= Length of the connecting rod,
l1= Length of the centre of gravity of the connecting rod from P.
Radial engines of a motor cycle.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  553
∴  Mass of the connecting rod at P,
1
Cllm
l−=×
The mass of the reciprocating parts ( mR) is also acting at P. Therefore,
Total equivalent mass of the reciprocating parts acting at P
1
RCllmml−=+ ×
∴  Total inertia force of the equivalent mass acting at P,
1
IR RCllFamml−=+×...(i)
where aR = Acceleration of the reciprocating parts
2 cos 2.cosr
∴2 1
I RCcos 2.cosllFrmmn l−θ  =ω θ+ +×   

and corresponding torque exerted on the crank shaft,
II I 22sin 2sin.
2s i nTFO MF r
nθ θ+ =× = −θ...(ii)
Note :  Usually the value of OM is measured by drawing the perpendicular from O on PO which intersects PC
produced at M.
4. In deriving the equation (ii) of the torque exerted on the crankshaft, it is assumed that one
of the two masses is placed at C and the other at P. This assumption does not satisfy the condition for

kinetically equivalent system of a rigid bar. Hence to compensate for it, a correcting torque is neces-sary whose value is given by
22
CP C C 1 P C1G.( ) () ( ) Tm m l l L kk ′=α = − α −
where L= Equivalent length of a simple pendulum when swung about an
axis through P
22
G1
1() ( )kl
l+=
αPC= Angular acceleration of the connecting rod PC.
2sin
n−ω θ= ...(From Art. 15.9)
The correcting torque T' may be applied to the system by two equal and opposite forces FY

acting through P and C. Therefore,
FY × PN=T'    or   FY  =  T'/PN
and corresponding torque on the crankshaft,
CYTT F NO NO
PN′=× = × ...(iii)
We know that, NO=OC cos θ = r cos θ
and PN=PC cos φ = l cos φ

554  /G6C   Theory of Machines
∴                       cos cos
cos cosNO r
PN l nθθ==φφ...lnr=/G51
                                               22 2
2cos cos
sin sin1nn
nθθ==
θ− θ−2
2sin...cos 1
nθφ= −/G51
Since sin2θ is very small as compared to n2, therefore neglecting sin2θ, we have
                                       cosNO
PN nθ=
Substituting this value in equation  (iii), we have
                                          CC 1 P Ccos cos() TT ml l L
nnθθ′=× = × −α×

                                               2
C1sin cos()ml l Lnnωθ θ=− × − ×        2
PCsin...
n −ω θα=/G51
                                               2
C1 2sin 2()
2ml l L
nωθ=− × −()...2s in c os si n2θθ = θ /G51
5.  The equivalent mass of the rod acting at C,
                                         1
2Clmm
l=×
∴  Torque exerted on the crank shaft due to mass m2 ,
                    11
W2 C C cosllTm g N O m g N O m g r
ll=− × × =− × × × =− × × × θ

...( cos )NO r=θ /G51
                                               1
C coslmg
n=− × × × θ ...( / )lr n= /G51
6. The total torque exerted on the crankshaft due to the inertia of the moving parts is the
algebraic sum of TI , TC and TW.
Example 15.19. The crank and connecting rod lengths of an engine are 125 mm and 500

mm respectively. The mass of the connecting rod is 60 kg and its centre of gravity is 275 mm from thecrosshead pin centre, the radius of gyration about centre of gravity being 150 mm.
If the engine speed is 600 r.p.m. for a crank position of 45° from the inner dead centre,
determine, using Klien’s or any other construction 1. the acceleration of the piston; 2. the magni-
tude, position and direction of inertia force due to the mass of the connecting rod.

Solution. Given : r = OC = 125 mm; l = PC = 500 mm; m
C = 60 kg ; PG = 275 mm ;
mC = 60 kg ; PG = 275 mm ; kG = 150 mm; N = 600 r.p.m. or ω = 2π × 600/60 = 62.84 rad/s ; θ = 45°
1. Acceleration of the piston
Let                          aP = Acceleration of the piston.
First of all, draw the configuration diagram OCP, as shown in Fig. 15.24, to some suitable
scale, such that
                           OC  = r = 125 mm ; PC = l = 500 mm ; and θ = 45°.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  555
Now, draw the Klien’s acceleration diagram OCQN, as shown in Fig. 15.24, in the same
manner as already discussed. By measurement,
                                    NO = 90 mm = 0.09 m
∴  Acceleration of the piston ,
                                               aP = ω2 × NO = (62.84)2  × 0.09 = 355.4 m/s  Ans.
Fig. 15.24
2.  The magnitude, position and direction of inertia force due to the mass of the connecting rod

The magnitude, postition and direction of the inertia force may be obtained as follows:
(i) Replace the connecting rod by dynamical equivalent system of two masses, assuming that
one of the masses is placed at P and the other mass at D. The position of the point D is obtained as
discussed in Art. 15.12.
(ii) Locate the points G and D on NC which is the acceleration image of the connecting rod.
Let these points are g and d on NC. Join gO and dO. By measurement,

gO =103 mm = 0.103 m
∴  Acceleration of G,   aG = ω2 × gO, acting in the direction from g to O.
(iii) From point D, draw DE parallel to dO. Now E is the point through which the inertia force
of the connecting rod passes. The magnitude of the inertia force of the connecting rod  is given by
                       FC  = mC × ω2 × gO = 60 × (62.84)2 × 0.103 = 24 400 N = 24.4 kN Ans.

(iv) From point E, draw a line parallel to gO, which shows the position of the inertia force of
the connecting rod and acts in the opposite direction of gO.
Example 15.20. The following data refer to a steam engine:
Diameter of piston = 240 mm; stroke = 600 mm ; length of connecting rod = 1.5 m ; mass of
reciprocating parts = 300 kg; mass of connecting r od = 250 kg; speed = 125 r. p.m ; centre of gravity

of connecting rod from crank pin = 500 mm ; radius of gyration of the connecting rod about an axisthrough the centre of gravity = 650 mm.

556  /G6C   Theory of Machines
Determine the magnitude and direction of the torque exerted on the crankshaft when the
crank has turned through 30° from inner dead centre.
Solution. Given:   D = 240 mm = 0.24 m ;  L = 600 mm  or  r = L/2 = 300 mm = 0.3 m ;
l = 1.5 m ; mR = 300 kg ;  mC = 250 kg;   N = 125 r.p.m.  or   ω = 2π × 125/60 = 13.1 rad/s ;
GC = 500 mm = 0.5 m;  kG = 650 mm = 0.65 m; θ = 30°
The inertia torque on the crankshaft may be determined by graphical method or analytical

method as discussed below:
First of all, draw the configuration diagram OCP, as shown in Fig. 15.25, to some suitable
                                     OC  = r = 300 mm; PC = l = 1.5 m; and angle POC = θ = 30°.
Fig. 15.25
Now draw the Klien’s acceleration diagram OCQN, as shown in Fig. 15.25, and complete the
figure in the similar manner as discussed in Art. 15.14.
By measurement; NO = 0.28 m ; gO = 0.28 m ; IP = 1.03 m ; I X = 0.38 m ; I Y = 0.98 m,

and IC = 1.7 m.
We know that inertia force of reciprocating parts,
22
IR 300 (13.1) 0.28 14 415 N Fm N O=× ω ×=× × =
and inertia force of connecting rod,
22
CC 250 (13.1) 0.28 12 013 N Fm g O=× ω ×=× × =
Let FT= Force acting perpendicular to the crank OC.
Taking moments about point I,
TICCF IC F IP W IY F IX×=×+ ×+×

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  557
T1.7 14 415 1.03 250 9.81 0.98 12013 0.38 21816F×= × + × × + × =
∴T2.816/1.7 12 833 NF== CC...( . )Wm g= /G51
We know that torque exerted on the crankshaft
T12 833 0.3 3850 N-m Fr=× = ×= Ans.
We know that the distance of centre of gravity ( G) of the connecting rod from P, i.e.,
l1 = l – GC = 1.5 – 0.5 = 1 m
∴  Inertia force due to total mass of the reciprocating parts at P,
2 1

2 1
I RCcos 2.cos5llFrmml−θ =ω θ+ +×   
                            2 1.5 1 cos 60(13.1) .0.3 19 064 N 300 250 cos 301.5 5−°   =× × =+× ° +    
1.5... 50.3lnr== =/G51
∴  Corresponding torque due to FI ,
nθ θ+ =× = −θ
     22sin 60sin 30 19 064 0.3
25 s i n3 0° °+ =× −°
     5719.2 0.587 3357 N-m (anticlockwise)=× =
Equivalent length of a simple pendulum when swung about an axis through P,
22 22
G1

22 22
G1
1() ( ) (0.65) 11.42 m1klLl+ +== =
∴  Correcting torque,
2
CC 1 2sin 2.( )
2Tm l l L
nωθ=− 

2
2(13.1) sin 60250 1(1.5 1.42) 59.5 N-m (anticlockwise)
25 °=× − = ×
Torque due to the weight of the connecting rod at C,
11
WC C cos cos
1250 9.81 cos 30 424.8 N-m (anticlockwise)5llTW mgnn=× ×θ =× × ×θ
=×× × ° =
∴  Total torque exerted on the crankshaft,
ICW
3357 59.5 424.8 3841.3 N-m (anticlockwise)TTT=++
=+ + =

=+ + =
Note: The slight difference in results arrived at by the above two methods is mainly due to error in measurement
in graphical method.Ans.

558  /G6C   Theory of Machines
Example 15.21.  A vertical engine running at 1200 r.p.m. with a stroke of 110 mm, has a
connecting rod 250 mm between centres and mass 1.25 kg. The mass centre of the connecting  rod is75 mm from the big end centre and when suspended as a pendulum from the gudgeon pin axis makes21 complete oscillations in 20 seconds.
1. Calculate the radius of gyration of the connecting rod about an axis through its mass
centre.

centre.
2. When the crank is at 40° from the top dead centre and the piston is moving downwards,
find analytically, the acceleration of the pi ston and the angular acceleration of the connecting rod.
Hence find the inertia torque exerted on the cranks haft. To make t he two-mass system to be dynami-
cally equivalent to the connecting rod, necessary correction torque has to be applied and since theengine is vertical, gravity effects are to be considered.

Solution. Given : N = 1200 r.p.m. or ω = 2π × 1200/60 = 125.7 rad/s ;
L = 110 mm or r = L/2 = 55 mm = 0.055 m;  l = PC = 250 mm = 0.25 m;
m
C = 1.25 kg; CG = 75 mm = 0.075 m ; θ = 40°
The configuration diagram of the engine is shown in Fig. 15.26.
1. Radius of gyration of the connecting rod about an axis through its mass
centre
Let kG= Radius of gyration of the connecting rod about
an axis through its mass centre,
l1= Distance of the centre of gravity from the point
of suspension = PG

= 250 – 75 = 175 mm = 0.175 m
Since the connecting rod makes 21 complete oscillations in 20 seco nds,
therefore frequency of oscillation,
Z211.05 H
20n==
We know that for a compound pendulum, frequency of oscillation,
           2 11
22 2 22
G1 G1.. 11or
2() ( ) 4 () ( )gl glnn
kl kl== ×
π +π +
...(Squaring both sides)
and                 22 2 2 1
G1 22 2 2. 9.81 0.175( ) ( ) (0.175) 0.0088 m
44 ( 1 . 0 5 )glkl
n×=− = − =
ππ ×
∴                    kG = 0.094 m = 94 mm Ans.

2.  Acceleration of the piston
We know that acceleration of the piston,
22
Pcos 80 cos 2cos 40 . (125.7) 0.055cos0.25/0.055ar
n°  θ °+ =ω = θ+  
= 698.7 m/s2 Ans.  ...( / )nl r=/G51
We know that mass of the connecting rod at P,
22
2
PCsin (125.7) sin402234.4 rad/s0.25/0.055 n−ω θ − °α= = = −  Ans.Fig. 15.26

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  559
Inertia torque exerted on the crankshaft
                          1
1C0.25 0.1751.25 0.375 kg
0.25llmm
l− −=× = × =
∴ Vertical inertia force,
                          I1 P. 0.375 698.7 262 NFm a==×=
and corresponding torque due to FI,
                           II 262 0.0425 11.135 N-m TF O M=− × =− × =−

=11.135 N-m (anticlockwise) ...(By measurement, OM = 0.0425 m)
We know that the equivalent length of a simple pendulum when swung about an axis passing
through P,
                            22 22
G1
1() ( ) (0.094) (0.175)0.225 m0.175klLl+ +== =
C1 P C. ( ) 1.25 0.175 (0.25 0.225) 2234.4 12.22 N-m Tm l l L′=− − α =− × − =−
Corresponding torque on the crankshaft,
       Ccos 12.22 cos 402.06 N-m 2.06 N-m (anticlockwise)
0.25/0.055TT

n′θ− × °== = − =
Torque due to the mass at P,
P1 0.375 9.81 0.0425 0.156 N-m (clockwise) Tmg O M=× × = × × =
Equivalent mass of the connecting rod at C,
                          1
2C0.1751.25 0.875 kg0.25lmml=× = × =
Torque due to mass at C,
W2 0.875 9.81 0.035 0.3 N-m (clockwise) Tm g N C=× × = × × =
...(By measurement, NC = 0.035 m)
∴  Inertia torque exerted on the crankshaft
                               ICPWTTTT=+−−

ICPWTTTT=+−−
                                11.135 2.06 0.156 0.3 12.739 N-m (anticlockwise)=+ − − = Ans.
Example 15.22.  The connecting rod of an internal combustion engine is 225 mm long and
has a mass 1.6 kg. The mass of the piston and gudgeon pin is 2.4 kg and the stroke is 150 mm. Thecylinder bore is 112.5 mm. The cent re of gravity of the connection rod is 150 mm from the small end.

Its radius of gyration about the centre of gravity for oscillations in the plane of swing of the connect-ing rod is 87.5 mm. Determine the magnitude and direction of the resultant force on the crank pinwhen the crank is at 40° and the piston is moving away from inner dead centre under an effective gaspresure of 1.8 MN/m
2. The engine speed is 1200 r.p.m.
Solution.  Given : l = PC = 225 mm = 0.225 m; mC = 1.6 kg; mR = 2.4 kg; L = 150 mm or

r = L/2 = 75 mm = 0.075 m ; D = 112.5 mm = 0.1125 m ;  PG = 150 mm ; kG = 87.5 mm = 0.0875 m ;
θ = 40° ; p = 1.8 MN/m2 = 1.8 × 106 N/m2; N = 1200 r.p.m. or ω = 2π × 1200/60 = 125.7 rad/s

560  /G6C   Theory of Machines
First of all, draw the configuration diagram OCP, as shown in Fig. 15.27 to some suitable
scale, such that OC = r = 75 mm ; PC = l = 225 mm ; and θ = 40°.
Fig. 15.27
Now, draw the Klien’s acceleration diagram OCQN. Complete the diagram in the same manner
as discussed earlier. By measurement,
NO = 0.0625 m ; gO = 0.0685 m ; IC = 0.29 m; IP = 0.24 m; IY = 0.148 m ; and IX = 0.08 m
22 6

22 6
L (0.1125) 1.8 10 17 895 N44FD pππ=× ×=× × × =
Inertia force due to mass of the reciprocating parts,
22
IR 2.4 (125.7) 0.0625 2370 N Fm N O=× ω ×= × =
∴  Net force on the piston,
PLI 17 895 2370 15 525 N FFF=−= − =
Inertia force due to mass of the connecting rod,
22
CC 1.6 (125.7) 0.0685 1732 N Fm g O=× ω ×=× × =
Let FT = Force acting perpendicular to the crank OC.
Now, taking moments about point I,
PCCT

PCCT
T 15 525 0.24 1.6 9.81 0.148 1732 0.08 0.29F IP W IY F IX F IC
F×=×+×+×
×= ×× +×+ ×
∴  T12 362 NF= CC...( . )Wm g= /G51
Let us now find the values of FN and FR in magnitude and direction. Draw the force polygon
as shown in Fig. 15.25.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  561
By measurement, FN = 3550 N; and FR = 7550 N
The magnitude and direction of the resultant force on the crank pin is given by FQ , which is
the resultant of FR and FT.
By measurement, FQ = 13 750 N  Ans.
EXERCISES
1.The crank and connecting rod of a reciprocating engine are 150 mm and 600 mm respectively. The
crank makes an angle of 60° with the inner dead centre and revolves at a uniform speed of 300 r.p.m.

Find, by Klein’s or Ritterhaus’s construction, 1. Velocity and acceleration of the piston, 2. Velocityand acceleration of the mid-point D of the connecting rod, and 3. Angular velocity and angular accel-
eration of the connecting rod. [Ans. 4.6 m/s, 61.7 m/s
2 ; 4.6 m/s, 93.8 m/s2 ; 4.17 rad/s, 214 rad/s2]
2.In a slider crank mechanism, the length of the crank and connecting rod are 100 mm and 400 mm

respectively. The crank rotates uniformly at 600 r.p.m. clockwise. When the crank has turned through45° from the inner dead centre, find, by analytical method : 1. Velocity and acceleration of the slider,2. Angular velocity and angular acceleration of the connecting rod. Check your result by Klein’s orBennett’s construction. [Ans.  5.2 m/s; 279 m/s
2; 11 rad/s; 698 rad/s2]
3.A petrol engine has a stroke of 120 mm and connecting rod is 3 times the crank length. The crank

rotates at 1500 r.p.m. in clockwise direction. Determine: 1. Velocity and acceleration of the piston,and 2. Angular velocity and angular acceleration of the connecting rod, when the piston had travelledone-fourth of its stroke from I.D.C. [Ans. 8.24 m/s, 1047 m/s
2; 37 rad/s, 5816 rad/s2]
4.The stroke of a steam engine is 600 mm and the length of connecting rod is 1.5 m. The crank rotates

at 180 r.p.m. Determine: 1. velocity and acceleration of the piston when crank has travelled throughan angle of 40° from inner dead centre, and 2. the position of the crank for zero acceleration of thepiston. [Ans. 4.2 m/s, 85.4 m/s
2; 79.3° from I.D.C]
5.The following data refer to a steam engine :
Diameter of piston = 240 mm; stroke = 600 mm; length of connecting rod = 1.5 m; mass of reciprocat-
ing parts = 300 kg; speed = 125 r.p.m.

Determine the magnitude and direction of the inertia force on the crankshaft when the crank has
turned through 30° from inner dead centre. [Ans. 14.92 kN]
6.A vertical petrol engine 150 mm diameter and 200 mm stroke has a connecting rod 350 mm long. The
mass of the piston is 1.6 kg and the engine speed is 1800 r.p.m. On the expansion stroke with crankangle 30° from top dead centre, the gas pressure is 750 kN/m
2. Determine the net thrust on the piston.

[Ans.  7535 N]
7.A horizontal steam engine running at 240 r.p.m. has a bore of 300 mm and stroke 600 mm. The
connecting rod is 1.05 m long and the mass of reciprocating parts is 60 kg. When the crank is 60° pastits inner dead centre, the steam pressure on the cover side of the piston is 1.125 N/mm
2 while that on
the crank side is 0.125 N/mm2. Neglecting the area of the piston rod, determine : 1. the force in the

piston rod ; and 2. the turning moment on the crankshaft. [Ans. 66.6 kN ; 19.86 kN-m]
8.A steam engine 200 mm bore and 300 mm stroke has a connecting rod 625 mm long. The mass of the
reciprocating parts is 15 kg and the speed is 250 r.p.m.When the crank is at 30° to the inner dead centre
and moving outwards, the difference in steam pressures is 840 kN/m2. If the crank pin radius is 30 mm,
determine:  1. the force on the crankshaft bearing; and 2. the t orque acting on the frame.

[Ans. 20.04 kN ; 2253 N-m]
9.A vertical single cylinder engine has a cylinder diameter of 250 mm and a stroke of 450 mm. The
reciprocating parts have a mass of 180 kg. The connecting rod is 4 times the crank radius and thespeed is 360 r.p.m. When the crank has turned through an angle of 45° from top dead centre, the netpressure on the piston is 1.05 MN/m
2. Calculate the effective turning moment on the crankshaft for
this position. [Ans. 2368 N-m]

562  /G6C   Theory of Machines
10.A horizontal, double acting steam engine has a stroke of 300 mm and runs at 240 r.p.m. The  cylinder
diameter is 200 mm, connecting rod is 750 mm long and the mass of the reciprocating parts is 70 kg.The steam is admitted at 600 kN/m
2 for one-third of the stroke, after which expansion takes place
according to the hyperbolic law p.V = constant. The exhaust pressure is 20 kN/m2. Neglecting the

effect of clearance and the diameter of the piston rod, find : 1. Thrust in the connecting rod, and 2.Effective turning moment on the crankshaft when the crank has turned through 120° from inner deadcentre. [Ans. 11.506 kN; 1322 N-m]
11.A horizontal steam engine running at 150 r.p.m. has a bore of 200 mm and a stroke of 400 mm. The

connecting rod is 1 m long and the reciprocating parts has a mass of 60 kg. When thecrank has turned through an angle of 30° from inner dead centre, steam pressure on the cover side is0.6 N/mm
2  while on the crankside is 0.1 N/mm2. Neglecting the area of the piston rod, determine:
1. turning moment on the crankshaft, 2. acceleration of the flywheel, if the mean resistance torque is600 N-m and the moment of inertia is 2.8 kg-m
2. [Ans. 1508 N-m; 324.3 rad/s2]

12.The ratio of the connecting rod length to crank length for a vertical petrol engine is 4:1. The bore /
stroke is 80/100 mm and mass of the reciprocating parts is 1 kg. The gas pressure on the piston is0.7N/mm
2 when it has moved 10 mm from T.D.C. on its power stroke. Determine the net load on the
gudgeon pin. The engine runs at 1800 r.p.m. At what engine speed will this load be zero?
[Ans. 1862.8 N; 2616 r.p.m.]

13.A petrol engine 90 mm in diameter and 120 mm stroke has a connecting rod of 240 mm length. The
piston has a mass of 1 kg and the speed is 1800 r.p.m. On the explosion stroke with the crank at 30°from top dead centre, the gas pressure is 0.5 N/mm
2. Find :
1. the resultant load on the gudgeon pin, 2. the thrust on the cylinder walls, and 3. the speed, above
which other things remaining same, the gudgeon pin load would be reserved in direction.

Also calculate the crank effort at the given position of the crank.
[Ans. 1078 N; 136 N ; 2212 r.p.m.; 39.4 N-m]
14.A single cylinder vertical engine has a bore of 300 mm, storke 360 mm and a connecting rod of length

720 mm. The mass of the reciprocating parts is 130 kg. When the piston is at quarter stroke from topdead centre and is moving downwards, the net pressure on it is 0.6 MPa. If the speed of the engine is250 r.p.m., calculate the turning moment on the crankshaft at the instant corresponding to the positionstated above. [Ans. 6295 N-m]
15.A horizontal, single cylinder, single acting, otto cycle gas engine has a bore of 300 mm and a stroke of

500 mm. The engine runs at 180 r.p.m. The ratio of compression is 5.5. The maximum explosionpressure is 3.2 N/mm
2 gauge and expansion follows the law p.V1.3 = constant. If the mass of the piston
is 150 kg and the connecting rod is 1.25 m long. Calculate the turning moment on the crankshaft whenthe crank has turned through 60° from the inner dead centre. The atmospheric pressure is 0.1 N/mm
2.
[Ans. 15.6 kN-m]

2.
16.A vertical single cylinder, diesel engine running at 300 r.p.m. has a cylinder diameter 250 mm
and stroke 400 mm. The mass of the reciprocating parts is 200 kg. The length of the connectingrod is 0.8 m. The ratio of compression is 14 and the pressure remains constant during injection ofoil for 1/10th of stroke. If the index of the law of expansion and compression is 1.35, find the torque

on the crankshaft when it makes an angle of 60° with the top dead centre during the expansion stroke.The suction pressure may be taken as 0.1 N/mm
2. [Ans. 7034 N-m]
17.A gas engine is coupled to a compressor, the two cylinders being horizontally opposed with the
pistons connected to a common crank pin. The stroke of each piston is 500 mm and the ratio of the

length of the connecting rod to the length of crank is 5. The cylinder diameters are 200 mm and 250 mmand the masses of reciprocating parts are 130 kg and 150 kg respectively. When the crank has movedthrough 60° from inner  dead centre on the firing stroke, the pressure of gas on the engine cylinder is1 N/mm
2 gauge and the pressure  in the compressor cylinder is 0.1 N/mm2 gauge. If the crank moves

with 200 r.p.m. and the flywheel of radius of gyration 1 m has a mass of 1350 kg, determine theangular acceleration of the flywheel. [Ans.  2.4 rad/s
2]
18.The length of a connecting rod of an engine is 500 mm measured between the centres and its mass is 18
kg. The centre of gravity is 125 mm from the crank pin centre and the crank radius is 100 mm.

Chapter 15 : Inertia Forces in Reciprocating Parts    /G6C  563
Determine the dynamically equivalent system keeping one mass at the small end. The frequency of
oscillation of the rod, when suspended from the centre of the small end is 43 vibrations per minute.
[Ans. 4.14 kg; 13.86 kg]
19.A small connecting rod 220 mm long between centres has a mass of 2 kg and a moment of inertia of
0.02 kg-m2 about its centre of gravity. The centre of gravity is located at a distance of 150 mm from

the small end centre. Determine the dynamically equivalent two mass system when one mass is lo-cated at the small end centre.
If the connecting rod is replaced by two masses located at the two centres, find the correction couple
that must be applied for complete dynamical equivalence of the system when the angular accelerationof the connecting rod is 20 000 rad/s
2 anticlockwise.
[Ans. 0.617 kg; 1.383 kg; 20 N-m (anticlockwise)]

20.The connecting rod of a horizontal reciprocating engine is 400 mm and length of the stroke is 200
mm. The mass of the reciprocating parts is 125 kg and that the connecting rod is 100 kg. The radiusof gyration of the connecting rod about an ax is through the centre of  gravity is 120 mm and the

distance of centre of gravity of the connecting rod from big end centre is 160 mm. The engine runs at750 r.p.m. Determine the torque exerted on the crankshaft when the crank has turned 30° from theinner dead centre. [Ans. 7078 N-m]
21.If the crank has turned through 135° from the inner dead centre in the above question, find the torque
on the crankshaft. [Ans. 5235 N-m]
1.Define ‘inertia force’ and ‘inertia torque’.

2.Draw and explain Klien’s construction for determining the velocity and acceleration of the piston in a
slider crank mechanism.
3.Explain Ritterhaus’s and Bennett’s constructions for determining the acceleration of the piston of a
reciprocating engine.
4.How are velocity and acceleration of the slider of a single slider crank chain determined analytically?

5.Derive an expression for the inertia force due to reciprocating mass in reciprocating engine, ne-
glecting the mass of the connecting rod.
6.What is the difference between piston effort, crank effort and crank-pin effort?
7.Discuss the method of finding the crank effort in a reciprocating single acting, single cylinder petrol
engine.
8.The inertia of the connecting rod can be replaced by two masses concentrated at two points and

connected rigidly together. How to determine the two masses so that it is dynamically equivalent tothe connecting rod ?  Show this.
9.Given acceleration image of a link. Explain how dynamical equivalent system can be used to
determine the direction of inertia force on it.
10.Describe the graphical and analytical method of finding the inertia torque on the crankshaft of a
horizontal reciprocating engine.

11.Derive an expression for the correction torque to be applied to a crankshaft if the connecting rod of
a reciprocating engine is replaced by two lumped masses at the piston pin and the crank pinrespectively.
1.When the crank is at the inner dead centre, in a horizontal reciprocating steam engine, then the velocity
of the piston will be

564  /G6C   Theory of Machines
2.The acceleration of the piston in a reciprocating steam engine is given by
(a)sin 2.sinr
nθ ωθ+(b)cos 2.cosr
nθ ωθ+
(c)2 sin 2.sinr
nθ ω θ+(d)2 cos 2.cosr
nθ ω θ+
where ω= Angular velocity of the crank,
r= Radius of the crank,
θ= Angle turned by the crank from inner dead centre, and
n= Ratio of length of connecting rod to crank radius.
3.A rigid body, under the action of external forces, can be replaced by two masses placed at a fixed

distance apart. The two masses form an equivalent dynamical system, if
(a) the sum of two masses is equal to the total mass of the body
(b) the centre of gravity of the two masses coincides with that of the body
(c) the sum of mass moment of inertia of the masses about their centre of gravity is equal to the  mass
moment of inertia of the body
(d) all of the above
4.The essential condition of placing the two masses, so that the system becomes dynamically equivalent
is

is
(a)l1 .l2 = kG2(b)l1 .l2 = kG(c)l1  = kG(d)l2 = kG
where l1 and l2  = Distance of two masses from the centre of gravity of the body, and
kG = Radius of gyration of the body.
5.In an engine, the work done by inertia forces in a cycle is
(a) positive ( b) zero ( c) negative ( d) none of these
ANSWERS
1.(a) 2.(d) 3.(d) 4.(a) 5.(a)

Chapter 16 : Turning Moment Diagrams and Flywheel            /G6C          565
16.1.16.1.16.1.16.1.16.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoduction
The turning moment diagram (also known as crank-
effort diagram ) is the graphical representation of the turning
moment or crank-effort for various positions of the crank. It is
plotted on cartesian co-ordinates, in which the turning momentis taken as the ordinate and crank angle as abscissa.

16.2.16.2.16.2.16.2.16.2. TTTTTurururururning Moment Diagram fning Moment Diagram fning Moment Diagram fning Moment Diagram fning Moment Diagram f or a Singleor a Singleor a Singleor a Singleor a Single
Cylinder Double Acting Steam EngineCylinder Double Acting Steam EngineCylinder Double Acting Steam EngineCylinder Double Acting Steam EngineCylinder Double Acting Steam Engine
A turning moment diagram for a single cylinder

double acting steam engine is shown in Fig. 16.1. The verticalordinate represents the turning moment and the horizontalordinate represents the crank angle.
We have discussed in Chapter 15 (Art. 15.10.) that
the turning moment on the crankshaft,
        
P
22sin 2sin
2s i nTF r
nθ =× θ +−θ
565TTTTTurururururningningningningning
MomentMomentMomentMomentMoment
DiagramsDiagramsDiagramsDiagramsDiagrams

and Flywheeland Flywheeland Flywheeland Flywheeland Flywheel16
2. Turning Moment Diagram
for a Single Cylinder
Double Acting SteamEngine.
3. Turning Moment Diagram
for a Four Stroke CycleInternal CombustionEngine.
4. Turning Moment Diagram
for a Multicylinder Engine.
5. Fluctuation of Energy.
6. Determination of Maximum
Fluctuation of Energy.
7. Coefficient of Fluctuation
of Energy.

of Energy.
8. Flywheel.9. Coefficient of Fluctuation of
Speed.
10. Energy Stored in a
Flywheel.
11. Dimensions of the Flywheel
Rim.
12. Flywheel in Punching Press.

566  /G6C   Theory of Machines
Fig. 16.1. Turning moment diagram for a single cylinder, double acting steam engine.
where                   FP= Piston effort,
r= Radius of crank,
n= Ratio of the connecting rod length and radius of crank, and
θ= Angle turned by the crank from inner dead centre.
From the above expression, we see
that the turning moment ( T) is zero, when the
crank angle ( θ) is zero. It is maximum when
the crank angle is 90° and it is again zero whencrank angle is 180°.

This is shown by the curve abc in
Fig. 16.1 and it represents the turningmoment diagram for outstroke. The curvecde is the turning moment diagram for
instroke and is somewhat similar to thecurve abc.
Since the work done is the product
of the turning moment and the angle turned,therefore the area of the turning momentdiagram represents the work done perrevolution. In actual practice, the engine isassumed to work against the mean resistingtorque, as shown by a horizontal  line AF.

The height of the ordinate a A represents the
mean height of the turning moment diagram.Since it is assumed that the work done bythe turning moment per revolution is equalto the work done against the mean resistingtorque, therefore the area of the rectangleaAFe is proportional to the work done against
the mean resisting torque.
Notes: 1.  When the turning moment is positive ( i.e. when the engine torque is more than the mean resisting

torque) as shown between points B and C (or D and E) in Fig. 16.1, the crankshaft  accelerates and the work
is done by the steam.
For flywheel, have a look at your tailor’s manual
sewing machine.

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  567
 2.When the turning moment is negative ( i.e. when the engine torque is less than the mean resisting
torque) as shown between points C and D in Fig. 16.1, the crankshaft retards and the work is done on the
steam.
 3.If                       T = Torque on the crankshaft at any instant, and
Tmean= Mean resisting torque.
Then accelerating torque on the rotating parts of the engine
=T – Tmean

=T – Tmean
4.If (T –Tmean) is positive, the flywheel accelerates and if ( T – Tmean) is negative, then the flywheel retards.
16.3. Turning Moment Diagram for a Four Stroke Cycle Internal
Combustion Engine
A turning moment diagram for a four stroke cycle internal combustion engine is shown in
Fig. 16.2. We know that in a four stroke cycle internal combustion engine, there is one workingstroke after the crank has turned through two revolutions,  i.e. 720° (or 4 π radians).

Fig. 16.2. Turning moment diagram for a four stroke cycle internal combustion engine.
Since the pressure inside  the engine cylinder is less than the atmospheric pressure during

the suction stroke, therefore a negative loop is formed as shown in Fig. 16.2. During the compressionstroke, the work is done on the gases, therefore a higher negative loop is obtained. During theexpansion or working stroke, the fuel burns and the gases expand, therefore a large positive loop isobtained. In this stroke, the work is done by the gases. During exhaust stroke, the work is done onthe gases, therefore a negative loop is formed. It may be noted that the effect of the inertia forces

be noted that the effect of the inertia forces onthe piston is taken into account in Fig. 16.2.

16.4. Turning Moment Diagram for a Multi-cylinder Engine
A separate turning moment diagram for a compound steam engine having three cylinders

and the resultant turning moment diagram is shown in Fig. 16.3. The resultant turning  momentdiagram is the sum of the turning moment diagrams for the three cylinders. It may be noted that thefirst cylinder is the high pressure cylinder, second cylinder is the intermediate cylinder and the thirdcylinder is the low pressure cylinder. The cranks, in case of three cylinders, are usually placed at120° to each other.

568  /G6C   Theory of Machines
Fig. 16.3. Turning moment diagram for a multi-cylinder engine.
16.5. Fluctuation of Energy
The fluctuation of energy may be determined by the turning moment diagram for one complete
cycle of operation. Consider the turning moment diagram for a single cylinder double acting steamengine as shown in Fig. 16.1. We see that the mean resisting torque line AF cuts the turning moment

diagram at points B, C, D and E. When the crank moves from a to p, the work done by the engine is
equal to the area aBp, whereas the energy required is represented by the area aABp. In other words,
the engine has done less work (equal to the area a AB) than the requirement. This amount of energy
is taken from the flywheel and hence the speed of the flywheel decreases. Now the crank movesfrom p to q, the work done by the engine is equal to the area pBbCq, whereas the requirement of

energy is represented by the area pBCq. Therefore, the engine has done more work than the
requirement. This excess work (equal to the area  BbC) is stored in the flywheel and hence the speed
of the flywheel increases while the crank moves from p to q.
Similarly, when the crank moves from q to r, more work is taken from the engine than is
developed. This loss of work is represented by the area  C c D. To supply this loss, the flywheel gives

up some of its energy and thus the speed decreases while the crank moves from q to r. As the crank
moves from r to s, excess energy is again developed given by the area  D d E and the speed again
increases. As the piston moves from s to e, again there is a loss of work and the speed decreases. The
variations of energy above and below the mean resisting torque line are called  fluctuations of
energy. The areas BbC, CcD, DdE, etc. represent fluctuations of energy.

A little consideration will show that the engine has a maximum speed either at q or at s. This
is due to the fact that the flywheel absorbs energy while the crank moves from p to q and from r to s.
On the other hand, the engine has a minimum speed either at p or at r. The reason is that the flywheel
gives out some of its energy when the crank moves from a to p and q to r. The difference between the
maximum and the minimum energies is known as  maximum fluctuation of energy .

16.6. Determination of Maximum Fluctuation of Energy
A turning moment diagram for a multi-cylinder engine is shown by a wavy curve in Fig.
16.4. The horizontal line AG represents the mean torque line. Let a1, a3, a5 be the areas above the
mean torque line and a2, a4 and  a6 be the areas below the mean torque line. These areas represent
some quantity of energy which is either added or subtracted from the energy of the moving parts ofthe engine.

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  569
Let the energy in the flywheel at A = E,
then from Fig. 16.4, we have
Energy at B  = E + a1
Energy at C  =  E + a1– a2
Energy at D  =  E + a1 – a2 + a3
Energy at E  =  E + a1 – a2 + a3 – a4
Energy at F  = E + a1 – a2 + a3 – a4 + a5
Energy at G =  E + a1 – a2 + a3 – a4 + a5 – a6
                    = Energy at A (i.e. cycle
                        repeats after G)
Let us now suppose that the greatest of

these energies is at B and least at E. Therefore,
Maximum energy in flywheel
=E + a1
Minimum energy in the flywheel
=E + a1 – a2 + a3 – a4
∴  Maximum fluctuation of energy,
∆ E=Maximum energy – Minimum energy
=(E + a1) – (E + a1 – a2 + a3 – a4) = a2 – a3 + a4
Fig. 16.4.  Determination of maximum fluctuation of energy.
16.7. Coefficient of Fluctuation of Energy
It may be defined as the ratio of the maximum fluctuation of energy to the work done

per cycle.  Mathematically, coefficient of fluctuation of energy,
                                  EMaximum fluctuation of energy
Work done per cycleC=
The work done per cycle (in N-m or joules) may be obtained by using the following two
relations :
1.Work done per cycle = Tmean × θ
where Tmean= Mean torque, and
θ= Angle turned (in radians), in one revolution.
=2π, in case of steam engine and two stroke internal combustion
engines
=4π, in case of four stroke internal combustion engines.

A flywheel stores  energy when the supply
is in excess  and releases energy when
energy is in deficit.

570  /G6C   Theory of Machines
The mean torque ( Tmean) in N-m may be obtained by using the following relation :
                              60
2meanPPTN×==πω
where P= Power transmitted in watts,
N= Speed in r.p.m., and
ω=Angular speed in rad/s  = 2 πN/60
2.  The work done per cycle may also be obtained by using the following relation :
     Work done per cycle  60P
n×=
where n= Number of working strokes per minute,
=N, in case of steam engines and two stroke internal combustion
engines,

engines,
=N /2, in case of four stroke internal combustion engines.
The following table shows the values of coefficient of fluctuation of energy for steam engines
and internal combustion engines.
Table 16.1. Coefficient of fluctuation of energy ( CE) for steam and internal
combustion engines.
      S.No. Type of engine Coefficient of fluctuation
of energy (CE)
1. Single cylinder, double acting steam engine 0.21

2. Cross-compound steam engine 0.0963. Single cylinder, single acting, four stroke gas engine 1.934. Four cylinders, single acting, four stroke gas engine 0.066
5. Six cylinders, single acting, four stroke gas engine 0.031
16.8. Flywheel
A flywheel used in machines serves as a reservoir, which stores energy during the period
when the supply of energy is more than the requirement, and releases it during the period when therequirement of energy is more than the supply.

In case of steam engines, internal combustion engines, reciprocating compressors and pumps,

the energy is developed during one stroke and the engine is to run for the whole cycle on the energyproduced during this one stroke. For example, in internal combustion engines, the energy is developedonly during expansion or power stroke which is much more than the engine load and no energy isbeing developed during suction, compression and exhaust strokes in case of four stroke engines andduring compression in case of two stroke engines. The excess energy developed during power strokeis

The excess energy developed during power strokeis absorbed by the flywheel and releases it to the crankshaft during other strokes in which no energyis developed, thus rotating the crankshaft at a uniform speed. A little consideration will show thatwhen the flywheel absorbs energy, its speed increases and when it releases energy, the speed decreases.Hence a flywheel does not maintain a constant speed, it simply reduces the fluctuation of speed. Inother words,  a flywheel controls the speed

Inother words,  a flywheel controls the speed variations caused by the fluctuation of the engine

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  571
In machines where the operation is intermittent like *punching machines, shearing machines,
rivetting machines, crushers, etc., the flywheel stores energy from the power source during the greaterportion of the operating cycle and gives it up during a small period of the cycle. Thus, the energyfrom the power source to the machines is supplied practically at a constant rate throughout theoperation.

Note:  The function of a  **governor in an engine is entirely different from that of a flywheel. It
regulates the mean speed of an engine when there are variations in the load, e.g., when the load on the engine

increases, it becomes necessary to increase the supply of working fluid. On the other hand, when the loaddecreases, less working fluid is required. The governor automatically controls the supply of working fluid tothe engine with the varying load condition and keeps the mean speed of the engine within certain limits.
 As discussed above, the flywheel does not maintain a constant speed, it simply reduces the fluctuation

of speed. It does not control the speed variations caused by the varying load.
16.9. Coefficient of Fluctuation of Speed
The difference between the maximum and minimum speeds during a cycle is called the
maximum fluctuation of speed . The ratio of the maximum fluctuation of speed to the mean speed is
called the coefficient of fluctuation of speed .
Let N1 and N2= Maximum and minimum speeds in r.p.m. during the cycle, and
N= Mean speed in r.p.m. 12
2NN+
=
∴  Coefficient of fluctuation of speed,

2NN+
=
                                  ()12 12
122
sNN NN
CNN N− −
==+
                                        ()12 12
122ω− ω ω− ω
==ωω + ω...(In terms of angular speeds)
                                        ()12 12
122vv vv
vv v− −
==+...(In terms of linear speeds)
The coefficient of fluctuation of speed is a limiting factor in the design of flywheel. It varies
depending upon the nature of service to which the flywheel is employed.

Note.  The reciprocal of the coefficient of fluctuation of speed is known
as coefficient of steadiness  and is denoted by m.
∴                                
121
sNmCN N==−
16.10. Energy Stored in a Flywheel
A flywheel is shown in Fig. 16.5. We have discussed in
Art. 16.5 that when a flywheel absorbs energy, its speed increasesand when it gives up energy, its speed decreases.
Let m= Mass of the flywheel in kg,
k= Radius of gyration of the
flywheel in metres,
Fig. 16.5. Flywheel.

* See Art. 16.12.
** See Chapter 18 on Governors.

572  /G6C   Theory of Machines
I= Mass moment of inertia of t he flywheel about its axis of rotation
in kg-m2 = m.k2,
N1 and N2= Maximum and minimum speeds during the cycle in r.p.m.,
ω1 and ω2= Maximum and minimum angular speeds during the cycle in rad/s,
N= Mean speed during the cycle in r.p.m. 12,2NN+
=
ω= Mean angular speed during the cycle in rad/s 12,2ω+ ω
=
CS= Coefficient of fluctuation of speed, 12 12orNN
N−ω − ω
=ω
We know that the mean kinetic energy of the flywheel,

22 211.. .22EI m k=×ω =× ω (in N-m or joules)
As the speed of the flywheel changes from ω1 to ω2, the maximum fluctuation of energy,
∆E= Maximum K.E. – Minimum K.E.
                                        () () () ()2 22 2
12 1 2111
222III=×ω −×ω =× ω − ω
                                        () () ()12 12 121.2II= × ω+ ω ω− ω = ωω− ω ...(i)
12...2ω+ ωω=/G51
                                        212.Iω− ω

=ωω... (Multiplying and dividing by ω)
=I.ω2.CS = m.k2.ω2.CS... (∵ I = m.k2)    ...(ii)
=2 .E.CS (in N–m or joules)21... .2EI=×ω/G51  ... (iii)
The radius of gyration ( k) may be taken equal to the mean radius of the rim ( R), because the
thickness of rim is very small as compared to the diameter of rim. Therefore, substituting k = R, in
equation (ii), we have
∆E=m.R2.ω2.CS = m.v2.CS
where v= Mean linear velocity ( i.e. at the mean radius) in m/s = ω.R

Notes.  1.  Since ω = 2 π N/60, therefore equation  (i) may be written as
                                              ()2
12
122224
60 60 60 3600NNNEI IN N Nππππ∆=× − = ×× − 
                                                    ()2
2
12..900mk N N Nπ=× −
                                                    2
S2
2...900mk N Cπ=×12...sNN
CN− =/G51-

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  573
2.In the above expressions, only the mass moment of inertia of the flywheel rim ( I) is considered
and the mass moment of inertia of the hub and arms is neglected. This is due to the fact that the major portion
of the mass of the flywheel is in the rim and a small portion is in the hub and arms. Also the hub and arms are
nearer to the axis of rotation, therefore the mass moment of inertia of the hub and arms is small.

Example 16.1.  The mass of flywheel of an engine is 6.5 tonnes and the radius of gyration
is 1.8 metres. It is found from the turning moment diagram that the fluctuation of energy is56 kN-m. If the mean speed of the engine is 120 r. p.m., find the maximum and minimum speeds.
Solution. Given :m = 6.5 t = 6500 kg;  k = 1.8 m ; ∆ E = 56 kN-m = 56 × 10
3 N-m ;
N = 120 r.p.m.
Let N1 and N2= Maximum and minimum speeds respectively.
We know that fluctuation of energy ( ∆ E),

                         56 × 103 2
900π= × m.k2 . N (N1 – N2) 2
900π= × 6500 (1.8)2 120 (N1 – N2)
= 27715 ( N1 – N2)
∴ N1 – N2= 56 × 103 /27 715 = 2 r.p.m. ... (i)
We also know that mean speed ( N),
                                 12
12120 or 120 2 240 r.p.m.2NN
NN+
=+ = × = ...(ii)
N1= 121 r.p.m., and N2 = 119 r.p.m.     Ans.
Example 16.2.  The flywheel of a steam engine has a radius of gyration of  1 m and mass

2500 kg. The starting torque of the steam engine is  1500 N-m and may be assumed constant .
Determine : 1. the angular acceleration of the flywheel , and 2. the kinetic energy of the flywheel
after 10 seconds from the start .
Solution.  Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m
Let α= Angular acceleration of the flywheel.
I=m.k2 = 2500 ×  12 = 2500 kg-m2
∴ Starting torque of the engine ( T),

1500 =I.α = 2500 × α    or    α = 1500 / 2500 = 0.6 rad /s2  Ans.
2.  Kinetic energy of the flywheel
First of all, let us find out the angular speed of the flywheel after 10 seconds from the start
(i.e. from rest), assuming uniform acceleration.
Let ω1= Angular speed at rest = 0
ω2= Angular speed after 10 seconds, and
t= Time in seconds.
We know that ω2=ω1 + α t = 0 + 0.6 × 10 = 6 rad /s

574  /G6C   Theory of Machines
∴  Kinetic energy  of  the  flywheel
                                        ()22
2112500 6 45 000 N-m = 45 kN-m22I=×ω =× × =  Ans.
Example 16.3.   A horizontal cross compound steam engine develops  300 kW at 90 r.p.m.
The coefficient of fluctuation of energy as f ound from the turning moment diagram is to be  0.1 and
the fluctuation of speed is to be kept within  ± 0.5% of the mean speed . Find the weight of the

flywheel required, if the radius of gyration is 2 metres .
Solution.  Given : P = 300 kW = 300 × 103 W; N = 90 r.p.m.; CE = 0.1; k = 2 m
We know that the mean angular speed,
ω=2  π N/60 = 2 π × 90/60 = 9.426 rad/s
Let ω1 and ω2= Maximum and minimum speeds respectively.
Since the fluctuation of speed is ± 0.5% of mean speed, therefore total fluctuation of speed,
ω1 – ω2=1 %  ω = 0.01 ω
and coefficient of fluctuation of speed,
                                  120.01sCω− ω
==ω

==ω
We know that work done per cycle
=P × 60 / N = 300 × 103 × 60 / 90 = 200 × 103 N-m
∆E= Work done per cycle × CE = 200 × 103 × 0.1 = 20 × 103 N-m
Let m= Mass of the flywheel.
We know that maximum fluctuation of energy ( ∆E ),
20 × 103=m.k2.ω2.CS = m × 22 × (9.426)2 × 0.01 = 3.554 m
∴ m=20 × 103/3.554 = 5630 kg     Ans.
Example 16.4.   The turning moment diagram for a petrol engine is drawn to the following

scales : Turning moment,  1 mm = 5 N-m ; crank angle,  1 mm = 1°. The turning moment diagram
repeats itself at every half revolution of the engine and the areas above and below the mean turningmoment line taken in order  are 295, 685, 40, 340, 960, 270 mm
2. The rotating parts are  equivalent
to a mass of  36 kg at a radius of gyration of  150 mm. Determine the coefficient of fluctuation of
speed when the engine runs at  1800 r.p.m.

Solution.  Given : m = 36 kg ; k = 150 mm = 0.15 m ;  N = 1800 r.p.m. or ω = 2 π × 1800/60
= 188.52 rad /s
Fig. 16.6
The turning moment diagram is shown in Fig. 16.6.

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  575
Since the turning moment scale is 1 mm = 5 N-m and
crank angle scale is 1 mm = 1° = π /180 rad, therefore,
1 mm2 on turning moment diagram
                                        5N - m
180 36ππ=× =
Let the total energy at A = E, then referring to
Fig. 16.6,
Energy at B=E + 295
        ... (Maximum energy)
Energy at C=E + 295 – 685 = E – 390
Energy at D=E – 390 + 40 = E – 350
Energy at E=E – 350 – 340 = E – 690 ... (Minimum energy)

Energy at F=E – 690 + 960 = E + 270
Energy at G=E + 270 – 270 = E = Energy at A
We know that maximum fluctuation of energy,
∆ E= Maximum energy – Minimum energy
=(E + 295) – ( E – 690) = 985 mm2
                                        985 86 N-m 86J36π=× = =
Let CS= Coefficient of fluctuation of speed.
We know that maximum fluctuation of energy ( ∆ E),
86 =m.k2 ω2.CS = 36 × (0.15)2 × (188.52)2 CS = 28 787 CS
∴ CS= 86 / 28 787 = 0.003 or 0.3%     Ans.

Example 16.5.   The turning moment diagram for a multicylinder engine has been drawn to
a scale 1 mm = 600 N-m vertically and  1 mm = 3° horizontally . The intercepted areas between the
output torque curve and the mean resistance line , taken in order from one end , are as follows  :
+ 52, – 124, + 92, – 140, + 85, – 72 and + 107  mm2, when the engine is running at a speed

of 600 r.p.m. If the total fluctuation of speed is not to exceed  ± 1.5% of the  mean , find the necessary
mass of the flywheel of radius  0.5 m.
Solution.  Given : N = 600 r.p.m.  or  ω = 2 π × 600 / 60 = 62.84 rad / s ;  R = 0.5 m
Fig. 16.7
Since the total fluctuation of speed is not to exceed ± 1.5% of the mean speed, therefore
ω1 – ω2=3 %  ω = 0.03 ω
Flywheel of an electric motor.

576  /G6C   Theory of Machines
                                  120.03sCω− ω
==ω
The turning moment diagram is shown in Fig. 16.7.
Since  the  turning  moment  scale is  1 mm  =  600 N-m  and crank angle scale is 1 mm = 3°
= 3° × π/180 = π / 60  rad, therefore
= 600 × π/60 = 31.42 N-m
Let the total energy at A= E, then referring to Fig. 16.7,
Energy at B=E + 52 ...(Maximum energy)
Energy at C=E + 52 – 124 = E – 72

Energy at D=E – 72 + 92 = E + 20
Energy at E=E + 20 – 140 = E – 120 ...(Minimum energy)
Energy at F=E – 120 + 85 = E – 35
Energy at G=E – 35 – 72 = E – 107
Energy at H=E – 107 + 107 = E = Energy at A
=(E + 52) – ( E – 120) = 172 = 172 × 31.42 = 5404 N-m
Let m= Mass of the flywheel in kg.
We know that maximum fluctuation of energy ( ∆ E ),

5404 =m.R2.ω2.CS = m × (0.5)2 × (62.84)2 × 0.03 = 29.6  m
∴ m= 5404 / 29.6 = 183 kg     Ans.
Example 16.6.  A shaft fitted with a flywheel rotates at  250 r.p.m. and drives a machine .
The torque of machine varies in a cyclic manner over a period of  3 revolutions . The torque rises
from 750 N-m to 3000 N-m uniformly during  1/2 revolution and remains constant for the following
revolution . It then falls uniformly to  750 N-m during the next  1/2 revolution and remains constant

for one revolution , the cycle being repeated thereafter .
Determine the power required to drive the machine and percentage fluctuation in speed , if
the driving torque applied to the shaft is constant and the mass of the flywheel is  500 kg with radius
of gyration of  600 mm.
Solution.   Given : N = 250 r.p.m. or  ω = 2π × 250/60 = 26.2 rad/s; m = 500 kg;
k = 600 mm = 0.6 m

The turning moment diagram for the complete cycle is shown in Fig. 16.8.We know that the torque required for one complete cycle
= Area of figure OABCDEF
= Area OAEF + Area ABG + Area BCHG + Area CDH
                                        11
22OF OA AG BG GH CH HD CH=×+ ××+×+ ××

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  577
                                        () ()16 750 3000 750 2 3000 7502=π × + × π − +π −
                                             ()13000 7502+× π −
= 11 250 π N-m ... (i)
If Tmean is the mean torque in N-m, then torque required for one complete cycle
=Tmean × 6 π Ν-m  ... (ii)
                            Tmean = 11 250 π/6  π = 1875 N-m
Fig. 16.8
Power required to drive the machine

Fig. 16.8
We know that power required to drive the machine,
P=Tmean × ω = 1875 × 26.2 = 49 125 W = 49.125 kW   Ans.
Coefficient of fluctuation of speed
Let CS= Coefficient  of  fluctuation of  speed.
First of all, let us find the values of LM and NP. From similar triangles ABG and BLM,
                                   LM BM
AG BG=    or    3000 18750.53000 750LM −==π−    or    LM = 0.5 π
Now, from similar triangles CHD and CNP,

                                  NP CN
HD CH=    or    3000 18750.53000 750NP −==π−    or    NP = 0.5 π
From Fig. 16.8, we find that
BM=CN = 3000 – 1875 = 1125 N-m
Since the area above the mean torque line represents the maximum fluctuation of energy,
therefore, maximum fluctuation of energy,
∆E= Area LBCP = Area LBM + Area MBCN + Area PNC
                                        11
22LM BM MN BM NP CN= ××+×+ × ×

578  /G6C   Theory of Machines
                                        110.5 1125 2 1125 0.5 112522=× π × + π × +× π ×
                                        8837 N-m=
8837 =m.k2.ω2.CS = 500 × (0.6)2 × (26.2)2 × CS = 123 559 CS
S88370.071123559C== Ans.
Flywheel of a  pump run by a  diesel engine.
Example 16.7.  During forward stroke of the piston of the double acting steam engine , the

turning moment has the maximum value of  2000 N-m when the crank makes an angle of  80° with
the inner dead centre . During the backward stroke , the maximum turning moment is  1500 N-m
when the crank makes an angle of  80 ° with the outer dead centre . The turning moment diagram for
the engine may be assumed for simplicity to be represented by two triangles .
If the crank makes  100 r.p.m.  and the radius of gyration of the flywheel is  1.75 m, find the

coefficient of fluctuation of energy and the mass of the flywheel to keep the speed within  ±  0.75% of
the mean speed. Also determine the crank angle at which the speed has its minimum and maximumvalues.
Solution.  Given : N = 100 r.p.m. or ω = 2π × 100/60 = 10.47 rad /s; k = 1.75 m
Since the fluctuation of speed is ± 0.75% of mean speed, therefore total fluctuation of speed,
ω
1 – ω2=1 . 5 %  ω
                                  12
S–
1.5% 0.015 Cωω

S–
== =ω
Coefficient of fluctuation of energy
The turning moment diagram for the engine during forward and backward strokes is shown
in Fig. 16.9. The point  O represents the inner dead centre (I.D.C.) and point G represents the
outer dead centre (O.D.C). We know that maximum turning moment when crank makes anangle of 80° (or 80 × π / 180 = 4 π/9 rad) with I.D.C.,
∴ AB= 2000 N-m

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  579
and maximum turning moment when crank makes an angle of 80° with outer dead centre (O.D.C.) or
180° + 80° = 260° = 260 × π /180 = 13 π / 9 rad with I.D.C.,
LM= 1500 N-m
Let Tmean=EB = QM = Mean resisting torque.
Fig. 16.9
= Area of triangle OAG + Area of triangle GLS
                                        11
22OG AB GS LM=× × +× ×

                                        112000 1500 1750 N-m22= ×π× + ×π× = π ...(i)
We also know that work done per cycle
=Tmean × 2 π N-m  ... (ii)
Tmean= 1750 π / 2 π = 875 N-m
From similar triangles ACD and AOG,
                               CD OG
AE AB=
or                                 () () – 2000 875 1.764 rad2000OG OGCD AE AB EBAB ABπ=×= = − =

                               1 = Area of triangle2E ACD CD AE∆= × ×
                                        () ()11– 1.764 2000 875 992 N-m22CD AB EB=× =× − =
We know that coefficient of fluctuation of energy,
                                 EMax.fluctuation of energy 9920.18 or 18%Work done per cycle 1750C== =π  Ans.

580  /G6C   Theory of Machines
Mass of the flywheel
Let  m= Mass of the flywheel.
We know that maximum fluctuation of energy ( ∆E),
992 =m.k2.ω2.CS = m × (1.75)2 × (10.47)2 × 0.015 = 5.03 m
∴ m= 992 / 5.03 = 197.2 kg   Ans.
Crank angles for the minimum and
maximum speeds
We know that the speed of
the flywheel is minimum at point  C
and maximum at point D (See
Art. 16.5).
Let θC and θD = Crank angles  from
I.D.C., for the minimum and  maximumspeeds.
From similar triangles ACE and
AOB,

AOB,
                 CE AE
OB AB=
or                           – 2000 875 4rad
2000 9 4AE AB EBCE OB OB
AB AB−π π=×= ×= ×=
∴                               C4 7 180= rad 35°
9 4 36 36ππ π 7 πθ− == × =
π    Ans.
Again from similar triangles AED and ABG,
                                ED AE
BG AB=
or                                  ()– AE AB EBED BG OG OBAB AB=×= −
                                        2000 875 4 2.8rad2000 9 9−π π=π − = 

∴                              D4 2.8 6.8 6.8 180= rad 136°
99 9 9πππ πθ+ = = × =
πAns.
Example 16.8.  A three cylinder single acting engine has its cranks set equally at 120 ° and
it runs at 600 r.p.m. The torque-crank angle diagram for each cycle is a triangle for the power stroke
with a maximum torque of  90 N-m at 60° from dead centre of corresponding crank. The torque on the
return stroke is sensibly zero . Determine :  1. power developed.  2. coefficient of fluctuation of speed,

if the mass of the flywheel is  12 kg and has a radius of gyration of  80 mm, 3. coefficient of fluctuation
of energy, and  4. maximum angular acceleration of the flywheel .
Solution.   Given :  N = 600 r.p.m. or  ω = 2 π × 600/60 = 62.84 rad /s;  Tmax = 90 N-m;
m = 12 kg; k = 80 mm = 0.08 m
Flywheel of small steam engine.

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  581
The torque-crank angle diagram for the individual cylinders is shown in Fig. 16.10 ( a), and
the resultant torque-crank angle diagram for the three cylinders is shown in Fig. 16.10 ( b).
Fig. 16.10
1.  Power developed
We know that work done/cycle
                                     1Area of three triangles = 3 90 424 N-m2=× × π × =
and mean torque,             Work done/cycle 42467.5 N-mCrank angle/cycle 2meanT== =π

∴   Power developed   = Tmean × ω = 67.5 × 62.84 = 4240 W = 4.24 kW   Ans.
2.  Coefficient of fluctuation of speed
First of all, let us find the maximum fluctuation of energy ( ∆E).
From Fig. 16.10 ( b), we find that
                                11Area of triangle =2aA a B A B A a=× ×
                                     () 71=6 7 . 5 4 5 5 . 8 9 N - m26aπ×× − = = ...(∵ AB = 30° = π / 6 rad)

21Area of triangle = '2a BbC BC bb=× ×
                                     ()1= 90 67.5 11.78 N-m
23π×−= ...(∵BC = 60° = π/3 rad)
= a3 = a4 = a5 = a6
Now, let the total energy at A = E, then referring to Fig. 16.10 ( b),
Energy at  B=E – 5.89
Energy at  C=E – 5.89 + 11.78 = E + 5.89
Energy at  D=E + 5.89 – 11.78 = E – 5.89
Energy at  E=E – 5.89 + 11.78 = E + 5.89
Energy at  G=E + 5.89 – 11.78 = E – 5.89
Energy at  H=E – 5.89 + 11.78 = E + 5.89

                              Energy at J  = E + 5.89 – 5.89 = E = Energy at A

582  /G6C   Theory of Machines
From above we see that maximum energy
=E + 5.89
and minimum energy = E – 5.89
∴ * Maximum fluctuation of energy,
∆E=(E + 5.89) – ( E – 5.89) = 11.78 N-m
11.78 =m.k2.ω2.CS = 12 × (0.08)2 × (62.84)2 × CS = 303.3 CS
∴ CS= 11.78/303.3 = 0.04 or 4%   Ans.
3.  Coefficient of fluctuation of energy

EMax.fluctuation of energy 11.780.0278 2.78%Work done/cycle 424C== = = Ans.
4.  Maximum angular acceleration of the flywheel
Let α= Maximum angular acceleration of the flywheel.
Tmax – Tmean=I.α = m.k2.α
90 – 67.5 = 12 × (0.08)2 × α = 0.077 α
∴                                 2 90 67.5292rad/s
0.077−α= =  Ans.
Example 16.9.   A single cylinder , single acting , four stroke gas engine develops  20 kW at

300 r.p.m. The work done by the gases during the expansion stroke is three times the work done on
the gases during the compression stroke , the work done during the suction and exhaust strokes
being negligible . If the total fluctuation of speed is not to exceed ± 2 per cent of the mean speed and
the turning moment diagram during compression and expansion is assumed to be triangular inshape, find the moment of inertia of the flywheel .
Solution.  Given : P = 20 kW = 20 × 10

3 W; N = 300 r.p.m. or ω = 2π × 300/60 = 31.42  rad/s
Since the total fluctuation of speed ( ω1 – ω2) is not to exceed ± 2 per cent of the mean speed
(ω), therefore
ω1 – ω2=4 %  ω
                                  12
S 4% 0.04 Cω− ω== =ω
The turning moment-crank angle diagram for a four stroke engine is shown in Fig. 16.11. It

is assumed to be triangular during compression and expansion strokes, neglecting the suction andexhaust strokes.
*Since the area above the mean torque line represents the maximum fluctuation of energy, therefore maxi-
mum fluctuation of energy,
∆E = Area Bbc = Area DdE = Area Ggh
1( 9 0–6 7 . 5 ) 1 1 . 7 8N - m23π=× =

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  583
We know that for a four stroke engine, number of working strokes per cycle,
                            n  = N/2 = 300 / 2 = 150
∴ Work done/cycle = P × 60/n = 20 × 103 × 60/150 = 8000 N-m ... (i)
Fig. 16.11
Since the work done during suction and exhaust strokes is negligible, therefore net work
done per cycle (during compression and expansion strokes)
                               E
ECE E2––33WWWW W=== ... ( ∵ WE = 3WC) ...(ii)

Equating equations  (i) and (ii), work done during expansion stroke,
                       WE  = 8000 × 3/2 = 12 000 N-m
We know that work done during expansion stroke ( WE),
                   1112 000 Area of triangle22ABC BC AG AG == × × = × π ×
∴                   AG   = Tmax = 12 000 × 2/ π = 7638 N-m
and mean turning moment,
                   *Workdone/cycle 8000637 N-mCrank angle/cycle 4meanTF G== = =π
∴ Excess turning moment,

                  Texcess  = AF = AG – FG = 7638 – 637 = 7001 N-m
Now, from similar triangles ADE and ABC,
            DE AF
BC AG=    or    70012.88rad
7638AFDE BC
AG=×= × π =
Since the area above the mean turning moment line represents the maximum fluctuation of
energy, therefore maximum fluctuation of energy,
                        11Area of 2.88 7001 10081 N-m22EA D E D E A F∆= ∆ = × × = × × =

* The mean turning moment ( Tmean) may also be obtained by using the following relation :
P = Tmean × ω or Tmean = P/ω = 20 × 103/31.42 = 637 N-m

584  /G6C   Theory of Machines
Let                                 I = Moment of inertia of the flywheel in kg-m2 .
10 081 = I.ω2.CS = I × (31.42)2 × 0.04 = 39.5 I
∴ I= 10081/ 39.5 = 255.2 kg-m2    Ans.
Example 16.10.   The turning moment diagram for a four stroke gas engine may be assumed
for simplicity to be represented by four triangles, the areas of which from the line of zero pressureare as follows :
Suction stroke  = 0.45 × 10

–3 m2; Compression stroke = 1.7 × 10–3 m2; Expansion stroke
= 6.8 × 10–3 m2; Exhaust stroke = 0.65 × 10–3 m2. Each m2 of area represents 3 MN-m of energy.
Assuming the resisting torque to be uniform, find the mass of the rim of a flywheel required
to keep the speed between 202 and 198 r.p.m. The mean radius of the rim is 1.2 m.
Solution.  Given : a1 = 0.45 × 10–3 m2 ; a2  = 1.7 × 10–3 m2 ; a3  = 6.8 × 10–3 m2;

a4 = 0.65  × 10–3 m2; N1 = 202 r.p.m; N2 = 198 r.p.m.;  R = 1.2 m
The turning moment crank angle diagram for a four stroke engine is shown in Fig. 16.12.
The areas below the zero line of pressure are taken as negative while the areas above the zero line ofpressure are taken as positive.
∴ Net area = a
3 – (a1 + a2 + a4)
= 6.8 × 10–3 – (0.45 × 10–3 + 1.7 × 10–3 + 0.65 × 10–3) = 4 × 10–3 m2
Since the energy scale is 1 m2 = 3 MN-m = 3 × 106 N-m, therefore,

Net work done per cycle = 4 × 10–3 × 3 ×106 = 12 × 103 N-m  . . . (i)
We also know that work done per cycle,
=Tmean × 4π N-m  . . . (ii)
Tmean=FG = 12 × 103/4π = 955 N-m
Fig. 16.12
Work done during expansion stroke
=a3 × Energy scale = 6.8 × 10–3 × 3 × 106 = 20.4 × 103 N-m   ...(iii)

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  585
Also, work done during expansion stroke
= Area of triangle ABC
                                        111.57122BC AG AG AG= ××= × π ×= × . . . (iv)
AG= 20.4 × 103/1.571 = 12 985 N-m
∴  Excess torque,
Texcess=AF = AG – FG = 12 985 – 955 = 12030 N-m
Now from similar triangles ADE and ABC,
                                DE AF
BC AG=    or    12 0302.9 rad12 985AFDE BCAG=×= × π =

We know that the maximum fluctuation of energy,
                                 11Area of 2.9 12030 N-m22E ADE DE AF∆= ∆ =× × =× ×
= 17 444 N-m
Mass of the rim of a flywheel
Let m= Mass of the rim of a flywheel in kg, and
N= Mean speed of the flywheel
                                        12 202 198200 r.p.m.22NN++== =
We know that the maximum fluctuation of energy ( ∆E ),
                            () )22
2 2
1217 444 . . ( – ) 1.2 200 (202 – 198900 900mR N N Nππ=× =× ×
= 12.63 m

= 12.63 m
∴ m= 17444 /12.36 = 1381 kg   Ans.
Example 16.11.  The turning moment curve for an engine is represented by the equation,
T = (20 000 + 9500 sin 2 θ – 5700 cos 2 θ) N-m, where  θ is the angle moved by the crank from
inner dead centre. If the resisting torque is constant, find:
1. Power developed by the engine ; 2. Moment of inertia of flywheel in kg-m2, if the total
fluctuation of speed is not  exceed 1% of mean speed whi ch is 180 r. p.m; and 3. Angular acceleration

of the flywheel when the crank has turned through 45° from inner dead centre.
Solution.  Given : T = (20 000 + 9500 sin 2 θ – 5700 cos 2 θ) N-m; N = 180 r.p.m. or
ω = 2π × 180/60 = 18.85 rad/s
Since the total fluctuation of speed ( ω1 – ω2) is 1% of mean speed ( ω), therefore coefficient
of fluctuation of speed,
                                 12
S–
1% 0.01 Cωω
== =ω
1.  Power developed by the engine
We know that work done per revolution
                                        ()22

586  /G6C   Theory of Machines
                                        29500cos2 5700sin220 000 – –220πθθ =θ
= 20 000 × 2 π = 40 000 π N-m
and mean resisting torque of the engine,
            Workdone per revolution 40 00020000 N-m22meanT== =ππ
We know that power developed by the engine
=Tmean . ω = 20 000 × 18.85 = 377 000 W = 377 kW  Ans.
2.  Moment of inertia of the flywheel
Let I= Moment of inertia of the flywheel in kg-m2.

The turning moment diagram for one stroke ( i.e. half revolution of the crankshaft) is shown
in Fig. 16.13. Since at points B and D, the torque exerted on the crankshaft is equal to the mean
resisting torque on the flywheel, therefore,
T=Tmean
20 000 + 9500 sin 2 θ – 5700 cos 2 θ = 20000
or 9500 sin 2 θ= 5700 cos 2 θ
tan 2θ= sin 2θ/cos 2θ = 5700/9500 = 0.6
∴ 2θ= 31° or θ = 15.5°
∴θB= 15.5° and θD = 90° + 15.5° = 105.5°
Fig. 16.13
Maximum fluctuation of energy,

Fig. 16.13
                                 ()D
B–meanET Tdθ
θ∆= θ∫
                                        ()105.5
15.520 000 9500 sin2 – 5700cos2 – 20 000 d°
°=+ θ θ θ∫
                                        105.5
15.59500cos2 5700sin2– – 11078 N-m22°
°θθ ==

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  587
We know that maximum fluctuation of energy
(∆ E),
11 078 = I.ω2.CS = I × (18.85)2 × 0.01 = 3.55 I
∴     I =11078/3.55 = 3121 kg-m2 Ans.
3.  Angular acceleration of the flywheel
Let α  = Angular acceleration of the flywheel,
              and
      θ   = Angle turned by the crank from inner
              dead centre = 45°               . . . (Given)
The angular acceleration in the flywheel is

produced by the excess torque over the mean torque.We know that excess torque at any instant,
T
excess  = T – Tmean
           = 20000 + 9500 sin 2 θ – 5700 cos 2 θ
                 – 20000
           = 9500 sin 2 θ – 5700 cos 2 θ
∴  Excess torque at 45°
  = 9500 sin 90° – 5700 cos 90° = 9500 N-m . . . (i)
We also know that excess torque
 =I.α = 3121 × α . . . (ii)
 α  =  9500/3121 = 3.044 rad /s2 Ans.

α  =  9500/3121 = 3.044 rad /s2 Ans.
Example 16.12.  A certain machine requires a torque of (5000 + 500 sin θ ) N-m to drive it,
where θ is the angle of rotation of shaft measured from certain datum. The machine is directly
coupled to an engine which produces a torque of (5000 + 600 sin 2 θ) N-m. The flywheel and the

other rotating parts attached to the engine has a mass of 500 kg at a radius of gyration of 0.4 m. Ifthe mean speed is 150 r.p.m., find : 1. the fluctuation of energy, 2. the total percentage fluctuation of
speed, and 3. the maximum and minimum angular acceleration of the flywheel and the corresponding
shaft position.
Solution.   Given : T
1 = ( 5000 + 500 sin θ) N-m; T2 = (5000 + 600 sin 2 θ) N-m;
m = 500 kg; k = 0.4 m; N = 150 r.p.m. or  ω = 2 π × 150/60 = 15.71 rad/s
Fig. 16.14

Fig. 16.14
Nowadays steam turbines  like this can
be produced entirely by computer-
controlled machine tools, directly from the
engineer’s computer.
chapter.

588  /G6C   Theory of Machines
1.  Fluctuation of energy
We know that change in torque                                        = T
2 – T1 = (5000 + 600 sin 2 θ) – (5000 + 500 sin θ)
= 600 sin 2 θ – 500 sin θ
This change is zero when
600 sin 2θ= 500 sin θ    or    1.2 sin 2 θ = sin θ
          1.2 × 2 sin θ cos θ =s i n  θ    or    2.4 sin θ cos θ = sin θ. . . (∵sin 2θ = 2 sin θ cos θ)
∴  Either sin θ= 0  or   cos θ = 1/2.4 = 0.4167
when sin θ=0 ,  θ = 0°, 180° and 360°

i.e. θA=0 ° ,  θC = 180° and θE = 360°
when cos θ= 0.4167, θ = 65.4° and 294.6°
i.e. θB= 65.4° and θD = 294.6°
The turning moment diagram is shown in Fig. 16.14. The maximum fluctuation of energy
lies between C and D (i.e. between 180° and 294.6°), as shown shaded in Fig. 16.14.
                                 ()294.6
21
180– ET T d°
°∆= θ∫
                                        () ( )294.6

1805000 600sin2 – 5000 500sin d°
°=+ θ + θ θ∫
                                        294.6
180600cos2– 500cos 1204N-m2°
°θ =+ θ =Ans.
2.  Total percentage fluctuation of speed
Let CS= Total percentage fluctuation of speed.
1204 =m.k2.ω2.CS = 500 × (0.4)2 × (15.71)2 × CS = 19744 CS
∴ CS= 1204 /19744 = 0.061    or    6.1%  Ans.
3.  Maximum and minimum angular acceleration of the flywheel and the corresponding shaft
positions

positions
The change in torque must be maximum or minimum when acceleration is maximum or
minimum. We know that
Change in torque, T=T2 – T1 = (5000 + 600 sin 2 θ) – (5000 + 500 sin θ)
= 600 sin 2 θ – 500 sin θ ...(i)
Differentiating this expression with respect to θ and equating to zero for maximum or
minimum values.
∴    ()600sin2 – 500sin 0d
dθθ =θ     or    1200 cos 2 θ – 500 cos θ = 0
or                  12 cos 2 θ – 5 cos θ = 0

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  589
12 (2 cos2θ – 1) – 5 cos θ = 0 . . . (∵ cos 2θ = 2 cos2θ – 1)
24 cos2θ – 5 cos θ – 12 = 0
∴                           5 25 4 12 24 5 34.3cos22 4 4 8±+ × × ±θ= =×
= 0.8187    or    – 0.6104
∴θ = 35°    or    127.6°   Ans.
Substituting θ = 35° in equation (i), we have maximum torque,
Tmax= 600 sin 70° – 500 sin 35° = 277 N-m
Substituting θ =127.6° in equation (i), we have minimum torque,

Tmin= 600 sin 255.2° – 500 sin 127.6° = – 976 N-m
                             ()2
22773.46 rad/s
500 0.4max
maxT
Iα= = =
×    Ans. . . . (∵ I = m.k2)
and minimum acceleration (or maximum retardation),
                             ()2
297612.2 rad/s
500 0.4min
minT
Iα= = =
×  Ans.
Example 16.13.   The equation of the turning moment curve of a three crank engine is
(5000 + 1500 sin 3 θ) N-m, where θ is the crank angle in radians. The moment of inertia of the

flywheel is 1000 kg-m2 and the mean speed is 300 r.p.m. Calculate : 1. power of the engine, and 2.
the maximum fluctuation of the speed of the flywheel in percentage when  (i) the resisting torque is
constant, and (ii) the resisting torque is (5000 + 600 sin θ) N-m.
Solution.  Given : T = (5000 + 1500 sin 3 θ ) N-m; I = 1000 kg-m2; N = 300 r.p.m. or
ω = 2 π × 300/60 = 31.42 rad /s
1.  Power of the engine
                                        ()2

()2
0 01500cos35000 1500sin3 5000 –3dπ πθ =+θ θ = θ∫ 
= 10000 π N-m
∴  Mean resisting torque,
                            Workdone/rev 100005000 N-m22meanTπ== =ππ
We know that power of the engine,
P=Tmean . ω = 5000 × 31.42 = 157 100 W = 157.1 kW  Ans.
2.  Maximum fluctuation of the speed of the flywheel
Let CS= Maximum or total fluctuation of speed of the flywheel.

590  /G6C   Theory of Machines
(i)  When resisting torque is constant
The turning moment diagram is shown in Fig. 16.15. Since the resisting torque is constant,
therefore the torque exerted on the shaft is equal to the mean resisting torque on the flywheel.
Fig. 16.15
∴                                  T = Tmean
5000 + 1500 sin 3 θ= 5000
1500 sin 3 θ= 0    or    sin 3 θ = 0
∴ 3θ= 0°    or    180°
θ= 0°    or    60°
                                 () ( )60 60

() ( )60 60
00– 5000 1500sin3 – 5000mean ET T d d°°
∆= θ = + θ θ∫∫
                                        60 60
0 01500cos31500sin3 – 1000N-m3d° °θ =θ θ = =∫ 
1000 =I.ω2.CS = 1000 × (31.42)2 × CS = 987 216 CS
∴ CS= 1000/ 987216 = 0.001 or 0.1%  Ans.
(ii)  When resisting torque is  (5000 + 600 sin θθθθθ ) N-m
The turning moment diagram is shown in Fig. 16.16. Since at points B and C, the torque

exerted on the shaft is equal to the mean resisting torque on the flywheel, therefore
Fig. 16.16

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  591
         5000 + 1500 sin 3 θ  = 5000 + 600 sin θ    or    2.5 sin 3 θ = sin θ
       2.5 (3 sin θ – 4 sin3 θ) =sin θ  ...(∵ sin 3θ = 3 sin θ – 4 sin3θ)
         3  – 4 sin2θ  =  0.4...(Dividing by 2.5 sin θ)
                 23–0 . 4 sin 0.654θ= =     or    sin θ = 0.8062
∴                    θ  = 53.7°     or    126.3°    i.e.    θB = 53.7°, and θC = 126.3°
                    * () ( )126.3

* () ( )126.3
53.75000 1500sin3 – 5000 600sin Ed°
°∆= + θ + θ θ ∫
                                    ()126.3 126.3
53.7 53.71500cos31500sin3 – 600sin – 600cos3d° °
° °θ =θ θ θ =+ θ∫
                            = – 1656 N-m
                 1656  =  I.ω2.CS = 1000 × (31.42)2 × CS = 987 216 CS
∴                     CS = 1656/ 987 216 = 0.00 168    or    0.168%  Ans.
16.11.  Dimensions of the Flywheel Rim

Consider a rim of the flywheel as shown in Fig. 16.17.
LetD= Mean diameter of rim in metres,
R= Mean radius of rim in metres,
A= Cross-sectional area of rim in m2,
ρ=Density of rim material in kg/m3,
N= Speed of the flywheel in r.p.m.,
ω= Angular velocity of the flywheel in rad/s,
v= Linear velocity at the mean radius in m/s
= ω .R = π D.N/60, and
σ= Tensile stress or hoop stress in N/m2 due to the centrifugal force.

Consider a small element of the rim as shown shaded in Fig. 16.17. Let it subtends an angle
δθ at the centre of the flywheel.
Volume of the small element                            =  A × R.δθ
∴ Mass of the small element
                      dm   = Density × volume = ρ.A.R.δθ
and centrifugal force on the element, acting radially outwards,
                       dF   = dm.ω
2.R = ρ.A.R2.ω2.δθFig. 16.17.  Rim of a flywheel.

*Since the fluctuation of energy is negative, therefore it is shown below the mean resisting torque curve, in
Fig. 16.16.

592  /G6C   Theory of Machines
*Superfluous data.Vertical component of dF
=dF.sin θ = ρ.A.R2.ω2.δθ.sin θ
∴  Total vertical upward force tending to burst the rim across the diameter X Y.
                                        []22 22
0
0.. . s i n. .. . –c o sAR d ARπ
π=ρ ω θ θ=ρ ω θ∫
=2ρ.A.R2.ω2. . . (i)
This vertical upward force will produce tensile stress or hoop stress (also called centrifugal
stress or circumferential stress), and it is resisted by 2 P, such that
2P=2  σ.A  . . . (ii)

Equating equations ( i) and (ii),
2.ρ.A.R2.ω2=2σ.A
or σ=ρ.R2.ω2 = ρ.v2....(∵ v = ω.R)
∴                                 vσ=ρ...(iii)
We know that mass of the rim,
m= Volume × density = π D.A.ρ
∴                                ..mAD=πρ...(iv)
From equations (iii) and (iv), we may find the value of the mean radius and cross-sectional
area of the rim.
Note:  If the cross-section of the rim is a rectangular, then
A=b × t
where b= Width of the rim, and
t= Thickness of the rim.

Example 16.14.   The turning moment diagram for a multi-cylinder engine has been drawn
to a scale of 1 mm to 500 N-m torque and 1 mm to 6° of crank displacement. The intercepted areasbetween output torque curve and mean resistance line taken in order from one end, in sq. mm are
– 30, + 410, – 280, + 320, – 330, + 250, – 360, + 280, – 260 sq. mm, when the engine is
running at 800 r.p.m.

The engine has a stroke of 300 mm and the fluctuation of speed is not to exceed ± 2% of the
mean speed. Determine a suitable diameter and cross-section of the flywheel rim for a limitingvalue of the safe centrifugal stress of 7 MPa. The material density may be assumed as 7200 kg/m
3.
The width of the rim is to be 5 times the thickness.
Solution.   Given : N = 800 r.p.m. or ω = 2π × 800 / 60 = 83.8 rad/s; *Stroke = 300 mm ;
σ = 7 MPa = 7 × 106 N/m2 ; ρ = 7200 kg/m3

Since the fluctuation of speed is ± 2% of mean speed, therefore total fluctuation of
speed,
                          ω1 – ω2 = 4% ω = 0.04 ω

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  593
12
S/GF9/G960.04 Cω==ω
Diameter of the flywheel rim
Let D= Diameter of the flywheel rim in metres, and
v= Peripheral velocity of the flywheel rim in m/s.
We know that centrifugal stress ( σ),
7 × 106=ρ.v2 = 7200 v2   or   v2 = 7 × 106/7200 = 972.2
∴ v= 31.2 m/s
We know that v=π D.N/60
∴ D=v × 60 / π N = 31.2 × 60/ π × 800 = 0.745 m  Ans.
Cross-section of the flywheel rim

Let t= Thickness of the flywheel rim in metres, and
b= Width of the flywheel rim in metres = 5 t ...(Given)
∴  Cross-sectional area of flywheel rim,
A=b.t = 5 t × t = 5 t2
First of all, let us find the mass ( m) of the flywheel rim. The turning moment diagram is
shown in Fig 16.18.
Fig. 16.18
Since the turning moment scale is 1 mm = 500 N-m  and  crank  angle  scale is 1 mm = 6°
= π /30 rad, therefore
1 mm2 on the turning moment diagram

= 500 × π / 30 = 52.37 N-m
Let the energy at A = E, then referring to Fig. 16.18,
Energy at B=E – 30 . . . (Minimum energy)
Energy at C=E – 30 + 410 = E + 380
Energy at D= E + 380 – 280 = E + 100
Energy at E=E + 100 + 320 = E + 420 . . . (Maximum energy)
Energy at F=E + 420 – 330 = E + 90
Energy at G=E + 90 + 250 = E + 340
Energy at H=E + 340 – 360 = E – 20

594  /G6C   Theory of Machines
Energy at K=E – 20 + 280 = E + 260
Energy at L=E + 260 – 260 = E = Energy at A
∆E= Maximum energy – Minimum energy
=(E + 420) – ( E – 30) = 450 mm2
= 450 × 52.37 = 23 566 N-m
We also know that maximum fluctuation of energy ( ∆E),
23566 = m.v2.CS = m × (31.2)2 × 0.04 = 39 m
∴ m= 23566 /39 = 604 kg
We know that mass of the flywheel rim ( m),
604 = Volume × density = π D.A.ρ
=π × 0.745 × 5 t2 × 7200 = 84 268 t2

∴ t2= 604 /84268 = 0.007 17 m2  or  t = 0.085 m = 85 mm  Ans.
and b=5t = 5 × 85 = 425 mm   Ans.
Example 16.15.   A single cylinder double acting steam engine develops 150 kW at a mean
speed of 80 r.p.m. The coefficient of flu ctuation of energy is 0.1 and the fluctuation of speed is ± 2%

of mean speed. If the mean diameter of the flywheel rim is 2 metre and the hub and spokes provide5% of the rotational inertia of the flywheel, find the mass and cross-sectional area of the flywheelrim. Assume the density of the flywheel material (which is cast iron ) as 7200 kg/m
3.
Solution.  Given : P = 150 kW = 150 × 103 W; N = 80 r.p.m. or ω = 2 π × 80 /60 = 8.4 rad/s;
CE = 0.1; D = 2 m or R = 1 m ; ρ = 7200 kg/m3

Since the fluctuation of speed is ± 2% of mean speed, therefore total fluctuation of speed,
                          ω1 – ω2 =  4% ω  = 0.04 ω
                                  12
S–
0.04 Cωω
==ω
Mass of the flywheel rim
Let m= Mass of the flywheel rim in kg, and
I= Mass moment of inertia of the flywheel in kg-m2.
=P × 60/N = 150 × 103 × 60/ 80 = 112.5 × 103 N-m

and maximum fluctuation of energy,
∆E= Work done /cycle × CE = 112.5 × 103 × 0.1 = 11 250 N-m
11250 = I.ω2.CS = I × (8.4)2 × 0.04 = 2.8224 I
∴ I= 11250 /2.8224 = 3986 kg-m2
Since the hub and spokes provide 5% of the rotational inertia of the flywheel, therefore,
mass moment of inertia of the flywheel rim ( Irim) will be 95% of the flywheel, i.e.
Irim= 0.95 I = 0.95 × 3986 = 3787 kg-m2

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  595
and                             Irim = m.k2    or    *2237873787 kg
1rimI
m
k== =  Ans. . . . (∵ k = R)
Cross-sectional area of the flywheel rim
Let A= Cross-sectional area of flywheel rim in m2.
We know that the mass of the flywheel ( m),
3787 = 2 π R × A × ρ = 2 π × 1 × A × 7200 = 45 245 A
∴ A= 3787/45 245 = 0.084 m2 Ans.
Example 16.16.   A multi-cylinder engine is to run at a speed of 600 r.p.m. On drawing the

turning moment diagram to a scale of 1 mm = 250 N-m and 1 mm = 3°, the areas above and belowthe mean torque line in mm
2 are : + 160, – 172, + 168, – 191, + 197, – 162
The speed is to be kept within ± 1% of the mean speed of the engine. Calculate the necessary
moment of inertia of the flywheel. Determine the suitable dimensions of a rectangular flywheel rimif the breadth is twice its thickness. The density of the cast iron is 7250 kg/m
3 and its hoop stress is

6 MPa. Assume that the rim contributes 92% of the flywheel effect.
Solution.  Given : N = 600 r.p.m. or ω = 2π × 600/60 = 62.84 rad /s; ρ = 7250 kg/m3;
σ = 6 MPa = 6 × 106 N/m2
Fig. 16.19
Since the fluctuation of speed is ± 1% of mean speed, therefore, total fluctuation of speed,
ω1 – ω2=2 %  ω = 0.02 ω
                                  12
S–0.02 Cωω==
ω
Moment of inertia of the flywheel

ω
The turning moment diagram is shown in Fig. 16.19. The turning moment scale is 1 mm =
250 N-m and crank angle scale is 1 mm = 3° = π /60 rad, therefore,
1 mm2 of turning moment diagram
= 250 × π /60 = 13.1 N-m
*The mass of the flywheel rim ( m) may also be obtained by using the following relation:
∆Erim=0.95 (∆E) = 0.95 × 11 250 = 10687.5 N-m

and ∆Erim=m.k2.ω2.CS = m (1)2 × (8.4)2 × 0.04 = 2.8224 m
∴ m=(∆E)rim/2.8224 = 10 687.5/ 2.8224 = 3787 kg

596  /G6C   Theory of Machines
Let the total energy at A=E. Therefore from Fig. 16.19, we find that
Energy at B=E + 160
Energy at C= E + 160 – 172 = E – 12
Energy at D= E – 12 + 168 = E + 156
Energy at E= E + 156 – 191 = E – 35 . . . (Minimum energy)
Energy at F= E – 35 + 197 = E + 162 . . . (Maximum energy)
Energy at G= E + 162 – 162 = E = Energy at A
=(E + 162) – ( E – 35) = 197 mm2
= 197 × 13.1 = 2581 N-m

We also know that maximum fluctuation of energy ( ∆E ),
2581 =I.ω2.CS = I × (62.84)2 × 0.02 = 79 I
∴ I= 2581/79 = 32.7 kg-m2 Ans.
Dimensions of the flywheel rim
Let t= Thickness of the flywheel rim in metres,
b= Breadth of the flywheel rim in metres = 2  t . . . (Given)
D= Mean diameter of the flywheel in metres, and
v= Peripheral velocity of the flywheel in m/s.
We know that hoop stress ( σ),
6 × 106=ρ.v2 = 7250 v2    or    v2 = 6 × 106/7250 = 827.6
∴ v= 28.8 m/s

We know that v=π DN/60,    or    D = v × 60 / π N = 28.8 × 60/ π × 600 = 0.92 m
Now, let us find the mass ( m) of the flywheel rim. Since the rim contributes 92% of the
flywheel effect, therefore maximum fluctuation of energy of rim,
∆Erim= 0.92 × ∆E = 0.92 × 2581 = 2375 N-m
We know that maximum fluctuation of energy of rim ( ∆Erim),
2375 =m.v2.CS = m × (28.8)2 × 0.02 = 16.6 m
∴ m= 2375/16.6 = 143 kg
Also m= Volume × density = π D.A.ρ = π D.b.t.ρ

∴ 143 =π × 0.92 × 2 t × t × 7250 = 41914 t2
t2= 143 /41914 = 0.0034 m2
or t= 0.0584 m = 58.4 mm  Ans.
and b=2  t = 116.8 mm  Ans.
Example 16.17.   The turning moment diagram of a four stroke engine may be assumed for
the sake of simplicity to be represented by four triangles in each stroke. The areas of these trianglesare as follows:

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  597
Suction stroke = 5 × 10–5 m2; Compression stroke = 21 × 10–5 m2; Expansion stroke =
85 × 10–5 m2; Exhaust stroke = 8 × 10–5 m2.
All the areas excepting expression stroke are negative. Each m2 of area represents 14 MN-m
of work.
Assuming the resisting torque to be constant, determine the moment of inertia of the flywheel

to keep the speed between 98 r.p.m. and 102 r.p.m. Also find the size of a rim-type flywheel based onthe minimum material criterion, given that density of flywheel material is 8150 kg/m
3 ; the allowable
tensile stress of the flywheel material is 7.5 MPa. The rim cross-section is rectangular, one sidebeing four times the length of the other.
1 = 5 × 10–5 m2; a2 = 21 × 10–5 m2; a3 = 85 × 10–5 m2; a4 = 8 × 10–5 m2;

N2 = 98 r.p.m.; N1 = 102 r.p.m.; ρ = 8150 kg/m3; σ = 7.5 MPa = 7.5 × 106 N/m2
Fig. 16.20
The turning moment-crank angle diagram for a four stroke engine is shown in Fig. 16.20.
∴Net area = a
= 85 × 105 – (5 × 105 + 21 × 10–5 + 8 × 10–5) = 51 × 10–5 m2
Since  1m2 = 14 MN-m = 14 × 106 N-m of work, therefore
Net work done per cycle

= 51 × 10–5 × 14 × 106 = 7140 N-m ... (i)
= Tmean × 4π N-m ... (ii)
Tmean = FG = 7140 /4 π = 568 N-m
= a3 × Work scale = 85 × 10–5 × 14 × 106 = 11900 N-m ... (iii)

598  /G6C   Theory of Machines
1
2= × BC × AG = 1
2= × π × AG = 1.571 AG ...(iv)
AG  = 11 900/1.571 = 7575 N-m
∴ Excess torque = AF = AG – FG = 7575 – 568 = 7007 N-m
DE AF
BC AG= or70072.9rad
7575AFDE BC
AG=×= × π =
∆E = Area of ∆ ADE = 1
2 × DE × AF
= 1
2 × 2.9 × 7007 = 10160 N-m
Moment of Inertia of the flywheel

Let I = Moment of inertia of the flywheel in kg-m2.
We know that mean speed during the cycle
12 102 98100r.p.m.22NNN++===
∴  Corresponding angular mean speed,
ω = 2πN/60 = 2π × 100/60 = 10.47 rad/s
12
S102 980.04
100NNC
N− −===
10 160 = I.ω2.CS = I (10.47)2 × 0.04 = 4.385 I
∴ I = 10160/ 4.385 = 2317 kg-m2 Ans.
Size of flywheel

Let t = Thickness of the flywheel rim in metres,
b = Width of the flywheel rim in metres = 4 t ...(Given)
D = Mean diameter of the flywheel in metres, and
v = Peripheral velocity of the flywheel in m/s.
7.5 × 106 = ρ .  v2 = 8150 v2
∴6
27.5 10920 or 30.3m/s8150vv×===
and v = πDN/60  or  D = v × 60/πN = 30.3 × 60/ π × 100 = 5.786 m

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  599
*The work done per cycle for a four stroke engine is also given by
360 60 50 10 60Work done per cycle 40000 N-m
Number of explosions/min 75PP
n×× × ×== = =Now let us find the mass ( m) of the flywheel rim. We know that maximum fluctuation of
energy (∆E),
10 160 = m.v2 CS = m × (30.3)2  × 0.04 = 36.72 m
∴ m = 10160/36.72 = 276.7 kg
Also m = Volume × density = DA Dbtggρρπ××= π×× ×
276.7 = π × 5.786 × 4 t × t × 4
42 81 059.3 109.81t×=×

∴ t2 = 276.7/59.3 × 104 = 0.0216 m or 21.6 mm Ans.
and b = 4t = 4 × 21.6 = 86.4 mm Ans.
Example 16.18.   An otto cycle engine develops 50 kW at 150 r.p.m. with 75 explosions per

minute. The change of speed from the commencement to the end of power stroke must not exceed0.5% of mean on either side. Find the mean diameter of the flywheel and a suitable rim cross-section having width four times the depth so that the hoop stress does not exceed 4 MPa. Assumethat the flywheel stores 16/15 times the energy stored by the rim and the work done during powerstroke is 1.40 times the work done during the cycle. Density of rim material is 7200 kg/m
3.

3.
Solution.  Given : P = 50 kW = 50 × 103 W; N = 150 r.p.m. or ω = 2 π × 150/60 = 15.71 rad/s;
n = 75; σ = 4 MPa = 4 × 106 N/m2; r = 7200 kg/m3
First of all, let us find the mean torque ( Tmean) transmitted by the engine or flywheel. We
know that the power transmitted ( P),
                         50 × 103 = Tmean × ω = Tmean × 15.71
∴ Tmean= 50 × 103/15.71 = 3182.7 N-m
Since the explosions per minute are equal to N/2, therefore, the engine is a four stroke cycle

engine. The turning moment diagram of a four stroke engine is shown in Fig. 16.21.
Fig. 16.21
We know that *work done per cycle
=Tmean × θ = 3182.7 × 4 π = 40 000 N-m

600  /G6C   Theory of Machines
∴ Workdone during power or working stroke
= 1.4 × work done per cycle ....(Given)
= 1.4 × 40 000 = 56 000 N-m ... (i)
The workdone during power stroke is shown by a triangle ABC in Fig. 16.20, in which base
AC = π radians and height BF = Tmax.
∴  Work done during working stroke
                                        11.5712max maxTT =× π × =  . . . (ii)
From equations  (i) and (ii), we have
Tmax= 56 000/1.571 = 35646 N-m
We know that the excess torque,

Texcess=BG = BF – FG = Tmax – Tmean = 35646 – 3182.7 = 32 463.3 N-m
Now, from similar triangles BDE and ABC,
                                DE BG
AC BF=    or    32463.30.910735646BGDE ACBF=×= × π = π
∆E= Area of triangle BDE 1
2DE BG=× ×
                                            10.9107 32463.3 46 445N-m2=× π × =
Mean diameter of the flywheel
Let D= Mean diameter of the flywheel in metres, and

4 × 106=ρ.v2 = 7200 v2    or    v2 = 4 × 106/7200 = 556
∴ v= 23.58 m/s
We know that v= π DN/60    or    D = v × 60/N = 23.58 × 60/ π × 150 = 3 m  Ans.
Cross-sectional dimensions of the rim
Let t= Thickness of the rim in metres, and
b= Width of the rim in metres = 4 t ...(Given)
∴  Cross-sectional area of the rim,
A=b × t = 4 t × t = 4 t2
First of all, let us find the mass of the flywheel rim.

E= Total energy of the flywheel in N-m.
Since the fluctuation of speed is 0.5% of the mean speed on either side, therefore total
fluctuation of speed,
N2 – N1= 1% of mean speed = 0.01 N
                                  12
S–
0.01NN
CN==

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  601
We know that the maximum fluctuation of energy ()E∆,
                             S 4 6 4 4 5 2 20 . 0 10 . 0 2 ECE E=× =×× =
∴                                 346 445/0.02 2322 10 N-mE== ×
Since the energy stored by the flywheel is 16
15 times the energy stored by the rim, therefore,
the energy of the rim,
                                33 15 15232 10 2177 10 N-m
16 16rimEE== × × =×
We know that energy of the rim ()rimE,

                     ()32212177 10 278 23.582mv m m ×= × × = =
∴                                 32177 10 /278 7831kgm=× =
We also knonw that mass of the flywheel rim ( m),
                                227831 3 4 7200 271469 DA t t=π × ×ρ=π× × × =
∴                                 2831/271469 0.0288 or 0.17m 170 mmtt== = = Ans.
and b=4 t = 4 × 170 = 680 mm Ans.
16.12. Flywheel in Punching Press

We have discussed in Art. 16.8 that the function of a flywheel in an engine is to reduce the
fluctuations of speed, when the load on the crankshaftis constant and the input torque varies during thecycle. The flywheel can also be used to perform the
same function when the torque is constant and the

load varies during the cycle. Such an application isfound in punching press  or in a rivetting machine.A punching press is shown diagrammatically in Fig.16.22. The crank is driven by a motor which suppliesconstant torque and the punch is at the position of
the slider in a slider-crank mechanism. From Fig.
16.22, we see that the load acts only during therotation of the crank from 
12to =θ=θ θ θ , when

the actual punching takes place and the load is zerofor the rest of the cycle. Unless a flywheel isused, the speed of the crankshaft will increase too
much during the rotation of crankshaft will
increase too much during the rotation of crankfrom
2θ=θto 2θ= π or 0θ= and again from
0θ= to 1θ=θ, because there is no load  while
input energy continues to be supplied. On the otherhand,  the drop in speed of the crankshaft is

very large during the rotation of crank fromFig. 16.22. Operation of flywheel in a
punching press.

602  /G6C   Theory of Machines
12to =θ=θ θ θ  due to much more load than the
energy supplied. Thus the flywheel has to absorbexcess energy available at one stage and has to make
up the deficient energy at the other stage to keep to
fluctuations of speed within permissible limits. Thisis done by choosing the suitable moment of inertia ofthe flywheel.
Let 
1E be the energy required for punching a
hole. This energy is determined by the size of the hole

punched, the thickness of the material and the physi-
cal properties of the material.
Let   1d =Diameter of the hole punched,
         1t  = Thickness of the plate, and
         uτ=  Ultimate shear stress for the plate
                 material.
∴ Maximum shear force required for punching,
        S1 1Area sheared × Ultimate shear stress .u Fd t== π τ
It is assumed that as the hole is punched, the shear force decreases uniformly from maximum
value to zero.

∴ Work done or energy required for punching a hole,
                                   11
2s EF t=××
Assuming one punching operation per revolution, the energy supplied to the shaft per revolu-
tion should also be equal to 1E. The energy supplied by the motor to the crankshaft during actual
punching operation,
                                   21
212EEθ− θ=π
∴ Balance energy required for punching
                                         21 21

21 21
1211 1 – 122EEEE Eθ− θ θ− θ  =−= = −   ππ  
This energy is to be supplied by the flywheel by  the decrease in its kinetic energy when its
speed falls from maximum to minimum. Thus maximum fluctuation of energy,
                                  21
121 12EE E Eθ− θ∆= − = −π 
The values of 1θ and 2θ may be determined only if the crank radius ( r), length of connecting

rod (l) and the relative position of the job with respect to the crankshaft axis are known. In the
absence of relevant data, we assume that
                          21
22 4tt
srθ− θ==π
Punching press and flywheel.

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  603
where                             t   = Thickness of the material to be punched,
                             s = Stroke of the punch = 2 Crank radius 2 r ×= .
By using the suitable relation for the maximum fluctuation of energy ( ∆E) as discussed in the
previous articles, we can find the mass and size of the flywheel.
Example 16.19.  A punching press is driven by a constant torque electric motor. The press is

provided with a flywheel that rotates at maximum speed of 225 r.p.m. The radius of gyration of theflywheel is 0.5 m. The press punches 720 holes per hour; each punching operation takes 2 secondand requires 15 kN-m of energy. Find the power of  the motor and the minimum mass of the flywheel
if speed of the same is not to fall below 200 r. p. m.
Solution.  Given 
1225N= r.p.m ; k = 0.5 m ; Hole punched = 720 per hr; 1E = 15 kN-m
= 15 × 103 N-m; 2200N= r.p.m.
Power of the motor

We know that the total energy required per second
                               = Energy required / hole × No. of holes/ s                               =15 × 10
3 × 720/3600 = 3000 N-m/s
        ∴ Power of the motor = 3000 W = 3 kW Ans. ( /G51 1 N-m/s = 1 W)
Minimum mass of the flywheel
Let                     m  =  Minimum mass of the flywheel.
Since each punching operation takes 2 seconds, therefore energy supplied by the motor in 2
seconds,

seconds,
                         23000 2 6000 N-mE=× =
∴ Energy to be supplied by the flywheel during punching or maximum fluctuation of energy,
                        3
12 15 10 6000 9000N-m EE E∆= − = × − =
Mean speed of the flywheel,
                          12 225 200212.5 r.p.m22NNN++== =
                     2
2
12 9000 . . ( )900mk N N Nπ=× −
                               () ()2
2212.5225 200 0.5900mπ=× × × × − =  14.565 m

    ∴                   m   = 9000/14.565 = 618 kg Ans.
Example 16.20.  A machine punching 38 mm holes in 32 mm thick plate requires 7 N-m of
energy per sq. mm of sheared area, and punches one hole in every 10 seconds. Calculate the power
of the motor required. The mean speed of the flywheel is 25 metres per second. The punch has astroke of 100 mm.
Find the mass of the flywheel required, if the total fluctuation of  speed is not to exceed 3%

of the mean speed. Assume that the motor su pplies energy to the machine at uniform rate.
Solution.  Given : d = 38 mm ;  t = 32 mm ; 
2
17N - m / m mE=  of sheared area; 25 m/sv= ;
s = 100 mm; 123% 0.03 vv v v−= =

604  /G6C   Theory of Machines
Power of the motor required
We know that sheared area,                                     
2. 38 32 3820 mmAd t=π =π× × =
Since the energy required to punch a hole is 7 N-m/mm2 of sheared area, therefore total
energy required per hole,
                                   17 3820 26 740 N-mE=× =
Also the time required to punch a hole is 10 second, therefore energy required for punching
work per second
= 26 740/10  =  2674  N-m/s
∴ Power of the motor required

= 2674 W = 2.674 kW Ans.
Mass of the flywheel required
 Let m= Mass of the flywheel in kg.
Since the stroke of the punch is 100 mm and it punches one hole in every 10 seconds, there-
fore the time required to punch a hole in a 32 mm thick plate
                                         1032 1.6 s2 100=× =×
∴  Energy supplied by the motor in 1.6 seconds,
                                   22674 1.6 4278 N-mE=× =

Energy to be supplied by the flywheel during punching or the maximum fluctuation of energy,
                                   12 26 740 4278 22 462 N-m EE E∆= − = − =
Coefficient of fluctuation of speed,
                                   12
S 0.03vvC
v−==
We know that maximum fluctuation of energy ()E∆,
                             ()2222 462 . . 0.03 18.75 25 Smv C m m== × × =
∴  m= 22462 /18.75 = 1198 kg Ans.

Note : The value of maximum fluctuation of energy ( ∆E) may also be determined as discussed in Art. 16.12. We
know that energy required for one punch,
                                    126740 N-mE=
and                                          21
11 12 2tEEs θ− θ− ∆= = − π 21......
22t
sθ−θ= π/G51
                                         32126740 22 462 N-m21 0 0− ==×

Example 16.21.  A riveting machine is driven by a constant torque 3 kW motor. The moving
parts including the flywheel are equivalent to 150 kg at 0.6 m radius. One riveting operation takes1 second and absorbs 10 000 N-m of energy . The speed of the flywheel is 300 r.p.m. before riveting.
Find the speed immediately after riveting. How many rivets can be closed per minute?
Solution. Given : P = 3 kW; m = 150 kg; k = 0.6 m; 
1N = 300 r.p.m. or
ω1 = 2π × 300/60 = 31.42 rad/s

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  605
Speed of the flywheel immediately after riveting
Let2ω= Angular speed of the flywheel immediately after riveting.
We know that energy supplied by the motor,                                   
23kW 3000 W 3000N-m/sE== = (∵ 1 W = 1 N-m/s)
But energy absorbed during  one riveting operation which takes 1 second,
                                   1=1000 0N - mE

1=1000 0N - mE
∴ Energy to be supplied by the flywheel for each riveting operation per second or the
maximum fluctuation of energy,
                                   12 10 000 3000 7000 N-m EE E∆= − = − =
                         () () ()() ()22 22 2 2
12 2117000 . 150 0.6 31.4222mk   =× =× × × −−ωω ω  
                                         ()2
227987.2= −ω
∴                          ()2

∴                          ()2
2987.2 7000/27 728=− =ω  or 226.98rad/sω=
Corresponding speed in r.p.m.,
                                 226.98 60/2N=×π  = 257.6 r.p.m. Ans.
Number of rivets that can be closed per minute
Since the energy absorbed by each riveting operation which takes 1 second is 10 000 N-m,
therefore, number of rivets that can be closed per minute,
                                        2
1300060 60 1810 000E
E=×= × =  rivets Ans.

Example 16.22.   A  punching press is required to punch 40 mm diameter holes in a plate of
15 mm thickness at the rate of 30 holes per minute. It requires 6 N-m  of energy per mm2 of sheared
area. If the punching takes 1/10 of a second and the r.p.m. of the flyw heel varies from 160 to 140,
determine the mass of the flywheel having radius of gyration of 1 metre.
Solution.  Given: d = 40 mm; t = 15  mm; No. of holes = 30 per min.; Energy required

= 6 N-m/mm2; Time = 1/10 s = 0.1 s; N1 = 160 r.p.m.; N2 =  140 r.p.m.; k = 1m
We know that sheared area per hole
                                        2. 40 15 1885 mmdt=π =π× × =
∴ Energy required to punch a hole,
                                   1= 6×1885 = 11310 N-mE
and energy required for punching work per second
= Energy required per hole × No. of holes per second= 11 310  × 30/60 = 5655 N-m/s
Since the punching takes 1/10 of a second, therefore, energy supplied by the motor in 1/10

second,
                                   
2= 5655×1/10 = 565.5 N-mE
∴ Energy to be supplied by the flywheel during punching a hole or maximum fluctuation of
energy of the flywheel,
                                  12 11310 565.5 10 744.5 N-m EE E∆= − = − =

606  /G6C   Theory of Machines
                                      12 160 140150 r.p.m.22NNN++== =
                              ()2
2
12 10 744.5 .900mk NNNπ=× −
   ()20.011 1 150 33 160 140 mm =× × × = −
∴                                m  = 10744.5 / 33 = 327 kg Ans.
Example 16.23.  A punching machine makes 25 working strokes per minute and is capable

of punching 25 mm diameter holes in 18 mm thick steel plates having an ultimate shear strength 300MPa. The punching operation takes place during 1/10th of a revolution of the crankshaft.
Estimate the power needed for the driving motor, assuming a mechanical ef ficiency of 95

percent. Dtetermine suitable dimensions for the rim cross-section of the flywheel, having width equalto twice thickness. The flywheel is to revolve at 9 times the speed of the crankshaft. The permissiblecoefficient of fluctuation of speed is 0.1.
The flywheel is to be made of cast iron having a working stress (tensile) of 6 MPa and density
of 7250 kg/m
3. The diameter of the flywheel must not exceed 1.4 m owing to space restrictions. The

hub and the spokes may be assumed to provide 5% of the rotational inertia of the wheel.
Solution.  Given :  n = 25; d1 = 25 mm =  0.025 m; t1 = 18 mm = 0.018 m; uτ = 300 MPa
= 300 × 106 N/m2; mη = 95% = 0.95 ; SC = 0.1; σ= 6 MPa = 6 × 106 N/m2; ρ = 7250 kg/m3;
D = 1.4 m or   R =   0.7 m
Power needed for the driving motor
We know  that the area of plate sheared ,
                                    62
s1 1 0.025 0.018 1414 10 m Ad t−=π × =π× × = ×
∴  Maximum shearing force required for punching,

                                   66
SS 1414 10 300 10 424 200Nu FA−=× τ = × × × =
and energy required per stroke
= Average shear force × Thickness of plate
                                         S111424200 0.018 3817.8 N-m22Ft=××=× × =
∴  Energy required per min
= Energy/stroke × No. of working strokes/min= 3817.8 × 25 = 95 450 N-m
We know that the power needed for the driving motor

= Energy required per min 95 450=1675 W =1.675 kW60× /GE7/G36 /G30 /G30 /G2E /G39 /G35m=×Ans.
Dimensions for the rim cross-section
Let t= Thickness of rim in metres, and
b= Width of rim in metres = 2 t ... ( Given)
∴ Cross-sectional area of rim,
                                    222 Abt tt t=×= ×=

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  607
Since the punching operation takes place ( i.e. energy is consumed) during 1/10th of a
revolution of the crankshaft, therefore during 9/10th of the revolution of a crankshaft, the energyis stored in the flywheel.
∴ Maximum fluctuation of energy,
                                  99Energy/stroke = × 3817.8 = 3436N-m
10 10E∆= ×

10 10E∆= ×
Since the hub and the spokes provide 5% of the rotational inertia of the wheel, therefore the
maximum fluctuation of energy provided by the flywheel by the rim will be 95%.
∴  Maximum fluctuation of energy provided by the rim,
                             0.95 0.95 3436 3264 N-mrimEE∆=× ∆ =× =
Since the flywheel is to revolve at 9 times the speed of the crankshaft and there are 25 work-
ing strokes per minute, therefore, mean speed of the flywheel,

9 25 225 r.p.mN=× = .
and mean angular speed,
                                     2 225/60 23.56rad/sω= π× =
We know that maximum fluctuation of energy ()rimE∆,
3264 = ()( )22 2 2. . . 0.1 27.2 0.7 23.56 s mR C m mω= × × × =
∴ m= 3264/27.2 = 120 kg
We also know that mass of the flywheel ( m),
                                 22120 1.4 2 7250 63782DA t t=π × ×ρ=π× × × =
∴                               2120/63782 0.00188t==  or t = 0.044 m = 44 mm Ans.

and b=2 t = 2 × 44 = 88 mm  Ans.
EXERCISES
1.An engine flywheel has a mass of 6.5 tonnes and the radius of gyration is 2 m. If the maximum and
minimum speeds are 120 r. p. m. and 118 r. p. m. respectively, find maximum fluctuation of energy.
[Ans. 67. 875 kN-m]
2.A vertical double acting steam engine develops 75 kW at 250 r.p.m. The maximum fluctuation of

energy is 30 per cent of the work done per stroke.  The maximum and minimum speeds are not to varymore than 1 per cent on either side of the mean speed. Find the mass of the flywheel required, if theradius of gyration is 0.6 m. [Ans. 547 kg]
3.In a turning moment diagram, the areas above and below the mean torque line taken in order are 4400,
1150, 1300 and 4550 mm
2 respectively. The scales of the turning moment diagram are:

Turning moment, 1 mm = 100 N-m; Crank angle, 1 mm = 1°Find the mass of the flywheel required to keep the speed between 297 and 303 r.p.m., if the radius of
gyration is 0.525 m. [Ans. 417 kg]
4.The turning moment diagram for a multicylinder engine has been drawn to a scale of 1 mm =
4500  N-m vertically and 1 mm = 2.4° horizontally. The intercepted areas between output torque curveand mean resistance line taken in order from one end are 342, 23, 245, 303, 115, 232, 227, 164 mm
2,

2,
when the engine is running at 150 r.p.m. If the mass of the flywheel is 1000 kg and the total fluctuationof speed does not exceed 3% of the mean speed, find the minimum value of the radius of gyration.
[Ans. 1.034 m]

608  /G6C   Theory of Machines
5.An engine has three single-acting cylinders whose cranks are spaced at 120° to each other. The turn-
ing moment diagram for each cylinder consists of a triangle having the following values:
Angle 0° 60° 180° 180° – 360°
Torque (N-m) 0 200 0 0
Find the mean torque and the moment of inertia of the flywheel to keep the speed within 180 ± 3 r.p.m.
[Ans. 150 N-m; 1.22 kg-m2]

6.The turning moment diagram for a four stroke gas engine may be assumed for simplicity to be repre-
sented by four triangles, the areas of which from the line of zero pressure are as follows:
Expansion stroke = 3550 mm2; exhaust stroke = 500 mm2; suction stroke  = 350 mm2; and compres-
sion stroke = 1400 mm2. Each mm2 represents 3 N-m.
Assuming the resisting moment to be uniform, find the mass of the rim of a flywheel required to keep

the mean speed 200 r.p.m. within ± 2%. The mean radius of the rim may be taken as 0.75 m. Alsodetermine the crank positions for the maximum and minimum speeds.
[Ans. 983 kg; 4° and 176° from I. D. C]
7.A single cylinder, single acting, four stroke cycle gas engine develops 20 kW at 250 r.p.m. The work
done by the gases during the expansion stroke is 3 times the work done on the gases during thecompression stroke. The work done on the suction and exhaust strokes may be neglected. If the

flywheel has a mass of 1.5 tonnes and has a radius of gyration of 0.6m, find the cyclic fluctuation of
energy and the coefficient of fluctuation of speed.
[Ans. 12.1 kN-m; 3.26%]
8.The torque exerted on the crank shaft of a two stroke engine is given by the equation:
() 14500 2300sin2 1900cosN-mT =+ θ − 2 θ
where θ is the crank angle displacement from the inner dead centre. Assuming the resisting torque to

be constant, determine: 1. The power of the engine when the speed is 150 r.p.m. ; 2. The moment ofinertia of the flywheel if the speed variation is not to exceed 
0.5%± of the mean speed; and 3. The
angular acceleration of the flywheel when the crank has turned through 30° from the inner dead
centre. [Ans. 228 kW; 1208 kg-m2; 0.86 rad/s2]
9.A certain machine requires a torque of (2000 + 300 sin θ) N-m to drive it, where θ is the angle of

rotation of its shaft measured from some datum. The machine is directly coupled to an electric motordeveloping uniform torque. The mean speed of the machine is 200 r.p.m.
Find: 1. the power of the driving electric motor, and 2. the moment of inertia of the flywheel required
to  be used if the fluctuation of speed is limited to 
2%±.
[Ans. 41.9 kW;  34.17 kg-m2]
10.The equation of the turning moment diagram for the three crank engine is given by:
() 25000 7500sin3N-mT =− θ

where θradians is the crank angle from inner dead centre. The moment of inertia of the flywheel is
400 kg-m2 and the mean engine speed is 300 r.p.m. Calculate the power of the engine and the total
percentage fluctuation of speed of the flywheel, if 1. The resisting torque is constant, and  2. Theresisting torque is (25 000  + 3600 
sinθ) N-m.
[Ans. 785 kW; 1.27%; 2.28%]

sinθ) N-m.
11.A single cylinder double acting steam engine delivers 185 kW at 100 r.p.m. The maximum fluctuation
of energy per  revolution  is 15 per cent of the energy developed per  revolution. The speed variationis limited to 1 per cent either way from the mean. The mean diameter of the rim is 2.4 m. Find the massand cross-sectional dimensions of the flywheel rim when width of rim is twice the thickness. Thedensity of flywheel material is 7200 kg/m
3.

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  609
12.A steam engine runs at 150 r.p.m. Its turning moment diagram gave the following area measurements
in mm2 taken in order above and below the mean torque line:
500, – 250, 270, – 390, 190, – 340, 270, – 250The scale for the turning moment is 1 mm = 500 N-m, and for crank angle is 1mm = 5°.The fluctuation of speed is not to exceed 
±1.5% of the mean, determine the cross-section of the rim

of the flywheel assumed rectangular with axial dimension equal to 1.5 times the radial dimension. The
hoop stress is limited to 3 MPa and the density of the material of the flywheel is 7500 kg/m3.
[Ans. 222 mm; 148 mm]
13.The turning moment diagram for the engine is drawn to the following scales:
Turning moment, 1 mm = 1000 N-m and crank angle, 1 mm = 6°.The areas above and below the mean turning moment line taken in order are : 530, 330, 380, 470, 180,
360, 350 and 280 mm
2.

2.
The mean speed of the engine is 150 r.p.m. and the total fluctuation of speed must not exceed 3.5% of
mean speed. Determine the diameter and mass of the flywheel rim, assuming that the total energy ofthe flywheel to be 15/14 that of rim. The peripheral velocity of the flywheel is 15 m/s. Find also thesuitable cross-scetional area of the rim of the flywheel. Take density of the material of the rim as 7200kg/m
3.
[Ans. 1.91 m; 8063 kg; 0.1866 m2]

3.
14.A single cylinder internal combustion engine working on the four stroke cycle develops 75 kW at 360
r.p.m. The fluctuation of energy can be assumed to be 0.9 times the energy developed per cycle. If the
fluctuation of speed is not to exceed 1 per cent and the maximum centrifugal stress in the flywheel is
to be 5.5 MPa, estimate the mean diameter and the cross-sectional area of the rim. The material of therim has a density of 7.2 Mg/m
3.

3.
[Ans. 1.47 m; 0.088 m2]
15.A cast iron flywheel used for a four stroke I.C. engine is developing 187.5 kW at 250 r.p.m. The hoop
stress developed in the flywheel is 5.2 MPa. The total fluctuation of speed is to be limited to 3% of the
mean speed. If the work done during the power stroke is 1/3 times more than the average workdone
during the whole cycle, find:
1. mean diameter of the flywheel, 2. mass of the flywheel and 3. cross-sectional dimensions of the rim

when the width is twice the thickness. The density of cast iron may be taken as 7220 kg/m3.
[Ans 2.05m; 4561 kg; 440 mm, 220 mm]
16.A certain machine tool does work intermittently. The machine is fitted with a flywheel of mass 200 kg
and radius of gyration of 0.4 m. It runs at a speed of 400 r.p.m. between the operations. The machine

is driven continuously by a motor and each operation takes 8 seconds. When the machine is doing itswork, the speed drops from 400 to 250 r.p.m. Find 1. minimum power of the motor, when there are 5
operations performed per minute, and 2. energy expanded in performing each operation.
[Ans. 4.278 kW; 51.33 kN-m]
17.A constant torque 4 kW motor drives a riveting machine. A flywheel of mass 130 kg and radius of

gyration 0.5 m is fitted to the riveting machine. Each riveting operation takes 1 second and requires 9000
N-m of energy. If the speed of the flywheel is 420 r.p.m. before riveting, find: 1. the fall in speed of the
flywheel after riveting; and 2. the number of rivets fitted per hour.
[Ans. 385.15 r.p.m.; 1600]
18.A machine has to carry out punching operation at the rate of 10 holes per minute. It does 6 kN-m of
work per mm

work per mm
2 of the sheared area in cutting 25 mm diameter holes in 20 mm thick plates. A flywheel
is fitted to the machine shaft which is driven by a constant torque. The fluctuation of speed is between
180 and 200 r.p.m. The actual punching takes 1.5 seconds. The frictional  losses are equivalent to
1/6 of the work done during punching. Find: 1. Power required to drive the punching  machine, and 2.Mass of the flywheel, if the radius of gyration of the wheel is 0.5 m. [Ans. 1.588 W; 686 kg]

610  /G6C   Theory of Machines
19.The crankshaft of a punching machine runs at a speed of 300 r.p.m. During punching of 10 mm
diameter holes in mild steel sheets, the torque required by the machine increases uniformly from 1000
N-m to 4000 N-m while the shaft turns through 40°, remains constant for the next 100°,  decreases
uniformly to 1000 N-m for the next 40° and remains constant for the next 180°. This cycle is repeated

during each revolution. The power is supplied  by a constant torque motor and the fluctuation of speed
is to be limited to ± 3% of  the mean speed. Find the power of the motor and the moment of inertia ofthe flywheel fitted to the machine.
[Ans. 68 kW; 67.22 kg-m
2]
20.A punching press pierces 35 holes per minute in a plate using 10 kN-m of energy per hole during each
revolution. Each piercing takes 40 per cent of the time needed to make one revolution. A cast iron

flywheel used with the punching machine is driven by a constant torque electric motor. The flywheel
rotates at a mean speed of 210 r.p.m. and the fluctuation of speed is not to exceed 1± per cent of the
mean speed. Find : 1. power of the electric motor, 2. mass of the flywheel, and 3. cross-sectional
dimensions of the rim when the width is twice its thickness. Take hoop stress for cast iron = 4 MPa
and density of cast iron = 7200 kg/m3.
[Ans. 5.83 kW; 537 kg; 148 mm, 74 mm]

1.Draw the turning moment diagram of a single cylinder double acting steam engine.
2.Explain precisely the uses of turning moment diagram of reciprocating engines.
3.Explain the turning moment diagram of a four stroke cycle internal combustion engine.
4.Discuss the turning moment diagram of a multicylinder engine.
5.Explain the terms ‘fluctuation of energy’ and ‘fluctuation of speed’ as applied to flywheels.

6.Define the terms ‘coefficient of fluctuation of energy’ and ‘coefficient of fluctuation of speed’, in the
case of flywheels.
7.What is the function of a flywheel? How does it differ from that of a governor?
8.Prove that the maximum fluctuation of energy,
                                                    S2EE C∆=×
where                          E  =  Mean kinetic energy of the flywheel, and
CS= Coefficient of fluctuation of speed.

1.The maximum fluctuation of energy is the
(a) sum of maximum and minimum energies
(b) difference between the maximum and minimum energies
(c) ratio of the maximum energy and minimum energy
(d) ratio of the mean resisting torque to the work done per cycle
2.In a turning moment diagram, the variations of energy above and below the mean resisting torque line
is called
(a) fluctuation of energy
(b) maximum fluctuation of energy
(c) coefficient of fluctuation of energy

Chapter 16 : Turning Moment Diagrams and Flywheel    /G6C  611
3.The ratio of the maximum fluctuation of speed  to the mean speed is called
(a) fluctuation of speed ( b) maximum fluctuation of speed
(c) coefficient of  fluctuation of speed ( d) none of these
4.The ratio of the maximum fluctuation of energy to the, ......... is called coefficient of fluctuation  of
energy.
(a) minimum fluctuation of energy ( b) work done per cycle
5.The maximum fluctuation of energy in a flywheel is equal to

(a)()12.Iωω− ω (b)2
S..ICω
(c) S2.EC (d) all of these
where I = Mass moment of inertia of the flywheel,
E = Mean kinetic energy of the flywheel,
CS = Coefficient of fluctuation of speed, and
ω= Mean angular speed 12
2ω+ ω= .
ANSWERS
1.(b) 2.(a) 3.(c) 4.(b) 5.(d)

612  /G6C   Theory of Machines
Steam Engine
Valves and
Reversing
Gears17
Features
2. D-slide V alve.
3. Piston Slide Valve.4. Relative Positions of Crank
and Eccentric Centre Lines.
5. Crank Positions for
Admission, Cut off, Releaseand Compression.
6. Approximate Analytical
Method for Crank Positions.
7. Valve Diagram.8. Zeuner Valve Diagram.9. Reuleaux Va lve Diagram.
10. Bilgram Valve Di agram.
11. Effect of the Early Point of
Cut-off.
12. Meyer’s Expansion V alve.

Cut-off.
13. Virtual or Equivalent
Eccentric for the Meyer’sExpansion Valve.
14. Minimum Width and Best
Setting of the ExpansionPlate.
15. Reversing Gears.16. Principle of Link Motions.17. Stephenson Link Motion.18. Virtual or Equivalent
Eccentric for StephensonLink Motion.
19. Radial Valve Gears.
20. Hackworth Va lve Gear.
21. Walschaert Valve Gear.17.1. Introduction
The valves are

used to control the steamwhich drives the pistonof a reciprocating steamengine. The valves haveto perform the four dis-tinct operations on thesteam used on one side(i.e. cover end) of the pis-
ton, as shown by the in-dicator diagram (alsoknown as pressure-vol-ume diagram) in Fig.17.1. These operationsare as follows:
1. Admission  or opening of inlet valve for admission
of steam to the cylinder. The point A represents the point for

admission of steam just before the end of return stroke and itis continued up to the point B.
Fig. 17.1. Indicator diagram of a
reciprocating steam engine.
612

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  613
2. Cut-off or closing of inlet valve in order to stop the admission of steam prior to expansion.
The point B represents the cut-off point of steam. The curve BC represents the expansion of steam in
the engine cylinder.
3. Release or opening of exhaust valve to allow the expanded steam to escape from the
cylinder to the atmosphere or to the condenser or to a larger cylinder. The point C represents the

opening of the valve for releasing the steam. The exhaust continues during the return stroke uptopoint D.
4. Compression  or closing of exhaust valve for stopping the release of steam from the cylinder
prior to compression. The point D represents the closing of exhaust valve. The steam which remains
in the cylinder is compressed from D to A and acts as a cushion for the reciprocating parts.
The same operations, as discussed above, are performed on steam in the same order on the

other side (or crank end) of the piston for each cycle or each revolution of the crank shaft. In otherwords, for a double acting piston, there are eight valve operations per cycle. All these eight opera-tions may be performed
(i)by a single slide valve such as D-slide valve,
(ii)by two piston valves, one for either end of cylinder, and
(iii)by two pairs of valves (one pair for each end of the cylinder), such as corliss valves or

drop valves. One valve at each end of the cylinder performs the operations of admission and cut-offwhile the other valve performs the operations of release and compression.
The engine performance depends upon the setting of the valves. In order to set a valve at a
correct position, a valve diagram is necessary.
17.2. D-slide Valve
The simplest type of the slide valve, called the D-slide valve, is most commonly used to

control the admission, cut-off, release and compression of steam in the cylinder of reciprocatingsteam engine. The usual arrangement of the D-slide valve, valve chest and cylinder for a double
acting steam engine is shown in Fig. 17.2 ( a).
Sectional view of a steam engine.Regulator valve
Fire
tubes Smokebox
Cylinder
valves
Cylinder PistonBoilerFireboxDriver’s
cab

614  /G6C   Theory of Machines
The steam from the boiler is admitted to the steam chest through a steam pipe. The recess R
in the valve is always open to the exhaust port which, in turn, is open either to atmosphere or to thecondenser. The ports P
1 and P2 serve to admit steam into the cylinder or to pass out the steam from the

cylinder. The valve is driven from an eccentric keyed to the crankshaft. It reciprocates across theports and opens them alternately to admit high pressure steam from the steam chest and to exhaust theused steam, through recess R to exhaust port.
The D-slide valve in its mid position relative to the ports is shown in Fig. 17.2 ( a). In this
position, the outer edge of the valve overlaps the steam port by an amount s. This distance s (i.e.

lapping on the outside of steam port) is called the steam lap  or outside lap . The inner edge of the
valve, also, overlaps the steam port by an amount e. This distance e (i.e. overlapping on the inside) is
called the exhaust lap  or inside lap .
Fig. 17.2.  D-slide valve.
The displacement of the valve may be assumed to take place with simple harmonic motion,
since the *obliquity of the eccentric rod is very small. Thus the eccentric centre line OE will be at

right angles to the line of stroke when the valve is in its mid-position. This is shown in Fig. 17.2 ( b)
for clockwise rotation of the crank.
Note: Since the steam is admitted from outside the steam chest, therefore the D-slide valve is also known as
Outside admission valve .
17.3. Piston Slide Valve
The piston slide valve, as shown in Fig. 17.3 ( a), consists of two rigidly connected pistons.

These pistons reciprocate in cylindrical liners and control the admission to, and exhaust from the twoends of the cylinder. In this case, high pressure superheated steam is usually admitted to the spacebetween the two pistons through O and exhaust takes place from the ends of the valve chest through
E. This type of valve is mostly used for locomotives and high pressure cylinders of marine engines.
The piston slide valve has the following advantages over the D-slide valve :

*Since the length of the eccentric rod varies from 15 to 20 times the eccentricity (also known as throw of the
eccentric), therefore the effect of its obliquity is very small. The eccentricity  or the throw of eccentric  is
defined as the distance between the centre of crank shaft O and the centre of eccentric E. Thus the distance OE
is the eccentricity.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  615
1.  Since there is no unbalanced steam thrust between the valve and its seat as the pressure
on the two sides is same, therefore the power absorbed in operating the piston valve is less than theD-slide valve.
2.  The wear of the piston valve is less than the wear of the D-slide valve.
3.   Since the valve spindle packing is subjected to the relatively low pressure and tempera-

ture of the exhaust steam, therefore the danger of leakage is less.
Fig. 17.3.  Piston slide valve.
The position of the steam lap ( s) and exhaust lap ( e) for the piston valve in its mid position is
shown in Fig. 17.3 ( a). The eccentric position for the clockwise rotation of the crank is shown in Fig.
17.3(b).
Note: Since the steam enters from the inside of the two pistons, therefore the piston valve is also known as
inside admission valve .

17.4. Relative Positions of Crank and Eccentric Centre Lines
Fig. 17.4.  Relative positions of crank and eccentric centre lines for D-slide valve.
We have discussed the D-slide valve (also known as outside admission valve) and piston
slide valve (also known as inside admission valve) in Art. 17.2 and Art. 17.3 respectively. Now weshall discuss the relative positions of the crank and the eccentric centre lines for these slide valves.

616  /G6C   Theory of Machines
1. D-slide or outside admission valve . The D-slide valve in its mid position is shown in
Fig. 17.2 ( a). At the beginning of the stroke of the piston from left to right as shown in Fig. 17.4 ( a),
the crank OC is at its inner dead centre position as shown in Fig. 17.4 ( b). A little consideration will
show that the steam will only be admitted to the cylinder if the D-slide valve moves from its mid

position towards the right atleast by a distance equal to the steam lap ( s). It may be noted that if
only this minimum required distance is moved by the valve, then the steam admitted to the cylinderwill be subjected to severe throttling or wire drawing. Therefore, in actual practice, the displace-ment of the D-slide valve is greater than the steam lap ( s) by a distance l which is known as the lead
of the valve.

In order to displace the valve from its mid position by a distance equal to steam lap plus lead
(i.e. s + l ), the eccentric centre line must be in advance of the 90° position by an angle α , such that
sinsl
OE+α=
The angle α is known as the angle of advance  of the eccentric. The relative positions of the
crank OC and the eccentric centre line OE remain unchanged during rotation of the crank OC, as
shown in Fig. 17.4 ( b).

Note : The eccentricity (or throw of the eccentric) OE is equal to half of the valve travel. The valve travel is the
distance moved by the valve from one end to the other end.
2. Piston slide valve or inside admission valve.  The piston slide valve in its mid position is
shown in Fig. 17.3 ( a). At the beginning of the outward stroke of the piston, from left to right as
shown in Fig. 17.5 ( a), the crank OC is at its inner dead centre as shown in Fig. 17.5 ( b). In the

similar way as discussed for D-slide valve, the valve should be displaced from its mean position by
a distance equal to the steam lap plus lead ( i.e. s + l) of the valve. The relative positions of the crank
OC and the eccentric centre line OE are as shown in Fig. 17.5 ( b). In this case, the angle of advance
is (180° + α ), and
sinsl
OE+α=
Fig. 17.5.  Relative positions of crank and eccentric centre lines for piston slide valve.

17.5. Crank Positions for Admission, Cut-off, Release and Compression
In the previous article, we have discussed the relative positions of the crank and eccentric
centre lines for both the D-slide valve and piston slide valve. Here we will discuss only the D-slide
valve to mark the positions of crank for admission, cut-off, release and compression. The sameprinciple may be applied to obtain the positions of crank for piston slide valve.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  617
1. Crank position for admission . At admission for the cover end of the cylinder, the outer
edge of the D-slide valve coincides with the outer edge of the port P1. The valve moves from its mid
position towards right, as shown by arrow A in Fig. 17.6 ( a), by an amount equal to steam lap s. At the
same time, the piston moves towards left as shown by thick lines in Fig. 17.6 ( a). The corresponding

position of the crank is OC1 and the eccentric centre line is shown by OE1 in Fig. 17.6 ( b), such that
∠ C1 OE1= 90° + α
Fig. 17.6.  Crank positions for admission and cut-off.
2. Crank position for cut-off . A little consideration will show that the cut-off will occur on
the cover end of the cylinder when the outer edge of the D-slide valve coincides with the outer edge
of the port P1 while the valve moves towards left as shown by arrow B. The piston now occupies the

position as shown by dotted lines in Fig. 17.6 ( a). The corresponding position of the crank is OC2 and
the eccentric centre line is shown by OE2 in Fig. 17.6 ( c), such that
∠ C2OE2=∠ C1OE1 = 90° + α
3. Crank position for release . At release for the cover end of the cylinder, the inner edge of
the D-slide valve coincides with the inner edge of the port P1. The valve moves from its mid position

towards left, as shown by arrow C in Fig. 17.7 ( a), by a distance equal to the exhaust lap e. Thus the
Fig. 17.7.  Crank positions for release and compression.

618  /G6C   Theory of Machines
 valve opens the port to exhaust. At the same time, the piston
moves towards right as shown by thick lines in Fig. 17.7 ( a).
The corresponding positions of crank and eccentric centre lineare shown by OC
3 and OE3 in Fig. 17.7 ( b), such that
∠ C3OE3= 90° + α
4. Crank position for compression . At compression,
for the cover end of the cylinder, the inner edge of the valvecoincides with the inner edge of the port P
1. The valve moves

from its mid position towards right, as shown by arrow D in Fig.
17.7 (a), by a distance equal to the exhaust lap e. The valve now
closes the port to exhaust. The piston moves towards left asshown by dotted lines in Fig. 17.7 ( a). The corresponding posi-
tions of crank and eccentric centre line are shown by OC
4 and
OE4 in Fig. 17.7 ( c), such that
∠ C4OE4=∠ C3OE3 = 90° + α
The positions of crank and eccentric centre line for all

the four operations may be combined into a single diagram, asshown in Fig. 17.8 ( a). Since the ideal indicator diagram, as
shown in Fig. 17.8 ( b), is drawn by taking projections from the
crank positions, therefore the effect of the obliquity of the con-necting rod is neglected. Fig. 17.8. Combined diagram
of crank positions.
17.6. Approximate Analytical Method for Crank Positions at Admission,
Cut-off, Release and Compression

The crank positions at which admission, cut-off, release and compression occur may be
obtained directly by analytical method as discussed below :(a)
(b)

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  619
Let x= Displacement of the valve from its mid-position,
θ= Crank angle,
 r= Eccentricity or throw of eccentric 1
2Travel of valve,=×
α= Angle of advance of eccentric.
Since the displacement of the valve may be assumed to take place with simple harmonic
motion, therefore
x=r sin (θ + α)  . . . (i)
But at admission and cut-off,
x= Steam lap, s
∴ s=r sin (θ + α) ...[From equation (i)]
 or                  –1 –1sin , and sin –ss

or                  –1 –1sin , and sin –ss
rr θ+α= θ= α   . . . (ii)
The two values of θ which satisfy the equation (ii) give the crank positions for admission and
cut-off.
Similarly, at release and compression, x = exhaust lap ( e) and is negative as measured from
the origin O.
∴ – e=r sin  (θ + α) . . . [From equation  (i)]
or    –1 –1––sin , and sin –ee
rr θ+α= θ= α    . . . (iii)

The two values of θ which satisfy the equation (iii) give the crank positions for release and
compression.
Example 17.1.  The D-slide valve taking steam on its outside edges has a total travel of 150

mm. The steam and exhaust laps for the cover end of the cylinder are 45 mm and 20 mm respectively.If the lead for the cover end is 6 mm, calculate the angle of advance and determine the main crankangles at admission, cut-of f, release and compression respectively for the cover end. Assume the
motion of the valve as simple harmonic.
Solution.  Given : 2 r = 2 OE = 150 mm or r = OE = 75 mm ; s = 45 mm ; e = 20 mm ;
l = 6 mm
Angle of advance
Let                      α  = Angle of advance.

45 6sin 0.6875sl
OE++α= = =
or                                   42.8α= °  Ans.
Crank angles at admission, cut-off, release and
compression
Let    θ
1, θ2, θ3 and θ4 = Crank angles at admission,
  cut-off, release and
                                 compression  respectively.
We know that for admission and cut-off,
–1 –1 –1 45sin sin sin (0.6) 36.87 or143.13
75s
r  θ+α= = = = ° °   

75s
Prototype of an industrial steam engine.

620  /G6C   Theory of Machines
∴θ1= 36.87° – α = 36.87° – 42.8° = – 5.93° Ans.
and θ2= 143.13° – α = 143.13° – 42.8° = 100.33° Ans.
We know that for release and compression,
–1 –1 –1–– 2 0sin sin sin (– 0.2667)
75e
r  θ+α= = =   
= 195.47° or 344.53°
∴θ3= 195.47° – α = 195.47° – 42.8° = 152.67° Ans.
and θ4= 344.53° – α = 344.53° – 42.8° = 301.73° Ans.
17.7. Valve Diagram
The crank positions for admission, cut-off, release and compression may be easily deter-

mined by graphical constructions known as valve diagrams . There are various methods of drawing
the valve diagrams but the following three are important from the subject point of view :
1. Zeuner valve diagram. 2. Reuleaux valve diagram, and 3. Bilgram valve diagram.
We shall discuss these valve diagrams, in detail as follows.
17.8. Zeuner Valve Diagram
The Zeuner’s valve diagram, as shown in Fig. 17.9, is drawn as discussed in the  following
steps :

steps :
1. First of all, draw AB equal to the travel of the valve to some suitable scale. This diameter
AB also represents the stroke of the piston to a different scale.
Fig. 17.9.  Zeuner valve diagram.
2. Draw a circle on the diameter AB such that OA = OB = eccentricity or throw of the eccen-
tric. The circle ACBD is known as the valve travel circle , where diameter CD is perpendicular to AB.
3. Draw EOF making an angle α, the angle of advance of the eccentric, with CD. It may be

noted that the angle α is measured from CD in the direction opposite to the rotation of the crank and
eccentric as marked by an arrow in Fig. 17.9.
4. In case the angle of advance ( α ) is not given, then mark OJ = steam lap ( s), and JK = lead
(l) of the valve. Draw KE perpendicular to AB which intersects the valve travel circle at E. The angle
EOC is now the angle of advance.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  621
5. Draw circles on OE and OF as diameters. These circles are called valve circles .
6. With O as centre, draw an arc of radius OG equal to steam lap ( s) cutting the valve circle
at M and N. Join OM and ON and produce them to cut the valve travel circle at C1 and C2 respectively.
Now OC1 and OC2 represent the positions of crank at admission and cut-off respectively.

Note : The circle with centre A and radius equal to lead ( l) will touch the line C1 C2.
7. Again with O as centre, draw an arc of radius OH equal to exhaust lap ( e) cutting the valve
circle at P and Q. Join OP and OQ and produce them to cut the valve travel circle at C3 and C4
respectively. Now OC3 and OC4 represent the position of crank at release and compression respec-
tively.
8. For any position of the crank such as OC', as shown in Fig. 17.9, the distance OX1

represents the displacement of the valve from its mid position and the distance X1 X2 (the point X1
is on the valve circle and the point X2 is on the arc JGN) gives the opening of the port to steam. The
distance X3 X4 ( point X3 is on the arc QHP and point X4 is on the valve circle) obtained by produc-
ing the crank OC',  gives the opening of the port to exhaust for the crank position OC'. The proof of
the diagram is as follows :
Join EX1. Now angle EX1O = 90°.

∴∠ OEX1 + ∠ X1OE= 90° = ∠ AOC = θ + ∠ X1OE + α
or ∠ OEX1=θ + α
Now from triangle OEX1,
OX1=OE sin (θ + α)
or x=r sin (θ + α)
where x= Displacement of the valve from its mid position, and
r= Eccentricity or throw of eccentric.
Now OX1=OX2 + X1 X2
∴ Opening of port to steam when the valve has moved a distance x from its mid-position,
X1 X2=OX1 – OX2 = r sin (θ + α) – s ...(∵ OX2 = Steam lap, s)

9. Mark HR = width of the steam port. Now with O as centre, draw an arc through R intersect-
ing the valve circle at X and Y. The lines OS and OW through X and Y respectively determines the
angle WOS through  which the crank turns while the steam port is full open to exhaust. The maximum
opening of port to exhaust is HF. A similar construction on the other valve circle will determine the

angle through which the crank turns while the steam port is not full open to steam. Fig. 17.9 showsthat the steam port is not full open to steam and the maximum opening of the port to steam is GE.
Note : The valve diagram, as shown in Fig. 17.9, is for the steam on the cover end side of the piston or for one-
half of the D-slide valve. In order to draw the valve diagram for the crank end side of the piston (or the other half

of the valve), the same valve circles are used but the two circles and the lines associated with them change
places. For the sake of clearness, the valve diagrams for the two ends of the piston is drawn separately.
17.9. Reuleaux’s Valve Diagram
The Reuleaux’s valve diagram is very simple to draw as compared to the Zeuner’s valve

diagram. Therefore it is widely used for most problems on slide valves. The Reuleaux’s valve dia-gram, as shown in Fig. 17.10, is drawn as discussed in the following steps :

622  /G6C   Theory of Machines
tric. This  circle ACBD is known as valve travel circle
where diameter CD is perpendicular to AB.
3. Draw EOF making an angle α, the angle of
advance of eccentric, with CD. It may be noted that the
angle α  is measured from CD in the direction opposite
to the rotation of crank and eccentric as marked by anarrow in Fig. 17.10.
4. Draw GOH perpendicular to EOF. Now draw
chords C
1 C2 and C3C4 parallel to GH and at distances

equal to steam lap ( s) and exhaust lap ( e) from GH
5. Now OC1, OC2, OC3 and OC4 represent the
positions of crank at admission, cutt-off, release and com-pression respectively.
The proof of the diagram is as follows :Let OC' be any crank position making an angle θ with the inner dead centre, as shown
in Fig.17.10. Draw C' J perpendicular to GH. From right angled triangle OC' J,
C' J=C' O sin C' OJ = r sin (θ + α) ... (i)

where r=C' O = Eccentricity or throw of eccentric.
But the displacement of the valve from its mid-position corresponding to crank angle θ is
given by
x=r sin (θ + α) ... (ii)
C' J=x
It, therefore, follows that the length of the perpendicular from C' to the diameter GH is equal
to the displacement of the valve from mid-position when the crank is in the position OC'.

We see from Fig. 17.10, that when the crank is in position OC
1 or OC2, the length of the
perpendicular from C1 or C2 on GH is equal to the steam lap ( s). Therefore OC1 and OC2 must
represent the crank positions at admission and cut-off respectively. Similarly, when the crank is inposition OC
3 and OC4, the length of perpendicular from C3 or  C4 on GH is equal to the exhaust lap
(e). Therefore OC3 and OC4 must represent the crank positions at release and compression respec-
tively.

tively.
Opening of the port to steam
We see from Fig. 17.10, that when the crank is in position OC', the displacement of the valve
C'J, from its mid-position,  exceeds the steam lap ( s) by a distance C' K. The distance C' K represents
the amount of port opening to steam. Therefore when the crank is in position OA (i.e. at the inner dead
centre), the perpendicular distance from A to GH, i.e. AP represents the displacement of the valve

from its mid-position. The distance AP exceeds the steam lap ( s) by a distance AQ which is equal to
the lead of valve and represents the amount of port opening to steam.
The maximum possible opening of the port to steam is equal to NE i.e. (r–s) where r is the
throw of the eccentric or half travel of the valve and s is the steam lap. Similarly, the maximum
possible opening of the port to exhaust is equal to MF i.e. (r–e) where e is the exhaust lap. The

difference ( r–e) may exceed the width of the actual port through which the steam is admitted to and
exhausted from the cylinder. In that case, the port will remain fully open for a certain period of crankFig. 17.10.  Reuleaux’s valve diagram.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  623
rotation. In order to find the duration of this period, draw a chord SW parallel to C3C4 at a distance
equal to  the width of the steam port ( w). The port will remain fully open to exhaust when the crank
rotates from the positon OW to OS.
17.10. Bilgram Valve Diagram
The Bilgram valve diagram, as shown in Fig. 17.11, is drawn as discussed in the following steps :

tric. This circle is known as valve travel circle.
3. Draw diameter GOH making an angle α, the angle of advance, with AB. The angle α is
measured from AB in the direction opposite to the rotation of crank and eccentric as marked by an

arrow in Fig. 17.11.
4. Draw two circles with centres G and H and radii equal to steam lap ( s) and exhaust lap ( e)
respectively as shown in Fig. 17.11.
5. The lines OC1 and OC2 are tangential to the steam lap circles and they represent the crank
positions for admission and cut-off respectively. Similarly OC3 and OC4 are tangential to the exhaust
lap circles and represent the crank positions for release and compression.
The proof of the diagram is as follows :

Let OC' be any crank position making an angle θ with the inner dead centre, as shown in
Fig. 17.11. Draw perpendiculars GE on OC' and HF on OC' produced.
Since the triangles OGE and OHF are similar, therefore
GE=HF = OG sin (θ + α) = r sin (θ + α) ... (i)
where r=OG = Eccentricity or throw of eccentric.
But the displacement of the valve from its mid-position corresponding to crank angle θ, is given by
A 1930’s Steam locomotive.

624  /G6C   Theory of Machines
GE=x
It, therefore, follows that the length of the perpendicular from G (or H) on OC' (or OC' pro-
duced) is equal to the displacement of the valve from mid-position when the crank is in the position OC'.
Fig. 17.11.  Bilgram valve diagram.
We see from Fig. 17.11, that the length of the perpendiculars from G and H on OC1 and OC2
respectively are equal to steam lap ( s). Therefore OC1 and OC2 must represent the crank positions at

admission and cut-off respectively, Similarly, the length of perpendiculars from H and G on OC3 and
OC4 respectively are equal to exhaust lap ( e). Therefore OC3 and OC4 must represent the crank
positions at release and compression respectively.
We see from Fig. 17.11, that when the  crank is in position OC', the displacement of the valve
GE, from its mid-position, exceeds the steam lap ( s) by a distance DE.  The distance DE represents

the amount of port opening to steam. Therefore, when the crank is in position OA (i.e. at the inner
dead centre), the perpendicular distance from G to OA (i.e. GL) represents the displacement of the
valve from its mid-position. The distance GL exceeds the steam lap ( s) by a distance ML which is
equal to lead of the valve and represents the amount of port opening to steam.
The maximum opening of the port to steam is equal to OP or (r – s) where r is the throw of

eccentric or half travel of the valve and s is the steam lap. Similarly the maximum opening of the port
to exhaust is OQ or (r – e) where e is the exhaust lap.
Notes :  1. The point G lies on the intersection of the bisectors of angles C1OT and RST.
2. The Bilgram valve diagram is usually used to determine throw of the eccentric, valve travel, angle of
advance of the eccentric, steam and exhaust laps when the crank positions at cut-off and release, lead of valve

and width of steam port are known.
Example 17.2.  The following particulars refer to a D-slide valve :
Total valve travel = 150 mm ; Steam lap = 45  mm ; Exhaust lap = 20 mm ; Lead = 6 mm.
Draw the Zeuner’s valve diagram for the cover end and determine the angle of advance of

the eccentric, main crank angles at admission, cut-off, release and compression, opening of port tosteam for 30° of crank rotation and maximum opening of port to steam. If the width of the port is40 mm, determine the angle through which the crank turns so that the exhaust valve is full open.
Solution. Given : AB = 150 mm ; s = 45 mm ; e = 20 mm ; l = 6 mm

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  625
Angle of advance of eccentric and main crank angles at admission, cut-off, release and compression
The Zeuner’s valve diagram for the cover end is drawn as discussed in the following steps :
1. First of all draw AB = 150 mm, to some suitable scale, to represent the total valve travel.
Draw a circle on this diameter AB such that OA = OB = throw of the eccentric. This circle is known as

valve travel circle. Draw COD perpendicular to AB.
2. Mark OJ = steam lap = 45 mm, and JK = lead = 6 mm. Through K, draw a perpendicular
on AB which intersects the valve travel circle at E. Join OE. Now angle COE represents the angle of
advance of the eccentric ( α) in a direction opposite to the direction of rotation of crank and eccentric
(shown clockwise in Fig. 17.12). By measurement, we find that
α=∠ COE = 42.5° Ans.
Fig. 17.12

Fig. 17.12
3. Draw a circle on OE as diameter. This circle is known as valve circle. Now with O as
centre, draw an arc of radius equal to steam lap ( i.e. 45 mm) which intersects the valve circle at M and
N. Join OM and ON and produce them to intersect the valve travel circle at C1 and  C2 respectively.
The lines OC1 and OC2 represent the crank positions at admission and cut-off respectively. By mea-
surement, we find that
Crank angle at admission from inner dead centre A

=∠ AOC1 = – 6° Ans.
and crank angle at cut-off from inner dead centre A
=∠ AOC2 = 101° Ans.
4. Now draw the diameter EOF. On OF draw a valve circle as shown in Fig. 17.12. With O as
centre, draw an arc of radius equal to exhaust lap ( i.e. 20 mm) which intersects the valve circle at P
and Q. Join OP and OQ and produce them to intersect the valve travel circle at C3 and C4 respectively.

Now OC3 and OC4 represent the crank positions at release and compression respectively. By mea-
Crank angle at release from inner dead centre A
=∠ AOC3 = 153° Ans.
and crank angle at compression from inner dead centre A
=∠ AOC4 = 302° Ans.
Opening of port to steam for 30° of crank rotation and maximum opening of port to steam
Let OC' be the crank at θ = 30° from the inner dead centre, as shown in Fig. 17.12. The crank

626  /G6C   Theory of Machines
OC' intersects the valve circle at X1 and arc MGN at X2. Now X1 X2 represents the opening of port to
steam for 30° of crank rotation. By measurement, we find that
X1X2= 27 mm Ans.
and maximum opening of port to steam,
GE= 30 mm Ans.
Angle through which the crank turns so that the exhaust valve is full open
Mark HR = width of port = 40 mm as shown in Fig. 17.12. Now with O as centre, draw an arc

passing through R which intersects the valve circle at X and Y. The angle XOY represents the crank
angle at which the exhaust valve is full open. By measurement, we find that
∠ XOY= 72° Ans.
Example 17.3.  The following data refer to a D-slide valve :
Total valve travel = 120 mm ; Angle of advance = 35° ; Steam lap = 25 mm ; Exhuast
lap = 8 mm.
If the length of the connecting rod is four times the crank radius , determine the positions of

the piston as percentage of the stroke for adm ission, cut-off, release and compression for both ends
of the piston.
Solution. Given : AB = 120 mm ; α = 35°, s = 25 mm ; e = 8 mm
Positions of the piston as a percentage of the stroke for admission, cut-off, release and compres-
sion for cover end of the piston.
First of all, determine the crank positions for the cover end either by Zeuner’s or Reuleaux

valve diagram. The Reuleaux valve diagram for the cover end, as shown in Fig. 17.13, is drawn asfollows :
1. Draw AB = 120 mm to some suitable scale to represent the total valve travel. This diameter
AB also represents the stroke of the piston. Draw a circle on diameter AB such that OA = OB  = Throw
of the eccentric or radius of crank.
Fig. 17.13
2. Draw a line GOH making an angle of 35°, the angle of advance, in a direction opposite to

the rotation of crank which is clockwise as shown in Fig. 17.13.
3. Draw EOF perpendicular to GOH and mark ON = steam lap =  25 mm and OM = exhaust
lap = 8 mm. Through N and M draw lines parallel to GOH which intersect the valve travel circle at C1,
C2, C3 and C4. Now OC1, OC2, OC3 and OC4 represent the crank positions at admission, cut-off
release and compression respectively.
At inner dead centre A and outer dead centre B, the connecting rod is in line with the crank.

Since the connecting rod is 4 times the crank radius OA, therefore mark AA1 = BB1 = 4 × OA = 4 × 60
= 240 mm. Now A1B1 = AB = 120 mm and represents the stroke of the piston. With C1, C2, C3, C4 as

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  627
centres and radius equal to length of connecting rod i.e. 240 mm, mark the corresponding positions of
piston as shown by points P1, P2, P3 and P4 in Fig. 17.13. By measurement, the piston position as
percentage of stroke is given by :
At admission11
11100BP
BA=× of return stroke119100 99.17%120=×= Ans.
At cut-off12
11100AP
AB=× of forward stroke96100 80%
120=×= Ans.
At release13
11100AP
AB=×  of forward stroke116100 96.67%

11100AP
At compression14
11100BP
BA=× of return stroke100100 83.33%
Note : The points p1, p2, p3 and p4 on AB are the corresponding points of P1, P2, P3 and P4 respectively. These
points may be obtained by drawing the arcs through C1, C2, C3 and C4 with their centres at P1, P2, P3 and P4 and
radius equal to the length of connecting rod. Now the piston positions as percentage of stroke is given by :
At admission 1100Bp

BA=×  of return stroke ; At cut-off 2100Ap
AB=× of forward stroke
     At release 3100Ap
AB=× of forward stroke ; At compression 4100Bp
BA=×  of return stroke
sion for crank end of the piston.
Fig. 17.14
The valve diagram for the crank end is drawn by rotating the valve diagram for the cover end

through 180° in the direction of rotation of the crank, as shown in Fig. 17.14. By measurement, thepiston positions as percentrage of stroke is given by :
At admission
11
11100BP
BA=× of forward stroke 1100Bp
BA=× of forward stroke
119100 99.17%120=×=  Ans.

628  /G6C   Theory of Machines
11100AP
AB=× of return stroke2100Ap
AB=× of return stroke
85100 70.83%
120=×=  Ans.
11100AP
AB=× of return stroke3100Ap
112100 93.33%120=×=  Ans.
11100BP
BA=× of forward stroke4100Bp
106100 88.33%
Example 17.4.  A slide valve has a travel of 125 mm. The angle of advance of the eccentric is

35°. The cut-off and release takes place at 75 per cent and 95 per cent of the stroke at each end of the
cylinder. If the connecting rod is 4 times the crank length, find steam lap, exhaust lap and lead foreach end of the valve.
Solution. Given : AB = 125 mm ; α = 35°
Piston position at cut-off for both ends ( i.e. cover and crank end)
= 75% of stroke
Piston position at release for both ends
= 95% of stroke
Steam lap , exhaust lap and lead for the
cover end
The Reuleaux’s diagram for the

cover end
cover end, as shown in Fig. 17.15, is drawnas discussed below :
1. First of all, draw AB = 125 mm
to some suitable scale, to represent the valvetravel. This diameter AB also represents the
piston stroke. On this diameter AB draw a
valve travel circle such that OA = OB =
Throw of eccentric. The radius OA or OB
also represents the crank radius.
2. Draw a line GOH making an

angle of 35°, the angle of advance, in a di-rection opposite to the rotation of the crankwhich is clockwise as shown in Fig. 17.15.
3. At inner dead centre A and outer
dead centre B, the connecting rod is in line
with the crank. Since the connecting rod is4 times the crank radius, therefore mark AA
1
= 4 × OA = 4 × 125 /2 = 250 mm. Now
A1B1 = AB = 125 mm and represents the
stroke of the piston.
Broaching machine. Broaching is a process of machin-

ing through holes of any cross sectional shape, straightand helical slots, external surfaces of various shapes,external and internal toothed gears, splines, keywaysand rifling.
Note : This picture is given as additional information and is
not a direct example of the current chapter.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  629
4. Since the cut-off takes place at 75 percent of the stroke, therefore
12 2
110.75AP Ap
AB AB==
∴ A1P2=A p2 = 0.75 × A1B1 = 0.75 × 125 = 94 mm
...(∵ A1B1 = AB = 125 mm)
With P2 as centre and radius equal to P2 p2 (i.e. length of connecting rod), draw an arc
through p2 which intersects the valve travel circle at C2. This point C2 represents the crank-pin
position at cut-off.
Fig. 17.15

Fig. 17.15
5.  From C2 draw C2C1 parallel to GH which intersects a line EOF (perpendicular to GOH)
at N. The point C1 represents the crank-pin position at admission and ON is the steam lap. By
measurement,
Steam lap = ON = 32 mm Ans.
6. Since the release takes place at 95 % of the stroke, therefore
13 3
110.95AP Ap
AB AB==
∴∴∴∴∴ A1P3=Ap3 = 0.95 × A1B1 = 0.95 × 125 = 118.8 mm
With P3 as centre and radius equal to P3 p3 (i.e. length of connecting rod), draw an arc

through p3 which intersects the valve travel circle at C3. This point C3 represents the crank-pin posi-
tion at release.
7. From C3 draw C3C4 parallel to GH which intersects a line EOF at M. The point C4 repre-
sents the crank-pin position at compression and OM is the exhaust lap. By measurement,
Exhaust lap = OM = 8 mm  Ans.
8. In order to find the lead, draw a circle with centre A such that C1C2 is tangential to this

circle (or draw AL perpendicular to C1C2). The perpendicular AL represents the lead of the valve.
By measurement,
Lead =AL = 6 mm Ans.
Steam lap, exhaust lap and lead for the crank end
Fig. 17.16

630  /G6C   Theory of Machines
The Reuleaux’s valve diagram for the crank end, as shown in Fig. 17.16, is drawn by rotating
the valve diagram for the cover end, through 180°. By measurement,
Steam lap = ON = 20 mm  Ans.
Exhaust lap = OM = 12 mm Ans.
and Lead = AL = 16 mm  Ans.
Example 17.5.  The following data refer to a D-slide valve for the cover end :
Position of the crank at cut off = 0.7 of stroke ; Lead = 6 mm ; Maximum opening of port to

steam = 45 mm ; connecting rod length = 4 times crank length.
Find the travel of valve, angle of advance and steam lap.Solution. Given : Position of Crank at cut-off = 0.7 of stroke ; Lead = 6 mm ; Maximum
opening of port to steam = 45 mm ; Connecting rod length = 4 times the crank length.
The travel of valve, angle of advance and steam lap may be obtained by using Bilgram valve
diagram as discussed below :

1. Draw A'B' of any convenient length, as shown in Fig. 17.17, to represent the assumed valve
travel. Draw the assumed valve travel circle on this diameter A' B' which also represents the piston
stroke (assumed).
Fig. 17.17
2. Since the length of connecting rod is 4 times the crank OA', therefore mark A'A'1, = B' B'1
= 4 × OA'. Now A'1 B'1 represents the piston stroke.
3. The cut-off for the cover end takes place at 0.7 of the stroke, therefore mark
12 2
110.7AP Ap

12 2
110.7AP Ap
AB AB′′ ′′==′′′′
4. Now P'2 as centre and radius P'2 p'2 (i.e. length of connecting rod), draw an arc p'2 C'2.
Now OC'2 represents the crank position at cut-off.
5.  Draw a line RS parallel to A' B' and at a distance equal to the lead i.e. 6 mm, to some
suitable scale. The point S lies on the line C'2 OT.
6. With O as centre, draw an arc of radius equal to the maximum opening of port to steam ( i.e.
45 mm) which intersects RS at X and OT at Y.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  631
7. Draw the bisector of angle RST. The point G on this bisector is obtained by hit and trial
such that the circle with centre G touches the maximum opening arc at P, the lines RS and ST. The
point G is a point on the actual valve travel circle and represents the centre for steam lap circle.
Travel of valve = 2 AO = 2 GO = 216 mm Ans.
Angle of advance = ∠ AOG = 40° Ans.
Steam lap = GP = 63 mm Ans.

Example 17.6.  In a steam engine , the D-slide valve has a cut-off at 70 per cent of the stroke
at each end of the cylinder. The steam lap and the lead for the cover end are 20 mm and 6 mmrespectively. If the length of the c onnecting rod is 4 times the crank length , find : valve travel , and
angle of advance of the eccentric. Determine also the steam lap and lead of the crank end.

Solution.   Given : Position of piston at cut-off on both sides of the cylinder  = 70% of stroke ;
Steam lap = 20 mm ; L ead = 6 mm ; Connecting rod length = 4 × crank length.
Valve travel and angle of advance of the eccentric
Since we have to find the valve travel, therefore the Bilgram valve diagram is used. The
position of the crank OC'
2 for cut-off at 70 per cent of stroke is obtained in the similar manner as

discussed in the previous example.The Bilgram valve diagram is now completed as follows :
1. Draw RS parallel to A' B' and at a distance equal to the lead i.e. 6 mm, to some suitable
scale as shown in Fig. 17.18. The point S lies on the line C'2 OT.
Fig. 17.18
2. Draw the bisector of the angle RST. Obtain the *point G on this bisector, by hit and
trial, such that a circle with centre G and radius equal to the steam lap i.e. 20 mm touches the
lines RS and ST.

3. Now with O as centre and radius equal to OG draw the actual valve travel circle.
Valve travel = AB = 76 mm Ans.
and angle of advance of the eccentric = ∠AOG = 42° Ans.
*The point G may also be obtained by drawing a perpendicular from A'O such that LG = steam lap
+ lead = 20 + 6 = 26 mm.

632  /G6C   Theory of Machines
Steam lap and lead for the crank end
Since the valve travel and angle of advance of the eccentric is known, therefore the steam lap
and lead for the crank end may be easily found by using Reuleaux valve diagram as discussed below:
1. Draw AB = 76 mm to some suitable scale, as shown in Fig. 17.19, to represent the valve
travel. This diameter AB also represents the piston stroke and OA = OB = Throw of eccentric or
radius of crank.

2. Since cut-off for the crank end takes place at 70 per cent of the stroke, threfore mark
20.7Ap
AB=or Ap2=0 . 7  ×  AB = 0.7 × 76 = 53.2 mm
3. Now with centre P2 and radius equal to P2 p2 (i.e. length of connecting rod), draw an arc
p2C2. Now OC2 represents the position of crank at cut-off for crank end.
4. Draw GOH at an angle of 42°, the advance of the eccentric, with AB in the direction
opposite to the rotation of crank which is shown clockwise in Fig. 17.19.
Fig. 17.19

Fig. 17.19
5. From C2 draw C2C1 parallel to GH. Now the perpendicular, ON and AL on C2C1 represent
the steam lap and lead respectively. By measurement
Steam lap = ON = 11 mm Ans.
Lead =AL = 14 mm Ans.
17.11. Effect of the Early Point of Cut-off with a simple Slide Valve
We have seen in the previous articles that the point of cut-off occurs very late i.e. when the

crank makes an angle greater than 90° with the inner dead centre (or when the piston moves greaterthan 50 per cent of the stroke), as shown in Fig. 17.20 ( a). We shall now consider the effect of the
early point of cut-off on the points of admission, release and compression. The early point of cut-off(considering the crank at 90°) may be obtained by the following three methods :
Fig. 17.20.  Effect of early point of cut-off with a simple slide valve.

Chapter 17 : Steam Engine Valves and Reversing Gears            /G6C          633
First Method
The simplest method of obtaining the earlier
cut-off is by increasing the angle of advance of the
eccentric while the throw of the eccentric, steam lapand exhaust lap are kept constant. We see from Fig.17.20 ( b), that by increasing the angle of advance for
the earlier cut-off will also make admission, release andcompression earlier than as shown in Fig. 17.20 ( a).

This will, obviously reduce the length of effective strokeof the piston.
Second Method
Fig. 17.20 ( c) shows that the earlier cut-off
may also be obtained by increasing the angle of
advance of the eccentric but reducing the throw of the
eccentric (or the valve travel) in order to retain the sametiming for admission as in the normal diagram shownin Fig. 17.20 ( a). The steam lap and exhaust lap are
constant. We see from Fig. 17.20 ( c) that the release

and compression occur earlier but not so early as inFig. 17.20 ( b). The objection to this method is that the
maximum opening of the port to steam and exhaust isreduced due to the shortening of valve travel. This willcause withdrawing or throttling of steam.
Third Method
Another method for obtaining the earlier cut-off is to increase the steam lap and the angle of
advance of the eccentric, as shown in Fig. 17.20 ( d), but keeping constant the travel and lead of the

valve [ i.e. same as in Fig. 17.20 ( a)]. The advantage of this method is that there will be a normal
timing of the admission and a smaller reduction in the maximum opening of the port to steam. But the
necessity of increasing the steam lap of the valve makes it unsuitable from practical point of view.
17.12.17.12.17.12.17.12.17.12. MeMeMeMeMeyyyyyer’er’er’er’er’s Expansion s Expansion s Expansion s Expansion s Expansion VVVVValvalvalvalvalveeeee

We have seen in the previous article that in order to obtain earlier cut-off, other operations
such as admission, release and compression also take place earlier which is undesirable. The Meyer’sexpansion valve not only enables the cut-off to take place early in the stroke with normal timing foradmission, release and compression, but it also enables the cut-off to be varied while the engine isrunning. There are two valves known as main valve  and expansion valve which are driven by sepa-

rate eccentric from the main crankshaft as shown in Fig. 17.21.
The main valve, is similar to the ordinary slide valve, except that it is provided with exten-
sions and the steam passes from the steam chest through the ports P
1 or P2. The admission of steam
to the main valve is controlled by the expansion valve which slides on the back of the main valve.The  expansion valve consists of two blocks or plates E
1 and E2 mounted on a spindle. It may be

noted that in order to admit steam into the cylinder, not only the ports P1 or P2 in the main valve are
in communication with the main ports P'1 or P'2 but at the same time these must be uncovered by the
expansion plates E1 or E2 as the case may be.
In order to obtain variable cut-off according to the requirement, the position of the plates E1
and E2 is varied by means of a spindle having right and left hand threads. The spindle extends to the

engine room so that the operator can vary  the position of  E1 and E2 while the engine is running . Thus
Blades of the helicopter propeller push the
air downwards and the resultant reaction
gives helicopter the necessary lifting power.

634  /G6C   Theory of Machines
the variable cut-off is achieved by the expansion valve without the change of lead, maximum opening
to steam or points of admission, release and compression.
Fig. 17.21.  Meyer’s expansion valve.
Fig. 17.21 shows that both the valves are in mid-position. A little consideration will show
that steam lap ( s) and exhaust lap ( e) for the main valve are positive whereas the steam lap ( a) for the

expansion valve is negative ( i.e. the port P1 instead of being covered by the expansion valve in
mid-position, is open to steam by a distance a ). The points of admission, release, compression and
the least point of cut-off may be obtained in the usual manner by the Reuleaux or Bilgram valvediagrams.
17.13. Virtual or Equivalent Eccentric for the Meyer’s Expansion Valve
In order to obtain the setting of the expansion valve for the predetermined cut-off or vice-

versa,the Reuleaux or Bilgram valve diagram is drawn from the virtual or equivalent eccentric. It is
defined as an eccentric having such a length and angle of advance that will cause cut-off to take placeat the same position, as is caused by the combined effect of main eccentric and expansion eccentric.
Fig. 17.22.  Virtual or equivalent eccentric for the Meyer’s expansion valve.
In the Meyer’s expansion valve, the main valve is driven by an eccentric having an angle of

advance of 25° to 30° and the expansion valve is driven by an eccentric having an angle of advance80° to 90°. If the engine has to be reversible, the angle of advance must be 90° so that the cut-off takesplace at the same fraction of the stroke for the same setting of the expansion valve whatever may bethe direction of rotation of the crank. Fig. 17.22 ( a) shows the relative positions of the crank OC,

main eccentric OM, and expansion eccentric OE, when the crank is at the inner dead centre. The
angle of advance of the main eccentric  OM is denoted  by α and the angle of advance of the expansion
eccentric OE is taken as β = 90°. When the crank OC has tuned through an angle θ from the inner
dead centre, the corresponding  positions of the main eccentric and the expansion eccentric are shownin Fig. 17.22 ( b). The displacement of the main valve from its mid-position is represented by OP, the

projection of OM' on the line of stroke. Similarly the displacement of expansion valve from its mid-

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  635
position is represented by OQ, the projection of OE' on the line of stroke. In both the cases, the
obliquity of the eccentric rod is neglected. The difference between OQ and OP (i.e. PQ) is the
displacement of the expansion valve relative to the main valve, in this case towards the right. Thedisplacement PQ may also be obtained by drawing a line OV parallel and equal to M'E' and then OR,

the projection of OV on the line of stroke, will be equal to PQ. Thus we see that for the given positions
of the main eccentric OM' and expansion eccentric OE', the displacement PQ is equal to the displace-
ment  given by a single eccentric OV. This eccentric OV is termed as virual or equivalent  eccentric.

Since the throw and the angle of advance are referred with respect to inner dead centre , the virtual orequivalent eccentric may be obtained by drawing  OV parallel and equal to ME in Fig. 17.22 ( a). Now
γ is the angle of advance and OV is the throw for the virtual or equivalent eccentric. The cut-off will
take place for the crank position in which  R lies at a distance ‘ a’ to the left of O. This position is most

easily found by applying the Reuleaux valve diagram to the virtual or equivalent eccentric OV.
Fig. 17.23 ( a) shows the Reuleaux valve diagram for the main eccentric OM in order to
determine the crank positions for admission, release, compression and for the latest possible cut-off.
(a) For main eccentric OM.( b) For virtual eccentric OV.
Fig. 17.23.  Reuleaux valve diagram.
The Reuleaux valve diargam for the virtual eccentric OV as shown in Fig. 17.23 ( b) is drawn
as follows :

1. First of all, draw a circle on diameter AB = 2 OV.
2. Draw G' H' making an angle α with AB in the direction opposite to the rotation of crank.
3. Draw C'1C'2 parallel to G' H' and at a distance equal to the steam lap ( a) for the expansion valve.
4. Now OC '2 represents the crank position at which the cut-off takes place. An increase or
decrease of the steam lap ( a) gives respectively a later or earlier cut-off. The steam lap ( a) is altered

by means of a right and a left hand threaded spindle.
17.14. Minimum Width and Best Setting of the Expansion Plate for Meyer’s
Expansion Valve
The minimum width and best setting of the expansion plate E1 or E2 for the Meyer’s
expansion valve may be obtained as discussed below :
1. Minimu m width of the expansion plate E1 or E2
Let OV= Throw of the virtual eccentric,
a= Steam lap of the expansion valve, and
p= Width of the port P1 or P2 in the main valve as shown in Fig. 17.21.

Since the maximum displacement, from the mid-position, of the expansion valve relative to
the main valve is equal to the throw of the virtual eccentric OV, therefore
Maximum overlap of the expansion valve and the port
=OV – a
∴  Minimum width of the expansion plate E1 or E2 (Fig. 17.21) required to prevent steam
from being re-admitted past the inner edge of the plate
=OV – a + p

636  /G6C   Theory of Machines
2. Best setting of the expansion plate E1 or E2
Due to the obliquity of the connecting rod, the steam lap ( a) for the two expansion plates, for
the same point of cut-off, must be different.
Let a1= Steam lap for the expansion plate E1 on the cover side, and
a2= Steam lap for the expansion plate E2 on the crank side.
Generally, the difference between the two steam laps, i.e. (a1 – a2) is different with the change

of point of cut-off. In actual practice, when the expansion plates are assembled on the valve spindle,they may be given different laps. But the difference ( a
1 – a2) once fixed will remain constant for all
values of the steam lap. Therefore for the best results, it is necessary to use such a value of ( a1 – a2)
which gives as nearly as possible equal cut-off on both strokes over the full range of cut-off required.
Example 17.7.  The following particulars refer to a Meyer’s expansion valve :

Throw of main eccentric = 50 mm ; angle of advance of main eccentric = 30° ; Throw of
expansion eccentric = 55 mm ; Angle of advance of expansion eccentric = 90° ; Ratio of connecting
rod length to crank length = 5.
Find : 1. Steam laps required on the expansion plates in order to give cut-off at 0.2 , 0.3, 0.4,
0.5, and 0.6 of the stroke on both strokes , 2. The best setting of the expansion plates , and 3. The
minimum width of the plate , if the width of the steam port in main valve is 28 mm.

Solution. Given : OM = 50 mm ; α = 30° ; OE = 55 mm ; β = 90° ; Ratio of connecting rod
length to crank length = 5
1. Steam lap required on the expansion plates
First of all, determine the throw and angle of advance of the
virtual eccentric OV as shown in Fig. 17.24. By measurement, throw of
virtual eccentric,
OV= 53 mm
and angle of advance, γ= 143°
Now draw the Reuleaux valve diagram, as shown in Fig. 17.25,
for the virtual eccentric OV as discussed below :

1. Draw a cirlce on the diameter AB, such that
AB=2  ×  OV = 2 × 53 = 106 mm
2. Draw C'1 OC'2 making an angle γ = 143° with AB, where γ is the angle of advance for the
virtual eccentric.
3. Since the length of connecting rod is 5 times the crank length, therefore draw AA1 = BB1 =
5 × OA. The distance A1B1 = AB represents the stroke length.
Fig. 17.25
4. Mark the points P1, P2, P3, P4 and P5 corresponding to 0.2, 0.3, 0.4, 0.5 and 0.6 of the

stroke respectively for the cover end and crank end as shown in Fig. 17.25.Fig. 17.24

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  637
5. Now with centres P1, P2, P3, P4 and P5 and radius equal to 5 times OA, obtain the crank-pin
positions P'1, P'2, P'3, P'4, and P'5 corresponding to 0.2, 0.3, 0.4, 0.5 and 0.6 of the stroke at which
cut-off is required to take place for both the ends.
6. The  perpendicular distances for P'1, P'2 ..... etc. on C'1 OC'2 for both  the ends represent

the steam lap ( a) for the virtual eccentric. By measurement, the required values of the steam lap ( a)
are tabulated below :
Cut off 0.2 0.3 0.4 0.5 0.6
Steam lap (cover end) in mm 12 24 32 39 46 Ans.
Steam lap (crank end) in mm 19 30 39 46 50 Ans.
Difference in mm 7 6 7 7 4
2. Best setting of the expansion plates
From the above table, we see that if the steam lap on the end is kept 7mm more than the steam

lap on the cover end, then the cut off will occur at approximately the same fraction of the stroke forboth ends of the cylinder.
Therefore, for best results, the expansion plates may be set with steam lap at the crank end
7mm greater than that at cover end. Ans.
3. Minimum width of expansion plate
From the above table, we see that the minimum steam lap on the cover end is 12 mm and on the
crank end is 19 mm. Since the width of the steam port in the main valves is 28 mm ( i.e. p = 28 mm),

therefore
Minimum width of the expansion plate on the cover end
=OV – a + p = 53 – 12 + 28 = 69 mm  Ans....
(Substituting a = 12 mm)
and minimum width of the expansion plate on the crank end
=OV – a + p = 53 – 19 + 28 = 62 mm  Ans....(Substituting, a = 19 mm)
17.15. Reversing Gears
The primary function of the reversing gear is to reverse the direction of motion of the crank-

shaft in steam engines. It also enables to vary the power developed by the engine by altering the pointof cut-off while the engine is running. Following two types of the reversing gears are generally used :
1. Link motions, and 2. Radial valve gears.
In link motions,  two eccentrics are keyed to the crankshaft, one for forward motion and the

other for backward motion. A suitable link mechanism is introduced between the eccentrics and thevalve rod so that the valve may receive its motion either wholly from one of the two eccentrics orpartly from one and partly from the other. The examples of link motions are Stephenson link motion,Gooch link motion  and Allan link motion. The  Stephenson link motion is most widely used.
In radial valve gears , a single eccentric or its equivalent is used which serves the same object

as two separate eccentrics of link motions. The examples of radial valve gears are Hackworth gearand Walschaert gear.
Notes : 1.  In order to determine the approximate piston position at which admission, cut-off, release and com-

pression takes place for a given setting of the gear, a simplified graphical method may be used. The  methodconsists in finding the throw and angle of advance of a single eccentric (known as virtual or equivalent eccen-tric) which gives approximately the same motion as obtained from a reversing gear. The method of finding the

638  /G6C   Theory of Machines
throw and angle of advance differs for the two types of the reversing gears and is discussed in the following
pages.
2. After determining the equivalent eccentric, the Reuleaux or Bilgram valve diagram is drawn to
determine the piston positions at which admission, cut-off, release and compression take place for a given
setting of the gear.
17.16.  Principle of Link Motions–Virtual Eccentric for a Valve With an Offset
  Line of Stroke

Line of Stroke
Let OC be the crank making an angle θ with the inner dead centre as shown in Fig. 17.26. The
corresponding position of one of the eccentrics is represented by OE making an angle ( θ + α) with the
vertical, where α is the angle of advance. As the crank OC revolves, the end A of the eccentric rod EA
reciprocates along the line PA (i.e. in the direction of the path of the valve rod connected to the valve).

The line of stroke of the valve is off-set by OP. It is required to find the throw and angle of advance of
an eccentric with axis at P, which will give to A the same motion as it receives from the actual
eccentric OE.
Fig. 17.26.  Principle of link motions.
A little consideration will show that when the line AE is produced to cut the vertical line OP
at M, then the triangle OEM represents the velocity triangle for the mechanism OEA, with all its sides

perpendicular to those of a usual velocity triangle.
Let ω= Angular speed of crank or eccentric in rad/s,
vA= Velocity of the point A,
vE= Velocity of the point E, and
β= Angle of inclination of AE with AP.
A
Esin
sinvO M O E M
vO E O M E∠==∠... In any triangle,
sin sinab
AB=/G51

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  639
sin [180 – ( ) – (90 )] cos ( )
sin (90 ) cos°θ + α ° + β θ + α + β==°+β β
∴AEcos ( )cos ( )cos cosOEvvθ+α+β=× = ω × × θ + α + βββ...(∵ vE = ω. OE)
Thus the velocity of A of the eccentric EA is same as can be obtained from a virtual eccentric
with centre P having throw equal to ( OE/ cos β) and the angle of advance ( α + β). Since the position

of eccentric OE changes with the rotation of crank , therefore the inclination of the eccentric EA with
the horizontal ( i.e. angle β) also changes. This change of angle β is very small because the eccentric
rod length EA is 10 to 20 times the throw of eccentric OE. If γ is taken as the mean inclination of the
eccentric rod EA, then the throw of virtual eccentric will be OE/ cos γ with an angle of advance ( α + γ).
Such an arrangement of the eccentric rod is called open rod arrangement.

If the eccentric rod EA instead of lying above the line of stroke of the piston ( i.e. open rod
arrangement), it is in crossed position by crossing the line of stroke ( i.e. crossed-rod arrangement) as
shown by EA1 in Fig. 17.26, then the velocity triangle for the mechanism OEA1 will be triangle
OEM1. In this case
A1 1 1
Esin
sinvO M O E M
vO E O M E∠==∠
11
11sin [180 – ( ) – (90 – )] cos ( – )
sin (90 – ) cos°θ + α ° β θ + α β==°β β
∴1
A1 E 1

∴1
A1 E 1
11cos ( – )cos ( – )cos cosOEvvθ+α β=× = ω × × θ + α βββ
cos ( – )
cosOE=ω× × θ+α γ
γ...(Taking mean β1 = γ)
From the above expression, we see that for the crossed rod arrangement, the throw for the
virtual eccentric with centre P1 is same i.e. OE/ cos γ but the angle of advance is ( α – γ).
The throw and angle of advance of the virtual eccentric may be determined by the simple
graphical construction as discussed below :

1. For A, draw PF parallel and equal to OE. From F draw FG perpendicular to PF. The angle
FPG is equal to γ. Now PG is the virtual eccentric.
2. Similarly for A1, P1G1 is the virtual eccentric.
(a) For open rod arrangement ( b) For crossed rod arrangement.
Fig. 17.27.  Determination of virtual eccentrics.

640  /G6C   Theory of Machines
We have already discussed that in link motions, there are two eccentrics. These two eccen-
trics are connected to their respective eccentric rods. The ends of these eccentric rods are connected
to a slotted link in which a die block slides. The die block is connected to the valve through a valve

rod. The resultant graphical construction for the open rod and crossed rod arrangements is shown inFig. 17.27 ( a) and (b) respectively. The determination of virtual eccentrics OG and OG' for the two
eccentrics of the open rod arrangement is shown in Fig. 17.27 ( a) whereas the determination of
virtual eccentrics OG
1 and OG'1 for the eccentrics of the crossed rod arrangement is shown in
Fig. 17.27 ( b). The straight lines GG' in Fig. 17.27 ( a) and G1G'1 in Fig. 17.27 ( b) are divided at Z and

Z' respectively in the same ratio in which the die block, for a given setting of the link motion, divides
the slotted link in which it slides. Now  OZ is the throw of the virtual eccentric and δ is the angle of
advance for the open rod arrangement. Similarly OZ1 is the throw of the virtual eccentric and δ' is the
angle of advance for the crossed rod arrangement.
17.17. Stephenson Link Motion
The Stephenson link motion, as shown in Fig. 17.28, is the most commonly used reversing

gear in steam engines. It is simple in construction and gives a good steam distribution. Fig. 17.28
shows the arrangement of the gear in mid-position, where OC is the crank and OE and OE1 are the
two eccentrics fixed on the driving shaft or axle in case of a locomotive. The eccentric OE is for
Fig. 17.28.  Stephenson link motion.
forward running and OE1 is for backward running. The motion of these eccentrics is transmitted to

the *curved slotted link AB by means of eccentric rods EA and E1B respectively. The link AB can also
slide on the die block D. The end A on the slotted link is connected to the controlling rod in the engine
cabin through the link AP and the bell crank lever RQP which is pivoted at the fixed fulcrum Q. By
moving the lever, the curved link AB is made to slide through the block D and enables the latter to

derive its motion either from B or A. In this way, the point of cut-off may be changed and the direction
of motion of the engine may be reversed. The valve receives its motion from the block D and the
valve rod is guided horizontally.
It may be noted that when the eccentrics OE and OE1 drives the eccentric rods EA and E1B
respectively, then the link motion is said to have an open rod arrangement. On the other hand, if the

eccentrics OE and OE1 drives the eccentric rods EB and E1A respectively, then the link motion is said
to have crossed rod arrangement. This arrangement gives different steam distribution.
*The radius of curvature of the link AB with either open or crossed rod arrangement is generally equal to the
length of the eccentric rod EA or E1B.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  641
The link motion is said to be in full forward gear position , when the curved link is lowered
so that A and D coincides. In this position, the valve receives its motion entirely from the eccentric
OE. When the curved link is raised so that B and D coincide, it is said to be in full backward gear
position. In this position, the valve recieves its motion entirely from the eccentric OE1. Similarly,

when D lies in the middle of AB, the link motion is said to be in mid-gear position. In this position (or
for any other position of D between AB), the valve recieves its motion partly from the eccentric OE
and partly from the eccentric OE1.
17.18. Virtual or Equivalent Eccentric for Stephenson Link Motion
The Stephenson link motion in an intermediate position is shown in Fig. 17.29. Let us now
find out the equivalent eccentric for the intermediate positions of the die block D. If we assume that

ends A and B of the curved link AB move along a straight path parallel to the line of stroke of the
valve, the equivalent eccentric for the ends A and B and for the die block D may be determined in the
similar manner as discussed in Art 17.16. Fig. 17.30 ( a) and (b) has been reproduced for the two
positions ( i.e. when D is in mid-position and in intermediate position) of the open rod arrangement. In

Fig. 17.30 ( a), OH is the equivalent eccentric for the mid-gear whereas in Fig. 17.30 ( b), OZ is the
equivalent eccentric for any other position. If the construction is repeated for different positions of
the die block D, various points similar to Z may be obtained. Now a curve is drawn through the
various positions of Z. A close approximation to this curve may be obtained by drawing a circular arc

through the points E, H and E1 as shown in Fig. 17.30 ( a). Now the equivalent eccentric for the gear
position, as shown in Fig. 17.29, may be determined by dividing the *arc EHE1 at Z in the same ratio
as D divides AB.
Fig. 17.29.  Stephenon link motion in an intermediate position.
Let R be the radius of the arc of the circle representing the locus of the points similar to Z as
shown in Fig. 17.30 ( a).
∴ R2=(OJ)2 + (EJ)2 = (OH – HJ)2 + (EJ)2 = (R – HJ)2 + (EJ)2
=R2 + (HJ)2 – 2R × HJ + (EJ)2

or22() ()
2HJ EJRHJ+= ...(i)
Now EJ=OE cos α ...(ii)
*A very close result can be obtained by dividing GH at Z in the same ratio as D divides AB.

642  /G6C   Theory of Machines
and HJ=GD = EG cos α = (OE tan γ) cos α ...(iii)
Fig. 17.30. Equivalent eccentric for the two positions of the open rod arrangement.
Substuting the values of EJ and HJ from equations (ii) and (iii) in equation  (i),
22 2( tan ) cos ( cos )
2( tan )cosOE OEROEγα + α=γα
22cos (1 tan ) cos sec
2t a n 2t a nOE OEα+ γ α × γ==γγ    ...(∵ 1 + tan2γ = sec2γ)
cos cos
2sin cos sin 2OE OEαα==γγ γ...(iv)

cos cos
where γ = Mean inclination of the eccentric rod to the line of stroke of the valve.
Since γ is very small, therefore sin 2 γ = 2γ in radians. From Fig. 17.29,
arc arcsin 2 2AB AB
OA AEγ= γ= =
Now equation (iv) may be written as
cosarcEARO EAB=α ×...(v)
The equivalent eccentric for the two positions of the crossed rod arrangement, as shown in
Fig. 17.31 ( a) and (b), may be determined in the similar manner as discussed above.

Fig. 17.31.  Equivalent eccentric for the two positions of the crossed rod arrangement.
After finding the equivalent eccentric for a given setting of the gear, the corresponding Reul eaux
of Bilgram diagram is drawn to determine the crank positions at admission, cut-off, release andcompression.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  643
Note: Comparing Fig. 17.30 ( a) for open rod arrangement and Fig. 17.31 ( a) for crossed rod arrangement, we
see that the projection of the virtual eccentric on the line of stroke for any given setting of the gear when itmoves from full gear to mid gear position,
1.increases in case of open rod arrangement, and
2.decreases in case of crossed rod arrangement.

This projection is equal to steam lap plus lead. The steam lap being constant, it follows that during
linking up the gear (i.e. when the gear moves from full gear position to mid-gear position), the lead increases in
open rod arrangement, while it decreases in crossed rod arrangement.
Example 17.8. A stephenson link motion with open rods has a throw of each eccentric 75

mm and an angle of advance 18º. The length of the curved slotted link is 400 mm and its radius ofcurvature is equal to the length of the eccentric rod which is 1.15 m. Determine the throw and angleof advance of the equivalent eccentric when 1.  the gear is in the mid-position and 2. the gear is in the
middle of full-gear and mid-gear.
Solution. Given OE = 75 mm ; α = 18º; Arc AB = 400 mm; EA = 1.15 m = 1150 mm

1. Throw and angle of advance of the equivalent eccentric when the gear is in mid-position
Let R= Radius of the arc or the locus of points similar to Z, as shown in Fig.
17.30 (a).
cosarcEARO EAB=α ×
115075 cos18º 205 mm400=× × =
Now draw OE and OE1 equal to 75 mm and at an
angle of 18º to vertical YY1 as shown in Fig. 17.32. The
point P on the line of stroke is found by drawing an arc
either from E or E1 such that EP = E1P = R = 205

mm. With P as centre and radius 205 mm draw an arc
EHE1. Now OH represents the equivalent eccentric and
angle YOH is its angle of advance when the gear is in
mid-position. By measurement, throw of equivalenteccentric,
      OH  = 38 mm Ans.
and angle of advance =
90YOH∠=  Ans.
2. Throw and angle of advance of the equivalent eccentric when the gear is in the middle of the
full-gear and mid-gear

In Fig. 17.32, the point E represents the full-gear position and H the mid-gear position. When
the gear is in the middle of the full-gear and mid-gear positions, i.e. in the middle of E and H, divide
the arc EH such that EZ = ZH. Now OZ represents the equivalent eccentric and angle YOZ is its angle
of advance. By measurement, throw of equivalent eccentric,
OZ= 50 mm  Ans.
and angle of advance = 45ºYOH∠= Ans.
17.19. Radial Valve Gears

We have already discussed that in radial valve gears, only one eccentric or its equivalent is
used. The principle on which the radial valve gears operate is discussed below :Fig. 17.32

644  /G6C   Theory of Machines
Let OC be the crank and OE the eccentric for a D-slide valve as shown in Fig. 17.33 ( a). OX
and OY are the projections of OE along OC and perpendicular to OC respectively. When the crank
turns through an angle θ from the inner dead centre, the distance moved by the valve from its mid-
position is given by OM which is the projection of the eccentric OE on the line of stroke, as shown in

Fig. 17.33 ( b). OX and OY are the projections of the eccentric OE along the crank OC and perpen-
dicular to OC. OP and OL are the projections of OX and OY on the line of stroke. From Fig. 17.33 ( b),
OM=OL + LM = OL + *OP
From the above expression, it follows that a motion given to the valve by the eccentric OE
(i.e. displacement OM) may be obtained by combining the displacements OL and OP obtained from

two separate eccentrics OY and OX. The eccentric OY is 90º out of phase with the engine crank OC
and is known as 90º component eccentric. The eccentric OX is 180º out of phase with the engine
crank OC and is known as 180º component eccentric.
Fig. 17.33. Radial valve gear.
The critical examination of Fig. 17.33 shows that
1. The throw of the 180º component eccentric ( i.e. OX) is equal to the sum of steam lap and

lead. If lead is kept constant for all settings of the gear, the throw of the 180º component eccentric willalso be constant.
2. If the throw of the 90º component eccentric ( i.e. OY) is reduced, the eccentric OE will have
larger angle of advance α (∵ tan α = OX/OY). The increase of angle of advance will cause cut-off to
take place earlier in the stroke of the piston.
3. In order to reverse the direction of rotation of the crank, the direction of 90º component

eccentric must be reversed as shown by OY' in Fig. 17.33 ( a).
17.20. Hackworth Valve Gear
This is the earliest of the radial valve gears in which the eccentric OE is placed directly
opposite to the main crank OC, as shown in Fig. 17.34. The eccentric centre E is coupled to a sliding
or die block D which reciprocates along the slotted bar GH which is pivoted to the frame at F. The
slotted bar GH is inclined to OF which is perpendicular to the line of stroke. The inclination of GH

(i.e. angle β) is fixed for a given setting of the gear and is a maximum for the full gear positions. In
order to reverse the direction of rotation of the engine, the slotted bar GH is tilted into the dotted
position as shown in Fig. 17.34. In mid-gear position,the slotted bar occupies the vertical position OZ
so that the motion of die block D is then perpendicular to the line of stroke of the engine. For constant

lead of the valve for all settings, the length of eccentric rod ED must be such that D and F coincide,
when the crank is in either dead centre positions. The valve is driven by a connecting link AB from a
point A on the eccentric rod ED.
*LM = OP, being the projection of two equal and parallel lines OX and EY respectively.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  645
The throw of the virtual or equivalent eccentric and its angle of advance may be determined
Fig. 17.34.  Hackworth valve gear.
The virtual eccentric OV is assumed to be equivalent to two eccentrics, i.e. 180º compo-
nent eccentric OX and 90º component eccentric OY, as shown in Fig. 17.35 ( a). First of all, let us
find the values of OX and OY. During the motion of crank OC from one dead centre to another

dead centre, the point E on the eccentric as well as link DAE moves through a distance equal to
 2 OE along the line of stroke, while the distance moved perpendicular to the line of stroke is zero
because D occupies the position F at both the dead centres. Now the displacement of the valve
during the motion of crank from one dead centre to another will be AA' or 2 OX. This may be
clearly understood from Fig. 17.35 ( b).
∴2or2OX DA DAOX OEOE DE DE== × ...(i)

This equation shows that the throw of 180º component eccentric is independent of the setting
of the gear. This throw ( i.e. OX) is equal to steam lap plus lead.
Fig. 17.35. Determination of throw and angle of advance of the virtual or equivalent eccentric.
Now considering the motion of the crank OC from one vertical position to another vertical
position. The point E on the eccentric as well as the link DAE moves through a vertical distance equal

to 2 OE, while the horizontal distance moved by the point E is zero. Since the slotted bar GH is
inclined at an angle β, therefore from Fig. 17.35 ( c),
Horizontal distance moved by D = DD' = 2OE tan β
Now the displacement of the valve during the motion of crank from one vertical position to
another will be 2 OY. Therefore from Fig. 17.35 ( d),

646  /G6C   Theory of Machines
2or tan2t a nOY EA EAOY OEOE ED ED== × ββ...(ii)
This equation shows that the throw of 90º component eccentric varies with the angle β i.e.
with the particular setting of the gear.
From Fig. 17.35 ( a), the throw of the virtual eccentric,
22()()OV OX OY=+ ...(iii)
and the angle of advance of the virtual eccentric,
1tanOX
OY−α= ...(iv)
The equivalent eccentric for a given setting of the gear may be
determined graphically as discussed below :

1. Draw OE, to some suitable scale, to represent the throw of
the actual eccentric as shown in Fig. 17.36.
2. Through E, draw ED inclined at angle β to OE so that OD
represents OE tan β to scale. The angle β is drawn upwards when slotted
bar GH (Fig. 17.34) is in full line position and it is drawn downwards
when GH is tilted to the dotted position, as shown in Fig. 17.36.
3. Divide OE at X in the same proportion as A divides ED in

Fig. 17.34. Through X draw a line perpendicualr to OE to meet ED at V.
Now OV is the equivalent eccentric for the motion of the valve.
Example 17.9. In a Hackworth radial valve gear , as shown in Fig. 17.37 , the dimensions of
various link are as follows :
OC = 225 mm; CP = 800 mm; DE = 625 mm and AE = 300 mmFig. 17.36Inside view of a factory.

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  647
If the lead is constant at 3 mm , the steam lap is 18 mm and the angle β is 20º, find the length
of eccentric OE , the distance OF , the effective valve travel and the effective angle of advance.
Fig. 17.37
Solution. Given : l = 3 mm; s = 18 mm; β = 20º
Length of eccentric OE
We know that in case of a Hackworth radial valve gear, the virtual eccentric OV may be

assumed to be equivalent to two eccentrics, i.e. 180º component eccentric and 90º component eccen-
tric as shown in Fig. 17.38. The throw of the 180º component eccentric is given by
625 300
625DE AE DAOX OE OE OEDE DE−−=×= ×= ×
= 0.52 OE ... (i)
Also OX=s + l = 18 + 3 = 21 mm ... (ii)
OE= 21/ 0.52 = 40.4 mm
Distance OF
When the eccentric OE is along the line of stroke, the point D coincides with F and the angle
FOE = 90º.
∴ (OF)2 + (OE)2=(DE)2

FOE = 90º.
or22 2 2( ) ( ) (625) (40.4) 623.7 mmOF DE OE=− =− = Ans.
Effective valve travel
We know that the 90º component eccentric,
300tan 40.4 tan 20º 7.06625EAOY OEED=× β =× = mm
∴  Throw of the virtual eccentric,
22 2 2( ) ( ) (21) (7.06) 22.15 mmOV OX OY=+ = + =
and effective valve travel = 2 × OV = 2 × 22.15 = 44.3 mm Ans.Fig. 17.38

648  /G6C   Theory of Machines
Effective angle of advance
We know that effective angle of advance,
()()11 1 21tan tan tan 2.9745 71.4º7.06OX
OY−− −α= = = = Ans.
17.21. Walschaert’s Valve Gear
The Walschaert’s valve gear, as shown in Fig. 17.39, is the most extensively used of all
reversing gears on modern locomotives. In this gear, a single eccentric OE is used and is set at 90º to
the main crank OC. The eccentric rod EG oscillates the curved slotted link GH about the fulcrum F

which is fixed to the frame of the engine. The relative motion of the sliding or die block D in the
curved slotted link GH is due to the link PQ which is operated by means of a bell crank lever SRQ and
a rod from the engine room. The die block D is capable of movement along the whole length of the
link GH. The pin K receives its motion from the die block D and ultimately from the eccentric OE,
while the pin L receives its motion from a point B on the main crosshead A, where AC is the connect-

ing rod of the engine.
When the gear is in mid-position, the block D is at F. The radius of link GH is such that when
the crank OC is at the inner dead centre position and the gear is reversed, the point K remains at rest.
This characteristic gives constant lead during all conditions of running.
Fig. 17.39. Walschaert’s valve gear.
Neglecting obliquities of all the rods, the throw of the virtual eccentric and its angle of advance

may be determined in the similar manner as discussed in the previous article. The virtual eccentricOV is assumed to be equivalent to two eccentrics i.e. 180º component eccentric OX and 90º component
eccentric OY, as shown in Fig. 17.40 ( a).
First of all, let us find the values of OX and OY . Considering the motion of crank OC from one
dead centre to another dead centre, the crosshead A and hence the point L on the link JKL moves through

a distance equal to 2 OC. During this motion of the crank, the point G on the slotted link and hence the
point K on the link JKL occupy the same position as at start. In other words, the distance moved by K is
zero. Now the displacement of the valve during the motion of crank from one dead centre to anotherdead centre will be JJ' or 2 OX.
From Fig. 17.40 ( b),
2or2OXJK JKOX OCOC KL KL== ×        ...(i)

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  649
Thus OX is constant for all positions of the block D on the link GH and the lead remains
unchanged during all conditions of running. In case the point J lies on the same side of K as L, the
motion of J and L will be in phase. In this position, OX is termed as 0º component eccentric.
Fig. 17.40. Determination of throw and angle of advance of the virtual eccentric.

Now considering the motion of the eccentric OE from one dead centre position to another
dead centre, the crank pin C moves from one vertical position to another, thus not traversing any
horizontal distance. During this motion of the eccentric, the distance moved by the crosshead A and
thus the point L on the link JKL will be zero. At the same time, the point E on the eccentric OE and the
point G on the curved slotted link GH moves through a distance 2 OE. Since the curved slotted link

GH is hinged at F, therefore the die block D moves through a distance DD' which is given by
or 22DD FD FDDD OEOE FG FG′′ == ×...[From Fig. 17.40 ( c)]
Since point K lies on the link DK, therefore point K will move through the same distance as
that of D, i.e.
2FDKK DD OEFG′′==×
Now the displacement of the valve during the motion of the crank from one vertical position
to another will be JJ' or 2 OY. From Fig. 17.40 ( d),
2or2OY FG JJ JL JL
KK KL FD OE KL× ′==′ ×
∴JL FDOY OGKL FG=×× ...(ii)

The position of D on the curved slotted link GH may be varied by operating the bell crank
lever from the rod in the engine room in order to suit load conditions or to effect the reversal ofdirection or rotation.
From Fig. 17.40 ( a), the throw of the virtual eccentric,
1tanOX
Example 17.10. In a Walschaert valve gear, as shown in Fig. 17.39, the engine crank is 300

mm long. The least cut-off in the head end of the cylinder is at 120º. At this, the maximum opening to
steam is 45 mm and the lead is 6 mm. If the length of the eccentric is 115 mm, find the ratios JL
KL and
FGFD of the gear. Neglect the obliquities of all the rods.
Solution. Given : OC = 300 mm; crank angle at cut-off = 120º; Maximum opening to steam
= 45 mm; l = 6 mm; OE = 115 mm

650  /G6C   Theory of Machines
First of all, draw the Bilgram valve diagram as discussed below :
1.Draw OC2, the position of crank at cut-off, at 120º to the line of stroke OV as shown in
Fig. 17.41.
Fig. 17.41
2.Draw a line parallel to OV and at a distance equal to the lead (6 mm) which intersects C2O
produced at S.
3.With centre O, draw an arc with radius OP = 45 mm, the maximum opening of steam.
4.Draw the bisector of the angle RST. On this bisector, obtain a point G such that a circle

drawn with centre G touches the lines SR, ST and the point P. Join OG.
By measurement, we find that throw of the virtual eccentric,
OV = OG = 81.5 mm
and angle of advance of the virtual eccentric
α= 32º
The virtual eccentric OV is assumed to be equivalent to two eccen-
trics, i.e. 180º component eccentric OX and 90º component eccentric OY, as
shown in Fig. 17.42.
∴  180º component eccentric
=OX = VY = OV sin α = 81.5 sin 32º = 43.2 mm
and 90º component eccentric,

OY=OV cos 32º = 81.5 × 0.848 = 69.1 mm
We know thatJKOX OCKL=× ...(Refer Fig. 17.39)
∴43.20.144300JK OX
KL OC== =
Now 1 1 0.144 1.144KL JKJL JK
KL KL KL+== + = + =  Ans.
Again we know that
JL FDOY OEKL FG=××
∴1151.144 1.969.1FG JL OE
FD KL OY=×= × =  Ans.Fig. 17.42

Chapter 17 : Steam Engine Valves and Reversing Gears    /G6C  651
EXERCISES
1.A D-slide valve has a travel of 100 mm, angle of advance 30º and the exhaust lap at both ends of the
valve is 12 mm. The cut-off takes place at 85% of the stroke on the cover end of the cylinder. Deter-
mine the lead and the point of release for this stroke.If the lead is same for both the ends, find the point of cut-off on the crank end and the steam lap. The
length of the connecting rod is 4 times the crank length.

[Ans. 7.2 mm; 98% of stroke, 78% of stroke, 18 mm]
2.The travel of a slide valve is 100 mm and the lead at the crank end is 6 mm. If the length of connecting
rod is 4.5 times the crank length, find the angle of advance, steam lap and exhaust lap to give cut-off
and release at 65% and 95% of the stroke respectively. [Ans. 38°, 24 mm, 12 mm]
3.The following data refer to a D-slide valve :
Valve travel = 150 mm; Lead at cover end = 6 mm ; Connecting rod length = 5 times crank length;

Cut-off at both ends of the piston = 0.7 strokeDetermine the angle of advance of the eccentric and maximum opening of port to steam and steam lap
for the cover end. Find also the steam lap and lead for the crank end of the valve.
[Ans. 40°, 35 mm; 40 mm; 26 mm; 22 mm]
4.A simple slide valve with outside admission provides a lead of 3 mm, a maximum port opening of 18
mm and a cut-off at 62% at the out-stroke ( i.e. at the crank end). The ratio of the connecting rod length

to the crank radius is 3.7. Find the valve travel, steam lap and the angle of advance. If the exhaust lapis 12.5 mm, find the percentage of stroke at which release and compression will occur.
[Ans.  78 mm, 20 mm, 37º; 92.5% of forward stroke, 88% of return stroke]
5.A steam engine fitted with a D-slide valve gives a cut-off at 65% of the stroke and release at 90% of
the stroke for both ends of the cylinder. The width of the ports is 25 mm, maximum opening of port to

steam is 18 mm and the valve lead at the cover end is 6 mm. If the length of the connecting rod is 5
times the crank length, find the total valve travel and the angle of advance of the eccentric. Determinealso the valve lead at the crank end and the steam and exhaust laps  for both ends of the cylinder.
[Ans. 88 mm, 45.5; l (crank end) = 14 mm,  s (cover end) = 25 mm,
(crank end) = 17.5 mm, e (cover end) = 6.5 mm, e (crank end) = 10.8 mm]

6.The following particulars refer to a Meyer’s expansion valve :
Angle of advance of main eccentric = 35º; Travel of main valve = 150 mm; Angle of advance of
expansion eccentric = 90º; Throw of expansion eccentric = 75 mm; Ratio of connecting rod length tocrank length = 4.
Find the steam laps required at the two ends of the expansion valve in order to give cut-off at 0.2, 0.4
and 0.6 of the stroke on both strokes. Determine also the best setting of the expansion plates.

[Ans. 24 mm, 48 mm, 64 mm for cover end; 38 mm, 60 mm, 68 mm for crank end]
7.The dimensions of a Hackworth valve gear, as shown in Fig. 17.34 (Page 645) are as follows :
OC = 300 mm; CP = 700 mm; OE = 145 mm; ED = 800 mm and EA = 550 mm.
The die block D coincides with F when the crank is at dead centres. Find the throw and the angle of
advance of the equivalent eccentric when the inclination of the slotted link is 30º with the vertical.
[Ans. 73.3 mm; 38.2º]

8.The Walschaert radial valve gear of a locomotive engine in which the slide valve has inside admission,
is shown in Fig. 17.43. The main crank OC is 250 mm and the crank OE is 75 mm.
Fig. 17.43

652  /G6C   Theory of Machines
If DE = 0.6 FG and JK = 0.15 JL, find the travel of the valve. When the cut off is to take place at 60%
of the stroke of the piston, find steam lap and lead of the valve. The motion of points K and L may each
be assumed simple harmonic along a horizontal straight line.
[Ans. 100 mm, 32.5 mm, 0.75 mm]
1.State the function of a valve in a steam engine. Name the types of valves commonly used to control the
various operations in a steam engine.

2.Describe the action of a D-slide valve and piston slide valve. Discuss the advantages of a  piston slide
valve over D-slide valve.
3.Define steam lap, exhaust lap and angle of advance for a simple slide valve with 1. out-side steam
admission, and 2. inside steam admission.
4.Explain how the ponts of admission, cut-off, release and compression are determined for a D-slide
valve using any one of the following constructions :

5.Discuss with the help of diagrams, the disadvantages of earlier cut-off with a simple slide valve.
6.Explain with the help of a neat sketch, the function of a Meyer’s expansion valve in a steam engine.
7.What do you understand by virtual or equivalent eccentric?  How it is obtained for the Meyer’s
expansion valve?

8.Why the reversing gears are used in steam engines? State the commonly used types of reversing gears
and how they differ from one another?
9.Describe, with the help of a line diagram, the working of a Stephenson link motion. How the virtual
eccentric and its angle of advance for any setting of this link motion is determined?
10.Discuss the principle underlying the use of a radial valve gear.

11.Explain, with the help of a line diagram, the working of a Hackworth valve gear. Discuss the method
to determine the virtual eccentric and its angle of advance for this gear.
12.Describe, with the help of a line diagram, the working of a Walschaert’s valve gear. How  will you
determine the virtual eccentric and its angle of advance for this gear?
1.In a steam engine, the distance by which the outer edge of the D-slide valve overlaps the steam port is

called(a) lead ( b) steam lap
(c) exhaust lap ( d) none of these
2.The D-slide valve is also known as
(a) inside admission valve ( b) outside admission valve
(c) piston slide valve ( d) none of these
3.In Meyer’s expansion valve, main valve is driven by an eccentric having an angle of advance
(a) 10º – 15º ( b) 15º – 25º
(c) 25º – 30º ( d) 30º – 40º
4.In Meyer’s expansion valve, the expansion valve is driven by an eccentric having an angle of advance
(a) 50º – 60º ( b) 60º – 70º

(c) 70º – 80º ( d) 80º – 90º
5.The function of a reversing gear in a steam engine is
(a) to control the supply of steam
(b) to alter the point of cut-off while the engine is running
(c) to reverse the direction of motion of the crankshaft
ANSWERS
1.(b) 2.(b) 3.(c) 4.(d) 5.(b),(c)

Chapter 18 : Governors    /G6C  653
Governors18
Features
2. Types of Governors.3. Centrifugal Governors.4. Terms Used in Governors.5. Watt Governor.6. Porter Governor.7. Proell Governor.8. Hartnell Governor.9. Hartung Governor.
10. Wilson-Hartnell Governor.
11. Pickering Governor.
12. Sensitiveness of Governors.13. Stability of Governors.14. Isochronous Governor.15. Hunting.
16. Effort and Power of a
Governor.
17. Effort and Power of a Porter
Governor.

Governor.
18. Controlling Force.19. Controlling Force Diagram
for a Porter Governor.
20. Controlling Force Diagram
for a Spring-controlledGovernor.
21. Coefficient of
Insensitiveness.18.1. Introduction
The function of a governor is to regulate the mean
speed of an engine, when there are variations in the load e.g.

when the load on an engine increases, its speed decreases,therefore it becomes necessary to increase the supply of work-ing fluid. On the other hand, when the load on the enginedecreases, its speed increases and thus less working fluid isrequired. The governor automatically controls the supply ofworking fluid to the engine with the varying load conditionsand keeps the mean speed within certain limits.
A little consideration will show, that when the load

increases, the configuration of the governor changes and avalve is moved to increase the supply of the working fluid ;conversely , when the load decreases, the engine speed in-
creases and the governor decreases the supply of workingfluid.
Note : We have discussed in Chapter 16 (Art. 16.8) that the func-

tion of a flywheel in an engine is entirely different from that of agovernor. It controls the speed variation caused by the fluctuationsof the engine turning moment during each cycle of operation. Itdoes not control the speed variations caused by a varying load. Thevarying demand for power is met by the governor regulating the
supply of working fluid.
18.2. Types of Governors
The governors may, broadly, be classified as
1. Centrifugal governors, and 2. Inertia governors.
653

654  /G6C   Theory of Machines
The centrifugal governors, may further be classified as follows :
18.3. Centrifugal  Governors
The centrifugal governors are based on the balancing of centrifugal force on the rotating balls
by an equal and opposite radial force, known as the controlling force *.It consists of two balls of
equal mass, which are attached to the arms as shown in Fig. 18.1. These balls are known as governor
balls or fly balls. The balls revolve with a spindle, which

is driven by the engine through bevel gears. The upperends of the arms are pivoted to the spindle, so that theballs may rise up or fall down as they revolve about thevertical axis. The arms are connected by the links to asleeve, which is keyed to the spindle. This sleeve re-volves with the spindle ; but can slide up and down.The balls and the sleeve rises when the spindle speedincreases, and falls when the speed decreases. In orderto limit the travel of the sleeve in upward and down-ward

the travel of the sleeve in upward and down-ward directions, two stops S, S are provided on the

spindle. The sleeve is connected by a bell crank leverto a throttle valve. The supply of the working fluid de-creases when the sleeve rises and increases when it falls.
When the load on the engine increases, the en-

gine and the governor speed decreases. This results inthe decrease of centrifugal force on the balls. Hencethe balls move inwards and the sleeve moves down-wards. The downward movement of the sleeve oper-ates a throttle valve at the other end of the bell cranklever to increase the supply of working fluid and thusthe engine speed is increased. In this case, the extrapower output is provided to balance the increased load.When the load on the engine decreases, the engine andthe governor speed

decreases, the engine andthe governor speed increases, which results in the in-crease of centrifugal force on the balls. Thus the ballsmove outwards and the sleeve rises upwards. This up-ward movement of the sleeve reduces the supply of theworking fluid and hence the speed is decreased. In thiscase, the power output is reduced.

*The controlling force is provided either by the action of gravity as in Watt governor or by a spring as in case
of Hartnell governor.A governor controls engine speed. As itrotates, the weights swing outwards, pullingdown a spindle that reduces the fuel supplyat high speed.Spring steel
strip
Spindle
controls fuel
supply
Rotating
weight

Chapter 18 : Governors    /G6C  655
Note :  When the balls rotate at uniform speed, controlling force is equal to the centrifugal force and they
balance each other.
Fig. 18.1. Centrifugal governor.
18.4. Terms Used in Governors
The following terms used in governors are
important from the subject point of view ;
1. Height of a governor . It is the vertical

distance from the centre of the ball to a point wherethe axes of the arms (or arms produced) intersecton the spindle axis. It is usually denoted by h.
2. Equilibrium speed . It is the speed at
which the governor balls, arms etc., are in completeequilibrium and the sleeve does not tend to moveupwards or downwards.
3. Mean equilibrium speed . It is the speed
at the mean position of the balls or the sleeve.
4. Maximum and minimum equilibrium

speeds. The speeds at the maximum and minimum
radius of rotation of the balls, without tending tomove either way are known as maximum and mini-mum equilibrium speeds respectively.
Note : There can be many equilibrium speeds between
the mean and the maximum and the mean and the mini-
mum equilibrium speeds.
5. Sleeve lift . It is the vertical distance which
the sleeve travels due to change in equilibriumspeed.
Centrifugal governor

656  /G6C   Theory of Machines
18.5. Watt Governor
The simplest form of a centrifugal governor is a Watt governor, as shown in Fig. 18.2. It is
basically a conical pendulum with links attached to a sleeve of negligible mass. The arms of thegovernor may be connected to the spindle in the following three ways :
1.The pivot P, may be on the spindle axis as shown in Fig. 18.2 ( a).
2.The pivot P, may be offset from the spindle axis and the arms when produced intersect at

O, as shown in Fig. 18.2 ( b).
3.The pivot P, may be offset, but the arms cross the axis at O, as shown in Fig. 18.2 ( c).
Fig. 18.2. Watt governor.
Let m= Mass of the ball in kg,
w= Weight of the ball in newtons  =  m.g,
T= Tension in the arm in newtons,
ω= Angular velocity of the  arm and  ball about  the spindle axis in
rad/s,
r= Radius of the path of  rotation of the ball i.e. horizontal distance
from the centre of  the ball to the  spindle axis in metres,

FC= Centrifugal force acting on the ball in newtons   =  m.ω2.r, and
h= Height of the governor in metres.
It is assumed that the weight of the arms, links and the sleeve are negligible as compared to
the weight of the balls. Now, the ball is in equilibrium under the action of
1. the centrifugal force ( FC) acting on the ball, 2.  the tension ( T) in the arm, and 3. the weight
(w) of the ball.
Taking moments about point O, we have
FC × h=w × r  =  m.g.r
or m.ω2.r.h=m.g.rorh = g/ω2. . . (i)

When g is expressed in m/s2 and ω in rad/s, then h is in metres. If N is the speed in r.p.m., then
ω=2πN/60
∴229.81 895metres
(2 /60)h
NN==
π. . . (∵ g = 9.81 m/s2) . . . (ii)
Note : We see from the above expression that the height of a governor h, is inversely proportional to N2.
Therefore at high speeds, the value of h is small. At such speeds, the change in the value of h corresponding to

a small change in speed is insufficient to enable a governor of this type to operate the mechanism to give thenecessary change in the fuel supply. This governor may only work satisfactorily at relatively low speeds i.e.
from 60 to 80 r.p.m.

Chapter 18 : Governors    /G6C  657
Example 18.1 . Calculate the vertical height of a Watt governor when it rotates at 60 r.p.m.
Also find the change in vertical height when its speed increases to 61 r.p.m .
Solution. Given : N1 = 60 r.p.m. ; N2 = 61 r.p.m.
Initial height
We know that initial height,
1 22
1895 8950.248 m
( ) (60)h
N===
Change in vertical height
We know that final height,
2 22
2895 8950.24 m
() ( 6 1 )h
N== =
∴ Change in vertical height

() ( 6 1 )h
N== =
=  h1 – h2 =  0.248 – 0.24  =  0.008 m  =  8  mm Ans.
18.6. Porter Governor
The Porter governor is a modification of a Watt’s governor, with central load attached to the
sleeve as shown in Fig. 18.3 ( a). The load moves up and down the central spindle. This additional
downward force increases the speed of revolution required to enable the balls to rise to any pre-determined level.

Consider the forces acting on one-half of the governor as shown in Fig. 18.3 ( b).
(a)( b)
Fig. 18.3.  Porter governor.
Let m= Mass of each ball in kg,
w= Weight of each ball in newtons = m.g,
M= Mass of the central load in kg,
W= Weight of the central  load in newtons = M.g,
r= Radius of rotation in metres,

658  /G6C   Theory of Machines
h= Height of governor in metres ,
N= Speed of the balls in r.p.m .,
ω= Angular speed of the  balls in  rad/s
= 2πN/60 rad/s,
FC= Centrifugal force acting on the  ball
in newtons = m.ω2.r,
T1= Force in the arm in  newtons,
T2= Force in the  link in newtons,
α= Angle  of  inclination  of the arm (or
upper  link) to the  vertical, and
β= Angle of  inclination of  the link
(or lower  link) to the vertical.
Though there are several ways of determining the

relation between the height of the governor ( h) and the
angular speed of the balls ( ω), yet the following two
methods are important from the subject point of view :
1. Method of resolution of forces ; and
2.  Instantaneous centre method.
1. Method of resolution of forces
Considering the equilibrium of the forces acting
at D, we have
2.cos22WM gTβ= =
or           2.
2c o sMgT=β. . . (i)

or           2.
Again, considering the equilibrium of the forces acting on B. The point B is in equilibrium
under the action of the following forces, as shown in Fig. 18.3 ( b).
(i)The weight of ball ( w = m.g),
(ii)The centrifugal force ( FC),
(iii)The tension in the arm ( T1), and
(iv)The tension in the link ( T2).
Resolving the forces vertically,
                                                         12.cos cos .2MgTTw m gα= β+ = + . . . (ii)
2.... cos

2.... cos
2MgTβ= /G51
T1 sin α + T2 sin β = FC
                             1C.sin sin2c o sMgTFα+ × β=β2....2cosMgT=β /G51
                                  1C.sin tan2MgTFα+ × β=
∴ 1C.sin – tan2MgTFα= × β . . . (iii)
A big hydel generator. Governors are
used to control the supply of working
fluid (water in hydel generators).
information and is not a direct example of
the current chapter.

Chapter 18 : Governors    /G6C  659
Dividing equation (iii) by equation (ii),
C1
1.–t a nsin 2
. cos.2MgFT
Mg Tmg×βα=α+
or C... tan – tan22Mg Mgmg F+α = × β
C .. t a n.–2 tan 2 tanF Mg Mgmgβ+= ×αα
Substitutingtan,tanqβ=αand tan ,r
hα= we have
2 .... .–22Mg h Mgmg m r qr+= ω × × . . . (∴ FC = m.ω2.r)
or2 ... . ( 1 )2Mgmh m g qω= + +
∴               22(1 ).1 2.( 1 )2 .MmqMg ghm g qm m++=+ + = ×ωω 
. . . (iv)
or               2(1 )1 2.( 1 )2.MmqMg gmg q
mh m h++ω= + + = ×

or               2 (1 )2 2
60MmqNg
mh++π=× 
∴                2
2(1 ) (1 )60 895 22
2MMmq mqgNmh m h++ ++=× =× π
. . . (v)
. . . (Taking g = 9.81 m/s2)
Notes : 1.  When the length of arms are equal to the length of links and the points P and D lie on the same vertical
line, then
             tan α = tan β        or   q =  tan α / tan β = 1
Therefore, the equation  (v) becomes
                2() 8 9 5mMNmh+=× . . . (vi)

2() 8 9 5mMNmh+=× . . . (vi)
2. When the loaded sleeve moves up and down the spindle, the frictional force acts on it in a direction
opposite to that of the motion of sleeve.
If F = Frictional force acting on the sleeve in newtons, then the equations (v) and (vi) may be written as
                       2..( 1 )895 2
.MgFmgq
Nmg h±++=× . . . (vii)
                      . ( . ) 895
.mg Mg F
mg h+±=×. . . (When q = 1) . . . (viii)

660  /G6C   Theory of Machines
The + sign is used when the sleeve moves upwards or the governor speed increases and negative sign is
used when the sleeve moves downwards or the governor speed decreases.
3.  On comparing the equation  (vi) with equation (ii) of Watt’s governor (Art. 18.5), we find that the
mass of the central load ( M) increases the height of governor in the ratio mM
m+
.
2. Instantaneous centre method
In this method, equilibrium of the forces acting on

the link BD are considered. The instantaneous centre I lies
at the point of intersection of PB produced and a line through
D perpendicular to the spindle axis, as shown in Fig. 18.4.
Taking moments about the point I,
C2WFB Mw I M I D×= × + ×
..2Mgm g IM ID=×+ ×
∴ C..2IM M g IDFm gBM BM=× + ×
..2IM M g IM MDmgBM BM+=× + 
..2IM M g IM MDmgBM BM BM=× + + 
.. tan (tan tan )2Mgmg=α + α + β
... tan ,and tanIM MD
BM BM=α =β/G51
Dividing throughout by tan α,
C .t a n ..1 . ( 1 )

tan 2 tan 2F Mg Mgmg mg q β=+ + =+ + αα tan...tanq β=α /G51
        We know that FC =  m.ω2.r,and tanr
hα=
∴ 2 ... . ( 1)2hM gmr m g qrω×= + +
or                   
22..( 1 ) ( 1 )1 22Mg Mmg q m qghmm++ + +
=× = ×
ωω
. . . (Same as before)
When tan α = tan β   or  q = 1, then
                          2mM ghm+=×
ωFig. 18.4.  Instantaneous centre
method.

Chapter 18 : Governors    /G6C  661
Example 18.2.  A Porter governor has equal arms each 250  mm long and pivoted on the axis
of rotation . Each ball has a mass of 5  kg and the mass of the central load on the sleeve is 25  kg. The
radius of rotation of the ball is 150  mm when the governor begins to lift and 200  mm when the
governor is at maximum speed . Find the minimum and maximum speeds and range of speed of the
governor.

governor.
Solution.  Given : BP = BD = 250 mm = 0.25 m ; m = 5 kg ; M = 15 kg ; r1 = 150 mm
= 0.15m; r2 = 200 mm = 0.2 m
Fig. 18.5
The minimum and maximum positions of the governor are shown in Fig. 18.5 ( a) and (b)
Minimum speed when r1 = BG = 0.15 m
Let N1 = Minimum speed.
From Fig. 18.5 ( a), we find that height of the governor,
22 2 2
1 ( ) – ( ) (0.25) – (0.15) 0.2 m hP G P B B G== = =
2
1
1895 5 15 895( ) 17 90050 . 2mMNmh++=× = × =
∴ N1 = 133.8 r.p.m.  Ans.

Maximum speed when   r2 = BG = 0.2 m
Let N2 = Maximum speed.
From Fig. 18.5 ( b), we find that height of the governor,
22 22
2 ( ) – ( ) (0.25) – (0.2) 0.15 m hP G P B B G== = =
                                  2
2
2895 5 15 895( ) 2386750 . 1 5mMNmh++=× = × =
∴                                   N2    = 154.5 r.p.m. Ans.

662  /G6C   Theory of Machines
Range of speed
We know that range of speed
= N2 – N1 = 154.4 – 133.8 = 20.7 r.p.m. Ans.
Example 18.3.  The arms of a Porter governor are each 250  mm long and pivoted on the
governor axis . The mass of each ball is  5  kg and the mass of the central sleeve is 30  kg. The radius
of rotation of the balls is 150  mm when the sleeve begins to rise and reaches a value of 200  mm for

maximum speed . Determine the speed range of the governor . If the friction at the sleeve is equivalent
of 20 N of load at the sleeve , determine how the speed range is modified .
Solution. Given : BP = BD = 250 mm ; m = 5 kg ; M = 30 kg ; r1 = 150 mm ; r2 = 200 mm
First of all, let us find the minimum and maximum speed of the governor. The minimum and
maximum position of the governor is shown in Fig. 18.6 ( a) and (b) respectively.
Let N1 = Minimum  speed  when r1 = BG = 150 mm, and

N2 = Maximum  speed  when r2 = BG = 200 mm.
Fig. 18.6
Speed range of the governor
From Fig. 18.6 ( a), we find that height of the governor,
22 22
1 ( ) – ( ) (250) – (150) 200 mm 0.2 m hP G P B B G== = = =
2
1
1895 5 30 895( ) 3132550 . 2mMNmh++=× = × =
∴ N1 =  177 r.p.m.
From Fig. 18.6 ( b), we find that height of the governor,
22 22
2 ( ) – ( ) (250) – (200) 150 mm 0.15 m hP G P B B G== = = =
2
2
2895 5 30 895() 4 1 7 6 750 . 1 5mMNmh++=× = × =
∴ N2 = 204.4 r.p.m.

Chapter 18 : Governors    /G6C  663
We know that speed range of the governor
= N2 – N1 = 204.4 – 177 = 27.4 r.p.m. Ans.
Speed range when friction at the sleeve is equivalent of 20
N of load (i.e. when F = 20 N)
We know that when the sleeve moves downwards,
the friction force ( F) acts upwards and the minimum speed is
given by
     2
1
1. ( . – ) 895().mg Mg FNmg h+=×
                5 9.81 (30 9.81 – 20) 8952950059 . 8 1 0 . 2×+ ×=× =×
∴ N1 = 172 r.p.m.

We also know that when the sleeve moves upwards,
the frictional force ( F) acts downwards and the maximum
speed is given by
2
2
2. ( . ) 895().mg Mg FNmg h++=×
                5 9.81 (30 9.81 20) 8954420059 . 8 1 0 . 1 5×+ ×+=× =×
∴ N2 = 210 r.p.m.
= N2 – N1 = 210 – 172 = 38 r.p.m. Ans.
Example 18.4.  In an engine governor of the Porter type , the upper and lower arms are 200  mm

and 250 mm respectively and pivoted on the axis of rotation . The mass of the central load is 15  kg,
the mass of each ball is 2  kg and friction of the sleeve together with the resistance of the operating
gear is equal to a load of 25  N at the sleeve . If the limiting inclinations of the upper arms to the
vertical are 30° and 40°, find, taking friction into account , range of speed of the governor .
Solution .  Given :  BP = 200 mm = 0.2 m ; BD = 250 mm = 0.25 m ; M = 15 kg ; m = 2 kg ;

F = 25 N ; α1 = 30°; α2 = 40°
First of all, let us find the minimum and maximum speed of the governor.The minimum and maximum position of the governor is shown Fig. 18.7 ( a) and (b)
Let N
1 = Minimum  speed,  and
N2 = Maximum  speed.
From Fig. 18.7 ( a), we find that minimum radius of rotation,
r1 = BG = BP sin 30° = 0.2 × 0.5 = 0.1 m
Height of the governor,
h1 = PG = BP cos 30° = 0.2 × 0.866 = 0.1732 m
A series of hydel generators.

664  /G6C   Theory of Machines
and22 2 2( ) – ( ) (0.25) – (0.1) 0.23DG BD BG=== m
∴ tan β1 = BG/DG = 0.1/0.23 = 0.4348
and tan α1 = tan 30° = 0.5774
∴1
1
1tan 0.43480.753tan 0.5774qβ===α
Fig. 18.7
We know that when the sleeve moves downwards, the frictional force ( F) acts upwards and
the minimum speed is given by
1
2
1
1.–.( 1 )895 2().Mg Fmg q
Nmg h++=×
15 9.81 – 2429 . 8 1 ( 10 . 7 5 3 )895 2335962 9.81 0.1732××+ +=× =×
∴ N1 = 183.3 r.p.m.

Now from Fig. 18.7 ( b),we find that maximum radius of rotation,
r2 = BG = BP sin 40° =  0.2 × 0.643 = 0.1268 m
h2 = PG = BP cos 40° = 0.2 × 0.766 = 0.1532 m
and22 2 2( ) – ( ) (0.25) – (0.1268) 0.2154 mDG BD BG== =
∴ tan β2 = BG/DG = 0.1268 / 0.2154 = 0.59
and tan α2 = tan 40° = 0.839
∴2
2
2tan 0.590.703tan 0.839qβ== =α

Chapter 18 : Governors    /G6C  665
We know that when the sleeve moves upwards, the frictional force ( F) acts downwards and
the maximum speed is given by
                                     2
2
2
2..( 1 )895 2().mg Fmg q
Nmg h+++=×
15 9.81 242 9.81 (1 0.703)895 249 2362 9.81 0.1532×+×+ +=× =×
∴ N2 = 222  r.p.m.
= N2 – N1 = 222 – 183.3 = 38.7 r.p.m. Ans.
Example 18.5.  A Porter governor has all four arms 250  mm long. The upper arms are

attached on the axis of rotation and the lower arms are attached to the sleeve at a distance of 30  mm
from the axis . The mass of each ball is 5  kg and the sleeve has a mass of 50  kg. The extreme radii of
rotation are 150  mm and 200  mm. Determine the range of speed of the governor .
Solution.  Given : BP = BD = 250 mm ; DH = 30 mm ; m = 5 kg ; M = 50 kg ;
r1 = 150 mm ; r2 = 200 mm

maximum position of the governor is shown in Fig. 18.8 ( a) and (b) respectively.
Fig. 18.8
Let N1 = Minimum  speed  when r1 = BG = 150 mm ; and
From Fig. 18.8 ( a), we find that height of the governor,
22 22
1 ( ) – ( ) (250) – (150) 200 mm 0.2 m hP G B P B G== = = =
BF  = BG – FG = 150 – 30 = 120 mm . . . (/G8C FG = DH)

666  /G6C   Theory of Machines
and22 22( ) – ( ) (250) – (120) 219 mmDF DB BF===
∴ tan α1 = BG/PG = 150 / 200 = 0.75
and tan β1 = BF/DF = 120/219 = 0.548
∴1
1
1tan 0.5480.731tan 0.75qβ== =α
We know that  12
1
150(1 ) 5 (1 0.731)895 895 22( ) 4320650 . 2Mmq
N
mh++ ++
=× = × =
∴ N1 = 208 r.p.m.
From Fig. 18.8( b), we find that height of the governor,
22 22
2 ( ) ( ) (250) (200) 150mm 0.15m hP G B P B G== − = − = =
BF = BG – FG = 200 – 30 = 170 mm

and                                          22 22( ) ( ) (250) (170) 183mmDF DB BF=− =− =
∴                           tan α2= BG/PG = 200/150 = 1.333
and                                      tan β2= BF/DF = 170/183 = 0.93
2
2
2tan 0.930.7tan 1.333qβ∴= = =α
                                   22
2
250(1 ) 5 (1 0.7)895 895 22( ) 56 68350 . 1 5Mmq
N
mh++ ++
=× =× =
∴  N2 = 238 r.p.m.

      = N2 – N1 = 238 – 208 = 30 r.p.m. Ans.
Example 18.6.  The arms of a Porter governor are 300  mm long. The upper arms are pivoted
on the axis of rotation . The lower arms are attached to a sleeve at a distance of 40 mm from the axis
of rotation . The mass of the load on the sleeve is 70  kg and the mass of each ball is 10  kg. Determine
the equilibrium speed when the radius of rotation of the balls is 200  mm. If the friction is equivalent

to a load of 20  N at the sleeve , what will be the range of speed for this position ?
Solution. Given : BP = BD = 300 mm ;  DH = 40 mm ;  M = 70 kg ;  m = 10 kg ;  r = BG = 200 mm
Equilibrium speed when the radius of rotation r = BG = 200 mm
Let N  = Equilibrium  speed.
The equilibrium position of the governor is shown in Fig. 18.9. From the figure, we find that
height of the governor,
                                       22 22( ) – ( ) (300) – (200) 224 mm hP G B P B G== = =

Chapter 18 : Governors    /G6C  667
     ∴                                BF  = BG – FG = 200 – 40 = 160
. . . (/G51 FG = DH)
and      22 22( ) – ( ) (300) – (160) 254 mmDF DB BF===
∴ tan α = BG/PG = 200 / 224 = 0.893
and           tan β = BF/DF = 160 / 254 = 0.63
∴       tan 0.630.705tan 0.893qβ===α
                     2(1 )895 2Mmq
Nmh++
=×
   7010 (1 0.705)895 227 84010 0.224++
=× =
∴      N2  = 167 r.p.m. Ans.
Range of speed when friction is equivalent to load of 20 N

at the sleeve ( i.e. when F = 20 N)
Let     N1 = Minimum  equilibrium  speed,  and
         N2 = Maximum  equilibrium  speed.
the frictional force ( F) acts upwards and the minimum equi-
librium speed is given by
                 2
                            70 9.81 – 2010 9.81 (1 0.705)895 22714410 9.81 0.224××+ +=× =×
∴      N1 = 164.8 r.p.m.

We also know that when the sleeve moves upwards, the frictional force ( F) acts downwards
and the maximum equilibrium speed is given by
         2
2..( 1 )895 2().Mg Fmg q
                    70 9.81 2010 9.81 (1 0.705)895 22853310 9.81 0.224×+×+ +=× =×
∴     N2 = 169 r.p.m.Fig. 18.9
An 18th century governor.

668  /G6C   Theory of Machines
= N2 – N1 = 169 – 164.8 = 4.2 r.p.m. Ans.
Example 18.7.  A loaded Porter governor has four links each 250  mm long, two revolving
masses each of 3  kg and a central dead weight of mass 20  kg. All the links are attached to respective
sleeves at radial distances of 40  mm from the axis of rotation . The masses revolve at a radius of 150
mm at minimum speed and at a radius of 200  mm at maximum speed . Determine the range of speed .

Solution.  Given : BP = BD = 250 mm ; m = 3 kg ;  M = 20 kg ;  PQ = DH = 40 mm ;
First of all, let us find the minimum and maximum speed of the governor.The minimum and maximum position of the governor is shown in Fig. 18.10 ( a) and (b) respectively.
Let N
1 =   Minimum  speed  when r1 = BG = 150 mm, and
N2 = Minimum  speed  when r2 = BG = 200 mm.
From Fig. 18.10 ( a), we find that
BF = BG – FG = 150 – 40 = 110 mm
and sin α1 =  BF / BP = 110 / 250 = 0.44 or α1 = 26.1°

∴ Height of the governor,
h1 = OG  = BG / tan α1 = 150 / tan 26.1° = 306 mm = 0.306 m
Fig. 18.10
Since all the links are attached to respective sleeves at equal distances ( i.e.40 mm) from the
axis of rotation, therefore
tan α1 = tan β1orq = 1
1
1895 3 20 895( ) 224243 0.306mMNmh++=× = × =
        N1 = 150 r.p.m.

Chapter 18 : Governors    /G6C  669
Now from Fig. 18.10 ( b), we find that
BF = BG – FG = 200 – 40   =   160  mm
and sin α2 = BF/BP = 160 / 250 = 0.64 or β2  =  39.8°
h2 = OG = BG / tan α2 = 200 / tan 39.8° = 240 mm = 0.24 m
In this case also,
tan α2 = tan β2      orq = 1
We know that            2
2
2895 3 20 895( ) 28 59030 . 2 4mMNmh++=× = × =
∴                                    N2 = 169 r.p.m.

                                                 = N2 – N1 = 169 – 150 = 19 r.p.m. Ans.
Example 18.8.  All the arms of a Porter governor are 178  mm long and are hinged at a
distance of 38  mm from the axis of rotation . The mass of each ball is 1.15  kg and mass of the sleeve
is 20 kg. The governor sleeve begins to rise at 280  r.p.m. when the links are at an angle of 30 ° to the
vertical. Assuming the friction force to be constant , determine the mini-

mum and maximum speed of rotation when the inclination of the armsto the vertical is 45 °.
Solution.  Given : BP = BD = 178 mm ; PQ = DH = 38 mm ;
m = 1.15 kg ; M = 20 kg ; N = 280 r.p.m. ; α = β = 30°
First of all, let us find the friction force ( F). The equilibrium
position of the governor when the lines are at 30° to vertical, is shownin Fig. 18.11. From the figure, we find that radius of rotation,
r = BG = BF + FG = BP × sin α + FG
= 178 sin 30° + 38 = 127 mm
and height of the governor,

h = BG / tan α
= 127 / tan 30° = 220 mm = 0.22 m
2.( ) 8 9 5
.mg M g FNmg h+±=×
. . . (∴ tan α = tan β or q = 1)
21.15 9.81 20 9.81 895(280)1.15 9.81 0.22F ×+ ×±=××
or2(280) 1.15 9.81 0.22–1.15 9.81 – 20 9.81895F×××±= × ×
= 217.5 – 11.3 – 196.2 = 10 N
We know that radius of rotation when inclination of the arms to the vertical is 45 ( i.e. when
 α = β = 45°),
= 178 sin 45° + 38 = 164 mmFig. 18.11

670  /G6C   Theory of Machines
h = BG / tan α = 164 / tan 45° = 164 mm = 0.164 m
Let N1 = Minimum speed of rotation, and
N2 = Maximum speed of rotation.
2
1.15 9.81 (20 9.81 –10) 89595 3821.15 9.81 0.164×+ ×=× =×
∴ N1 = 309 r.p.m. Ans.
and2
1.15 9.81 (20 9.81 10) 895105 0401.15 9.81 0.164×+ ×+=× =×
N2 = 324 r.p.m. Ans.
18.7. Proell Governor

The Proell governor has the balls fixed at B and C to the extension of the links DF and EG, as
shown in Fig. 18.12 ( a). The arms FP and GQ are pivoted at P and Q respectively.
Consider the equilibrium of the forces on one-half of the governor as shown in Fig. 18.12 ( b).
The instantaneous centre ( I) lies on the intersection of the line PF produced and the line from D
drawn perpendicualr to the spindle axis. The prependicular BM is drawn on ID.

Fig. 18.12.  Proell governor.
Taking moments about I, using the same notations as discussed in Art. 18.6 (Porter governor),
C..22WM gF BM w IM ID m g IM ID× = ×+ ×= ×+ × . . . (i)
∴ C..2IM M g IM MDFm gBM BM+=×+ . . . ( /G51  ID = IM + MD)

Chapter 18 : Governors    /G6C  671
Multiplying and dividing by FM, we have
C..2FM IM M g IM MDFm gBM FM FM FM =× + +   
.. tan (tan tan )2FM M gmgBM=× α + α + β
.t a ntan . 12t a nFM M gmgBM β=× α + +  α 
We know that   FC = m.ω2 r ; tanr
hα= and tan
tanqβ=α
∴2 ... . ( 1 )2FM r M gmr m g qBM hω= × + +
and2(1 )2MmqFM g
BM m h++ω= 
. . . (ii)
Substituting ω = 2π N/60, and g = 9.81 m/s2, we get
2(1 )895 2MmqFMNBM m h++=
. . . (iii)

Notes : 1.  The equation (i) may be applied to any given configuration of the governor.
2.  Comparing equation (iii) with the equation  (v) of the Porter governor (Art. 18.6), we see that the
equilibrium speed reduces for the given values of m, M and h. Hence in order to have the same equilibrium
speed for the given values of m, M and h, balls of smaller masses are used in the Proell governor than in the
Porter governor.

3.  When α = β, then q = 1. Therefore equation (iii) may be written as
2 895 FM m MNBM m h+=(h being in metres ) ...(iv)
Example 18.9.  A Proell governor has equal arms of length 300  mm. The upper and lower
ends of the arms are pivoted on the axis of the governor . The extension arms of the lower links are
each 80 mm long and parallel to the axis when the radii of rotation of the balls  are 150 mm and

200 mm. The mass of each ball is 10  kg and the mass of the central load is 100  kg. Determine the
range of speed of the governor .
Solution.  Given : PF = DF = 300 mm ; BF = 80 mm ; m = 10 kg ; M = 100 kg ;
r1 = 150 mm; r2 = 200 mm
maximum position of the governor is shown in Fig. 18.13.
Let N1 = Minimum  speed  when  radius  of  rotation, r1 = FG = 150 mm ; and

N2 = Maximum  speed  when  radius  of  rotation , r2 = FG = 200 mm.
From Fig. 18.13 ( a), we find that height of the governor,
22 22
1 ( ) – ( ) (300) – (150) 260 mm 0.26 m hP G P F F G== = = =

672  /G6C   Theory of Machines
and FM = GD = PG = 260 mm = 0.26 m
∴ BM = BF + FM = 80 + 260 = 340 mm = 0.34 m
1
1895()FM m MNBM m h+=. . . (∴ α = β or q = 1)
           0.26 10 100 895289560.34 10 0.26+== or   N1 = 170 r.p.m.
Fig. 18.13
Now from Fig. 18.13 ( b), we find that height of the governor,
22 22
2 ( ) – ( ) (300) – (200) 224 mm 0.224 m hP G P F F G== = = =
and FM = GD = PG = 224 mm = 0.224 m
∴ BM = BF +  FM = 80 + 224 = 304 mm = 0.304 m
We know that 2
2

2
2895()FM m MNBM m h+=. . . (/G51 α = β or q = 1)
0.224 10 100 89532 3850.304 10 0.224+==orN2 = 180 r.p.m.
= N2 – N1 = 180 – 170 = 10 r.p.m. Ans.
Note : The example may also be solved as discussed below :
From Fig. 18.13 ( a), we find that
sin α= sin β = 150 / 300 = 0.5 or α = β = 30°
and MD=FG = 150 mm = 0.15 m
FM=FD cos β = 300 cos 30° = 260 mm = 0.26 m

Chapter 18 : Governors    /G6C  673
IM=FM tan α = 0.26 tan 30° = 0.15 m
BM=BF + FM = 80 + 260 = 340 mm = 0.34 m
ID=IM + MD = 0.15 + 0.15 = 0.3 m
 2
22 1
C1 1 12( ) 10 0.15 0.0165( )60NFm r Nπ=ω = = 
Now taking moments about point I,
   C..2MgFB M m g I M I D×=× + ×
or2
1100 9.810.0165 ( ) 0.34 10 9.81 0.15 0.32N×=× × + ×
0.0056 (N1)2 = 14.715 + 147.15 = 161.865
∴2
1161.865( ) 28904
0.0056N==    orN1 = 170 r.p.m.
Similarly N2 may be calculated.

Example 18.10.  A governor of the Proell type has each arm 250  mm long. The pivots of the
upper and lower arms are 25  mm from the axis . The central load acting on the sleeve has a mass of
25 kg and the each rotating ball has a mass of 3.2  kg. When the governor sleeve is in mid-position ,
the extension link of the lower arm is vertical and the radius of the path of rotation of the masses is175 mm. The vertical height of the governor is 200 mm.

If the governor speed is 160 r.p.m. when in mid-position, find : 1. length of the extension
link; and 2. tension in the upper arm.
Note : This picture is given as additional information and is not a direct example of the current chapter.An overview of a combined cycle power plant. Governors are used in power plants to control the
flow of working fluids.

674  /G6C   Theory of Machines
Solution. Given : PF = DF = 250 mm = 0.25 m ; PQ = DH = KG = 25 mm = 0.025 m ;
M = 25 kg ; m = 3.2 kg ; r = FG = 175 mm = 0.175 m ; h = QG = PK = 200 mm = 0.2 m ; N = 160 r.p.m.
1. Length of the extension link
Let BF= Length  of  the  extension  link.
The Proell governor in its mid-position is shown
in Fig. 18.14.
From the figure, we find that
FM=GH = QG = 200 mm = 0.2 m
2 895 FM m MNBM m h+=
. . . (/G51 α = β or q = 1)

20.2 3.2 25 895 7887(160)3.2 0.2BM BM+==
∴ BM= 7887/(160)2 = 0.308 m
From Fig. 18.14,
BF=BM – FM = 0.308 – 0.2 = 0.108 m = 108 mm   Ans.
2. Tension in the upper arm
Let T1= Tension  in  the  upper  arm.
                       22 2 2() – () () – ( – )PK PF FK PF FG KG==
22(250) – (175 – 25) 200 mm==
cos α=PK/PF  = 200/250 = 0.8
and 125 9.81cos 3.2 9.81 154 N22MgTm g×α= + = × + =
∴1154 154192.5 Ncos 0.8T== =αAns.

Example 18.11.  The following particulars refer to a
Proell governor with open arms :
Length of all arms = 200 mm ; distance of pivot of arms
from the axis of rotation = 40 mm ; length of extension of lower
arms to which each ball is attached = 100 mm ; mass of each ball
= 6 kg and mass of the central load = 150 kg. If the radius of
rotation of the balls is 180  mm when the arms are inclined at an

angle of 40 ° to the axis of rotation , find the equilibrium speed for
the above configuration .
Solution.  Given : PF = DF = 200 mm ; PQ = DK =
HG = 40 mm ; BF = 100 mm ; m = 6 kg; M = 150 kg ;
r = JG = 180 mm = 0.18 m ; α = β = 40°
Let N= Equilibrium  speed.Fig. 18.15. All dimensions in mm.Fig. 18.14.  All dimensions in mm.

Chapter 18 : Governors    /G6C  675
From the equilibrium position of the governor, as shown in Fig. 18.15, we find that
PH = PF × cos 40°
= 200 × 0.766 = 153.2 mm
= 0.1532 m
and FH=PF × sin 40°= 200 × 0.643 = 128.6 mm
∴ JF=JG – HG – FH = 180 – 40 – 128.6 = 11.4 mm
and22() – ()BJ BF JF=22(100) – (11.4) 99.4 mm==
We know that BM = BJ + JM = 99.4 + 153.2 = 252.6 mm . . . (/G51JM = HD = PH)
IM = IN – NM = FH – JF = 128.6 – 11.4 =  117.2 mm
. . . (/G51 IN = ND = FH)

and ID = IN + ND = 2 × IN = 2 × FH = 2 × 128.6 = 257.2 mm
Now taking moments about the instantaneous centre I,
C..2MgFB M m g I M I D×=× + ×
C150 9.81252.6 6 9.81 117.2 257.2 196 1252F××= × ×+ ×=
∴ C196125776.4 N
252.6F==
We know that centrifugal force ( FC),
              2
222776.4 . . 6 0.18 0.01260Nmr Nπ=ω= = 
∴2776.4647000.012N== orN = 254 r.p.m. Ans.
Example 18.12.  A Proell governor has all four arms of length 305  mm. The upper arms are

pivoted on the axis of rotation and the lower arms are attached to a sleeve at a distance of 38 mmfrom the axis . The mass of each ball is 4.8  kg and are attached to the extension of the lower arms
which are 102  mm long. The mass on the sleeve is 45  kg. The minimum and maximum radii of
governor are 165  mm and 216  mm. Assuming that the extensions of the lower arms are parallel to the
governor axis at the minimum radius , find the corresponding equilibrium speeds .

Solution. Given : PF = DF = 305 mm ; DH = 38 mm ; BF = 102 mm ; m = 4.8 kg ; M = 54 kg
Equilibrium speed at the minimum radius of governor
The radius of the governor is the distance of the point of intersection of the upper and lower
arms from the governor axis. When the extensions of the lower arms are parallel to the governor axis,then the radius of the governor ( FG) is equal to the radius of rotation ( r
1).

676  /G6C   Theory of Machines
The governor configuration at the minimum radius
(i.e. when FG = 165 mm) is shown in Fig. 18.16.
Let                  N1 = Equilibrium  speed  at the
    minimum radius i.e. when
   FG =  r1 = 165 mm.
From Fig. 18.16, we find that
                        165sin 0.541305FG
FPα= = =
∴                   α = 32.75°
and                       tan α = tan 32.75° = 0.6432
   –sin
165 – 380.4164305FK FG KG
DF DFβ= =
==
∴                    β = 24.6°

DF DFβ= =
==
∴                    β = 24.6°
and                        tan β = tan 24.6° = 0.4578
                             tan 0.45780.712tan 0.6432qβ== =α
From Fig. 18.16, we find that height of the governor,
                                        22 22( ) – ( ) (305) – (165) 256.5 mm 0.2565 m hP G P F F G== = = =
                       MD  = FK = FG – KG = 165 – 38 = 127 mm
∴    22 2 2( ) – ( ) (305) – (127) 277 mm 0.277 mFM DF MD=== =

and                            BM = BF + FM = 102 + 277 = 379 mm = 0.379 m
                      2
1(1 )895 2()MmqFMNBM m h++=

                                 544.8 (1 0.712)0.277 895 2271090.379 4.8 0.2565++==

∴                  N1 = 165 r.p.m . Ans.
Note : The valve of N1 may also be obtained by drawing the governor configuration to some suitable scale and
measuring the distances BM, IM and ID. Now taking moments about point I,

C..,2MgFB M m g I M I D×=× + ×
where                          2
2 1
C1 1 12Centrifugal force ( )60NFm r m rπ== ω = 
Equilibrium speed at the maximum radius of governor
Let                 N2 = Equilibrium speed at the maximum radius of governor, i.e. when F1G1
                                    = r2 = 216 mm.Fig. 18.16Also

Chapter 18 : Governors    /G6C  677
First of all, let us find the values of BD and γ in Fig. 18.16. We know that
22 22( ) ( ) (397) (127) 400 mmBD BM MD=+ =+ =
and tan γ = MD/BM = 127/379 = 0.335 or γ = 18.5°
The governor configuration at the maximum ra-
dius of F1G1 = 216 mm is shown in Fig. 18.17. From the
geometry of the figure,
11
1
11216sin 0.7082305FG
PFα= = =
∴α1 = 45.1°
1 1 11 11
1
11 11–sin
216 – 380.5836305FK FG KG
FD FDβ= =
==
∴β1 = 35.7°

FD FDβ= =
==
∴β1 = 35.7°
Since the extension is rigidly connected to the
lower arm ( i.e. DFB or D1F1B1 is one continuous link)
therefore B1D1 and angle B1D1F1 do  not change. In other
words,
B1D1 = BD = 400 mm
and γ – β = γ1 – β1 or γ1 = γ – β + β1
= 18.5° – 24.6° + 35.7° = 29.6°
∴ Radius of rotation,
r2 =  M1D1 + D1H1 = B1D1 × sin γ1 + 38 mm
= 400 sin 29.6° + 38 = 235.6 mm = 0.2356 m
From Fig. 18.17, we find that
B1M1 = B1D1 × cos γ1 = 400 × cos 29.6° = 348 mm = 0.348 m

F1N1=F1D1 × cos β1 = 305 × cos 35.7° = 248 mm = 0.248 m
I1N1=F1N1 × tan α1 = 0.248 × tan 45.1° = 0.249 m
N1D1=F1D1 × sin β1 = 305 × sin 35.7 = 178 mm = 0.178 m
∴ I1D1=I1N1 + N1D1 = 0.249 + 0.178 = 0.427 m
M1D1=B1D1 sin γ1 = 400 sin 29.6° = 198 mm = 0.198 m
∴ I1M1=I1D1 − M1D1 = 0.427 – 0.198 = 0.229 m
2
22 2
C2 2 22( ) 4.8 0.2356 0.0124 ( )60NFm r Nπ=ω = = 
Now taking moments about point I1,
C1 1 1 1 1 1..2MgFB Mm g I M I D×= ×+×
      2

      2
254 9.810.0124 ( ) 0.348 4.8 9.81 0.229 0.4272N××= × ×+ ×
0.0043 (N2)2 = 10.873 + 113.1 = 123.883Fig. 18.17

678  /G6C   Theory of Machines
∴2
2123.883( ) 28810
0.0043N== orN2 = 170 r.p.m. Ans.
Note : The value of N2 may also be obtained by drawing the governor configuration to some suitable scale and
measuring the distances B1M1, I1M1 and I1D1.
18.8. Hartnell Governor
A Hartnell governor is a spring loaded governor as shown in Fig. 18.18. It consists of two
bell crank levers pivoted at the points O,O to the frame. The frame is attached to the governor spindle

and therefore rotates with it. Each lever carries a ball at the end of the vertical arm OB and a roller at
the end of the horizontal arm OR. A helical spring in compression provides equal downward forces
on the two rollers through a collar on the sleeve. The spring force may be adjusted by screwing a nutup or down on the sleeve.
Letm= Mass of each ball in kg,
M= Mass of sleeve in kg,
r
1= Minimum radius of rotation in
metres,
r2= Maximum radius of rotation in
metres,

metres,
metres,
ω1= Angular speed of the governor at
minimum radius in rad/s,
ω2= Angular speed of the governor at
maximum radius in rad/s,
S1= Spring force exerted on the sleeve
at ω1 in newtons,
S2= Spring force exerted on the sleeve
at ω2 in newtons,
FC1= Centrifugal force at ω1 in newtons = m (ω1)2 r1,
FC2= Centrifugal force at ω2 in newtons = m (ω2)2 r2,
s= Stiffness of the spring or the force required to compress the spring by one mm,

x= Length of the vertical or ball arm of the lever in metres,
y= Length of the horizontal or sleeve arm of the lever in metres, and
r= Distance of fulcrum O from the governor axis or the radius of rotation when the
governor is in mid-position, in metres.
Consider the forces acting at one bell crank lever. The minimum and maximum position is
shown in Fig. 18.19. Let h be the compression of the spring when the radius of rotation changes from
r1 to r2.

r1 to r2.
For the minimum position i.e. when the radius of rotation changes from r to r1, as shown in
Fig. 18.19 ( a), the compression of the spring or the lift of sleeve h1 is given by
11 1 – har r
yx x==. . . (i)
Similarly, for the maximum position i.e. when the radius of rotation changes from r to r2, as
shown in Fig. 18.19 ( b), the compression of the spring or lift of sleeve h2 is given by
22 2 – harr
yx x== . . . (ii)Fig. 18.18. Hartnell governor.

Chapter 18 : Governors    /G6C  679
Adding equations  (i) and (ii),
122 1 – hh rr
yx+= or21–rrh
yx= . . . (/G51 h = h1 + h2)
∴ 21(–)yhrr
x= . . . (iii)
Fig. 18.19
Now for minimum position, taking moments about point O, we get
1
1C 1 1 1..2Mg SyF xm g a+×= ×− ×
or1C 1 1 1
12.( – . )Mg S F x mg ay+= × × . . . (iv)
Again for maximum position, taking moments about point O, we get
2
2C 22 2..2Mg SyF xm g a+×= ×+ ×
or 2C 2 2 2
22.(. )Mg S F x mg ay+= ×+ × . . . (v)

Subtracting equation  (iv) from equation  (v),
21 C 22 2 C 1 1 1
2122–( . ) – (– . )SS F xm g a Fxm g ayy=× + × × ×
S2 – S1 = h.s, and21(–)yhrrx=
∴21 21
21––
–SS SS xshr r y== 

Neglecting the obliquity effect of the arms ( i.e. x1 = x2 = x, and y1 = y2 = y) and the moment
due to weight of the balls ( i.e. m.g), we have for minimum position,
1
C1.
2Mg SyF x+×= × or1C 1.2xMg S Fy+= × . . . (vi)

680  /G6C   Theory of Machines
Similarly for maximum position,
2
C2.
2Mg SyF x+×= × or2C 2.2xMg S Fy+= × . . . (vii)
Subtracting equation (vi) from equation (vii),
21 C 2C 1–2 ( – )xSS F Fy= ...(viii)
S2 – S1 = h.s, and21(–)yhrr
x=
∴2
C2 C1 21
21– –2–FF SS xshr r y ==   . . . (ix)
Notes : 1. Unless otherwise stated, the obliquity effect of the arms and the moment due to the weight of the balls
is neglected, in actual practice.

2. When friction is taken into account, the weight of the sleeve ( M.g) may be replaced by ( M.g. ± F).
3. The centrifugal force ( FC) for any intermediate position ( i.e. between the minimum and maximum
position) at a radius of rotation ( r) may be obtained as discussed below :
Since the stiffness for a given spring is constant for all positions, therefore for minimum and interme-
diate position,
2
CC 1
1–2–FF xsrr y =  . . . (x)

2
CC 1
and for intermediate and maximum position,
2
C2 C
2–2–FFxsrr y =  . . . (xi)
∴        From equations (ix), (x) and (xi),
C2 C1 C C1 C2 C
21 1 2–– –
–––FFF FFF
rr r r rr==
or12
CC 1 C 2C 1 C 2 C 2C 1
21 21––(–) – (–)––rr r rFF F F F F Frr rr =+ =  
 
Example 18.13.  A Hartnell governor having a central sleeve spring and two right-angled

bell crank levers moves between 290  r.p.m. and 310 r.p.m. for a sleeve lift of 15  mm. The sleeve arms
and the ball arms are 80  mm and 120  mm respectively . The levers are pivoted at 120  mm from the
governor axis and mass of each ball is 2.5  kg. The ball arms are parallel to the governor axis at the
lowest equilibrium speed . Determine : 1. loads on the spring at the lowest and the highest equilib-
rium speeds , and 2. stiffness of the spring .

Solution.  Given : N1 = 290 r.p.m. or ω1 = 2 π × 290/60 =  30.4 rad/s ; N2 = 310 r.p.m. or
ω2 = 2 π × 310/60 = 32.5 rad/s ; h = 15 mm = 0.015 m ; y = 80 mm = 0.08 m ; x = 120 mm =
0.12 m ; r = 120 mm = 0.12 m ; m = 2.5 kg
1. Loads on the spring at the lowest and highest equilibrium speeds
Let S  = Spring  load  at  lowest  equilibrium  speed, and
S2 = Spring load at highest  equilibrium speed.

Since the ball arms are parallel to governor axis at the lowest equilibrium speed ( i.e. at
N1 = 290 r.p.m.), as shown in Fig. 18.20 ( a), therefore
r = r1 = 120 mm = 0.12 m

Chapter 18 : Governors    /G6C  681
We know that centrifugal force at the minimum speed,
FC1 = m (ω1)2 r1 =  2.5 (30.4)2  0.12 = 277 N
Now let us find the radius of rotation at the highest equilibrium speed, i.e. at N2 = 310 r.p.m.
The position of ball arm and sleeve arm at the highest equilibrium speed is shown in Fig. 18.20 ( b).
Let r2 = Radius of rotation at N2 = 310 r.p.m.
We know that21(–)yhrr
x=
or 210.120.12 0.015 0.1425 m
0.08xrrh
y =+ = + =  

0.08xrrh
∴  Centrifugal force at the maximum speed,
FC2=m (ω2)2 r2 = 2.5 × (32.5)2 × 0.1425 = 376 N
Fig. 18.20
Neglecting the obliquity effect of arms and the moment due to the weight of the balls, we
have for lowest position,
1C 10.12. 2 2 277 831 N0.08xMg S Fy+= ×= × × =
∴ S2= 831 N Ans.             (/G51 M = 0)
and for highest position,
2C 20.12. 2 2 376 1128 N0.08xMg S Fy+= × = × × =
∴ S1= 1128 N Ans.             (/G51 M = 0)
2. Stiffness of the spring

We know that stiffness of the spring,
21–1128 – 83119.8 N/mm15SSsh== =  Ans.
Example 18.14.  In a spring loaded Hartnell type governor , the extreme radii of rotation of
the balls are 80  mm and 120  mm. The ball arm and the sleeve arm of the bell crank lever are equal
in length. The mass of each ball is 2  kg. If the speeds at the two extreme positions are 400  and
420 r.p.m., find : 1. the initial compression of the central spring , and 2. the spring constant .

682  /G6C   Theory of Machines
Solution. Given : r1 = 80 mm = 0.08 m ; r2 = 120 mm = 0.12 m ; x = y ; m = 2 kg ; N1 = 400
r.p.m. or ω = 2 π × 400/60 = 41.9 rad/s ; N2 = 420 r.p.m. or ω2 = 2 π × 420/60 = 44 rad/s
Initial compression of the central spring
We know that the centrifugal force at the minimum speed,
FC1=m (ω1)2 r1 = 2 (41.9)2  0.08 = 281 N
and centrifugal force at the maximum speed,
FC2=m (ω2)2 r2 = 2 (44)2  0.12 = 465 N
Let S1= Spring force at the minimum speed, and

S2= Spring force at the maximum speed.
We know that for minimum position,
1C 1.2xMg S Fy+= ×
∴ S1=2  FC1 = 2 × 281 = 562 N . . . (/G51M = 0 and x = y)
2C 2.2xMg S Fy+= ×
∴ S2=2 FC2 = 2 × 465 = 930 N
We know that lift of the sleeve,
21 21( – ) – 120 – 80 40 mmyhrr rrx== = =         . . . ( /G51x  = y)
∴   Stiffness of the spring,
21–930 – 5629.2 N/mm40SSsh== =
We know that initial compression of the central spring

156261mm9.2S
s== =  Ans.
2. Spring constant
We have calculated above that the spring constant or stiffness of the spring,
s= 9.2 N/mm Ans.
Example 18.15.  A spring loaded governor of the Hartnell type has arms of equal length . The
masses rotate in a circle of 130  mm diameter when the sleeve is in the mid position and the ball arms
are vertical . The equilibrium speed for this position is 450  r.p.m., neglecting friction . The maximum

sleeve movement is to be 25  mm and the maximum variation of speed taking in account the friction to
be 5 per cent of the mid position speed . The mass of the sleeve is 4  kg and the friction may be
considered equivalent to 30  N at the sleeve . The power of the governor must be sufficient to over-
come the friction by one per cent change of speed either way at mid-position . Determine , neglecting

obliquity effect of arms ; 1. The value of each rotating mass : 2. The spring stiffness in N/mm ; and
3. The initial compression of spring .
Solution. Given : x = y ; d = 130 mm or r = 65 mm = 0.065 m ; N = 450 r.p.m. or
ω = 2 π × 450/60 = 47.23 rad/s ; h = 25 mm = 0.025 m ; M = 4 kg ; F = 30 N
1. Value of each rotating mass
Let m= Value of each rotating mass in kg,  and
S= Spring force on the sleeve at mid position in newtons.

Chapter 18 : Governors    /G6C  683
Since the change of speed at mid position to overcome friction is 1 per cent either way
(i.e. ± 1%), therefore
Minimum speed at mid position,
ω =ω – 0.01ω = 0.99ω = 0.99 × 47.13 =  46.66 rad/s
and maximum speed at mid-position,
ω2=ω + 0.01ω = 1.01ω = 1.01 × 47.13 = 47.6 rad/s
∴   Centrifugal force at the minimum speed,
FC1=m (ω1 )2 r = m (46.66)2  0.065 = 141.5 m N
FC2=m (ω2)2 r = m (47.6)2  0.0065 = 147.3 m N

We know that for minimum speed at mid-
position,
C1 (. )2xSM g F Fy++ =×
or S + (4 × 9.81 – 30) = 2 × 141.5 m × 1
. . . (/G51 x = y)
∴ S + 9.24 = 283 m. . . (i)
and for maximum speed at mid-position,
C2 (. )2xSM g F Fy++ =×
S + (4 × 9.81 + 30) = 2 × 147.3 m × 1
. . . (/G51x = y)
∴ S + 69.24 = 294.6 m
. . . (ii)
m=5 . 2  k g Ans.
2. Spring stiffness in N/mm
Let s= Spring stiffness in N/mm.

Since the maximum variation of speed, considering friction is ± 5% of the mid-position
speed, therefore,
Minimum speed considering friction,
ω1'=ω – 0.05ω = 0.95ω = 0.95 × 47.13 = 44.8 rad/s
and maximum speed considering friction,
ω2'=ω + 0.05ω = 1.05ω = 1.05 × 47.13 = 49.5 rad/s
We know that minimum radius of rotation considering friction,
110.025– 0.065 – 0.0525 m2xrrhy=× = =
1 ... , and
2hxy h==/G51
A steam turbine used in thermal power
stations.

stations.

684  /G6C   Theory of Machines
and maximum radius of rotation considering friction,
220.0250.065 0.0775 m2xrr hy=+ ×= + =
2 ... ,and2hxy h==/G51
∴   Centrifugal force at the minimum speed considering friction,
FC1'=m (ω′1)2 r1 = 5.2 (44.8)2  0.0525 = 548 N
and centrifugal force at the maximum speed considering friction,
FC2'=m (ω2')2 r2 = 5.2 (49.5)2  0.0775 = 987 N
Let S1= Spring force at minimum speed considering friction, and
S2= Spring force at maximum speed considering  friction.

We know that for minimum speed considering friction,
1C 1(.–)2xSM g F Fy′ += ×
S1 + (4 × 9.81 – 30) = 2 × 548 × 1 . . . (/G51 x = y)
∴ S1 + 9.24 = 1096 or S1 = 1096 – 9.24 = 1086.76 N
and for maximum speed considering friction,
2C 2(. )2xSM g FFy′ ++ =×
S2 + (4 × 9.81 + 30) = 2 × 987 × 1 . . . (/G51 x = y)
∴ S2 + 69.24 = 1974 or S2 = 1974 – 69.24 = 1904.76 N
21–1904.76 –1086.7632.72 N/mm25SSsh== = Ans.
3. Initial compression of the spring

We know that initial compression of the spring
11086.7633.2 mm
32.72S
s== = Ans.
Example 18.16.  In a spring loaded governor of the Hartnell type , the mass of each ball is
1kg, length of vertical arm of the bell crank lever is 100  mm and that of the horizontal arm is 50  mm.
The distance of fulcrum of each bell crank lever is 80  mm from the axis of rotation of the governor .

The extreme radii of rotation of the balls are 75  mm and 112.5  mm. The maximum equilibrium speed
is 5 per cent greater than the minimum equilibrium speed which is 360  r.p.m. Find, neglecting obliq-
uity of arms , initial compression of the spring and equilibrium speed corresponding to the radius of
rotation of 100  mm.
Solution.  Given : m = 1 kg ; x = 100 mm = 0.1 m ; y = 50 mm = 0.05 m ; r = 80 mm
= 0.08 m ; r1 = 75 mm = 0.075 m ; r2 = 112.5 mm = 0.1125 m ; N1 = 360 r.p.m. or

ω1   =  2 π × 360/60  =  37.7 rad/s
Since the maximum equilibrium speed is 5% greater than the minimum equilibrium speed
(ω1), therefore maximum equilibrium speed,
ω2= 1.05 × 37.7 = 39.6 rad/s
We know that centrifugal force at the minimum equilibrium speed,
FC1=m (ω1)2 r1 = 1 (37.7)2 0.075 = 106.6 N

Chapter 18 : Governors    /G6C  685
 and centrifugal force at the maximum equilibrium speed,
FC2=m (ω2)2 r2 = 1 (39.6)2 0.1125 = 176.4 N
Initial compression of the spring
Let S1= Spring force corresponding to ω1, and
S2= Spring force corresponding to ω2.
Since the obliquity of arms is neglected, therefore for minimum equilibrium position,
1C 10.1. 2 2 106.6 426.4 N0.05xMg S Fy+= ×= × × =
∴ S1= 426.4 N           ...(/G51M = 0)
and for maximum equilibrium position,

2C 20.1. 2 2 176.4 705.6 N0.05xMg S Fy+= × = × × =
∴ S2= 705.6 N           ...(/G51M = 0)
210.05( – ) (0.1125 – 0.075) 0.018 75 m0.1yhrrx== =
and stiffness of the spring  21– 705.6 – 426.414 890 N/m 14.89 N/mm0.018 75SSsh== = =
∴   Initial compression of the spring
1426.428.6 mm14.89S
s== = Ans.
Equilibrium speed corresponding to radius of rotation r = 100 mm = 0.1 m
Let N= Equilibrium speed in r.p.m.

Since the obliquity of the arms is neglected, therefore the centrifugal force at any instant,
1
CC 1 C 2C 1
21–(–)–rrFF F Frr=+ 

0.1 – 0.075106.6 (176.4 –106.6) 153 N
0.1125 – 0.075=+ = 

We know that centrifugal force ( FC ),
2
222153 . . 1 0.1 0.0011
60Nmr Nπ=ω= = 
∴ N2= 153 / 0.0011 = 139 090     or N = 373 r.p.m. Ans.
Example 18.17.  In a spring loaded governor of the Hartnell type , the mass of each ball is

5 kg and the lift of the sleeve is 50  mm. The speed at which the governor begins to float is 240  r.p.m.,
and at this speed the radius of the ball path is 110  mm. The mean working speed of the governor is 20
times the range of speed when friction is neglected . If the lengths of ball and roller arm of the bell
crank lever are 120  mm and 100  mm respectively and if the distance between the centre of pivot of

bell crank lever and axis of governor spindle is 140  mm, determine the initial compression of the
spring taking into account the obliquity of arms .
If friction is equivalent to a force of 30  N at the sleeve , find the total alteration in speed before
the sleeve begins to move from mid-position .

686  /G6C   Theory of Machines
Solution. Given : m = 5 kg ; h = 50 mm = 0.05 m ; N1 = 240 r.p.m. or ω1 = 2 π × 240/60 =
25.14 rad/s ; r1 = 110 mm = 0.11 m ; x = 120 mm = 0.12 m ; y = 100 mm = 0.1 m ;
r = 140 mm = 0.14 m ; F = 30 N
Initial compression of the spring taking into account the obliquity of arms
First of all, let us find out the maximum speed of rotation ( ω2) in rad/s.
We know that mean working speed,
12
2ω+ ωω=
and range of speed, neglecting friction
= ω2 – ω1

= ω2 – ω1
Since the mean working speed is 20 times the range of speed, therefore
ω= 20 (ω2 – ω1)
or12
2120 ( – )2ω+ ω=ω ω
25.14 + ω2= 40 (ω2 – 25.14) = 40 ω2 – 1005.6
∴ 40 ω2 – ω2= 25.14 + 1005.6 = 1030.74 or ω2 = 26.43 rad/s
The minimum and maximum position of the governor balls is shown in Fig. 18.21 ( a) and (b)
Let r2= Maximum radius of rotation.
                           21(–)yhrr
x=

21(–)yhrr
x=
or                           210.120.11 0.05 0.17 m0.1xrrhy=+×= + × =
FC1=m (ω1)2 r1 = 5 (25.14)2 0.11 = 347.6 N
FC2=m (ω2)2 r2 = 5 (26.43)2 0.17 = 593.8 N
Fig. 18.21
Since the obliquity of arms is to be taken into account, therefore from the minimum position
as shown in Fig. 18.21 ( a),
a1=r – r1 = 0.14 – 0.11 = 0.03 m

Chapter 18 : Governors    /G6C  687
22 2 2
11– ( ) (0.12) – (0.03) 0.1162 m xxa== =
and22 2 2
11 – ( ) (0.1) – (0.025) 0.0986 m yyh== =
. . . (/G51 h1 = h / 2 = 0.025 m)
Similarly, for the maximum position, as shown in Fig. 18.21 ( b),
a2=r2 – r = 0.17 – 0.14 = 0.03 m
∴ x2=x1 = 0.1162 m . . . (/G51 a2 = a1)
and y2=y1 = 0.0986 m . . . (/G51 h2 = h1)
Now taking moments about point O for the minimum position as shown in Fig. 18.21 ( a),
1
1C 1 1 1.–.2Mg SyF xm g a+×= × ×

1
10.0968 347.6 0.1162 – 5 9.81 0.03 38.9 N2S×= × × × = . . . (/G51 M = 0)
∴                   S1 = 2 × 38.9/0.0968 = 804 N
Similarly, taking moments about point O for the maximum position as shown in Fig. 18.21 ( b),
2
20.0968 593.8 0.1162 5 9.81 0.03 70.47 N2S×= ×+ × × = . . . (/G51M = 0)
∴ S2= 2 × 70.47/0.0968 = 1456 N
We know that stiffness of the spring
21–1456 – 80413.04 N/mm50SSsh== =

180461.66 mm
13.04S
s== =  Ans.
Total alternation in speed when friction is taken into account
We know that spring force for the mid-position,
                                         S = S1 + h1.s =8.4 + 25 × 13.04 = 1130  N   . . . (/G51 h1 = h / 2 = 25 mm)
and mean angular speed,12 25.14 26.4325.785 rad/s22ω+ ω +ω= = =
or N=ω × 60 / 2π = 25.785 × 60 / 2 π = 246.2 r.p.m.
∴   Speed when the sleeve begins to move downwards from the mid-position,

– 1130 – 30246.2 243 r.p.m.1130SFNNS′== =
and speed when the sleeve begins to move upwards from the mid-position,
1130 30246.2 249 r.p.m.1130SFNNS++′′== =
∴  Alteration in speed        = N'' – N' = 249 – 243 = 6 r.p.m. Ans.

688  /G6C   Theory of Machines
Example 18.18.  Fig. 18.22 shows diagram-
matically a centrifugal governor . The masses ‘m’ are
directly connected to one another by two parallel andidentical close coiled springs , one on either side . In
the position shown , with the mass arms parallel to
the axis of rotation , the equilibrium speed is 900
r.p.m. Given ball circle radius = 70 mm ; length of
ball arm = 85 mm and length of sleeve arm = 50
mm.
1. When the speed is increased by 1 % without

mm.
any change of radius for the given position , an axial
force of 30  N is required at the sleeve to maintain
equilibrium . Determine the mass of each ball .
2. Find the stiffness and initial extension of each
spring, if the rate of sleeve movement , when in mid position is 20  mm for 480  r.p.m.   change of speed .
Solution.  Given : N = 900 r.p.m. or ω = 2 π × 900/60 = 94.26 rad/s ; r = 70 mm

= 0.07 m; x = 85 mm = 0.085 m ; y = 50 mm = 0.05 m ; W = 30 N
1. Mass of each ball
Let m = Mass of each ball in kg.
We know that centrifugal force at the equilibrium speed,
F
C = m.ω2.r = m (94.26)2 0.07 = 622 mN
Since the speed is increased by 1% without any change of radius, therefore increased speed,
ω1 = ω + 0.01 ω = 1.01 ω = 1.01 × 94.26 = 95.2 rad/s
and centrifugal force at the increased speed,
FC1 = m (ω1)2 r = m (95.2)2 0.07 = 634.4 mN

Now taking moments about point O as shown in Fig. 18.23, we get
C1 C( – ) 0.085 0.052WFF =×
30(634.4 – 622 ) 0.085 0.05 0.752mm =× =
1.054 m=0 . 7 5
or m= 0.75/1.054 = 0.7 kg Ans.
2. Stiffness and initial extension of each spring
Let s= Stiffness of each spring.
We know that centrifugal force at the equilibrium speed, i.e. at 900 r.p.m.
FC= 622 m = 622 × 0.7 = 435.4 N
Since the change of speed is 480 r.p.m., therefore increased speed,

N2= 900 + 480 = 1380 r.p.m.
∴  Angular increased speed,
ω2=2  π × 1380/60 = 144.5 rad/sFig. 18.23Fig. 18.22.

Chapter 18 : Governors    /G6C  689
Also, it is given that for 480 r.p.m. change of speed, the rate of sleeve movement is 20 mm, i.e.
                              h = 20 mm = 0.02 m
Let                         r  = Radius of rotation at 900 r.p.m. = 0.07 m . . . (Given)
                              r2 = Radius of rotation at 1380 r.p.m.
We know that for the radius of rotation to change from r  to r2 , the increase in length of radius of
rotation is

                            20.085– 0.02 0.034 m0.05xrr hy=× = × =
∴                        r2 = r + 0.034 = 0.07 + 0.034 = 0.104 m
and centrifugal force at the increased speed ( ω2),
                          FC2= m (ω2)2 r2 = 0.7 (144.5)2  0.104 = 1520 N
∴   Stiffness of each spring,
C2 C
2– Increase in force for one ball 1520 – 435.4=Increase in length for each spring 2 ( – ) 2 (0.104 – 0.07)FFsrr==
= 15 950 N/m = 15.95 N/mm Ans.
and initial extension of each spring

                                       C435.427.3 mm
15.95F
s== =  Ans.
Example 18.19.  In a spring controlled governor
of the type , as shown in Fig . 18.24, the mass of each ball is
1.5 kg and the mass of the sleeve is 8  kg. The two arms of
the bell crank lever are at right angles and their lengths
are OB = 100 mm and OA = 40 mm. The distance of
the fulcrum O of each bell crank l ever from the axis of

rotation is 50  mm and minimum radius of rotation of the
governor balls is also 50  mm. The corresponding
equilibrium speed is 240  r.p.m. and the sl eeve is required
to lift 10 mm for an increase in speed of 5  per cent. Find
the stiffness and initial compression of the spring .
Solution. Given : m = 1.5 kg ; M = 8 kg ; OB = x
= 100 mm = 0.1 m ; OA = y = 40 mm = 0.04 m ; r = 50 mm
= 0.05 m; r1 = 50 mm = 0.05 m ; N1 = 240 r.p.m. or
ω1 = 2 π × 240/60 = 25.14 rad/s ; h = 10 mm = 0.01 m ;

Increase in speed = 5%
Stiffness of the spring
The spring controlled governor of the type, as shown in Fig. 18.24, has the pivots for the bell
crank lever on the moving sleeve. The spring is compressed between the sleeve and the cap which isfixed to the end of the governor shaft. The simplest way of analysing this type of governor is by takingmoments about the instantaneous centre of all the forces which act on one of the bell crank levers.

The minimum position of the governor is shown in Fig. 18.25 ( a).
We know that the centrifugal force acting on the ball at the minimum equilibrium speed,
                            F
C1 = m (ω1)2 r1 = 1.5 (25.14)2 0.05 = 47.4 NFig. 18.24

690  /G6C   Theory of Machines
Let                           S1 = Spring force at the minimum equilibrium speed.
The instantaneous centre I for the bell crank lever coincides with the roller centre A. Taking
moments about A,
                            1
C1..2Mg SFxm g O A+ ×= +
                        1
189 . 8 147.4 0.1 1.5 9.81 0.04 0.6 1.57 0.022SS×+ ×= × + =+ +
                                 4.74 = 2.17 + 0.02 S1    or  14.74 – 2.17128.5 N
0.02S==
Fig. 18.25

0.02S==
Fig. 18.25
The maximum position of the governor is shown in Fig. 18.25 ( b). From the geometry of the
figure,
                 21–rrh
xy=     or    210.10.05 0.01 0.075 m0.04xrrhy=+×= + × =
Since the increase in speed is 5%, therefore the maximum equilibrium speed of rotation,
                                 N2= N1 + 0.05 N1 = 1.05 N1 = 1.05 × 240 = 252 r.p.m.
or                                  ω2= 2 π × 252/60 = 26.4 rad/s

∴  Centrifugal force acting on the ball at the maximum equilibrium speed,
                                FC2= m (ω2)2 r2 = 1.5 (26.4)2 0.075 = 78.4 N
Let                         S2 = Spring force at the maximum equilibrium speed.
The instantaneous centre in this case lies at I as shown in Fig. 18.25 ( b). From the geometry
                       22 2 2 2 2( ) – ( ) – (0.04) – (0.01) 0.0387 mOI OA IA y h== = =
                                   22 2 2

22 2 2
21 () – () – (– )BD OB OD x r r==
                                          22(0.1) – (0.075 – 0.05) 0.097 m==
                            ID  = OI + OD = 0.0387 + (0.075 – 0.05) = 0.0637 m

Chapter 18 : Governors    /G6C  691
Now taking moments about I,
2
C2..2Mg SFB D m g I D O I+×= × + ×
2 89 . 8 178.4 0.097 1.5 9.81 0.0637 0.03872S×+×= × × + 
7.6 = 0.937 + 1.52 + 0.02 S2 = 2.457 + 0.019 S2
∴27.6 – 2.457270.7 N0.019S==
21–270.7 –128.514.22 N/mm
10SSs
h== =  Ans.
1128.59.04 mm14.22S
s== = Ans.
18.9. Hartung Governor

s== = Ans.
A spring controlled governor of the Hartung type is shown in Fig. 18.26 ( a). In this type of
governor, the vertical arms of the bell crank levers are fitted with spring balls which compress againstthe frame of the governor when the rollers at the horizontal arm press against the sleeve.
An overview of a thermal power station.

692  /G6C   Theory of Machines
Let S= Spring force,
FC= Centrifugal force,
M= Mass on the sleeve, and
x and y= Lengths of the vertical and horizontal arm of the bell crank
lever respectively.
Fig. 18.26. Hartung governor.
Fig. 18.26 ( a) and (b) show the governor in mid-position. Neglecting the effect of obliquity
of the arms, taking moments about the fulcrum O,
C.
2MgFx S x y×=×+ ×
Example 18.20. In a spring-controlled governor of the Hartung type , the length of the ball

and sleeve arms are 80  mm and 120  mm respectively . The total travel of the sleeve is 25  mm. In the
mid position , each spring is compressed by 50  mm and the radius of rotation of the mass centres is
140 mm. Each ball has a mass of 4  kg and the spring has a stiff ness of 10  kN/m of compression . The
equivalent mass of the governor gear at the sleeve is 16  kg. Neglecting the moment due to the revolving

masses when the arms are inclined , determine the ratio of the range of speed to the mean speed of the
governor. Find, also, the speed in the mid-position .
Solution. Given : x = 80 mm = 0.08 mm ; y = 120 mm = 0.12 m ; h = 25 mm = 0.025 m ;
r = 140 mm = 0.14 m ; m = 4 kg ; s = 10 kN/m = 10 × 103 N/m ; M = 16 kg ; Initial compression
= 50 mm = 0.05 m
Mean speed of the governor
First of all, let us find the mean speed of the governor i.e. the speed when the governor is in

Chapter 18 : Governors    /G6C  693
Let ω= Mean angular speed in rad/s, and
N= Mean speed in r.p.m.
We know that the centrifugal force acting on the ball spring,
FC=m.ω2.r = 4 × ω2 × 0.14 = 0.56 ω2 N
and Spring force, S= Stiffness × Initial compression = 10 × 103 × 0.05 = 500 N
Now taking moments about point O, neglecting the moment due to the revolving masses, we
have
C.
2 16 9.810.56 0.08 500 0.08 0.12 40 9.42 49.422×ω× = × + × = + =
∴249.4211030.56 0.08ω= =×orω = 33.2 rad/s

and33.23 60317 r.p.m.2N×==π Ans.
Ratio of range of speed to mean speed
Let ω1= Minimum angular speed in rad/s, at the minimum radius of
rotation r1,
ω2= Maximum angular speed in rad/s, at the maximum radius of
rotation r2,
N1 and N2= Corresponding minimum and maximum speeds in r.p.m.
The minimum and maximum position is shown in Fig. 18.27 ( b) and (c) respectively. First of
all, let us find the minimum speed N1.
From the geometry of the Fig. 18.27 ( b),

1
1–rr x
hy=or 110.025 0.08– 0.14 – 0.132 m20 . 1 2xrrhy=× = × =
. . . (/G51 h1 = h/2)
We know that centrifugal force at the minimum position,
FC1=m (ω1)2 r1 = 4 (ω1)2 0.132 = 0.528 ( ω1)2 N
and spring force at the minimum position,
S1= [Initial compression – ( r – r1)] × Stiffness
= [0.05 – (0.14 – 0.132)] 10 × 103 = 420 N
Now taking moments about the fulcrum O, neglecting the obliquity of arms ( i.e. taking x1 = x
and y1 = y),
C1 1.
2MgFx S x y×= ×+ ×

and y1 = y),
C1 1.
2
116 9.810.528 ( ) 0.08 420 0.08 0.12 33.6 9.42 43.022×ω = ×+ ×=+=
∴2
143.02( ) 10190.528 0.08ω= =×orω1 = 32 rad/s
and132 60305.5 r.p.m.2N×==π

694       /G6C               Theory of Machines
Now let us find the maximum speed N2. From the geometry of the Fig. 18.27 ( c),
                          2
2–rrx
hy=    or   220.025 0.080.14 0.148 m20 . 1 2xrr hy=+ × = + × =
. . . (∵ h2 = h/2)
We know that centrifugal force at the maximum position,
FC2=m (ω2)2 r2 = 4 (ω2)2 0.148 = 0.592 ( ω2)2 N
and spring force at the maximum position,
S2= [Initial compression + ( r2 – r) × Stiffness
= [0.05 + (0.148 – 0.14)] × 10 × 103 = 580 N

Now taking moments about the fulcrum O, neglecting obliquity of arms ( i.e. taking x2 = x and
y2 = y),
C2 2.
2
216 9.810.592 ( ) 0.08 580 0.08 0.12 46.4 9.42 55.822×ω = ×+ ×=+=
∴2
255.82( ) 11780.592 0.08ω= =×orω2 = 34.32 rad/s
and 234.32 60327.7 r.p.m.2N×==π
= N2 – N1 = 327.7 – 305.5 = 22.2 r.p.m.
∴ Ratio of range of speed to mean speed
21– 22.20.07
317NN
N== = or 7% Ans.

317NN
18.10.18.10.18.10.18.10.18.10. WWWWWilson-Harilson-Harilson-Harilson-Harilson-Har tnell Gotnell Gotnell Gotnell Gotnell Go vvvvvererererernornornornornor
A Wilson-Hartnell governor is a governor in which the balls are connected by a spring in
tension as shown in Fig. 18.28. An auxiliary spring is attached to the sleeve mechanism through a

lever by means of which the equilibrium speed for a given radius may be adjusted. The main springmay be considered of two equal parts each belonging to both the balls. The line diagram of a Wilson-Hartnell governor is shown in Fig. 18.29.
       
Fig. 18.28 Fig. 18.29

Chapter 18 : Governors    /G6C  695
Let P= Tension in the main spring or ball spring A,
S= Tension in the auxiliary spring B,
m= Mass of each ball,
M= Mass of sleeve,
sb= Stiffness of each ball spring,
sa= Stiffness of auxiliary spring,
FC= Centrifugal force of each ball, and
r= Radius of rotation of balls,
Now total downward force on the sleeve
= M.g + S × b/a
Taking moments about O and neglecting the effect of the pull of gravity on the ball,
C./(– )2Mg S baFP x y+×=×

Let suffixes 1 and 2 be used to denote the values at minimum and maximum equilibrium
speeds respectively.
∴  At minimum equilibrium speed,
1
C1 1./(– )2Mg S baFP x y+×=× . . . (i)
and at maximum equilibrium speed,
2
C2 2./(– )2Mg S baFP x y+×=× . . . (ii)
Subtracting equation (i) from equation (ii), we have
   C2 C1 2 1 2 1[( – ) – ( – )] ( – )
2byFF P P x S S
a=× . . . (iii)

When the radius increases from r1 to r2, the ball springs extend by the amount 2 ( r2 – r1) and
the auxiliary spring extend by the amount 21(–)ybrrxa×
∴ P2 – P1=2  sb × 2 (r2 – r1) = 4 sb (r2 – r1)
and 21 2 1–( – )aybSSs rrxa=×
Substituting the values of ( P2 – P1) and (S2 – S1) in equation  (iii),
     C 2 C 1 21 21[( – ) – 4 ( – )] ( – )
2baybbyFF s r r x s r r
xaa=× × ×
          2
C 2 C 1 21 21(–) – 4 ( – ) ( – )2a
bs ybFF s r r r rxa=× 
∴2
C2 C1

∴2
C2 C1
21–42–a
bsF Fybsxa r r+× =
Note : When the auxiliary spring is not used, then sa = 0.
∴C2 C1
21–4–bFFsrr= orC2 C1
21–
4( – )bFFs
rr=

696  /G6C   Theory of Machines
Example 18.21.  The following particulars refer to  a Wilson-Hartnell governor :
Mass of each ball = 2 kg ; minimum radius = 125 mm ; maximum radius = 175 mm ; minimum
speed = 240 r.p.m. ; maximum speed = 250 r.p.m. ; length of the ball arm of each bell crank lever = 150
mm; length of the sleeve arm of each bell crank lever = 100 mm ; combined stiffness of the two ball

springs = 0.2 kN/m. Find the equivalent stiffness of the auxiliary spring referred to the sleeve .
Solution.  Given : m = 2 kg ; r1 = 125 mm = 0.125 m ; r2 = 175 mm = 0.175 m ;
N1 = 240 r.p.m. or ω1 = 2 π × 240/60 = 25.14 rad/s ; N2 = 250 r.p.m. or ω2 = 2 π = 250/60
= 26.2 rad/s ; x = 150 mm = 0.15 m; y = 100 mm = 0.1 m ; sb = 0.2 kN/m = 200 N/m
Let s= Equivalent stiffness of the auxiliary spring referred to the sleeve
2
absa=

FC1=m (ω1)2 r1 = 2 (25.14)2 0.125 = 158 N
FC2=m (ω2)2 r2 = 2 (26.2)2 0.175 = 240 N
2
C2 C1
21–42–a
20.1 240 –1584 200 16402 0.15 0.175 – 0.125bs b
a×+ ×= = 
2
800 0.22 1640absa+= or2
0.22 1640 – 800 840absa==
∴2
800/0.22 3818 N/m = 3.818 kN/mabsa== Ans.
Example 18.22.  A spring loaded governor is shown
in Fig. 18.30. The two balls , each of mass 6 kg, are connected

across by two springs . An auxiliary spring B provides an
additional force at the sleeve through the medium of a leverwhich pivots about a fixed centre at its left hand end . In the
mean position , the radius of the governor balls is 120  mm
and the speed is 600  r.p.m. The tension in each spring is
then 1 kN. Find the tension in the spring B for this position .
When the sleeve moves up 15  mm, the speed is to be
630 r.p.m. Find the necessary stiff ness of the spring B , if the

stiffness of each spring A is 10  kN/m. Neglect the moment
produced by the mass of the balls .
Solution. Given : m = 6 kg ; r = r
1 = 120 mm =  0.12 m ; N = N1 = 600 r.p.m. or
ω1 = 2π × 600/60 = 62.84 rad/s
Tension in spring  B
Let SB1= Spring force or tension in spring B, and
M.g= Total load at the sleeve.Fig. 18.30

Chapter 18 : Governors    /G6C  697
FC1=m (ω1)2 r1 = 6 (62.84)2 0.12 = 2843 N
Since the tension in each spring A is 1 kN and there are two springs, therefore
Total spring force in spring A,
SA1= 2 × 1 = 2 kN = 2000 N
Taking moments about the pivot P (neglecting the moment produced by the mass of balls) in
order to find the force Mg on the sleeve, in the mean position as shown in Fig. 18.31 ( a),
C1 A1.90 90 902MgFS×= ×+ × or C1 A1.
2MgFS=+

2MgFS=+
∴ M.g=2  FC1 – 2 SA1 = 2 × 2843 – 2 × 2000 = 1686 N
Now taking moments about point Q,
SB1 × 160 = M.g. (80 + 160) = 1686 × 240 = 404 640
∴ SB1= 404 640/160 = 2529 N Ans.
Fig. 18.31
Stiffness of the spring B
Given : h = 15 mm = 0.015 m ; N2 = 630 r.p.m. or ω2 = 2 π × 630/60 = 66 rad/s;
SA = 10 kN/m = 10 × 103 N/m
Let sB= Stiffness of spring B.
The maximum position is shown in Fig. 18.31 ( b).

First of all, let us find the maximum radius of rotation ( r2) when the sleeve moves up by
0.015 m. We know that
                                         21(–)yhrrx=    or  21 0.12 0.015 0.135 mxrrhy=+×= + =
. . . (/G51 x = y = 90 mm = 0.09 m)
FC2=m (ω2)2 r2 = 6 (66)2 0.135 = 3528 N
We know that extension of the spring A,
= 2 (r2 – r1) × No. of springs = 2 (0.135 – 0.12) 2 = 0.06 m

698  /G6C   Theory of Machines
∴   Total spring force in spring A,
SA2=SA1 + Extension of springs × Stiffness of springs ( sA)
= 2000 + 0.06 × 10 × 103 = 2600 N
Now taking moments about P, neglecting the obliquity of arms,
C2 A2.90 90 902MgFS×= ×+ × or C2 A2.
2MgFS=+
∴ M.g=2  FC2 – 2 SA2 = 2 × 3528 – 2 × 2600 = 1856 N
Again taking moments about point Q (neglecting the moment produced by the mass of balls)
in order to find the spring force ( SB2) when sleeve rises as shown in Fig. 18.31 ( b),

SB2 × 160 = M.g (80 + 160) = 1856 × 240 = 445 440
∴ SB2= 445 440 /160 = 2784 N
When the sleeve rises 0.015 m, the extension in spring B
1600.015 0.01m
80 160==+
∴  Stiffness of spring B,
B2 B1
B– 2784 – 252925500 N/mExtension of spring 0.01SSsB== =
= 25.5 N/mm Ans.
18.11. Pickering Governor
A Pickering governor is mostly used for driving gramophone. It consists of *three straight

leaf springs arranged at equal angular intervals round the spindle. Each spring carries a weight at thecentre. The weights move outwards and the springs bend as they rotate about the spindle axis withincreasing speed.
Fig. 18.32. Pickering governor.
In Fig. 18.32 ( a), the governor is at rest. When the governor rotates, the springs together with
the weights are deflected as shown in Fig. 18.32 ( b). The upper end of the spring is attached by a
*Only two leaf springs are shown in Fig. 18.32.

Chapter 18 : Governors    /G6C  699
screw to hexagonal nut fixed to the governor spindle. The lower end of the spring is attached to a
sleeve which is free to slide on the spindle. The spindle runs in a bearing at each end and is driventhrough gearing by the motor. The sleeve can rise until it reaches a stop, whose position is adjustable.
Let                     m   =  Mass attached at the centre of the leaf spring,

a    = Distance from the spindle axis  to the centre of gravity of the mass,
                                    when the governor is at rest,
                          ω  = Angular speed of the governor spindle,
                          δ   = Deflection of the centre of the leaf spring at angular speed ω,
a + δ= Distance from the spindle axis to the centre of gravity of  the mass,
when the governor is rotating, and

λ= Lift of the sleeve corresponding to the deflection δ.
We know that the maximum deflection of a leaf spring with both ends fixed and carrying a
load (W) at the centre is,
3.
192Wl
EIδ= . . . (i)
where l= Distance between the fixed ends of the spring,
E= Young’s modulus of the material of the spring, and
I= Moment of inertia of its cross-section about the neutral axis 3.
12bt=
(where b and t are width and thickness of spring).

In case of a Pickering governor, the central load is the centrifugal force.∴ W=F
C = m.ω2 (a + δ) . . . (ii)
Substituting the value of W in equation ( i), we have
23.( )
192 .mal
EIω+ δδ=
Note: The empirical relation between the lift of the sleeve and the deflection δ is, 22.4
lδλ=  approximately.
Example 18.23.  A gramophone is driven by a Pickering governor . The mass of each disc
attached to the centre of a leaf spring is 20  g. The each spring is 5 mm wide and 0.125  mm thick . The

effective length of each spring is 40  mm. The distance from the spindle axis to the centre of gravity of
the mass when the governor is at rest , is 10 mm. Find the speed of the turntable when the sleeve has
risen 0.8 mm and the ratio of the governor speed to the turntable speed is 10.5 . Take E = 210 kN/mm2.
Solution. Given: m =20 g =  0.02 kg ; b = 5 mm ; t = 0.125 mm ; a = 10 mm = 0.01 m ;
E = 210 kN/mm2 = 210 × 103 N/mm2
We know that moment of inertia of the spring about its neutral axis,

33
–3 4 . 5 (0.125)0.8 10 mm12 12btI== = ×
Since the effective length of each spring is 40 mm and lift of sleeve ( λ) = 0.8 mm, therefore
Length of spring between fixed ends,
l= 40 – 0.8 = 39.2 mm
We know that the central deflection ( λ),
22
2 2.4 2.40.8 0.0639.2lδδ=== δ

700  /G6C   Theory of Machines
∴δ2= 0.8/0.06 = 13.3      or δ = 3.65 mm
Let N= Speed of the governor, and
N1= Speed of the turntable.
∴ N/N1= 10.5
We know that23.( )
192 .mal
EIω+ δδ=
23 2
2
3– 30.02 (10 3.65) (39.2) 16 4453.65 0.5132256 192 210 10 0.8 10ω+ ω== = ω
××× ×
∴23.657.156
0.51ω= = orω = 2.675 rad/s
and N=ω × 60/2π = 2.675 × 60/2 π = 25.5 r.p.m. Ans.
∴ N1=N/10.5 = 25.5/10.5 = 2.43 r.p.m.  Ans.
18.12. Sensitiveness of Governors

Consider two governors A and B running at the same speed. When this speed increases or
decreases by a certain amount, the lift of the sleeve of governor A is greater than the lift of the sleeve
of governor B. It is then said that the governor A is more sensitive than the governor B.
In general, the greater the lift of the sleeve corresponding to a given fractional change in

speed, the greater is the sensitiveness of the governor. It may also be stated in another way that for agiven lift of the sleeve, the sensitiveness of the governor increases as the speed range decreases. Thisdefinition of sensitiveness may be quite satisfactory when the governor is considered as an independentmechanism. But when the governor is fitted to an engine, the practical requirement is simply that thechange of equilibrium speed from the full load to the no load position of the sleeve

full load to the no load position of the sleeve should be as smalla fraction as possible of the mean equilibrium speed. The actual displacement of the sleeve is immaterial,provided that it is sufficient to change the energy supplied to the engine by the required amount. Forthis reason, the sensitiveness is defined as the ratio of the difference between the maximum and

minimum equilibrium speeds to the mean equilibrium speed.
Let N
1= Minimum equilibrium speed,
N2= Maximum equilibrium speed, and
N= Mean equilibrium speed 12.2NN+=
∴     Sensitiveness of the governor
21 21
122( – ) NN NN
NN N−==+
21
122( – )ωω=ω+ ω. . . (In terms of angular speeds)
18.13. Stability of Governors
A governor is said to be stable when for every speed within the working range there is a

definite configuration i.e. there is only one radius of rotation of the governor balls at which the
governor is in equilibrium. For a stable governor, if the equilibrium speed increases, the radius ofgovernor balls must also increase.
Note : A governor is said to be unstable, if the radius of rotation decreases as the speed  increases....(Given)

Chapter 18 : Governors    /G6C  701
18.14. Isochronous Governors
A governor is said to be  isochronous  when the equilibrium speed is constant ( i.e. range of
speed is zero) for all radii of rotation of the balls within the working range, neglecting friction. Theisochronism is the stage of infinite sensitivity.
Let us consider the case of a Porter governor running at speeds N
1 and N2 r.p.m. We have
discussed in Art. 18.6 that
2
1
1(1 )895 2()Mmq
N
mh++
=× . . . (i)
and2
2
2(1 )895 2()Mmq
N
mh++

N
mh++
=× . . . (i)
and2
2
N
mh++
=× . . . (ii)
For isochronism, range of speed should be zero i.e. N2 – N1 = 0 or N2 = N1. Therefore from
equations (i) and (ii), h1 = h2, which is impossible in case of a Porter governor. Hence a Porter
governor cannot be isochronous.
Now consider the case of a Hartnell governor running at speeds N1 and N2 r.p.m. We have
discussed in Art. 18.8 that
2
1
1C 1 12.2 260N xxMg S F m ryyπ+= ×= × × . . . (iii)
and2
2

and2
2
2C 2 22.2 260N xxMg S F m ryyπ+= × = × × . . . (iv)
For isochronism, N2 = N1. Therefore from equations (iii) and (iv),
11
22.
.Mg S r
Mg S r+=+
Note : The isochronous governor is not of practical
use because the  sleeve will move to one of its extremepositions immediately the speed deviates from the
isochronous speed.
18.15. Hunting
A governor is said to be hunt if the speed

of the engine fluctuates continuously above andbelow the mean speed. This is caused by a toosensitive governor which changes the fuel supplyby a large amount when a small change in thespeed of rotation takes place. For example, whenthe load on the engine increases, the engine speeddecreases and, if the governor is very sensitive,the governor sleeve immediately falls to its lowestposition. This will result in the opening of thecontrol valve wide which will supply the fuel to the engine in excess

will supply the fuel to the engine in excess of its requirement so that theengine speed rapidly increases again and the governor sleeve rises to its highest position. Due to thismovement of the sleeve, the control valve will cut off the fuel supply to the engine and thus the enginespeed begins to fall once again. This cycle is repeated indefinitely.

Such a governor may admit either the maximum or the minimum amount of fuel. The effect
of this will be to cause wide fluctuations in the engine speed or in other words, the engine will hunt.
A forklift is used to carry small loads from one
place to the other inside a factory.

702  /G6C   Theory of Machines
18.16. Effort and Power of a Governor
The effort of a governor  is the mean force exerted at the sleeve for a given percentage
change of speed * (or lift of the sleeve). It may be noted that when the governor is running steadily,

there is no force at the sleeve. But, when the speed changes, there is a resistance at the sleeve whichopposes its motion. It is assumed that this resistance which is equal to the effort, varies uniformlyfrom a maximum value to zero while the governor moves into its new position of equilibrium.
The power of a governor  is the work done at the sleeve for a given percentage change of

speed. It is the product of the mean value of the effort and the distance through which the sleevemoves. Mathematically,
Power = Mean effort × lift of sleeve
18.17. Effort and Power of a Porter Governor
The effort and power of a Porter governor may be determined as discussed below.Let N= Equilibrium speed corresponding to the configuration as shown in
Fig. 18.33 ( a), and
c= Percentage increase in speed.
∴Increase in speed = c.N
and increased speed = N + c.N = N (1 + c)

The equilibrium position of the governor at the increased speed is shown in Fig. 18.33 ( b).
Fig. 18.33
We have discussed in Art. 18.6 that when the speed is N r.p.m., the sleeve load is M.g.
Assuming that the angles α and β are equal, so that q = 1, then the height of the governor,
2895(in metres)mMh
mN+=× . . . (i)
When the increase of speed takes place, a downward force P will have to be exerted on the

sleeve in order to prevent the sleeve from rising. If the speed increases to (1 + c) N r.p.m. and the
height of the governor remains the same, the load on the sleeve increases to M1.g. Therefore
1
22895(in metres)
(1 )mMhm cN+=×
+. . . (ii)
Equating equations (i) and (ii), we have
1
2(1 )mMmM
c++=
+orM1 = (m + M) (1 + c2) – m
and M1 – M=(m + M) (1 + c)2 – m – M = (m + M) [(1 + c)2 – 1] . . . (iii)

*In comparing different types of governors, it is convenient to take the change of speed as one per cent.

Chapter 18 : Governors    /G6C  703
A little consideration will show that ( M1 – M)g is the downward force which must be applied
in order to prevent the sleeve from rising as the speed increases. It is the same force which acts on thegovernor sleeve immediately after the increase of speed has taken place and before the sleeve beginsto move. When the sleeve takes the new position as shown in Fig. 18.33 ( b), this force gradually
diminishes to zero.

Let P= Mean force exerted on the sleeve during the increase in speed or
the effort of the governor.
∴
2
1(– ) ( ) [ ( 1) – 1 ]
22MM gm M c gP++==
2() [ 1 2 – 1 ]()2mM c c gcm Mg++ +== + . . . (iv)
. . . (Neglecting c2, being very small)
If F is the frictional force (in newtons) at the sleeve, then
P=c (m.g + M.g ± F)
We have already discussed that the power of a governor is the product of the governor effort
and the lift of the sleeve.
Let x= Lift of the sleeve.

∴Governor power = P × x . . . (v)
If the height of the governor at speed N is h and at an increased speed (1 + c) N is h1, then
x=2  (h – h1)
As there is no resultant force at the sleeve in the two equilibrium positions, therefore
2895,mMhmN+=× and 1 22895,
(1 )mMh
m cN+=×
+
∴1
21
(1 )h
h c=
+or1 2(1 )hh
c=
+
We know that 1 2212( – ) 2 – 2 1–
(1 ) (1 )hxh h h h
cc == =  ++ 
2
212 – 1 22212 12cc chhc cc  ++== + ++  . . . (vi)

Substituting the values of P and x in equation  (v), we have
Governor power         224() 2 () .12 12cccm Mg h m Mg hcc=+ × = + ++. . . (vii)
Notes : 1.  If α is not equal to β, i.e. tan β / tan α = q, then the equations  (i) and (ii) may be written as
2(1 )895 2Mmq
hm N++
=× . . . (viii)
When speed increases to (1 + c) N and height of the governor remains the same, then
1
2(1 )895 2
(1 )Mmq
h
m cN++
=×
+. . . (ix)

704  /G6C   Theory of Machines
From equations (viii) and (ix), we have
1
2(1 )2(1 )
2 (1 )MmqMmq
c++
++ =
+
or2 1(1 ) (1 ) (1 ) –
22MMqm q cm+= + + +
∴2
2 1(1 )(1 ) –21 2 1Mm c M mcqq+=+ +++
or2
2 1 (1 )–( 1 ) – –22 1 2 1 2MM m c M mMcqq+=+ +++
22[ ( 1) – 1 ] [ ( 1) – 1 ]
12mMcc
q=+ + +
+
2[( 1 ) –1]12mMcq=+ ++
∴      Governor effort , 2 1–[1 2 – 1]21 2MM m MPg c c gq == + + +  +  
2(2 ) .12 1mM mcg Mc gqq =+ =+ ++ . . . (Neglecting c2)

The equation (vi) for the lift of the sleeve becomes,
2(1 )12cxq hc=+ +
∴        22Governor power . (1 )11 2mcPx Mc g q hqc  =×= + +   ++  
[]22242 ( 1) . ( 1).12 12 2cc MmM qg h m qg hcc=+ + = + +++ 
2. The above method of determining the effort and power of a Porter governor may be followed for any
other type of the governor.
Example 18.24.  A Porter governor has equal arms each 250  mm long and pivoted on the

axis of rotation . Each ball has a mass of 5  kg and the mass of the central load on the sleeve is 25  kg.
The radius of rotation of the ball is 150  mm when the governor begins to lift and 200  mm when the
governor is at maximum speed . Find the range of speed , sleeve lift , governor effort and power of the
governor in the following cases :
1.When the friction at the sleeve is neglected , and
2.When the friction at the sleeve is equivalent to 10  N.

Solution.  Given : BP = BD = 250 mm ; m = 5 kg ; M = 25 kg ; r1 =150 mm ; r2 = 200 mm ;
F = 10 N
1. When the friction at the sleeve is neglected
First of all, let us find the minimum and maximum speed of rotation. The minimum and
maximum position of the governor is shown in Fig. 18.34 ( a) and (b) respectively.
Let N1= Minimum speed, and
N2= Maximum speed.
From Fig. 18.34 ( a),
          22 22

Chapter 18 : Governors    /G6C  705
From Fig. 18.34 ( b),
22 22
2 ( ) – ( ) (250) – (200) 150 mm 0.15 m hP G B P B G== = = =
1
1895 5 25 895() 2 6 8 5 050 . 2mMNmh++=× = × =
∴ N1= 164 r.p.m.
and                                  2
2
2895 5 25 895( ) 3580050 . 1 5mMNmh++=× = × =
/G51                           N2 = 189 r.p.m.
Fig. 18.34
= N2 – N1 = 189 – 164 = 25 r.p.m. Ans.
Sleeve lift
We know that sleeve lift,

x=2  (h1 – h2) = 2 (200 – 150) = 100 mm = 0.1 m Ans.
Governor effort
Let c= Percentage increase in speed.
We know that increase in speed or range of speed,
c.N1=N2 – N1 = 25 r.p.m.
∴ c= 25/N1 = 25/164 = 0.152
We know that governor effort
P=c (m + M) g = 0.152 (5 + 25) 9.81 = 44.7 N Ans.

706  /G6C   Theory of Machines
Power of the governor
We know that power of the governor
= P.x = 44.7 × 0.1 = 4.47 N-m Ans.
2. When the friction at the sleeve is taken into account
1
5 9.81 (25 9.81 –10) 89525 93859 . 8 1 0 . 2×+ ×=× =×
∴                           N1 = 161 r.p.m.
and                                 2
2
                                          5 9.81 (25 9.81 10) 89537 01659 . 8 1 0 . 1 5×+ ×+=× =×

∴                            N2 = 192.4 r.p.m.
= N2 – N1 = 192.4 – 161 = 31.4 r.p.m. Ans.
The sleeve lift ( x) will be same as calculated above.
∴   Sleeve lift, x= 100 mm = 0.1 m Ans.
c.N1=N2 – N1 = 31.4 r.p.m.
∴ c= 31.4/N1 = 31.4/161 = 0.195
We know that governor effort,

P=c (m.g + M.g + F) = 0.195 (5 × 9.81 + 25 × 9.81 + 10) N
= 57.4 N Ans.
= P.x = 57.4 × 0.1 = 5.74 N-m Ans.
Example 18.25. The upper arms of a Porter governor has lengths 350  mm and are pivoted
on the axis of rotation . The lower arms has lengths 300  mm and are attached to the sleeve at a
distance of 40  mm from the axis . Each ball has a mass of 4  kg and mass on the sleeve is 45  kg.

Determine the equilibrium speed for a radius of rotation of 200  mm and find also the effort and
power of the governor for 1  per cent speed change .
Solution. Given : PB = 350 mm = 0.35 m ; BD = 300 mm = 0.3 m ; DE = 40 mm = 0.04 m ;
m = 4 kg ; M = 45 kg ; r = BG = 200 mm = 0.2 m ; c = 1% = 0.01

Chapter 18 : Governors    /G6C  707
Equilibrium speed
Let      N = Equilibrium speed.
The equilibrium position of the governor is shown in Fig.
18.35. From the geometry of the figure,
     22() – () hP G P B B G==
      22(0.35) – (0.2) 0.287 m==
      0.2tan 0.697
0.287BG
PGα= = =
∴ BH = BG – HG = 0.2 – 0.04 = 0.16 m
. . . (/G51 HG = DE)
and22() – ()DH BD BH=
                                     22(0.3) – (0.16) 0.254 m==
∴ tan β = BH/DH = 0.16 / 0.254 = 0.63

and     tan 0.630.904tan 0.697qβ===α
245(1 ) 4 (1 0.904)895 895 2236 5174 0.287Mmq
Nmh++ ++
=× = × =
∴ N = 191 r.p.m. Ans.
Effort of the governor
We know that effort of the governor,
22 40.01 45 9.81 4.8 N
11 0 . 9 0 4mPc Mg
q   ×=+ = +=  ++   Ans.
24(1 ) .12 2cMmq g hc=+ ++
24 (0.01) 454 (1 0.904) 9.81 0.287 0.052 N-m120 . 0 1 2=+ + × =+× 
= 52 N-mm Ans.

Example 18.26.  The radius of rotation of the balls of a Hartnell governor is 80  mm at the
minimum speed of 300  r.p.m. Neglecting gravity effect , determine the speed after the sleeve  has
lifted by 60  mm. Also determine the initial compression of the spring , the governor effort and the
power.
The particulars of the governor are given below :
Length of ball arm = 150 mm ; length of sleeve arm = 100 mm ; mass of each ball = 4 kg ; and

708  /G6C   Theory of Machines
Solution. Given : r1 = 80 mm = 0.08 m ; N1 = 300 r.p.m. or ω1 = 2π × 300/60 = 31.42 rad/s ;
h = 60 mm = 0.06 m ; x = 150 mm = 0.15 m ; y = 100 mm = 0.1 m ; m = 4 kg ; s = 25 N/mm
Fig. 18.36
The minimum and maximum position of the governor is shown in Fig. 18.36 ( a) and (b)
respectively. First of all, let us find the maximum radius of rotation ( r2). We know that lift of the
sleeve,
21(–)yhrr
x=
or 210.150.08 0.06 0.17 m0.1xrrhy=+×= + × =. . . (/G51h = h1 + h2)

Maximum speed of rotation
Let N2= Maximum speed of rotation, and
S1 and S2= Spring force at the minimum and maximum speed respectively, in
newtons.
We know centrifugal force at the minimum speed,
FC1=m (ω1)2 r1 = 4 (31.42)2 0.08 = 316 N
Now taking moments about the fulcrum O of the bell crank lever when in minimum position
as shown in Fig 18.36 ( a). The gravity effect is neglected, i.e. the moment due to the weight of balls,
sleeve and the bell crank lever arms is neglected.
∴1
C1.

∴1
C1.
2Mg SFx y+×= × or 1C 10.152 2 316 948 N0.1xSFy=× = × ×=
            ...(/G51M =0)
We know that  S2 – S1 = h.s   or    S2 = S1 + h.s = 948 + 60 × 25 = 2448 N
We know that centrifugal force at the maximum speed,
        22
2 2 22
C1 2 2 2 222( ) 0.17 0.00746 ( )60 60NNFm r r m Nππ  =ω = = =     
Now taking moments about the fulcrum O when in maximum position, as shown in Fig. 18.36 ( b),
2
C2.
2Mg SFx y+×= ×
2

2
C2.
2
224480.00746 ( ) 0.15 0.12N =× or 0.001 12 ( N2)2 = 122.4 . . . (/G51M = 0)
2
2122.4( ) 109 2860.00112N== orN2 = 331 r.p.m. Ans.

Chapter 18 : Governors    /G6C  709
        194837.92 mm
25S
s== =  Ans.
We know that the governor effort,
        21– 2448 – 948750 N22SSP== =  Ans.
Governor power
We know that the governor power
      = P × h = 750 × 0.06 = 45 N-m Ans.
Example 18.27. In a Hartnell governor , the lengths of ball and sleeve arms of a bell crank

lever are 120  mm and 100  mm respectively . The distance of the fulcrum of the bell crank lever from
the governor axis is 140  mm. Each governor ball has a mass of 4  kg. The governor runs at a mean
speed of 300  r.p.m. with the ball arms vertical and sleeve arms horizontal . For an increase of speed
of 4 per cent, the sleeve moves 10  mm upwards . Neglecting friction , find :
1. the minimum equilibrium speed if the total sleeve movement is limited to 20 mm, 2. the

spring stiffness , 3. the sensitiveness of the governor , and 4. the spring stiffness if the governor is to be
isochronous at 300  r.p.m.
Solution.  Given : x = 120 mm = 0.12 m ; y = 100 mm = 0.1 m ; r = 140 mm = 0.14 m ;
m = 4 kg ; N = 300 r.p.m. or ω = 2π × 300/60 = 31.42 rad/s ; h1 = 10 mm = 0.01 m ; h = 20 mm = 0.02 m
1. Minimum equilibrium speed
Let N1= Minimum equilibrium speed,
r1= Radius of rotation in the minimum position, i.e. when the sleeve
moves downward, and

r2= Radius of rotation in the maximum position, i.e. when the sleeve
moves upward.
Since the increase in speed is 4%, therefore maximum speed,
N2=N + 0.04 N = 1.04 N = 1.04 × 300 = 312 r.p.m.
or ω2=2π × 312 / 60 = 32.7 rad/s
We know that lift of the sleeve for the maximum position,
h2=h – h1 = 0.02 – 0.01 = 0.01 m
Now for the minimum position,
11–hr r
yx= or110.12– 0 . 1 4–0 . 0 1 0 . 1 2 8m0.1xrrhy=× = × =
Similarly for the maximum position,
22–hrr

22–hrr
yx= or220.120.14 0.01 0.152 m0.1xrr hy=+ ×= + × =
We know that centrifugal force in the mean position,
FC=m.ω2.r = 4 (31.42)2 0.14 = 553 N
Centrifugal force in the minimum position,
2
22 1
C1 1 2 12( ) 4 0.128 0.0056 ( )
60NFm r Nπ=ω = = . . . (i)

710  /G6C   Theory of Machines
and centrifugal force in the maximum position,
FC2=m.(ω2)2 r2 = 4 (32.7)2 0.152 = 650 N
We know that centrifugal force at any instant,
1
CC 1 C 2C 1

C1 C1 10.14 – 0.128553 (650 – ) 0.5 325
0.152 – 0.128C FF F=+ = + 

∴ C1553 – 325456 N
0.5F== ...(ii)
2
1456() 8 1 4 2 80.0056N== orN1 = 285.4 r.p.m. Ans.
2. Spring stiffness
LetS1 and S2= Spring force at the minimum and maximum position.

Neglecting the effect of obliquity of arms, we have for the minimum position,
1
C1.
2Mg SyF x+×= × or1C 10.122 2 456 1094.4 N0.1xSFy=× = × ×=
. . . (/G51M = 0)
and for the maximum position,
2
C2.
2Mg SyF x+×= × or 2C 20.122 2 650 1560 N0.1xSFy=× = × ×=
We know that spring stiffness,
21–1560 –1064.423.28 N/mm
20SSs
h== =  Ans.
3. Sensitiveness of the governor
We know that sensitiveness of the governor
21
122 ( – ) 2 (312 – 285.4)0.089 or 8.9%285.4 312NN
NN== =++ Ans.

4. Spring stiffness for the governor to be isochronous at 300 r.p.m.
The governor is isochronous, when N = N1 = N2 = 300 r.p.m. or ω = ω1 = ω2 =  31.42 rad/s
∴ FC1 = m.ω2.r1 = 4 (31.42)2 0.128 = 505.5 N
and FC2 = m.ω2.r2 = 4 (31.42)2 0.152 = 600 N
We know that 1C 10.122 2 505.5 1213 N0.1xSFy=× = ×× =
and2C 20.122 2 600 1440 N0.1xSFy=× = × × =
∴ Spring stiffness,21–1440 – 121311.35 N/mm
20SSs
h== =  Ans.

Chapter 18 : Governors    /G6C  711
18.18. Controlling Force
We have seen earlier that when a body rotates in a circular
path, there is an inward radial force or centripetal force acting on it.In case of a governor running at a steady speed, the inward forceacting on the rotating balls is known as controlling force . It is equal
and opposite to the centrifugal reaction.
∴   Controlling force,  F
C  =  m.ω2.r
The controlling force is provided by the weight of the sleeve

and balls as in Porter governor and by the spring and weight as inHartnell governor (or spring controlled governor).
When the graph between the controlling force ( F
C) as
ordinate and radius of rotation of the balls ( r) as abscissa is drawn,
then the graph obtained is known as controlling force diagram . This diagram enables the stability
and sensitiveness of the governor to be examined and also shows clearly the effect of friction.
18.19. Controlling Force Diagram for Porter Governor

The controlling force diagram for a Porter governor is a curve as shown in Fig. 18.37. We
know that controlling force,
2
2
C2..60NFmr m rπ=ω= 
or22
2 C 16 0 16 0(tan )22FNmr m == φ ππ C... tanF
r=φ/G51
∴1/260 tan
2Nmφ=π. . . (i)
where φ is the angle between the axis of radius of rotation and a line joining a given point (say A) on
the curve to the origin O.

Notes : 1.  In case the governor satisfies the condition for stability, the angle φ must increase with radius of
rotation of the governor balls. In other words, the equilibrium speed must increase with the increase of radius ofrotation of the governor balls.
2. For the governor to be more sensitive, the change in the value of φ over the change of radius of
rotation should be as small as possible.

3. For the isochronous governor, the controlling force curve is a straight line passing through the
origin. The angle φ will be constant for all values of the radius of rotation of the governor. From equation  (i)
2 2
22 C.. 2tan . .N60Fmr Nmm Crrωπ φ= = = ω = = 
where22constant60Cmπ==
Using the above relation, the angle φ may be determined for different values of N and the lines are drawn from

the origin*. These lines enable the equilibrium speed corresponding to a given radius of rotation to be determined.
Alternatively, the same results may be obtained more simply by setting-off a speed scale along any arbitrarilychosen ordinate. The controlling force is calculated for one constant radius of rotation and for different arbitrarilychosen values of speed. The values thus obtained are set-off along the ordinate that corresponds to the chosen

radius and marked with the appropriate speeds.Fig. 18.37.  Controlling
force diargram.
*See Example 18.28, Fig. 18.39.

712  /G6C   Theory of Machines
Example 18.28.  In a Porter governor , the length of each arm is 300  mm and all the arms are
pivoted on the axis of rotation . The mass of each ball is 7.5  kg and the mass of the sleeve  is 45  kg.
The extreme radii of rotation are 150  mm and 225  mm. Draw the controlling force curve and set-off
a speed scale along the ordinate corresponding to a radius of 250  mm.
Solution. Given : l = 300 mm = 0.3 m ; m = 7.5 kg ; r1 = 150 mm = 0.15 m ; r2 = 225 mm
= 0.225m

= 0.225m
Let FC = Controlling force.
We have discussed in Art 18.6 that
C...t a n – t a n22Mg Mgmg F+α = × β
∴ C...t a n t a n22Mg MgFm g=+ α + × β
=.. t a n.t a n
22 t a nMg Mgmg β++ × αα 
tanSubstituting , and tan , we gettanrqhβ=α =α
C.(1 ) .2Mg rFq m gh=+ +
Since α = β as shown in Fig. 18.38, therefore q = 1.
∴                      C22(. .) ( )
–rrFm g M g g m Mhlr=+ =+ ×22... –hlr=/G51
The following table shows the values of FC for different values of r.

r (in metres) 0.025 0.05 0.075 0.1 0.125 0.15 0.175 0.2 0.225 0.25
22–hlr= 0.2894 0.2958 0.2905 0.2828 0.2727 0.2598 0.2437 0.2236 0.1985 0.1658
FC (in newtons) 44.5 87 133 182 236 297 370 461 584 776
These values are plotted to draw the controlling force curve as shown in Fig. 18.39. In order
to set-off the speed scale along the ordinate through r = 250 mm = 0.25 m, we have
2
22
C2. . 7.5 0.25 0.02
60NFmr Nπ=ω= = 

2
22
The values of FC for different values of N are given in the following table.
N 100 125 150 160 170 180 190 200
(in r.p.m.)
FC200 312.5 450 512 578 648 722 800
(in newtons)
The speed scale is now marked on the graph as shown in Fig. 18.39.Fig. 18.38

Chapter 18 : Governors    /G6C  713
Fig. 18.39
The range of equilibrium speeds for the governor is obtained by drawing lines from the origin
(shown dotted in Fig. 18.39 ) through the two points A (when r = 0.15 m ) and B ( when r = 0.225 m)
on the controlling force curve.
From the graph, we see that these lines intersect the speed scale at approximately 160 r.p.m.
and 180 r.p.m. Ans.
18.20. Controlling Force Diagram for Spring-controlled Governors

The controlling force diagram for the spring controlled governors is a straight line, as shown
in Fig. 18.40. We know that controlling force,
FC = m.ω2.rorFC /r = m.ω2
The following points, for the stability of spring-controlled governors, may be noted :
1. For the governor to be stable, the controlling force ( FC) must increase as the radius of rotation
(r) increases, i.e. FC/r must increase as r increases. Hence the
controlling force line AB when produced must intersect the

controlling force axis below the origin, as shown in Fig. 18.40.
The relation between the controlling force ( FC) and the
radius of rotation ( r) for the stability of spring controlled
governors is given by the following equation
FC = a.r – b . . . (i)
where a and b are constants.
2. The value of b in equation ( i) may be made either
zero or positive by increasing the initial tension of the spring. Ifb is zero, the controlling force line CD passes through the origin

and the governor becomes isochronous  because F
C /r will
remain constant for all radii of rotation.
The relation between the controlling force and the radius of rotation, for an isochronous
governor is, therefore,
FC = a.r . . . (ii)
3. If b is greater than zero or positive, then FC /r decreases as  r  increases, so that the equilibrium
speed of the governor decreases with an increase of the radius of rotation of balls, which is impracticable.Fig. 18.40

714  /G6C   Theory of Machines
Such a governor is said to be unstable and the relation between the controlling force and the radius of
rotation is, therefore
FC = a.r + b . . . (iii)
Example 18.29.  The particulars of a governor of
the type as shown in Fig. 18.41, are as follows :
The mass of each ball is 1.5  kg and the mass of the
sleeve is 7.5  kg. The lengths of the ball arm and sleeve arm
of the bell crank lever are 112.5  mm and 50  mm respectively

and are at right angles to each other . The extreme radii of
rotation are 62.5  mm and 112.5  mm. At the minimum radius ,
the ball arm is vertical and the spring load is 160  N. The
spring stiffness is 10.5  N/mm. Draw the controlling force
curve and mark the speed scale along the ordinate through125 mm.
Solution. Given : m = 1.5 kg ; M = 7.5 kg ;
x = 112.5 mm = 0.1125 m ;  y = 50 mm = 0.05 m ; r
1 = 62.5 mm
= 0.0625 m ; r2 = 112.5 mm = 0.1125 m ; S1 = 160 N ;
s = 10.5 N/mm = 10 500 N/m

Let S1 and S2 be  the  spring  loads  at  the minimum  and  maximum  radius  of  rotation.
The minimum and maximum position of the balls is shown in Fig. 18.42 ( a) and (b)
respectively. Taking moments about the instantaneous centre I, for the maximum position as shown
in Fig. 18.42 ( b),
Fig. 18.42
2
C2..2Mg SFB D m g I D I O+×= × + ×
or2
C2.cos . ( sin cos ) cos2Mg SFx m g x y y+×θ = θ +θ + ×θ
∴2
C2..t a n2Mg SyyFm gxx+ =θ + + ×. . . (i)

S2 – S1=h.s = y sin θ × s
∴ S2=S1 + y sin θ × s = 160 + 0.05 × sin θ × 10 500
= 160 + 525 sin θ . . . (ii)Fig. 18.41

Chapter 18 : Governors    /G6C  715
Now the equation  (i) may be written as
           C20.05 7.5 9.81 160 525 sin 0.051.5 9.81 tan0.1125 2 0.1125F×++ θ =× θ + + × 
                              = 14.7 tan θ + 116.5 sin θ + 58.5      ... (iii)
From the geometry of Fig. 18.42 ( b), we find that
                         r2 = r1 + OD = r1 + x sin θ = 0.0625 + 0.1125 sin θ     ...(iv)
In order to determine the controlling force and the radius of rotation of the ball for different

values of θ, the angle θ is treated as variable. From equations  (iii) and (iv), the values of controlling
force (FC) and radius of rotation ( r) for different values of θ are tabulated below :
θ° 0 51 0 1 52 0 2 53 0
sin θ 0 0.0871 0.1736 0.2588 0.342 0.4226 0.5
tan θ 0 0.0875 0.1763 0.2679 0.364 0.4663 0.5773
FC (N) 58.5 70 81.3 92.6 103.7 114.6 125.2
r (m) 0.0625 0.0723 0.082 0.092 0.101 0.11 0.1187
The graph between  FC and r is plotted as shown in Fig. 18.43. It may be seen that the

716  /G6C   Theory of Machines
In order to set off the speed scale along the ordinate through r = 125 mm = 0.125 m,
2
22
C2. . 1.5 0.125 0.0020660NFmr Nπ=ω= = 
The corresponding values of FC and N are given in the following table :
N (r.p.m.) 200 210 220 230 240 250 260
FC (N) 82.4 90.8 99.7 109 118.6 128.7 139.2
The speed scale is now marked on the graph as shown in Fig. 18.43. The range of equilibrium

speeds for the governor is obtained by drawing lines from the origin (shown dotted in Fig. 18.43)through the two points A (when r = 0.0625 m) and B (when r = 0.1125 m) on the controlling force curve.
From the graph we see that these lines intersect the speed scale at approximately 238 r.p.m.
and 252 r.p.m. Ans.
18.21. Coefficient of Insensitiveness
In the previous articles, we have assumed the governor to be frictionless. In actual practice,

there is always friction in the joints and operating mechanism of the governor. Since the frictionalforce always acts in the opposite direction to that of motion, therefore, when the speed of rotationdecreases, the friction prevents the downward movement of the sleeve and the radial inward movementof the balls. On the other hand, when the speed of rotation increases, the friction prevents the upwardmovement of the sleeve and radial outward movement of the balls.
Fig. 18.44

Fig. 18.44
Let FS= Force required at the sleeve to overcome friction,
FB= Corresponding radial force required at each ball,
FC= Controlling force on each ball, and
W= Total load on the sleeve = M.g.
∴ For decrease in speed, sleeve load (taking friction into account),
W1=W – FSorM1.g = M.g – FS
and for increase in speed, sleeve load (taking friction into account),
W2=W + FSorM1.g = M.g + FS
Similarly, for decrease in speed, controlling force,
FC1=FC – FB

FC1=FC – FB
and for increase in speed, controlling force,
FC2=FC + FB

Chapter 18 : Governors    /G6C  717
Thus for a Porter governor, as shown Fig. 18.44 ( a), the relation between FS and FB may be
obtained by taking moments about the instantaneous centre I.
∴              S
B ()2FFB M I MM D×= +
orSS S
Btan(tan tan ) 1 tan
22 2 t a nFF FIM MDF
BM +β== α + β = + α  α  
                                         SS(1 ) tan (1 )
22FF rqq
h=+ α =+ . . . (i)
tan... , and tantanrqh β=α =α /G51

Similarly, for spring loaded governors as shown in Fig. 18.44 ( b), taking moments about the
fulcrum O of the bell crank lever,
S
B2FFx y×= × or BS2yFFx=× . . . (ii)
Fig. 18.45 shows the effect of friction on the
controlling force diagram. We see that for one value of theradius of rotation ( i.e. OA), there are three values of controlling
force as discussed below:
1.  For speed decreasing, the controlling force reduces
to F

to F
C1 (or AD) and the corresponding speed on the speed scale
is N'.
2.  For speed increasing, the controlling force
increases to FC2 (or AC) and the corresponding speed on the
speed scale is N'.
3.  For friction neglected, the controlling force is
FC ( or AB) and the corresponding speed on the speed scale
is N.
From above, it is concluded that when the radius of rotation is OA, the speed of rotation may

vary within the limits N' and N'' without causing any displacement (up or down) of the governor
sleeve. The governor is said to be  insensitive  if the speed fluctuates over this range.
The ratio –NN
N′′ ′
 is called the coefficient of insensitiveness  of the governor.
Since the controlling force is proportional to the square of the speed at a given radius, therefore
for a governor speed N,
FC  ∝ N2orFC=C.N2...(iii)
Similarly, for speed N',
FC1=C (N')2...(iv)

and for speed N'', FC2=C (N'')2...(v)
Subtracting equation (iv) from equation  (v), we have
FC2 – FC1=C [(N'')2 – (N')2]
or ( FC + FB) – (FC – FB)=C [(N'')2 – (N')2]
2FB=C [(N'')2 – (N')2] ... (vi)Fig. 18.45.  Effect of friction
on controlling force.

718  /G6C   Theory of Machines
Dividing equation  (vi) by equation  (iii)
            22
B
22
C2( ) – ( ) ( ) ( – ) –FN N N N N NN N N N
FN N NN′′ ′ ′′ ′ ′′ ′ ′′ ′ ′′ ′ ++== = ×
Since 
2NN′′ ′+ is approximately equal to N, therefore
B
C2–2FN N
FN′′ ′=×
∴  Coefficient of insensitiveness
B
C–F NN
NF′′ ′==. . . (vii)
Notes : 1 In case of a Porter governor, as shown in Fig. 18.44 ( a), (i.e. when the lower arm is not attached on the
governor axis),
S
B(1 )2FrFqh=+
∴   Coefficient of insensitiveness

BS
CC–(1 )2NNF F rqNFF h′′ ′== = +. . . (viii)
2. When all the arms of a Porter governor are attached to the governor axis, then q = 1. In that case,
BSrFFh=×
BS
CC–NNFFr
NF F h′′ ′== = ×. . . (ix)
3. In case of a Porter governor when all the arms are attached to the governor axis, the coefficient of
insensitiveness may also be determined as discussed below :

Let h= Height of the governor at the mean speed N, when friction is neglected,
F= Frictional force on the sleeve,
N' and N''= Minimum and maximum speed when friction is taken into account.
We have discussed above that the governor is insensitive when the sleeve does not move downwards
when the speed falls to N' or upwards when the speed rises to N'' . In other words, the height of the governor ( h)
remains the same for minimum and maximum speeds N' and N'' respectively. We know that

2 895 mMNmh+=×
Similarly,2. ( . – ) 895().mg Mg FNmg h+′=×
and2.( . ) 8 9 5()
.mg Mg FN
mg h++′′=×
Now ( N'')2 – (N')2 = (N'' + N') (N'' – N') = 2 N (N'' – N') ...2NNN′′ ′+ =/G51
∴22() – ()–2NNNNN′′ ′′′ ′=

Chapter 18 : Governors    /G6C  719
and coefficient of insensitiveness
                                      22
2.( . ) .( . – )––( ) – ( ) ..
22mg Mg F mg Mg F
NN N N mg mg
mM N N
m++ +
′′ ′ ′′ ′== =+
                                              12
2( ) ( )FF
mM g mM g==++        ...(x)
4. In case of a Porter governor, when the upper arms are pivoted to the governor axis and the lower

arms are  at a certain distance from the governor axis i.e. when α is not equal to β (Refer Art. 18.6), then it may
be proved that
Coefficient of insensitiveness
                               –( 1 )
2. .( 1 )NN F q
Nm g M g q′′ ′ +=
++
5. In case of a Hartnell governor,
                                        BS2yFF
x=×
∴  Coefficient of insensitiveness BS
CC–
2NNFFy
NF F x′′ ′== = × ...(xi)
Example 18.30.  A Porter governor has equal arms 200  mm long pivoted on the axis of

rotation. The mass of each ball is 3  kg and the mass on the sleeve is 15  kg. The ball path is 120  mm
when the governor begins to lift and 160  mm at the maximum speed . Determine the range of speed .
If the friction at the sleeve is equivalent to a force of 10  N, find the coefficient of insensitiveness .
Solution. Given : BP = BD = 200 mm = 0.2 m ; m = 3 kg ; M = 15 kg ; r1 = 120 mm = 0.12 m ;
r2 = 160 mm = 0.16 m ; F = 10 N

First of all, let us find the minimum and maximum speed of rotation.
Fig. 18.46
The minimum and maximum position of the balls is shown in Fig 18.46 ( a) and (b) respectively.
Let N1 = Minimum speed, and
N2 = Maximum speed.

720  /G6C   Theory of Machines
From Fig. 18.46 ( a),22 2 2
1( ) –( ) ( 0 . 2 ) –( 0 . 1 2 ) 0 . 1 6mhB PB G== =
and from Fig. 18.46 ( b),22 2 2
2( ) – ( ) (0.2) – (0.16) 0.12 mhB PB G== =
We know that  2
1
1895 3 15 895( ) 3356330 . 1 6mMNmh++=× = × =
∴ N1 = 183.2 r.p.m.
Similarly2
2
1895 3 15 895() 4 4 7 5 030 . 1 2mMNmh++=× = × =
∴ N2 = 211.5 r.p.m.
= N2 – N1 = 211.5 – 183.2 = 28.3 r.p.m. Ans.

We know that coefficient of insensitiveness,
                   –1 00.0566 5.66%() ( 3 1 5 ) 9 . 8 1NN F
Nm M g′′ ′== = =++ Ans.
Example 18.31.  The following particulars refer to a Proell governor with open arms :
Length of all arms = 200 mm, distance of pivot of arms from the axis of rotation = 40 mm,
length of extension of lower arms to which the ball is attached = 100 mm, mass of each ball = 6 kg

and mass of the central load = 150 kg. If the radius of rotation of the balls is 180  mm when the arms
are inclined at 40 ° to the axis of rotation , find :
1. the equilibrium speed for the above configuration , 2. the coefficient of insensitiveness if
the friction of the governor mechanism is equivalent to a force of 20  N at the sleeve , and 3. the range
of speed between which the governor is inoperative .
Solution. Given : PF = FD = 200 mm = 0.2 m ; DK = 40 mm = 0.04 m ; BF = 100 mm

= 0.1 m ; m = 6 kg ; M = 150 kg ; r = JG = 180 mm = 0.18 m ; F = 20 N
1. Equilibrium speed
Let N= Equilibrium speed.
From the equilibrium position, as shown in Fig 18.47, we find that the height of the governor,
h=PH = PF cos 40°
= 0.2 × 0.766 = 0.1532 m
and FH=PF sin 40°
= 0.2 × 0.643 = 0.1286 m
∴ JF=JG – HG – FH
= 0.18 – 0.04 – 0.1286
= 0.0114 m
and22() – ()BJ BF JF=
22(0.1) – (0.0114)=
= 0.0993 m
BM=BJ + JM

= 0.0993 m
BM=BJ + JM
= 0.0993 + 0.1532 = 0.2525 m . . . ( /G51 JM = MD = PH)Fig. 18.47All dimensions in mm.

Chapter 18 : Governors    /G6C  721
 IM = IN – MN = FH – JF = 0.1286 – 0.0114 = 0.1172 m
  ID = 2 × IN = 2 × FH = 2 × 0.1286 = 0.2572 m
        2
22
C2. . 6 0.18 0.01260NFmr Nπ=ω= = 
        C..2MgFB M m g I M I D×=× + ×
                  2 150 9.810.012 0.2525 6 9.81 0.1172 0.25722N××= × ×+ ×
or                        0.003 03 N2 = 6.9 + 189.2 = 196.1

∴                          N2 = 196.1/0.003 03 = 64 720 or N = 254.4 r.p.m. Ans.
Let N' and N'' = Minimum and maximum speed considering friction.
2
22
C2( ) 6 0.18 0.012 ( )60NFm r N′π′′′=ω = = 
2
22
C2( ) 6 0.18 0.012 ( )60NFm r N′′π ′′ ′′ ′′=ω = = 
Taking moments about I, when sleeve moves downwards,
C..2Mg FFB M m g I M I D−′×=× + ×

               2 150 9.81 200.012( ) 0.2525 6 9.81 0.1172 0.25722N×−′= × × + ×
0.003 03 ( N')2= 6.9 + 186.7 = 193.6
∴ (N')2= 193.6/0.003 03 = 63 894 or N' = 252.8 r.p.m.
Again taking moments about I, when the sleeve moves upwards,
C..2Mg FFB M m g I M I D+′′×=× + ×
2 150 9.81 200.012 ( ) 0.2525 6 9.81 0.1172 0.25722N×+′′ =× × + ×
0.003 03 ( N'')2= 6.9 + 191.8 = 198.7
∴ (N'')2= 198.7/0.003 03 = 65 578 or N'' = 256 r.p.m.

– 256 – 252.80.0126 or 1.26%
254.4NN
N′′ ′==  Ans.
3. Range of speed
= N'' – N' = 256 – 252.8 = 3.2 r.p.m. Ans.

722  /G6C   Theory of Machines
Example 18.32.  A spring controlled governor is shown in Fig . 18.48. The central spindle
does not move axially . The mass of the sleeve is 20  kg and the frictional resistance to its movement is
equivalent to 20  N. The balls attached to the right angled bell crank levers have mass 4  kg each. The
stiffness of the spring is 40  N/mm compression . The radius of rotation of the balls is 125  mm when the

sleeve is in its lowest position , and the ball arms are vertical and the spring exerts a force of 600  N.
Determine :
1.  the speed at which the sleeve will begin to rise from its lowest position,
2.  the range of speed when the sleeve is 12.5 mm above its lowest position, and
3.  the coefficient of insensitiveness at higher speed .
Solution. Given : M = 20 kg ; F = 20 N ; m = 4 kg ; s = 40 N/mm ; r1 = 125 mm = 0.125 m ;
S1 = 600 N

S1 = 600 N
1. Speed at which sleeve will begin to rise from its lowest position
Let N1= Required speed.
The lowest position is shown in Fig. 18.49 ( a).
We know that the centrifugal force at the lowest position,
2
22 1
C1 1 1 12( ) 4 0.125 0.0055 ( )60NFm r Nπ=ω = = 
Fig. 18.48 Fig. 18.49
Since the sleeve is about to rise, therefore frictional resistance is taken positive. Also the
central spindle is stationary, therefore all the forces are transferred to both the pivots of the bell crank

lever i.e. 1.
2Mg S F++ at each pivot, as shown in Fig. 18.49 ( a).
Since the pivot O moves vertically and the roller A moves horizontally, therefore A is the
instantaneous centre of the bell crank lever.
Now taking moments about A,
1
C1..2Mg S FFO Bm g O A++ ×= +
2
120 9.81 600 200.0055 ( ) 0.19 4 9.81 0.0852N×++ =× +
0.001 05 ( N1)2= 3.3 + 34.7 = 38 or ( N1)2 = 38 / 0.001 05 = 36 190
∴ N1= 190 r.p.m. Ans.

Chapter 18 : Governors    /G6C  723
2. Range of speed
The highest position is shown in Fig. 18.49 ( b).
Let                       N2 =Maximum speed,
                            h  = Lift of the sleeve = 12.5 mm = 0.0125 m              . . . (Given)
                              r2 = Maximum radius of rotation of the balls, and
                             S2 =  Maximum spring force.
We know that lift of the sleeve
                                   21 21(–) (–)yO Ahrr rr

xO B=× =×  . . (Here x = OB, and y = OA)
∴                          r2 = r1 + h × OB/OA = 0.125 + 0.0125 × 0.19/0.085 = 0.153 m
We know that centrifugal force at the highest position,
                                22 2
C2 2 2 22( ) 4 0.153 0.0067 ( )60NFm r Nπ=ω = = 
and                            S2 – S1  =h.sorS2 = S1 + h.s = 600 + 12.5 × 40 = 1100 N
From Fig 18.49 ( b), we find that
                                 22 2 2( ) – ( ) (85) – (12.5) 84 mm 0.084 mOI OA AI=== =

. . . (/G51 AI = h)
22 2 2( ) – ( ) (190) – (153 – 125) 188 mm 0.188 mBC OB OC== = =
. . . (/G51OC = r2 – r1)
and                                IC  = OI + OC = 84 + (153 – 125) = 112 mm = 0.112 m
                      2
C2..2Mg S FFB C m g I C O I+±×= × + ×
The ± sign denotes that at the highest position, the sleeve may either rise or fall. Therefore
     2
220 9.81 1100 200.0067 ( ) 0.188 4 9.81 0.112 0.0842N×+ ±=× × + 

0.001 26 ( N2)2 = 4.4 + 54.4 ± 0.84 = 58.8 ± 0.84
Taking – ve sign, when the sleeve is about to fall,
              0.001 26 ( N2)2 = 58.8 – 0.84 = 57.96
                           ( N2)2 = 57.96/0.001 26 = 46 000 or N2 = 214.5 r.p.m.
Taking + ve sign, when the sleeve is about to lift,
            0.001 26 ( N2')2 = 58.8 + 0.84 = 59.64
                           ( N2')2 = 59.64/0.001 26 = 47 333 or N2' = 217.5 r.p.m.
∴  Range of speed at the maximum radius

                                     = N2' – N2 = 217.5 – 214.5 = 3 r.p.m. Ans.

724  /G6C   Theory of Machines
3. Coefficient of insensitiveness at higher speed
We know that coefficient of insensitiveness
22
222( – ) 2 (217.5 – 214.5)0.014 1.4%217.5 214.5NN
NN′
== = =+ ′+ Ans.
Example 18.33.  In a spring controlled governor , the curve of controlling force is a straight
line. When balls are 400  mm apart , the controlling force is 1200  N and when 200  mm apart , the
controlling force is 450  N. At what speed will the governor run when the balls are 250  mm apart ?

What initial tension on the spring would be required for isochronism and what would then be thespeed ? The mass of each ball is 9  kg.
Solution. Given : When balls are 400 mm apart, i.e. when the radius of rotation ( r
2) is 200
mm, the controlling force,
FC2= 1200 N
When balls are 200 mm apart i.e. when the radius of rotation ( r1) is 100 mm, the controlling
force,
FC1= 450 N
Mass of each ball, m=9  k g

force,
FC1= 450 N
Speed of the governor when the balls are 250 mm apart, i.e. when radius of rotation (r) is 125 mm
Let N= Required speed.
We know that for the stability of the spring controlled governors, the controlling force ( FC) is
expressed in the form
* FC=a.r – b . . . (i)
When r=r1 = 100 mm = 0.1 m, then
450 =a × 0.1 – b = 0.1 a – b . . .(ii)
and when r=r2 = 200 mm = 0.2 m, then
1200 =a × 0.2 – b = 0.2 a – b . . . (iii)

From equations (ii) and (iii), we find that
a= 7500, and b = 300
Now the equation (i) may be written as
FC= 7500 r – 300 . . . (iv)
Substituting  r= 125 mm = 0.125 m, in equation (iv), we get
FC= 7500 × 0.125 – 300 = 637.5 N
2
2
2 2637.5 9 0.125 0.012 34
60NNπ==
∴ N2= 637.5/0.012 34  = 51 661 or N = 227.3 r.p.m. Ans.
*We find that C1 C2
12450 12004500 and 6000.0.1 0.2FF
rr== = =

rr== = =
Since FC/r  increases as r increases, therefore for stability
FC = a.r – b(See Art. 18.20)

Chapter 18 : Governors    /G6C  725
Initial tension on the spring for isochronism
We have discussed in Art. 18.20 that for an isochronous governor, the controlling force line
passed through the origin ( i.e. b = 0). The value of b is made zero by increasing the initial tension of
the spring to 300 N.
∴   Initial tension on the spring for isochronism = 300 N Ans.
Isochronous speed
Let N'= Isochronous speed, and
FC'= Controlling force at the isochronous speed.
We know that for isochronism,

FC'=a.rorm (ω')2 r = a.rorm (ω')2 = a
∴22
60Nma′π=      or 9 × 0.011 ( N')2 = 7500
(N)2= 7500 / 0.099 = 75 758 or N' = 275 r.p.m. Ans.
Example 18.34.  The controlling force (FC) in newtons and the radius of rotation (r) in
metres for a spring controlled governor is given by the expression
FC=2800 r – 76
The mass of the ball is 5 kg and the extreme radii of rotation of the balls are 100  mm and

175 mm. Find the maximum and minimum speeds of equilibrium . If the friction of the governor
mechanism is equivalent to a force of 5  N at each ball , find the coefficient of insensitiveness of the
governor at the extreme radii .
Solution. Given : m = 5 kg ; r1 = 100 mm = 0.1 m ; r2 = 175 mm = 0.175 m
Maximum and minimum speeds of equilibrium
Let N2 and N1= Maximum and minimum speeds of equilibrium respectively.
The controlling force is given by the expression,
FC= 2800 r – 76

∴  Controlling force at the minimum radius of rotation ( i.e. at r1 = 0.1 m),
FC1= 2800 × 0.1 – 76 = 204 N
and controlling force at the maximum radius of rotation ( i.e. at r2 = 0.175 m),
FC2= 2800 × 0.175 – 76 = 414 N
2 1
C1 1 1 12()
60NFm r m rπ=ω = 
or2
2 1
12204 5 0.1 0.0055 ( )60NNπ==
∴ (N1)2= 204 / 0.0055 = 37 091 or N1 = 192.6 r.p.m. Ans.
2 2
C2 2 2 22()
or2
2 2
22414 5 0.175 0.0096 ( )60NNπ==

or2
2 2
∴ (N2)2= 414 / 0.0096 = 43 125 or N2 = 207.6 r.p.m. Ans.

726  /G6C   Theory of Machines
Let N1' and N2'= Minimum and maximum speeds of the governor considering
   friction.
We know that frictional force at each ball
= 5 N . . . (Given)
∴2
2 1
C1 1 1 12–( )60NFF m r m r′π′  =ω =
2
2 1
12204 – 5 5 0.1 0.0055 ( )60NN′π′ ==
∴2
11204 – 5( ) 36182 or 190.2 r.p.m.0.0055NN′′== =
2 2
C2 2 2 22()60NFF m r m r′π′  += ω =
2
2 2
22414 5 5 0.175 0.0096 ( )60NN′π′ += =
∴2

∴2
22414 5( ) 43 646 or 209 r.p.m.
0.0096NN+′′== =
21
212( – ) 2 (209 – 190.2)0.094 or 9.4%209 190.2NN
NN′′
== =+ ′′+ Ans.
EXERCISES
1.The length of the upper arm of a Watt governor is 400 mm and its inclination to the vertical is 30°.
Find the percentage increase in speed, if the balls rise by 20 mm. [Ans. 3%]

2.A Porter governor has two balls each of mass 3 kg and a central load of mass 15 kg. The arms are all
200 mm long, pivoted on the axis. If the maximum and minimum radii of rotation of the balls are 160mm and 120 mm respectively, find the range of speed. [Ans. 28.3 r.p.m.]
3.In a Porter governor, the mass of the central load is 18 kg and the mass of each ball is 2 kg. The top

arms are 250 mm while the bottom arms are each 300 mm long. The friction of the sleeve is 14 N. Ifthe top arms make 45° with the axis of rotation in the equilibrium position, find the range of speed ofthe governor in that position. [Ans. 15 r.p.m.]
4.A loaded governor of the Porter type has equal arms and links each 250 mm long. The mass of each

ball is 2 kg and the central mass is 12 kg. When the ball radius is 150 mm, the valve is fully open andwhen the radius is 185 mm, the valve is closed. Find the maximum speed and the range of speed. If themaximum speed is to be increased 20% by an addition of mass to the central load, find what additionalmass is required. [Ans. 193 r.p.m. ; 16 r.p.m.; 6.14 kg]
5.The arms of a Porter governor are 300 mm long. The upper arms are pivoted on the axis of rotation

and the lower arms are attached to the sleeve at a distance of 35 mm from the axis of rotation. The loadon the sleeve is 54 kg and the mass of each ball is 7 kg. Determine the equilibrium speed when the

Chapter 18 : Governors    /G6C  727
radius of the balls is 225 mm. What will be the range of speed for this position, if the frictional
resistances to the motion of the sleeve are equivalent to a force of 30 N?
[Ans. 174.3 r.p.m. ; 8.5 r.p.m.]
6.In a Porter governor, the upper and lower arms are each 250 mm long and are pivoted on the axis of

rotation. The mass of each rotating ball is 3 kg and the mass of the sleeve is 20 kg. The sleeve is in itslowest position when the arms are inclined at 30° to the governor axis. The lift of the sleeve is 36 mm.Find the force of friction at the sleeve, if the speed at the moment it rises from the lowest position isequal to the speed at the moment it falls from the highest position. Also, find the range of speed of thegovernor. [Ans. 9.8 N ; 16 r.p.m.]

7.A Porter governor has links 150 mm long and are attached to pivots at a radial distance of 30 mm from
the vertical axis of the governor. The mass of each ball is 1.75 kg and the mass of the sleeve is 25 kg.The governor sleeve begins to rise at 300 r.p.m. when the links are at 30° to the vertical. Assuming thefriction force to be constant, find the minimum and maximum speed of rotation when the inclinationof the links is 45° to the vertical. [Ans. 284 r.p.m. ; 347 r.p.m.]

8.A Proell governor has all the four arms of length 250 mm. The upper and lower ends of the arms are
pivoted on the axis of rotation of the governor. The extension arms of the lower links are each 100 mmlong and parallel to the axis when the radius of the ball path is 150 mm. The mass of each ball is 4.5kg and the mass of the central load is 36 kg. Determine the equilibrium speed of the governor.
[Ans. 164 r.p.m.]

9.A Proell governor has arms of 300 mm length. The upper arms are hinged on the axis of rotation,

whereas the lower arms are pivoted at a distance of 35 mm from the axis of rotation. The extension oflower arms to which the balls are attached are 100 mm long. The mass of each ball is 8 kg and the masson the sleeve is 60 kg. At the minimum radius of rotation of 200 mm, the extensions are parallel to thegovernor axis. Determine the equilibrium speed of the governor for the given configuration. Whatwill be the equilibrium speed for the maximum radius of 250 mm?

[Ans. 144.5 r.p.m. ; 158.2 r.p.m.]
10.A spring controlled governor of the Hartnell type with a central spring under compression has balls

each of mass 2 kg. The ball and sleeve arms of the bell crank levers are respectively 100 mm and 60mm long and are at right angles. In the lowest position of the governor sleeve, the radius of rotation ofthe balls is 80 mm and the ball arms are parallel to the governor axis. Find the initial load on the springin order that the sleeve may begin to lift at 300 r.p.m. If the stiffness of the spring is 30 kN/m, what isthe equilibrium speed corresponding to a sleeve lift of 10 mm? [Ans. 527 N ; 342

11.In a governor of the Hartnell type, the mass of each ball is 1.5 kg and the lengths of the vertical and

horizontal arms of the bell crank lever are 100 mm and 50 mm respectively. The fulcrum of the bellcrank lever is at a distance of 90 mm from the axis of rotation. The maximum and minimum radii ofrotation of balls are 120 mm and 80 mm and the corresponding equilibrium speeds are 325 and  300r.p.m. Find the stiffness of the spring and the equilibrium speed when the radius of rotation is 100 mm.
[Ans. 18 kN/m, 315 r.p.m.]

12.A governor of the Hartnell type has equal balls of mass 3 kg, set initially at a radius of 200 mm. The
arms of the bell crank lever are 110 mm vertically and 150 mm horizontally. Find :  1. the initialcompressive force on the spring, if the speed for an initial ball radius of 200 mm is 240 r.p.m. ; and  2.the stiffness of the spring required to permit a sleeve movement of 4 mm on a fluctuation of 7.5 percent in the engine speed. [Ans. 556 N ; 23.75 N/mm]

13.A spring controlled governor of the Hartnell type has the following data :
Mass of the ball = 1.8 kg ; Mass of the sleeve = 6 kg ; Ball and sleeve arms of the bell crank lever = 150

mm and 120 mm respectively. The equilibrium speed and radius of rotation for the lowest position ofthe sleeve are 400 r.p.m. and 150 mm respectively. The sleeve lift is 10 mm and the change in speedfor full sleeve lift is 5%. During an overhaul, the spring was compressed 2 mm more than the correctcompression for the initial setting. Determine the stiffness of the spring and the new equilibriumspeed for the lowest position of the sleeve. [Ans. 28.96 N/mm ; 472 r.p.m.]

728  /G6C   Theory of Machines
14.A spring controlled governor of the Hartnell type has two rotating balls of mass 1.35 kg each. The ball
arm is 75 mm and the sleeve arm is 62.5 mm. In the mid position of the sleeve, the sleeve arm is
horizontal and the balls rotate in a circle of 100 mm radius. The total sleeve movement is 30 mm.
Due to maladjustment of the spring, it is found that the equilibrium speed at the topmost position

of the sleeve is 420 r.p.m. and that corresponding to the lowest position is 435 r.p.m.
Determine : 1. stiffness and initial compression of th e spring, and 2. the required initial compression
of the spring to give an equilibrium speed at the topmost position w hich is 12 r.p.m. m ore than at the
lowest position. Neglect the moment due to mass of the balls.
[Ans. 6.3 N/mm, 87.54 mm ; 53.5 mm]

15.A Hartnell governor has two rotating balls, of mass 2.7 kg each. The ball radius is 125 mm in the mean

position when the ball arms are vertical and the speed is 150 r.p.m. with the sleeve rising. The lengthof the ball arms is 140 mm and the length of the sleeve arms 90 mm. The stiffness of the spring is7 kN/m and the total sleeve movement is 12 mm from the mean position. Allowing for a constantfriction force of 14  N acting at the sleeve, determine the speed range of the governor in the lowest and
highest sleeve positions. Neglect the obliquity of the ball arms. [Ans. 10.7 r.pm., 6.6 r.pm.]

16.The spring controlled governor of the Hartung type has two rotating masses each of 2.5 kg and the
limits of their radius of rotation are 100 mm and 125 mm. The each mass is directly controlled by aspring attached to it and to the inner casing of the governor as shown in Fig 18.26 ( a). The stiffness of

the spring is 8 kN/m and the force on each spring, when the masses are in their mid-position, is 320 N.In addition, there is an equivalent constant inward radial force of 80 N acting on each revolving massin order to allow for the dead weight of the mechanism. Neglecting friction, find the range of speed of
the governor. [Ans. 51 r.p.m.]
17.In a spring controlled governor of the Hartung type, the lengths of the horizontal and vertical arms of

the bell crank levers are 100 mm and 80 mm respectively. The fulcrum of the bell crank lever is at adistance of 120 mm from the axis of the governor. The each revolving mass is 9 kg. The stiffness of thespring is 25 kN/m. If the length of each spring is 120 mm when the radius of rotation is 70 mm and theequilibrium speed is 360 r.p.m., find the free length of the spring. If the radius of rotation increases to
120 mm, what will be the corresponding percentage increase in speed ?

[Ans 145.75 mm ; 10.83%]
[Hint. Free length of the spring = Length of the spring + compression of the spring]
18.The following particulars refer to a Wilson-Hartnell governor :
Mass of each ball = 4 kg ; minimum radius = 80 mm ; maximum radius = 90 mm ; minimum speed =
240 r.p.m.; maximum speed = 252 r.p.m.; length of the ball arm of each bell crank lever   = 80 mm ;
length of sleeve arm of each bell crank lever = 60 mm ; combined stiffness of the two ball springs =750 N/m.

Find the required stiffness of the auxiliary spring, if the lever is pivoted at the mid-point.
[Ans. 6.786 kN/m]
19.A spring loaded governor of the Wilson-Hartnell type
is shown in Fig 18.50. Two balls each of mass 4 kg are
connected across by two springs A. The stiffness of

each spring is 750 N/m and a free length of 100 mm.The length of ball arm of each bell crank lever is 80mm and that of sleeve arm is 60 mm. The lever ispivoted at its mid-point. The speed of the governor is240 r.p.m. in its mean position and the radius of rotation
of the ball is 80 mm. If the lift of the sleeve is 7.5 mm
for an increase of speed of 5%, find the requiredstiffness of the auxiliary spring B.
                                                   [Ans. 6.756 kN/m]
Fig. 18.50

Chapter 18 : Governors    /G6C  729
20.A Porter governor has all four arms 200 mm long. The upper arms are pivoted on the axis of rotation

and the lower arms are attached to a sleeve at a distance of 25 mm from the axis. Each ball has a massof 2 kg and the mass of the load on the sleeve is 20 kg. If the radius of rotation of the balls at a speedof 250 r.p.m. is 100 mm, find the speed of the governor after the sleeve has lifted 50 mm. Alsodetermine the effort and power of the governor. [Ans. 275.6 r.p.m.; 22.4 N ; 1.12 N-m]
21.A Porter governor has arms 250 mm each and four rotating flyballs of mass 0.8 kg each. The sleeve

movement is restricted to ± 20 mm from the height when the mean speed is 100 r.p.m. Calculate thecentral dead load and sensitivity of the governor neglecting friction when the flyball exerts a centrifugalforce of 9.81 N. Determine also the effort and power of the governor for 1 percent speed change.
[Ans. 11.76 N; 11.12; 0.196 N; 7.7 N-mm]
22.The upper arms of a Porter governor are pivoted on the axis of rotation and the lower arms are pivoted

to the sleeve at a distance of 30 mm from the axis of rotation. The length of each arm is 300 mm andthe mass of each ball is 6 kg. If the equilibrium speed is 200 r.p.m. when the radius of rotation is 200mm, find the required mass on the sleeve. If the friction is equivalent to a force of 40 N at the sleeve,find the coefficient of insensitiveness at 200 mm radius. [Ans. 61.1 kg. ; 6%]
23.In a spring controlled governor, the radial force acting on the balls was 4500 N when the centre of

balls was 200 mm from the axis and 7500 N when at 300 mm. Assuming that the force varies directlyas the radius, find the radius of the ball path when the governor runs at 270 r.p.m. Also find whatalteration in spring load is required in order to make the governor isochronous and the speed at which
it would then run. The mass of each ball is 30 kg. [Ans. 250 mm ; 1500 N ; 301.5 r.p.m.]
1.What is the function of a governor ? How does it differ from that of a flywheel ?

2.State the different types of governors. What is the difference between centrifugal and inertia type
governors ? Why is the former preferred to the latter ?
3.Explain the term height of the governor. Derive an expression for the height in the case of a Watt
governor. What are the limitations of a Watt governor ?
4.What are the effects of friction and of adding a central weight to the sleeve of a Watt governor ?

5.Discuss the controlling force and stability of a governor and show that the stability of a governor
depends on the slope of the curve connecting the controlling force ( FC) and radius of rotation ( r) and
the value ( FC /r).
6.What is stability of a governor ? Sketch the controlling force  versus radius diagrams for a stable,
unstable and isochronous governor. Derive the conditions for stability.
7.Explain clearly how would you determine from the controlling force curve whether a governor is

stable, unstable or isochronous. Show also how the effect of friction may be indicated on the curve.
8.Define and explain the following terms relating to governors :
1. Stability, 2 .Sensitiveness, 3. Isochronism, and 4. Hunting.
9.Explain the terms and derive expressions for ‘effort’ and ‘power’ of a Porter governor.
10.Prove that the sensitiveness of a Proell governor is greater than that of a Porter governor.
11.Write short note on ‘coefficient of insensitiveness’ of governors.

1.The height of a Watt’s governor (in metres) in equal to
(a) 8.95/N2(b) 89.5/N2(c) 895/N2(d) 8950/N2
where    N = Speed of the arm and ball about the spindle axis.

730  /G6C   Theory of Machines
2.The ratio of the height of a Porter governor (when the length of arms and links are equal) to the height
of a Watt’s governor is
(a)m
mM+(b)M
mM+(c)mM
m+(d)mM
M+
where          m = Mass of the ball, and
                    M = Mass of the load on the sleeve.
3.When the sleeve of a Porter governor moves upwards, the governor speed
(a) increases ( b) decreases ( c) remains unaffected
4.A Hartnell governor is a
(a) pendulum type governor ( b) spring loaded governor

(c) dead weight governor ( d) inertia governor
5.Which of the following governor is used to drive a gramophone ?
(a) Watt governor ( b) Porter governor
(c) Pickering governor ( d) Hartnell governor
6.Which of the following is a spring controlled governor?
(a) Hartnell ( b) Hartung ( c) Pickering ( d) all of these
7.For two governors A and B, the lift of sleeve of governor A is more than that of governor B, for a given
fractional change in speed. It indicates that

(a) governor A is more sensitive than governor B
(b) governor B is more sensitive than governor A
(c) both governors A and B are equally sensitive
8.The sensitiveness of a governor is given by
(a)
21meanω
ω− ω(b)21
meanω− ω
ω(c)21
2meanω− ω
ω(d) none of these
where           ω1 and ω2 = Minimum and maximum angular speed, and
                     ωmean = Mean angular speed.

ωmean = Mean angular speed.
9.In a Hartnell governor, if a spring of greater stiffness is used, then the governor will be
(a) more sensitive ( b) less sensitive ( c) isochronous
10.A governor is said to be hunting, if the speed of the engine
(a) remains constant at the mean speed
(b) is above the mean speed
(c) is below the mean speed
(d) fluctuates continuously above and below the mean speed.
11.A hunting governor is

(a) more stable ( b) less sensitive ( c) more sensitive ( d) none of these
12.Isochronism in a governor is desirable when
(a) the engine operates at low speeds
(b) the engine operates at high speeds
(c) the engine operates at variable speeds
(d) one speed is desired under one load

Chapter 18 : Governors    /G6C  731
13.The power of a governor is equal to
(a)2
()12cmM hc++(b)22()12cmM hc++
(c)23()12cmM hc++(d)24()12cmM hc++
where   c = Percentage increase in speed.
14.When the relation between the controlling force ( FC ) and radius of rotation ( r) for a spring controlled
governor is FC  = a.r + b, then the governor will be
(a) stable ( b) unstable ( c) isochronous

15.For a governor, if FC is the controlling force, r is the radius of rotation of the balls, the stability of the
governor will be ensured when
(a)CCdF F
dr r> (b)CCdF F
dr r< (c)C0dF
dr= (d) none of these
ANSWERS
1.(c) 2.(c) 3.(a) 4.(b) 5.(c)
6.(d) 7.(a) 8.(b) 9.(b) 10.(d)
11.(c) 12.(d) 13.(d) 14.(b) 15.(a)

732  /G6C   Theory of Machines
Brakes and
Dynamometers19
Features
2. Materials for Brake Lining.3. Types of Brakes.4. Single Block or Shoe Brake.5. Pivoted Block or Shoe Brake.6. Double Block or Shoe Brake.7. Simple Band Brake.8. Differential Band Brake.9. Band and Block Brake.
10. Internal Expanding Brake.
11. Braking of  a Vehicle.
12. Dynamometer.13. Types of Dynamometers.14. Classification of Absorption
Dynamometers.

15. Prony Brake Dynamometer.16. Rope Brake Dynamometers.17. Classification of Transmission
18. Epicyclic-train
19. Belt Transmission
Dynamometer-Froude orThroneycraft T ransmission
Dynamometer.
20. Torsion Dynamometer.21. Bevis Gibson Flash Light
Torsion Dynamometer.19.1. Introduction
A brake is a device by means of which artificial
frictional resistance is applied to a moving machine member,

in order to retard or stop the motion of a machine. In the process
of performing this function, the brake absorbs either kinetic
energy of the moving member or potential energy given up by
objects being lowered by hoists, elevators etc. The energy
absorbed by brakes is dissipated in the form of heat. This heat
is dissipated in the surrounding air (or water which is circulated
through the passages in the brake drum) so that excessive

heating of the brake lining does not take place. The capacity of
a brake depends upon the following factors :
1.The unit pressure between the braking surfaces,
2.The coefficient of friction between the braking
surfaces,
3.The peripheral velocity of the brake drum,
4.The projected area of the friction surfaces, and
5.The ability of the brake to dissipate heat equivalent
to the energy being absorbed.
The major functional difference between a clutch and

a brake is that a clutch is used to keep the driving and driven
member moving together, whereas brakes are used to stop a
moving member or to control its speed.
19.2. Materials for Brake Lining
The material used for the brake lining should have the
following characteristics :
732

Chapter 19 : Brakes and Dynamometers            /G6C          733
1.It should have high coefficient of friction with minimum fading. In other words, the coeffi-
cient of friction should remain constant with change in temperature.
2.It should have low wear rate.
3.It should have high heat resistance.
4.It should have high heat dissipation capacity.
5.It should have adequate mechanical strength.
6.It should not be affected by moisture and oil.

The materials commonly used for facing or lining of brakes and their properties are shown in
the following table.
Table 19.1. Properties of materials for brake lining.Table 19.1. Properties of materials for brake lining.Table 19.1. Properties of materials for brake lining.Table 19.1. Properties of materials for brake lining.Table 19.1. Properties of materials for brake lining.
Coefficient of friction (µ) Allowable

Material for braking lining  pressure (p)
Dry Greasy Lubricated N/mm2
Cast iron on cast iron 0.15 – 0.2 0.06 – 0.10 0.05 – 0.10 1.0 – 1.75
Bronze on cast iron – 0.05 – 0.10 0.05 – 0.10 0.56 – 0.84
Steel on cast iron 0.20 – 0.30 0.07 – 0.12 0.06 – 0.10 0.84 – 1.40
Wood on cast iron 0.20 – 0.35 0.08 – 0.12 – 0.40 – 0.62Fibre on metal – 0.10 – 0.20 – 0.07 – 0.28Cork on metal 0.35 0.25 – 0.30 0.22 – 0.25 0.05 – 0.10

Leather on metal 0.30 – 0.5 0.15 – 0.20 0.12 – 0.15 0.07 – 0.28
Wire asbestos on metal 0.35 – 0.5 0.25 – 0.30 0.20 – 0.25 0.20 – 0.55Asbestos blocks on metal 0.40 – 0.48 0.25 – 0.30 – 0.28 – 1.1Asbestos on metal (Short – – 0.20 – 0.25 1.4 – 2.1
action)
Metal on cast iron (Short – – 0.05 – 0.10 1.4 – 2.1
action)
19.3.19.3.19.3.19.3.19.3. Types of BrakesTypes of BrakesTypes of BrakesTypes of BrakesTypes of Brakes

The brakes, according to the means used for transforming the energy by the braking  elements,
are classified as :
1.  Hydraulic brakes e.g. pumps or hydrodynamic brake
and fluid agitator,
2.  Electric brakes e.g. generators and eddy current
brakes, and
3.  Mechanical brakes.
The hydraulic and electric brakes cannot bring the
member to rest and are mostly used where large amounts of
energy are to be transformed while the brake is retarding the

load such as in laboratory dynamometers, high way trucks andelectric locomotives. These brakes are also used for retardingor controlling the speed of a vehicle for down-hill travel.
The mechanical brakes, according to the direction of
acting force, may be divided into the following two groups :
(a) Radial brakes.  In these brakes, the force acting on
the brake drum is in radial direction. The radial brakes may be
Simple bicycle brakes.

734  /G6C   Theory of Machines
sub-divided into external brakes  and internal brakes . According to the shape of the friction ele-
ments, these brakes may be block or shoe brakes  and band brakes .
(b) Axial brakes . In these brakes, the force acting on the brake drum is in axial direction. The
axial brakes may be disc brakes and cone brakes. The analysis of these brakes is similar to clutches.
Since we are concerned with only mechanical brakes, therefore, these are discussed, in detail,

in the following pages.
19.4. Single Block or Shoe Brake
A single block or shoe brake is shown in Fig. 19.1. It consists of a block or shoe which is

pressed against the rim of a revolving brake wheel drum. The block is made of a softer material thanthe rim of the wheel. This type of a brake is commonly used on railway trains and tram cars. Thefriction between the block and the wheel causes a tangential braking force to act on the wheel, whichretard the rotation of the wheel. The block is pressed against the wheel by a force applied to one endof a lever to which the block is rigidly fixed as shown in Fig. 19.1. The other end of the lever

as shown in Fig. 19.1. The other end of the lever ispivoted on a fixed fulcrum O.

(a) Clockwise rotation of brake wheel                             ( b) Anticlockwise rotation of brake wheel.
Fig. 19.1. Single block brake. Line of action of tangential force passes through the fulcrum of the lever.
Let             P = Force applied at the end of the lever,
          RN= Normal force pressing the brake block on the wheel,
              r  = Radius of the wheel,
          2θ= Angle of contact surface of the block,
           µ = Coefficient of friction, and

µ = Coefficient of friction, and
             Ft = Tangential braking force or the frictional force acting at the contact
      surface of the block and the wheel.
If the angle of contact is less than 60°, then it may
be assumed that the normal pressure between the block andthe wheel is uniform. In such cases, tangential braking forceon the wheel,
            F
t = µ.RN ...(i)
and the braking torque,   TB = Ft.r = µ.RN.r ... (ii)

Let us now consider the following three cases :Case 1. When the line of action of tangential brak-
ing force ( F
t ) passes through the fulcrum O of the lever,
and the brake wheel rotates clockwise as shown in Fig. 19.1(a), then for equilibrium, taking moments about the fulcrum
O, we have
  
NRx P l×=×  or NPlR
x×=
∴ Braking torque,
                       BN.. . ...Pl PlrTR r r
xxµ=µ =µ× × =
When brakes are on, the pads grip the

wheel rim from either side, frictionbetween the pads and the rim convertsthe cycle's kinetic energy into heat asthey reduce its speed.

Chapter 19 : Brakes and Dynamometers    /G6C  735
It may be noted that when the brake wheel rotates anticlockwise as shown in Fig. 19.1 ( b),
then the braking torque is same, i.e.
          BN.. ...PlrTR r
xµ=µ =
Case 2. When the line of action of the tangential braking force ( Ft) passes through a distance
‘a’ below the fulcrum O, and the brake wheel rotates clockwise as shown in Fig. 19.2 ( a), then for
equilibrium, taking moments about the fulcrum O,

      RN × x + Ft × a = P.l or   RN × x + µ RN × a = P.l     or   RN = .
.Pl
xa+µ
and braking torque,       BN.. ...plrTR rxaµ=µ =+µ
(a)  Clockwise rotation of brake wheel.       ( b) Anticlockwise rotation of brake wheel.
Fig. 19.2.  Single block brake. Line of action of Ft passes below the fulcrum.
When the brake wheel rotates anticlockwise, as shown in Fig. 19.2 ( b), then for equilibrium,
         RN.x = P.l + Ft.a = P.l + µ.RN.a ...(i)

RN.x = P.l + Ft.a = P.l + µ.RN.a ...(i)
or         RN (x – µ.a) = P.l   orRN = .
.Pl
xa−µ
and braking torque,        BN.. ....PlrTR rxaµ=µ =−µ
Case 3. When the line of action of the tangential braking force ( Ft) passes through a distance
‘a’ above the fulcrum O, and the brake wheel rotates clockwise as shown in Fig. 19.3 ( a), then for
equilibrium, taking moments about the fulcrum O, we have
         RN.x = P.l + Ft.a = P.l + µ.RN.a . . . (ii)

or          RN (x – µ.a) = P.l        or           RN = .
.Pl
xa−µ
(a) Clockwise rotation of brake wheel.                   (b) Anticlockwise rotation of brake wheel.
Fig. 19.3.  Single block brake. Line of action of Ftpasses above the fulcrum.
and braking torque,       TB = µ.RN.r = .. .
.Plr
xaµ
−µ

736  /G6C   Theory of Machines
When the brake wheel rotates anticlockwise as shown in Fig. 19.3 ( b), then for equilibrium,
taking moments about the fulcrum O, we have
    RN × x + Ft × a = P.l     or     RN × x + µ.RN × a = P.l    or    RN = .
.Pl
xa+µ
.Plr
xaµ
+µ
Notes : 1. From above we see that when the brake wheel rotates anticlockwise in case 2 [Fig. 19.2 ( b)] and when

it rotates clockwise in case 3 [Fig. 19.3 ( a)], the equations (i) and (ii) are same, i.e.
    RN × x = P.l + µ.RN.a
From this we see that
the moment of frictional force(µ.R
N.a) adds to the moment
of force ( P.l). In other words,
the frictional force helps toapply the brake. Such type ofbrakes are said to be  self ener-
gizing brakes . When the fric-
tional force is great enough toapply the brake with no exter-nal force, then the brake is saidto be self-locking brake .
From the above ex-

pression, we see that if
.xa≤µ, then P will be negative or equal to zero. This means no external force is needed to apply the brake and
hence the brake is self locking. Therefore the condition for the brake to be self locking is
.xa≤µ
The self locking brake is used only in back-stop applications.
2.The brake should be self energizing and not the self locking.
3.In order to avoid self locking and to prevent the brake from grabbing, x is kept greater than µ.a.

4.If Ab is the projected bearing area of the block or shoe, then the bearing pressure on the shoe,
             pb = RN / Ab
We know that       Ab = Width of shoe × Projected length of shoe = (2 sin )wrθ
5.When a single block or shoe brake is applied to a rolling wheel, an additional load is thrown on the
shaft bearings due to heavy normal force ( RN) and produces bending of the shaft.
In order to overcome this drawback, a double block or shoe brake is used, as discussed in Art. 19.6.

19.5. Pivoted Block or Shoe Brake
We have discussed in the previous article that when the angle of contact is less than 60°, then

it may be assumed that the normal pressure between the block and the wheel is uniform. Butwhen the angle of contact is greater than 60°, then the unitpressure normal to the surface of contact is less at the endsthan at the centre. In such cases, the block or shoe is pivotedto the lever, as shown in Fig. 19.4, instead of being rigidlyattached to the lever. This gives uniform wear of the brakelining in the direction of the applied force. The braking torque

for a pivoted block or shoe brake ( i.e. when 2
θ > 60°) is
given by
        BN ..tTF r R r ′ =× = µ
where         ′µ= Equivalent coefficient of friction = 4s i n
2s i n 2µθ
θ+ θ, and
           µ = Actual coefficient of friction.
These brakes have more life and may provide a higher braking torque.Fig. 19.4.  Pivoted block or shoe brake.Shoe brakes of a racing car

Chapter 19 : Brakes and Dynamometers    /G6C  737
Example 19.1. A single block brake is shown in Fig. 19.5.
The diameter of the drum is 250 mm and the angle of contact is90°. If the operating force of 700 N is applied at the end of a leverand the coefficient of friction between the drum and the lining is0.35, determine the torque that may be transmitted by the blockbrake.
Solution. Given : d = 250 mm or r = 125 mm ; 
2θ= 90°
 = /2π rad ; P = 700 N ; µ = 0.35

2θ= 90°
Since the angle of contact is greater than 60°, therefore
equivalent coefficient of friction,
                4 sin 4 0.35 sin45
2 sin2 /2 sin90µθ × × °′µ= =θ+ θ π + °= 0.385
Let                  RN = Normal force pressing the block to the brake drum, and
                  Ft = Tangential braking force = N.R′µ
Taking moments about the fulcrum O, we have
            N 700(250 200) 50 200 200 200 5200.385tt
ttFFFR F+ + × =×= ×= ×=′µ

or  520 Ft – 50Ft = 700 × 450    or     Ft = 700 × 450/470 = 670 N
We know that torque transmitted by the block brake,
               TB = Ft × r = 670 × 125 = 8 3750 N-mm =  83.75N-m Ans.
Example 19.2. Fig. 19.6 shows a brake shoe

applied to a drum by a lever AB which is pivoted at a fixed point A and rigidly fixed to the shoe.The radius of the drum is 160 mm. The coefficient offriction at the brake lining is 0.3. If the drum rotatesclockwise, find the braking torque due to the horizon-tal force of 600 N at B.
Solution. Given : r = 160 mm = 0.16 m ;
µ = 0.3 ; P = 600 N
Since the angle subtended by the shoe at the
centre of drum is 40°, therefore we need not to calcu-

late the equivalent coefficient of friction 
.′µ
Let RN = Normal force pressing the
block to the brake drum, and
 Ft = Tangential braking force = µ. RN
Taking moments about point A,
RN × 350 + Ft (200 – 160) = 600 (400 + 350)
350 40 600 7500.3t
tFF ×+ =×  or 1207 Ft = 450 × 103
∴                 Ft = 450 × 103/1207 = 372.8 N
We know that braking torque,
      TB = Ft × r = 372.8 × 0.16 = 59.6 N-m Ans.All dimensions in mm.
Fig. 19.5
Fig. 19.6

738  /G6C   Theory of Machines
Example 19.3. A bicycle and rider of mass 100 kg are travelling at the rate of 16 km/h on a
level road. A brake is applied to the rear wheel which is 0.9 m in diameter and this is the only
resistance acting. How far will the bicycle travel and how many turns will it make before it comes to
rest ? The pressure applied on the brake is 100 N and µ = 0.05.
Solution. Given : m = 100 kg, v = 16 km / h = 4.44 m/ s ; D = 0.9 m ; RN = 100 N ; µ = 0.05

Distance travelled by the bicycle before it comes to rest
Let      x = Distance travelled (in metres) by the bicycle before it comes to rest.
We know that tangential braking force acting at the point of contact of the brake and wheel,
                Ft = µ.RN = 0.05 × 100 = 5 N
and work done        = Ft × x = 5  × x = 5x N-m . . . (i)
We know that kinetic energy of the bicycle
                      22. 100(4.44)
22
986N-mmν==
=
In order to bring the bicycle to rest, the work done

against friction must be equal to kinetic energy of the bi-cycle. Therefore equating equations  (i) and (ii),
   5 x = 986  or   x = 986/5 =  197.2 m  Ans.
Number of revolutions made by the bicycle before itcomes to rest
Let     N = Required number of revolutions.
We know that distance travelled by the bicycle ( x),
            
197.2 0.9 2.83=π =π× =DN N N
∴              N = 197.2 / 2.83 = 70 Ans.
Example 19.4.  A braking system has its braking lever inclined at an angle of 30° to the

horizontal plane, as shown in Fig. 19.7. The mass and diameter of the brake drum are 218 kg and0.54 m respectively.
Fig. 19.7
At the instant the lever is pressed on the brake drum with a vertical force of 600 N, the drum
is found to rotate at 2400 r.p.m. clockwise. The coefficient of friction between the brake shoe and the
brake drum is 0.4. Assume that the lever and brake shoe are perfectly rigid and possess negligibleweight. Find :
Shoe brake.. . . (ii)

Chapter 19 : Brakes and Dynamometers    /G6C  739
1. Braking torque,  2. Number of revolutions the drum will make before coming to rest from
the instant of pressing the lever, and 3. Time taken for the drum to come to rest from the instant of
pressing the lever.
Solution. Given : m = 218 kg ; d = 0.54 m   or   r = 0.27 m ;  P = 600 N ;  N = 2400 r.p.m.;
µ = 0.41. Braking torque
Let R
N = Normal force pressing the block to the brake drum, and
Ft = Tangential braking force.

The various forces acting on the braking system are shown in Fig. 19.8.
Fig. 19.8
Taking moments about the fulcrum O,
   600 cos 30° × 1.2 = RN × 0.4      or      623.5 = 0.4 RN
∴            RN =  623.5/0.4 = 1560 N
and Ft  = µ.RN = 0.4 × 1560 = 624 N
TB = Ft  × r = 624 × 0.27 = 168.5 N-m  Ans.
2. Number of revolutions the drum will make before coming to rest
Let  n = Required number of revolutions.

We know that kinetic energy of the brake drum
   = 22 2. 218 . 0.54 2400109 N-m
2 2 60 60mv dN ππ × × ==  
   = 502 × 103 N-m . . . (i)
and work done by the brake drum due to braking torque
    = B2 168.5 2 1060 N-mTn n n×π= ×π= . . . (ii)
Since the kinetic energy of the brake drum is used to overcome the work done due to braking
torque, therefore equating equations (i) and (ii),
 n = 502 × 103/1060 = 474 Ans.

n = 502 × 103/1060 = 474 Ans.
3. Time taken for the drum to come to rest
We know that time taken for the drum to come to rest i.e. time required for 474 revolutions,
  4740.2min
2400== =nt
N= 12 s Ans.

740  /G6C   Theory of Machines
19.6. Double Block or Shoe Brake
When a single block brake is applied to a rolling wheel, an additional load is thrown on the
shaft bearings due to the normal force ( RN). This produces

bending of the shaft. In order to overcome this drawback, a doubleblock or shoe brake, as shown in Fig. 19.9, is used. It consists oftwo brake blocks applied at the opposite ends of a diameter ofthe wheel which eliminate or reduces the unbalanced force onthe shaft. The brake is set by a spring which pulls the upper endsof the brake arms together. When a force P is applied to the bell

crank lever, the spring is compressed and the brake is released.This type of brake is often used on electric cranes and the forceP is produced by an electromagnet or solenoid. When the current
is switched off, there is no force on the bell crank lever and thebrake is engaged automatically due to the spring force and thusthere will be no downward movement of the load.
In a double block brake, the braking action is doubled

by the use of two blocks and these blocks may be operatedpractically by the same force which will operate one. In case ofdouble block or shoe brake, the braking torque is given by
      T
B = (Ft1 + Ft2) r
where Ft1 and Ft2 are the braking forces on the two blocks.
Example 19.5. A double shoe brake, as shown in Fig. 19.10,

is capable of absorbing a torque of 1400 N-m. The diameter of thebrake drum is 350 mm and the angle of contact for each shoe is 100°.If the coefficient of friction between the brake drum and lining is0.4 ; find 1. the spring force necessary to set the brake ; and 2. the
width of the brake shoes, if the bearing pressure on the liningmaterial is not to exceed 0.3 N/mm
2.
Solution. Given :  TB = 1400 N-m = 1400 × 103 N-mm ;
d = 350 mm or r = 175 mm ; 2θ = 100° = 100 × π/180 = 1.75rad;

µ = 0.4 ; pb = 0.3 N/mm2
1. Spring force necessary to set the brake
Let            S  = Spring force necessary to
          set the brake.
RN1 and Ft1 = Normal reaction and the
        braking force on the right           hand side shoe, and
 R
N2 and Ft2 = Corresponding values on
           the left hand side shoe.
Since the angle of contact is greater than
60°, therefore equivalent coefficient of friction,
         4s i n 40 . 4s i n 5 00.45
2 sin2 1.75 sin100µθ ×× °µ′= = =

θ+ θ + °Fig. 19.9.  Double block or shoe
 brake.
Fig. 19.10
Brakes on a railway coach.

Chapter 19 : Brakes and Dynamometers    /G6C  741
Taking moments about the fulcrum O1, we have
         1
N1 1 1 1450 200 (175 40) 200 135 579.4
0.45ttt
tFSR F F F×=×+ −= ×+ ×=
. . . 1
N1 Substituting tFR= ′µ 
∴   Ft1 = S × 450 / 579.4 = 0.776 S
Again taking moments about O2, we have
            2
2N 2 2450 (175 40) 200 200 444.40.45t
ttFSF R F×+ −=×= ×=
. . .2
N2 Substituting tFR= ′µ 
      444.4 Ft2 – 135Ft2 = S × 450   or   309.4 Ft2 = S × 450
∴  Ft2 = S × 450 / 309.4 = 1.454 S

We know that torque capacity of the brake ( TB),
  1400 × 103  = (Ft1 + Ft2) r = (0.776 S + 1.454 S) 175 = 390.25 S
∴     S = 1400 × 103/390.25 = 3587 N Ans.
2. Width of the brake shoes
Let     b = Width of the brake shoes in mm.
We know that projected bearing area for one shoe,
2(2 sin ) (2 175sin50 ) 268 mmbAb r b b=θ = × ° =
Normal force on the right hand side of the shoe,
1
N10.776 0.776 35876186 N0.45 0.45tF SR××== = =′µ

and normal force on the left hand side of the shoe,
             2
N21.454 1.454 358711590N0.45 0.45tF SR××== = =′µ
We see that the maximum normal force is on the left hand side of the shoe. Therefore we shall
find the width of the shoe for the maximum normal force i.e. RN2.
We know that the bearing pressure on the lining material ( pb),
 N211590 43.250.3268bR
Ab b== =
∴                  b = 43.25 / 0.3 = 144.2 mm  Ans.
19.7. Simple Band Brake

A band brake consists of a flexible band of leather, one or more ropes,or a steel lined with
friction material, which embraces a part of the circumference of the drum. A band brake, as shown inFig. 19.11, is called a  simple band brake  in which one end of the band is attached to a fixed pin or
fulcrum of the lever while the other end is attached to the lever at a distance b from the fulcrum.

When a force P is applied to the lever at C, the lever turns about the fulcrum pin O and tightens
the band on the drum and hence the brakes are applied. The friction between the band and the drumprovides the braking force. The force P on the lever at C may be determined as discussed below :
Let T
1 = Tension in the tight side of the band,
T2 = Tension in the slack side of the band,

742  /G6C   Theory of Machines
θ = Angle of lap (or embrace) of the band on the drum,
 µ = Coefficient of friction between the band and the drum,
  r = Radius of the drum,
  t = Thickness of the band, and
re = Effective radius of the drum = 2+tr
(a) Clockwise rotation of drum. ( b) Anticlockwise rotation of drum.
Fig. 19.11.  Simple band brake.
We know that limiting ratio of the tensions is given by the relation,
             1
2µθ=TeTor1
22.3log .T
T=µθ


1
22.3log .T
T=µθ

and braking force on the drum = T1 – T2
∴ Braking torque on the drum,
              TB = (T1 – T2) r . . . (Neglecting thickness of band)
    = (T1 – T2) re . . . (Considering thickness of band)
Now considering the equilibrium of the lever OBC. It may be noted that when the drum
rotates in the clockwise direction, as shown in Fig. 19.11 ( a), the end of the band attached to the

fulcrum O will be slack with tension T2 and end of the band attached to B will be tight with tension T1.
On the other hand, when the drum rotates in the anticlockwise direction, as shown in Fig. 19.11 ( b),
the tensions in the band will reverse, i.e. the end of the band attached to the fulcrum O will be tight
with tension T1 and the end of the band attached to B will be slack with tension T2. Now taking
moments about the fulcrum O, we have

           P.l = T1.b . . . (For clockwise rotation of the drum)
and            P.l = T2.b  . . . (For anticlockwise rotation of the drum)Band brake Bands of a brake shown separately

Chapter 19 : Brakes and Dynamometers    /G6C  743
where   l = Length of the lever from the fulcrum ( OC), and
 b = Perpendicular distance from O to the line of action of T1  or T2.
Notes : 1.  When the brake band is attached to the lever, as shown in Fig. 19.11 ( a) and (b), then the force ( P)
must act in the upward direction in order to tighten the band on the drum.
2. If the permissible tensile stress ( σ) for the material of the band is known, then maximum tension in
the band is given by

T1 = ..σwt
where w = Width of the band, and
  t = thickness of the band.
Example 19.6.  A band brake acts on the 3/4th of circumference of a drum of 450 mm diam-
eter which is keyed to the shaft. The band brake provides a braking torque of 225 N-m. One end ofthe band is attached to a fulcrum pin of the lever and the other end to a pin 100 mm from the fulcrum.If the operating force is applied at 500 mm from the fulcrum and the coefficient  of friction is 0.25,

find the operating force when the drum rotates in the (a) anticlockwise direction, and (b) clockwise
Solution. Given : d = 450 mm or r = 225 mm = 0.225 m ; T
B = 225 N-m ; b = OB = 100 mm
= 0.1 m ; l = 500 mm = 0.5 m ; µ = 0.25
Let    P = Operating force.
(a) Operating force when drum rotates in anticlockwisedirection
The band brake is shown in Fig. 19.11. Since one
end of the band is attached to the fulcrum at O, therefore the
operating force P will act upward and when the drum ro-

tates anticlockwise, as shown in Fig. 19.11 ( b), the end of
the band attached to O will be tight with tension T
1 and the
end of the band attached to B will be slack with tension T2.
First of all, let us find the tensions T1  and T2.
We know that angle of wrap,
                                33th of circumference = 360 27044θ= × °= °
     270 /180 4.713rad=× π =
and                1
12.3log . 0.25 4.713 1.178T

∴                 1
21.178log 0.51232.3==
T
T or 1
23.253=T

T
T or 1
23.253=T
T. . . (i)
. . . (Taking antilog of 0.5123)
We know that braking torque ( TB),
             225 = ( T1 – T2) r = (T1 – T2) 0.225
∴                   T1 – T2 = 225/ 0.225 = 1000 N . . . (ii)
  T1 = 1444 N;  and      T2 = 444 N
Now taking moments about the fulcrum O, we have
          P × l = T2.b       or        P × 0.5 = 444 × 0.1 = 44.4
∴                 P = 44.4 / 0.5 = 88.8 N Ans.
Drums for band brakes.

744  /G6C   Theory of Machines
(b) Operating force when drum rotates in clockwise direction
When the drum rotates in clockwise direction, as shown in Fig.19.11 ( a), then taking mo-
ments about the fulcrum O, we have
        P × l = T1. b        or       P × 0.5 = 1444 × 0.1 = 144.4
∴ P = 144.4 / 0.5 = 288.8 N  Ans.
Example 19.7.  The simple band brake, as shown in Fig. 19.12, is applied to a shaft carrying

a flywheel of mass 400 kg. The radius of gyration of the flywheel is 450 mm and runs at 300 r.p.m.
If the coefficient of friction is 0.2 and the brake drum
diameter is 240 mm, find :
1.the torque applied due to a hand load of 100 N,
2.the number of turns of the wheel before it is brought to
rest, and
3.the time required to bring it to rest, from the moment of
the application of the brake.
Solution. Given :  m = 400 kg ; k = 450 mm = 0.45 m ;

N = 300 r.p.m. or  23 0 0 / 6 0ω= π×  = 31.42 rad/s  ; µ = 0.2 ;
d = 240 mm = 0.24 m or r = 0.12 m
1. Torque applied due to hand load
First of all, let us find the tensions in the tight and slack sides of the band i.e. T1 and T2
From the geometry of the Fig. 19.12, angle of lap of the band on the drum,
         360 150 210 210 3.666rad
180πθ= °− °= °= × =
     1
22.3log . 0.2 3.666 0.7332=µθ= × =
T
T
           1
20.7332log 0.31882.3==
T
T       or1

T
T       or1
22.08=T
T. . . (i)
... (Taking antilog of 0.3188)
           T2 × 120 = 100 × 300 = 30 000 or  T2 = 30 000/120 = 250 N
∴                     T1 = 2.08T2 = 2.08 × 250 = 520 N  . . . [From equation  (i)]
We know that torque applied,
       TB = (T1 – T2 ) r =  (520 – 250) 0.12 = 32.4 N-m  Ans.
2. Number of turns of the wheel before it is brought to rest

Let          n = Number of turns of the wheel before it is brought to rest.
We know that kinetic energy of rotation of the drum
           22 2 2 211 1. . . 400(0.45) (31.42)22 2=×ω =× ω =×Im k = 40 000 N-m
This energy is used to overcome the work done due to the braking torque ( TB).
∴             40 000 = TB × 2πn = 32.4 × 2πn = 203.6 n
or          n = 40 000 / 203.6 = 196.5 Ans.All dimensions in mm.
Fig. 19.12

Chapter 19 : Brakes and Dynamometers    /G6C  745
3. Time required to bring the wheel to rest
We know that the time required to bring the wheel to rest
= n / N = 196.5 / 300 = 0.655 min =  39.3 s Ans.
Example 19.8. A simple band brake operates on a drum of 600 mm in diameter that is
running at 200 r.p.m. The coefficient of friction is 0.25. The brake band has a contact of 270°, one

end is fastened to a fixed pin and the other end to the brake arm 125 mm from the fixed pin. Thestraight brake arm is 750 mm long and placed perpendicular to the diameter that bisects the angle ofcontact.
1. What is the pull necessary on the end of the brake
arm to stop the wheel if 35 kW is being absorbed ? What is thedirection for this minimum pull ?
2. What width of steel band of 2.5 mm thick is required
for this brake if the maximum tensile stress is not to exceed50 N/mm
2 ?

2 ?
Solution. Given : d = 600 mm or r = 300 mm ;
N = 200 r.p.m. ; µ = 0.25 ; 270 270 /180θ= °= ×π = 4.713 rad ;
Power = 35 kW = 35 × 103 W ; t = 2.5 mm ; σ = 50 N/mm2
1. Pull necessary on the end of the brake arm to stop the wheel
Let P = Pull necessary on the end of the brake arm to
stop the wheel.
The simple band brake is shown in Fig. 19.13. Since one end of the band is attached to the
fixed pin O, therefore the pull P on the end of the brake arm will act upward and when the wheel

rotates anticlockwise, the end of the band attached to O will be tight with tension T1 and the end of the
band attached to B will be slack with tension T2. First of all, let us find the tensions T1 and T2. We
know that
     1
22.3log . 0.25 4.713 1.178T

∴      1
21.178log 0.51222.3T
T==
 or  1
23.25T
T=     ... (Taking antilog of 0.5122)  ... (i)
Let  TB = Braking torque.
We know that power absorbed,
       3 BB
B2.2 2 0 035 10 2160 60NT TTπ× π× ××= = =

∴                33
B35 10 /21 1667 N-m 1667 10 N-mmT=× = = ×
We also know that braking torque ( TB),
  1667 × 103 = (T1 – T2)  r = (T1 – T2) 300
∴                    T1 – T2 = 1167 × 103/300 = 5556 N ... (ii)
From equations (i) and (ii), we find that
  T1 = 8025 N; and T2 = 2469 NFig. 19.13All dimensions in mm

746  /G6C   Theory of Machines
Now taking moments about O, we have
              P × 750 = T2 × *OD = T2 × 62.52 = 2469 × 88.4 = 218 260
∴                      P =218260/ 750 = 291 N  Ans.
2.  Width of steel band
Let         w = Width of steel band in mm.
We know that maximum tension in the band ( T1),
    8025 = ..wtσ = 50 × w × 2.5 = 125 w
∴          w= 8025 / 125 = 64.2 mm Ans.
19.8. Differential Band Brake

In a differential band brake, as shown in Fig. 19.14, the ends of the band are joined at A and
B to a lever AOC pivoted on a fixed pin or fulcrum O. It may be noted that for the band to tighten, the
length OA must be greater than the length OB.
    (a) Clockwise rotation of the drum. ( a) Anticlockwise rotation of the drum.
Fig. 19.14.   Differential band brake.
The braking torque on the drum may be obtained

in the similar way as discussed in simple band brake. Nowconsidering the equilibrium of the lever AOC. It may be
noted that when the drum rotates in the clockwise direc-tion, as shown in Fig. 19.14 ( a), the end of the band
attached to A will be slack with tension T
2 and end of the
band attached to B will be tight with tension T1. On the

other hand, when the drum rotates in the anticlockwisedirection, as shown in Fig. 19.14 ( b), the end of the band
attached to A will be tight with tension T
1 and end of the
band attached to B will be slack with tension T2. Now
 P.l + T1.b = T2.a
... (For clockwise rotation of the drum )
or                     P.l = T2.a – T1.b  ... (i)
and        P.l + T2.b = T1.a
... (For anticlockwise rotation of the drum )
or         P.l = T1.a – T2.b ... (ii)

*OD = Perpendicular distance from O to the line of action of tension T2.
OE = EB = OB/2 = 125/2 = 62.5 mm, and ∠DOE = 45°
 ∴OD = OE sec 45° = 62.5 2 mm
Tractors  are specially made to move on
rough terrain and  exert high power at
low speeds.

Chapter 19 : Brakes and Dynamometers    /G6C  747
We have discussed in block brakes (Art. 19.4), that when the frictional force helps to apply
the brake, it is said to be self energizing brake. In case of differential band brake, we see from equa-tions (i) and (ii) that the moment T
1.b and T2.b helps in applying the brake (because it adds to the
moment P.l) for the clockwise and anticlockwise rotation of the drum respectively.

We have also discussed that when the force P is negative or zero, then brake is self locking.
Thus for differential band brake and for clockwise rotation of the drum, the condition for self lockingis
21..Ta Tb≤ or21//TT b a≤
and for anticlockwise rotation of the drum, the condition for self locking is
12..Ta Tb≤ or12//TT ba≤
Notes : 1.  The condition for self locking may also be written as follows :
    For clockwise rotation of the drum,
12..Tb Ta≥ or 12//TT ab≥

and for anticlockwise rotation of the drum,
21..Tb Ta≥ or 12//TT a b≥
2. When in Fig. 19.14 ( a) and (b), the length OB is greater than OA, then the force P must act in the
upward direction in order to apply the brake. The tensions in the band, i.e. T1 and T2 will remain unchanged.
Example 19.9.  In a winch, the rope supports a load W and is wound round a barrel 450 mm

diameter. A differential band brake acts on a drum 800 mm diameter which is keyed to the same shaftas the barrel. The two ends of the bands are attached to pins on opposite sides of the fulcrum of thebrake lever and at distances of 25 mm and 100 mm from the fulcrum. The angle of lap of the brakeband is 250° and the coefficient of friction is 0.25. What is the maximum load W which can besupported by the brake when a force of 750 N is applied to the lever at a distance of 3000 mm fromthe fulcrum

Solution. Given : D = 450 mm or R = 225 mm ; d = 800 mm or r = 400 mm ; OB = 25 mm ;
OA = 100 mm ; 
θ = 250° = 250 × π/180 = 4.364 rad ;
µ = 0.25 ; P = 750 N ; l = OC = 3000 mm
Since OA is greater than OB, therefore the
operating force ( P = 750 N) will act downwards.
First of all, let us consider that the drum rotates
in clockwise direction.
We know that when the drum rotates in clock-
wise direction, the end of band attached to A will be
slack with tension T2 and the end of the band attached

to B will be tight with tension T1, as shown in Fig. 19.15.
Now let us find out the values of tensions T1 and T2. We
know that
1
22.3log . 0.25 4.364 1.091T

∴   1
21.091log 0.47432.3T
T==
 or 1
22.98T
T= ... (Taking antilog of 0.4743)
and                        T1 = 2.98 T2... (i)
Now taking moments about the fulcrum O,
    750 × 3000 + T1 × 25 = T2 × 100All dimensions in mm.
Fig. 19.15

748  /G6C   Theory of Machines
orT2 × 100 – 2.98 T2 × 25 = 2250 × 103 ... (/G51 T1 = 2.98 T2)
 25.5 T2= 2250 × 103orT2 = 2250 × 103/25.5 = 88 × 103 N
and         T1= 2.98T2 = 2.98 × 88 × 103 = 262 × 103 N
       TB= (T1 – T2) r
= (262 × 103 – 88 × 103) 400 = 69.6 × 106 N-mm ... (i)
and the torque due to load W newtons,
      TW = W.R = W × 225 = 225 W N-mm ... (ii)
Since the braking torque must be equal to the torque due to load W  newtons, therefore from

equations  (i) and (ii),
      W = 69.6 × 106/225 = 309 × 103 N = 309 kN
Now let us consider that the drum rotates in
anticlockwise direction. We know that when the drum rotatesin anticlockwise direction, the end of the band attached to A
will be tight with tension T
1 and end of the band attached to
B will be slack with tension T2, as shown in Fig. 19.16. The
ratio of tensions T1 and T2 will be same as calculated above,
i.e.
       1
22.98T
T= or T1 = 2.98 T2

i.e.
       1
22.98T
       750 × 3000 + T2 × 25 = T1 × 100
or2.98 T2 × 100 – T2× 25 = 2250 × 103 ... (/G51 T1 = 2.98 T2)
273 T2 = 2250 × 103      or      T2 = 2250 × 103/273 = 8242 N
and        T1 = 2.98 T2 = 2.98 × 8242 = 24 561 N
∴       Braking torque,  TB = (T1 × T2) r
           = (24 561 – 8242)400 = 6.53 × 106 N-mm ... (iii)
From equations  (ii) and (iii),
W = 6.53 × 106/225 = 29 × 103 N = 29 kN

From above, we see that the maximum load ( W) that can be supported by the brake is 309 kN,
when the drum rotates in clockwise direction.  Ans.
Example 19.10.  A differential band brake, as shown in Fig. 19.17, has an angle of contact of
225°. The band has a compressed woven lining and bears against a cast iron drum of 350 mmdiameter. The brake is to sustain a torque of 350 N-m and the coefficient of friction between the band

and the drum is 0.3. Find : 1. The necessary force (P) for the clockwise and anticlockwise rotation of
the drum; and  2. The value of ‘OA’ for the brake to be self locking, when the drum rotates clockwise.
Solution. Given:
θ= 225° = 225 × π/180 = 3.93 rad ; d = 350 mm    or   r = 175 mm ;
T = 350 N-m = 350 × 103 N-mm
1. Necessary force (P) for the clockwise and anticlockwise rotation of the drum
When the drum rotates in the clockwise direction, the end of the band attached to A will be

slack with tension T2 and the end of the band attached to B will be tight with tension T1, as shown in
Fig. 19.18. First of all, let us find the values of tensions T1 and T2.All dimensions in mm.
Fig. 19.16

Chapter 19 : Brakes and Dynamometers    /G6C  749
     Fig. 19.17   Fig. 19.18
      1
22.3log . 0.3 3.93 1.179T

∴        1
21.179log 0.51262.3T
T==
     or    1
23.255T
T= ... (Taking antilog of 0.5126 ) ... (i)
and braking torque ( TB),
          350 × 103 = (T1 – T2)r = (T1 – T2) 175
∴          T1 – T2 = 350 × 103/175 = 2000 N  ... (ii)
        T1 = 2887 N ; and T2 = 887 N

T1 = 2887 N ; and T2 = 887 N
            P  × 500 = T2 × 150 – T1 × 35 = 887 × 150 – 2887 × 35 = 32 ×103
∴          P = 32 × 103/500 = 64 N Ans.
When the drum rotates in the anticlockwise
direction, the end of the band attached to A will be tightwith tension T
1 and end of the band attached to B will
be slack with tension T2, as shown in Fig. 19.19. Taking
             P  × 500 = T1 × 150 – T2 × 35

P  × 500 = T1 × 150 – T2 × 35
            = 2887 × 150 – 887 × 35            = 402 × 10
3
          P = 402 × 103/500 = 804 N   Ans.
2. Value of ‘OA’ for the brake to be self locking, when
the drum rotates clockwise
The clockwise rotation of the drum is shown in Fig 19.18.For clockwise rotation of the drum, we know that
            P × 500 = T
2 × OA – T1 × OB
For the brake to be self locking, P must be equal to zero. Therefore
             T2 × OA = T1 × OB
and      1
22887 35

750  /G6C   Theory of Machines
19.9. Band and Block Brake
The band brake may be lined with blocks of wood or other material, as shown in Fig. 19.20
(a). The friction between the blocks and the drum provides braking action. Let there are ‘ n’ number
of blocks, each subtending an angle 2θ at the centre and the drum rotates in anticlockwise direction.
(a)( b)
Fig. 19.20.  Band and block brake.
Let T1 = Tension in the tight side,
T2 = Tension in the slack side,

 µ = Coefficient of friction between the blocks and drum,
1T′= Tension in the band between the first and second block,
               23,TT′′ etc.=Tensions in the band between the second and third block,
         between the third and fourth block etc.
Consider one of the blocks (say first block) as shown in Fig. 19.20 ( b). This is in equilibrium
under the action of the following forces :
1. Tension in the tight side ( T1),

2. Tension in the slack side (1T′) or tension in the band between the first and second block,
3. Normal reaction of the drum on the block ( RN), and
4. The force of friction ( µ. RN ).
Resolving the forces radially, we have
11 N() s i nTT R′+θ =  ... (i)
Resolving the forces tangentially, we have
11 N() c o s .TT R′+θ = µ  ... (ii)
Dividing equation  (ii) by (i), we have
N 11
N 11. () c o s
() s i nR TT
R TT′ µ −θ=
′+θ
or11 11() t a n ()TT TT′′−= µθ +
∴     1

′+θ
∴     1
11t a n
1t a nT
T+µ θ=−µ θ′
Similarly, it can be proved for each of the blocks that
                  31 12
2 2341t a n........1t a nn TT TT
T TTT−′ ′′ +µ θ=== =−µ θ ′′′

Chapter 19 : Brakes and Dynamometers    /G6C  751
∴             1 11 12
221231t a n.........1t a nn
nT TT TT
TTTTT−′′ +µ θ=××× × = −µ θ ′′′  ... (iii)
Braking torque on the drum of effective radius re,
   TB = (T1 – T2) re
        = (T1 – T2) r  ... [Neglecting thickness of band]
Note : For the first block, the tension in the tight side is T1 and in the slack side is 1T′ and for the second block,

the tension in the tight side is  1T′and in the slack side is 2T′. Similarly for the third block, the tension in the
tight side is 2T′ and in the slack side is 3T′and so on. For the last block, the tension in the tight side is
Tn-1 and in the slack side is T2.
Example 19.11. In the band and block brake

shown in Fig. 19.21, the band is lined with 12 blockseach of which subtends an angle of 15° at the centreof the rotating drum. The thickness of the blocks is 75mm and the diameter of the drum is 850 mm. If, whenthe brake is in action, the greatest and least tensionsin the brake strap are T
1 and T2, show that
12
1
21t a n 7 . 5
1t a n 7 . 5T
T+µ °=−µ °, where µ is the
coefficient of friction for the blocks.
With the lever arrangement as shown in

Fig.19.21, find the least force required at C for theblocks to absorb 225 kW at 240 r.p.m. The coefficient
of friction between the band and blocks is 0.4.
Solution. Given : n = 12;    
2θ = 15°  or  θ = 7.5°;   t = 75 mm = 0.075 m;   d = 850 mm
= 0.85m ;    Power = 225 kW = 225 × 103 W;    N = 240 r.p.m.;    µ = 0.4
Since OA > OB, therefore the force at C must act downward. Also, the drum rotates clock-

wise, therefore the end of the band attached to A will be slack with tension T2 (least tension) and the
end of the band attached to B will be tight with tension T1 (greatest tension).
Consider one of the blocks (say first block) as shown in Fig. 19.22. This is in equilibrium
under the action of the following four forces :
2. Tension in the slack side (1T′) or the tension in the band between the first and second block,

11 N() s i n 7 . 5TT R′+° = ... (i)
11 N() c o s 7 . 5 .TT R′−° = µ ... (ii)
Dividing equation (ii) by (i), we have
11
11() c o s 7 . 5
() s i n 7 . 5′−°=µ
′+°TT
TT    or   11
11tan7.5TT
TT′−=µ °
′+All dimensions in mm.
Fig. 19.21
Fig. 19.22

752  /G6C   Theory of Machines
∴  11 1 1 .t a n 7 . 5 .t a n 7 . 5 TT T T′′−=µ ° +µ °
or                11(1 tan7.5 ) (1 tan7.5 )TT ′ −µ ° = +µ °
∴                       1
11t a n 7 . 5
1t a n 7 . 5T
T+µ °=−µ °′
Similarly, for the other blocks, the ratio of tensions 12
23TT
TT′′
=
′′ etc. remains constant.
 Therefore for 12 blocks having greatest tension T1 and least tension T2 is
        12
1
21t a n 7 . 5
1t a n 7 . 5T
T+µ °=−µ °
Least force required at C

Let         P = Least force required at C.
We know that diameter of band,
         D = d + 2t = 0.85 + 2 × 0.075 = 1 m
∴ Power absorbed  = 12() .
60TTD N−π
or                3
12Power 60 225 10 60
1 240TTDN×× ×−= =ππ × ×= 17 900 N  ... (iii)
We have proved that
              12 12 12
1
21 tan7.5 1 0.4 0.1317 1.05273.55
1 tan7.5 1 0.4 0.1317 0.9473+µ ° + ×   == = =   −µ ° − ×    T
T
... (iv)
From equations (iii) and (iv), we find that

       T1 = 24  920 N, and T2 = 7020 N
              P  × 500 = T2 × 150 – T1 × 30 = 7020 × 150 – 24 920 × 30 = 305 400
∴           P = 305 400 / 500 = 610.8 N Ans.
Example 19.12.  A band and block brake, having 14 blocks each of which subtends an angle
of 15° at the centre, is applied to a drum of 1 m effective diameter. The drum and flywheel mounted

on the same shaft has a mass of 2000 kg and a combined radius of gyration of 500 mm. The two endsof the band are attached to pins on opposite sides of the brake lever at distances of 30 mm and 120mm from the fulcrum. If a force of 200 N is applied at a distance of 750 mm from the fulcrum, find:
1. maximum braking torque,  2. angular retardation of the drum, and 3. time taken by the
system to come to rest from the rated speed of 360 r.p.m.

The coefficient of friction between blocks and drum may be taken as 0.25.
Solution. Given : n = 14 ; 
2θ = 15°     or   θ = 7.5° ;   d = 1 m    or    r = 0.5 m ;    m = 2000kg;
k = 500 mm = 0.5 m ;    P = 200 N ;    N = 360 r.p.m. ;    l = 750 mm ;    µ = 0.25
1. Maximum braking torque
The braking torque will be maximum when OB > OA and the drum rotates anticlockwise as
shown in Fig. 19.23. The force P must act upwards and the end of the band attached to A is tight under

tension T1 and the end of the band attached to B is slack under tension T2.

Chapter 19 : Brakes and Dynamometers            /G6C          753
Taking moments about O,
           200 × 750 + T1 × 30 = T2 × 120
        12 T2 – 3T1 = 15 000              . . . (i)
We know that             1
21t a n
1t a nnT
T+µ θ=−µ θ
          = 141 0.25tan 7.5
1 0.25tan 7.5+°
−°
          = 141 0.25 0.1317
1 .025 0.1317+×−×

          = (1.068)14 = 2.512 . . . (ii)
                 T1 = 8440 N, and T2 = 3360 N

T1 = 8440 N, and T2 = 3360 N
We know that maximum braking torque,
                   B1 2( ) (8440 3360)0.5 2540 N-m TT T r=− = − = Ans.
2.  Angular retardation of the drum
Let                  α = Angular retardation of the drum.
We know that braking torque ( TB ),
              222540 . . . 2000(0.5) 500 Im k=α = α = α = α
∴                                2540/ 500α= = 5.08 rad/s2 Ans.
3.  Time taken by the system to come to rest
Let                     t  = Required time.

Since the system is to come to rest from the rated speed of 360 r.p.m., therefore
Initial angular speed, 12 360 / 60 37.7 rad/sω=π × =
and final angular speed,20ω=
We know that21 .t ω= ω− α  . . . (– ve sign due to retardation )
∴                   1/ 37.7 / 5.08 t=ω α=  = 7.42 s Ans.
19.10. Internal Expanding Brake19.10. Internal Expanding Brake19.10. Internal Expanding Brake19.10. Internal Expanding Brake19.10. Internal Expanding Brake

An internal expanding brake consists of two shoes S1 and S2 as shown in Fig. 19.24. The
outer surface of the shoes are lined with some friction material (usually with Ferodo) to increase the
coefficient of friction and to prevent wearing away of the metal. Each shoe is pivoted at one end abouta fixed fulcrum O
1 and O2 and made to contact a cam at the other end. When the cam rotates, the

shoes are pushed outwards against the rim of the drum. The friction between the shoes and the drumproduces the braking torque and hence reduces the speed of the drum. The shoes are normally held inoff position by a spring as shown in Fig. 19.24. The drum encloses the entire mechanism to keep outdust and moisture. This type of brake is commonly used in motor cars and light trucks.All dimensions in mm
Fig. 19.23

754       /G6C               Theory of Machines
Fig. 19.24.  Internal expanding brake. Fig. 19.25.  Forces on an internal expanding brake.
We shall now consider the forces acting on such a brake, when the drum rotates in the
anticlockwise direction as shown in Fig. 19.25. It may be noted that for the anticlockwise direction,
the left hand shoe is known as leading  or primary shoe  while the right hand shoe is known as  trailing
or secondary shoe .
Let  r = Internal radius of the wheel rim,

 b = Width of the brake lining,
p1 = Maximum intensity of normal
        pressure,
            pN = Normal pressure,
F1= Force exerted by the cam on
        the leading shoe, and
F2 = Force exerted by the cam on
        the trailing shoe.
Consider a small element of the brake lining
AC subtending an angle δθ at the centre. Let OA
makes an angle θ with OO1 as shown in Fig. 19.25. It
is assumed that the pressure distribution on the shoe

is nearly uniform, however the friction lining wearsout more at the free end. Since the shoe turns aboutO
1, therefore the rate of wear of the shoe lining at A
will be proportional to the radial displacement of that point. The rate of wear of the shoe lining variesdirectly as the perpendicular distance from O
1 to OA, i.e. O1B. From the geometry of the figure,
                 O1B = OO1 sinθ
and normal pressure at A,
         N1 Nsin or sin pp p∝θ = θ
∴     Normal force acting on the element,

 NRδ  = Normal pressure × Area of the element
          = 1 N(.. ) s i n (.. )pb r p b rδθ = θ δθ
and braking or friction force on the element,
    N 1.s i n ( . .) FR p b rδ= µ × δ = µ θ δ θ
∴  Braking torque due to the element about O,    
                 2
B1 1 .s i n ( . .) . ( s i n .) TF rp b r r p b rδ= δ × = µ θ δ θ = µ θ δ θ
Internal expanding brake.Loading ShoeReturn Spring
40 mm70 mm overall,
50 mm spring,
one on each
side
135 mm

one on each
side
135 mm
35 mm25 mmTrailing
Shoe
110 mm overall
(behind shoes)60 mm overall, 25 mm
springLever

Chapter 19 : Brakes and Dynamometers            /G6C          755
and total braking torque about O for whole of one shoe,
         []2
2
1
122
B1 1 sin cos T pbr d pbrθ
θ
θ
θ=µ θ θ=µ − θ∫
= 2
12 1(cos cos ) pb rµθ − θ
Moment of normal force NRδ  of the element about the fulcrum O1,
   NN 1 N 1 (s i n ) MR O B R O Oδ= δ ×= δ θ
= 12
11 1sin (.. )( sin ) sin (.. ) pb r O O p b r O Oθδ θ θ = θδ θ
∴  Total moment of normal forces about the fulcrum O1,
     22
1122

22
1122
N1 1 1 1s i n (.. ) ... s i n Mp b r O O p b r O O dθθ
θθ=θ δ θ = θ θ∫∫
= 2
11 11... ( 1 c o s 2)2pb r O O dθ
θ−θ θ∫        ...2 1sin (1 cos2 )2θ= − θ∵
= 2
11 11s i n 2...22pb r O Oθ
θθθ−
= 2112 1 1s i n2 s i n2 1...22 2pb r O Oθθ θ− − θ+
= 121 1 2 111. . . ( ) (sin 2 sin 2 )22pb r O Oθ− θ+ θ− θ
Moment of frictional force Fδ about the fulcrum O1,
      F1 ( cos ) MF A B F r O Oδ= δ ×= δ− θ ... (∵ AB = r – OO1 cosθ)
1 1sin ( . . ) ( cos )pb r r O O=µ θ δθ − θ

11.. . ( s i n s i n c o s )pb rr O O=µ θ− θ θδθ
= 11.. . s i n s i n 22OOpb r rµθ − θ δ θ... ( 2sin cos sin2 )θθ = θ∵
∴   Total moment of frictional force about the fulcrum O1,
         MF =2
11
1 sin sin 22OOpb r r dθ
θµθ − θ θ ∫
2
11
1 cos cos 24OOpbr rθ
θ=µ − θ+ θ
        = 1112 2 1 1 cos cos 2 cos cos 244OO OOpbr r rµ− θ + θ + θ − θ

= 111 2 2 1 (cos cos ) (cos 2 cos 2 )4OOpbr rµθ − θ + θ − θInternal exparding brake.

756       /G6C               Theory of Machines
Now for leading shoe, taking moments about the fulcrum O1,
   F1 × l = MN – MF
and for trailing shoe, taking moments about the fulcrum O2,
 F2 × l  = MN + MF
Note :  If MF > MN, then the brake becomes self locking.
Example 19.13. The arrangement of an internal expanding friction brake, in which the
brake shoe is pivoted at ‘C’ is shown in Fig. 19.26. The distance ‘CO’ is 75 mm, O being the centre

of the drum. The internal radius of the brake drum is100 mm. The friction lining extends over an arc AB, suchthat the angle AOC is 135° and angle BOC is 45°. Thebrake is applied by means of a force at Q, perpendicularto the line CQ, the distance CQ being 150 mm.
The local rate of wear on the lining may be taken as
proportional to the normal pressure on an element at an
angle of ‘
θ’ with OC and may be taken as equal to
 p1 sin θ, where p1 is the maximum intensity of normal
pressure.

pressure.
The coefficient of friction may be taken as 0.4 and
the braking torque required is 21 N-m. Calculate the force
Q required to operate the brake when 1. The drum rotates
clockwise, and 2. The drum rotates anticlockwise.
Solution.  Given : OC = 75 mm ; r = 100 mm ;
2θ = 135° = 135 × π /180 = 2.356 rad ; 1θ = 45° = 45 × π/180 = 0.786 rad ; l = 150 mm ;
µ = 0.4 ; TB = 21 N-m = 21 × 103 N-mm
1. Force ‘Q’ required to operate the brake when drum rotates clockwise

We know that total braking torque due to shoe ( TB ),
 132
12 21 10 . . . (cos cos ) pb r ×= µ θ −θ
12
1 0.4 (100) (cos 45 cos135 ) 5656 . pb p b=× × ° − ° =
∴3
1. 21 10 / 5656 3.7pb=× =
Total moment of normal forces about the fulcrum C,
     N1 2 1 1 211. . . ( ) (sin 2 sin 2 )22Mp b r O C=θ − θ + θ − θ
= 113.7 100 75 (2.356 0.786) (sin 90 sin 270 )22××× − + ° − °
= 13 875 (1.57 + 1) = 35 660 N-mm
and total moment of friction force about the fulcrum C,

F1 1 2 2 1 . . . (cos cos ) (cos 2 cos 2 )4OCMp b r r=µ θ − θ + θ − θ
= 0.4 × 3.7 × 100 75100 (cos 45 cos135 ) (cos 270 cos90 )4°− ° + °− °
= 148 × 141.4 = 20 930 N-mmAll dimensions in mm
Fig. 19.26

Chapter 19 : Brakes and Dynamometers            /G6C          757
Taking moments about the fulcrum C, we have
 Q × 150 = MN + MF = 35 660 + 20 930 = 56 590
∴                     Q = 56 590 / 150 = 377 N  Ans.
2. Force ‘Q’ required to operate the brake when drum rotates anticlockwise
 Q × 150 = MN – MF = 35 660 – 20 930 = 14 730
∴                    Q  = 14 730/150 =  98.2 N Ans.

19.11. Braking of a Vehicle19.11. Braking of a Vehicle19.11. Braking of a Vehicle19.11. Braking of a Vehicle19.11. Braking of a Vehicle
In a four wheeled moving vehicle, the brakes may be applied to
1. the rear wheels only,
2. the front wheels only, and
3. all the four wheels.
In all the above mentioned three types of

braking, it is required to determine the retardationof the vehicle when brakes are applied. Since thevehicle retards, therefore it is a problem ofdynamics. But it may be reduced to an equivalentproblem of statics by including the inertia force inthe system of forces actually applied to the vehicle.The inertia force is equal and opposite to thebraking force causing retardation.
Now, consider a vehicle moving up an
inclined plane, as shown in Fig. 19.27.

Let  
α  =  Angle of inclination of the plane to the horizontal,
m= Mass of the vehicle in kg (such that its weight is m.g newtons),
h= Height of the C.G. of the vehicle above the road surface in metres,
x= Perpendicular distance of C.G. from the rear axle in metres,
L=  Distance between the centres of the rear and front wheels (also called wheel
base) of the vehicle in metres,
RA= Total normal reaction between the ground and the front wheels in newtons,

RB= Total normal reaction between the ground and the rear wheels in newtons,
µ = Coefficient of friction between the tyres and road surface, and
a= Retardation of the vehicle in m/s2.
We shall now consider the above mentioned three cases of braking, one by one. In all these
cases, the braking force acts in the opposite direction to the direction of motion of the vehicle.
1. When the brakes are applied to the rear wheels only

It is a common way of braking the vehicle in which the braking force acts at the rear wheels
only.
Let   FB = Total braking force (in newtons) acting at the rear wheels due to the
         application of the brakes. Its maximum value is µ. RB.
The various forces acting on the vehicle are shown in Fig. 19.27. For the equilibrium of the
vehicle, the forces acting on the vehicle must be in equilibrium.
Resolving the forces parallel to the plane,

B .. s i n . Fm g m a+α =  . . . (i)Fig. 19.27.  Motion of vehicle up the inclined
plane and brakes are applied to rear wheels only.

758       /G6C               Theory of Machines
Resolving the forces perpendicular to the plane,
AB .c o s RRm g+= α  . . . (ii)
Taking moments about G, the centre of gravity of the vehicle,
FB × h + RB × x = RA (L – x)  . . . (iii)
Substituting the value of FB = µ. RB, and RA = m.g cosα – RB  [from equation (ii) ] in the
above expression, we have
 µ.RB × h + RB × x = (m.g cos α – RB) (L – x)
RB (L + µ. h) = m.g cos α (L – x)
∴ RB = .c o s( )
.mg L x
Lhα−
+µ

.mg L x
Lhα−
+µ
and RA = B.c o s( ).c o s .c o s.mg L xmg R mgLhα−α− = α−+µ
. cos ( . )
.mg x h
Lhα+ µ=+µ
We know from equation  (i),
            BB B .s i n .sin sin
mFm g F Rag g
mm+α µ== + α = + α
               .c o s ( )sin.gL xgLhµα −=+ α+µ . . . (Substituting the value of RB)
Notes : 1.  When the vehicle moves on a level track, then α = 0.
∴ BA.( ) .( .);..mg L x mg x hRRLh Lh−+ µ==+µ +µ  and  .( )
.gL xaLhµ−=+µ

2. If the vehicle moves down the plane, then equation (i) becomes
B .s i n . Fm g m a−α =
∴BB .. c o s ( ).sin .sin sin.FRg L xag g gmm L hµµ α −=− α = − α = − α+µ
2. When the brakes are applied to front wheels only
It is a very rare way of braking the
vehicle, in which the braking force acts at the
front wheels only.Let F
A = Total braking force (in newtons)
     acting at the front wheels due to   the application of brakes. Its    maximum value is µ. R
A.

A.
The various forces acting on the vehicle
are shown in Fig. 19.28.
Resolving the forces parallel to the
plane,
A .s i n . Fm g m a+α = . . . (i)
Resolving the forces perpendicular to
the plane,
    AB .c o s RRm g+= α  . . . (ii)Fig. 19.28.  Motion of the vehicle up the inclined
plane and brakes are applied to front wheels only.

Chapter 19 : Brakes and Dynamometers            /G6C          759
FA × h + RB × x = RA (L – x)
Substituting the value of FA = µ.RA and RB = m.g cosα– RA [from equation (ii) ] in the above
expression, we have
µ.RA × h + (m.g cos α– RA) x = RA (L – x)
µ.RA × h + m.g cos α× x = RA × L
∴A.c o s
.mg xRLhα×=−µ
andBA.c o s.c o s .c o s.mg xRm g Rm gLhα×=α − =α −−µ
    ..c o s 1 .c o s
..xL h xmg mg
Lh Lh  −µ −=α − =α  −µ −µ 

We know from equation (i),
  AA .s i n . .s i n Fm g Rm gamm+α µ +α==
     = ..c o s .s i n
(. )mg x mg
Lh m mµα × α+−µ . . . (Substituting the value of RA)
    = .c o ssin.gxgLhµα ×+α−µ
Notes :  1. When the vehicle moves on a level track, then α = 0.
∴ AB.. ( . );;
..mg x mg L h xRR
Lh Lh×− µ −==−µ −µ   and   .
.gxa
Lhµ⋅=−µ
2. When the vehicle moves down the plane, then equation  (i) becomes
  A .s i n . Fm g m a−α =

A .s i n . Fm g m a−α =
∴  AA .. c o s.sin .sin sinm.FRg xag g gmL hµµ α ×=− α = − α = − α−µ
3. When the brakes are applied to all the four
wheels
This is the most common way of braking
the vehicle, in which the braking force acts on
both the rear and front wheels.
Let  FA = Braking force provided by the
             front wheels = µ. RA, and
    FB = Braking force provided by the
              rear wheels = µ. RB.
A little consideration will show that when

the brakes are applied to all the four wheels, thebraking distance ( i.e. the distance in which the
vehicle is brought to rest after applying thebrakes) will be the least. It is due to this reasonthat the brakes are applied to all the four wheels.
are shown in Fig. 19.29.Fig. 19.29.  Motion of the vehicle up the inclined
plane and the brakes are applied to allthe four wheels.

760       /G6C               Theory of Machines
                      AB .s i n . FFm g m a++ α =  . . . (i)
           AB .c o s RRm g+= α  . . . (ii)
                     AB B A() ( )FF h Rx R L x++ × = −  . . . (iii)
Substituting the value of AA BB ., . FR FR=µ =µ  and BA .c o s Rm g R=α −  [From equation
(ii)] in the above expression,

AB A A() ( . c o s ) ( )RR h m g R x R L xµ+ + α − = −
AA A A( . cos ) ( . cos ) ( )Rm g R h m g R x R L xµ+ α − + α − = −
        A ..c o s .c o smg h mg x R Lµα × +α × = ×
∴               RA = . cos ( . )mg h x
Lαµ +
and             BAcos ( . ). cos . cosmg h xRm g Rm gLαµ +=α − =α −
                           = ... cos 1 . coshx L hxmg mgLLµ+ − µ−  α− = α   
   AB.. s i n .RR m g m aµ+ µ+ α =
  AB() . s i n .RR m g m aµ++ α =

AB() . s i n .RR m g m aµ++ α =
   ... c o s .s i n .mg mg maµα + α =  . . . [From equation (ii)]
∴          (. c o s s i n ) ag=µ α + α
∴                        AB.(. ) .;. ;mg h x L h xRR m g
LLµ+ − µ− ==  and a = g.µ
2. If the vehicle moves down the plane, then equation  (i) may be written as
            AB .s i n . FF m g m a+− α =
or                       AB() . s i n .RR m g m aµ+− α =

..c o s .s i n .mg mg maµα −α =
and                          (. c o s s i n ) ag=µ α − α
Example 19.14.  A car moving on a level road at a speed 50 km/h has a wheel base 2.8
metres, distance of C.G. from ground level 600 mm, and the distance of C.G. from rear wheels 1.2
metres. Find the distance travelled by the car before coming to rest when brakes are applied,
1. to the rear wheels, 2. to the front wheels, and  3. to all the four wheels.

The coefficient of friction between the tyres and the road may be taken as 0.6.
Solution. Given : u = 50 km/h = 1 3.89 m/s ; L = 2.8 m ; h = 600 mm = 0.6 m ; x = 1.2 m ; µ = 0.6
 Let      s  = Distance travelled by the car before coming to rest.
1. When brakes are applied to the rear wheels
Since the vehicle moves on a level road, therefore retardation of the car,
             2 . ( ) 0.6 9.81(2.8 1.2)2.98 m/s. 2.8 0.6 0.6gL xaLhµ −×−== =+µ + ×

Chapter 19 : Brakes and Dynamometers            /G6C          761
We know that for uniform retardation,
22(13.89)
2 2 2.98usa==×= 32.4 m  Ans.
2. When brakes are applied to the front wheels
2 . . 0.6 9.18 1.22.9 m/s.2 . 8 0 . 6 0 . 6gxaLhµ× ×== =−µ − ×
22(13.89)
2 2 2.9usa==× = 33.26 m  Ans.
3. When the brakes are applied to all the four wheels

2. 9.81 0.6 5.886 m/s ag=µ = × =
22(13.89)
2 2 5.886usa==× = 16.4 m  Ans.
Example 19.15.  A vehicle moving on a rough plane inclined at 10° with the horizontal at a
speed of 36 km/h has a wheel base 1.8 metres. The centre of gravity of the vehicle is 0.8 metre from

the rear wheels and 0.9 metre above the inclined plane. Find the distance travelled by the vehiclebefore coming to rest and the time taken to do so when 1. The vehicle moves up the plane, and 2. The
vehicle moves down the plane.
The brakes are applied to all the four wheels and the coefficient of friction is 0.5.
Solution. Given :  
 α = 10°;  u = 36 km / h = 10 m / s ; L = 1.8 m ; x = 0.8 m ; h = 0.9 m ; µ = 0.5

Let                s = Distance travelled by the vehicle before  coming to rest, and
              t = Time taken by the vehicle in coming to rest.
1.  When the vehicle moves up the plane and brakes are applied to all the four wheels
Since the vehicle moves up the inclined plane, therefore retardation of the vehicle,
(c o s s i n) ag=µ α +α
    = 9.81 (0.5cos10 sin10 ) 9.81(0.5 0.9848 0.1736) °+ ° = × + = 6.53 m/s2
22(10)
2 2 6.53usa==× = 7.657 m Ans.

22(10)
and final velocity of the vehicle ( v),
0 . 10 6.53ua t t=+ = − . . .(Minus sign due to retardation)
∴                 t = 10 / 6.53 = 1.53 s Ans.
2.  When the vehicle moves down the plane and brakes are applied to all the four wheels
Since the vehicle moves down the inclined plane, therefore retardation of the vehicle,
(c o s s i n) ag=µ α −α
   9.81(0.5cos10 sin10 ) 9.81(0.5 0.9848 0.1736)=° − ° = × − = 3.13 m/s2

762       /G6C               Theory of Machines
22(10)
2 2 3.13usa==×= 16 m  Ans.
0 = u + a.t = 10 – 3.13 t . . . (Minus sign due to retardation)
∴   t = 10/3.13 = 3.2 s Ans.
Example 19.16.  The wheel base of a car is 3 metres and its centre of gravity is 1.2 metres
ahead the rear axle and 0.75 m above the ground level. The coefficient of friction between the wheels

and the road is 0.5. Determine the maximum deceleration of the car when it moves on a level road,if the braking force on all the wheels is the same and no wheel slip occurs.
Solution. Given : L = 3 m ; x = 1.2 m ; h = 0.75 m ; µ = 0.5
Let a = Maximum deceleration of the car,
m = Mass of the car,
F
A and FB = Braking forces at
the front and
rear wheels
 respectively, and
 RAandRB = Normal reactions
   at the front and
 rear wheels
 respectively.
The various forces acting on the car are

shown in Fig. 19.30.
We shall consider the following two cases:
(a) When the slipping is imminent at the rear wheels
We know that when the brakes are applied to all the four wheels and the vehicle moves on a
level road, then
            B.3 0 . 5 0 . 7 5 1 . 2.9 . 8 1 4 . 6 6 N3Lh xRm g m mL−µ − − × − == × =  
and       FA + FB = m.a or 2µ.   RB = m.a . . . (∵ FB = FA and FB = µ. RB)
∴            2 × 0.5 × 4.66 m = m.a or a = 4.66 m/s2

(b) When the slipping is imminent at the front wheels
            A. ( . ) 9.81(0.5 0.75 1.2)5.15 N
3mg h x mRm
Lµ+ × × +== =
and                    FA + FB = m.a      or          2µ . RA = m.a . . . (∵  FA = FB and FA = µ . RA)
∴      2 × 0.5 × 5.15 m = m.a      or                 a = 5.15 m/s2
Hence the maximum possible deceleration is 4.66 m/s2 and slipping would occur first at the

rear wheels. Ans.
19.12. Dynamometer19.12. Dynamometer19.12. Dynamometer19.12. Dynamometer19.12. Dynamometer
A dynamometer is a brake but in addition it has a device to measure the frictional resistance.
Knowing the frictional resistance, we may obtain the torque transmitted and hence the power of the
engine.Fig. 19.30

Chapter 19 : Brakes and Dynamometers            /G6C          763
19.13. Types of Dynamometers19.13. Types of Dynamometers19.13. Types of Dynamometers19.13. Types of Dynamometers19.13. Types of Dynamometers
Following are the two types of
dynamometers, used for measuring the brake
power of an engine.
1. Absorption dynamometers, and
2. Transmission dynamometers.
In the absorption  dynamometers , the

entire energy or power produced by theengine is absorbed by the friction resistancesof the brake and is transformed into heat,during the process of measurement. But inthe transmission dynamometers , the energy
is not wasted in friction but is used for doingwork. The energy or power produced by theengine is transmitted through the dynamom-eter to some other machines where the powerdeveloped is suitably measured.

19.14. Classification of Absorption Dynamometers19.14. Classification of Absorption Dynamometers19.14. Classification of Absorption Dynamometers19.14. Classification of Absorption Dynamometers19.14. Classification of Absorption Dynamometers
The following two types of absorption dynamometers are important from the subject point of
view :
1. Prony brake dynamometer, and  2. Rope brake dynamometer.
These dynamometers are discussed, in detail, in the following pages.

19.15. Prony Brake Dynamometer19.15. Prony Brake Dynamometer19.15. Prony Brake Dynamometer19.15. Prony Brake Dynamometer19.15. Prony Brake Dynamometer
A simplest form of an absorption type dynamometer is a prony brake dynamometer, as shown

in Fig. 19.31. It consists of two wooden blocks placed around a pulley fixed to the shaft of an enginewhose power is required to be measured. The blocks are clamped by means of two bolts and nuts, asshown in Fig. 19.31. A helical spring is provided between the nut and the upper block to adjust thepressure on the pulley to control its speed. The upper block has a long lever attached to it and carriesa weight W at its outer end. A counter weight is placed at the other end of the lever which

brake when unloaded. Two stops S, S are provided to limit the motion of the lever.
Fig. 19.31.  Prony brake dynamometer.
When the brake is to be put in operation, the long end of the lever is loaded with suitable
weights W and the nuts are tightened until the engine shaft runs at a constant speed and the lever is in
horizontal position. Under these conditions, the moment due to the weight W must balance the mo-
ment of the frictional resistance between the blocks and the pulley.

764       /G6C               Theory of Machines
Let W= Weight at the outer end of the lever in newtons,
L= Horizontal distance of the weight W
from the centre of the pulley in metres,
F= Frictional resistance between the blocks
and the pulley in newtons,
R= Radius of the pulley in metres, and
N= Speed of the shaft in r.p.m.
We know that the moment of the frictional re-
sistance or torque on the shaft,
                T = W.L = F.R N-m
Work done in one revolution

       = Torque × Angle turned in radians
                     = 2N - mT×π
  ∴    Work done per minute
            = 2N - mTN×π
We know that brake power of the engine,
            Work done per min. 2 . 2.. w a t t s60 60 60TN W L NBP×π ×π== =
Notes : 1.  From the above expression, we see that while determining the brake power of engine with the help of
a prony brake dynamometer, it is not necessary to know the radius of the pulley, the coefficient of friction

between the wooden blocks and the pulley and the pressure exerted by tightening of the nuts.
2. When the driving torque on the shaft is not uniform, this dynamometer is subjected to severe oscil-
lations.
19.16. Rope Brake Dynamometer19.16. Rope Brake Dynamometer19.16. Rope Brake Dynamometer19.16. Rope Brake Dynamometer19.16. Rope Brake Dynamometer
It is another form of absorption type dynamometer which is most commonly used for measur-

ing the brake power of the engine. It consists of one, two or more ropes wound around the flywheel orrim of a pulley fixed rigidly to the shaft of an engine. The upper end of the ropes is attached to a springbalance while the lower end of the ropes is kept in position by applying a dead weight as shown in Fig.19.32. In order to prevent the slipping of the rope over the flywheel, wooden blocks are placed atintervals around the circumference of the flywheel.

In the operation of the brake, the engine is made to run at a constant speed. The frictional
torque, due to the rope, must be equal to the torque being transmitted by the engine.
Let W = Dead load in newtons,
S = Spring balance reading in newtons,
D = Diameter of the wheel in metres,
d = diameter of rope in metres, and
N = Speed of the engine shaft in r.p.m.
∴ Net load on the brake
   = ( W – S) N
We know that distance moved in one revolution
  = () mDdπ+
Another dynamo

Chapter 19 : Brakes and Dynamometers            /G6C          765
∴    Work done per revolution
= () () N - mWS D d−π+
and work done per minute
= () () N - mWS D d N−π+
Fig. 19.32.  Rope brake dynamometer.
∴    Brake power of the engine,
 Work done per min ( ) ( )B.P watts60 60WS D d N−π+==
If the diameter of the rope ( d) is neglected, then brake
power of the engine,
()B.P. watts60WS D N−π=
Note: Since the energy produced by the engine is absorbed by the

frictional resistances of the brake and is transformed into heat,
therefore it is necessary to keep the flywheel of the engine cool withsoapy water. The flywheels have their rims made of a channel sectionso as to receive a stream of water which is being whirled round bythe wheel. The water is kept continually flowing into the rim and isdrained away by a sharp edged scoop on the other side, as shown in
Fig. 19.32.
Example 19.17. In a laboratory experiment, the

following data were recorded with rope brake:
Diameter of the flywheel 1.2 m; diameter of the rope
12.5 mm; speed of the engine 200 r.p.m.; dead load on the
brake 600 N; spring balance reading 150 N. Calculate thebrake power of the engine.
Solution. Given : D = 1.2 m ; d = 12.5 mm
= 0.0125 m ; N = 200 r.p.m ; W = 600 N ; S = 150 N
 An engine is being readied for
testing on a dynamometer

766       /G6C               Theory of Machines
( ) ( ) (600 150) (1.2 0.0125)200B.P. 5715 W
60 60WS D d N−π+ − π +== =
       = 5.715 kW Ans.
19.17.19.17.19.17.19.17.19.17. Classification of Transmission DynamometersClassification of Transmission DynamometersClassification of Transmission DynamometersClassification of Transmission DynamometersClassification of Transmission Dynamometers

The following types of transmission dynamometers are important from the subject point of
view :
1. Epicyclic-train dynamometer,  2. Belt transmission dynamometer, and 3. Torsion dyna-
mometer.
We shall now discuss these dynamometers, in detail, in the following pages.
19.18.19.18.19.18.19.18.19.18. Epicyclic-train DynamometerEpicyclic-train DynamometerEpicyclic-train DynamometerEpicyclic-train DynamometerEpicyclic-train Dynamometer
Fig. 19.33.  Epicyclic train dynamometer.

An epicyclic-train dynamometer, as shown in Fig. 19.33, consists of a simple epicyclic train
of gears, i.e. a spur gear, an annular gear (a gear having internal teeth) and a pinion. The spur gear is
keyed to the engine shaft ( i.e. driving shaft) and rotates in anticlockwise direction. The annular gear
is also keyed to the driving shaft and rotates in clockwise direction. The pinion or the intermediate

gear meshes with both the spur and annular gears. The pinion revolves freely on a lever which ispivoted to the common axis of the driving and driven shafts. A weight w is placed at the smaller end
of the lever in order to keep it in position. A little consideration will show that if the friction of the pinon which the pinion rotates is neglected, then the tangential effort P exerted by the spur gear on the
pinion and the tangential reaction of the annular gear on the pinion are equal.

Since these efforts act in the upward direction as shown, therefore total upward force on the
lever acting through the axis of the pinion is 2 P. This force tends to rotate the lever about its fulcrum
and it is balanced by a dead weight W at the end of the lever. The stops S, S are provided to control the
movement of the lever.
For equilibrium of the lever, taking moments about the fulcrum F,
 2P × a = W.L or P = W.L /2a
Let   R = Pitch circle radius of the spur gear in metres, and

N  = Speed of the engine shaft in r.p.m.
∴   Torque transmitted,   T = P.R
and power transmitted                    2. 2watts60 60TN P R N×π ×π==

Chapter 19 : Brakes and Dynamometers            /G6C          767
19.19.  Belt Transmission Dynamometer-Froude or Throneycroft Transmission19.19.  Belt Transmission Dynamometer-Froude or Throneycroft Transmission19.19.  Belt Transmission Dynamometer-Froude or Throneycroft Transmission19.19.  Belt Transmission Dynamometer-Froude or Throneycroft Transmission19.19.  Belt Transmission Dynamometer-Froude or Throneycroft Transmission
DynamometerDynamometerDynamometerDynamometerDynamometer

When the belt is transmitting power from one pulley to another, the tangential effort on the
driven pulley is equal to the difference between the tensions in the tight and slack sides of the belt. A
belt dynamometer is introduced to measure directly the difference between the tensions of the belt,while it is running.
Fig. 19.34.  Froude or Throneycroft transmission dynamometer.
A belt transmission dynamometer, as shown in Fig. 19.34, is called a Froude or Throneycroft

transmission dynamometer. It consists of a pulley A (called driving pulley) which is rigidly fixed to
the shaft of an engine whose power is required to be measured. There is another pulley B (called
driven pulley) mounted on another shaft to which the power from pulley A is transmitted. The pulleys
A and B are connected by means of a continuous belt passing round the two loose pulleys C and D
which are mounted on a T-shaped frame. The frame is pivoted at E and its movement is controlled by

two stops S,S. Since the tension in the tight side of the belt ( T1) is greater than the tension in the slack
side of the belt ( T2), therefore the total force acting on the pulley C (i.e. 2T1) is greater than the total
force acting on the pulley D (i.e. 2T2). It is thus obvious that the frame causes movement about E in
the anticlockwise direction. In order to balance it, a weight W is applied at a distance L from E on the
frame as shown in Fig. 19.34.

Now taking moments about the pivot E, neglecting friction,
1222 .Ta T a W L×= ×+ or12.
2WLTT
a−=
Let D = diameter of the pulley A in metres, and
∴         Work done in one revolution = 12() N - mTT D−π
and workdone per minute                    = 12()N - mTT D N−π
∴   Brake power of the engine, 12()B.P. watts
60TT D N−π=
Example 19.18.  The essential features of a transmission dynamometer are shown in Fig.

19.35. A is the driving pulley which runs at 600 r.p.m. B and C are jockey pulleys mounted on ahorizontal beam pivoted at D, about which point the complete beam is balanced when at rest. E is thedriven pulley and all portions of the belt between the pulleys are vertical. A, B and C are each 300mm diameter and the thickness and weight of the belt are neglected. The length DF is 750 mm.
Find : 1. the value of the weight W to maintain the beam in a horizontal position when

4.5 kW is being transmitted, and 2. the value of W, when the belt just begins to slip on pulley A. The
coefficient of friction being 0.2 and maximum tension in the belt 1.5 kN.

768       /G6C               Theory of Machines
Fig. 19.35.  All dimensions in mm.
Solution. Given : NA = 600 r.p.m. : DA = DB = DC = 300 mm = 0.3 m
1. Value of the weight W to maintain the beam in a horizontal position
Given : Power transmitted ( P) = 4.5 kW = 4500 W
Let T1 = Tension in the tight side of the belt on pulley A, and
T2 = Tension in the slack side of the belt on pulley A.
∴  Force acting upwards on the pulley C = 2T1
and force acting upwards on the pulley B = 2T2

Now taking moments about the pivot D,
W × 750 = 2 T1 × 300 – 2 T2 × 300 = 600 ( T1 – T2)
∴  T1 – T2 = W × 750 / 600 = 1.25 W N
12 A A() 1.25 0.3 6004500 11.7860 60TT D N WW−π ×π× ×== =
∴    W = 4500 / 11.78 = 382 N  Ans.
2. Value of W, when the belt just begins to slip on A
Given :   µ = 0.2 ; T1 = 1.5 kN = 1500 N
1
22.3log . 0.2 0.6284T
T=µθ= ×π=
. . . (∵θ = 180° = π rad)
    1

    1
20.6284log 0.27322.3==
T
T    or   1
21.876T
T=   . . . (Taking antilog of 0.2732)
∴  T2 = T1/ 1.876 = 1500 / 1.876 = 800 N
               W × 750 = 2 T1 × 300 – 2 T2 × 300 = 2 × 1500 × 300 – 2 × 800 × 300
                = 420 × 103
∴            W = 420 × 103/ 750 = 560 N  Ans.

Chapter 19 : Brakes and Dynamometers            /G6C          769
19.20. Torsion Dynamometer19.20. Torsion Dynamometer19.20. Torsion Dynamometer19.20. Torsion Dynamometer19.20. Torsion Dynamometer
A torsion dynamometer is used for measuring large powers particularly the power transmit-
ted along the propeller shaft of a turbine or motor vessel. A little consideration will show that when

the power is being transmitted, then the driving end of the shaft twists through a small angle relativeto the driven end of the shaft. The amount of twist depends upon many factors such as torque actingon the shaft ( T), length of the shaft ( l), diameter of the shaft ( D) and modulus of rigidity ( C) of the
material of the shaft. We know that the torsion equation is
. TC
Jlθ=
where θ = Angle of twist in radians, and
 J = Polar moment of inertia of the shaft.

J = Polar moment of inertia of the shaft.
For a solid shaft of diameter D, the polar moment of inertia
4
32JDπ=×
and for a hollow shaft of external diameter D and internal diameter d, the polar moment of inertia,
44()32JD dπ=−
From the above torsion equation,
..CJTk
l=× θ = θ
where k = C.J/l is a constant for a particular shaft. Thus, the torque acting on the shaft is proportional

to the angle of twist. This means that if the angle of twist is measured by some means, then the torqueand hence the power transmitted may be determined.
We know that the power transmitted
2
60TNP×π=  watts, where N is the speed in r.p.m.
A number of dynamometers are used to measure the angle of twist, one of which is discussed

in Art. 19.21. Since the angle of twist is measured for a small length of the shaft, therefore somemagnifying device must be introduced in the dynamometer for accurate measurement.
Example 19.19.  A torsion dynamometer is fitted to a propeller shaft of a marine engine. It is

found that the shaft twists 2° in a length of 20 metres at 120 r.p.m. If the shaft is hollow with 400 mmexternal diameter and 300 mm internal diameter, find the power of the engine. Take modulus ofrigidity for the shaft material as 80  GPa.
Solution. Given : 
θ = 2° = 2 × π /180 = 0.035 rad ; l = 20 m ; N = 120 r.p.m. ; D = 400 mm
= 0.4 m ; d = 300 mm = 0.3 m ; C = 80 GPa =  80 × 109 N/m2
We know that polar moment of inertia of the shaft,
44 4 4 4( ) (0.4) (0.3) 0.0017m32 32JD dππ =− = −=

and torque applied to the shaft,
93 . 80 10 0.00170.035 238 10 N-m20CJTl××=× θ = × =×
32 238 10 2 120
60 60×π × ×π ×==TNP  = 2990 × 103 W = 2990 kW  Ans.

770  /G6C   Theory of Machines
19.21. Bevis-Gibson Flash Light Torsion Dynamometer
Fig. 19.36.  Bevis-Gibson flash light torsion dynamometer.
It depends upon the fact that the light travels in a straight line through air of uniform density
and the velocity of light is infinite. It consists of two discs A and B fixed on a shaft at a convenient
distance apart, as shown in Fig. 19.36 ( a). Each disc has a small radial slot and these two slots are in

the same line when no power is transmitted and there is no torque on the shaft. A bright electric lampL, behind the disc A, is fixed on the bearing of the shaft. This lamp is masked having a slot directly
opposite to the slot of disc A. At every revolution of the shaft, a flash of light is projected through the
slot in the disc A towards the disc B in a direction parallel to the shaft. An eye piece E is fitted behind

the disc B on the shaft bearing and is capable of slight circumferential adjustment.
When the shaft does not transmit any torque ( i.e. at rest), a flash of light may be seen after
every revolution of the shaft, as the positions of the slit do not change relative to one another as shownin Fig. 19.36 ( b). Now when the torque is transmitted, the shaft twists and the slot in the disc B
changes its position, though the slots in L, A and E are still in line. Due to this, the light does not reach

to the eye piece as shown in Fig. 19.36 ( c). If the eye piece is now moved round by an amount equal
to the lag of disc B, then the slot in the eye piece will be opposite to the slot in disc B as shown in Fig.
19.36 (d) and hence the eye piece receives flash of light. The eye piece is moved by operating a
micrometer spindle and by means of scale and vernier, the angle of twist may be measured upto1/100th of a degree.
The torsion meter discussed above gives the angle of twist of

the shaft, when the uniform torque is transmitted during each revolutionas in case of turbine shaft. But when the torque varies during each revo-lution as in reciprocating engines, it is necessary to measure the angle oftwist at several different angular positions. For this, the discs A and B

are perforated with slots arranged in the form of spiral as shown in Fig.19.37. The lamp and the eye piece must be moved radially so as to bringthem into line with each corresponding pair of slots in the discs.
EXERCISES
1.A single block brake, as shown in Fig. 19.38, has the drum diameter 250 mm. The angle of contact is

90° and the coefficient of friction between the drum and the lining is 0.35. If the operating force of650 N is applied at the end of the lever, determine the torque that may be transmitted by the blockbrake. [Ans. 65.6 N-m]Fig. 19.37.   Perforated
disc.

Chapter 19 : Brakes and Dynamometers    /G6C  771
                        Fig. 19.38                                                        Fig. 19.39
2.The layout and dimensions of a double shoe brake is shown in Fig. 19.39. The diameter of the brake

drum is 300 mm and the contact angle for each shoe is 90°. If the coefficient of friction for the brakelining and the drum is 0.4, find the spring force necessary to transmit a torque of 30 N-m. Alsodetermine the width of the brake shoes, if the bearing pressure on the lining material is not to exceed0.28 N/mm
2. [Ans. 98.4 N ; 5 mm]
3.The arrangements of a transmission brake is shown in Fig. 19.40. The arms are pivoted at O1 and O2

and when force is applied at the end of a hand lever, the screw AB rotates. The left and right hand

threads working in nuts on the ends of the arms move the arms together and thus apply the brake. Theforce on the hand lever is applied 400 mm from the axis of the screw.The drum is 240 mm in diameter and the angle subtended by each is 90°. The screw has six squarethreads with a mean diameter of 20 mm and a lead of 55 mm. Assuming a coefficient of friction for thebraking surface as 0.3 and for the threads 0.15, determine the force on the hand lever required to setthe brake when the torque on the

required to setthe brake when the torque on the drum is 245 N-m. [Ans. 86.5 N]

All dimensions in mm.          All dimensions in mm.
Fig. 19.40 Fig. 19.41
4.The layout and dimensions of the block brake are shown in Fig. 19.41. The diameter of the wheel is
300 mm and the contact angle for each block is 90°. If the coefficient of friction for the brake liningand wheel is 0.4 and the torque on the wheel is 30 N-m, find the force P on the operating arm required
to set the brake for anticlockwise rotation of the wheel. [Ans. 10 N]

5.A simple band brake is operated by a lever of length 500 mm. The brake drum has a diameter of 500
mm and the brake band embraces 5/8 of the circumference. One end of the band is attached to thefulcrum of the lever while the other end is attached to a pin on the lever 100 mm from the fulcrum. Ifthe effort applied to the end of the lever is 2 kN and the coefficient of friction is 0.25, find themaximum braking torque on the drum. [Ans. 4.2 kN-m]All dimensions in mm.

772  /G6C   Theory of Machines
6.A differential band brake acting on the 3/4 th of the circumference of a drum of 450 mm diameter, is

to provide a braking torque of 225 N-m. One end of the band is attached to a pin 100 mm from thefulcrum of the lever and the other end to another pin 25 mm from the fulcrum on the other side of itwhere the operating force is also acting. If the operating force is applied at 500 mm from the fulcrumand the coefficient of friction is 0.25, find the two values of the operating force corresponding to twodirections of rotation of the drum. [Ans. 16.6 N for clockwise ; 266.6 N for anticlockwise]

7.A differential band brake is shown in Fig. 19.42. The diameter of the drum is 800 mm. The coefficient
of friction between the band and the drum is 0.3 and the angle of embrace is 240°.
     Fig. 19.42      Fig. 19.43
When a force of 600 N is applied at the free end of the lever, find for clockwise and anticlockwise

rotation of the drum: 1. the maximum and minimum forces in the band ; and 2. the torque which canbe applied by the brake.   [Ans. 176 kN, 50 kN, 50.4 kN-m ; 6.46 kN, 1.835 kN, 1.85 kN-m]
8.A differential band brake is shown in Fig. 19.43. The diameter of the drum is 1 metre and rotates at
1200 r.p.m. in the anticlockwise direction. The angle of contact is 320°. The various lengths are :OA = 30 mm; AB = 150 mm and OC = 700 mm. Find the pull

required at the end C of a lever to absorb 40 kW. Also find the
length of AB for self locking. The coefficient of friction may
be taken as 0.2.                             [Ans. 25.7 N ; 91.8 mm]
9.In a band and block brake, the band is lined with 14 blocks,

each of which subtends an angle of 20° at the drum centre.One end of the band is attached to the fulcrum of the brakelever and the other to a pin 150 mm from the fulcrum. Find theforce required at the end of the lever 1 metre long from thefulcrum to give a torque of 4 kN-m. The diameter of the brakedrum is 1 metre and the coefficient of friction between theblocks and the drum is 0.25.                         [Ans. 1712 N]
10.Fig. 19.44 shows the particulars of two brake shoes which act

on the internal surface of a cylindrical brake drum. The brakingforces F
1 and F2 are applied as shown, and each shoe pivots on its fixed fulcrum O1 and O2.
The width of the brake lining is 35 mm. The intensity of pressure at any point A is 0.4 sin θ N/mm2,
where θ is measured as shown from either pivot. The coefficient of friction is 0.4. Determine the
braking torque and the magnitude of the forces F1 and F2. [Ans. 373 N-m ; 685 N, 2323 N]

11.A lorry is moving on a level road at a speed of 36 km/h. Its centre of gravity lies at a distance of 0.6 m
from the ground level. The wheel base is 2.4 metres and the distance of C.G. from the rear wheels is0.9 m. Find the distance travelled by the car before coming to rest when brakes are applied,(a)   to the rear wheels, ( b) to the front wheels, and ( c) to all the four wheels.
The coefficient of friction between the tyres and the road surface is 0.45.
[Ans. 21.55 m; 26.82 m; 11.36 m]

12.A torsion dynamometer is fitted on a turbine shaft to measure the angle of twist. It is observed that the
shaft twists 1.5° in a length of 5 metres at 500 r.p.m. The shaft is solid and has a diameter of 200 mm.If the modulus of rigidity for the shaft material is 85 GPa, find the power transmitted by the turbine.
[Ans. 3662 kW]    Fig. 19.44All dimensions in mm.

Chapter 19 : Brakes and Dynamometers    /G6C  773
1.Distinguish between brakes and dynamometers.
2.Discuss the various types of the brakes.
3.Show that, in a band and block brake, the ratio of the maximum and minimum tensions in the brake
straps is
01t a n
1t a nn
nT
where  T0 = Maximum tension,
 Tn = Minimum tension
  µ = Coefficient of friction between the blocks and drum, and
             2θ = Angle subtended by each block at the centre of the drum.

4.Describe with the help of a neat sketch the principles of operation of an internal expanding shoe.
Derive the expression for the braking torque.
5.What are the leading and trailing shoes of an internal expanding shoe brake ?
6.What is the difference between absorption and transmission dynamometers ? What are torsion dyna-
mometers ?
7.Describe the construction and operation of a prony brake or rope brake absorption dynamometer.

8.Describe with sketches one form of torsion dynamometer and explain with detail the calculations
involved in finding the power transmitted.
9.Explain with neat sketches the Bevis-Gibson flash light dynamometer.
1.The brakes commonly used in railway trains is
(a) shoe brake ( b) band brake
(c) band and block brake ( d) internal expanding brake
2.The brake commonly used in motor cars is
(c) band and block brake ( d) internal
expanding brake

3.In a differential band brake, as shown in Fig. 19.45, the length
OA is greater than OB. In order to apply the brake, the force P
at C should
(a)  be zero ( b) act in upward direction
(c) act in downward direction
4.For the brake to be self locking, the force P at C as shown in
Fig. 19.45, should
(a) be zero
(b)  act in upward direction
5.When brakes are applied to all the four wheels of a moving car, the distance travelled by the car before

it is brought to rest, will be
 (a) maximum  ( b) minimum
6.Which of the following is an absorption type dynamometer ?
(a) prony brake dynamometer ( b) rope brake dynamometer
(c) epicyclic-train dynamometer ( d) torsion dynamometer
ANSWERS
1.(a) 2.(d) 3.(c) 4.(a) 5.(b) 6. (a), (b)Fig. 19.45

774       /G6C               Theory of Machines
CamsCamsCamsCamsCams20
2. Classification of Followers.3. Classification of Cams.4.Terms used in Radial cams.
5. Motion of the Follower .6. Displacement, Velocity and
Acceleration Diagramswhen the Follower Moveswith Uniform Velocity.
7. Displacement, Velocity and
Acceleration Diagramswhen the Follower Moveswith Simple HarmonicMotion.
8. Displacement, Velocity and

Acceleration Diagramswhen the Follower Moveswith Uniform Accelerationand Retardation.
9. Displacement, Velocity and
Acceleration Diagramswhen the Follower Moveswith Cycloidal Motion.
10. Construction of Cam
Profiles.
1 1. Cams with Specified
Contours.
12. Tangent Cam with
Reciprocating RollerFollower.
13. Circular Arc Cam with Flat-
faced Follower.20.1.20.1.20.1.20.1.20.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoduction

A cam is a rotating machine element which gives
reciprocating or oscillating motion to another element knownas follower . The cam and the follower have a line contact

and constitute a higher pair. The cams are usually rotated atuniform speed by a shaft, but the follower motion is pre-determined and will be according to the shape of the cam.The cam and follower is one of the simplest as well as oneof the most important mechanisms found in modernmachinery today. The cams are widely used for operatingthe inlet and exhaust valves of internal combustion engines,automatic attachment of machineries, paper cutting machines,spinning and weaving textile machineries,

and weaving textile machineries, feed mechanismof automatic lathes etc.

20.2.20.2.20.2.20.2.20.2. Classification of FollowersClassification of FollowersClassification of FollowersClassification of FollowersClassification of Followers
The followers may be classified as discussed below :
1. According to the surface in contact . The followers,
according to the surface in contact, are as follows :
(a)Knife edge follower . When the contacting end of
the follower has a sharp knife edge, it is called aknife edge follower, as shown in Fig. 20.1 ( a). The

sliding motion takes place between the contactingsurfaces ( i.e. the knife edge and the cam surface). It
is seldom used in practice because the small area ofcontacting surface results in excessive wear. In knifeedge followers, a considerable side thrust existsbetween the follower and the guide.
774

Chapter 20 : Cams            /G6C          775
(b)Roller follower . When the contacting end of the follower is a roller, it is called a roller
follower, as shown in Fig. 20.1 ( b). Since the rolling motion takes place between the
contacting surfaces ( i.e. the roller and the cam), therefore the rate of wear is greatly reduced.

In roller followers also the side thrust exists between the follower and the guide. Theroller followers are extensively used where more space is available such as in stationarygas and oil engines and aircraft engines.
(c)Flat faced or mushroom follower . When the contacting end of the follower is a perfectly
flat face, it is called a flat-faced follower, as shown in Fig. 20.1 ( c). It may be noted that

the side thrust between the follower and the guide is much reduced in case of flat facedfollowers. The only side thrust is due to friction between the contact surfaces of the followerand the cam. The relative motion between these surfaces is largely of sliding nature butwear may be reduced by off-setting the axis of the follower, as shown in Fig. 20.1 ( f ) so

that when the cam rotates, the follower also rotates about its own axis. The flat facedfollowers are generally used where space is limited such as in cams which operate thevalves of automobile engines.
Note :  When the flat faced follower is circular, it is then called a mushroom follower.
(d)Spherical faced follower . When the contacting end of the follower is of spherical shape,
it is called a spherical faced follower, as shown in Fig. 20.1 ( d). It may be noted that when

a flat-faced follower is used in automobile engines, high surface stresses are produced. Inorder to minimise these stresses, the flat end of the follower is machined to a sphericalshape.
(a) Cam with knife     ( b) Cam with roller ( c) Cam with flat
    edge follower. follower .      faced follower.
(d) Cam with spherical           ( e) Cam with spherical ( f) Cam with offset
     faced follower.     faced follower.     follower.
Fig. 20.1.  Classification of followers.

2. According to the motion of the follower . The followers, according to its motion, are of the
following two types:

776       /G6C               Theory of Machines
(a)Reciprocating or translating follower . When the follower reciprocates in guides as the
cam rotates uniformly, it is known as reciprocating or translating follower. The followersas shown in Fig. 20.1 ( a) to ( d) are all reciprocating or translating followers.
(b)Oscillating or rotating follower . When the uniform rotary motion of the cam is converted

into predetermined oscillatory motion of the follower, it is called oscillating or rotatingfollower. The follower, as shown in Fig 20.1 ( e), is an oscillating or rotating follower.
3. According to the path of motion of the follower . The followers, according to its path of
motion, are of the following two types:
(a)Radial follower . When the motion of the follower is along an axis passing through the
centre of the cam, it is known as radial follower. The followers, as shown in Fig. 20.1 ( a)

to (e), are all radial followers.
(b)Off-set follower . When the motion of the follower is along an axis away from the axis of
the cam centre, it is called off-set follower. The follower, as shown in Fig. 20.1 ( f), is an
off-set follower.
Note :  In all cases, the follower must be constrained to follow the cam. This may be done by springs, gravity
or hydraulic means. In some types of cams, the follower may ride in a groove.

20.3.20.3.20.3.20.3.20.3. Classification of CamsClassification of CamsClassification of CamsClassification of CamsClassification of Cams
Though the cams may be classified in many ways, yet the following two types are important
(a) Cylindrical cam with reciprocating     ( b) Cylindrical cam with oscillating follower.
follower.
Fig. 20.2.  Cylindrical cam.
1. Radial or disc cam . In radial

cams, the follower reciprocates oroscillates in a direction perpendicular tothe cam axis. The cams as shown in Fig.20.1 are all radial cams.
2. Cylindrical cam . In cylindrical
cams, the follower reciprocates oroscillates in a direction parallel to the camaxis. The follower rides in a groove at itscylindrical surface. A cylindrical groovedcam with a reciprocating and an oscillatingfollower is shown in Fig. 20.2 ( a) and ( b)

Note : In actual practice, radial cams are widely
used. Therefore our discussion will be only
confined to radial cams.
In IC engines, cams are widely used to
operate valves.

Chapter 20 : Cams            /G6C          777
20.4.20.4.20.4.20.4.20.4. TTTTTerererererms Used in Radial Camsms Used in Radial Camsms Used in Radial Camsms Used in Radial Camsms Used in Radial Cams
Fig. 20.3 shows a radial cam with reciprocating roller follower. The following terms are
important in order to draw the cam profile.
1. Base circle . It is the smallest circle that can be drawn to the cam profile.

2. Trace point . It is a reference point on the follower and is used to generate the  pitch curve .
In case of knife edge follower, the knife edge represents the trace point and the pitch curvecorresponds to the cam profile. In a roller follower, the centre of the roller represents the trace point.
3. Pressure angle . It is the angle between the direction of the follower motion and a normal

to the pitch curve. This angle is very important in designing a cam profile. If the pressure angle istoo large, a reciprocating follower will jam in its bearings.
4. Pitch point . It is a point on the pitch curve having the maximum pressure angle.
5. Pitch circle . It is a circle drawn from the centre of the cam through the pitch points.
6. Pitch curve . It is the curve generated by the trace point as the follower moves relative to

the cam. For a knife edge follower, the pitch curve and the cam profile are same whereas for aroller follower, they are separated by the radius of the roller.
7. Prime circle . It is the smallest circle that can be drawn from the centre of the cam and
tangent to the pitch curve. For a knife edge and a flat face follower, the prime circle and the basecircle are identical. For a roller follower, the prime circle is larger than the base circle by the radiusof the roller.

8. Lift or stroke . It is the maximum travel of the follower from its lowest position to the
topmost position.
Fig. 20.3.  Terms used in radial cams.
20.5.20.5.20.5.20.5.20.5. Motion of the FollowerMotion of the FollowerMotion of the FollowerMotion of the FollowerMotion of the Follower
The follower, during its travel, may have one of the following motions.
1. Uniform velocity, 2. Simple harmonic motion,  3. Uniform acceleration and retardation,
and 4. Cycloidal motion.

778       /G6C               Theory of Machines
We shall now discuss the displacement, velocity and acceleration diagrams for the cam when
the follower moves with the above mentioned motions.

20.6.20.6.20.6.20.6.20.6. Displacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when the
Follower Moves with Uniform VelocityFollower Moves with Uniform VelocityFollower Moves with Uniform VelocityFollower Moves with Uniform VelocityFollower Moves with Uniform Velocity

The displacement, velocity and acceleration diagrams when a knife-edged follower moves
with uniform velocity are shown in Fig. 20.4 ( a), (b) and ( c) respectively. The abscissa (base)
represents the time ( i.e. the number of seconds required for the cam to complete one revolution) or
it may represent the angular displacement of the cam in degrees. The ordinate represents the dis-placement, or velocity or acceleration of the follower.

Since the follower moves with uniform velocity during its rise and return stroke, therefore
the slope of the displacement curves must be constant. In other words, AB
1 and C1D must be
straight lines. A little consideration will show that the follower remains at rest during part of thecam rotation. The periods during which the follower remains at rest are known as dwell periods , as
shown by lines B
1C1 and DE in Fig. 20.4 ( a). From Fig. 20.4 ( c), we see that the acceleration or

retardation of the follower at the beginning and at the end of each stroke is infinite. This is due tothe fact that the follower is required to start from rest and has to gain a velocity within no time.This is only possible if the acceleration or retardation at the beginning and at the end of each strokeis infinite. These conditions are however, impracticable.
Fig. 20.4.  Displacement, velocity and Fig. 20.5.  Modified displacement, velocity and

acceleration diagrams when the acceleration diagrams when the  follower
      follower moves with uniform velocity.  moves with uniform velocity.
In order to have the acceleration and retardation within

the finite limits, it is necessary to modify the conditions whichgovern the motion of the follower. This may be done byrounding off the sharp corners of the displacement diagramat the beginning and at the end of each stroke, as shown inFig. 20.5 ( a). By doing so, the velocity of the follower
increases gradually to its maximum value at the beginningof each stroke and decreases gradually to zero at the end ofeach stroke as shown in Fig. 20.5 ( b). The modified
Camshaft of an IC engine.

Chapter 20 : Cams            /G6C          779
displacement, velocity and acceleration diagrams are shown in Fig. 20.5. The round corners of the
displacement diagram are usually parabolic curves because the parabolic motion results in a verylow acceleration of the follower for a given stroke and cam speed.

20.7.20.7.20.7.20.7.20.7. Displacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when the

Follower Moves with Simple Harmonic MotionFollower Moves with Simple Harmonic MotionFollower Moves with Simple Harmonic MotionFollower Moves with Simple Harmonic MotionFollower Moves with Simple Harmonic Motion
The displacement, velocity and acceleration diagrams when the follower moves with simple
harmonic motion are shown in Fig. 20.6 ( a), (b) and ( c) respectively. The displacement diagram is
drawn as follows :
1.Draw a semi-circle on the follower stroke as diameter.

2.Divide the semi-circle into any number of even equal parts (say eight).
3.Divide the angular displacements of the cam during out stroke and return stroke into the
same number of equal parts.
4.The displacement diagram is obtained by projecting the points as shown in Fig. 20.6 ( a).
The velocity and acceleration diagrams are shown in Fig. 20.6 ( b) and ( c) respectively. Since

the follower moves with a simple harmonic motion, therefore velocity diagram consists of a sinecurve and the acceleration diagram is a cosine curve. We see from Fig. 20.6 ( b) that the velocity of
the follower is zero at the beginning and at the end of its stroke and increases gradually to amaximum at mid-stroke. On the other hand, the acceleration of the follower is maximum at thebeginning and at the ends of the stroke and diminishes to zero at mid-stroke.

Fig. 20.6.  Displacement, velocity and acceleration diagrams when the follower
moves with simple harmonic motion.
Let S= Stroke of the follower,
     Oθand Rθ = Angular displacement of the cam during out stroke and return stroke of the
follower respectively, in radians, and
ω= Angular velocity of the cam in rad/s.

780       /G6C               Theory of Machines
∴Time required for the out stroke of the follower in seconds,
OO / t=θ ω
Consider a point P moving at a uniform speed Pω radians per sec round the circumference
of a circle with the stroke S as diameter, as shown in Fig. 20.7.
The point ′P (which is the projection of a point P on the diam-
eter) executes a simple harmonic motion as the point P rotates.
The motion of the follower is similar to that of point P′.
∴ Peripheral speed of the point P′,
P

P
OO1
22SSvtππ ω=× =×θ
and maximum velocity of the follower on the outstroke,
P O
OO.
22SSvvπω π ω== ×=θθ
We know that the centripetal acceleration of the point P,
        222 2
P
P2O O() .2 .
2 2( )v SSa
OP Sπω π ω== × = θ θ 
∴  Maximum acceleration of the follower on the outstroke,
       22
O P 2
O.
2( )Saaπω==
θ
Similarly, maximum velocity of the follower on the return stroke,
      R
R.
2Svπω=θ

R
R.
2Svπω=θ
and maximum acceleration of the follower on the return stroke,
       22
R 2
R.
2( )Saπω=
θ
20.8.20.8.20.8.20.8.20.8. Displacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when the

Follower Moves with Uniform Acceleration and RetardationFollower Moves with Uniform Acceleration and RetardationFollower Moves with Uniform Acceleration and RetardationFollower Moves with Uniform Acceleration and RetardationFollower Moves with Uniform Acceleration and Retardation
The displacement, velocity and acceleration diagrams when the follower moves with uniform
acceleration and retardation are shown in Fig. 20.8 ( a), (b) and ( c) respectively. We see that the

displacement diagram consists of a parabolic curve and may be drawn as discussed below :
1.Divide the angular displacement of the cam during outstroke (Oθ) into any even number
of equal parts (say eight) and draw vertical lines through these points as shown in Fig.
20.8 ( a).
2.Divide the stroke of the follower ( S) into the same number of equal even parts.
3.Join Aa to intersect the vertical line through point 1 at B. Similarly, obtain the other points

C, D etc. as shown in Fig. 20.8 ( a). Now join these points to obtain the parabolic curve
for the out stroke of the follower.
4.In the similar way as discussed above, the displacement diagram for the follower during
return stroke may be drawn.
Since the acceleration and retardation are uniform, therefore the velocity varies directly with
the time. The velocity diagram is shown in Fig. 20.8 ( b).
Let         S  = Stroke of the follower,Fig. 20.7.  Motion of a point.

Chapter 20 : Cams            /G6C          781
      Oθ and Rθ = Angular displacement of the cam during out stroke and return stroke
   of the follower respectively, and
        ω = Angular velocity of the cam.
We know that time required for the follower during outstroke,
        OO / t=θ ω
and time required for the follower during return stroke,
        RR / t=θ ω
Mean velocity of the follower during outstroke
           = S/tO
and mean velocity of the follower during return stroke

= S/tR
Fig. 20.8.  Displacement, velocity and acceleration diagrams when the follower moves
with uniform acceleration and retardation.
Since the maximum velocity of follower is equal to twice the mean velocity, therefore maxi-
mum velocity of the follower during outstroke,
        O
OO22 .SSvtω==θ
Similarly, maximum velocity of the follower during return stroke,
       R
R2.Svω=θ

782       /G6C               Theory of Machines
We see from the acceleration diagram, as shown in Fig. 20.8 ( c), that during first half of the
outstroke there is uniform acceleration and during the second half of the out stroke there is uniformretardation. Thus, the maximum velocity of the follower is reached after the time t
O/ 2 (during out
stroke) and tR /2 (during return stroke).
∴  Maximum acceleration of the follower during outstroke,
       2
O
O 2OO O O22. 4 .
/2 . ()v SSatt×ω ω== =

O
θθ. . . (∵  OO / t=θ ω )
Similarly, maximum acceleration of the follower during return stroke,
       2
R2
R4.
()Saω=
θ
20.9.20.9.20.9.20.9.20.9. Displacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when theDisplacement, Velocity and Acceleration Diagrams when the

Follower Moves with Cycloidal MotionFollower Moves with Cycloidal MotionFollower Moves with Cycloidal MotionFollower Moves with Cycloidal MotionFollower Moves with Cycloidal Motion
Fig. 20.9.  Displacement, velocity and acceleration diagrams when the
follower moves with cycloidal motion.
The displacement, velocity and acceleration diagrams when the follower moves with cycloidal
motion are shown in Fig. 20.9 ( a), (b) and ( c) respectively. We know that cycloid is a curve traced

by a point on a circle when the circle rolls without slipping on a straight line.
In case of cams, this straight line is a stroke of the follower which is translating and the
circumference of the rolling circle is equal to the stroke ( S) of the follower. Therefore the radius of

Chapter 20 : Cams            /G6C          783
the rolling circle is /2Sπ . The displacement diagram is drawn as discussed below :
1.Draw a circle of radius /2Sπ with A as centre.
2.Divide the circle into any
number of equal even parts(say six). Project these pointshorizontally on the verticalcentre line of the circle.
These points are shown by
a′
and b′ in Fig. 20.9 ( a).
3.Divide the angular displace-

ment of the cam during out-stroke into the same numberof equal even parts as thecircle is divided. Draw verti-cal lines through these points.
4.Join AB which intersects the
vertical line through  3 ′ at c.
From 
a′ draw a line parallel
to AB intersecting the verti-
cal lines through 1′ and 2′
at a and b respectively.
5.Similarly, from b′ draw a
line parallel to AB intersect-
ing the vertical lines through
4′ and 5′ at d and e respec-
tively.

tively.
6.Join the points A a b c d e B
by a smooth curve. This is the required cycloidal curve for the follower during outstroke.
Let           θ = Angle through which the cam rotates in time t seconds, and
         ω = Angular velocity of the cam.
We know that displacement of the follower after time t seconds,
          
OO12sin2xS  θπ θ=− θπθ  . . . (i)
∴  Velocity of the follower after time t seconds,

OO O12 2cos2dx d dSdt dt dt  θπ π θ θ=× − θπ θ θ  
 . . . [Differentiating equation  (i)]
OO O O2. 21c o s 1c o sSd S
dt  θπ θ ω π θ=× − = −   θθ θ θ   . . . (ii)
Cams are used in Jet and aircraft engines. The
above picture shows an aircraft engine.

784       /G6C               Theory of Machines
The velocity is maximum, when
O2cos 1πθ=−θ     or     
O2πθ=πθ     or    O/2θ=θ
Substituting O/2θ=θ in equation (ii), we have maximum velocity of the follower during
outstroke,
         O
OO.2 .(1 1)SSvωω=+ =θθ
R
R2.Svω=θ
Now, acceleration of the follower after time t sec,
         2
2OO O.2 2sindx S d
dt dt ωπ π θ θ=θθ θ . . . [Differentiating equation (ii)]

2
2O O2. 2sin
()S πω πθ= θ θ . . . d
dtθ=ω∵     . . .  (iii)
The acceleration is maximum, when
  
O2sin 1πθ=θ    or     
O2
2πθ π=θ    or    O/4θ=θ
Substituting O/4θ=θ in equation ( iii), we have maximum acceleration of the follower dur-
ing outstroke, 2
O 2
O2.
()Saπω=
θ
2
R2
R2.
()Saπω=
θ
The velocity and acceleration diagrams are shown in Fig. 20.9 ( b) and ( c) respectively.

20.10.20.10.20.10.20.10.20.10. Construction of Cam Profile for a Radial CamConstruction of Cam Profile for a Radial CamConstruction of Cam Profile for a Radial CamConstruction of Cam Profile for a Radial CamConstruction of Cam Profile for a Radial Cam
In order to draw the cam profile for a radial cam, first of all the displacement diagram for the

given motion of the follower is drawn. Then by constructing the follower in its proper position ateach angular position, the profile of the working surface of the cam is drawn.
In constructing the cam profile, the principle of kinematic inversion is used, i.e. the cam is
imagined to be stationary and the follower is allowed to rotate in the  opposite direction  to the cam
rotation .
The construction of cam profiles for different types of follower with different types of

motions are discussed in the following examples.
Example 20.1.  A cam is to give the following motion to a knife-edged follower :
1. Outstroke during 60° of cam rotation ; 2. Dwell for the next 30° of cam rotation ;
3. Return stroke during next 60° of cam rotation, and 4. Dwell for the remaining 210° of cam
rotation.
The stroke of the follower is 40 mm and the minimum radius of the cam is 50 mm. The

follower moves with uniform velocity during both the outstroke and return strokes. Draw the pro-file of the cam when (a) the axis of the follower passes through the axis of the cam shaft, and
(b) the axis of the follower is offset by 20 mm from the axis of the cam shaft .

Chapter 20 : Cams            /G6C          785
Construction
Fig. 20.10
First of all, the displacement diagram, as shown in Fig. 20.10, is drawn as discussed in the
following steps :
1.Draw a horizontal line AX = 360° to some suitable scale. On this line, mark AS = 60° to
represent outstroke of the follower, ST = 30° to represent dwell, TP = 60° to represent
return stroke and PX = 210° to represent dwell.
2.Draw vertical line AY equal to the stroke of the follower ( i.e. 40 mm) and complete the

rectangle as shown in Fig. 20.10.
3.Divide the angular displacement during outstroke and return stroke into any equal number
of even parts (say six) and draw vertical lines through each point.
4.Since the follower moves with uniform velocity during outstroke and return stroke, there-
fore the displacement diagram consists of straight lines. Join AG and HP.
5. The complete displacement diagram is shown by AGHPX  in Fig. 20.10.

(a) Profile of the cam when the axis of follower passes through the axis of cam shaft
The profile of the cam when the axis of the follower passes through the axis of the cam shaft,
as shown in Fig. 20.11, is drawn as discussed in the following steps :
Fig. 20.11

786       /G6C               Theory of Machines
1.Draw a base circle with radius equal to the minimum radius of the cam ( i.e. 50 mm) with
O as centre.
2.Since the axis of the follower passes through the axis of the cam shaft, therefore mark
trace point A, as shown in Fig. 20.11.
3.From OA, mark angle AOS = 60° to represent outstroke, angle SOT = 30° to represent
dwell and angle TOP = 60° to represent return stroke.

4.Divide the angular displacements during outstroke and return stroke ( i.e. angle AOS and
angle TOP) into the same number of equal even parts as in displacement diagram.
5.Join the points 1, 2, 3 ...etc. and 0′,1′, 2′, 3′,  ... etc. with centre O and produce beyond
the base circle as shown in Fig. 20.11.
6.Now set off 1 B, 2C, 3D ... etc. and 0′H,1′J ... etc. from the displacement diagram.
7.Join the points A, B, C,... M, N, P with a smooth curve. The curve AGHPA  is the complete

profile of the cam.
Notes :  The points B, C, D .... L, M, N may also be obtained as follows :
1. Mark AY = 40 mm on the axis of the follower, and set of Ab, Ac, Ad... etc. equal to the distances 1 B,
2C, 3D... etc. as in displacement diagram.
2. From the centre of the cam O, draw arcs with radii Ob, Oc, Od etc. The arcs intersect the produced
lines O1, O2... etc. at B, C, D ... L, M, N.
(b)Profile of the cam when the axis of the follower is offset by 20 mm from the axis of the cam
shaft

shaft
The profile of the cam when the axis of the follower is offset from the axis of the cam shaft,
as shown in Fig. 20.12, is drawn as discussed in the following steps :
Fig. 20.12
2.Draw the axis of the follower at a distance of 20 mm from the axis of the cam, which
intersects the base circle at A.
3.Join AO and draw an offset circle of radius 20 mm with centre O.

4.From OA, mark angle AOS = 60° to represent outstroke, angle SOT = 30° to represent

Chapter 20 : Cams            /G6C          787
5.Divide the angular displacement during outstroke and return stroke ( i.e. angle AOS and
6.Now from the points 1, 2, 3 ... etc. and 0, 1,2,3′′′′ ... etc. on the base circle, draw tangents
to the offset circle and produce these tangents beyond the base circle as shown in Fig.
20.12.
7.Now set off 1 B, 2C, 3D ... etc. and  0′H,1′J ... etc. from the displacement diagram.

8.Join the points A, B, C ...M, N, P with a smooth curve. The curve AGHP A  is the complete
Example 20.2. A cam is to be designed for a knife edge follower with the following data :
1.Cam lift = 40 mm during 90° of cam rotation with simple harmonic motion.
2.Dwell for the next 30°.
3.During the next 60° of cam rotation, the follower returns to its original position with
simple harmonic motion.
4.Dwell during the remaining 180°.

Draw the profile of the cam when(a) the line of stroke of the follower passes through the axis of the cam shaft, and
(b) the line of stroke is offset 20 mm from the axis of the cam shaft.
The radius of the base circle of the cam is 40 mm. Determine the maximum velocity and
acceleration of the follower during its ascent and descent, if the cam rotates at 240 r.p.m.
Solution.  Given : S = 40 mm = 0.04 m; 
Oθ = 90° = π/2 rad = 1.571 rad ; Rθ = 60° =

π/3 rad = 1.047 rad ; N = 240 r.p.m.
Fig. 20.13
First of all, the displacement diagram, as shown in Fig 20.13, is drawn as discussed in the
1.Draw horizontal line AX = 360° to some suitable scale. On this line, mark AS = 90° to
represent out stroke ; SR = 30° to represent dwell ; RP = 60° to represent return stroke
and PX = 180° to represent dwell.

2.Draw vertical line AY = 40 mm to represent the cam lift or stroke of the follower and
complete the rectangle as shown in Fig. 20.13.
3.Divide the angular displacement during out stroke and return stroke into any equal num-
ber of even parts (say six) and draw vertical lines through each point.
4.Since the follower moves with simple harmonic motion, therefore draw a semicircle with
AY as diameter and divide into six equal parts.

5.From points a, b, c ... etc. draw horizontal lines intersecting the vertical lines drawn through
1, 2, 3 ... etc. and 0′,1′,2′ ...etc. at B, C, D ... M, N, P.
6. Join the points A, B, C ... etc. with a smooth curve as shown in Fig. 20.13. This is the
required displacement diagram.

788       /G6C               Theory of Machines
(a)Profile of the cam when the line of stroke of the follower passes through the axis of the cam
shaft
The profile of the cam when the line of stroke of the follower passes through the axis of the
cam shaft, as shown in Fig. 20.14, is drawn in the similar way as is discussed in Example 20.1.
Fig. 20.14
(b)Profile of the cam when the line of stroke of the follower is offset 20 mm from the axis
of the cam shaft

The profile of the cam when the line of stroke of the follower is offset 20 mm from the axis
of the cam shaft, as shown in Fig. 20.15, is drawn in the similar way as discussed in Example 20.1.
Fig. 20.15

Chapter 20 : Cams            /G6C          789
Maximum velocity of the follower during its ascent and descent
We know that angular velocity of the cam,
22 2 4 025.14
60 60ππ ×ω= = =N rad/s
We also know that the maximum velocity of the
follower during its ascent,
       O
O. 25.14 0.04
2 2 1.571Svπω π× ×==θ× = 1 m/s  Ans.
and maximum velocity of the follower during its
descent,
 R
R. 25.14 0.04
2 2 1.047Svπω π× ×==θ× = 1.51 m/s  Ans.
Maximum acceleration of the follower during its

ascent and descent
We know that the maximum acceleration of the
22 2 2
O22
O. (25.14) 0.04
2( ) 2(1.571)Saπω π==
θ = 50.6 m/s2 Ans.
and maximum acceleration of the follower during its descent,
  22 2 2
R22
R. (25.14) 0.04
2( ) 2(1.047)Saπω π==
θ = 113.8 m/s2 Ans.
Example 20.3.  A cam, with a minimum radius of 25 mm, rotating clockwise at a uniform speed

is to be designed to give a roller follower, at the end of a valve rod, motion described below :
1. To raise the valve through 50 mm during 120° rotation of the cam ;
2. To keep the valve fully raised through next 30°;
3. To lower the valve during next 60°; and
4. To keep the valve closed during rest of the revolution i.e. 150° ;

The diameter of the roller is 20 mm and the diameter of the cam shaft is 25 mm.Draw the profile of the cam when (a) the line of stroke of the valve rod passes through the
axis of the cam shaft, and (b) the line of the stroke is offset 15 mm from the axis of the cam shaft .
The displacement of the valve, while being raised and lowered, is to take place with simple
harmonic motion. Determine the maximum acceleration of the valve rod when the cam shaft rotatesat 100 r.p.m.

Draw the displacement, the velocity and the acceleration diagrams for one complete revolu-
tion of the cam.
Solution. Given : S = 50 mm = 0.05 m ; 
Oθ = 120° = 2 π/3 rad = 2.1 rad ; Rθ = 60° =
π/3 rad = 1.047 rad ; N = 100 r.p.m.
Since the valve is being raised and lowered with simple harmonic motion, therefore the dis-
placement diagram, as shown in Fig. 20.16 ( a), is drawn in the similar manner as discussed in the
previous example.
Role of cams in piston movement.

790       /G6C               Theory of Machines
(((((a)Profile of the cam when the line of stroke of the valve rod passes through the axis of
the cam shaft
The profile of the cam, as shown in Fig. 20.17, is drawn as discussed in the following steps :
1.Draw a base circle with centre O and radius equal to the minimum radius of the cam
( i.e. 25 mm ).
Fig. 20.16
2.Draw a prime circle with centre O and radius,
         −OA = Min. radius of cam + 1
2 Dia. of roller = 125 20 352+× =  mm

3.Draw angle AOS = 120° to represent raising or out stroke of the valve, angle SOT = 30° to
represent dwell and angle TOP = 60° to represent lowering or return stroke of the valve.
4.Divide the angular displacements of the cam during raising and lowering of the valve ( i.e.
angle AOS and TOP ) into the same number of equal even parts as in displacement diagram.
5.Join the points 1, 2, 3, etc. with the centre O and produce the lines beyond prime circle as

shown in Fig. 20.17.
6.Set off 1 B, 2C, 3D etc. equal to the displacements from displacement diagram.
7.Join the points A, B, C ... N, P, A. The curve drawn through these points is known as pitch
curve.

Chapter 20 : Cams            /G6C          791
8.From the points A, B, C ... N, P, draw circles of radius equal to the radius of the roller.
9.Join the bottoms of the circles with a smooth curve as shown in Fig. 20.17. This is the
required profile of the cam.Fig. 20.17
Fig. 20.18

792       /G6C               Theory of Machines
(b) Profile of the cam when the line of stroke is offset 15 mm from the axis of the cam shaft
The profile of the cam when the line of stroke is offset from the axis of the cam shaft, as
shown in Fig. 20.18, may be drawn as discussed in the following steps :
1.Draw a base circle with centre O and radius equal to 25 mm.
2.Draw a prime circle with centre O and radius OA = 35 mm.
3.Draw an off-set circle with centre O and radius equal to 15 mm.

4.Join OA. From OA draw the angular displacements of cam i.e. draw angle AOS = 120°,
angle SOT = 30° and angle TOP = 60°.
5.Divide the angular displacements of the cam during raising and lowering of the valve into
the same number of equal even parts ( i.e. six parts ) as in displacement diagram.
6.From points 1, 2, 3 .... etc. and 0′,1′,3′, ...etc. on the prime circle, draw tangents to the
offset circle.
7.Set off 1 B, 2C, 3D... etc. equal to displacements as measured from displacement diagram.

8.By joining the points A, B, C ... M, N, P, with a smooth curve, we get a pitch curve .
9.Now A, B, C...etc. as centre, draw circles with radius equal to the radius of roller.
10.Join the bottoms of the circles with a smooth curve as shown in Fig. 20.18. This is the
required profile of the cam.
Maximum acceleration of the valve rod
We know that angular velocity of the cam shaft,
2 2 10010.47
We also know that maximum velocity of the valve rod to raise valve,
           O

O
O. 10.47 0.050.392 2 2.1Svπω π× ×== =θ× m/s
and maximum velocity of the valve rod to lower the valve,
         R
R. 10.47 0.050.7852 2 1.047Svπω π× ×== =θ× m/s
The velocity diagram for one complete revolution of the cam is shown in Fig. 20.16 ( b).
We know that the maximum acceleration of the valve rod to raise the valve,
           22 2 2
O22
0. (10.47) 0.056.13
2( ) 2(2.1 )Saπω π== =
θ m/s2 Ans.
and maximum acceleration of the valve rod to lower the valve,
         22 2 2
R22

22 2 2
R22
R. (10.47) 0.0524.67
2( ) 2(1.047)Saπω π== =
θ m/s2 Ans.
The acceleration diagram for one complete revolution of the cam is shown in Fig. 20.16 ( c).
Example 20.4. A cam drives a flat reciprocating follower in the following manner :
During first 120° rotation of the cam, follower moves outwards through a distance of 20 mm

with simple harmonic motion. The follower dwells during next 30° of cam rotation. During next120° of cam rotation, the follower moves inwards  with simple harmonic motion . The follower
dwells for the next 90° of cam rotation.
The minimum radius of the cam is 25 mm. Draw the profile of the cam.

Chapter 20 : Cams            /G6C          793
Since the follower moves outwards and inwards with simple harmonic motion, therefore the
displacement diagram, as shown in Fig. 20.19, is drawn in the similar manner as discussed earlier.
Fig. 20.19
Now the profile of the cam driving a flat reciprocating follower, as shown in Fig. 20.20, is
drawn as discussed in the following steps :
1.Draw a base circle with centre O and radius OA equal to the minimum radius of the cam
(i.e. 25 mm).

2.Draw angle AOS = 120° to represent the outward stroke, angle SOT = 30° to represent
dwell and angle TOP = 120° to represent inward stroke.
3.Divide the angular displacement during outward stroke and inward stroke ( i.e. angles AOS
and TOP ) into the same number of equal even parts as in the displacement diagram.
Fig. 20.20

794       /G6C               Theory of Machines
4.Join the points 1, 2, 3 . . . etc. with centre O and produce beyond the base circle.
5.From points 1, 2, 3 . . . etc., set off 1 B, 2C, 3D . . . etc. equal to the distances measured
from the displacement diagram.
6.Now at points B, C, D . . . M, N, P, draw the position of the flat-faced follower. The axis
of the follower at all these positions passes through the cam centre.

7.The curve drawn tangentially to the flat side of the follower is the required profile of the
cam, as shown in Fig. 20.20.
Example 20.5. Draw a cam profile to drive an oscillating roller follower to the specifica-
tions given below :
(a) Follower to move outwards through an angular displacement of 20° during the first 120°
rotation of the cam ;
(b) Follower to return to its initial position during next 120° rotation of the cam ;
(c) Follower to dwell during the next 120° of cam rotation.

The distance between pivot centre and roller centre = 120 mm ; distance between pivot
centre and cam axis = 130 mm ; minimum radius of cam = 40 mm ; radius of roller = 10 mm ;inward and outward strokes take place with simple harmonic motion.
ConstructionWe know that the angular displacement
of the roller follower
20 20 /180 /9=° =× π = π  rad
Since the distance between the pivot
centre and the roller centre ( i.e. the radius
A1 A) is 120 mm, therefore length of the arc

AA2, as shown in Fig. 20.21, along which the
displacement of the roller actually takes place
120 /9 41.88=× π =  mm
. . . (∵   Length of arc = Radius of arc × Angle subtended by the arc at the centre in radians)
Since the angle is very small, therefore length of chord AA2 is taken equal to the length of arc
AA2. Thus in order to draw the displacement diagram, we shall take lift of the follower equal to
length of chord AA2 i.e. 41.88 mm.
Fig. 20.22

Fig. 20.22
The outward and inward strokes take place with simple harmonic motion, therefore the dis-
placement diagram, as shown in Fig. 20.22, is drawn in the similar way as discussed in Example20.4.Fig. 20.21

Chapter 20 : Cams            /G6C          795
The profile of the cam to drive an oscillating roller follower, as shown in Fig. 20.23, is drawn
as discussed in the following steps :
1.First of all, draw a base circle with centre O and radius equal to the minimum radius of
the cam ( i.e. 40 mm)
2.Draw a prime circle with centre O and radius OA
= Min. radius of cam + radius of roller = 40 + 10 = 50 mm
3.Now locate the pivot centre A1 such that OA1  = 130 mm and AA1 = 120 mm. Draw a

pivot circle with centre O and radius OA1 = 130 mm.
Fig. 20.23
4.Join OA1. Draw angle A1OS = 120° to represent the outward stroke of the follower, angle
SOT = 120° to represent the inward stroke of the follower and angle TOA1 = 120° to
represent the dwell.
5.Divide angles A1OS and SOT into the same number of equal even parts as in the displace-
ment diagram and mark points 1, 2, 3 . . . 4′,5′,6′  on the pivot circle.

6.Now with points 1, 2, 3 . . . 4′,5′,6′(on the pivot circle) as centre and radius equal to
A1A (i.e. 120 mm) draw circular arcs to intersect the prime circle at points 1, 2, 3 . . .
4′,5′,6′.

796       /G6C               Theory of Machines
7.Set off the distances 1 B, 2C, 3D... 4,′L 5′Malong the arcs drawn equal to the distances
as measured from the displacement diagram.
8.The curve passing through the points A, B,  C....L, M, N is known as pitch curve.
9.Now draw circles with A, B, C, D....L, M, N as centre and radius equal to the radius of
roller.
10.Join the bottoms of the circles with a smooth curve as shown in Fig. 20.23. This is the

Example 20.6. A cam, with a minimum radius of 50 mm, rotating clockwise at a uniform
speed, is required to give a knife edge follower the motion as described below :
1. To move outwards through 40 mm during 100° rotation of the cam ; 2. To dwell for next
80° ; 3. To return to its starting position during next 90°, and 4. To dwell for the rest period of a
revolution i.e. 90°.

Draw the profile of the cam(i) when the line of stroke of the follower passes through the centre of the cam shaft, and
(ii) when the line of stroke of the follower is off-set by 15 mm.
The displacement of the follower is to take place with uniform acceleration and uniform
retardation. Determine the maximum velocity and acceleration of the follower when the cam shaftrotates at 900 r.p.m.
Draw the displacement, velocity and acceleration diagrams for one complete revolution of

the cam.
oθ=100° = 100 × π/180 = 1.745 rad ; Rθ = 90° =
π/2 = 1.571 rad ; N = 900 r.p.m.
First of all, the displacement diagram, as shown in Fig. 20.24 ( a), is drawn as discussed in
the following steps :
1.Draw a horizontal line ASTPQ  such that AS represents the angular displacement of the
cam during outward stroke ( i.e. 100° ) to some suitable scale. The line ST represents the

dwell period of 80° after outward stroke. The line TP represents the angular displacement
of the cam during return stroke ( i.e. 90°) and the line PQ represents the dwell period of
90° after return stroke.
2.Divide AS and TP into any number of equal even parts (say six).
3.Draw vertical lines through points 0, 1, 2, 3 etc. and equal to the lift of the valve i.e. 40
mm.
4.Divide the vertical lines 3- f and 3-f′′into six equal parts as shown by points a, b, c . . .

and ′a,′b,′c. . . in Fig. 20.24 ( a).
5.Since the follower moves with equal uniform acceleration and uniform retardation, there-
fore the displacement diagram of the outward and return stroke consists of a double pa-rabola.
6.Join Aa, Ab and Ac intersecting the vertical lines through 1, 2 and 3 at B, C and D respec-
tively.
7.Join the points B, C and D with a smooth curve. This is the required parabola for the half

outstroke of the valve. Similarly the other curves may be drawn as shown in Fig. 20.24.
8.The curve A B C  . . . N P Q  is the required displacement diagram.

Chapter 20 : Cams            /G6C          797
Fig. 20.24
Fig. 20.25

798       /G6C               Theory of Machines
(i)Profile of the cam when the line of stroke of the follower passes through the centre of
The profile of the cam when the line of stroke of the follower passes through the centre of
cam shaft, as shown in Fig. 20.25, may be drawn as discussed in the following steps :
1.Draw a base circle with centre O and radius 50 mm (equal to minimum radius of the
cam).

cam).
2.Divide the base circle such that angle AOS = 100° ; angle SOT = 80° and angle TOP =
90°.
3.Divide angles AOS and TOP into the same number of equal even parts as in displacement
diagram ( i.e. six parts).
4.Join the points 1, 2, 3 . . . and 1′,2′,3′, . . . with centre O and produce these lines beyond
the base circle.
5.From points 1, 2, 3 . . . and 1′,2′,3′, . . . mark the displacements 1 B, 2C, 3D . . . etc. as
measured from the displacement diagram.

6.Join the points A, B, C . . . M, N, P with a smooth curve as shown in Fig. 20.25. This is
the required profile of the cam.
(ii) Profile of the cam when the line of stroke of the follower is offset by 15 mm
The profile of the cam when the line of stroke of the follower is offset may be drawn as
discussed in Example 20.2. The profile of cam is shown in Fig. 20.26.
Fig. 20.26
Maximum velocity of the follower during out stroke and return stroke

          22 9 0 094.2660 60ππ ×ω= = =N rad/s

Chapter 20 : Cams            /G6C          799
We also know that the maximum velocity of the follower during out stroke,
        O
O2 . 2 94.26 0.044.321.745Svω××== =θ m/s Ans.
and maximum velocity of the follower during return
stroke,
        R
R2 . 2 94.26 0.044.81.571Svω××== =θ m/s Ans.
The velocity diagram is shown in Fig. 20.24 ( b).
Maximum acceleration of the follower during out
stroke and return stroke
We know that the maximum acceleration of
the follower during out stroke,
22
O22

22
O22
O4 . 4(94.26) 0.04467
( ) (1.745)Saω== =
θm/s2 Ans.
and maximum acceleration of the follower during return stroke,
    22
R22
R4 . 4(94.26) 0.04
() ( 1 . 5 7 1 )Saω== =
θ 576 m/s2 Ans.
The acceleration diagram is shown in Fig. 20.24 ( c).
Example 20.7. Design a cam for operating the exhaust valve of an oil engine. It is required

to give equal uniform acceleration and retardation during opening and closing of the valve eachof which corresponds to 60° of cam rotation. The valve must remain in the fully open position for20° of cam rotation.
The lift of the valve is 37.5 mm and the least radius of the cam is 40 mm. The follower is
provided with a roller of radius 20 mm and its line of stroke passes through the axis of the cam.

First of all, the displacement diagram, as shown in Fig. 20.27, is drawn as discussed in the
Fig. 20.27
1.Draw a horizontal line ASTP  such that AS represents the angular displacement of the cam
during opening ( i.e. out stroke ) of the valve (equal to 60°), to some suitable scale. The
line ST represents the dwell period of 20° i.e. the period during which the valve remains
A type of roller follower.

800       /G6C               Theory of Machines
fully open and TP represents the angular displacement during closing ( i.e. return stroke)
of the valve which is equal to 60°.
3.Draw vertical lines through points 0, 1, 2, 3 etc. and equal to lift of the valve i.e. 37.5
mm.
4.Divide the vertical lines 3 f and 3′′finto six equal parts as shown by the points a, b, c . ..
and ′a, ′b, ′c. . . in Fig. 20.27.

5.Since the valve moves with equal uniform acceleration and retardation, therefore the dis-
placement diagram for opening and closing of a valve consists of double parabola.
6.Complete the displacement diagram as shown in Fig. 20.27.
Now the profile of the cam, with a roller follower when its line of stroke passes through the
axis of cam, as shown in Fig. 20.28, is drawn in the similar way as discussed in Example 20.3.
Fig. 20.28

Fig. 20.28
Example 20.8.  A cam rotating clockwise at a uniform speed of 1000 r.p.m. is required to
give a roller follower the motion defined below :
1. Follower to move outwards through 50 mm during 120° of cam rotation,
2. Follower to dwell for next 60° of cam rotation,
3. Follower to return to its starting position during next 90° of cam rotation,
4. Follower to dwell for the rest of the cam rotation.
The minimum radius of the cam is 50 mm and the diameter of roller is 10 mm. The line of

stroke of the follower is off-set by 20 mm from the axis of the cam shaft. If the displacement of the

Chapter 20 : Cams            /G6C          801
follower takes place with uniform and equal acceleration and retardation on both the outward and
return strokes, draw profile of the cam and find the maximum velocity and acceleration during outstroke and return stroke.
Solution.  Given : N = 1000 r.p.m. ; S = 50 mm = 0.05 m ; 
Oθ = 120° = 2 π/3 rad = 2.1 rad ;
Rθ = 90° = π/2 rad = 1.571 rad
Since the displacement of the follower takes place with uniform and equal acceleration and

retardation on both outward and return strokes, therefore the displacement diagram, as shown inFig. 20.29, is drawn in the similar manner as discussed in the previous example. But in this case,the angular displacement and stroke of the follower is divided into eight equal parts.
Fig. 20.29
Now, the profile of the cam, as shown in Fig. 20.30, is drawn as discussed in the following
steps :
(i.e. 50 mm).

802       /G6C               Theory of Machines
2.Draw a prime circle with centre O and radius
OA = Minimum radius of the cam + radius of roller = 50 + 5 = 55 mm
3.Draw an off-set circle with centre O and radius equal to 20 mm.
4.Divide the angular displacements of the cam during out stroke and return stroke into eight
equal parts as shown by points 0, 1, 2 . . . and 0′,1′,2′. . . etc. on the prime circle in Fig.
20.30.
5.From these points draw tangents to the off-set circle.

6.Set off 1 B, 2C, 3D . . . etc. equal to the displacements as measured from the displacement
diagram.
7.By joining the points A, B, C . . . T, U, A with a smooth curve, we get a pitch curve.
8.Now from points A, B, C . . . T, U, draw circles with radius equal to the radius of the
roller.
9.Join the bottoms of these circles with a smooth curve to obtain the profile of the cam as
shown in Fig. 20.30.

     2 2 1000104.760 60ππ ×ω= = =N rad/s.
follower during outstroke,
           O
O2 . 2 104.7 0.0552.1Svω××== =θ m/s Ans.
stroke,
           R
R2 . 2 104.7 0.056.661.571Svω××== =θ m/s Ans.
follower during out stroke,
        22
O22

        22
O22
O4 . 4(104.7) 0.05497.2
() ( 2 . 1 )Saω== =
θ m/s2 Ans.
        22
R22
R4 . 4(104.7) 0.05888
( ) (1.571)Saω== =
θ m/s2 Ans.
Example 20.9. Construct the profile of a cam to suit the following specifications :
Cam shaft diameter = 40 mm ; Least radius of cam = 25 mm ; Diameter of roller = 25 mm;

Angle of lift = 120° ; Angle of fall = 150° ; Lift of the follower = 40 mm ; Number of pauses aretwo of equal interval between motions.
A rocker using a cam.

Chapter 20 : Cams            /G6C          803
During the lift, the motion is S.H.M. During the fall the motion is uniform acceleration and
deceleration. The speed of the cam shaft is uniform. The line of stroke of the follower is off-set12.5 mm from the centre of the cam.
First of all the displacement diagram, as shown in Fig. 20.31, is drawn as discussed in the
1.Since the follower moves with simple harmonic motion during lift ( i.e. for 120° of cam

rotation), therefore draw the displacement curve ADG  in the similar manner as discussed
in Example 20.2.
2.Since the follower moves with uniform acceleration and deceleration during fall ( i.e. for
150° of cam rotation), therefore draw the displacement curve HLP consisting of double
parabola as discussed in Example 20.6.
Fig. 20.31
Now the profile of the cam, when the line of stroke of the follower is off-set 12.5 mm from

the centre of the cam, as shown in Fig. 20.32, is drawn as discussed in the following steps :
1.Draw a base circle with centre O and radius equal to the least radius of cam ( i.e. 25 mm).
Fig. 20.32

804       /G6C               Theory of Machines
OA = Least radius of cam + radius of roller = 25 + 25/2 = 37.5 mm
3.Draw a circle with centre O and radius equal to 20 mm to represent the cam shaft.
4. Draw an offset circle with centre O and radius equal to 12.5 mm.
5. Join OA. From OA draw angular displacements of the cam, i.e. draw angle AOS = 120° to
represent lift of the follower, angle SOT = 45° to represent pause, angle TOP = 150° to

represent fall of the follower and angle POA  = 45° to represent pause.
Note.  Since the number of pauses are two of equal interval between motions ( i.e. between lift and
fall of the follower), therefore angular displacement of each pause
 360 (120 150 )452°− °+ °== °
6.Divide the angular displacements during lift and fall ( i.e. angle AOS and TOP) into the
same number of equal even parts ( i.e. six parts) as in the displacement diagram.

7.From points 1, 2, 3 . . . etc. and  0,1,2,3′′′′ . . . etc. on the prime circle, draw tangents to
the off-set circle.
8.Set off 1 B, 2C, 3D . . . etc. equal to the displacements as measured from the displacement
diagram.
9.By joining the points A, B, C . . . M, N, P with a smooth curve, we get a pitch curve.
10.Now with A, B, C . . . etc. as centre, draw circles with radius equal to the radius of roller.
11.Join the bottoms of the circles with a smooth curve as shown in Fig. 20.32. This is the

Example 20.10.  It is required to set out the profile of a cam to give the following motion to
the reciprocating follower with a flat mushroom contact face :
(i) Follower to have a stroke of 20 mm during 120° of cam rotation ;
(ii) Follower to dwell for 30° of cam rotation ;
(iii) Follower to return to its initial position during 120° of cam rotation ; and
(iv) Follower to dwell for remaining 90° of cam rotation.

The minimum radius of the cam is 25 mm. The out stroke of the follower is performed with
simple harmonic motion and the return stroke with equal uniform acceleration and retardation.
ConstructionSince the out stroke of the follower is performed with simple harmonic motion and the return
stroke with uniform acceleration and retardation, therefore the displacement diagram, as shown inFig. 20.33, is drawn in the similar manner as discussed in the previous example.
Fig. 20.33

Chapter 20 : Cams            /G6C          805
The profile of the cam with a flat mushroom contact face reciprocating follower, as shown in
Fig. 20.34, is drawn in the similar way as discussed in Example 20.4.
Fig. 20.34
Example 20.11.  It is required to set out the profile of a cam with oscillating follower for the
following motion :
(a) Follower to move outward through an angular displacement of 20° during 90° of cam

rotation ; (b) Follower to dwell for 45° of cam rotation ; (c) Follower to return to its original
position of zero displacement in 75° of cam rotation ; and (d) Follower to dwell for the remaining
period of the revolution of the cam.
The distance between the pivot centre and the follower roller centre is 70 mm and the roller

diameter is 20 mm. The minimum radius of the cam corresponds to the starting position of thefollower as given in (a). The location of the pivot point is 70 mm to the left and 60 mm above the
axis of rotation of the cam. The motion of the follower is to take place with S.H.M. during outstroke and with uniform acceleration and retardation during return stroke.
ConstructionWe know that the angular displacement of the
roller follower,

Since the distance between the pivot centre and
the roller centre ( i.e. radius A1A) is 70 mm, therefore
length of arc AA2, as shown in Fig. 20.35, along which
the displacement of the roller actually takes place
70 /9 24.5=× π=  mm Fig. 20.35

806       /G6C               Theory of Machines
AA2. Thus in order to draw the displacement diagram, we shall take lift of the follower equal to the
length of chord AA2 i.e. 24.5 mm.
Fig. 20.36
Fig. 20.37

Chapter 20 : Cams            /G6C          807
The follower moves with simple harmonic motion during out stroke and with uniform accel-
eration and retardation during return stroke. Therefore, the displacement diagram, as shown in Fig.20.36, is drawn in the similar way as discussed in the previous example.
The profile of the cam, as shown in Fig. 20.37, is drawn as discussed in the following steps :1.First of all, locate the pivot point A
1 which is 70 mm to the left and 60 mm above the axis

of the cam.
2.Since the distance between the pivot centre A1 and the follower roller centre A is 70 mm
and the roller diameter is 20 mm, therefore draw a circle with centre A and radius equal to
the radius of roller i.e. 10 mm.
3.We find that the minimum radius of the cam
      60 10 50=−=  mm
∴  Radius of the prime circle,
OA = Min. radius of cam + Radius of roller = 50 + 10 = 60 mm
4.Now complete the profile of the cam in the similar way as discussed in Example 20.5.

Example 20.12.  Draw the profile of the cam when the roller follower moves with cycloidal
motion during out stroke and return stroke, as given below :
1. Out stroke with maximum displacement of 31.4 mm during 180° of cam rotation,
2. Return stroke for the next 150° of cam rotation,
3. Dwell for the remaining 30° of cam rotation.
The minimum radius of the cam is 15 mm and the roller diameter of the follower is 10 mm.

The axis of the roller follower is offset by 10 mm towards right from the axis of cam shaft.
First of all, the displacement diagram, as shown in Fig. 20.38, is drawn as discussed in the
Fig. 20.38
1.Draw horizontal line ASP such that AS = 180° to represent the out stroke, SN = 150° to
represent the return stroke and NP = 30° to represent the dwell period.
2.Divide AS and SN into any number of even equal parts (say six).

3.From the points 1, 2, 3 . . . etc. draw vertical lines and set-off equal to the stroke of the
follower.
4.From a point G draw a generating circle of radius,
    Stroke 31.4
22==ππr  = 5 mm

808       /G6C               Theory of Machines
5.Divide the generating circle into six equal parts and from these points draw horizontal
lines to meet the vertical diameter at ′a, G and ′b.
6.Join AG and GN. From point ′a, draw lines parallel to AG and GN to intersect the verti-
cal lines drawn through 1, 2,  4′ and 5′at B, C, L and M respectively. Similarly draw
parallel lines from ′bintersecting the vertical lines through 4, 5, 1′ and 2′at E, F, H and
J respectively.
Fig. 20.39

Fig. 20.39
7.Join the points A, B, C . . . L, M, N with a smooth curve.
8.The curve A B C . . . L M N is the required displacement diagram.
Now the profile of the cam, as shown in Fig. 20.39, may be drawn in the similar way as
discussed in Example 20.9.
20.11.20.11.20.11.20.11.20.11. Cams with Specified ContoursCams with Specified ContoursCams with Specified ContoursCams with Specified ContoursCams with Specified Contours

In the previous articles, we have discussed about the design of the profile of a cam when the
follower moves with the specified motion. But, the shape of the cam profile thus obtained may bedifficult and costly to manufacture. In actual practice, the cams with specified contours (cam pro-files consisting of circular arcs and straight lines are preferred) are assumed and then motion of thefollower is determined.

20.12.20.12.20.12.20.12.20.12. Tangent Cam with Reciprocating Roller FollowerTangent Cam with Reciprocating Roller FollowerTangent Cam with Reciprocating Roller FollowerTangent Cam with Reciprocating Roller FollowerTangent Cam with Reciprocating Roller Follower
When the flanks of the cam are straight and tangential to the base circle and nose circle, then
the cam is known as a tangent cam , as shown in Fig. 20.40. These cams are usually symmetrical

about the centre line of the cam shaft. Such type of cams are used for operating the inlet andexhaust valves of internal combustion engines. We shall now derive the expressions for displace-ment, velocity and acceleration of the follower for the following two cases :

Chapter 20 : Cams            /G6C          809
1. When the roller has contact with the straight flanks ; and
2. When the roller has contact with the nose.
Let               r1 = Radius of the base circle or minimum radius of the cam,
                r2 = Radius of the roller,
     r3 = Radius of nose,
    α = Semi-angle of action of cam or angle of ascent,
                  θ= Angle turned by the cam from the beginning of the roller displacement,

φ = Angle turned by the cam for contact of roller with the straight flank, and
    ω = Angular velocity of the cam.
1. When the roller has contact with straight flanks . A roller having contact with straight
flanks is shown in Fig. 20.40. The point O is the centre of cam shaft and the point K is the centre
of nose. EG and PQ are straight flanks of the cam. When the roller is in lowest position, ( i.e. when

the roller has contact with the straight flank at E ), the centre of roller lies at B on the pitch curve.
Let the cam has turned through an angle * θ ( less than φ) for the roller to have contact at any point
(say F) between the straight flanks EG. The centre of roller at this stage lies at C. Therefore
displacement (or lift or stroke) of the roller from its lowest position is given by
       1c o s
cos cos−θ=−= −= θθ OBxO CO B O BO B

         121c o s()cos−θ=+ θrr . . . 12 ()OB OE EB r r=+= +∵  . . . (i)
Fig. 20.40.  Tangent cam with reciprocating roller follower having contact with straight flanks.
* Since the cam is assumed to be stationary, the angle θ is turned by the roller.

810       /G6C               Theory of Machines
Differentiating equation ( i) with respect to t, we have velocity of the follower,
       122sin()
cosθθ θ ==×=+ θ θdx dx d dvr rdt d dt dt
122sin()
cosθ=ω + θrr . . . (/ )θ= ω∵dd t  . . . (ii)
From equation (ii), we see
that when θ increases, sin θ
increases and cos θ decreases. In
other words, sin θ/ cos2
θincreases. Thus the velocity is
maximum where θ is maximum.
This happens when θ = φ i.e.
when the roller just leaves contact

with the straight flank at G or when
the straight flank merges into acircular nose.
∴  Maximum velocity of the
follower,
        122sin()
cosmaxvr rφ=ω + φ
Now differentiating equation
(ii) with respect to t, we have
acceleration of the follower,
θ==×θdv dv dadt d dt
   2
124cos .cos sin 2cos sin()
cosθθ −θ × θ × − θ θ=ω + θ drrdt
   222
123cos 2sin()
cosθ+ θ=ω + θ rr. . . d
dtθ=ω∵
   222
123cos 2 (1 cos )()
cosθ+ − θ=ω + 

222
θ rr
   22
1232c o s()
cos−θ=ω + θrr. . . (iii)
A little consideration will show that the acceleration is minimum when 2
32c o s
cos−θ
θ is mini-
mum. This is only possible when (22c o s−θ ) is minimum and cos3θ is maximum. This happensA car in the assembly line.

Chapter 20 : Cams            /G6C          811
when θ = 0°, i.e. when the roller is at the beginning of its lift along the straight flank (or when the
roller has contact with the straight flank at E ).
∴  Minimum acceleration of the follower,
        2
12()minar r=ω +
The acceleration is maximum when θ = φ, i.e. when the roller just leaves contact with the
straight flank at G or when the straight flank merges into a circular nose.
∴ Maximum acceleration of the follower,
    122
232c o s()

122
232c o s()
cosmaxar r−φ=ω + φ
2. When the roller has contact with the nose . A roller having contact with the circular nose
at G is shown in Fig 20.41. The centre of roller lies at D on the pitch curve. The displacement is
usually measured from the top position of the roller, i.e. when the roller has contact at the apex of
the nose (point H) and the centre of roller lies at J on the pitch curve.

Fig. 20.41.  Tangent cam with reciprocating roller follower having contact with the nose.
Let           1θ = Angle turned by the cam measured from the position when the roller
 is at the top of the nose.

812       /G6C               Theory of Machines
The displacement of the roller is given by
 () () () =−=− + = + − +xO JO DO J O AA D O KK J O AA D
Substituting  =OK r  and 32 =+= + =KJ KH HJ r r L , we have
 1 () ( c o s c o s )=+− × θ + βxr L O K D K
  1 () ( c o s c o s )=+− θ + βrL r L . . . 32 ()DK KJ r r L== + =∵
   1 cos cos =+− θ − βLrr L . . . (i)
Now from right angled triangles OAK  and DAK ,
          1 sin sin=β =θAK DK OK
or       1 sin sinβ= θ Lr
       22 22

       22 22
1 sin sinβ= θ Lr  or 22 2 2
1 (1 cos ) sin−β = θLr
222 22
1 cos sin−β =θLL r  or 2 2 222
1 cos sinβ= − θ LL r
∴       1
222 21 cos ( sin )β= − θ LL r
Substituting the value of cosβ L in equation  (i), we get
             1
222 211 cos ( sin ) =+− θ − − θxL r r L r. . . (ii)
Differentiating equation (ii) with respect to t, we have velocity of the follower,
 1
1θ== ×θd dx dxvdt d dt
   1

1
   1
222 2 11 211 1 11sin ( sin ) ( 2sin cos )2− θθ=− ×− θ× − − θ − × θ θddrL r rdt dt
   1
222 2 11 211 11sin ( sin ) sin 22− θθ=θ ×+ − θ ×θ ×ddrL r rdt dt
   1
11
222 21sin 2.s i n
2( sin )
 θ=ω θ +

−θ rr
Lr        . . . 1Substitutingd
dtθ =ω  . . . (iii)
Now differentiating equation  (iii) with respect to t, we have acceleration of the follower,
 1
1θ== ×θd dv dvadt d dt
   11
222 211
1
222 2 211 1 1
1222
1( sin ) ( 2cos 2

222 211
1
222 2 211 1 1
1222
1sin 2 ( sin ) ( 2sin cos )2.cos
2( sin )Lr r
drL r rdtr
Lr−
 −θ × θ +
θ θ× − θ × θ θ
=ω θ+−θ 

Chapter 20 : Cams            /G6C          813
Substituting 1θ=ωd
dtand multiplying the numerator and denominator of second term by
1
222 21 (s i n )−θLr , we have
     222 32
11 12
12223 / 2
11(s i n ) ( 2 c o s 2 ) s i n 22.c o s
2( sin )−θ θ + × θ=ω θ +−θ Lr r r
ar
Lr
        23 2 3 2
11 1 1 12
12223 / 2
112 cos 2 2 sin .cos 2 (2sin cos )2.c o s
2( sin )×θ − θ θ + × θ θ=ω θ +−θ Lr r r
r
Lr
        23 2 2 3 2 22 11 1 1 1
12223 / 2

r
Lr
12223 / 2
12 . cos 2 2 .sin (1 2sin ) 2 sin (1 sin ).c o s
2( sin ) θ− θ − θ + θ − θ=ω θ +
−θ Lr r rr
Lr
        23 42 11
12223 / 2
1.c o s 2 s i n.c o s
(s i n ) θ+ θ=ω θ +
−θ Lr rr
Lr
Notes :  1. Since 1θis measured from the top position of the roller, therefore for the roller to have contact at
the apex of the nose ( i.e. at point H),then 1θ = 0, and for the roller to have contact where straight flank

merges into a nose ( i.e. at point G), then 1θ= α − φ .
2. The velocity is zero at H and maximum at G.
3. The acceleration is minimum at H and maximum at G.
4. From Fig 20.41, we see that the distances OK and KD remains constant for all positions of the roller
when it moves along the circular nose. In other words, a tangent cam operating a roller follower and havingcontact with the nose is equivalent to a slider crank mechanism ( i.e. ODK  ) in which the roller is assumed

equivalent to the slider D, crank OK and connecting rod
DK. Therefore the velocity and acceleration of the roller
follower may be obtained graphically as discussed in
Chapters 7 and 8.
Example 20.13.  In a symmetrical tangent cam
operating a roller follower, the least radius of thecam is 30 mm and roller radius is 17.5 mm. The angleof ascent is 75° and the total lift is 17.5 mm. Thespeed of the cam shaft is 600 r.p.m. Calculate :  1. the
principal dimensions of the cam ;  2. the accelerations

of the follower at the beginning of the lift, wherestraight flank merges into the circular nose and atthe apex of the circular nose. Assume that there is nodwell between ascent and descent.
Solution.  Given : r
1 = 30 mm ; r2 = 17.5 mm ;
α= 75° ; Total lift = 17.5 mm ; N = 600 r.p.m. or
ω= 2π× 600/60 = 62.84 rad/s
Fig. 20.42

814       /G6C               Theory of Machines
1. Principal dimensions of the cam
Let                     r  = OK = Distance between cam  centre and nose centre,
                    r3 =  Nose radius, and
                    φ = Angle of contact of cam with straight flanks.
From the geometry of Fig. 20.42,
    31+=rr r  + Total lift
 = 30 + 17.5 = 47.5 mm
∴          3 47.5=−rr . . . (i)
Also,        =+OE OP PE        or          13=+rO P r
∴        13 3 30 =−= −OP r r r . . . (ii)

Now from right angled triangle OKP ,
       cos OP OK=×α . . . (c o s / ) OP OKα=∵
or                       33 3 3 30 (47.5 )cos75 (47.5 )0.2588 12.3 0.2588−= − ° = − = −rr r r
. . . ()OK r=∵
∴                  r3 = 23.88 mm  Ans.
and                      3 47.5 47.5 23.88 23.62==− =− =rO K r  mm Ans.
Again, from right angled triangle ODB ,
     
12sin 23.62sin 75tan 0.480330 17.5α°φ= = = = =++DB KP OK
OB OB r r
∴          25.6φ= ° Ans.

OB OB r r
2. Acceleration of the follower at the beginning of the lift
We know that acceleration of the follower at the beginning of the lift, i.e. when the roller has
contact at E on the straight flank,
     22 2
12( ) (62.84) (30 17.5) 187 600minar r=ω + = + =  mm/s2
            =  187.6 m/s2 Ans.
Acceleration of the follower where straight flank merges into a circular nose
We know that acceleration of the follower where straight flank merges into a circular nose

i.e. when the roller just leaves contact at G,
    2222
12332 cos 2 cos 25.6( ) (62.84) (30 17.5)
cos cos 25.6maxar r  −φ − °=ω + = +  φ°  
            20 . 8 1 3187600 303 8000.733−==  mm/s2 = 303.8 m/s2 Ans.

Chapter 20 : Cams            /G6C          815
Acceleration of the follower at the apex of the circular nose
We know that acceleration of the follower for contact with the circular nose,
      23 42 11
12223 / 2
−θ Lr rar
Lr
Since 1θ is measured from the top position of the follower, therefore for the follower to have
contact at the apex of the circular nose ( i.e. at point H ), 1θ = 0.

∴  Acceleration of the follower at the apex of the circular nose,
   222 2
323..1 .1 .1  =ω + =ω + =ω +    +   Lr r rar r rLr r L
       2 23.62(62.84) 23.62 117.5 23.88=+ + = 146 530 mm/s2    . . . 23 ()Lr r=+∵
         = 146.53 m/s2 Ans.
Example 20.14. A cam has straight working faces which are tangential to a base circle of

diameter 90 mm. The follower is a roller of diameter 40 mm and the centre of roller moves alonga straight line passing through the centre line of the cam shaft. The angle between the tangentialfaces of the cam is 90° and the faces are joined by a nose circle of 10 mm radius. The speed ofrotation of the cam is 120 revolutions per min.
Find the acceleration of the roller centre 1. when during the lift, the roller is just about to

leave the straight flank ; and 2. when the roller is at the outer end of its lift.
1 = 90 mm  or  r1 = 45 mm ; d2 = 40 mm  or  r2 = 20 mm ; 2 α = 90° or
α = 45° ; r3 = 10 mm ; N = 120 r.p.m. or ω = 2π × 120/60 = 12.57 rad/s
The tangent cam operating a roller follower is shown in Fig. 20.43.
Fig. 20.43
First of all, let us find the *angle turned by the cam ( φ) when the roller is just about to leave
the straight flank at G. The centre of roller at this position lies at D.

*Since the cam is assumed to be stationary, φ is the angle turned by the roller when it is just about to leave
the straight flank at G.

816       /G6C               Theory of Machines
From the geometry of the figure,
                BD PK OP OE PE===−
                     
13 45 10 35 mmOE KG
rr=−
=−= −=
Now from triangle OBD ,
               
12tan
350.538545 20BD BD
OB OE EB
BD
rrφ= =+
== =++
∴    28.3φ= °
1. Acceleration of the roller centre when roller is just about to leave the straight flank
We know that acceleration of the roller centre when the roller is just about to leave the
straight flank,
                2222

                2222
12332 cos 2 cos 28.3( ) (12.57) (45 20)
cos cos 28.3  −φ − °=ω + = +     φ°  ar r
     = 18 500 mm/s2 = 18.5 m/s2 Ans.
2. Acceleration of the roller centre when the roller is at the outer end of the lift
First of all, let us find the values of OK and KD. From the geometry of the figure,
22()() 2== + = ×OK r OP PK OP . . . ()OP PK=∵
       2( ) 2(45 10) 49.5=− = − = OE EP  mm
 32 10 20 30 == + =+=+ =KD L KG GD r r  mm

32 10 20 30 == + =+=+ =KD L KG GD r r  mm
We know that acceleration of the roller centre when the roller is at the outer end of the lift,
i.e. when the roller has contact at the top of the nose,
   23 4
22 11
12223 / 2
1.c o s 2 s i n.c o s .1
(s i n ) θ+ θ =ω θ + =ω +  −θ Lr r rar rL Lr
. . . (∵  At the outer end of the lift, 10θ= )
      2 49.5(12.57) 49.5 1 20 73030=+ =   mm/s2 = 20.73 m/s2 Ans.

20.13. Circular Arc Cam with Flat-faced Follower20.13. Circular Arc Cam with Flat-faced Follower20.13. Circular Arc Cam with Flat-faced Follower20.13. Circular Arc Cam with Flat-faced Follower20.13. Circular Arc Cam with Flat-faced Follower
When the flanks of the cam connecting the base circle and nose are of convex circular arcs,
then the cam is known as circular arc cam . A symmetrical circular arc cam operating a flat-faced

follower is shown in Fig. 20.44, in which O and Q are the centres of cam and nose respectively. EF
and GH are two circular flanks whose centres lie at P and P′  respectively. The centres  * P and P′
*The centres P and P′ may also be obtained by drawing arcs with centres O and Q and radii equal to OP
and PQ respectively. The circular flanks EF and GH are now drawn with centres P and P′ and radius
equal to PE.
In aircraft engines roller followers are
widely used.

Chapter 20 : Cams            /G6C          817
lie on lines EO and GO produced.
Let r1 = Minimum radius of the cam or radius of the base circle = OE,
r2 = Radius of nose,
          R  = Radius of circular flank = PE,
          2α = Total angle of action of cam = angle EOG ,
α = Semi-angle of action of cam or angle of ascent = angle EOK , and
           φ = Angle of action of cam on the circular flank.

Fig. 20.44. Circular arc cam with flat face of the follower having contact with the circular flank.
       We shall consider the following two cases :
1. When the flat face of the follower has contact on the circular flank, and
2. When the flat face of the follower has contact on the nose.
In deriving the expressions for displacement, velocity and acceleration of the follower for the

above two cases, it is assumed that the cam is fixed and the follower rotates in the opposite senseto that of the cam. In Fig. 20.44, the cam is rotating in the clockwise direction and the followerrotates in the counter-clockwise direction.
1. When the flat face of the follower has contact on the circular flank . First of all, let us
consider that the flat face of the follower has contact at E (i.e. at the junction of the circular flank
and base circle). When the cam turns through an angle

θ (less than φ) relative to the follower, the
contact of the flat face of the follower will shift from E to C on the circular flank, such that flat face
of the follower is perpendicular to PC. Since OB is perpendicular to BC, therefore OB is parallel to
PC. From O, draw OD perpendicular to PC.
From the geometry of the figure, the displacement or lift of the follower ( x) at any instant for
contact on the circular flank, is given by
            ==−=−xB AB OA OC D E O . . . (i)

        cos =−=− θCD PC PD PE OP
   cos (1 cos ) =+− θ =+ −θOP OE OP OE OP

818       /G6C               Theory of Machines
Substituting the value of CD in equation (i),
(1 cos ) (1 cos ) xO EO P E OO P=+ −θ −= −θ
  1 ( )(1 cos ) ( ) (1 cos )PE OE R r= − −θ =− −θ . . . (ii)
dx dx d dxv
dt d dt dθ==×=× ω
θθ . . . substitutingd
   11 () s i n () s i nRr Rr=− θ × ω = ω− θ  . . . (iii)
From the above expression, we see that at the beginning of the ascent ( i.e. when θ = 0 ), the

velocity is zero (because sin 0 = 0 ) and it increases as θ increases. The velocity will be maximum
when θ = φ, i.e. when the contact of the follower just shifts from circular flank to circular nose.
Therefore maximum velocity of the follower,
        1 () s i n maxvR r=ω − φ
Now differentiating equation (iii) with respect to t, we have acceleration of the follower,
           dv dv d dvadt d dt dθ==×=× ωθθ
   2
11 () c o s () c o sRr Rr=ω − θ×ω=ω − θ  . . . (iv)

acceleration is maximum (because cos 0 = 1 ) and it decreases as θ increases. The acceleration
will be minimum when θ = φ.
∴   Maximum acceleration of the follower,
       2
1 ()maxaR r=ω −
and minimum acceleration of the follower,
        2
1 () c o sminaR r=ω − φ
2. When the flat face of the follower has contact on the nose . The flat face of the follower

having contact on the nose at C is shown in Fig. 20.45. The centre of curvature of the nose lies at
Q. In this case, the displacement or lift of the follower at any instant when the cam has turned
through an angle θ (greater than φ ) is given by
 xA BO B O A C D O A==−=− . . .( ∵  OB = CD ) . . . (i)
But         cos ( ) CD CQ QD CQ OQ=+=+ α − θ
Substituting the value of CD in equation (i), we have
 cos ( ) xC QO Q O A=+ α − θ −  . . . (ii)

cos ( ) xC QO Q O A=+ α − θ −  . . . (ii)
The displacement or lift of the follower when the contact is at the apex K of the nose i.e.
when α−θ  = 0 is
*21 x C QO QO A r O Qr=+−= +−
*From the geometry of Fig. 20.45, we also find that lift of the follower when the contact is at the apex K
of the nose is
x = JK = OQ + QK – OJ = OQ + r2 – r1

Chapter 20 : Cams            /G6C          819
    dx dx d dxv
θθ
        sin ( ) sin ( )OQ OQ=α − θ ω = ω ×α − θ  . . . (iii)
. . .(∵  CQ, OQ, OA and α are constant)
From the above expression, we see that the velocity is zero when α−θ  = 0 or α=θ  i.e.
when the follower is at the apex K of the nose. The velocity will be maximum when ( α−θ ) is

maximum. This happens when the follower changes contact from circular flank to circular nose at
point F, i.e. when (α−θ ) = φ.
    dv dv d dva
θθ
        2cos ( ) cos ( )OQ OQ=−ω× α−θω=−ω × α−θ     . . . (iv)
The negative sign in the above expression shows that there is a retardation when the
follower is in contact with the nose of the cam.

From the above expression, we see that retardation is maximum when α−θ  = 0 or θ=α ,
i.e. when the follower is at the apex K of the nose.
∴ Maximum retardation = 2OQω×
The retardation is minimum when α−θ  is maximum. This happens when the follower changes
contact from circular flank to circular nose at point F i.e.  when θ=φ .
∴ Minimum retardation = 2cos ( )OQω× α − φ

Fig. 20.45.  Circular arc cam with flat face of the follower having contact on the nose.

820       /G6C               Theory of Machines
Example 20.15. A symmetrical circular cam operating a flat-faced follower has the
following particulars :
Minimum radius of the cam = 30 mm ; Total lift = 20 mm ; Angle of lift = 75° ; Nose radius
= 5 mm ; Speed = 600 r.p.m. Find : 1. the principal dimensions of the cam, and 2. the accelera-

tion of the follower at the beginning of the lift, at the end of contact with the circular flank , at thebeginning of contact with nose and at the apex of the nose.
Solution. Given : r
1 = OE = 30 mm ; x = JK = 20 mm ; α = 75° ; r2 = QF = QK = 5 mm ;
N = 600 r.p.m. or ω =2 600 / 60 62.84π× =  rad/s
A symmetrical circular cam operating a flat faced follower is shown in Fig. 20.46.
Let         OQ  = Distance between cam centre and nose centre,

R  = PE = Radius of circular flank, and
         φ = Angle of contact on the circular flank.
We know that lift of the follower ( x),
        21 20 5 30 25=+ − =+ − =−OQ r r OQ OQ
∴         OQ  = 20 + 25 = 45 mm  Ans.
We know that    PQ PF FQ PE FQ OP OE FQ=−=−=+−
             30 5 ( 25) OP OP=+ − = +  mm
Now from a triangle OPQ ,
             222() ()()2 c o sPQ OP OQ OP OQ=+− × ×β
       22 2( 25) ( ) 45 2 45cos (180 75 )OP OP OP+= + − × × ° − °

22( ) 50 625 ( ) 2025 23.3OP OP OP OP++ =+ +
   50 23.3 2025 625OP OP−= −
or   26.7 OP = 1400
and        OP  = 1400/26.7 = 52.4 mm
∴ Radius of circular flanks,
          RP EO P O E==+ = 52.4 + 30
  = 82.4 mm  Ans.
and         PQ  = OP + 25 = 52.4 + 25
 = 77.4 mm  Ans.
In order to find angle φ, consider a triangle OPQ . We
know that
     sin sinOQ PQ=φβ
or        sin 45 sin (180 75 )sin 0.561677.4OQ
PQ×β × ° − °φ= = =
∴            34.2φ= ° Ans.Fig. 20.46

Chapter 20 : Cams            /G6C          821
2. Acceleration of the follower
We know that acceleration of the follower at the beginning of the lift,
22
11 () c o s () aR r R r=ω − θ=ω −   . . .  (∵ At the beginning of lift, 0θ= ° )
= (62.84)2 (82.4 – 30) = 206 930 mm/s = 206.93 m/s2 Ans.
Acceleration of the follower at the end of contact with the circular flank,
      22
11 () c o s () c o s aR r R r=ω − θ=ω − φ
. . . (∵  At the end of contact with the circular flank, θ=φ  )

= – (62.84)2 (82.4 – 30) cos 34.2° = 171 130 mm/s2 = 171.13  m/s2 Ans.
Acceleration of the follower at the beginning of contact with nose,
      22cos ( ) cos ( ) aO Q O Q=−ω × α−θ =−ω × α−φ
 . . . (∵  At the beginning of contact with nose, θ=φ  )
= – (62.84)2 45 cos (75°– 34.2°) = – 134 520 mm/s2 = – 134.52 m/s2
= 134.52 m/s2 (Retardation) Ans.
and acceleration of the follower at the apex of nose,
        22cos ( ) aO Q O Q=−ω × α−θ =−ω ×  ... (∵At the apex of nose, α−θ = 0 )

= – (62.84)2 45 = – 177 700 mm/s2 = – 177.7 m/s2
          = 177.7 m/s2 (Retardation) Ans.
Example 20.16. A symmetrical cam with convex flanks operates a flat-footed follower. The
lift is 8 mm, base circle radius 25 mm and the nose radius 12 mm. The total angle of the camaction is 120°.
1. Find the radius of convex flanks,  2. Draw the profile of the cam, and 3. Determine the
maximum velocity and the maximum acceleration when the cam shaft rotates at 500 r.p.m.

Solution.  Given : x = JK = 8 mm ; r
1 = OE = OJ = 25 mm ; r2 = QF = QK = 12 mm ;
2 120 EOGα=∠ = °   or 60 EOKα=∠ = ° ; N = 500 r.p.m. or 2ω= π  × 500/60 = 52.37 rad/s
1. Radius of convex flanks
Let           R = Radius of convex flanks = PE = PG′
A symmetrical cam with convex flanks operating a flat footed follower is shown in Fig.
20.47. From the geometry of the figure,
     OQ OJ JK QK=+−12rx r=+−
 = 25 + 8 – 12 = 21 mm
       PQ PF QF PE QF=−=−   = (R – 12) mm

PQ PF QF PE QF=−=−   = (R – 12) mm
and        (2 5 ) OP PE OE R=−= −  mm

822       /G6C               Theory of Machines
Now consider the triangle OPQ . We know that
 222() ()()2 c o sPQ OP OQ OP OQ=+− × × β
        22 2( 12) ( 25) (21) 2( 25)21cos(180 60 )RR R−= −+ −− ° − °
  2224 144 50 625 441 21 525 RR RR R−+ = −+ + +−
     – 24 R + 144 = – 29 R + 541 or 5 R = 397
∴                    R  = 397/5 = 79.4 mm  Ans.
Fig. 20.47
2. Profile of the cam
The profile of the cam, as shown in Fig. 20.47, is drawn as discussed in the following steps :

(a)First of all, draw a base circle with centre O and radius OE = r1 = 25 mm.
(b)Draw angle EOK  = 60° and angle KOG  = 60° such that the total angle of cam action is
120°.
(c)On line OK mark OQ = 21 mm (as calculated above). Now Q as centre, draw a circle of
radius equal to the nose radius r2 = QK = QF = 12 mm. This circle cuts the line OK at J.
Now JK represents the lift of the follower ( i.e. 8 mm).
(d)Produce EO and GO as shown in Fig. 20.47. Now with Q as centre and radius equal to

PQ = R – r2 = 79.4 – 12 = 67.4 mm, draw arcs intersecting the lines EO and GO produced
at P and P′ respectively. The centre  P′ may also be obtained by drawing arcs with
centres O and Q and radii OP and PQ respectively.
(e)Now with P and P′as centres and radius equal to R = 79.4 mm, draw arcs EF and GH
which represent the convex flanks. EFKHGAE  is the profile of the cam.
3. Maximum velocity and maximum acceleration
First of all, let us find the angle φ. From triangle OPQ ,
   sin sinOQ PQ=φβ

Chapter 20 : Cams            /G6C          823
or                  21sin sin sin (180 60 ) 0.269879.4 12OQ
PQφ= × β= × °− ° =−
. . . (∵  PQ = R – 12 )
∴          15.65φ= °
We know that maximum velocity,
    1 ( )sin 52.37(79.4 25)sin15.65 770 maxvR r=ω − φ= − °=  mm/s
= 0.77 m/s Ans.
and maximum acceleration,
    22
1 ( ) (52.37) (79.4 25) 149200maxaR r=ω − = − =  mm/s2= 149.2 m/s2 Ans.
Example 20.17. The following particulars relate to a symmetrical circular cam operating a

flat faced follower :
Least radius = 16 mm, nose radius = 3.2 mm, distance between cam shaft centre and nose
centre = 25 mm, angle of action of cam = 150°, and cam shaft speed = 600 r.p.m.
Assuming that there is no dwell between ascent or descent, determine the lift of the valve,
the flank radius and the acceleration and retardation of the follower at a point where circularnose merges into circular flank.
Solution.   Given : r
1 = OE = OJ = 16 mm ; r2 = QK = QF = 3.2 mm ; OQ = 25 mm ;

2 α = 150° or α = 75° ; N = 600 r.p.m. or ω = 2π× 600/60 = 62.84 rad/s
Lift of the valve
A symmetrical circular cam operating a flat faced
follower is shown in Fig. 20.48.
We know that lift of the valve,
         xJ KO K O J==−
            21 OQ QK OJ OQ r r=+−=+ −
         = 25 + 3.2 – 16 = 12.2 mm  Ans.
Flank radius
Let         R  = PE = Flank radius.
First of all, let us find out the values of OP and
PQ. From the geometry of Fig. 20.48,
       16 OP PE OE R=−= −

16 OP PE OE R=−= −
and        3.2 PQ PF FQ R=−= −
  222() ()()2 c o sPQ OP OQ OP OQ=+− × × β
Substituting the values of OP and PQ in the above expression,
           22 2( 3.2) ( 16) (25) 2( 16) 25cos(180 75 )RR R−= − +− − × ° − °Fig. 20.48

824       /G6C               Theory of Machines
  226.4 10.24 32 256 625 (50 800)( 0.2588) RR R R R−+= − + + −−−
      6.4 10.24 19.06 673.96RR−+= − +   or  12.66 R = 663.72
∴            R  = 52.43 mm  Ans.
Acceleration and retardation of the follower at a point where circular nose merges into circular
flank
From Fig. 20.48 we see that at a point F, the circular nose merges into a circular flank. Let φ
be the angle of action of cam at point F. From triangle OPQ ,
       sin sinOQ PQ=φβ

sin sinOQ PQ=φβ
or         sin sin (180 75 ) sin105OQ OQ
PQ PF FQφ= × °− ° = × °−
   250.966 0.4907
52.43 3.2=× =
−
∴  29.4φ= °
We know that acceleration of the follower,
22
1 cos ( ) cos aO P R r=ω × × θ=ω − φ . . . (∵    θ=φ )
   = (62.84)2 (52.43 – 16) cos 29.4° = 125 330 mm/s2
   = 125.33 m/s2 Ans.
We also know that retardation of the follower,
22cos ( ) cos ( ) aO Q O Q=ω × α−θ =ω × α−φ . . . (∵    θ=φ )
   = (62.84)2 25 cos (75° – 29.4°) = 69 110 mm/s2
    = 69.11 m/s2 Ans.

= 69.11 m/s2 Ans.
Example 20.18.  A flat ended valve tappet is operated by a symmetrical cam with circular
arc for flank and nose. The straight line path of the tappet passes through the cam axis. Totalangle of action = 150°. Lift = 6 mm. Base circle diameter = 30 mm. Period of acceleration is halfthe period of retardation during the lift. The cam rotates at 1250 r.p.m. Find : 1. flank and nose
radii ; 2. maximum acceleration and retardation during the lift.
Solution.  Given : 2

α = 150° or α = 75° ; x = JK = 6 mm ;
d1 = 30 mm or r1 = OE = OJ = 15 mm ; N = 1250 r.p.m. or
ω = 2π × 1250/60 = 131 rad/s
1. 1. 1. 1. 1. Flank and nose radii
The circular arc cam operating a flat ended valve tappet
is shown in Fig. 20.49.
Let R = PE = Flank radius, and
           2rQ F Q K==  = Nose radius.Fig. 20.49

Chapter 20 : Cams            /G6C          825
First of all, let us find the values of OP, OQ and PQ. The acceleration takes place while the
follower is on the flank and retardation while the follower is on nose. Since the period of accelera-tion is half the period of retardation during the lift, therefore
1
2φ= γ  . . . (i)
We know that         180 180 75 105β= °−α= °− °= °
∴        75 180 180 105 75φ+γ= °= °−β= °− °= ° . . . (ii)
25φ= ° ,     and       50γ= °

Now from the geometry of Fig. 20.49,
         12 2 2 15 6 21 OQ OJ JK QK r x r r r=+−= + − = + − = −  . . . (iii)
and          2 () 1 5 PQ PF FQ PE FQ OP OE FQ OP r=−=−=+ −=+ − . . . (iv)
Now from triangle OPQ ,
       sin sin sinOP OQ PQ==γφβ
or     22 21 15
sin 50 sin 25 sin105rO P r OP −+ −==
°° °
∴         22
221 21sin 50 0.766 38 1.8
sin 25 0.4226rrOP r−−=× ° =× = −
° . . . (v)
Also          22 15 15sin 50 0.766sin105 0.966OP r OP rOP+− +−=× ° =×°

= 0.793 × OP + 11.9 – 0.793  r2
∴   0.207 OP = 11.9 – 0.793 r2     or         OP  =  57.5 – 3.83 r2. . . (vi)
From equations  (v) and (vi),
38 – 1.8 r2 = 57.5 – 3.83 r2      or     2.03 r2 = 19.5
∴           r2 = 9.6 mm  Ans.
We know that      2 38 1.8 38 1.8 9.6 20.7 OP r=− =−×=  mm . . . [From equation (v)]
∴ R = PE = OP + OE = 20.7 + 15 = 35.7 mm  Ans.
2. Maximum acceleration and retardation during the lift
We know that maximum acceleration
 22 2

 22 2
1 ( ) (131) 20.7 355230Rr O P=ω − =ω × = =  mm/s2
   = 355.23 m/s2 Ans.
and maximum retardation,  22
2 (21 ) OQ r=ω × =ω − . . . [From equation (iii) ]
   = (131)2 (21 – 9.6) = 195 640 mm/s2 = 195.64 m/s2 Ans.

826       /G6C               Theory of Machines
Example 20.19. A cam consists of a circular disc of diameter
75 mm with its centre displaced 25 mm from the camshaft axis. Thefollower has a flat surface (horizontal) in contact with the cam andthe line of action of the follower is vertical and passes through theshaft axis as shown in Fig. 20.50. The mass of the follower is 2.3 kgand is pressed downwards by a spring which has a stiffness of 3.5N/mm. In the lowest position the spring force is 45 N.

1. Derive an expression for the acceleration of the follower in
terms of the angle of rotation from the beginning of the lift.
2. As the cam shaft speed is gradually increased, a value is
reached at which the follower begins to lift from the cam surface.Determine the camshaft speed for this condition .
Solution.  Given : d = 75 mm   or   r = OA = 37.5 mm ;
OQ = 25 mm ;   m = 2.3 kg ;   s = 3.5 N/mm ;  S = 45 N
1. Expression for the acceleration of the follower

The cam in its lowest position is shown by full lines in Fig. 20.51 and by dotted lines when
it has rotated through an angle 
θ.
From the geometry of the figure, the displacement of the follower,
         xA BO SO Q Q S===−
  cos OQ PQ=− θ
  cos OQ OQ=− θ       . . . ()PQ OQ=∵
 ( 1 c o s) 2 5 ( 1 c o s)OQ=− θ = − θ . . . (i)
Differentiating equation  (i) with respect to t, we
get velocity of the follower,
dx dx d dxv
θθ
 . . .(Substituting /dd tθ= ω )

θθ
 25sin 25 sin=θ × ω = ω θ . . . (ii)
Now differentiating equation  (ii) with respect to
t, we get acceleration of the follower,
           25 cosdv dv dadt d dtθ==×= ωθ × ωθ
 225 cos=ω θ  mm/s2 = 20.025 cosωθ   m/s2 Ans.
2. Cam shaft speed
Let            N  = Cam shaft speed in r.p.m.
We know that accelerating force
 22. 2.3 0.025 cos 0.0575 cosma== × ωθ = ωθ  N
Now for any value of θ, the algebraic sum of the spring force, weight of the follower and the

accelerating force is equal to the vertical reaction between the cam and follower. When this reac-
tion is zero, then the follower will just begin to leave the cam.Fig. 20.50
Fig. 20.51

Chapter 20 : Cams            /G6C          827
∴     .. . 0Ss xm gm a++ + =
245 3.5 25(1 cos ) 2.3 9.81 0.0575 cos 0+× −θ +× + ω θ =
245 87.5 87.5cos 22.56 0.0575 cos 0+− θ + + ωθ =
2155.06 87.5cos 0.0575 cos 0−θ + ω θ =
22697 1522cos cos 0−θ + ω θ =  . . . ( Dividing by 0.0575 )
          2cos 1522cos 2697ωθ = θ −  or 21522 2697secω= − θ
Since sec 1θ≥+  or , 1≤− , therefore the minimum value of 2ωoccurs when θ= 180° there-
fore

fore
                21522 ( 2697) 4219ω= −− =  . . .  [Substituting sec θ = – 1 ]
∴        65ω=  rad/s
and maximum allowable cam shaft speed,
     60 65 60
22Nω× ×==ππ = 621 r.p.m.  Ans.
EXERCISESEXERCISESEXERCISESEXERCISESEXERCISES
1. A disc cam is to give uniform motion to a knife edge follower during out stroke of 50 mm during

the first half of the cam revolution. The follower again returns to its original position with uniformmotion during the next half of the revolution. The minimum radius of the cam is 50 mm and thediameter of the cam shaft is 35 mm. Draw the profile of the cam when 1. the axis of follower passesthrough the axis of cam shaft, and 2. the axis of follower is offset by 20 mm from the axis of thecam shaft.
2. A cam operating a knife-edged follower has the following data :

(a) Follower moves outwards through 40 mm during 60° of cam rotation.
(b) Follower dwells for the next 45°.
(c) Follower returns to its original position during next 90°.
(d) Follower dwells for the rest of the rotation.
The displacement of the follower is to take place with simple harmonic motion during both the

outward and return strokes. The least radius of the cam is 50 mm. Draw the profile of the cam when1. the axis of the follower passes through the cam axis, and 2. the axis of the follower is offset 20mm towards right from the cam axis. If the cam rotates at 300 r.p.m., determine maximum velocityand acceleration of the follower during the outward stroke and the return stroke.
[Ans. 1.88 m/s, 1.26 m/s ; 177.7 m/s
2, 79 m/s2]

2, 79 m/s2]
3. A disc cam rotating in a clockwise direction is used to move a reciprocating roller with simple
harmonic motion in a radial path, as given below :
(i) Outstroke with maximum displacement of 25 mm during 120° of cam rotation,
(ii)  Dwell for 60° of cam rotation,
(iii)  Return stroke with maximum displacement of 25 mm during 90° of cam rotation, and
(iv)  Dwell during remaining 90° of cam rotation.

The line of reciprocation of follower passes through the camshaft axis. The maximum radius of cam
is 20 mm. If the cam rotates at a uniform speed of 300 r.p.m. find the maximum velocity andacceleration during outstroke and return stroke. The roller diameter is 8 mm.

828       /G6C               Theory of Machines
Draw the profile of the cam when the line of reciprocation of the follower is offset by 20 mm
towards right from the cam shaft axis. [Ans. 0.59 m/s, 0.786 m/s ; 27.8 m/s2, 49.4 m/s2]
4. Design a cam to raise a valve with simple harmonic motion through 50 mm in 1/3 of a revolution,

keep if fully raised through 1/12 revolution and to lower it with harmonic motion in 1/6 revolution.The valve remains closed during the rest of the revolution. The diameter of the roller is 20 mm andthe minimum radius of the cam is 25 mm. The diameter of the camshaft is 25 mm. The axis of thevalve rod passes through the axis of the camshaft. If the camshaft rotates at uniform speed of 100r.p.m. ; find the maximum velocity and acceleration of a valve during raising and lowering.

[ Ans. 0.39 m/s, 0.78 m/s ; 6.17 m/s
2, 24.67 m/s2]
5. A cam rotating clockwise with a uniform speed is to give the roller follower of 20 mm diameter with
the following motion :
(a) Follower to move outwards through a distance of 30 mm during 120° of cam rotation ;
(b) Follower to dwell for 60° of cam rotation ;
(c) Follower to return to its initial position during 90° of cam rotation ; and
(d) Follower to dwell for the remaining 90° of cam rotation.

The minimum radius of the cam is 45 mm and the line of stroke of the follower is offset 15 mm
from the axis of the cam and the displacement of the follower is to take place with simple harmonicmotion on both the outward and return strokes. Draw the cam profile.
6. A cam rotating clockwise at a uniform speed of 100 r.p.m. is required to give motion to knife-edge
follower as below :
(a) Follower to move outwards through 25 mm during 120° of cam rotation,

(b) Follower to dwell for the next 60° of cam rotation,
(c) Follower to return to its starting position during next 90° of cam rotation, and
(d) Follower to dwell for the rest of the cam rotation.
The minimum radius of the cam is 50 mm and the line of stroke of the follower passes through the

axis of the cam shaft. If the displacement of the follower takes place with uniform and equal accel-eration and retardation on both the outward and return strokes, find the maximum velocity andacceleration during outstroke and return stroke. [Ans. 0.25 m/s, 0.33 m/s ; 2.5 m/s
2 , 4.44 m/s2]
7. A cam with 30 mm as minimum diameter is rotating clockwise at a uniform speed of 1200 r.p.m.
and has to give the following motion to a roller follower 10 mm in diameter:

(a) Follower to complete outward stroke of 25 mm during 120° of cam rotation with equal uniform
acceleration and retardation ;
(c) Follower to return to its initial position during 90° of cam rotation with equal uniform accelera-
tion and retardation ;

Draw the cam profile if the axis of the roller follower passes through the axis of the cam.Determine the maximum velocity of the follower during the outstroke and return stroke and
also the uniform acceleration of the follower on the out stroke and the return stoke.
[Ans. 3 m/s , 4 m/s ; 360.2 m/s
2, 640.34 m/s2]
8. A cam rotating clockwise at a uniform speed of 200 r.p.m. is required to move an offset roller

follower with a uniform and equal acceleration and retardation on both the outward and returnstrokes. The angle of ascent, the angle of dwell (between ascent and descent) and the angle ofdescent is 120°, 60° and 90° respectively. The follower dwells for the rest of cam rotation. The leastradius of the cam is 50 mm, the lift of the follower is 25 mm and the diameter of the roller is 10mm. The line of stroke of the follower is offset by 20 mm from the axis of the cam. Draw the camprofile and find

the axis of the cam. Draw the camprofile and find the maximum velocity and acceleration of the follower during the outstroke.

9. A flat faced reciprocating follower has the following motion :

Chapter 20 : Cams            /G6C          829
(i) The follower moves out for 80° of cam rotation with uniform acceleration and retardation, the
acceleration being twice the retardation.
(ii) The follower dwells for the next 80° of cam rotation.
(iii) It moves in for the next 120° of cam rotation with uniform acceleration and retardation, the
retardation being twice the acceleration.
(iv) The follower dwells for the remaining period.

The base circle diameter of the cam is 60 mm and the stroke of the follower is 20 mm. The line
of movement of the follower passes through the cam centre.
Draw the displacement diagram and the profile of the cam very neatly showing all construc-
tional details.
10. From the following data, draw the profile of a cam in which the follower moves with simple har-
monic motion during ascent while it moves with uniformly accelerated motion during descent :

Least radius of cam = 50 mm ; Angle of ascent = 48° ; Angle of dwell between ascent and
descent = 42° ; Angle of descent = 60° ; Lift of follower = 40 mm ; Diameter of roller = 30 mm ;Distance between the line of action of follower and the axis of cam = 20 mm.
If the cam rotates at 360 r.p.m. anticlockwise, find the maximum velocity and acceleration of
the follower during descent. [Ans. 2.88 m/s ; 207.4 m/s
2]

2]
11. Draw the profile of a cam with oscillating roller follower for the following motion :
(a) Follower to move outwards through an angular displacement of 20° during 120° of cam rota-
tion.
(b) Follower to dwell for 50° of cam rotation.
(c) Follower to return to its initial position in 90° of cam rotation with uniform acceleration and
retardation.
(d) Follower to dwell for the remaining period of cam rotation.

The distance between the pivot centre and the roller centre is 130 mm and the distance between
the pivot centre and cam axis is 150 mm. The minimum radius of the cam is 80 mm and thediameter of the roller is 50 mm.
12. Draw the profile of the cam when the roller follower moves with cycloidal motion as given below :
(a) Outstroke with maximum displacement of 44 mm during 180° of cam rotation.
(b) Return stroke for the next 150° of cam rotation.
(c) Dwell for the remaining 30° of cam rotation.

The minimum radius of the cam is 20 mm and the diameter of the roller is 10 mm. The axis of
the roller follower passes through the cam shaft axis.
13. A symmetrical tangent cam operating a roller follower has the following particulars :
Radius of base circle of cam = 40 mm, roller radius = 20 mm, angle of ascent = 75°, total lift = 20
mm, speed of cam shaft = 300 r.p.m.

Determine : 1. the principal dimensions of the cam, 2. the equation for the displacement curve,
when the follower is in contact with the straight flank, and 3. the acceleration of the follower whenit is in contact with the straight flank where it merges into the circular nose.
[Ans. r
3 = 33 mm ;  θ θ θ θ θ  = 23.5° ; 89.4 m/s2]
14. A cam profile consists of two circular arcs of radii 24 mm and 12 mm, joined by straight lines,

giving the follower a lift of 12 mm. The follower is a roller of 24 mm radius and its line of actionis a straight line passing through the cam shaft axis. When the cam shaft has a uniform speed of 500rev/min, find the maximum velocity and acceleration of the follower while in contact with the straightflank of the cam. [Ans. 1.2 m/s ; 198 m/s
2]
15. The following particulars relate to a symmetrical tangent cam operating a roller follower :-

Least radius = 30 mm, nose radius = 24 mm, roller radius = 17.5 mm, distance between cam shaft
and nose centre = 23.5 mm, angle of action of cam = 150°, cam shaft speed = 600 r.p.m.
Assuming that there is no dwell between ascent and descent, determine the lift of the valve and the

830       /G6C               Theory of Machines
acceleration of the follower at a point where straight flank merges into the circular nose.
[Ans. 17.5 mm ; 304.5 m/s2]
16. Following is the data for a circular arc cam working with a flat faced reciprocating follower :
Minimum radius of the cam = 30 mm ; Total angle of cam action = 120° ; Radius of the circular arc
= 80 mm ; Nose radius = 10 mm.
1. Find the distance of the centre of nose circle from the cam axis ; 2. Draw the profile of the cam

to full scale; 3. Find the angle through which the cam turns when the point of contact moves fromthe junction of minimum radius arc and circular arc to the junction of nose radius arc and circulararc ; and 4. Find the velocity and acceleration of the follower when the cam has turned through an
angle of 
θ = 20°. The angle θ is measured from the point where the follower just starts moving
away from the cam. The angular velocity of the cam is 10 rad/s.
[Ans. 30 mm ; 22°; 68.4 mm/s ; 1880 mm/s2]

17. The suction valve of a four stroke petrol engine is operated by a circular arc cam with a flat faced

follower. The lift of the follower is 10 mm ; base circle diameter of the cam is 40 mm and the noseradius is 2.5 mm. The crank angle when suction valve opens is 4° after top dead centre and whenthe suction valve closes, the crank angle is 50° after bottom dead centre. If the cam shaft rotates at600 r.p.m., determine: 1. maximum velocity of the valve, and 2. maximum acceleration and retarda-tion of the valve.
[Ans. 1.22 m/s ; 383 m/s
2, 108.6 m/s2]

[Hint.  Total angle turned by the crankshaft when valve is open
= 180° – 4° + 50° = 226°
Since the engine is a four stroke cycle, therefore speed of cam shaft is half of the speed of the crank
shaft.
        ∴Total angle turned by the cam shaft during opening of valve, 2 α = 226/2 = 113° or
α = 56.5°].
18. The following particulars relate to a symmetrical circular cam operating a flat-faced follower :

Least radius = 25 mm ; nose radius = 8 mm, lift of the valve = 10 mm, angle of action of cam =
120°, cam shaft speed = 1000 r.p.m.
Determine the flank radius and the maximum velocity, acceleration and retardation of the follower. If
the mass of the follower and valve with which it is in contact is 4 kg, find the minimum force to beexerted by the spring to overcome inertia of the valve parts.
[Ans. 88 mm ; 1.93 m/s, 690.6 m/s
2, 296 m/s2 ; 1184 N]

DO YOU KNOW ?DO YOU KNOW ?DO YOU KNOW ?DO YOU KNOW ?DO YOU KNOW ?
1. Write short notes on cams and followers.
2. Explain with sketches the different types of cams and followers.
3. Why a roller follower is preferred to that of a knife-edged follower ?
4. Define the following terms as applied to cam with a neat sketch :-
(a)  Base circle,    ( b)   Pitch circle,    ( c)  Pressure angle, and   ( d)   Stroke of the follower.

5. What are the different types of motion with which a follower can move ?
6. Draw the displacement, velocity and acceleration diagrams for a follower when it moves with simple
harmonic motion. Derive the expression for velocity and acceleration during outstroke and return
stroke of the follower.
7. Draw the displacement, velocity and acceleration diagrams for a follower when it moves with uni-
form acceleration and retardation. Derive the expression for velocity and acceleration during out-

stroke and return stroke of the follower.
8. Derive expressions for displacement, velocity and acceleration for a tangent cam operating on a
radial-translating roller follower :

Chapter 20 : Cams            /G6C          831
(i) when the contact is on straight flank, and
(ii) when the contact is on circular nose.
9. Derive the expressions for displacement, velocity and acceleration for a circular arc cam operating a
flat-faced follower
(i) when the contact is on the circular flank, and
OBJECTIVE TYPE QUESTIONSOBJECTIVE TYPE QUESTIONSOBJECTIVE TYPE QUESTIONSOBJECTIVE TYPE QUESTIONSOBJECTIVE TYPE QUESTIONS

1. The size of a cam depends upon
(a) base circle ( b) pitch circle ( c) prime circle ( d) pitch curve
2. The angle between the direction of the follower motion and a normal to the pitch curve is called
(a) pitch angle ( b)  prime angle
(c) base angle ( d)  pressure angle
3. A circle drawn with centre as the cam centre and radius equal to the distance between the cam
centre and the point on the pitch curve at which the pressure angle is maximum, is called
(a) base circle ( b)  pitch circle

(c) prime circle ( d)  none of these
4. The cam follower generally used in automobile engines is
(a) knife edge follower ( b)  flat faced follower
(c) spherical faced follower  ( d)  roller follower
5. The cam follower extensively used in air-craft engines is
(c) spherical faced follower ( d)  roller follower
6. In a radial cam, the follower moves
(a) in a direction perpendicular to the cam axis

(b) in a direction parallel to the cam axis
(c) in any direction irrespective of the cam axis
(d) along the cam axis
7. A radial follower is one
(a) that reciprocates in the guides ( b)  that oscillates
(c) in which the follower translates along an axis passing through the cam centre of rotation.
8. Offet is provided to a cam follower mechanism to
(a) minimise the side thrust ( b)  accelerate

(c) avoid jerk ( d)  none of these
9. For low and moderate speed engines, the cam follower should move with
(a) uniform velocity ( b)  simple harmonic motion
(c) uniform acceleration and retardation ( d)  cycloidal motion
10. For high speed engines, the cam follower should move with

11. Which of the following displacement diagrams should be chosen for better dynamic performance of
a cam-follower mechanism ?
(a) simple hormonic motion ( b)  parabolic motion
(c) cycloidal motion ( d)  none of these

832       /G6C               Theory of Machines
12. For a given lift of the follower of a cam follower mechanism, a smaller base circle diameter is
desired.
(a) because it will give a steeper cam and higher pressure angle.
(b) because it will give a profile with lower pressure angle
(c) because it will avoid jumping
(d) none of the above.
13. The linear velocity of the reciprocating roller follower when it has contact with the straight flanks of
the tangent cam, is given by

(a)12() s i nrrω− θ (b) 12() c o srrω− θ
(c)122() s i n s e crrω+ θ θ (d) 122( )cos cosecrrω+ θ θ
where       ω = Angular velocity of the cam shaft,
 r1 = Minimum radius of the cam,
 r2 = Radius of the roller, and
θ = Angle turned by the cam from the beginning of the displacement for contact
                of roller with the straight flanks.
14. The displacement of a flat faced follower when it has contact with the flank of a circular arc cam, is
given by

given by
(a)(1 cos )−θR          ( b)  (1 sin )−θR
(c) 1 () ( 1 c o s )Rr−− θ                      ( d)  1 () ( 1 s i n )Rr−− θ
where  R = Radius of the flank,
 r1 = Minimum radius of the cam, and
 θ = Angle turned by the cam for contact with the circular flank.
15. The retardation of a flat faced follower when it has contact at the apex of the nose of a circular arc
cam, is given by
(a)  2ω× OQ (b) 2sinω× θ OQ
(c)  2cosω× θ OQ                                ( d) 2tanω× θ OQ

where OQ = Distance between the centre of circular flank and centre of nose.
ANSWERSANSWERSANSWERSANSWERSANSWERS
1.  (a) 2.  (d) 3.  (b) 4.  (c) 5.  (d)
6.  (a) 7.  (a) 8.  (a) 9. (b) 10. (d)
11.  (c) 12.  (d) 13.  (c) 14. (c) 15. (a)

Chapter 21  : Balancing of Rotating Masses            /G6C          833
21.1.21.1.21.1.21.1.21.1. IntroductionIntroductionIntroductionIntroductionIntroduction
The high speed of engines and other machines is a
common phenomenon now-a-days. It is, therefore, very

essential that all the rotating and reciprocating parts shouldbe completely balanced as far as possible. If these parts arenot properly balanced, the dynamic forces are set up. Theseforces not only increase the loads on bearings and stressesin the various members, but also produce unpleasant andeven dangerous vibrations. In this chapter we shall discussthe balancing of unbalanced forces caused by rotating masses,in order to minimise pressure on the main bearings when anengine is running.

21.2.21.2.21.2.21.2.21.2. Balancing of Rotating MassesBalancing of Rotating MassesBalancing of Rotating MassesBalancing of Rotating MassesBalancing of Rotating Masses
We have already discussed, that whenever a certain

mass is attached to a rotating shaft, it exerts some centrifu-gal force, whose effect is to bend the shaft and to producevibrations in it. In order to prevent the effect of centrifugalforce, another mass is attached to the opposite side of theshaft, at such a position so as to balance the effect of thecentrifugal force of the first mass. This is done in such a
833BalaBalaBalaBalaBala ncing ofncing ofncing ofncing ofncing of
RotatingRotatingRotatingRotatingRotating

MassMassMassMassMass eseseseses21
2. Balancing of Rotating
Masses.
3. Balancing of a Single
Rotating Mass By a Single
Mass Rotating in the SamePlane.
4. Balancing of a Single
Rotating Mass By TwoMasses Rotating inDifferent Planes.
5. Balancing of Several
Masses Rotating in theSame Plane.
6. Balancing of Several
Masses Rotating inDifferent Planes .

834       /G6C               Theory of Machines
way that the centrifugal force of both the masses are made to be equal and opposite. The process of
providing the second mass in order to counteract the effect of the centrifugal force of the first mass,is called balancing of rotating masses .
The following cases are important from the subject point of view:
1.Balancing of a single rotating mass by a single mass rotating in the same plane.

2.Balancing of a single rotating mass by two masses rotating in different planes.
3.Balancing of different masses rotating in the same plane.
4.Balancing of different masses rotating in different planes.
We shall now discuss these cases, in detail, in the following pages.

21.3.21.3.21.3.21.3.21.3. Balancing of a Single Rotating Mass By a Single Mass Rotating inBalancing of a Single Rotating Mass By a Single Mass Rotating inBalancing of a Single Rotating Mass By a Single Mass Rotating inBalancing of a Single Rotating Mass By a Single Mass Rotating inBalancing of a Single Rotating Mass By a Single Mass Rotating in
the Same Planethe Same Planethe Same Planethe Same Planethe Same Plane

Consider a disturbing mass m1 attached to a shaft rotating at ω rad/s as shown in Fig. 21.1.
Let r1 be the radius of rotation of the mass m1 (i.e. distance between the axis of rotation of the shaft
and the centre of gravity of the mass m1).
We know that the centrifugal force exerted by the mass m1 on the shaft,
=⋅ ω ⋅2
Cl 1 1Fm r . . . (i)
This centrifugal force acts radially outwards and thus produces bending moment on the

shaft. In order to counteract the effect of this force, a balancing mass ( m2) may be attached in the
same plane of rotation as that of disturbing mass ( m1) such that the centrifugal forces due to the
two masses are equal and opposite.
Fig. 21.1. Balancing of a single rotating mass by a single mass rotating in the same plane.
Let          r2 = Radius of rotation of the balancing mass m2 (i.e. distance between the
 axis of rotation of the shaft and the centre of gravity of mass m2 ).

∴  Centrifugal force due to mass m2,
     =⋅ ω ⋅2
C2 2 2Fm r . . . (ii)
         22
11 2 2.mr m rω⋅ = ⋅ ω⋅    or   11 2 2mr mr⋅= ⋅
Notes : 1.  The product m2.r2 may be split up in any convenient way. But the radius of rotation of the
balancing mass ( m2) is generally made large in order to reduce the balancing mass m2.
  2 . The centrifugal forces are proportional to the product of the mass and radius of rotation of

Chapter 21  : Balancing of Rotating Masses            /G6C          835
21.4.21.4.21.4.21.4.21.4. Balancing of a Single Rotating Mass By Two Masses Rotating inBalancing of a Single Rotating Mass By Two Masses Rotating inBalancing of a Single Rotating Mass By Two Masses Rotating inBalancing of a Single Rotating Mass By Two Masses Rotating inBalancing of a Single Rotating Mass By Two Masses Rotating in
Different PlanesDifferent PlanesDifferent PlanesDifferent PlanesDifferent Planes

We have discussed in the previous article that by introducing a single balancing mass in the
same plane of rotation as that of disturbing mass, the centrifugal forces are balanced. In other

words, the two forces are equal in magnitude and opposite in direction. But this type of arrange-ment for balancing gives rise to a couple which tends to rock the shaft in its bearings. Therefore inorder to put the system in complete balance, two balancing masses are placed in two differentplanes, parallel to the plane of rotation of the disturbing mass, in such a way that they satisfy thefollowing two conditions of equilibrium.

1.The net dynamic force acting on the shaft is equal to zero. This requires that the line of
action of three centrifugal forces must be the same. In other words, the centre of themasses of the system must lie on the axis of rotation. This is the condition for static
balancing .
2.The net couple due to the dynamic forces acting on the shaft is equal to zero. In other
words, the algebraic sum of the moments about any point in the plane must be zero.

The conditions  (1) and (2) together give dynamic balancing . The following two possibili-
ties may arise while attaching the two balancing masses :
1.The plane of the disturbing mass may be in between the planes of the two balancing
masses, and
2.The plane of the disturbing mass may lie on the left or right of the two planes containing
the balancing masses.
 We shall now discuss both the above cases one by one.

1.  When the plane of the disturbing mass lies in between the planes of the two balancing
   masses
Consider a disturbing mass m lying in a plane A to be balanced by two rotating masses m
1
and m2 lying in two different planes L and M as shown in Fig. 21.2. Let r, r1 and r2 be the radii of
rotation of the masses in planes A, L and M respectively.
The picture shows a diesel engine. All diesel, petrol and steam engines have reciprocating and
rotating masses inside them which need to be balanced.

836       /G6C               Theory of Machines
Let       l1 = Distance between the planes A and L,
                  l2 = Distance between the planes A and M, and
                   l  = Distance between the planes L and M.
Fig. 21.2.  Balancing of a single rotating mass by two rotating masses in different planes when the
     plane of single rotating mass lies in between the planes of two balancing masses.
We know that the centrifugal force exerted by the mass m in the plane A,
    =⋅ ω ⋅2

=⋅ ω ⋅2
CFm r
Similarly, the centrifugal force exerted by the mass m1 in the plane L,
  =⋅ ω ⋅2
C1 1 1Fm r
and, the centrifugal force exerted by the mass m2 in the plane M,
  =⋅ ω ⋅2
C2 2 2Fm r
Since the net force acting on the shaft must be equal to zero, therefore the centrifugal force
on the disturbing mass must be equal to the sum of the centrifugal forces on the balancing masses,
therefore
    CC 1 C 2FF F=+          or      ⋅ω ⋅ = ω ⋅ + ⋅ω ⋅22 2
11 2 2. mr mr m r

∴              11 2 2 mr m r m r⋅= ⋅+ ⋅  . . . (i)
Now in order to find the magnitude of balancing force in the plane L (or the dynamic force
at the bearing Q of a shaft), take moments about P which is the point of intersection of the plane M
and the axis of rotation. Therefore
           C1 C 2Fl F l×= ×      or    22
11 2mr l m r l⋅ω ⋅ × = ⋅ω ⋅ ×
∴          11 2mrl m r l⋅⋅=⋅⋅      or    2
11lmr m rl⋅=⋅ ×  . . . (ii)

Similarly, in order to find the balancing force in plane M (or the dynamic force at the
bearing P of a shaft), take moments about Q which is the point of intersection of the plane L and
the axis of rotation. Therefore
          C2 C 1Fl F l×= ×    or   22
22 1mr l m r l⋅ω ⋅ × = ⋅ω ⋅ ×
∴          22 1mrl m r l⋅⋅ =⋅ ⋅    or   1
22lmr m rl⋅=⋅ ×  . . . (iii)

Chapter 21  : Balancing of Rotating Masses            /G6C          837
It may be noted that equation (i) represents the condition for static balance, but in order to
achieve dynamic balance, equations (ii) or (iii) must also be satisfied.
2.  When the plane of the disturbing mass lies on one end of the planes of the balancing
    masses
Fig. 21.3. Balancing of a single rotating mass by two rotating masses in different planes, when the

plane of single rotating mass lies at one end of the planes of balancing masses.
In this case, the mass m lies in the plane A and the balancing masses lie in the planes L and
M, as shown in Fig. 21.3. As discussed above, the following conditions must be satisfied in order
to balance the system, i.e.
      C2 C1 CFF F+= or     22 2
22 1 1 mr m r m r⋅ω ⋅ + ⋅ω ⋅ = ⋅ω ⋅
∴    22 1 1 . mr m r mr⋅+ ⋅ =  . . . (iv)

Now, to find the balancing force in the plane L (or the dynamic force at the bearing Q of a
shaft), take moments about P which is the point of intersection of the plane M and the axis of
rotation. Therefore
         C1 C 2Fl F l×= ×   or   22
∴                   11 2mrl m r l⋅⋅=⋅⋅   or   2
11lmr m r
l⋅=⋅ ×  . . . (v)
. . . [Same as equation (ii)]

Similarly, to find the balancing force in the plane M (or the dynamic force at the bearing P
of a shaft), take moments about Q which is the point of intersection of the plane L and the axis of
          C2 C 1Fl F l×= ×   or   22
         22 1mrl m r l⋅⋅ =⋅ ⋅  or   1
22lmr m r
l⋅=⋅ × . . . (vi)
. . . [Same as equation  (iii)]

21.5.21.5.21.5.21.5.21.5. Balancing of Several Masses Rotating in the Same PlaneBalancing of Several Masses Rotating in the Same PlaneBalancing of Several Masses Rotating in the Same PlaneBalancing of Several Masses Rotating in the Same PlaneBalancing of Several Masses Rotating in the Same Plane
Consider any number of masses (say four) of magnitude m1, m2, m3 and m4 at distances of

r1, r2, r3 and r4 from the axis of the rotating shaft. Let 123 4,,a n dθθθ θ be the angles of these
masses with the horizontal line OX, as shown in Fig. 21.4 ( a). Let these masses rotate about an axis
through O and perpendicular to the plane of paper, with a constant angular velocity of ω rad/s.

838       /G6C               Theory of Machines
The magnitude and position of the balancing mass may be found out analytically or
graphically as discussed below :
(a) Space diagram.         ( b) V ector diagram.
Fig. 21.4. Balancing of several masses rotating in the same plane.
1. Analytical method
The magnitude and direction of the balancing mass may be obtained, analytically, as
1.First of all, find out the centrifugal force * (or the product of the mass and its radius of

rotation) exerted by each mass on the rotating shaft.
* Since ω2 is same for each mass, therefore the magnitude of the centrifugal force for each mass is propor-
tional to the product of the respective mass and its radius of rotation.
A car assembly line.

Chapter 21  : Balancing of Rotating Masses            /G6C          839
2.Resolve the centrifugal forces horizontally and vertically and find their sums, i.e. HΣ
and VΣ. We know that
Sum of horizontal components of the centrifugal forces,
   11 1 2 2 2cos cos . ..... Hm r m rΣ= ⋅ θ + ⋅ θ +
and sum of vertical components of the centrifugal forces,
  11 1 2 2 2sin sin . ..... Vm r m rΣ= ⋅ θ + ⋅ θ +
3.Magnitude of the resultant centrifugal force,
    =Σ + Σ22
C ()() FH V

=Σ + Σ22
C ()() FH V
4.If θ is the angle, which the resultant force makes with the horizontal, then
 tan / VHθ=Σ Σ
5.The balancing force is then equal to the resultant force, but in opposite direction .
6.Now find out the magnitude of the balancing mass, such that
    =⋅CFm r
where       m  = Balancing mass, and
       r  = Its radius of rotation.
2. Graphical method
The magnitude and position of the balancing mass may also be obtained graphically as

1.First of all, draw the space diagram with the positions of the several masses, as shown in
Fig. 21.4 ( a).
2.Find out the centrifugal force (or product of the mass and radius of rotation) exerted by
each mass on the rotating shaft.
3.Now draw the vector diagram with the obtained centrifugal forces (or the product of the
masses and their radii of rotation), such that ab represents the centrifugal force exerted by

the mass m1 (or m1.r1) in magnitude and direction to some suitable scale. Similarly, draw
bc, cd and de to represent centrifugal forces of other masses m2, m3 and m4 (or m2.r2,
m3.r3 and m4.r4).
4.Now, as per polygon law of forces, the closing side ae represents the resultant force in
magnitude and direction, as shown in Fig. 21.4 ( b).
5.The balancing force is, then, equal to the resultant force, but in opposite direction.

6.Now find out the magnitude of the balancing mass ( m) at a given radius of rotation ( r),
such that
        2mr⋅ω ⋅  = Resultant centrifugal force
or     m.r = Resultant of m1.r1, m2.r2, m3.r3 and m4.r4
Example 21.1. Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg and 260 kg
respectively. The corresponding radii of rotation are 0.2 m, 0.15 m, 0.25 m and 0.3 m respectively

and the angles between successive masses are 45°, 75° and 135°. Find the position and magnitudeof the balance mass required, if its radius of rotation is 0.2 m.
1 = 200 kg ; m2 = 300 kg ; m3 = 240 kg ; m4 = 260 kg ; r1 = 0.2 m ;
r2 = 0.15 m ; r3= 0.25 m ; r4 = 0.3 m ; 1θ = 0° ; 2θ = 45° ; 3θ = 45° + 75° = 120° ; 4θ = 45° + 75°
+ 135° = 255° ; r= 0.2 m

840       /G6C               Theory of Machines
Let               m  = Balancing mass, and
                 θ= The angle which the balancing mass makes with m1.
Since the magnitude of centrifugal forces are
proportional to the product of each mass and its radius,
therefore
            ⋅= × =11 200 0.2 40 kg-m mr
           ⋅= × =22 300 0.15 45kg-m mr
                        ⋅= × =33 240 0.25 60 kg-m mr
           ⋅= × =44 260 0.3 78 kg-m mr
The problem may, now, be solved either analytically

or graphically. But we shall solve the problem by both themethods one by one.
The space diagram is shown in Fig. 21.5.
Resolving m
1.r1, m2.r2, m3.r3 and m4.r4 horizontally,
   11 1 2 2 2 33 3 4 4 4cos cos cos cos Hm r m r m r m rΣ =⋅ θ +⋅ θ +⋅ θ +⋅ θ
        40 cos 0 45cos45 60 cos120 78 cos255=° + ° + ° + °
       40 31.8 30 20.2 21.6 kg-m=+ −− =
Now resolving vertically,
   Σ =⋅ θ +⋅ θ +⋅ θ +⋅ θ1 1 1 2 22 3 33 4 44sin sin sin sin Vm r m r m r m r

40 sin0 45sin45 60 sin120 78 sin255=° + ° + ° + °
       0 31.8 52 75.3 8.5 kg-m=+ + − =
∴Resultant,   R =  22 2 2( ) ( ) (21.6) (8.5) 23.2 kg-mHVΣ+ Σ= + =
 23.2 mr R⋅= =    or  == =23.2 / 23.2 / 0.2 116 kgmr  Ans.
and  tan / 8.5/ 21.6 0.3935 VH′θ=Σ Σ = =    or  ′θ = 21.48°
Since ′θ is the angle of the resultant R from the horizontal mass of 200 kg, therefore the
angle of the balancing mass from the horizontal mass of 200 kg,
     θ = 180° + 21.48° = 201.48° Ans.

θ = 180° + 21.48° = 201.48° Ans.
The magnitude and the position of the balancing mass may also be found graphically as
1.First of all, draw the space diagram showing the positions of all the given masses as
shown in Fig 21.6 ( a).
2.Since the centrifugal force of each mass is proportional to the product of the mass and
radius, therefore
  m1.r1 = 200 × 0.2 =  40 kg-m
 m2.r2 = 300 × 0.15 = 45 kg-mFig. 21.5

Chapter 21  : Balancing of Rotating Masses            /G6C          841
             m3.r3 = 240 × 0.25 = 60 kg-m
   m4.r4 = 260 × 0.3 =  78 kg-m
3.Now draw the vector diagram with the above values, to some suitable scale, as shown in
Fig. 21.6 ( b). The closing side of the polygon ae represents the resultant force. By mea-
surement, we find that ae = 23 kg-m.
(a) Space diagram. ( b) Vector diagram
Fig. 21.6

Fig. 21.6
4.The balancing force is equal to the resultant force, but opposite  in direction as shown in
Fig. 21.6 ( a). Since the balancing force is proportional to m.r, therefore
          m × 0.2 = vector ea = 23 kg-m   or   m = 23/0.2 = 115 kg Ans.
By measurement we also find that the angle of inclination of the balancing mass ( m) from
the horizontal mass of 200 kg,
    θ = 201°  Ans.

21.6.21.6.21.6.21.6.21.6. Balancing of Several Masses Rotating in Different PlanesBalancing of Several Masses Rotating in Different PlanesBalancing of Several Masses Rotating in Different PlanesBalancing of Several Masses Rotating in Different PlanesBalancing of Several Masses Rotating in Different Planes
When several masses revolve in different planes, they
may be transferred to a reference plane  (briefly written as

R.P.), which may be defined as the plane passing through a
point on the axis of rotation and perpendicular to it. The

effect of transferring a revolving mass (in one plane) to areference plane is to cause a force of magnitude equal to thecentrifugal force of the revolving mass to act in the referenceplane, together with a couple of magnitude equal to theproduct of the force and the distance between the plane ofrotation and the reference plane. In order to have a completebalance of the several revolving masses in different planes,the following two conditions must be satisfied :

1.The forces in the reference plane must balance,  i.e.
the resultant force must be zero.
2.The couples about the reference plane must balance,
i.e. the resultant couple must be zero.
Let us now consider four masses m
1, m2, m3 and m4
revolving in planes 1, 2, 3 and 4 respectively as shown inDiesel engine.

842       /G6C               Theory of Machines
Fig. 21.7 ( a). The relative angular positions of these masses are shown in the end view [Fig. 21.7
(b)]. The magnitude of the balancing masses mL and mM in planes L and M may be obtained as
1.Take one of the planes, say L as the reference plane ( R.P.). The distances of all the other
planes to the left of the reference plane may be regarded as  negative , and those to the
right as  positive .

2.Tabulate the data as shown in Table 21.1. The planes are tabulated in the same order in
which they occur, reading from left to right.
Table 21.1Table 21.1Table 21.1Table 21.1Table 21.1
Plane Mass (m) Radius(r) Cent.force 2÷ω Distance from Couple2÷ω
(m.r) Plane L  (l) (m.r.l)
(1) (2) (3) (4) (5) (6)
1 m1r1m1.r1–l1– m1.r1.l1
L(R.P.) mLrLmL.rL00
2 m2r2m2.r2l2m2.r2.l2
3 m3r3m3.r3l3m3.r3.l3
MmMrMmM.rMlMmM.rM.lM
4 m4r4m4.r4l4m4.r4.l4

(a) Position of planes of the masses.   ( b) Angular position of the masses.
(c) Couple vector.      ( d) Couple vectors turned  ( e) Couple polygon.   ( f) Force polygon.
     counter clockwise through
a right angle.
Fig. 21.7. Balancing of several masses rotating in different planes.
3.A couple may be represented by a vector drawn perpendicular to the plane of the couple.

The couple C1 introduced by transferring m1 to the reference plane through O is propor-

Chapter 21  : Balancing of Rotating Masses            /G6C          843
tional to m1.r1.l1 and acts in a plane through Om1 and perpendicular to the paper. The
vector representing this couple is drawn in the plane of the paper and perpendicular to
Om1 as shown by OC1 in Fig. 21.7 ( c). Similarly, the vectors OC2, OC3 and OC4 are
drawn perpendicular to Om2, Om3 and Om4 respectively and in the plane of the paper.

4.The couple vectors as discussed above, are turned counter clockwise through a right angle
for convenience of drawing as shown in Fig. 21.7 ( d). We see that their relative positions
remains unaffected. Now the vectors OC2, OC3 and OC4 are parallel and in the same
direction as Om2, Om3 and Om4, while the vector OC1 is parallel to Om1 but in *opposite
direction. Hence the couple vectors are drawn radially outwards for the masses on one

side of the reference plane and radially inward for the masses on the other side of thereference plane .
5.Now draw the couple polygon as shown in Fig. 21.7 ( e). The vector 
do′′ represents the
balanced couple. Since the balanced couple CM is proportional to mM.rM.lM, therefore
   MM M M vector Cm r l d o ′′ =⋅ ⋅ =     or    M
MMvector domrl′′=⋅
From this expression, the value of the balancing mass mM in the plane M may be obtained,

and the angle of inclination φ of this mass may be measured from Fig. 21.7 ( b).
6.Now draw the force polygon as shown in Fig. 21.7 ( f ). The vector eo (in the direction
from e to o ) represents the balanced force. Since the balanced force is proportional to
mL.rL, therefore,
           LL vector mr e o⋅=      or     L
Lvector eomr=
From this expression, the value of the balancing mass mL in the plane L may be obtained

and the angle of inclination α of this mass with the horizontal may be measured from Fig. 21.7 ( b).
Example 21.2. A shaft carries four masses A, B, C and D of magnitude 200 kg, 300 kg,
400 kg and 200 kg respectively and revolving at radii 80 mm, 70 mm, 60 mm and 80 mm in planes
measured from A at 300 mm, 400 mm and 700 mm. The angles between the cranks measuredanticlockwise are A to B 45°, B to C 70° and C to D 120°. The balancing masses are to be placed

in planes X and Y. The distance between the planes A and X is 100 mm, between X and Y is 400
mm and between Y and D is 200 mm. If the balancing masses revolve at a radius of 100 mm, findtheir magnitudes and angular positions .
Solution.  Given : m
A = 200 kg ; mB = 300 kg ; mC = 400 kg ; mD = 200 kg ; rA = 80 mm
= 0.08m ; rB = 70 mm = 0.07 m ; rC = 60 mm = 0.06 m ; rD = 80 mm = 0.08 m ; rX = rY = 100 mm
= 0.1 m
Let     mX = Balancing mass placed in plane X, and

mY = Balancing mass placed in plane Y.
The position of planes and angular position of the masses (assuming the mass A as
horizontal) are shown in Fig. 21.8 ( a) and ( b) respectively.
Assume the plane X as the reference plane ( R.P.). The distances of the planes to the right of
plane X are taken as + ve while the distances of the planes to the left of plane X are taken as – ve.
The data may be tabulated as shown in Table 21.2.

*  From Table 21.1 (column 6) we see that the couple is – m1,r1.l1.

844       /G6C               Theory of Machines
Table 21.2Table 21.2Table 21.2Table 21.2Table 21.2
Plane Mass (m) Radius (r) Cent.force ÷ω2Distance from Couple ÷ω2
kg  m (m.r) kg-m Plane x(l) m (m.r.l) kg-m2
A 200 0.08 16 – 0.1 – 1.6
X(R.P.) mX0.1 0.1 mX00
B 300 0.07 21 0.2 4.2
C 400 0.06 24 0.3 7.2
YmY0.1 0.1 mY0.4 0.04 mY
D 200 0.08 16 0.6 9.6
The balancing masses mX and mY and their angular positions may be determined graphi-
cally as discussed below :

1.First of all, draw the couple polygon from the data given in Table 21.2 (column 6) as
shown in Fig. 21.8 ( c) to some suitable scale. The vector  do′′ represents the balanced
couple. Since the balanced couple is proportional to 0.04 mY, therefore by measurement,
Y . vector . kg-mmd o ′′ ==200 4 73      or    mY = 182.5 kg Ans.
(a) Position of planes. ( b) Angular position of masses.
          ( c) Couple polygon. ( d) Force polygon.
Fig. 21.8

Chapter 21  : Balancing of Rotating Masses            /G6C          845
The angular position of the mass mY is obtained by drawing OmY in Fig. 21.8 ( b), parallel
to vector do′′ . By measurement, the angular position of mY is Y12θ= °  in the clockwise
direction from mass mA (i.e. 200 kg ). Ans.
2.Now draw the force polygon from the data given in Table 21.2 (column 4) as shown in
Fig. 21.8 ( d). The vector eo represents the balanced force. Since the balanced force is

proportional to 0.1 mX, therefore by measurement,
X 0.1 vector 35.5 kg-mme o==     or    mX = 355 kg Ans.
The angular position of the mass mX is obtained by drawing OmX in Fig. 21.8 ( b), parallel
to vector eo. By measurement, the angular position of mX is X145θ= ° in the clockwise
Example 21.3. Four masses A, B, C and D as shown below are to be completely balanced.
ABCD
Mass (kg) — 30 50 40
Radius (mm) 180 240 120 150

The planes containing masses B and C are 300 mm apart. The angle between planes
containing B and C is 90°. B and C make angles of 210° and 120° respectively with D in the same
sense. Find :
1.The magnitude and the angular position of mass A ; and
2.The position of planes A and D.
Solution.  Given : rA = 180 mm = 0.18 m ; mB = 30 kg ; rB = 240 mm = 0.24 m ;
mC = 50 kg ; rC = 120 mm = 0.12 m ; mD = 40 kg ; rD = 150 mm = 0.15 m ;  ∠BOC  = 90° ;

∠BOD  = 210° ;  ∠COD  = 120°
1. The magnitude and the angular position of mass A
Let            mA = Magnitude of Mass A,
  x = Distance between the planes B and D, and
              y  = Distance between the planes A and B.
The position of the planes and the angular position of the masses is shown in Fig. 21.9 ( a)
and ( b) respectively.
Assuming the plane B as the reference plane ( R.P.) and the mass B (mB) along the horizon-

tal line as shown in Fig. 21.9 ( b), the data may be tabulated as below :
Table 21.3Table 21.3Table 21.3Table 21.3Table 21.3
Plane Mass Radius Cent.force ÷ω2Distance from Couple ÷ω2
(m) kg (r) m (m.r) kg-m plane B (l) m (m.r.l) kg-m2
AmA0.18 0.08 mA– y – 0.18 mA.y
B (R.P) 30 0.24 7.2 0 0
C 50 0.12 6 0.3 1.8
D 40 0.15 6 x 6x
The magnitude and angular position of mass A may be determined by drawing the force

polygon from the data given in Table 21.3 (Column 4), as shown in Fig. 21.9 ( c), to some suitable

846       /G6C               Theory of Machines
scale. Since the masses are to be completely balanced, therefore the force polygon must be a closed
figure. The closing side ( i.e. vector do) is proportional to 0.18 mA. By measurement,
     0.18 mA = Vector do = 3.6 kg-m   or   mA = 20 kg Ans.
In order to find the angular position of mass A, draw OA in Fig. 21.9 ( b) parallel to vector
do. By measurement, we find that the angular position of mass A from mass B in the anticlockwise

direction is  ∠AOB = 236° Ans .
(c) Force polygon.                    ( d) Couple polygon.
Fig. 21.9.
2. Position of planes A and D
The position of planes A and D may be obtained by drawing the couple polygon, as shown
in Fig. 21.9 ( d), from the data given in Table 21.3 (column 6). The couple polygon is drawn as
1.Draw vector oc′′  parallel to OC and equal to 1.8 kg-m2, to some suitable scale.

2.From points c′ and o′, draw lines parallel to OD and OA respectively, such that they
intersect at point d′. By measurement, we find that
6x = vector cd′′ = 2.3 kg-m2 or x = 0.383 m
We see from the couple polygon that the direction of vector  cd′′ is opposite to the
direction of mass D. Therefore the plane of mass D is 0.383 m or 383 mm towards left of plane B

and not towards right of plane B as already assumed. Ans.(a) Position of planes.                 ( b) Angular position of masses.All dimensions in mm.

Chapter 21  : Balancing of Rotating Masses            /G6C          847
Again by measurement from couple polygon,
           – 0.18 mA.y = vector ′′od  = 3.6 kg-m2
         – 0.18 × 20 y = 3.6     or   y = – 1 m
The negative sign indicates that the plane A is not towards left of B as assumed but it is
1 m or 1000 mm towards right of plane B. Ans.
Example 21.4.  A, B, C and D are four masses carried by a rotating shaft at radii 100,

125, 200 and 150 mm respectively. The planes in which the masses revolve are spaced 600 mm
apart and the mass of B, C and D are 10 kg, 5 kg, and 4 kg respectively.
Find the required mass A and the relative angular settings of the four masses so that the
shaft shall be in complete balance.
Solution.  Given : rA = 100 mm = 0.1 m ; rB = 125 mm = 0.125 m ; rC = 200 mm = 0.2 m ;
rD = 150 mm = 0.15 m ; mB = 10 kg ; mC = 5 kg ; mD = 4 kg

The position of planes is shown in Fig. 21.10 ( a). Assuming the plane of mass A as the
reference plane ( R.P.), the data may be tabulated as below :
Table 21.4Table 21.4Table 21.4Table 21.4Table 21.4
Plane Mass (m) Radius (r) Cent. Force ÷ω2Distance from Couple ÷ω2
kg m (m.r)kg-m plane A (l)m (m.r.l) kg-m2
A(R.P .) mA0.1 0.1 mA00
B 10 0.125 1.25 0.6 0.75
C 5 0.2 1 1.2 1.2
D 4 0.15 0.6 1.8 1.08

First of all, the angular setting of masses C and D is obtained by drawing the couple
polygon from the data given in Table 21.4 (column 6). Assume the position of mass B in the
horizontal direction OB as shown in Fig. 21.10 ( b). Now the couple polygon as shown in Fig.
21.10 ( c) is drawn as discussed below :
1.Draw vector o′ b′  in the horizontal direction ( i.e. parallel to OB) and equal to 0.75 kg-m2,
to some suitable scale.

2.From points o′ and b′, draw vectors  o′ c′ and b′ c′  equal to 1.2 kg-m2 and 1.08 kg-m2
respectively. These vectors intersect at c′.
3.Now in Fig. 21.10 ( b), draw OC parallel to vector  o′ c′ and OD parallel to vector b′ c′.
By measurement, we find that the angular setting of mass C from mass B in the anticlockwise
direction, i.e.
   BOC∠  = 240°  Ans.
and angular setting of mass D from mass B in the anticlockwise direction, i.e.
    BOD∠  = 100° Ans.

BOD∠  = 100° Ans.
In order to find the required mass A (mA) and its angular setting, draw the force polygon to
some suitable scale, as shown in Fig. 21.10 ( d), from the data given in Table 21.4 (column 4).
Since the closing side of the force polygon (vector do) is proportional to 0.1 mA, therefore
by measurement,
      0.1 mA = 0.7 kg-m2   or   mA = 7 kg Ans.

848       /G6C               Theory of Machines
Now draw OA in Fig. 21.10 ( b), parallel to vector do. By measurement, we find that the
angular setting of mass A from mass B in the anticlockwise direction, i.e.
    BOA∠  = 155°  Ans.
(c) Couple polygon. ( d) Force polygon.
Fig. 21.10
Example 21.5. A shaft carries four masses in parallel planes A, B, C and D in this order
along its length. The masses at B and C are 18 kg and 12.5 kg respectively, and each has an

eccentricity of 60 mm. The masses at A and D have an eccentricity of 80 mm. The angle betweenthe masses at B and C is 100° and that between the masses at B and A is 190°, both beingmeasured in the same direction. The axial distance between the planes A and B is 100 mm and
that between B and C is 200 mm. If the shaft is in complete dynamic balance, determine :
1. The magnitude of the masses at A and D ; 2. the distance between planes A and D ; and
3. the angular position of the mass at D.

B = 18 kg ; mC = 12.5 kg ; rB = rC = 60 mm = 0.06 m ; rA = rD = 80 mm
= 0.08 m ;  ∠BOC  = 100° ;  ∠ BOA  = 190°
1. Magnitude of the masses at A and D
Let            MA = Mass at A,
          MD = Mass at D, and
  x = Distance between planes A and D.All dimensions in mm

Chapter 21  : Balancing of Rotating Masses            /G6C          849
The position of the planes and angular position of the masses is shown in Fig. 21.11 ( a)
and ( b) respectively. The position of mass B is assumed in the horizontal direction, i.e. along OB.
Taking the plane of mass A as the reference plane, the data may be tabulated as below :
Table 21.5Table 21.5Table 21.5Table 21.5Table 21.5
Plane Mass Eccentricity Cent. force ÷ω2Distance from Couple ÷ω2

(m) kg (r) m (m.r) kg-m plane A(l)m (m.r.l) kg-m2
A (R.P.) mA0.08 0.08 mA00
B 18 0.06 1.08 0.1 0.108
C 12.5 0.06 0.75 0.3 0.225
DmD0.08 0.08 mDx 0.08 mD . x
Fig. 21.11
First of all, the direction of mass D is fixed by drawing the couple polygon to some suit-
able scale, as shown in Fig. 21.11 ( c), from the data given in Table 21.5 (column 6). The closingAll dimensions in mm.

850       /G6C               Theory of Machines
side of the couple polygon (vector co′′ ) is proportional to 0.08 mD.x. By measurement, we find that
0.08 mD.x = vector co′′ = 0.235 kg-m2 . . . (i)
In Fig. 21.11 ( b), draw OD parallel to vector co′′ to fix the direction of mass D.
Now draw the force polygon, to some suitable scale, as shown in Fig. 21.11 ( d), from the
data given in Table 21.5 (column 4), as discussed below :
1.Draw vector ob parallel to OB and equal to 1.08 kg-m.

2.From point b, draw vector bc parallel to OC and equal to 0.75 kg-m.
3.For the shaft to be in complete dynamic balance, the force polygon must be a closed
figure. Therefore from point c, draw vector cd parallel to OA and from point o draw
vector od parallel to OD. The vectors cd and od intersect at d. Since the vector cd is
proportional to 0.08 mA, therefore by measurement
   0.08 mA = vector cd = 0.77 kg-m   or   mA = 9.625 kg Ans.

and vector do is proportional to 0.08 mD, therefore by measurement,
 0.08 mD = vector do = 0.65 kg-m   or    mD = 8.125 kg Ans.
2. Distance between planes A and D
0.08 mD.x = 0.235 kg-m2
   0.08 × 8.125 × x = 0.235 kg-m2   or   0.65 x = 0.235
∴         0.2350.36150.65x== m = 361.5 mm  Ans.
3. Angular position of mass at D
By measurement from Fig. 21.11 ( b), we find that the angular position of mass at D from
mass B in the anticlockwise direction, i.e.  ∠BOD  = 251°  Ans.

Example 21.6. A shaft has three eccentrics, each 75 mm diameter and 25 mm thick,
machined in one piece with the shaft. The central planes of the eccentric are 60 mm apart. The
distance of the centres from the axis of rotation are 12 mm, 18 mm and 12 mm and their angular
positions are 120° apart. The density of metal is 7000 kg/m3. Find the amount of out-of-balance

force and couple at 600 r.p.m. If the shaft is balanced by adding two masses at a radius 75 mmand at distances of 100 mm from the central plane of the middle eccentric, find the amount of themasses and their angular positions.
Solution. Given : D = 75 mm = 0.075 m ; t = 25 mm = 0.025 m ; r
A = 12 mm = 0.012 m ;
rB = 18 mm = 0.018 m ; rC = 12 mm = 0.012 mm ; ρ = 7000 kg/m3 ; N = 600 r.p.m. or
2ω= π × 600/60 = 62.84 rad/s ; rL = rM = 75 mm = 0.075 m
We know that mass of each eccentric,

        mA = mB = mC = Volume × Density = 2
4Dtπ×× × ρ
 = 2(0.075) (0.025)7000 0.77 kg4π=
Let L and M be the planes at distances of 100 mm from the central plane of middle
eccentric. The position of the planes and the angular position of the three eccentrics is shown inFig. 21.12 ( a) and ( b) respectively. Assuming L as the reference plane and mass of the eccentric A
in the vertical direction, the data may be tabulated as below :

Chapter 21  : Balancing of Rotating Masses            /G6C          851
Table 21.6.Table 21.6.Table 21.6.Table 21.6.Table 21.6.
Plane Mass Radius Cent. force 2÷ω Distance from Couple 2÷ω
(m) kg (r) m (m.r) kg-m plane L.(l)m (m.r.l) kg-m2
L (R.P.) mL0.075 75 × 10–3 mL00
A 0.77 0.012 9.24 × 10–30.04 0.3696 × 10–3
B 0.77 0.018 13.86 × 10–30.1 1.386 × 10–3
C 0.77 0.012 9.24 × 10–30.16 1.4784 × 10–3
MmM0.075 75 × 10–3 mM0.20 15 × 10–3 mM
Out-of-balance force

The out-of-balance force is obtained by drawing the force polygon, as shown in Fig. 21.12
(c), from the data given in Table 21.6 (column 4). The resultant oc represents the out-of-balance
force.
39.24 10 oa−=×30.3696 10 oa−′′=×39.24 10 oa−=×
313.86 10 ab−=×31.386 10 ab−′′=×313.86 10 ab−=×
39.24 10 bc−=×31.4784 10 bc−′′=×39.24 10 bc−=×
3
M 75 10 cd m−=×
(c) Force polygon.  ( d) Couple polygon. ( e) Force polygon.

852       /G6C               Theory of Machines
Since the centrifugal force is proportional to the product of mass and radius ( i.e. m.r),
therefore by measurement.
 Out-of-balance force = vector oc = 4.75 × 10–3 kg-m
  = 4.75 × 10–3 × 2ω = 4.75 × 10–3 (62.84)2 = 18.76 N  Ans.
Out-of-balance couple
The out-of-balance couple is obtained by drawing the couple polygon from the data given
in Table 21.6 (column 6), as shown in Fig. 21.12 ( d). The resultant oc′′ represents the out-of-

balance couple. Since the couple is proportional to the product of force and distance ( m.r.l), there-
fore by measurement,
Out-of-balance couple = vector oc′′ = 1.1 × 10–3 kg-m2
  32 3 21.1 10 1.1 10 (62.84)−−=× × ω =×  = 4.34 N-m  Ans.
Amount of balancing masses and their angular positions
The vector co′′ (in the direction from c′ to o′), as shown in Fig. 21.12 ( d) represents the
balancing couple and is proportional to 15 × 10–3 mM, i.e.
       15 × 10–3 mM = vector co′′ = 1.1 × 10–3 kg-m2

or           mM = 0.073 kg Ans.
Draw OM in Fig. 21.12 ( b) parallel to vector co′′. By measurement, we find that the angular
position of balancing mass ( mM) is 5° from mass A in the clockwise direction. Ans.
Ship powered by a diesel engine.

Chapter 21  : Balancing of Rotating Masses            /G6C          853
In order to find the balancing mass ( mL), a force polygon as shown in Fig. 21.12 ( e) is
drawn. The closing side of the polygon i.e. vector do (in the direction from d to o) represents the
balancing force and is proportional to 75 × 10–3 mL. By measurement, we find that
         75 × 10–3 mL = vector do = 5.2 ×10–3 kg-m
or           mL = 0.0693 kg Ans.

Draw OL in Fig. 21.12 ( b), parallel to vector do. By measurement, we find that the angular
position of mass ( mL) is 124° from mass A in the clockwise direction . Ans.
Example 21.7. A shaft is supported in bearings 1.8 m apart and projects 0.45 m beyond
bearings at each end. The shaft carries three pulleys one at each end and one at the middle of its

length. The mass of end pulleys is 48 kg and 20 kg and their centre of gravity are 15 mm and 12.5mm respectively from the shaft axis. The centre pulley has a mass of 56 kg and its centre of gravityis 15 mm from the shaft axis. If the pulleys are arranged so as to give static balance, determine :1. relative angular positions of the pulleys, and 2. dynamic forces produced on the bearings when
the shaft rotates at 300 r.p.m .

A = 48 kg ; mC = 20 kg ;  rA = 15 mm = 0.015 m ; rC = 12.5 mm =
0.0125 m ; mB = 56 kg ; rB = 15 mm = 0.015 m ; N = 300 r.p.m . or ω= 2 π × 300/60
= 31.42 rad/s
1. Relative angular position of the pulleys
The position of the shaft and pulleys is shown in Fig. 21.13 ( a).
Let          mL and mM = Mass at the bearings L and M, and
    rL and rM = Radius of rotation of the masses at L and M respectively.

Assuming the plane of bearing L as reference plane, the data may be tabulated as below :
Table 21.7.Table 21.7.Table 21.7.Table 21.7.Table 21.7.
(m) kg (r) m (m.r) kg-m plane L(l)m (m.r.l) kg-m2
A 48 0.015 0.72 – 0.45 – 0.324
L(R.P) mLrLmL.rL00
B 56 0.015 0.84 0.9 0.756
MmMrMmM.rM1.8 1.8 mM.rM
C 20 0.0125 0.25 2.25 0.5625

First of all, draw the force polygon to some suitable scale, as shown in Fig. 21.13 ( c), from
the data given in Table 21.7 (column 4). It is assumed that the mass of pulley B acts in vertical
direction. We know that for the static balance of the pulleys, the centre of gravity of the system

must lie on the axis of rotation. Therefore a force polygon must be a closed figure. Now in Fig.21.13 ( b), draw OA parallel to vector bc and OC parallel to vector co. By measurement, we find
that
Angle between pulleys B and A  = 161° Ans.
Angle between pulleys A and C  = 76°  Ans.
and Angle between pulleys C and B  = 123°  Ans.
2. Dynamic forces at the two bearings
In order to find the dynamic forces (or reactions) at the two bearings L and M, let us first
calculate the values of m

854       /G6C               Theory of Machines
(a) Position of shaft and pulleys. ( b) Angular position of pulleys.
(c) Force polygon. ( d) Couple polygon. ( e) Force polygon.
Fig. 21.13
1.Draw the couple polygon to some suitable scale, as shown in Fig. 21.13 ( d), from the data
given in Table 21.7 (column 6). The closing side of the polygon (vector  co′′ ) represents
the balanced couple and is proportional to 1.8 mM.rM. By measurement, we find that

1.8 mM.rM = vector co′′ = 0.97 kg-m2        or          mM.rM = 0.54 kg-m
 ∴ Dynamic force at the bearing M
  = 22
MM. . 0.54 (31.42)mrω=  = 533 N  Ans.
2.Now draw the force polygon, as shown in Fig. 21.13 ( e), from the data given in Table
21.7 (column 4) and taking mM.rM = 0.54 kg-m. The closing side of the polygon (vector
do) represents the balanced force and is proportional to mL.rL. By measurement, we find
that
      mL.rL = 0.54 kg-m
∴ Dynamic force at the bearing L
              = 2

Chapter 21  : Balancing of Rotating Masses            /G6C          855
Notes : 1.  The dynamic force at the two bearings are equal in
magnitude but opposite in direction.
2. The dynamic force at the two bearings may also be ob-
tained as discussed below :
From couple polygon as shown in Fig. 21.13 ( d), we see
that the vector ′′oc   in the direction from o′ to c′ represents the
out-of-balance couple.
        =  vector oc′′ = 0.97 kg-m2

=  vector oc′′ = 0.97 kg-m2
        = 0.97 × ω2 = 0.97  (31.42)2 = 957.6 N-m
Since the shaft is in static balance, therefore it is only sub-
jected to an unbalanced couple which is same about all planes and the
bearing reactions are then equal and opposite. We know that
Dynamic force on each bearing
          Out-of-balance couple 957.6
Distance between bearings 1.8=== 532 N  Ans.

1. Four masses A, B, C and D are attached to a shaft and revolve in the same plane. The masses are 12
kg, 10 kg, 18 kg and 15 kg respectively and their radii of rotations are 40 mm, 50 mm, 60 mm and
30 mm. The angular position of the masses B, C and D are 60°, 135° and 270° from the mass A.
Find the magnitude and position of the balancing mass at a radius of 100 mm.
[Ans. 7.56 kg ; 87° clockwise from A]

2. Four masses A, B, C and D revolve at equal radii and are equally spaced along a shaft. The mass B
is 7 kg and the radii of C and D make angles of 90° and 240° respectively with the radius of B. Find
the magnitude of the masses A, C and D and the angular position of A so that the system may be
completely balanced.
[Ans. 5 kg ; 6 kg ; 4.67 kg ; 205° from mass B in anticlockwise direction]

3. A rotating shaft carries four masses A, B, C and D which are radially attached to it. The mass
centres are 30 mm, 38 mm, 40 mm and 35 mm respectively from the axis of rotation. The masses
A, C and D are 7.5 kg, 5 kg and 4 kg respectively. The axial distances between the planes of
rotation of A and B is 400 mm and between B and C is 500 mm. The masses A and C are at right
angles to each other. Find for a complete balance,
1. the angles between the masses B and D from mass A,

2. the axial distance between the planes of rotation of C and D,
3. the magnitude of mass B. [Ans. 162.5°, 47.5° ; 511 mm : 9.24 kg]
4. A rotating shaft carries four unbalanced masses 18 kg, 14 kg, 16 kg and 12 kg at radii 50 mm, 60
mm, 70 mm and 60 mm respectively. The 2nd, 3rd and 4th masses revolve in planes 80 mm, 160
mm and 280 mm respectively measured from the plane of the first mass and are angularly located at

60°, 135° and 270° respectively measured clockwise from the first mass looking from this mass endof the shaft. The shaft is dynamically balanced by two masses, both located at 50 mm radii and
revolving in planes mid-way between those of 1st and 2nd masses and midway between those of
3rd and 4th masses. Determine, graphically or otherwise, the magnitudes of the masses and theirrespective angular positions.
[Ans. 13.3 kg and 10.4 kg at 25° and 275° from mass A in anticlockwise direction]

A spiral elevator conveyor for
material handling.
Note : This picture is given as
additional information and is not a

856       /G6C               Theory of Machines
5. A shaft carries five masses A, B, C, D and E which revolve at the same radius in planes which are
equidistant from one another. The magnitude of the masses in planes A, C and D are 50 kg, 40 kg
and 80 kg respectively. The angle between A and C is 90° and that between C and D is 135°.
Determine the magnitude of the masses in planes B and E and their positions to put the shaft in
complete rotating balance.

[Ans. 12 kg, 15 kg ; 130° and 24° from mass A in anticlockwise direction]
6. A shaft with 3 metres span between two bearings carries two masses of 10 kg and 20 kg acting at
the extremities of the arms 0.45 m and 0.6 m long respectively. The planes in which these masses
rotate are 1.2 m and 2.4 m respectively from the left end bearing supporting the shaft. The angle

between the arms is 60°. The speed of rotation of the shaft is 200 r.p.m. If the masses are balancedby two counter-masses rotating with the shaft acting at radii of 0.3 m and placed at 0.3 m from each
bearing centres, estimate the magnitude of the two balance masses and their orientation with respect
to the X-axis, i.e. mass of 10 kg.
[Ans. 10 kg and 41 kg at 190° and 235° from X-axis in the anticlockwise direction]

7. A, B, C and D are four masses carried by a rotating shaft at radii 100 mm, 150 mm, 150 mm and
200 mm respectively. The planes in which the masses rotate are spaced at 500 mm apart and themagnitude of the masses B, C and D are 9 kg, 5 kg and 4 kg respectively. Find the required mass A
and the relative angular settings of the four masses so that the shaft shall be in complete balance.
[Ans. 10 kg ; Between B and A 165°, Between B and C 295°, Between B and D 145°]

8. A 3.6 m long shaft carries three pulleys, two at its two ends and third at the mid-point. The two endpulleys has mass of 79 kg and 40 kg and their centre of gravity are 3 mm and 5 mm respectively
from the axis of the shaft. The middle pulley mass is 50 kg and its centre of gravity is 8 mm from
the shaft axis. The pulleys are so keyed to the shaft that the assembly is in static balance. The shaftrotates at 300 r.p.m. in two bearings 2.4m apart with equal overhang on either side. Determine :

1. the relative angular positions of the pulleys, and 2. dynamic reactions at the two bearings.
9. The camshaft of high speed pump consists of a parallel shaft 25 mm diameter and 480 mm long. It
carries three eccentrics, each of diameter 60 mm and a uniform thickness of 18 mm. The assembly is
symmetrical as shown in Fig. 21.14 and the bearings are at A and B. The angle between the eccentrics
is 120° and the eccentricity of each is 12.5 mm. The material density is 7000 kg/m
3, and the speed of

rotation is 1430 r.p.m.
Fig. 21.14
Find :  1. dynamic load on each bearing due to the out-of-balance couple ; and 2. kinetic energy of
the complete assembly. [Ans. 6.12 kg ; 8.7 N-m]
1. Why is balancing of rotating parts necessary for high speed engines ?
2. Explain clearly the terms ‘static balancing’ and ‘dynamic balancing’. State the necessary conditions
to achieve them.

3. Discuss how a single revolving mass is balanced by two masses revolving in different planes.

Chapter 21  : Balancing of Rotating Masses            /G6C          857
4. Explain the method of balancing of different masses revolving in the same plane.
5. How the different masses rotating in different planes are balanced ?
1. The balancing of rotating and reciprocating parts of an engine is necessary when it runs at
(a) slow speed ( b)  medium speed        ( c) high speed

2. A disturbing mass m1 attached to a rotating shaft may be balanced by a single mass m2 attached in
the same plane of rotation as that of m1 such that
(a)m1.r2 = m2.r1(b)  m1.r1 = m2.r2       ( c)m1. m2 = r1.r2
3. For static balancing of a shaft,(a) the net dynamic force acting on the shaft is equal to zero
(b) the net couple due to the dynamic forces acting on the shaft is equal to zero
(c) both ( a) and ( b)

4. For dynamic balancing of a shaft,(a) the net dynamic force acting on the shaft is equal to zero
(b) the net couple due to dynamic forces acting on the shaft is equal to zero
5. In order to have a complete balance of the several revolving masses in different planes(a) the resultant force must be zero
(b) the resultant couple must be zero
(c) both the resultant force and couple must be zero

1. (c) 2. (b) 3. (a) 4. (c) 5. (c)

858       /G6C               Theory of Machines
858Balancing ofBalancing ofBalancing ofBalancing ofBalancing of
ReciprocatingReciprocatingReciprocatingReciprocatingReciprocating
MassesMassesMassesMassesMasses22
2. Primary and Secondary
Unbalanced Forces of
Reciprocating Masses.
3. Partial Balancing of
Unbalanced Primary Forcein a Reciprocating Engine.
4. Partial Balancing of
Locomotives.
5. Effect of Partial Balancing of

Reciprocating Parts of TwoCylinder Locomotives.
6. Variation of Tractive Force.
7. Swaying Couple.8. Hammer Blow.
9. Balancing of Coupled
10. Balancing of Primary Forces
of Multi-cylinder In-line
Engines.
1 1. Balancing of Secondary
Forces of Multi-cylinder In-line Engines.
12. Balancing of Radial Engines
(Direct and Reverse CrankMethod).

13. Balancing of V-engines .22.1.22.1.22.1.22.1.22.1. IntroductionIntroductionIntroductionIntroductionIntroduction
We have discussed in Chapter 15 (Art. 15.10), the
various forces acting on the reciprocating parts of an en-gine. The resultant of all the forces acting on the body of theengine due to inertia forces only is known as unbalanced
force or shaking force . Thus if the resultant of all the forces

due to inertia effects is zero, then there will be no unbal-anced force, but even then an unbalanced couple or shakingcouple will be present.
Consider a horizontal reciprocating engine mecha-
nism as shown in Fig. 22.1.
Fig. 22.1. Reciprocating engine mechanism.
Let FR = Force required to accelerate the
        reciprocating parts,

Chapter 22 : Balancing of Reciprocating Masses            /G6C          859
FI = Inertia force due to reciprocating parts,
           FN = Force on the sides of the cylinder walls or normal force acting on
       the cross-head guides, and
FB = Force acting on the crankshaft bearing or main bearing.
Since FR and FI are equal in magnitude but opposite in direction, therefore they balance
each other. The horizontal component of FB (i.e. FBH) acting along the line of reciprocation is also

equal and opposite to FI. This force FBH = FU is an unbalanced force or shaking force and required
to be properly balanced.
The force on the sides of the cylinder walls ( FN) and the vertical component of FB
(i.e. FBV) are equal and opposite and thus form a shaking couple of magnitude FN × x or FBV × x.
 From above we see that the effect of the reciprocating parts is to produce a shaking force
and a shaking couple. Since the shaking force and a shaking couple vary in magnitude and direc-

tion during the engine cycle, therefore they cause very objectionable vibrations.
Thus the purpose of balancing the reciprocating masses is to eliminate the shaking force
and a shaking couple. In most of the mechanisms, we can reduce the shaking force and a shakingcouple by adding appropriate balancing mass, but it is usually not practical to eliminate themcompletely. In other words, the reciprocating masses are only partially balanced.

Note : The masses rotating with the crankshaft are normally balanced and they do not transmit any unbalanced
or shaking force on the body of the engine.
22.2.22.2.22.2.22.2.22.2. Primary and Secondary Unbalanced Forces of Reciprocating MassesPrimary and Secondary Unbalanced Forces of Reciprocating MassesPrimary and Secondary Unbalanced Forces of Reciprocating MassesPrimary and Secondary Unbalanced Forces of Reciprocating MassesPrimary and Secondary Unbalanced Forces of Reciprocating Masses

Consider a reciprocating engine mechanism as shown in Fig. 22.1.
Let         m = Mass of the reciprocating parts,
         l = Length of the connecting rod PC,
         r  = Radius of the crank OC,
        θ = Angle of inclination of the crank with the line of stroke PO,
       ω = Angular speed of the crank,
         n  = Ratio of length of the connecting rod to the crank radius = l / r.
We have already discussed in Art. 15.8 that the acceleration of the reciprocating parts is

approximately given by the expression,
       θ =ω ⋅ θ+2
Rcos 2cos arn
∴ Inertia force due to reciprocating parts or force required to accelerate the reciprocating
parts,
        FI = FR = Mass × acceleration = 2 cos 2cos mrnθ ⋅ω ⋅ θ+
We have discussed in the previous article that the horizontal component of the force exerted
on the crank shaft bearing ( i.e. FBH) is equal and opposite to inertia force ( FI). This force is an
unbalanced one and is denoted by FU.
∴ Unbalanced force,

       22 2
Ucos 2 cos2cos . cos Fm r m r m rnnθθ = ⋅ ω⋅ θ + = ω⋅ θ + ⋅ ω⋅×  = FP + FS
The expression 2(c o s )mr⋅ω ⋅ θ  is known as primary unbalanced force  and
2 cos 2mrnθ ⋅ω ⋅ ×is called secondary unbalanced force .

860       /G6C               Theory of Machines
∴  Primary unbalanced force, =⋅ ω ⋅ θ2
P cos Fm r
and secondary unbalanced force,        θ=⋅ ω ⋅ ×2
Scos2Fm r
n
Notes:  1. The primary unbalanced force is maximum, when θ = 0° or 180°. Thus, the primary force is
maximum twice in one revolution of the crank. The maximum primary unbalanced force is given by
           =⋅ ω ⋅2
P( )maxFm r
2. The secondary unbalanced force is maximum, when θ = 0°, 90°,180° and 360°. Thus, the second-

ary force is maximum four times in one revolution of the crank. The maximum secondary unbalanced force is
given by
           =⋅ ω ×2
S( )maxrFmn
3. From above we see that secondary unbalanced force is 1/ n times the maximum primary
unbalanced force.
4. In case of moderate speeds, the secondary unbalanced force is so small that it may be neglected as
compared to primary unbalanced force.
5. The unbalanced force due to reciprocating masses varies in magnitude but constant in direction

while due to the revolving masses, the unbalanced force is constant in magnitude but varies in direction.
22.3.22.3.22.3.22.3.22.3. Partial Balancing of Unbalanced Primary Force in a ReciprocatingPartial Balancing of Unbalanced Primary Force in a ReciprocatingPartial Balancing of Unbalanced Primary Force in a ReciprocatingPartial Balancing of Unbalanced Primary Force in a ReciprocatingPartial Balancing of Unbalanced Primary Force in a Reciprocating
EngineEngineEngineEngineEngine

The primary unbalanced force 2(c o s )mr⋅ω ⋅ θ  may be considered as the component of the
centrifugal force produced by a rotating mass m placed at the crank radius r, as shown in Fig. 22.2.
Fig. 22.2. Partial balancing of unbalanced primary force in a reciprocating engine.
The primary force acts from O to P along the line of stroke. Hence, balancing of primary
force is considered as equivalent to the balancing of mass m rotating at the crank radius r. This is

balanced by having a mass B at a radius b, placed diametrically opposite to the crank pin C.
We know that centrifugal force due to mass B,
                 2Bb=⋅ ω ⋅
and horizontal component of this force acting in opposite direction of primary force
                 2cos Bb=⋅ ω ⋅ θ
The primary force is balanced, if
             22cos cos Bb mr⋅ω ⋅ θ= ⋅ω ⋅ θ    or    B.b = m.r

Chapter 22 : Balancing of Reciprocating Masses            /G6C          861
A little consideration will show,
that the primary force is completely
balanced if B.b = m.r, but the centrifugal
force produced due to the revolving massB, has also a vertical component
(perpendicular to the line of stroke) of
magnitude2sin Bb⋅ω ⋅ θ . This force
remains unbalanced. The maximum valueof this force is equal to 
2Bb⋅ω ⋅  when θ
is 90° and 270°, which is same as the
maximum value of the primary force

2mr⋅ω ⋅ .
From the above discussion, we see
that in the first case, the primary unbalanced

force acts along the line of stroke whereasin the second case, the unbalanced force actsalong the perpendicular to the line of stroke.The maximum value of the force remainssame in both the cases. It is thus obvious,that the effect of the above method ofbalancing is to change the direction of themaximum unbalanced force from the lineof stroke to the perpendicular of line ofstroke. As a compromise let a fraction ‘ c’

of the reciprocating masses is balanced,such that
                c .m.r = B.b
∴ Unbalanced force along the line
of stroke
            22cos cos mr Bb=⋅ ω ⋅ θ −⋅ ω ⋅ θ
            22cos cos mr c mr= ⋅ ω⋅ θ −⋅ ⋅ ω⋅ θ ... (∵ B.b = c.m.r)
            2(1 ) coscm r=− ⋅ ω ⋅ θ
and unbalanced force along the perpendicular to the line of stroke
           = ⋅ ω⋅ θ =⋅ ⋅ ω⋅ θ22sin sin Bb c mr
∴ Resultant unbalanced force at any instant

            2222(1 ) cos sincm r c m r  = − ⋅ω ⋅ θ + ⋅ ⋅ω ⋅ θ  
           22 2 2 2(1 ) cos sin mr c c=⋅ ω ⋅ − θ + θ
Note :  If the balancing mass is required to balance the revolving masses as well as reciprocating masses, then
        = ⋅+⋅⋅= +⋅11 .( )B b m r cmr m cmr
where                     m1 = Magnitude of the revolving masses, and
          m = magnitude of the reciprocating masses.
Operating handle
Carrying handle

Bin where largerdust particlesaccumulate.
Cleaner leadcontaining brush
Cable winderHose for use inawkward placeCable winderfor stowingflex
Cyclone cleaner.

862       /G6C               Theory of Machines
Example 22.1. A single cylinder reciprocating engine has speed 240 r.p.m., stroke 300
mm, mass of reciprocating parts 50 kg, mass of revolving parts at 150 mm radius 37 kg. If two-
third of the reciprocating parts and all the revolving parts are to be balanced, find : 1. The balance
mass required at a radius of 400 mm, and 2. The residual unbalanced force when the crank has
rotated 60° from top dead centre .

Solution.  Given : N = 240 r.p.m. or 22 4 0 / 6 0ω= π×  = 25.14 rad/s ; Stroke = 300 mm
= 0.3 m; m = 50 kg ; m1 = 37 kg ; r = 150 mm = 0.15 m ; c = 2/3
1. Balance mass required
Let          B = Balance mass required, and
         b = Radius of rotation of the balance mass = 400 mm = 0.4 m
. . . (Given)
      B .b = (m1 + c.m) r
 B × 0.4 = 237 50 0.15 10.553+× =   or   B = 26.38 kg  Ans.
2. Residual unbalanced force

Let         θ = Crank angle from top dead centre = 60° . . . (Given)
We know that residual unbalanced force
          22 2 2 2(1 ) cos sin mr c c=⋅ ω ⋅ − θ + θ
          22
22 2 2250(25.14) 0.15 1 cos 60 sin 60 N33  =− ° + °     
         = 4740 × 0.601 = 2849 N Ans.
22.4.22.4.22.4.22.4.22.4. Partial Balancing of LocomotivesPartial Balancing of LocomotivesPartial Balancing of LocomotivesPartial Balancing of LocomotivesPartial Balancing of Locomotives

The locomotives, usually, have two cylinders with cranks placed at right angles to each
other in order to have uniformity in turning moment diagram. The two cylinder locomotives may
be classified as :
1. Inside cylinder locomotives ; and 2. Outside cylinder locomotives.
In the  inside  cylinder locomotives , the two cylinders are placed in between the planes of
two driving wheels as shown in Fig. 22.3 ( a) ; whereas in the outside cylinder locomotives , the two

cylinders are placed outside the driving wheels, one on each side of the driving wheel, as shown inFig. 22.3 ( b). The locomotives may be
(a) Single or uncoupled locomotives ; and (b) Coupled locomotives.
        ( a) Inside cylinder locomotives. ( b) Outside cylinder locomotives.
Fig. 22.3

Chapter 22 : Balancing of Reciprocating Masses            /G6C          863
A single or uncoupled locomotive is one, in which the effort is transmitted to one pair of
the wheels only ; whereas in coupled locomotives, the driving wheels are connected to the leading
and trailing wheel by an outside coupling rod.

22.5.22.5.22.5.22.5.22.5. Effect of Partial Balancing of Reciprocating Parts of Two CylinderEffect of Partial Balancing of Reciprocating Parts of Two CylinderEffect of Partial Balancing of Reciprocating Parts of Two CylinderEffect of Partial Balancing of Reciprocating Parts of Two CylinderEffect of Partial Balancing of Reciprocating Parts of Two Cylinder
LocomotivesLocomotivesLocomotivesLocomotivesLocomotives

We have discussed in the previous article that the reciprocating parts are only partially
balanced. Due to this partial balancing of the reciprocating parts, there is an unbalanced primary
force along the line of stroke and also an unbalanced primary force perpendicular to the line ofstroke. The effect of an unbalanced primary force along the line of stroke is to produce;
1. Variation in tractive force along the line of stroke ; and 2. Swaying couple.

The effect of an unbalanced primary force perpendicular to the line of stroke is to produce
variation in pressure on the rails, which results in hammering action on the rails. The maximummagnitude of the unbalanced force along the perpendicular to the line of stroke is known as ahammer blow . We shall now discuss the effects of an unbalanced primary force in the following
articles.

articles.
22.6.22.6.22.6.22.6.22.6. Variation of Tractive ForceVariation of Tractive ForceVariation of Tractive ForceVariation of Tractive ForceVariation of Tractive Force
The resultant unbalanced force due to the two cylinders, along the line of stroke, is known
as tractive force . Let the crank for the first cylinder be inclined at an angle θ with the line of
stroke, as shown in Fig. 22.4. Since the crank for the second cylinder is at right angle to the first

crank, therefore the angle of inclination for the second crank will be (90° + θ).
Let          m = Mass of the reciprocating parts per cylinder, and
          c = Fraction of the reciprocating parts to be balanced.
We know that unbalanced force along the line of stroke for cylinder 1
          = 2(1 – ) . . coscm rωθ
Similarly, unbalanced force along the line of stroke for cylinder 2,
           2(1 ) . cos(90 )cm r=− ω ⋅ ° + θ
∴ As per definition, the tractive force,

         FT = Resultant unbalanced force
   along the line of stroke
          = 2(1 ) . . coscm r−ω θ
     + 2(1 ) . . cos(90 )−ω ° + θcm r
          = 2(1 ) . . (cos sin )cm r−ω θ − θ
The tractive force is maximum or minimum when (cos θ – sin θ) is maximum or mini-
mum. For (cos θ – sin θ) to be maximum or minimum,
   (cos sin ) 0d
dθ− θ =
θ or       sin cos 0−θ −θ =     or    sin cos−θ =θ
∴               tan 1θ=−      or        135θ= °     or     315°

Thus, the tractive force is maximum or minimum when θ = 135° or 315°.
∴ Maximum and minimum value of the tractive force or the variation in tractive force
           = 22(1 ) . . (cos135 sin135 ) 2 (1 ) . .cm r cm r±− ω ° − ° = ± − ωFig. 22.4.  Variation of tractive force.

864       /G6C               Theory of Machines
22.7.22.7.22.7.22.7.22.7. Swaying CoupleSwaying CoupleSwaying CoupleSwaying CoupleSwaying Couple
The unbalanced forces along the line of stroke for the two cylinders constitute a couple
about the centre line YY between the cylinders as shown in Fig. 22.5.
This couple has swaying effect about a vertical axis, and tends to sway the engine alternately
in clockwise and anticlockwise directions. Hence the couple is known as swaying couple.

Let            a  = Distance between the centre lines of the two cylinders.
∴ Swaying couple
           = 2(1 ) . . cos2acm r−ω θ ×
       2(1 ) . . cos (90 )2acm r−− ω ° + θ
             = 2(1 ) . . (cos sin )2acm r−ω × θ + θ
The swaying couple is maximum or minimum when
(cos sin )θ+ θ  is maximum or minimum. For (cos sin )θ+ θ  to
be maximum or minimum,
dθ+ θ =θ      or     sin cos 0−θ +θ =   or   sin cos−θ = −θ
∴     tan 1θ=        or       45θ= °   or    225°

Thus, the swaying couple is maximum or minimum when θ = 45° or 225°.
∴ Maximum and minimum value of the swaying couple
 = 22(1 ) . . (cos 45 sin 45 ) (1 ) . .2 2aacm r cm r±− ω × ° + ° = ± − ω
Note :  In order to reduce the magnitude of the swaying couple, revolving balancing masses are introduced.
But, as discussed in the previous article, the revolving balancing masses cause unbalanced forces to act at

right angles to the line of stroke. These forces vary the downward pressure of the wheels on the rails andcause oscillation of the locomotive in a vertical plane about a horizontal axis. Since a swaying couple is moreharmful than an oscillating couple, therefore a value of ‘ c’ from 2/3 to 3/4, in two-cylinder locomotives with
two pairs of coupled wheels, is usually used. But in large four cylinder locomotives with three or more pairs
of coupled wheels, the value of ‘ c’ is taken as 2/5.

22.8.22.8.22.8.22.8.22.8. Hammer BlowHammer BlowHammer BlowHammer BlowHammer Blow
We have already discussed that the maximum magnitude of the unbalanced force along the
perpendicular to the line of stroke is known as hammer blow .
We know that the unbalanced force along the perpendicular to the line of stroke due to the
balancing mass B, at a radius b, in order to balance reciprocating parts only is B.ω2.b sin θ. This
force will be maximum when sin θ is unity, i.e. when θ = 90° or 270°.

∴     Hammer blow = B.ω2.b (Substituiting sin θ = 1)
The effect of hammer blow is to cause the variation in pressure between the wheel and the
rail. This variation is shown in Fig. 22.6, for one revolution of the wheel.
Let P be the downward pressure on the rails (or static wheel load).Fig. 22.5.  Swaying couple.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          865
∴ Net pressure between the wheel and the rail
          = 2.. PB b±ω
Fig. 22.6. Hammer blow.
If (P–B.ω2.b) is negative , then the wheel will be lifted from the rails. Therefore the limiting
condition in order that the wheel does not lift from the rails is given by
       2.. PB b=ω
and the permissible value of the angular speed,
        .P
Bbω=

.P
Bbω=
Example 22.2.  An inside cylinder locomotive has its cylinder centre lines 0.7 m apart and
has a stroke of 0.6 m. The rotating masses per cylinder are equivalent to 150 kg at the crank pin,
and the reciprocating masses per cylinder to 180 kg. The wheel centre lines are 1.5 m apart. Thecranks are at right angles.
The whole of the rotating and 2/3 of the recipro-

cating masses are to be balanced by masses placed at aradius of 0.6 m. Find the magnitude and direction of thebalancing masses.
Find the fluctuation in rail pressure under one
wheel, variation of tractive effort and the magnitude ofswaying couple at a crank speed of 300 r.p.m.
Solution. Given : a = 0.7 m; l
B = lC = 0.6 m or
rB = rC = 0.3 m; m1 = 150 kg; m2 = 180 kg;
c = 2/3; rA = rD = 0.6 m; N = 300 r.p.m. or
23 0 0 / 6 0ω= π× = 31.42 rad/s

We know that the equivalent mass of the rotating
parts to be balanced per cylinder at the crank pin,
 m = mB = mC = m1 + c.m2 = 150 + 2
3× 180 = 270 kg
Magnitude and direction of the balancing masses
Let        mA and mD = Magnitude of the balancing
     masses
          Aθand Dθ= Angular position of the
     balancing masses mA
     and mD from the first
     crank B.
This Brinel hardness testing machine is
used to test the hardness of the metal.

866       /G6C               Theory of Machines
The magnitude and direction of the balancing masses may be determined graphically as
1.First of all, draw the space diagram to show the positions of the planes of the wheels and
the cylinders, as shown in Fig. 22.7 ( a). Since the cranks of the cylinders are at right
angles, therefore assuming the position of crank of the cylinder B in the horizontal direc-

tion, draw OC and OB at right angles to each other as shown in Fig. 22.7 ( b).
2.Tabulate the data as given in the following table. Assume the plane of wheel A as the
reference plane.
Table 22.1Table 22.1Table 22.1Table 22.1Table 22.1
Plane mass. Radius Cent. force ÷÷÷÷÷ 2ω Distance from Couple ÷÷÷÷÷ 2ω
(m) kg (r)m (m.r) kg-m plane A (l)m (m.r.l) kg-m2
A (R.P.) mA0.6 0.6 mA00
B 270 0.3 81 0.4 32.4
C 270 0.3 81 1.1 89.1
DmD0.6 0.6 mD1.5 0.9 mD

3.Now, draw the couple polygon from the data given in Table 22.1 (column 6), to some
suitable scale, as shown in Fig 22.7 ( c). The closing side co′′ represents the balancing
couple and it is proportional to 0.9 mD. Therefore, by measurement,
0.9 mD = vector c ′o′ = 94.5 kg-m2   or   mD = 105 kg  Ans.
            
               ( a) Position of planes.  ( b) Angular position of masses.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          867
4.To determine the angular position of the balancing mass D, draw OD in Fig. 22.7 ( b)
parallel to vector co′′. By measurement,
     Dθ = 250° Ans.
5.In order to find the balancing mass A, draw the force polygon from the data given in
Table 22.1 (column 4), to some suitable scale, as shown in Fig. 22.7 ( d), The vector do
represents the balancing force and it is proportional to 0.6 mA. Therefore by measurement,

0.6 mA = vector do = 63 kg-m or mA = 105 kg  Ans.
6. To determine the angular position of the balancing mass A, draw OA in Fig. 22.7 ( b)
parallel to vector do. By measurement,
     Aθ = 200° Ans.
Fluctuation in rail pressure
We know that each balancing mass
 = 105 kg
∴  Balancing mass for rotating masses,
          D = 1 150105 105 58.3 kg270m
m×= ×=
and balancing mass for reciprocating masses,
         2. 2 180105 105 46.6 kg3 270cmBm=× = × × =

This balancing mass of 46.6 kg for reciprocating masses gives rise to the centrifugal force.
∴ Fluctuation in rail pressure or hammer blow
 = 22. . 46.6 (31.42) 0.6Bbω= = 27 602 N. Ans. ... (∵ b = rA = rD)
Variation of tractive effort
We know that maximum variation of tractive effort
 = 22
222(1 ) . . 2 1 180(31.42) 0.3N3cm r±−ω = ±− 
 = ± 25 127 N  Ans. ... (∵ r = rB = rC)
Swaying couple
We know that maximum swaying couple
 = 22
220.7 1(1 ) 3. . 180(31.42) 0.3 N-m
22acmr−− ×ω= ×

 = 8797 N-m  Ans.
Example 22.3 The three cranks of a three cylinder locomotive are all on the same axle
and are set at 120°. The pitch of the cylinders is 1 metre and the stroke of each piston is 0.6 m. The
reciprocating masses are 300 kg for inside cylinder and 260 kg for each outside cylinder and theplanes of rotation of the balance masses are 0.8 m from the inside crank.
If 40% of the reciprocating parts are to be balanced, find :

1. the magnitude and the position of the balancing masses required at a radius of 0.6 m ;
and
2. the hammer blow per wheel when the axle makes 6 r.p.s.

868       /G6C               Theory of Machines
Solution.  Given : ∠AOB = ∠BOC = ∠COA = 120° ; lA = lB = lC = 0.6 m or rA = rB
= rC = 0.3 m ; mI = 300 kg ; mO = 260 kg ; c = 40% = 0.4 ; b1 = b2 = 0.6 m ; N = 6 r.p.s.
= 6 × 2π = 37.7 rad/s
Since 40% of the reciprocating masses are to be balanced, therefore mass of the reciprocat-
ing parts to be balanced for each outside cylinder,
       mA = mC = c × mO = 0.4 × 260 = 104 kg
and mass of the reciprocating parts to be balanced for inside cylinder,

mB = c × m1 = 0.4 × 300 = 120 kg
1. Magnitude and position of the balancing masses
Let        B1 and B2 = Magnitude of the balancing masses in kg,
      1θ and 2θ = Angular position of the balancing masses B1 and B2 from crank A.
The magnitude and position of the balancing masses may be determined graphically as
1.First of all, draw the position of planes and cranks as shown in Fig. 22.8 ( a) and ( b)

respectively. The position of crank A is assumed in the horizontal direction.
2.Tabulate the data as given in the following table. Assume the plane of balancing mass B1
(i.e. plane 1) as the reference plane.
Table 22.2Table 22.2Table 22.2Table 22.2Table 22.2
(m)kg (r) m (m.r) kg-m plane1 (l)m (m.r.l.) kg-m2
(1)  (2) (3) (4) (5) (6)
A 104 0.3 31.2 – 0.2 – 6.24
1 (R.P.) B10.6 0.6 B100
B 120 0.3 36 0.8 28.8
2 B20.6 0.6 B21.6 0.96 B2

C 104 0.3 31.2 1.8 56.16
3.Now draw the couple polygon with the data given in Table 22.2 (column 6), to some
suitable scale, as shown in Fig. 22.8 ( c). The closing side co′′ represents the balancing
couple and it is proportional to 0.96 B2. Therefore, by measurement,
  0.96 B2 = vector co′′ = 55.2 kg-m2 or B2 = 57.5 kg  Ans .
4.To determine the angular position of the balancing mass B2, draw OB2 parallel to vector

co′′ as shown in Fig. 22.8 ( b). By measurement,
       2θ = 24° Ans.
5.In order to find the balance mass B1, draw the force polygon with the data given in Table
22.2 (column 4 ), to some suitable scale, as shown in Fig. 22.8 ( d). The closing side co
represents the balancing force and it is proportional to 0.6 B1. Therefore, by measurement,
    0.6 B1 = vector co = 34.5 kg-m   or   B1 = 57.5 kg  Ans.
6.To determine the angular position of the balancing mass B1, draw OB1 parallel to vector

co, as shown in Fig. 22.8 ( b). By measurement,
        1θ = 215°  Ans.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          869
        ( a) Position of planes.      ( b) Position of cranks.
               ( c) Couple polygon.            ( d) Force polygon.
Fig. 22.8
2. Hammer blow per wheel
We know that hammer blow per wheel
            = 2
11..Bbω  = 57.5 (37.7)2 20.6 = 49 035 N  Ans.
Example 22.4. The following data refer to two cylinder locomotive with cranks at 90° :

Reciprocating mass per cylinder = 300 kg ; Crank radius = 0.3 m ; Driving wheel
diameter = 1.8 m ; Distance between cylinder centre lines = 0.65 m ; Distance between the driving
wheel central planes = 1.55 m.
Determine :  1. the fraction of the reciprocating masses to be balanced, if the hammer blow
is not to exceed 46 kN at 96.5 km. p.h. ;  2. the variation in tractive effort ; and 3. the maximum
swaying couple.This chamber is used to test the
acoustics of a vehicle so that the

noise it produces can be reduced.
The panels in the walls and ceiling
of the room absorb the sound which
is monitored (above)

870       /G6C               Theory of Machines
Solution. Given : m = 300 kg  ; r = 0.3 m ; D = 1.8 m or R = 0.9 m ; a = 0.65 m ; Hammer
blow = 46 kN = 46 × 103 N ; v = 96.5 km/h = 26.8 m/s
1. Fraction of the reciprocating masses to be balanced
Let         c  = Fraction of the reciprocating masses to be balanced, and
          B  = Magnitude of balancing mass placed at each of the driving wheels at
   radius b.
We know that the mass of the reciprocating parts to be balanced
. 300 kgcm c==

(a) Position of planes.          ( b) Position of cranks.
Fig. 22.9
The position of planes of the wheels and cylinders is shown in Fig. 22.9 ( a), and the
position of cranks is shown in Fig 22.9 ( b). Assuming the plane of wheel A as the reference plane,
the data may be tabulated as below :
Table 22.3Table 22.3Table 22.3Table 22.3Table 22.3
Plane Mass Radius Cent. force2÷ω Distance from Couple 2÷ω
(m) kg (r) m (m.r) kg-m plane A (l)m (m.r.l.) kg-m2

A (R.P.) B b B.b 00
B 300 c 0.3 90 c 0.45 40.5 c
C 300 c 0.3 90 c 1.1 99 c
DB b B .b 1.55 1.55 B.b
Now the couple polygon, to some suitable scale, may be drawn with the data given in
Table 22.3 (column 6), as shown in Fig. 22.10. The closing side of the polygon (vector co′′)
represents the balancing couple and is proportional to 1.55 B.b.
From the couple polygon,
            221.55 . (40.5 ) (99 ) 107 Bb c c c=+ =
∴                B .b = 107 c / 1.55 = 69 c

We know that angular speed,
        ω= v/R = 26.8/0.9 = 29.8 rad/s
∴ Hammer blow,
            46 × 103 = B. 2ω.b
  = 69 c (29.8)2 = 61 275 c
∴           c  = 46 × 103/61 275 = 0.751 Ans.Fig. 22.10

Chapter 22 : Balancing of Reciprocating Masses            /G6C          871
2. Variation in tractive effort
We know that variation in tractive effort
 = 22( 1 ) . . cm r±−ω = 22(1 0.751) 300(29.8) 0.3±−
           = 28 140 N = 28.14 kN  Ans.
Maximum swaying couple
We know the maximum swaying couple
          = 2 (1 )..
2acmr−×ω  = 2 0.65(1 0.751)300(29.8) 0.3 9148 N-m
2−×=
          = 9.148 kN-m  Ans.
Example 22.5. The following data apply to an outside cylinder uncoupled locomotive :

Mass of rotating parts per cylinder = 360 kg ; Mass of reciprocating parts per cylinder
= 300 kg ; Angle between cranks = 90° ; Crank radius = 0.3 m ; Cylinder centres = 1.75 m ;
Radius of balance masses = 0.75 m ; Wheel centres = 1.45 m.
If whole of the rotating and two-thirds of reciprocating parts are to be balanced in planes
of the driving wheels, find :
1. Magnitude and angular positions of balance masses,

2. Speed in kilometres per hour at which the wheel will lift off the rails when the load on
each driving wheel is 30 kN and the diameter of tread of driving wheels is 1.8 m, and
3. Swaying couple at speed arrived at in (2) above.
Solution :  Given : m1 = 360 kg ; m2 = 300 kg ; ∠ AOD = 90° ; rA = rD = 0.3 m ;
a = 1.75 m ; rB = rC = 0.75 m ; c = 2 / 3.
We know that the equivalent mass of the rotating parts to be balanced per cylinder,
         m  = mA = mD = m1 + c.m2 = 360 + 2
3 × 300 = 560 kg

1. Magnitude and angular position of balance masses
Let       mB and mC = Magnitude of the balance masses, and
        Bθ and Cθ = angular position of the balance masses mB and mC from the crank A.
The magnitude and direction of the balance masses may be determined, graphically, as
1.First of all, draw the positions of the planes of the wheels and the cylinders as shown in

Fig. 22.11 ( a). Since the cranks of the two cylinders are at right angles, therefore assum-
ing the position of the cylinder A in the horizontal direction, draw OA and OD at right
angles to each other as shown in Fig. 22.11 ( b).
2.Assuming the plane of wheel B as the reference plane, the data may be tabulated as be-
low:

872       /G6C               Theory of Machines
Table 22.4Table 22.4Table 22.4Table 22.4Table 22.4
(m) kg (r) m (m.r) kg-m plane B(l) m (m.r.l) kg-m2
A 560 0.3 168 – 0.15 – 25.2
B (R.P) mB0.75 0.75 mB00
CmC0.75 0.75 mC1.45 1.08 mC
D 560 0.3 168 1.6 268.8
3.Now draw the couple polygon with the data given in Table 22.4 column (6), to some

suitable scale as shown in Fig. 22.11( c). The closing side do′′ represents the balancing
couple and it is proportional to 1.08 mC. Therefore, by measurement,
 1.08 mC = 269.6 kg-m2    or    mC = 249 kg   Ans.
        
       ( a) Position of planes. ( b) Position of masses.
       ( c) Couple polygon.          ( d) Force polygon.
Fig. 22.11
4.To determine the angular position of the balancing mass C, draw OC parallel to vector
do′′ as shown in Fig. 22.11 ( b). By measurement,

      Cθ = 275° Ans.
5.In order to find the balancing mass B, draw the force polygon with the data given in Table
22.4 column (4), to some suitable scale, as shown in Fig. 22.11 ( d). The vector co represents

Chapter 22 : Balancing of Reciprocating Masses            /G6C          873
the balancing force and it is proportional to 0.75 mB. Therefore, by measurement,
  0.75 mB = 186.75 kg-m   or    mB = 249 kg  Ans.
6.To determine the angular position of the balancing mass B, draw OB parallel to vector oc
as shown Fig. 22.11 ( b). By measurement,
       Bθ = 174.5° Ans.
2. Speed at which the wheel will lift off the rails
Given :          P = 30 kN = 30 × 103 N ; D = 1.8 m

Let          ω = Angular speed at which the wheels will lift off the rails in rad/s, and
         v  = Corresponding linear speed in km/h.
We know that each balancing mass,
        mB = mC = 249 kg
∴  Balancing mass for reciprocating parts,
          2. 2 300249 249 89 kg3 560cmBm=× = × × =
We know that    330 1021.2 rad/s.8 9 0 . 7 5P
Bb×ω= = =×...( ∵ b = rB = rC)
and           / 2 21.2 1.8/ 2 19.08 m/s vD=ω× = × =
 = 19.08 × 3600/ 1000 = 68.7 km/h Ans.

= 19.08 × 3600/ 1000 = 68.7 km/h Ans.
3. Swaying couple at speed ω = 21.1 rad/s
We know that the swaying couple
            2
2(1 )..
2acmr−=× ω 221.75 13300(21.2) 0.3
2−=× N-m
 = 16 687 N-m = 16.687 kN-m  Ans.
22.9.22.9.22.9.22.9.22.9. Balancing of CoupledBalancing of CoupledBalancing of CoupledBalancing of CoupledBalancing of Coupled
The uncoupled locomotives as
discussed in the previous article, are

obsolete now-a-days. In a coupledlocomotive, the driving wheels areconnected to the leading and trailingwheels by an outside coupling rod. Bysuch an arrangement, a greater portionof the engine mass is utilised by tractivepurposes. In coupled locomotives, thecoupling rod cranks are placeddiametrically opposite to the adjacentmain cranks ( i.e. driving cranks). The
coupling rods together with cranks andpins may be treated as rotating masses

A dynamo converts mechanical energy into electrical
energy.

874       /G6C               Theory of Machines
and completely balanced by masses in the respective wheels. Thus in a coupled engine, the rotating
and reciprocating masses must be treated separately and the balanced masses for the two systemsare suitably combined in the wheel.
It may be noted that the variation of pressure between the wheel and the rail ( i.e. hammer
blow) may be reduced by equal distribution of balanced mass ( B) between the driving, leading and
trailing wheels respectively.

Example 22.6. The following particulars relate to a two-cylinder locomotive with two
coupled wheels on each side :
Stroke = 650 mm
Mass of reciprocating parts per cylinder = 240 kg
Mass of revolving parts per cylinder = 200 kgMass of each coupling rod = 250 kgRadius of centre of coupling rod pin = 250 mmDistances between cylinders = 0.6 mDistance between wheels = 1.5 m
Distance between coupling rods = 1.8 m

The main cranks are at right angles and the coupling rod pins are at 180° to their respec-
tive main cranks. The balance masses are to be placed in the wheels at a mean radius of 675 mm

in order to balance whole of the revolving and 3/4th of the reciprocating masses. The balancemass for the reciprocating masses is to be divided equally between the driving wheels and thecoupled wheels. Find : 1. The magnitudes and angular positions of the masses required for the
driving and trailing wheels, and 2. The hammer blow at 120 km/h, if the wheels are 1.8 metre
diameter.
Solution. Given : L
C = LD = 650 mm or rC = rD = 325 mm = 0.325 m ; m1 = 240 kg ;

m2 = 200 kg ; m3 = 250 kg ; rA = rF = 250 mm = 0.25 m ; CD = 0.6 m ; BE = 1.5 m ; AF = 1.8 m ;
rB = rE = 675 mm = 0.675 m ; c = 3/4
The position of planes for the driving wheels B and E, cylinders C and D, and coupling
rods A and F, are shown in Fig. 22.12 ( a).
The angular position of cranks C and D and coupling pins A and F are shown in Fig.
22.12( b).
We know that mass of the reciprocating parts per cylinder to be balanced
            = 13. 240 180 kg4cm=× =

= 13. 240 180 kg4cm=× =
Since the reciprocating masses are to be divided equally between the driving wheels and
trailing wheels, therefore 90 kg is taken for driving wheels and 90 kg for trailing wheels. Now foreach driving wheel, the following masses are to be balanced :
1.Half of the mass of coupling rod i.e. 12502× 125 kg= . In other words, the masses at the
coupling rods A and F to be balanced for each driving wheel are
      AF 125 kg mm==

Chapter 22 : Balancing of Reciprocating Masses            /G6C          875
2.  Whole of the revolving mass i.e. 200 kg and the mass of the reciprocating parts i.e.
90 kg. In other words, total mass at the cylinders C and D to be balanced for each driving wheel are
      mC = mD = 200 + 90 = 290 kg
(a) Position of planes. ( b) Angular position of cranks and coupling pins.
   (c) Couple polygon : Driving wheel E.       ( d) Force polygon : Driving wheel B.
Fig. 22.12

Fig. 22.12
Balanced masses in the driving wheels
Let mB and mE be the balance masses placed in the driving wheels B and E respectively.
Taking the plane of B as reference plane, the data may be tabulated as below :
Table 22.5. (For driving wheels)Table 22.5. (For driving wheels)Table 22.5. (For driving wheels)Table 22.5. (For driving wheels)Table 22.5. (For driving wheels)
(m) kg (r) m (m.r) kg-m Plane B(l) m (m.r.l) kg-m2

A 125 0.25 31.25 – 0.15 – 4.7
B (R.P.) mB0.675 0.675 mB00
C 290 0.325 94.25 0.45 42.4
D 290 0.325 94.25 1.05 99
EmE0.675 0.675 mE1.5 1.01 mE
F 125 0.25 31.25 1.65 51.6
In order to find the balance mass mE in the driving wheel E, draw a couple polygon from the
data given in Table 22.5 (column 6), to some suitable scale as shown in Fig 22.12 ( c). The closing
side of polygon as shown dotted is proportional to 1.01 mE, Therefore by measurement, we find that

876       /G6C               Theory of Machines
 1.01 mE = 67.4 kg-m2  or  mE = 66.7 kg  Ans.
and         θ = 45°  Ans.
Now draw the force polygon from the data given in Table 22.5 (column 4), to some suit-
able scale, as shown in Fig. 22.12 ( d). The closing side of the polygon as shown dotted is propor-
tional to 0.675 mB. Therefore by measurement, we find that
           0.675 mB = 45 kg-m    or   mB = 66.7 kg Ans.
and         φ = 45°  Ans.
Balance masses in the trailing wheels

For each trailing wheel, the following masses are to be balanced :
1. Half of the mass of the coupling rod i.e. 125 kg. In other words, the masses at the cou-
pling rods A and F to be balanced for each trailing wheel are
       mA = mF = 125 kg
2. Mass of the reciprocating parts i.e. 90 kg. In other words, the mass at the cylinders C and
D to be balanced for each trailing wheel are
      mC = mD = 90 kg

mC = mD = 90 kg
Let Bm′ and Em′ be the balanced masses placed in the trailing wheels. Taking the plane of
B as the reference plane, the data may be tabulated as below :
Table 22.6. (For trailing wheels)Table 22.6. (For trailing wheels)Table 22.6. (For trailing wheels)Table 22.6. (For trailing wheels)Table 22.6. (For trailing wheels)

B (R.P.)Bm′ 0.675 0.675 Bm′ 00
C 90 0.325 29.25 0.45 13.2
D 90 0.325 29.25 1.05 30.7
EEm′ 0.675 0.675 Em′ 1.5 1.01 Em′
In order to find the balance mass Em′ in the trailing wheel E, draw a couple polygon from the
data given in Table 22.6 (column 6), to some suitable scale, as shown in Fig. 22.13 ( a). The closing
side of the polygon as shown dotted is proportional to 1.01 Em′. Therefore by measurement, we find
that

that
       1.01 m′E = 27.5 m2    or    m′E = 27.5 kg Ans.
and   α = 40° Ans.
Now draw the force polygon from the data given in Table 22.6 (column 4), to some suit-
able scale, as shown in Fig. 22.13 ( b). The closing side of the polygon as shown dotted is propor-
tional to 0.675 Bm′. Therefore by measurement, we find that
B0.675 18.35 kg-m m′=   or  Bm′ = 27.2 kg Ans.
and          β = 50°  Ans.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          877
Fig. 22.14 shows the balance masses in the four wheels and it will be seen that the balance
masses for the driving wheels are symmetrical about the axis X-X [Fig. 22.12 ( b)]. Similarly the
balance masses for the trailing wheels are symmetrical about the axis X-X.
   (a) Couple polygon : Trailing wheel E.( b) Force polygon : Trailing wheel B.
Fig. 22.13

Fig. 22.13
Driving wheel E.        Trailing wheel E.       Driving wheel B.  Trailing wheel B.
          ( a)   ( b)   ( c)        ( d)
Fig. 22.14
Hammer blow
In order to find the hammer blow, we must find the balance mass required for reciprocating
masses only. For this, the data may be tabulated as below. Let Bm′′ and Em′′ be the balanced
masses required for the reciprocating masses.

Table 22.7. (For hammer blow)Table 22.7. (For hammer blow)Table 22.7. (For hammer blow)Table 22.7. (For hammer blow)Table 22.7. (For hammer blow)
B(R.P .)Bm′′ 0.675 0675 Bm′′ 00
EEm′′ 0.675 0.675Em′′ 1.5 1.01Em′′

Now the couple polygon and the force polygon may be drawn, but due to symmetry we
shall only draw the couple polygon from the data given in Table 22.7 (column 6), to some suitable
scale as shown in Fig 22.15.
From Fig. 22.15,
22
E1.01 (30.7) (13.2) 33.4 m′′=+ =
∴                   Em′′ = 33 kg

878       /G6C               Theory of Machines
We know that linear speed of the wheel,
           v  = 120 km/h = 33.33 m/s
and diameter of the wheel, D = 1.8 m
∴ Angular speed of the wheel
            33.3337rad/s
/2 1 . 8/2v
Dω= = =
We know that hammer blow
             22. . 33(37) 0.675Bb=± ω =  = ± 30.494 N Ans.
. . . (  E Bm′′=∵ , and b = rB = rE)

22.10.22.10.22.10.22.10.22.10. Balancing of Primary Forces of Multi-cylinder In-line EnginesBalancing of Primary Forces of Multi-cylinder In-line EnginesBalancing of Primary Forces of Multi-cylinder In-line EnginesBalancing of Primary Forces of Multi-cylinder In-line EnginesBalancing of Primary Forces of Multi-cylinder In-line Engines
The multi-cylinder engines with the cylinder centre lines in the same plane and on the same

side of the centre line of the crankshaft, are known as In-line engines . The following two conditions
must be satisfied in order to give the primary balance of the reciprocating parts of a multi-cylinder
engine :
1.The algebraic sum of the primary forces must be equal to zero. In other words, the pri-
mary force polygon must *close ; and
2.The algebraic sum of the couples about any point in the plane of the primary forces must

be equal to zero. In other words, the primary couple polygon must close.
We have already
discussed, that the primary
unbalanced force due to the
reciprocating masses is equal tothe component, parallel to the lineof stroke, of the centrifugal forceproduced by the equal massplaced at the crankpin andrevolving with it. Therefore, inorder to give the  primary balance
of the reciprocating parts of amulti-cylinder engine, it isconvenient to imagine thereciprocating masses to be

transferred to their respective
crankpins and to treat theproblem as one of revolvingmasses .
Notes : 1.  For a two cylinder engine
with cranks at 180°, condition (1) may
be satisfied, but this will result in an
unbalanced couple. Thus the abovemethod of primary balancing cannot be applied in this case.
2. For a three cylinder engine with cranks at 120° and if the reciprocating masses per cylinder are

same, then condition (1) will be satisfied because the forces may be represented by the sides of an equilateral
triangle. However, by taking a reference plane through one of the cylinder centre lines, two couples with non-
parallel axes will remain and these cannot vanish vectorially. Hence the above method of balancing fails in
this case also.

* The closing side of the primary force polygon gives the maximum unbalanced primary force and theclosing side of the primary couple polygon gives the maximum unblanced primary couple.Fig. 22.15
The speedometer is an instrument which shows how fast a
car is moving. It works with a magnet that spins around as
the car moves.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          879
Fig. 22.16.  Secondary force.  3. For a four cylinder engine, similar reasoning will show that complete primary balance is pos-
sible and it follows that
‘For a multi-cylinder engine, the primary forces may be completely balanced by suitably ar-
ranging the crank angles, provided that the number of cranks are not less than four’.

22.11.22.11.22.11.22.11.22.11. Balancing of Secondary Forces of Multi-cylinder In-line EnginesBalancing of Secondary Forces of Multi-cylinder In-line EnginesBalancing of Secondary Forces of Multi-cylinder In-line EnginesBalancing of Secondary Forces of Multi-cylinder In-line EnginesBalancing of Secondary Forces of Multi-cylinder In-line Engines
When the connecting rod is not too long ( i.e. when the obliquity of the connecting rod is

considered), then the secondary disturbing force due to the reciprocating mass arises.
We have discussed in Art. 22.2, that the secondary force,
      2
 Scos 2.. Fm rnθ=ω×
This expression may be written as
      2
 S .(2 ) cos 24rFmn=ω × ×θ
As in case of primary forces, the secondary forces may be considered to be equivalent to
the component, parallel to the line of stroke, of the centrifugal force produced by an equal mass

placed at the imaginary crank of length r/4n and revolving at twice the speed of the actual crank
(i.e. 2ω) as shown in Fig. 22.16.
Thus, in multi-cylinder in-line engines, each imagi-

nary secondary crank with a mass attached to the crankpinis inclined to the line of stroke at twice the angle of theactual crank. The values of the secondary forces and couplesmay be obtained by considering the revolving mass. This isdone in the similar way as discussed for primary forces.The following two conditions must be satisfied in order togive a complete secondary balance of an engine :
1.The algebraic sum of the secondary forces must be equal to zero. In other words, the

secondary force polygon must close, and
2.The algebraic sum of the couples about any point in the plane of the secondary forces
must be equal to zero. In other words, the secondary couple polygon must close.
Note : The closing side of the secondary force polygon gives the maximum unbalanced secondary force and
the closing side of the secondary couple polygon gives the maximum unbalanced
secondary couple.
Example 22.7. A four cylinder vertical engine has cranks 150 mm long. The planes of

rotation of the first, second and fourth cranks are 400 mm, 200 mm and 200 mm respectively from
the third crank and their reciprocating masses are 50 kg, 60 kg and 50 kg respectively. Find themass of the reciprocating parts for the third cylinder and the relative angular positions of thecranks in order that the engine may be in complete primary balance.
Solution. Given r
1 = r2 = r3 = r4 = 150 mm = 0.15 m ; m1 = 50 kg ; m2 = 60 kg ;
m4 = 50 kg

m4 = 50 kg
We have discussed in Art. 22.10 that in order to give the primary balance of the reciprocat-
ing parts of a multi-cylinder engine, the problem may be treated as that of revolving masses withthe reciprocating masses transferred to their respective crank pins.
The position of planes is shown in Fig. 22.17 ( a). Assuming the plane of third cylinder as
the reference plane, the data may be tabulated as given in Table 22.8.

880       /G6C               Theory of Machines
Table 22.8Table 22.8Table 22.8Table 22.8Table 22.8
(m) kg (r) m (m.r) kg-m plane 3(l) m (m.r.l) kg-m2
1 50 0.15 7.5 – 0.4 – 3
2 60 0.15 9 – 0.2 – 1.8
3(R.P.) m30.15 0.15 m300
4 50 0.15 7.5 0.2 1.5
First of all, the angular position of cranks 2 and 4 are obtained by drawing the couple

polygon from the data given in Table 22.8 (column 6). Assume the position of crank 1 in the
horizontal direction as shown in Fig 22.17 ( b), The couple polygon, as shown in Fig. 22.17 ( c), is
1.Draw vector oa′′ in the horizontal direction ( i.e. parallel to O1) and equal to – 3 kg-m2,
2.From point o′ and a′, draw vectors ob′′ and ab′′ equal to – 1.8 kg-m2 and 1.5 kg-m2
respectively. These vectors intersect at b′.

      ( a) Position of planes.       ( b) Angular position of cranks.
      ( c) Couple polygon.      ( d) Force polygon.
Fig. 22.17
3.Now in Fig. 22.17 ( b), draw O2 parallel to vector ob′′ and O4 parallel to vector ab′′.
By measurement, we find that the angular position of crank 2 from crank 1 in the
anticlockwise direction is
2θ = 160° Ans.
and the angular position of crank 4 from crank 1 in the anticlockwise direction is
 4θ= 26° Ans.

4θ= 26° Ans.
In order to find the mass of the third cylinder ( m3) and its angular position, draw the force
polygon, to some suitable scale, as shown in Fig. 22.17 ( d), from the data given in Table 22.8
(column 4). Since the closing side of the force polygon (vector co) is proportional to 0.15 m3,
therefore by measurement,
     0.15 m3 = 9 kg-m   or   m3 = 60 kg Ans.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          881
Now draw O3 in Fig 22.17 ( b), parallel to vector co. By measurement, we find that the
angular position of crank 3 from crank 1 in the anticlockwise direction is
            3θ = 227°  Ans.
Example 22.8.  A four crank engine has the two outer cranks set at 120° to each other,
and their reciprocating masses are each 400 kg. The distance between the planes of rotation of

adjacent cranks are 450 mm, 750 mm and 600 mm. If the engine is to be in complete primarybalance, find the reciprocating mass and the relative angular position for each of the inner cranks.
If the length of each crank is 300 mm, the length of each connecting rod is 1.2 m and the
speed of rotation is 240 r.p.m., what is the maximum secondary unbalanced force ?
1 = m4 = 400 kg ; r = 300 mm = 0.3 m ; l = 1.2 m ; N = 240 r.p.m. or
2 240/ 60ω= π× = 25.14 rad/s

Reciprocating mass and the relative angular position for each of the inner cranks
Let               m2 and m3 = Reciprocating mass for the inner cranks 2 and 3 respectively, and
              2θ and 3θ = Angular positions of the cranks 2 and 3 with respect to crank 1
                     respectively.
The position of the planes of rotation of the cranks and their angular setting are shown in

Fig. 22.18 ( a) and ( b) respectively. Taking the plane of crank 2 as the reference plane, the data may
be tabulated as below :
Table 22.9Table 22.9Table 22.9Table 22.9Table 22.9
(m) kg (r) m (m.r) kg-m plane (2) (l) m  (m.r.l.) kg-m2
1 400 0.3 120 – 0.45 – 54
2(R.P.) m20.3 0.3 m200
3 m30.3 0.3 m30.75 0.225 m3
4 400 0.3 120 1.35 162

Since the engine is to be in complete primary balance, therefore the primary couple poly-
gon and the primary force polygon must close. First of all, the primary couple polygon, as shownin Fig. 22.18 ( c), is drawn to some suitable scale from the data given in Table 22.9 (column 6), in
order to find the reciprocating mass for crank 3. Now by measurement, we find that
               
2
3 0.225 196 kg-m m=    or   m3 = 871 kg  Ans.

2
and its angular position with respect to crank 1 in the anticlockwise direction,
             3θ= 326°  Ans.
Now in order to find the reciprocating mass for crank 2, draw the primary force polygon,
as shown in Fig. 22.18 ( d), to some suitable scale from the data given in Table 22.9 (column 4).
Now by measurement, we find that
        0.3 m2 = 284 kg-m    or    m2 = 947 kg Ans.

            2θ = 168° Ans.
Maximum secondary unbalanced force
The secondary crank positions obtained by rotating the primary cranks at twice the angle,

882       /G6C               Theory of Machines
is shown in Fig. 22.18 ( e). Now draw the secondary force polygon, as shown in Fig. 22.18 (  f ), to
some suitable scale, from the data given in Table 22.9 (column 4). The closing side of the polygon
shown dotted in Fig. 22.18 (  f ) represents the maximum secondary unbalanced force. By measure-
ment, we find that the maximum secondary unbalanced force is proportional to 582 kg-m.
∴  Maximum secondary unbalanced force

         = 22582(25.14)582 91 960N1.2/ 0.3 nω×= =  = 91.96 kN Ans.     . . . (∵ n = l/r)
(a) Positions of planes. ( b) Primary crank positions.
(c) Primary couple polygon. ( d) Primary force polygon.
(e) Secondary crank positions. (  f ) Secondary force polygon.
Fig. 22.18
Example 22.9. The cranks and connecting rods of a 4-cylinder in-line engine running at

1800 r.p.m. are 60 mm and 240 mm each respectively and the cylinders are spaced 150 mm apart.If the cylinders are numbered 1 to 4 in sequence from one end, the cranks appear at intervals of90° in an end view in the order 1-4-2-3. The reciprocating mass corresponding to each cylinder is1.5 kg.
Determine :  1. Unbalanced primary and secondary forces, if any, and 2. Unbalanced
primary and secondary couples with reference to central plane of the engine.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          883
Solution. Given : N = 1800 r.p.m. or 2ω= π  × 1800/60 = 188.52 rad/s ; r = 60 mm
= 0.6 m ; l = 240 mm = 0.24 m ; m = 1.5 kg
1. Unbalanced primary and secondary forces
The position of the cylinder planes and cranks is shown in Fig.22.19 ( a) and ( b) respec-
tively. With reference to central plane of the engine, the data may be tabulated as below :
Table 22.10Table 22.10Table 22.10Table 22.10Table 22.10

Plane Mass Radius Cent. force 2÷ω Distance from ref. Couple 2÷ω
(m) kg (r) m (m.r) kg-m plane 3 (l) m (m.r.l.) kg-m2
1 1.5 0.6 0.9 – 0.225 – 0.2025
2 1.5 0.6 0.9 – 0.075 – 0.0675
3 1.5 0.6 0.9 + 0.075 + 0.0675
4 1.5 0.6 0.9 + 0.225 + 0.2025
(a) Cylinder plane positions. ( b) Primary crank positions.
(c) Primary force polygon. ( d) Primary couple polygon.
(e) Secondary crank           (  f) Secondary force ( g) Secondary couple
    positions.     polygon.      polygon.

884       /G6C               Theory of Machines
The primary force polygon from the data given in Table 22.10 (column 4) is drawn as
shown in Fig. 22.19 ( c). Since the primary force polygon is a closed figure, therefore there are no
unbalanced primary forces. Ans.
The secondary crank positions, taking crank 3 as the reference crank, is shown in Fig.
22.19 ( e). From the secondary force polygon as shown in Fig. 22.19 (  f ), we see that it is a closed

figure. Therefore there are no unbalanced secondary forces. Ans.
2. Unbalanced primary and secondary couples
The primary couple polygon from the data given in Table 22.10 (column 6) is drawn as
shown in Fig. 22.19 ( d). The closing side of the polygon, shown dotted in the figure, represents
unbalanced primary couple. By measurement, we find the unbalanced primary couple is propor-
tional to 0.19 kg-m2.
∴  Unbalanced primary couple,
   U.P .C = 0.19 × 2ω = 0.19 (188.52)2 = 6752 N-m Ans.

The secondary couple polygon is shown in Fig. 22.1 ( g). The unbalanced secondary couple
is shown by dotted line. By measurement, we find that unbalanced secondary couple is propor-tional to 0.54 kg-m
2.
∴  Unbalanced secondary couple,
  22(188.52)... 0 . 5 4 0 . 5 40.24 / 0.6USCnω=× =×  = 4798 N-m Ans. . . . (∵ n = l/r )
Example 22.10. Fig. 22.20 shows the arrangement of the cranks in a four crank symmetrical

engine in which the masses of the reciprocating parts at cranks 1 and 4 are each equal to m1 and
at cranks 2 and 3 are each equal to m2.
Fig. 22.20
Show that the arrangement is balanced for primary forces and couples and for secondary
forces provided that
      12 12
21 21cos tan;,cos tanma
maθθ==θθ    and   121cos .cos2θθ = .
Solution. Given : Mass of reciprocating parts at cranks 1 and 4 = m1 ; Mass of the
reciprocating parts at cranks 2 and 3 = m2

The position of planes and primary and secondary crank positions are shown in Fig. 22.21
(a), (b) and ( c) respectively. Assuming the reference plane midway between the planes of rotation
of cranks 2 and 3, the data may be tabulated as below :

Chapter 22 : Balancing of Reciprocating Masses            /G6C          885
Table 22.11Table 22.11Table 22.11Table 22.11Table 22.11
Plane Mass (m) Radius (r) Cent. force 2÷ω Distance from Couple 2÷ω
(m.r) ref. plane (l) (m.r.l)
1 m1rm1.r – a1– m1.r.a1
2 m2rm2.r – a2– m2.r.a2
3 m2rm2.r + a2+ m2.r.a2
4 m1rm1.r + a1+ m1.r.a1
 (a) Position of planes.      ( b) Primary crank          ( c) Secondary crank Positions.
           positions.

positions.
(d) Primary force         ( e) Primary couple (  f ) Secondary force
     polygon.  polygon.        polygon.
Fig. 22.21
In order to balance the arrangement for primary forces and couples, the primary force and
couple polygons must close. Fig. 22.21 ( d) and ( e) show the primary force and couple polygons,
which are closed figures. From Fig. 22.21 ( d),
      11 2 2.c o s .c o s PQ m r m r=θ =θ      or     12
21cos
cosm
mθ=θ 
  Ans.
From Fig. 22.21 ( e),

21cos
cosm
mθ=θ 
  Ans.
      11 1 2 2 2.. s i n .. s i n FG m r a m r a=θ = θ
or         11 1 2 2 2. sin . sinma m aθ= θ
            11 2
22 1sin
sinma
maθ×=θ     or     21 2
12 1cos sin
cos sina
aθθ×=θθ... 12
21cos
cosm
m θ=θ ∵

886       /G6C               Theory of Machines
∴        12 12
2121sin cos tan
sin cos tana
aθθθ=×=θθ θ Ans.
In order to balance the arrangement for secondary forces, the secondary force polygon
must close. The position of the secondary cranks is shown in Fig. 22.21 ( c) and the secondary force
polygon is shown in Fig. 22.21 ( f ).
Now from Fig. 22.21 (  f ),
                 11 2 2. cos 2 . cos(180 2 ) RS m r m r=θ = ° − θ
or                     12 22.cos2 .cos2mmθ= − θ
∴       2
12 2

∴       2
12 2
221 1cos 2 (2cos 1)
cos 2 2cos 1m
m−θ − θ −==θ θ− . . . ( 2cos 2 2cos 1θ= θ− ∵ )
             2
22
21 1cos (1 2cos )
cos 2cos 1θ−θ=θ θ−. . . 12
21cos
cosm
m θ=θ ∵
    22
12 2 1 21 2cos .cos cos cos 2cos .cosθθ − θ = θ − θθ
      1 2 12 12 2cos .cos (cos cos ) cos cosθ θ θ+ θ = θ+ θ
    12 2cos .cos 1θθ =     or   121cos .cos2θθ =  Ans.
Example 22.11. A four cylinder engine has cranks arranged symmetrically along the shaft

as shown in Fig. 22.22. The distance between the outer cranks A and D is 5.4 metres and that
between the inner cranks B and C is 2.4 metres. The mass of the reciprocating parts belonging toeach of the outer cylinders is 2 tonnes, and that belonging to each of the inner cylinders ism tonnes.
Fig. 22.22
If the primary and secondary forces are to be balanced and also the primary couples,

determine the crank angle positions and the mass of the reciprocating parts (m) corresponding tothe inner cylinders.
Find also the maximum value of the unbalanced secondary couple, if the stroke is 1 metre,
the connecting rod length 2 metres, and the speed of the engine is 110 r.p.m.
Solution. Given : AD = 5.4 m ; BC = 2.4 m ; m
A = mD = 2 t ; L = 1 m or r = L/ 2 = 0.5 m ;
l = 2 m ; N = 110 r.p.m. or 2ω= π  × 110/60 = 11.52 rad/s

Fig. 22.23 ( a) shows the position of planes and Fig. 22.23 ( b) shows the end view of the
cranks with primary crank angles α and φ which are to be determined. Assuming the reference

Chapter 22 : Balancing of Reciprocating Masses            /G6C          887
plane mid-way between the planes of rotation of cranks A and D, the data may be tabulated as
below :
Table 22.12Table 22.12Table 22.12Table 22.12Table 22.12
(m) t (r) m (m.r) t-m  plane (l) m (m.r.l) t-m2
(1) (2) (3) (3) (4) (5)
A 2 0.5 1 – 2.7 – 2.7
Bm 0.5 0.5 m – 1.2 – 0.6 m
Cm 0.5 0.5 m + 1.2 + 0.6 m
D 2 0.5 1 + 2.7 + 2.7

   (a) Positions of planes.        ( b) Primary crank positions.
    (c) Primary force polygon.        ( d) Primary couple polygon.
    (e) Secondary crank ( f) Secondary force ( g) Secondary couple
         positions.      polygon.       polygon.
Fig. 22.23

888       /G6C               Theory of Machines
Since the primary forces and couples are to be balanced, therefore the primary force and
couple polygons, drawn from the data given in Table 22.12 column (4) and (6) respectively, as
shown in Fig. 22.23 ( c) and ( d), must close.
From Fig. 22.23 ( c),
           1cos 0.5 cos PQ m=α = φ
∴                  1cos 2coscos0.5mmααφ= =  . . . (i)
From Fig. 22.23 ( d),
           2.7sin 0.6 sin FG m=α = φ
∴                  0.6 sin sinsin

2.7 4.5mmφφα= = . . . (ii)
Now draw the secondary crank positions as shown in Fig. 22.23 ( e). Let OP be the reference
line. The secondary crank angles are given below :
              OP  to OA = 2α
              OP to OC = 2 (180° – φ) = 360° – 2φ
              OP to OB = 2 (180° + φ) = 360° + 2φ
              OP to OD = 2 (360° – α) = 720° – 2α
Since the secondary forces are to be balanced, therefore the secondary force polygon, as

shown in Fig. 22.23 (  f ), must close. Now from Fig. 22.23 (  f ),
                        RS  = 1 cos 2α= 0.5 m cos (180° – 2φ)
or                    2
21c o s 2 ( 2 c o s 1 )
0.5 cos 2 2cos 1 m−φ − θ −==
α α−. . . 2( cos2 2cos 1)θ= θ−∵
A Steam-powered ship.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          889
          2
22 2cos2cos 1 0.5 ( 1 2cos ) 0.5 1 2 mmmαα− = − φ =  −      . . . [From equation (i)]
             22
28cos 4cos0.5 1 0.5mmm m αα=− = −

 22 4cos2cos 1 0.5 mmαα+ = +    or   2 24cos 1 0.5mmm+α= +
∴              2cos (1 0.5 )24 4mmmmα= + × =+ . . . (iii)
Now from equation (ii)
     2
2 sinsin4.5mφα=
or             2 22 2 222 sin 2 cos1c o s ( 1c o s ) 120.25 20.25 20.25mm m

mφα −α = = −φ =  −  
 . . . [From equations  (i)]
     22 2
24111 14 20.25 4 20.25 20.25 20.25mm m m m m
m m −= − × = − = −    
 . . . [From equation (iii) ]
or     2
1020.25 20.25 4mm m−+ − =  or 24.0625 20.25 0 mm+− =
∴             24.0625 (4.0625) 4 20.252.9 t2m−± + ×==
We know that 2 2.9cos 0.72544mα= = =
∴      cos 0.851α=  or α = 31.6°  Ans.
Also             2cos 2 0.851cos 0.5869
2.9 mα×φ= = =      or    φ = 54.06°  Ans.
Maximum unbalanced secondary couple

The secondary couple polygon is shown in Fig. 22.23 ( g). The maximum unbalanced sec-
ondary couple is shown by a dotted line. By measurement, we find that the maximum unbalanced
secondary couple is proportional to 8 t-m2.
∴  Maximum unbalanced secondary couple,
    22(11.52).. 8 82 / 0.5USCnω=× = ×  = 265.4 kN-m Ans. . . . (∵ n = l/r )
Example 22.12. A five cylinder in-line engine running at 750 r.p.m. has successive cranks

144° apart, the distance between the cylinder centre lines being 375 mm. The piston stroke is 225mm and the ratio of the connecting rod to the crank is 4. Examine the engine for balance ofprimary and secondary forces and couples. Find the maximum values of these and the position ofthe central crank at which these maximum values occur. The reciprocating mass for each cylinderis 15 kg.

890       /G6C               Theory of Machines
Solution.  Given : N = 750 r.p.m. or 2ω= π × 750/60 = 78.55 rad/s ; L = 225 mm = 0.225 m
or r = 0.1125 m ; n = l/r = 4 ; m = 15 kg
Assuming the engine to be a vertical engine, the positions of the cylinders and the cranks
are shown in Fig. 22.24 ( a), (b) and ( c). The plane 3 may be taken as the reference plane and the
crank 3 as the reference crank. The data may be tabulated as given in the following table.

Table 22.13Table 22.13Table 22.13Table 22.13Table 22.13
(m) kg (r) m (m.r) kg-m Plane 3 (l) m (m.r.l) kg-m2
1 15 0.1125 1.6875 – 0.75 – 1.265
2 15 0.1125 1.6875 – 0.375 – 0.6328
3(R.P.) 15 0.1125 1.6875 0 0
4 15 0.1125 1.6875 + 0.375 + 0.6328
5 15 0.1125 1.6875 + 0.75 + 1.265
Now, draw the force and couple polygons for primary and secondary cranks as shown in

Fig. 22.24 ( d), (e), ( f ), and ( g). Since the primary and secondary force polygons are close, there-
fore the engine is balanced for primary and secondary forces. Ans.
    (a) Position of planes. ( b) Primary crank positions. ( c) Secondary crank positions.
(d) Primary force polygon.    ( e) Primary couple polygon.
(f) Secondary force polygon.    ( g) Secondary couple polygon.
Fig. 22.24
Maximum unbalanced primary couple

Fig. 22.24
We know that the closing side of the primary couple polygon [shown dotted in Fig. 22.24
(e)] gives the maximum unbalanced primary couple. By measurement, we find that maximum un-
balanced primary couple is proportional to 1.62 kg-m2.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          891
∴  Maximum unbalanced primary couple,
U.P .C . = 1.62 × ω2 = 1.62 (78.55)2 = 9996 N-m  Ans.
We see from Fig. 22.24 ( e) [shown by dotted line] that the maximum unbalanced primary
couple occurs when crank 3 is at 90° from the line of stroke.
We know that the closing side of the secondary couple polygon [shown dotted in Fig.

22.24 ( g)] gives the maximum unbalanced secondary couple. By measurement, we find that maximum
unbalanced secondary couple is proportional to 2.7 kg-m2.
∴ Maximum unbalanced secondary couple.
 22(78.55).. 2 . 7 2 . 74USCnω=×=×  = 4165 N-m Ans.
We see from Fig. 22.24 ( g) that if the vector representing the unbalanced secondary couple
(shown by dotted line) is rotated through 90°, it will coincide with the line of stroke. Hence the

original crank will be rotated through 45°. Therefore, the maximum unbalanced secondary coupleoccurs when crank 3 is at 45° and at successive intervals of 90° ( i.e. 135°, 225° and 315°) from the
line of stroke.
Example 22.13. The firing order in a 6 cylinder vertical four stroke in-line engine is
1-4-2-6-3-5. The piston stroke is 100 mm and the length of each connecting rod is 200 mm. The
pitch distances between the cylinder centre lines are 100 mm, 100 mm, 150 mm, 100 mm, and 100

mm respectively. The reciprocating mass per cylinder is 1 kg and the engine runs at 3000 r.p.m.
Determine the out-of-balance primary and secondary forces and couples on this engine,
taking a plane midway between the cylinder 3 and 4 as the reference plane.
Solution. Given : L = 100 mm or r = L/ 2 = 50 mm = 0.05 m ; l = 200 mm ; m = 1 kg ;
N = 3000 r.p.m.
The position of the cylinders and the cranks are shown in Fig. 22.25 ( a), (b) and ( c). With

the reference plane midway between the cylinders 3 and 4, the data may be tabulated as given inthe following table :
Table 22.14Table 22.14Table 22.14Table 22.14Table 22.14
(m) kg (r) m (m.r) kg-m plane 3 (l)m (m.r.l) kg-m2
1 1 0.05 0.05 – 0.275 – 0.01375
2 1 0.05 0.05 – 0.175 – 0.008753 1 0.05 0.05 – 0.075 – 0.003754 1 0.05 0.05 + 0.075 + 0.003755 1 0.05 0.05 + 0.175 + 0.00875
6 1 0.05 0.05 + 0.275 + 0.01375

Now, draw the force and couple polygons for the primary and secondary cranks as shown
in Fig. 22.25 ( d), (e), ( f ) and ( g).

892       /G6C               Theory of Machines
(a) Positions of planes. ( b) Primary crank positions. ( c) Secondary crank positions.
 (d) Primary force polygon. ( e) Secondary force polygon.
  ( f ) Primary couple polygon. ( g) Secondary couple polygon.
Fig. 22.25
From Fig. 22.25 ( d) and ( e), we see that the primary and secondary force polygons are
closed figures, therefore there are no out-of-balance primary and secondary forces. Thus the engine

is balanced for primary and secondary forces. Also, the primary and secondary couple polygons, asshown in Fig. 22.25 (  f ) and ( g) are closed figures, therefore there are no out-of-balance primary
and secondary couples. Thus the engine is balanced for primary and secondary couples.  Ans.
Example 22.14.  In an in-line six cylinder engine working on two stroke cycle, the cylinder

centre lines are spaced at 600 mm. In the end view, the cranks are 60° apart and in the order1-4-5-2-3-6. The stroke of each piston is 400 mm and the connecting rod length is 1 metre. Themass of the reciprocating parts is 200 kg per cylinder and that of rotating parts 100 kg per crank.The engine rotates at 300 r.p.m. Examine the engine for the balance of primary and secondaryforces and couples. Find the maximum unbalanced forces and couples .

Solution. Given : L = 400 mm or r = L/2 = 200 mm = 0.2 m ; l = 1 m ; m
1 = 200 kg ;
m2 = 100 kg ; N = 300 r.p.m. or 2ω= π  × 300/60 = 31.42 rad/s
Assuming the engine to be a vertical engine, the position of planes of cylinders and the
angular position of primary and secondary cranks (assuming the crank 1 coinciding with the line ofstroke i.e. in the vertical direction ) are shown in Fig. 22.26 ( a), (b) and ( c) respectively. It may be
noted that the mass of rotating parts ( m

2) at each crank pin is included with the mass of reciprocating
parts ( m1) for primary forces and couples only. Taking the reference plane between the cylinders 3
and 4, the data may be tabulated as below:

Chapter 22 : Balancing of Reciprocating Masses            /G6C          893
Table 22.15. (For primary forces and couples only)Table 22.15. (For primary forces and couples only)Table 22.15. (For primary forces and couples only)Table 22.15. (For primary forces and couples only)Table 22.15. (For primary forces and couples only)
Plane Mass (m) kg Radius Cent. force 2÷ω Distance from Couple 2÷ω
m = m1 + m2(r) m (m.r)kg-m ref. plane (1) m (m.r .l) kg-m2
1 300 0.2 60 – 1.5 – 90

2 300 0.2 60 – 0.9 – 543 300 0.2 60 – 0.3 – 184 300 0.2 60 + 0.3 + 185 300 0.2 60 + 0.9 + 54
6 300 0.2 60 + 1.5 + 90
 (a) Positions of planes of cylinders. ( b) Primary crank positions. ( c) Secondary crank positions.
 (d) Primary force polygon.  ( e) Primary couple polygon.
( f ) Secondary force polygon.   ( g) Secondary couple polygon.     ( h) Secondary couple polygon.
Fig. 22.26

894       /G6C               Theory of Machines
Now draw the force polygon and couple polygon for primary cranks from the data given in
Table 22.15 (column 4 and 6) respectively, as shown in Fig. 22.26 ( d) and ( e). Since the force and
couple polygons are closed figures, therefore the engine is balanced for primary force and couple
(i.e. there is no unbalanced primary force and couple ).
The data for the secondary forces and couples, taking m = m1 = 200 kg, may be tabulated
as below :

as below :
Table 22.16. (For secondary forces and couples)Table 22.16. (For secondary forces and couples)Table 22.16. (For secondary forces and couples)Table 22.16. (For secondary forces and couples)Table 22.16. (For secondary forces and couples)
m = m1(r) m (m.r) kg-m ref. plane (l) m (m.r.l) kg-m2
1 200 0.2 40 – 1.5 – 60
2 200 0.2 40 – 0.9 – 363 200 0.2 40 – 0.3 – 124 200 0.2 40 + 0.3 + 125 200 0.2 40 + 0.9 + 36

6 200 0.2 40 + 1.5 + 60
First of all, draw the secondary force polygon for secondary cranks [the angular position of
which is shown in Fig. 22.26 ( c)] from the data given in Table 22.16 (column 4) as shown in Fig.
22.26 (  f ). Since the secondary force polygon is a closed figure, therefore the engine is balanced
for secondary forces ( i.e. there is no unbalanced secondary forces.) Now draw the secondary couple

polygon for the secondary cranks from the data given in Table 22.16 (column 6) as shown in Fig.
22.26 ( g). The closing side of the polygon as shown by dotted line represents the maximum unbal-
anced secondary couple. By measurement, we find that maximum unbalanced couple is propor-tional to 168 kg-m
2.
∴  Maximum unbalanced secondary couple
           22(31.42)168 168 33 170 N-m1/0.2 nω=×=× = = 33.17 kN-m Ans.
 . . .  ( ∵ n = l/r)

. . .  ( ∵ n = l/r)
Note : The secondary couple polygon may also be drawn as shown in Fig. 22.26 ( h).
22.12.22.12.22.12.22.12.22.12. Balancing of Radial Engines (Direct and Reverse Cranks Method )Balancing of Radial Engines (Direct and Reverse Cranks Method )Balancing of Radial Engines (Direct and Reverse Cranks Method )Balancing of Radial Engines (Direct and Reverse Cranks Method )Balancing of Radial Engines (Direct and Reverse Cranks Method )

The method of direct and reverse cranks is used in balancing of radial or V-engines, in
which the connecting rods are connected to a common crank. Since the plane of rotation of thevarious cranks (in radial or V-engines) is same, therefore there is no unbalanced primary or second-
ary couple.
Fig. 22.27. Reciprocating engine mechanism.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          895
Consider a reciprocating engine mechanism as shown in Fig. 22.27. Let the crank OC
(known as the direct crank) rotates uniformly at ω radians per second in a clockwise direction. Let
at any instant the crank makes an angle θ with the line of stroke OP. The indirect or reverse crank
OC′ is the image of the direct crank OC, when seen through the mirror placed at the line of stroke.

A little consideration will show that when the direct crank revolves in a clockwise direction, the
reverse crank will revolve in the anticlockwise direction. We shall now discuss the primary andsecondary forces due to the mass ( m) of the reciprocating parts at P.
Considering the primary forces
We have already discussed that primary force is 
2.. c o smrωθ . This force is equal to the
component of the centrifugal force along the line of stroke, produced by a mass ( m) placed at the

crank pin C. Now let us suppose that the mass ( m) of the reciprocating parts is divided into two
parts, each equal to m/2 .
Fig. 22.28. Primary forces on reciprocating engine mechanism.
It is assumed that m/ 2 is fixed at the direct crank  (termed as primary direct crank ) pin C
and m/ 2 at the reverse crank  (termed as primary reverse crank ) pin C′, as shown in Fig. 22.28.
We know that the centrifugal force acting on the primary direct and reverse crank
            2.2mr =× ω

2.2mr =× ω
∴  Component of the centrifugal force acting on the primary direct crank
           2.c o s2mr =× ω θ  . . . (in the direction from O to P)
and, the component of the centrifugal force acting on the primary reverse crank
            2.c o s2mr =× ω θ . . . (in the direction from O to P)
∴ Total component of the centrifugal force along the line of stroke
           222. c o s . . c o s2mrm r =× × ω θ = ω θ  = Primary force, FP

Hence, for primary effects the mass m of the reciprocating parts at P may be replaced
by two masses at C and ′C each of magnitude m/2.
Note : The component of the centrifugal forces of the direct and reverse cranks, in a direction perpendicular
to the line of stroke, are each equal to 2.s i n ,2mr×ω θ , but opposite in direction. Hence these components are
balanced.

896       /G6C               Theory of Machines
Considering secondary forces
We know that the secondary force
           22 cos 2(2 ) cos2 . .4rmm rnnθ=ω × θ = ω ×
In the similar way as discussed above, it will be seen that for the secondary effects, the
mass ( m) of the reciprocating parts may be replaced by two masses (each m/2) placed at D and D′
such that OD = OD′ = r/4n. The crank OD is the secondary direct crank  and rotates at 2ω rad/s in

the clockwise direction, while the crank OD′is the secondary reverse crank  and rotates at 2ω
rad/s in the anticlockwise direction as shown in Fig. 22.29.
Fig. 22.29. Secondary force on reciprocating engine mechanism.
Example 22.15. The three cylinders of an air compressor have their axes 120° to one
another, and their connecting rods are coupled to a single crank. The stroke is 100 mm and the

length of each connecting rod is 150 mm. The mass of the reciprocating parts per cylinder is 1.5kg. Find the maximum primary and secondary forces acting on the frame of the compressor whenrunning at 3000 r.p.m. Describe clearly a method by which such forces may be balanced.
A diesel train engine.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          897
Solution.  Given : L = 100 mm or r = L/ 2 = 50 mm = 0.05 m ; l = 150 mm = 0.15 m ;
m = 1.5 kg ; N = 3000 r.p.m. or 2ω= π  × 3000/60 = 314.2 rad/s
The position of three cylinders is shown in
Fig. 22.30. Let the common crank be along the inner
dead centre of cylinder 1. Since common crank rotates
clockwise, therefore θ is positive when measured
clockwise.
Maximum primary force acting on the frame of the
compressor

The primary direct and reverse crank positions
as shown in Fig. 22.31 ( a) and ( b), are obtained as
1.Since θ = 0° for cylinder 1, therefore both the
primary direct and reverse cranks will coincide
with the common crank.
2.Since θ =  ±120° for cylinder 2, therefore the
primary direct crank is 120° clockwise and theprimary reverse crank is 120° anti-clockwisefrom the line of stroke of cylinder 2.
3.Since

3.Since 
θ = ± 240° for cylinder 3, therefore the primary direct crank is 240° clockwise and
the primary reverse crank is 240° anti-clockwise from the line of stroke of cylinder 3.
From Fig. 22.31 ( b), we see that the primary reverse cranks form a balanced system. There-
fore there is no unbalanced primary force due to the reverse cranks. From Fig. 22.31 ( a), we see
that the resultant primary force is equivalent to the centrifugal force of a mass 3 m/2 attached to the
end of the crank.

∴ Maximum primary force  = 22 33 1 . 5. (314.2) 0.05 11 106 N22mr××ω = =  = 11.106 kN  Ans.
(a) Direct primary cranks. ( b) Reverse primary cranks.
Fig. 22.31
The maximum primary force may be balanced by a mass attached diametrically opposite to
the crank pin and rotating with the crank, of magnitude B1 at radius b1 such that
      113 3 1.5.0 . 0 522mBb r×=× = ×  = 0.1125 N-m  Ans.Fig. 22.30

898       /G6C               Theory of Machines
Maximum secondary force acting on the frame of the compressor
The secondary direct and reverse crank positions as shown in Fig. 22.32 ( a) and ( b), are
obtained as discussed below :
1.Since θ= 0° and 2 θ = 0° for cylinder 1, therefore both the secondary direct and reverse
cranks will coincide with the common crank.
2.Since θ =  ±120° and 2 θ =  ± 240° for cylinder 2, therefore the secondary direct crank

is 240° clockwise and the secondary reverse crank is 240° anticlockwise from the line of
stroke of cylinder 2.
3.Since θ = ± 240° and 2 θ =  ± 480°, therefore the secondary direct crank is 480° or 120°
clockwise and the secondary reverse crank is 480° or 120° anti-clockwise from the line ofstroke of cylinder 3.
(a) Direct secondary cranks. ( b) Reverse secondary cranks.
Fig. 22.32
From Fig. 22.32 ( a), we see that the secondary direct cranks form a balanced system.

Therefore there is no unbalanced secondary force due to the direct cranks. From Fig. 22.32 ( b),
we see that the resultant secondary force is equivalent to the centrifugal force of a mass 3 m/2
attached at a crank radius of r/4n and rotating at a speed of 2ω rad/s in the opposite direction to
the crank.
Submarines are powered by diesel or nuclear powered engines which have
reciprocating and rotating parts.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          899
∴  Maximum secondary force
           22 2 3 1.5 0.05(2 ) (2 314.2) N2 4 2 4 0.15/ 0.05mr
n×   =ω = ×  ×  
...( /)nl r=∵
= 3702 N  Ans.
This maximum secondary force may be balanced by a mass B2 at radius b2, attached dia-
metrically opposite to the crankpin, and rotating anti-clockwise at twice the crank speed, such that
   2233 1 . 5 0 . 0 5.
24 2 4 0 . 1 5 / 0 . 0 5mrBb
n×=×= × =
× 0.009 375 N-m Ans.

n×=×= × =
Notes : 1.  Proceeding in the same way as discussed in the above example, we may prove that in a radial
engine with an odd number of cylinders, the primary forces may be balanced by attaching single mass of
magnitude 1
2K.m (K being the number of cylinders), at crank radius diametrically opposite to the crank pin.
2. For a radial engine containing four or more cylinders, the secondary direct and reverse cranks

form a balanced system, i.e. the secondary forces are in complete balance.
22.13.22.13.22.13.22.13.22.13. Balancing of V-enginesBalancing of V-enginesBalancing of V-enginesBalancing of V-enginesBalancing of V-engines
Consider a symmetrical two cylinder V-engine as shown in Fig. 22.33, The common crank
OC is driven by two connecting rods PC and QC. The lines of stroke OP and OQ are inclined to
the vertical OY, at an angle α as shown in Fig 22.33.

Let          m = Mass of reciprocating parts per cylinder,
           l = Length of connecting rod,
          r = Radius of crank,
        n = Ratio of length of connecting rod to crank radius =  l / r
         θ = Inclination of crank to the vertical at any instant,
        ω = Angular velocity of crank.
Fig.22.33. Balancing of V-engines.
We know that inertia force due to reciprocating parts of cylinder 1, along the line of stroke
           2 cos 2( )..c o s ( )mrnα−θ =ω α − θ +

and the inertia force due to reciprocating parts of cylinder 2, along the line of stroke
            2 cos 2( )..c o s ( )mrnα+θ =ω α − θ +

900       /G6C               Theory of Machines
The balancing of V-engines is only considered for primary and secondary forces * as
Considering primary forces
We know that primary force acting along the line of stroke of cylinder 1,
      2
P1 .. c o s ( ) Fm r=ω α − θ
∴  Component of FP1 along the vertical line OY,
            = 2
P1cos . .cos( ) cosFm rα= ω α−θ α  . . . (i)
and component of FP1 along the horizontal line OX
= 2
P1sin . . cos( )sinFm rα= ω α−θ α . . . (ii)

= 2
Similarly, primary force acting along the line of stroke of cylinder 2,
    2
P2 .. c o s ( ) Fm r=ω α + θ
∴  Component of FP2 along the vertical line OY
= 2
P2cos . . cos( ) cosFm rα= ω α+θ α  . . . (iii)
and component of FP2 along the horizontal line OX′
                       =2
P2sin . . cos( )sinFm rα= ω α+θ α  . . . (iv)
Total component of primary force along the vertical line OY

PVF=(i) + (iii) 2. . cos [cos( ) cos( )]mr=ω α α − θ + α + θ
            = 2. . cos 2 cos cosmrωα × α θ
... [ cos( ) cos( ) 2cos cos ]α−θ + α+θ = α θ∵
= 222. . c o s . c o smrωα θ
and total component of primary force along the horizontal line OX
     PHF=(ii) – (iv) 2. . sin [cos( ) cos( )]mr=ω α α − θ − α + θ
= 2. . sin 2sin sinmrωα × α θ
...  [c o s ( ) c o s ( ) 2 s i n s i n ]α−θ − α+θ = α θ∵
= 222 . . sin .sinmrωα θ
∴  Resultant primary force,
       22
PP VP H () () FF F=+

22
= 22 2 2 22 . . (cos .cos ) (sin .sin )mrωα θ + α θ  . . . (v)
Notes : The following results, derived from equation (v), depending upon the value of α may be noted :
1. When 26 0α= °    or   30 ,α= °
     22 2 2 2P2 . . ( c o s 3 0c o s ) ( s i n 3 0s i n ) Fm r=ω °θ + ° θ
* Since the plane of rotation of the crank is same, therefore there are no unbalanced primary and secondary
couples.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          901
            = 22
2 312. . c o s s i n44mr ωθ + θ   = 22 2.9 c o s s i n2mr×ω θ+ θ      ... (vi)
2. When 29 0α= °    or  45α= °
      22 2 2 2
P2 . . ( c o s 4 5c o s ) ( s i n 4 5s i n ) Fm r=ω °θ + ° θ
            22
22 112. . c o s) s i n . .22mr mr  =ω θ+ θ = ω       . . . (vii)
3. When 21 2 0α= °    or   60α= ° ,
       22 2 2 2
P2 . . ( c o s 6 0c o s ) ( s i n 6 0s i n ) Fm r=ω °θ + ° θ
= 22

= 22
2 132. . c o s s i n44mr ωθ + θ   = 22 2.c o s 9 s i n2mr×ω θ+ θ ... (viii)
We know that secondary force acting along the line of stroke of cylinder 1,
     2
S1cos 2( ).. Fm r
nα−θ=ω×
∴  Component of FS1 along the vertical line OY
         2
S1cos 2( )cos . . cosFm r
nα−θ=α = ω × × α . . . (ix)
and component of FS1 along the horizontal line OX
          2
S1cos 2( )sin . . sinFm rnα−θ=α = ω × × α  . . . (x)

Similarly, secondary force acting along the line of stroke of cylinder 2,
      2
S2cos 2( ). Fm rnα+θ=ω×
∴  Component of FS2 along the vertical line OY
          2
S2cos 2( )cos . . cosFm r
nα+θ=α = ω × × α  . . . (xi)
and component of FS2 along the horizontal line OX′
           2
S2cos 2( )sin . . sinFm r
nα+θ=α = ω × × α  . . . (xii)
Total component of secondary force along the vertical line OY,
    SVF=(ix) + (xi) 2. cos [cos 2( ) cos 2( )]mrn= ×ω α α−θ + α+θ
=2. cos 2 cos 2 cos 2mr

n×ω α× α θ  = 2 2. cos .cos 2 cos 2mr
n×ω α α θ
and total component of secondary force along the horizontal line OX,
      FSH = (x) – (xii) 2. sin [cos 2( ) cos 2( )]mrn= ×ω α α−θ − α+θ
           2.s i n 2 s i n 2. s i n 2mr
n=× ω α × α θ
       2 2. sin .sin 2 .sin 2mr
n=× ω α α θ

902       /G6C               Theory of Machines
∴ Resultant secondary force,
      22
SS VS H () () FF F=+
          22 2 2. (cos .cos 2 .cos 2 ) (sin .sin 2 .sin 2 )mr
n= × ω ααθ + ααθ
 . . .(xiii)
Notes :  The following results, derived from equation (xiii),  depending upon the value of α, may be noted.
1. When 26 0α= °  or 30α= ° ,
       22 2
S2. (cos30 cos60 cos2 ) (sin30 sin 60 sin 2 )mFrn=× ω ° ° θ + ° °θ
= 22
2 23 1 1 3. cos2 sin 222 22mr
n ×ω × θ + × θ   

= 2 3.2mrn×× ω  . . . (xiv)
2. When 29 0α= °  or 45α= ° ,
       22 2
S2. (cos45 cos90 cos2 ) (sin 45 sin 90 sin 2 )mFrn=× ω ° °θ + ° °θ
            = 2
22 21 2.0 1 s i n 2 . s i n 2
2mmrr
nn×ω + × × θ = ×ω θ . . . (xv)
Note : This picture is given as additional information and is not a direct example of the current chapter.Automated Guided Vehicles,  AGVs, operate in many factories. They ferry goods and materials

along carefully marked routes. Many AGVs are guided by signals from electrical loops buried
under factory floors.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          903
        3.  When 2 120α= °  or 60α= °
        22 2
S2. (cos60 cos120 cos2 ) (sin 60 sin120 sin 2 )mFrn=× ω ° °θ + ° °θ
           2 22 21 1 3 3.c o s 2 s i n 2
22 22mr
n =× ω × − × θ + ××θ   
22 2 2.c o s 2 9 s i n 2mrn=× ω θ + θ            . . . (xvi)
Example 22.16.  A vee-twin engine has the cylinder axes at right angles and the connect-

ing rods operate a common crank. The reciprocating mass per cylinder is 11.5 kg and the crank
radius is 75 mm. The length of the connecting rod is 0.3 m. Show that the engine may be balancedfor primary forces by means of a revolving balance mass.
If the engine speed is 500 r.p.m. What is the value of maximum resultant secondary force ?
29 0α= °  or 45α= ° ; m = 11.5 kg ; r = 75 mm = 0.075 m ; l = 0.3 m ;
N = 500 r.p.m. or 25 0 0 / 6 0ω= π×  = 52.37 rad/s

We know that resultant primary force,
     22 2 22
 P2. . ( c o s c o s) ( s i n s i n) Fm r=ω αθ + αθ
          22 2 2 22. . ( c o s4 5 c o s) ( s i n4 5 s i n)mr=ω °θ + °θ
           22
22 cos sin2. . . .22mr mrθθ =ω + = ω  
Since the resultant primary force 2..mrω is the centrifugal force of a mass m at the crank
radius r when rotating at ω rad / s, therefore, the engine may be balanced by a rotating balance
mass.
Maximum resultant secondary force

mass.
We know that resultant secondary force,
      2
 S 2. s i n 2mFrn=× × ω θ  . . . ( When 2 α = 90°)
This is maximum, when sin 2 θ is maximum i.e. when sin 2 θ = ± 1 or θ = 45° or 135°.
∴ Maximum resultant secondary force,
  2
S 2.maxmFrn=× × ω  . . . (Substituting θ = 45° )
           2 11.52 (52.37) 0.075
0.3/ 0.075=× =  836 N Ans. . . . (∵ n = l / r)
Example 22.17.  The reciprocating mass per cylinder in a 60° V-twin engine is 1.5 kg. The

stroke and connecting rod length are 100 mm and 250 mm respectively. If the engine runs at 2500
r.p.m., determine the maximum and minimum values of the primary and secondary forces. Alsofind out the crank position corresponding these values.

904       /G6C               Theory of Machines
Solution.  Given 2α = 60°   or   α = 30°,   m = 1.5 kg ;  Stroke = 100 mm or r = 100/2
= 50 mm = 0.05 m ;  l = 250 mm = 0.25 m ;  N = 250 r.p.m.   or  ω = 2 π × 2500 / 60 = 261.8 rad/s
Maximum and minimum values of primary forces
We know that the resultant primary force,
        FP = 22 2 2 22. . ( c o s c o s) ( s i n s i n)mrωα ⋅ θ + α ⋅ θ
             = 22 2 2 22. . ( c o s3 0 c o s) ( c o s3 0 s i n)mrω° θ + ° θ
 = 22

= 22
2 312c o s s i n44mr ωθ + θ  
 = 22 29c os s in2mr×ω θ+ θ ...(i)
The primary force is maximum, when θ = 0°. Therefore substituting θ = 0° in equation (i),
we have maximum primary force,
  22
P( )1.53 (261.8) 0.05 3 7710.7 N22maxmFr=× ω × = × = Ans.
The primary force is minimum, when θ = 90°. Therefore substituting θ = 90° in equation
(i), we have minimum primary force,
            22
P( )1.5(261.8) 0.05 2570.2 N22minmFr=× ω= =  Ans.

Maximum and minimum values of secondary forces
We know that resultant secondary force.
        22 2
S2(cos cos 2 cos 2 ) (sin sin 2 sin 2 )mF
n=× ω α α θ +ααθ
 22 2 2(cos30 cos60 cos 2 ) (sin 30 sin 60 sin 2 )mr
n=× ω ° ° θ + ° °θ
 22
2 23 11 3cos 2 sin 222 22mrn =× ω × θ + × θ    
 2 3
2=× × ωmrn
 2 31 . 5(261.8) 0.052 0.25/ 0.05=× ... (∵ n = l/r)
 890.3 N=  Ans.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          905
1. A single cylinder horizontal engine runs at 120 r.p.m. The length of stroke is 400 mm. The mass of

the revolving parts assumed concentrated at the crank pin is 100 kg and mass of the reciprocatingparts is 150 kg. Determine the magnitude of the balancing mass required to be placed opposite tothe crank at a radius of 150mm which is equivalent to all the revolving and 2/3rd of the reciprocat-ing masses. If the crank turns 30° from the inner dead centre, find the magnitude of the unbalancedforce due to the balancing mass. [Ans. 212.4 kg]

2. A single cylinder engine runs at 250 r.p.m. and has a stroke of 180 mm. The reciprocating parts hasa mass of 120 kg and the revolving parts are equivalent to a mass of 70 kg at a radius of 90 mm. Amass is placed opposite to the crank at a radius of 150 mm to balance the whole of the revolvingmass and two-thirds of the reciprocating mass. Determine the magnitude of the balancing mass andthe resultant residual unbalance force when the crank has turned 30° from the inner dead centre,neglect the

turned 30° from the inner dead centre,neglect the obliquity of the connecting rod. [Ans. 90 kg ; 3.264 kN]

3. A two cylinder uncoupled locomotive has inside cylinders 0.6 m apart. The radius of each crank is300 mm and are at right angles. The revolving mass per cylinder is 250 kg and the reciprocatingmass per cylinder is 300 kg. The whole of the revolving and two-third of the reciprocating massesare to be balanced and the balanced masses are placed, in the planes of rotation of the drivingwheels, at a radius of 0.8 m. The driving wheels are 2 m in diameter and 1.5 m apart. If the speedof the engine

and 1.5 m apart. If the speedof the engine is 80 km. p.h. ; find hammer blow, maximum variation in tractive effort and maximumswaying couple. [Ans. 18.30 kN, 16.92 kN, 16.2 kN-m]

4. A two cylinder uncoupled locomotive with cranks at 90° has a crank radius of 325 mm. The distancebetween the centres of driving wheels is 1.5 m. The pitch of cylinders is 0.6 m. The diameter oftreads of driving wheels is 1.8 m. The radius of centres of gravity of balance masses is 0.65 m. Thepressure due to dead load on each wheel is 40 kN. The masses of reciprocating and rotating partsper cylinder are 330 kg and 300 kg respectively. The speed of the locomotive is 60 km. p.h. find :

1. The balancing masses both in magnitude and position required to be placed is the planes of
driving wheels to balance whole of the revolving and two-third of the reciprocating masses ; 2. Theswaying couple ; 3.The variation is tractive force ; 4. The maximum and minimum pressure onrails ; and 5. The maximum speed at which it is possible to run the locomotive, in order that thewheels are not lifted from the rails.
[Ans. 200 kg ; 13 kN-m ; 17.34 kN ; 58.86 kN, 21.14 kN ; 87.54 km/h]

5. Two locomotives are built with similar sets of reciprocating parts. One is an inside cylinder enginewith two cylinders with centre lines at 0.6 m apart. The other is an outside cylinder with centrelines at 1.98 m apart. The distance between the driving wheel centres is 1.5 m in both the cases.The inside cylinder locomotive runs at 0.8 times the speed of the outside cylinder locomotive andthe hammer blow of the inside cylinder locomotive is 1.2 times the hammer blow of the outsidecylinder

If the diameter of the driving wheel of the outside cylinder locomotive is 1.98 m, calculate the
diameter of the driving wheel of the inside cylinder locomotive. Compare also the variation in theswaying couples of the two engines. Assume that the same fraction of the reciprocating masses arebalanced in both the cases. [Ans. 1.184 m, 1.185]

6. An air compressor has four vertical cylinders 1,2,3 and 4 in line and the driving cranks at 90°intervals reach their upper most positions in this order. The cranks are of 150 mm radius, theconnecting rods 500 mm long and the cylinder centre line 400 mm apart. The mass of the recipro-cating parts for each cylinder is 22.5 kg and the speed of rotation is 400 r.p.m. Show that there areno out-of-balance primary or secondary forces and determine the corresponding couples, indicatingthe positions

corresponding couples, indicatingthe positions of No. 1 crank for maximum values. The central plane of the machine may be taken asreference plane. [Ans. Primary couple = 6.7 kN-m at 45° and 225° ;

Secondary couple = 1.4 kN-m at 0°, 90°, 180°, 270°]
7. A four cylinder engine has the two outer cranks at 120° to each other and their reciprocating massesare each 400 kg. The distance between the planes of rotation of adjacent cranks are 400 mm, 700mm, 700 mm and 500 mm. Find the reciprocating mass and the relative angular position for each ofthe inner cranks, if the engine is to be in complete primary balance. Also find the maximum

906       /G6C               Theory of Machines
unbalanced secondary force, if the length of each crank is 350 mm, the length of each connecting
rod 1.7 m and the engine speed 500 r.p.m.
[Ans. 800 kg at 163° counter clockwise from crank 1, 830 kg at
312° counter clockwise from crank 1 ; 397.3 kN ]

8. The reciprocating masses of the first three cylinders of a four cylinder engine are 4.1, 6.2 and 7.4tonnes respectively. The centre lines of the three cylinders are 5.2 m, 3.2 m and 1.2 m from thefourth cylinder. If the cranks for all the cylinders are equal, determine the reciprocating mass of thefourth cylinder and the angular position of the cranks such that the system is completely balancedfor the primary force and couple.

If the cranks are 0.8 m long, the connecting rods 3.8 m, and the speed of the engine 75 r.p.m. ; find
the maximum unbalanced secondary force and the crank angle at which it occurs.
[Ans. 6.19 t ; 7.5 kN, 33° clockwise from I.D.C.]

9. In a four cylinder petrol engine equally spaced, the cranks, numbered from the front end are 1,2,3,and 4. The cranks 1 and 4 are in phase and 180° ahead of cranks 2 and 3. The reciprocating massof each cylinder is 1 kg. The cranks are 50 mm radius and the connecting rod 200 mm long.
What are the resultant unbalanced forces and couples, primary and secondary, when cranks 1 and 4

are on top dead centre position ? The engine is rotating at 1500 r.p.m. in a clockwise directionwhen viewed from the front. Take the reference plane midway between cylinder 2 and 3.
10. A four cylinder inline marine oil engine has cranks at angular displacement of 90°. The outer
cranks are 3 m apart and inner cranks are 1.2 m apart. The inner cranks are placed symmetrically

between the outer cranks. The length of each crank is 450 mm. If the engine runs at 90 r.p.m. andthe mass of reciprocating parts for each cylinder is 900 kg, find the firing order of the cylinders for
the best primary balancing force of reciprocating masses. Determine the maximum unbalanced pri-
mary couple for the best arrangement. [Ans. 1-4-2-3 ; 45.7 kN-m]
11. In a four crank symmetrical engine, the reciprocating masses of the two outside cylinders A and D

are each 600 kg and those of the two inside cylinders B and C are each 900 kg. The distance
between the cylinder axes of A and D is 5.4 metres. Taking the reference line to bisect the angle
between the cranks A and D, and the reference plane to bisect the distance between the cylinder
axes of A and D, find the angles between the cranks and the distance between the cylinder axes of
B and C for complete balance except for secondary couples.

Determine the maximum value of the unbalanced secondary couple if the length of the crank is 425mm, length of connecting rod 1.8 m and speed is 150 r.p.m.
[Ans. A = 210°, B = 54.7°, C = 305.3°, D =150°; 2.2 m ; 67 N-m ]
12. In a four cylinder inline engine, the cylinders are placed symmetrically along the longitudinal axis,with a centre distance of 2.4 m between the outside cylinders and 0.6 m between the inside cylinders.

The cranks between the two inside cylinders are at 90° to each other and the mass of reciprocating
parts of each of these is 225 kg. All the four cranks are of 0.3 m radius. If the system is to becompletely balanced for the primary effects, determine 1. The mass of the reciprocating parts of
each of the outside cranks, and 2. The angular position of the outside cranks with reference to the

nearest inside cranks, measured in clockwise direction and draw an end view of the four primarycranks marking these angles therein.
With the above arrangement, evaluate the secondary unbalanced effects completely, with reference
to a plane through the centre line of cylinder no. 1 and show by means of an end view the angularposition of these with reference to secondary crank no. 1. The engine is running at 180 r.p.m. and
the length of each connecting rod is 1.2 m.

[ Ans. 164 kg each ; 128° and 148° ; 814 kN and 12.7 kN-m ]
13. A six-cylinder, single acting, two stroke Diesel engine is arranged with cranks at 60° for the firing
sequence 1-4-5-2-3-6. The cylinders, numbered 1 to 6 in succession are pitched 1.5 m apart, except
cylinders 3 and 4 which are 1.8 m apart. The reciprocating and revolving masses per line are 2.2

tonnes and 1.6 tonnes respectively. The crank length is 375 mm, the connecting rod length is 1.6 m,and the speed is 120 r.p.m.
Determine the maximum and minimum values of the primary couple due to the reciprocating and
revolving parts. Also find the maximum secondary couple and angular position relative to crankNo. 1. Take the plane between the cylinders 3 and 4 as the reference plane.

Chapter 22 : Balancing of Reciprocating Masses            /G6C          907
14. A three cylinder radial engine driven by a common crank has the cylinders spaced at 120°. The
stroke is 125 mm, length of the connecting rod 225 mm and the mass of the reciprocating parts per
cylinder 2 kg. Calculate the primary and secondary forces at crank shaft speed of 1200 r.p.m.
[Ans. 3000 N ; 830 N]
15. The pistons of a 60° twin V-engine has strokes of 120 mm. The connecting rods driving a common

crank has a length of 200 mm. The mass of the reciprocating parts per cylinder is 1 kg and the
speed of the crank shaft is 2500 r.p.m. Determine the magnitude of the primary and secondaryforces. [Ans. 6.3 kN ; 1.1 kN]
16. A twin cylinder V-engine has the cylinders set at an angle of 45°, with both pistons connected to

the single crank. The crank radius is 62.5 mm and the connecting rods are 275 mm long. Thereciprocating mass per line is 1.5 kg and the total rotating mass is equivalent to 2 kg at the crank
radius. A balance mass fitted opposite to the crank, is equivalent to 2.25 kg at a radius of 87.5 mm.
Determine for an engine speed of 1800 r.p.m. ; the maximum and minimum values of the primaryand secondary forces due to the inertia of reciprocating and rotating masses.

[ Ans. Primary forces : 3240 N (max.) and 1830 N (min.)
Secondary forces : 1020 N (max.) and 470 N (min.)]
1. Write a short note on primary and secondary balancing.
2. Explain why only a part of the unbalanced force due to reciprocating masses is balanced by revolving
mass.
3. Derive the following expressions, for an uncoupled two cylinder locomotive engine :

(a) Variation is tractive force ; ( b) Swaying couple ; and ( c) Hammer blow.
4. What are in-line engines ? How are they balanced ? It is possible to balance them completely ?
5. Explain the ‘direct and reverse crank’ method for determining unbalanced forces in radial engines.
6. Discuss the balancing of V-engines.

1. The primary unbalanced force is maximum when the angle of inclination of the crank with the line
of stroke is
(a)0 ° ( b) 90° ( c)  180° ( d)360°
2.  The partial balancing means
(a) balancing partially the revolving masses
(b) balancing partially the reciprocating masses
(c) best balancing of engines
3. In order to facilitate the starting of locomotive in any position, the cranks of a locomotive, with two
cylinders, are placed at . . . . . . to each other.

(a) 45° ( b) 90° ( c) 120° ( d) 180°
4. In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to
(a) minimise the effect of primary forces ( b) minimise the effect of secondary forces
(c) have perfect balancing ( d) start the locomotive quickly
5. If c be the fraction of the reciprocating parts of mass m to be balanced per cyclinder of a steam
locomotive with crank radius r, angular speed ω, distance between centre lines of two cylinders a,

then the magnitude of the maximum swaying couple is given by
(a)2 1
2cmr a−×ω (b)2 1
2cmr a−×ω
(c)22( 1 ) cm r a−ω (d) none of these

908       /G6C               Theory of Machines
6. The swaying couple is maximum or minimum when the angle of inclination of the crank to the line
of stroke ( θ) is equal to
(a) 45° and 135° ( b) 90° and 135°
(c) 135° and 225° ( d) 45° and 225°
7. The tractive force is maximum or minimum when the angle of inclination of the crank to the line of
stroke (θ) is equal to
(a) 90° and 225° ( b) 135° and 180° ( c) 180° and 225° ( d) 135° and 315°
8. The swaying couple is due to the

(a) primary unbalanced force ( b) secondary unbalanced force
(c) two cylinders of locomotive ( d) partial balancing
9. In a locomotive, the maximum magnitude of the unbalanced force along the perpendicular to the
line of stroke, is known as
(a ) tractive force ( b) swaying couple ( c) hammer blow ( d) none of these
10. The effect of hammer blow in a locomotive can be reduced by(a) decreasing the speed
(b) using two or three pairs of wheels coupled together

(c) balancing whole of the reciprocating parts
(d) both ( a) and ( b)
11. Multi-cylinder engines are desirable because(a) only balancing problems are reduced ( b) only flywheel size is reduced
(c) both ( a) and ( b)( d) none of these
12. When the primary direct crank of a reciprocating engine makes an angle 
θ with the line of stroke,
then the secondary direct crank will make an angle of . . . . . with the line of stroke.
(a)θ/2 ( b)θ (c)2  θ (d)4  θ

13. Secondary forces in reciprocating mass on engine frame are
(a) of same frequency as of primary forces
(b) twice the frequency as  of primary forces
(c) four times the frequency as of primary forces
14. The secondary unbalanced force produced by the reciprocating parts of a certain cylinder of a given
engine with crank radius r and connecting rod length l can be considered as equal to primary

unbalanced force produced by the same weight having
(a) an equivalent crank radius r2/4l and rotating at twice the speed of the engine
(b)r2/4l as equivalent crank radius and rotating at engine speed
(c) equivalent crank length of r2/4l and rotating at engine speed
15. Which of the following statement is correct?
(a) In any engine, 100% of the reciprocating masses can be balanced dynamically

(b) In the case of balancing of multicylinder engine, the value of secondary force is higher than
the value of the primary force
(c) In the case of balancing of multimass rotating systems, dynamic balancing can be directly
started without static balancing done to the system
    1.   (c) 2.(b) 3.(b) 4.(b) 5. (b)
    6.  (d) 7.(d) 8.(a) 9.(c) 10. (d)
   1 1.   (c) 12. (c) 13. (b) 14. (a) 15. (c)

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          909
909LongitudinalLongitudinalLongitudinalLongitudinalLongitudinal
andandandandand
TTTTTransvransvransvransvransv ererererersesesesese
VVVVVibraibraibraibraibrationstionstionstionstions23
FFFFFeaeaeaeaeaturturturturtures (Main)es (Main)es (Main)es (Main)es (Main)
2. Terms Used in Vibratory
Motion.
3. Types of Vibratory Motion.4. Types of Free Vibrations.5. Natural Frequency of Free

Longitudinal Vibrations.
6. Natural Frequency of Free
Transverse Vibrations.
7. Effect of Inertia of the
Constraint in Longitudinal and
8. Natural Frequency of Free
9. Natural Frequency of Free
10. Natural Frequency of Free
11. Natural Frequency of Free
12. Critical or Whirling Speed of
a Shaft.
13. Frequency of Free Damped
Vibrations(Viscous Damping).
14. Damping Factor or Damping

Ratio.
15. Logarithmic Decrement.
16. Frequency of Under Damped
Forced Vibrations.
17. Magnification Factor or
Dynamic Magnifier.
18. Vibration Isolation and
Transmissibility.23.1.23.1.23.1.23.1.23.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoduction
When elastic bodies such as a spring, a beam and a
shaft are displaced from the equilibrium position by the ap-

plication of external forces, and then released, they executea vibratory motion . This is due to the reason that, when a

body is displaced, the internal forces in the form of elasticor strain energy are present in the body. At release, theseforces bring the body to its original position. When the bodyreaches the equilibrium position, the whole of the elastic orstrain energy is converted into kinetic energy due to whichthe body continues to move in the opposite direction. Thewhole of the kinetic energy is again converted into strainenergy due to which the body again returns to the equilib-rium position. In this

returns to the equilib-rium position. In this way, the vibratory motion is repeatedindefinitely.

23.2.23.2.23.2.23.2.23.2. TTTTTerererererms Used in ms Used in ms Used in ms Used in ms Used in VVVVVibraibraibraibraibratortortortortory Motiony Motiony Motiony Motiony Motion
The following terms are commonly used in connec-
tion with the vibratory motions :

910       /G6C               Theory of Machines
1. Period of vibration or time period . It is the time interval after which the motion is
repeated itself. The period of vibration is usually expressed in seconds.
2. Cycle . It is the motion completed during one time period.
3. Frequency . It is the number of cycles described in one second. In S.I. units, the fre-
quency is expressed in hertz (briefly written as Hz) which is equal to one cycle per second.

23.3.23.3.23.3.23.3.23.3. Types of Vibratory MotionTypes of Vibratory MotionTypes of Vibratory MotionTypes of Vibratory MotionTypes of Vibratory Motion
The following types of vibratory motion are important from the subject point of view :
1. Free or natural vibrations . When no external force acts on the body, after giving it an
initial displacement, then the body is said to be under free or natural vibrations . The frequency of
the free vibrations is called free or natural frequency .

2. Forced vibrations . When the body vibrates under the influence of external force, then
the body is said to be under  forced vibrations . The external force applied to the body is a periodic
disturbing force created by unbalance. The vibrations have the same frequency as the applied force.
Note : When the frequency of the external force is same as that of the natural vibrations, resonance takes
place.
3. Damped vibrations . When there is a reduction in amplitude over every cycle of vibration,

the motion is said to be damped vibration . This is due to the fact that a certain amount of energy
possessed by the vibrating system is always dissipated in overcoming frictional resistances to the
motion.
23.4.23.4.23.4.23.4.23.4. Types of Free VibrationsTypes of Free VibrationsTypes of Free VibrationsTypes of Free VibrationsTypes of Free Vibrations
The following three types of free vibrations are important from the subject point of view :

1. Longitudinal vibrations, 2. Transverse vibrations, and 3. Torsional vibrations.
Consider a weightless constraint (spring or shaft) whose one end is fixed and the other end
carrying a heavy disc, as shown in Fig. 23.1. This system may execute one of the three above
mentioned types of vibrations.
B = Mean position ; A and C = Extreme positions.
(a) Longitudinal vibrations.  ( b) Transverse vibrations.     ( c) Torsional vibrations.
Fig. 23.1.  Types of free vibrations.

1. Longitudinal vibrations . When the particles of the shaft or disc moves parallel to the
axis of the shaft, as shown in Fig. 23.1 ( a), then the vibrations are known as longitudinal vibrations .
In this case, the shaft is elongated and shortened alternately and thus the tensile and compressivestresses are induced alternately in the shaft.

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          911
2. Transverse vibrations . When the particles of the shaft or disc move approximately
perpendicular to the axis of the shaft, as shown in Fig. 23.1 ( b), then the vibrations are known as
transverse vibrations . In this case, the shaft is straight and bent alternately and bending stresses are
induced in the shaft.
3. Torsional vibrations *. When the particles of the shaft or disc move in a circle about the

axis of the shaft, as shown in Fig. 23.1 ( c), then the vibrations are known as  torsional vibrations .
In this case, the shaft is twisted and untwisted alternately and the torsional shear stresses are in-
duced in the shaft.
Note : If the limit of proportionality ( i.e. stress proportional to strain) is not exceeded in the three types of

vibrations, then the restoring force in longitudinal and transverse vibrations or the restoring couple in torsionalvibrations which is exerted on the disc by the shaft (due to the stiffness of the shaft) is directly proportionalto the displacement of the disc from its equilibrium or mean position. Hence it follows that the accelerationtowards the equilibrium position is directly proportional to the displacement from that position and the vibration
is, therefore, simple harmonic.

23.5.23.5.23.5.23.5.23.5. Natural Frequency of Free Longitudinal VibrationsNatural Frequency of Free Longitudinal VibrationsNatural Frequency of Free Longitudinal VibrationsNatural Frequency of Free Longitudinal VibrationsNatural Frequency of Free Longitudinal Vibrations
The natural frequency of the free longitudinal vibrations may be determined by the following
three methods :
1. Equilibrium Method

Consider a constraint ( i.e. spring) of negligible mass in an unstrained position, as shown in
Fig. 23.2 ( a).
 Let  s= Stiffness of the constraint. It is the force required to produce
unit displacement in the direction of vibration. It is usually
expressed in N/m.
m= Mass of the body suspended from the constraint in kg,
*     The torsional vibrations are separately discussed in chapter 24.

912       /G6C               Theory of Machines
       δ= Static deflection of the spring in metres due to weight W
 newtons, and
             x  = Displacement given to the body by the external force, in metres.
Fig. 23.2.  Natural frequency of free longitudinal vibrations.
In the equilibrium position, as shown in Fig. 23.2 ( b), the gravitational pull W = m.g, is
balanced by a force of spring, such that W = s.δ.

Since the mass is now displaced from its equilibrium position by a distance x, as shown in
Fig. 23.2 ( c), and is then released, therefore after time t,
Restoring force           () .. Ws x Ws s x=−δ +=− δ −
          ... .sss xs x=δ −δ − = −  . . . (. )Ws=δ∵  . . . (i)
 . . . (Taking upward force as negative)
and          Accelerating force = Mass × Acceleration
2
2dxm
dt=× . . . (Taking downward force as positive)  . .  .  (ii)

Equating equations  (i) and (ii), the equation of motion of the body of mass m after time t is
          2
2.dxms x
dt×= −    or    2
2.0dxms x
dt×+ =
∴2
20dx sx
m dt+× =  . . . (iii)
We know that the fundamental equation of simple harmonic motion is
     22
2.0dxx
dt+ω =  . . . (iv)
Comparing equations (iii) and (iv), we have
       s
mω=
∴    Time period,      22pmtsπ== πω

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          913
and natural frequency,     11 1
22n
psgf
tm== =
ππ δ. . . (.. )mg s=δ∵
Taking the value of g as 9.81 m/s2 and δ in metres,
    1 9.81 0.4985Hz
2nf==
πδ δ
Note : The value of static deflection δmay be found out from the given conditions of the problem. For
longitudinal vibrations, it may be obtained by the relation,
Stress
StrainE=   or      WlEA×=δ    or      .
.Wl
EAδ=

.Wl
EAδ=
where                       δ  = Static deflection i.e. extension or compression of the constraint,
    W  = Load attached to the free end of constraint,
      l  = Length of the constraint,
    E  = Young’s modulus for the constraint, and
     A  = Cross-sectional area of the constraint.
2.  Energy method
We know that the kinetic
energy is due to the motion of the

body and the potential energy iswith respect to a certain datumposition which is equal to theamount of work required to movethe body from the datum position.In the case of vibrations, thedatum position is the mean orequilibrium position at which thepotential energy of the body or thesystem is zero.
In the free vibrations, no

energy is transferred to the systemor from the system. Therefore thesummation of kinetic energy andpotential energy must be aconstant quantity which is same atall the times. In other words,
∴  (. . . . ) 0dKE PE
dt+=
We know that kinetic en-
ergy,
    21..
2dxKE m
dt=×∵
This industrial compressor uses compressed air to power heavy-

duty construction tools. Compressors are used for jobs, suchas breaking up concrete or paving, drilling, pile driving, sand-blasting and tunnelling. A compressor works on the same prin-ciple as a pump. A piston moves backwards and forwards in-side a hollow cylinder, which compresses the air and forces itinto a hollow chamber. A pipe or hose connected to the cham-ber channels the compressed air to the tools.
Note : This picture is given as additional information and

914       /G6C               Theory of Machines
and potential energy,           2 0. 1.. .22sxPE x sx+== ×
. . . (∵P .E. = Mean force × Displacement )
∴          2
2 11.022dd xms xdt dt× +×  =
           2
21122 0
22dx d x dxms x
dt dt dt××× × +× × × =
or      2
2.0dxms x
dt×+ =     or     2
20dx sx
m dt+× =  . . . (Same as before)
The time period and the natural frequency may be obtained as discussed in the previous
method.
3. Rayleigh’s method

method.
In this method, the maximum kinetic energy at the mean position is equal to the maximum
potential energy (or strain energy) at the extreme position. Assuming the motion executed by the
vibration to be simple harmonic, then
       sin . xX t=ω . . . (i)
where          x  = Displacement of the body from the mean position after time t
   seconds, and
          X  = Maximum displacement from mean position to extreme position.
Now, differentiating equation  (i), we have

     cos .dxXt
dt=ω× ω
Since at the mean position, t = 0, therefore maximum velocity at the mean position,
        .dxvX
dt== ω
∴   Maximum kinetic energy at mean position
          22 2 11.. .22mv m X=× =×ω  . . . (ii)
and maximum potential energy at the extreme position
          2 0. 1.22sXXs X+== × . . . (iii)
Equating equations (ii) and (iii),
     22 2 11.. .22mX s X×ω =×     or   2s
mω=  , and s
mω=

mω=  , and s
mω=
∴   Time period,        22pstmπ== πω. . . (Same as before)

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          915
and natural frequency, 11
22n
psftmω===
ππ. . . (Same as before)
Note : In all the above expressions, ωis known as natural circular frequency  and is generally denoted by
ωn.

ωn.
23.6.23.6.23.6.23.6.23.6. Natural Frequency of Free Transverse VibrationsNatural Frequency of Free Transverse VibrationsNatural Frequency of Free Transverse VibrationsNatural Frequency of Free Transverse VibrationsNatural Frequency of Free Transverse Vibrations
Consider a shaft of negligible mass, whose one
end is fixed and the other end carries a body of weight
W, as shown in Fig. 23.3.
Let            s  = Stiffness of shaft,
         δ= Static deflection due to
    weight of the body,

weight of the body,
            x  = Displacement of body from
      mean position after time t.
           m  = Mass of body = W/g
As discussed in the previous article,
    Restoring force  = – s.x . . . (i)
  and accelerating force       2
2dxm
dt=× . . . (ii)
Equating equations (i) and (ii), the equation of motion becomes
   2
2.dxms x
dt×= −       or      2
2.0dxms x
dt×+ =
∴    2
20dx sx
m dt+× =                    . . . (Same as before )

Hence, the time period and the natural frequency of the transverse vibrations are same as
that of longitudinal vibrations. Therefore
Time period, 2pmts=π
and natural frequency,    11 1
22n
psgf
tm== =
ππ δ
Note :  The shape of the curve, into which the vibrating shaft deflects, is identical with the static deflection
curve of a cantilever beam loaded at the end. It has been proved in the text book on Strength of Materials,

that the static deflection of a cantilever beam loaded at the free end is
3
3Wl
EIδ=  (in metres)
where            W  = Load at the free end, in newtons,
l = Length of the shaft or beam in metres,
E = Young’s modulus for the material of the shaft or beam in
N/m2, and
I = Moment of inertia of the shaft or beam in m4.Fig. 23.3.  Natural frequency of free
       transverse vibrations.

916       /G6C               Theory of Machines
Example 23.1. A cantilever shaft 50 mm diameter and 300 mm long has a disc of mass
100 kg at its free end. The Young's modulus for the shaft material is 200 GN/m2. Determine the
frequency of longitudinal and transverse vibrations of the shaft.
Solution.  Given : d = 50 mm = 0.05 m ; l = 300 mm = 0.03 m ; m = 100 kg ;
E = 200 GN/m2 = 200 ×109 N/m2
We know that cross-sectional area of the shaft,
       22 3 2(0.05) 1.96 10 m44Ad− ππ=× = = ×

22 3 2(0.05) 1.96 10 m44Ad− ππ=× = = ×
and moment of inertia of the shaft,
      44 6 4(0.05) 0.3 10 m
64 64Id− ππ=×= =×
Frequency of longitudinal vibration
We know that static deflection of the shaft,
               6
39. 100 9.81 0.30.751 10. 1.96 10 200 10Wl
AE−
−××δ= = = ×
×× × m
(. )Wm g=…∵
∴Frequency of longitudinal vibration,
     
60.4985 0.4985575
0.751 10nf
−== =
δ× Hz  Ans.
Frequency of transverse vibration
      333

333
96. 100 9.81 (0.3)0.147 103. 3 200 10 0.3 10Wl
EI−
−××δ= = = ×
××× × m
∴    Frequency of transverse vibration,
     30.4985 0.4985
0.147 10nf
−==
δ× = 41 Hz Ans.

0.147 10nf
−==
23.7.23.7.23.7.23.7.23.7. Effect of Inertia of the Constraint in Longitudinal and TransverseEffect of Inertia of the Constraint in Longitudinal and TransverseEffect of Inertia of the Constraint in Longitudinal and TransverseEffect of Inertia of the Constraint in Longitudinal and TransverseEffect of Inertia of the Constraint in Longitudinal and Transverse
VibrationsVibrationsVibrationsVibrationsVibrations
In deriving the expressions for natural frequency of

longitudinal and transverse vibrations, we have neglected the inertia
of the constraint i.e. shaft. We shall now discuss the effect of the
inertia of the constraint, as below :
1. Longitudinal vibration
Consider the constraint whose one end is fixed and other end
is free as shown in Fig. 23.4.
Let       m1 = Mass of the constraint per unit length,
       l  = Length of the constraint,
     mC= Total mass of the constraint = m1. l, and

v  = Longitudinal velocity of the free end.Fig. 23.4.  Effect of inertia
of the constraint in
longitudinal vibrations.

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          917
Consider a small element of the constraint at a distance x from the fixed end and of
length xδ.
∴  Velocity of the small element
          xv
l=×
and kinetic energy possessed by the element
          1
2= × Mass (velocity)2
          2 22
1
1 2. 1.2 2mv x xmx v xl l=× δ × = × δ 
∴   Total kinetic energy possessed by the constraint,
         22 2 3
11
2200..
3 22llmvx mv xdx
ll=× = 
∫
23
22 2 C 11

ll=× = 
∫
23
22 2 C 11
1 2.. 11 1.32 3 23 2 3 2m mv ml llmv v v
l =× = × × = =   . . . (i)
. . . (Substituting  m1 . l = mC)
If a mass of C
3m
 is placed at the free end and the constraint is assumed to be of negligible
mass, then
Total kinetic energy possessed by the constraint
          2 C 1
23mv= . . . [Same as equation (i)] . . . (ii)
Hence the two systems are dynamically same. Therefore, inertia of the constraint may be

allowed for by adding one-third of its mass to the disc at the free end.
From the above discussion, we find that when the mass of the constraint mC and the mass
of the disc m at the end is given, then natural frequency of vibration,
      C1
2
3nsfmm=π+
2. Transverse vibration
Consider a constraint whose one end is fixed and the other
end is free as shown in Fig. 23.5.
Let       m1 = Mass of constraint per unit length,
        l  = Length of the constraint,

l  = Length of the constraint,
      mC = Total mass of the constraint = m1.l, and
        v  = Transverse velocity of the free end.
Consider a small element of the constraint at a distance x
from the fixed end and of length xδ. The velocity of this element isFig. 23.5.  Effect of inertia
transverse vibrations.

918       /G6C               Theory of Machines
given by 23
33.
2lx xv
l−× 
.
∴    Kinetic energy of the element
      223
1313 ..2 2lx xmx v
l−=× δ × 
and total kinetic energy of the constraint,
      22 23
24 5 6 1
136
00. 13 .(9 . 6 . )2 28llmv lx xm v dx l x l x x dx
ll−=× × = − + ∫∫
      2 25 6 7
1
6
0.9 . 6 .
56 7 8l
mv l x l x x
l
=− +

      22 77 7 7
11
66.. 96 3 3
56 7 3 5 88mv mv ll l l
ll  
=− + =    
      22 2

ll  
      22 2
11 C33 1 33 1 33.. .280 2 140 2 140ml v ml v m v = ×=× =×    . . . (i)
. . . (Substituting m1.l = mC)
If a mass of C33
140m is placed at the free end and the constraint is assumed to be of negli-
gible mass, then
      2 C331
2 140mv=. . . [Same as equation (i)]
Hence the two systems are dynamically same. Therefore the inertia of the constraint may
be allowed for by adding 33

140 of its mass to the disc at the free end.
of the disc m at the free end is given, then natural frequency of vibration,
 C1
33 2
140nsfmm=π+
Notes : 1.  If both the ends of the constraint are fixed, and the disc is situated in the middle of it, then
proceeding in the similar way as discussed above, we may prove that the inertia of the constraint may be
allowed for by adding 13

35 of its mass to the disc.
2. If the constraint is like a simply supported beam, then 17
35 of its mass may be added to the mass
of the disc.

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          919
23.8.23.8.23.8.23.8.23.8. Natural Frequency of Free Transverse Vibrations Due to a PointNatural Frequency of Free Transverse Vibrations Due to a PointNatural Frequency of Free Transverse Vibrations Due to a PointNatural Frequency of Free Transverse Vibrations Due to a PointNatural Frequency of Free Transverse Vibrations Due to a Point

Load Acting Over aLoad Acting Over aLoad Acting Over aLoad Acting Over aLoad Acting Over a Simply Supported ShaftSimply Supported ShaftSimply Supported ShaftSimply Supported ShaftSimply Supported Shaft
Consider a shaft AB of length l, carrying a point
load W at C which is at a distance of l1 from A and l2 from
B, as shown in Fig. 23.6. A little consideration will show
that when the shaft is deflected and suddenly released, it

will make transverse vibrations. The deflection of the shaftis proportional to the load W and if the beam is deflected
beyond the static equilibrium position then the load willvibrate with simple harmonic motion (as by a helical
spring). If 
δis the static deflection due to load W, then the
natural frequency of the free transverse vibration is
     1 0.4985
2ngf==
πδ δ Hz  . . . (Substituting, g = 9.81 m/s2)

Some of the values of the static deflection for the various types of beams and under various
load conditions are given in the following table.

Table 23.1. Values of static deflection (Table 23.1. Values of static deflection (Table 23.1. Values of static deflection (Table 23.1. Values of static deflection (Table 23.1. Values of static deflection ( δδδδδ) for the various types of beams) for the various types of beams) for the various types of beams) for the various types of beams) for the various types of beams

and under various load conditions.and under various load conditions.and under various load conditions.and under various load conditions.and under various load conditions.
S.No. Type of beam Deflection  (δδδδδ)
1. Cantilever beam with a point load W at the δ=3
3Wl
EI (at the free end)
free end.
2. Cantilever beam with a uniformly δ=4
8wl
distributed load of w per unit length.
3. Simply supported beam with an eccentric δ=22
3Wa b
EIl (at the point load)
point load W.

3Wa b
4. Simply supported beam with a central point δ=3
48Wl
EI (at the centre)
load W.Fig. 23.6.  Simply supported beam
with a point load.

920       /G6C               Theory of Machines
5. Simply supported beam with a uniformly δ= ×45
384wl
6. Fixed beam with an eccentric point load W. δ=33
3Wa b
7. Fixed beam with a central point load W. δ=3
192Wl
8. Fixed beam with a uniformly distributed δ=4
384wl
load of w per unit length.

Example 23.2. A shaft of length 0.75 m, supported freely at the ends, is carrying a body of
mass 90 kg at 0.25 m from one end. Find the natural frequency of transverse vibration. Assume
E = 200 GN/m2 and shaft diameter = 50 mm.
Solution.  Given : l = 0.75 m ; m = 90 kg ; a = AC = 0.25 m ; E = 200 GN/m2 = 200 × 109
N/m2; d = 50 mm = 0.05 m
The shaft is shown in Fig. 23.7.
We know that moment of inertia of the shaft,
       44 4(0.05) m
64 64Idππ=×=

44 4(0.05) m
          640.307 10 m−=×
and static deflection at the load point ( i.e. at point C),
       22 2 23
9690 9.81(0.25) (0.5)0.1 103 3 200 10 0.307 10 0.75Wa b
EIl−
−×δ= = = ×
××× × × m
. . . (∵b = BC = 0.5 m)∵S.No. Type of beam Deflection  (δδδδδ)
Fig. 23.7

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          921
We know that natural frequency of transverse vibration,
  30.4985 0.4985
0.1 10nf
−==
δ × = 49.85 Hz Ans.
Example 23.3. A flywheel is mounted on a vertical shaft as shown in Fig. 23.8. The both
ends of the shaft are fixed and its diameter is 50 mm. The flywheel has a mass of 500 kg. Find the
natural frequencies of longitudinal and transverse vibrations. Take E = 200 GN/m2.

Solution.  Given : d = 50 mm = 0.05 m ; m = 500 kg ; E = 200 GN/m2 = 200 × 109 N/m2
We know that cross-sectional area of shaft,
   22 3(0.05) 1.96 1044Ad− ππ=× = = ×  m2
and moment of inertia of shaft,
     44 6 4(0.05) 0.307 10 m64 64Id− ππ=×= = ×
Natural frequency of longitudinal vibration
Let     m1 = Mass of flywheel carried by the length l1.
∴           m  – m1= Mass of flywheel carried by length l2.
We know that extension of length l1
       11 1 1.. .
..Wl mg l
AE AE== . . . (i)

11 1 1.. .
..Wl mg l
Similarly, compression of length l2
       12 1 2 () ( ) .
..WWl mmg l
AE AE−−==  . . . (ii)
Since extension of length l1 must be equal to compression of length l2, therefore equating
 11 1 2.( )ml m m l=−
       11 10.9 (500 )0.6 300 0.6mm m×= − = −  or 1200 m=  kg
∴    Extension of length l1,
    6 11
39.. 200 9.81 0.94.5 10. 1.96 10 200 10mg l
AE−
−××δ= = = ×
×××m
We know that natural frequency of longitudinal vibration,

60.4985 0.4985
4.5 10nf
−==
δ × = 235 Hz Ans.
Natural frequency of transverse vibration
We know that the static deflection for a shaft fixed at both ends and carrying a point load
is given by
    33 3 3
3
39 6 3500 9.81(0.9) (0.6)1.24 10
3 3 200 10 0.307 10 (1.5)Wa b
EI l−
−×δ= = = ×
××× × m
. . . (Substituting W = m.g ; a = l1, and b = l2)Fig. 23.8

922       /G6C               Theory of Machines
    30.4985 0.4985
1.24 10nf
−==
δ × = 14.24 Hz    Ans.

1.24 10nf
−==
23.9.23.9.23.9.23.9.23.9. Natural Frequency of Free Transverse Vibrations Due to UniformlyNatural Frequency of Free Transverse Vibrations Due to UniformlyNatural Frequency of Free Transverse Vibrations Due to UniformlyNatural Frequency of Free Transverse Vibrations Due to UniformlyNatural Frequency of Free Transverse Vibrations Due to Uniformly

Distributed Load Acting Over a Simply Supported ShaftDistributed Load Acting Over a Simply Supported ShaftDistributed Load Acting Over a Simply Supported ShaftDistributed Load Acting Over a Simply Supported ShaftDistributed Load Acting Over a Simply Supported Shaft
Consider a shaft AB carrying a uniformly distributed load of w per unit length as shown in
Fig. 23.9.
Let     y1= Static deflection at the middle of the shaft,
     a1= Amplitude of vibration at the middle of the shaft, and

w1 = Uniformly distributed load per unit static deflection at the
 middle of the shaft = w/y1.
Fig. 23.9.  Simply supported shaft carrying a uniformly distributed load.
Now, consider a small section of the shaft at a distance x from A and length xδ.
Let        y = Static deflection at a distance x from A, and
     a = Amplitude of its vibration.
∴    Work done on this small section
         1
11 1
1111 1.. .22 2a wwa x a a x a w a xyy=× δ ×=× ×δ ×=×× ×× δ

Since the maximum potential energy at the extreme position is equal to the amount of work
done to move the beam from the mean position to one of its extreme positions, therefore
Maximum potential energy at the extreme position
         1
101
2.lawa d xy=× × ×∫  . . . (i)
Assuming that the shape of the curve of a vibrating shaft is similar to the static deflection
curve of a beam, therefore
     1
1aa
yy= = Constant, C      or     1
1aCy= and a = y.C

Substituting these values in equation  (i), we have maximum potential energy at the extreme
position
        2
0011.. . .22ll
wC y C d x w C y d x=× × × = ×∫∫ . . . (ii)

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          923
Since the maximum velocity at the mean position is 1.aω, where ωis the circular frequency
of vibration, therefore
Maximum kinetic energy at the mean position
          22 2 2
001.(.) .22llwd x waC y d xgg=× ω=× ω ×∫∫  . . . (iii)
. . .(Substituting a = y.C )
We know that the maximum potential energy at the extreme position is equal to the maximum

kinetic energy at the mean position, therefore equating equations (ii) and (iii),
           22 2 2
001..22llwwC y d x C yd xg×× = × ω ×∫∫
∴                  2 0
2
0.
.l
lgy d x
yd xω=∫
∫      or       0
2
0.
.l
lgy d x
yd xω=∫
∫. . . (iv)
When the shaft is a simply supported, then the static deflection at a distance x from A is
      *43 3(2 )24wyx l x l xEI=− + . . . (v)
where        w  = Uniformly distributed load unit length,
       E  = Young’s modulus for the material of the shaft, and

I  = Moment of inertia of the shaft.
* It has been proved in books on ‘Strength of Materials’ that maximum bending moment at a distance x
from A is
        22
2(. . )22maxdy w x w l xBM E I
dx== −
Integrating this expression,
 32
1..23 22dy wx wl xEI Cdx=− +××
On further integrating,
  43
12...234 223=−+ +×× ××wx wl xEIy Cx C
          43
1224 12=−+ +wx wlxCx C
where C1 and C2 are the constants of integration and may be determined from the given conditions of
the problem. Here

        when          x  = 0, y = 0 ;              ∴      C2 = 0
        and when          x  = l, y = 0 ;    ∴      C1 = 3
24wl
Substituting the value of C1, we get
       43 3(2 )
24=− +wyx l x l x
EI

924       /G6C               Theory of Machines
Now integrating the above equation (v) within the limits from 0 to l,
 54 3 2
43 3
0 002(2 )24 24 5 4 2l l lww x l x l xyd x x l x l x d xEI EI=− + = − + 
∫∫
           55 5 5 52.
24 5 4 2 24 5 120 .wl l l w l w l
EI EI E I=− + = × =
 . . . (vi)
Now2
24 3 3
00(2 )24llwyd x x l x lx d xEI=− +∫∫
           2
82 6 6 2 74 43 5
0(4 4 4 2)24lwxl x l xl xl xl x d xEI=+ + − − +∫
           2 92 7 6 3 84 53 6
22
044 4 2

2 92 7 6 3 84 53 6
22
044 4 2
97 38 5 6 576
=⋅ + + − − + 
l
w x lx lx l x lx lx
EI
           2 99 9999
2244 4 2
9738 5 6 576w ll llll
EI= ++ −−+ 

           29
2231
630 576wl
EI=× . . . (vii)
Substituting the value in equation (iv) from equations (vi) and (vii), we get circular frequency
due to uniformly distributed load,
        52 2
29576 630
120 31wl E IgEI wl ×ω= ×× 
A railway bridge.

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          925
          2
4424 630
155EI EI gg
wl wl=× = π  . . . (viii)
∴   Natural frequency due to uniformly distributed load,
     2
44 22 2nEIg EIgf
wl wlωπ π== =ππ. . . (ix)
We know that the static deflection of a simply supported shaft due to uniformly distributed
load of w per unit length, is
      4
S5
384wl
EIδ=       or     4S5
384EI
wl=δ
Equation (ix) may be written as
     
S S50 . 5 6 1 5
23 8 4ngfπ==

S S50 . 5 6 1 5
δδ Hz . . . (Substituting, g = 9.81 m/s2)
23.10. Natural Frequency of Free Transverse Vibrations of a Shaft Fixed at23.10. Natural Frequency of Free Transverse Vibrations of a Shaft Fixed at23.10. Natural Frequency of Free Transverse Vibrations of a Shaft Fixed at23.10. Natural Frequency of Free Transverse Vibrations of a Shaft Fixed at23.10. Natural Frequency of Free Transverse Vibrations of a Shaft Fixed at

Both Ends  Carrying a Uniformly Distributed Load Both Ends  Carrying a Uniformly Distributed Load Both Ends  Carrying a Uniformly Distributed Load Both Ends  Carrying a Uniformly Distributed Load Both Ends  Carrying a Uniformly Distributed Load
Consider a shaft AB fixed at both ends
and carrying a uniformly distributed load of w
per unit length as shown in Fig. 23.10.
We know that the static deflection at a
distance x from A is given by
      *42 2 3(2 )24wyx l x l xEI=+ −  . . . (i)

*42 2 3(2 )24wyx l x l xEI=+ −  . . . (i)
*      It has been proved in books on ‘Strength of Materials’ that the bending moment at a distance x from A is
       22 2
2 12 2 2== + −d y wl wx wlxME I
dx
Integrating this equation,
 23 2
112 2 3 2 2=+ − +××dy wl wx wlxEI x Cdx
where C1 is the constant of integration. We know that when 0, 0==dyx
dx. Therefore C1 = 0.
or23 2
12 6 4=+ −dy wl wx wlxEI xdx
Integrating the above equation,
   22 4 3 22 4 3

   22 4 3 22 4 3
2 .12 2 6 4 4 3 24 24 12=+ − × + =+ − +××wl x wx wl x wl x wx wlxEI y C C
where C2 is the constant of integration. We know that when x = 0, y = 0. Therefore C2 = 0.
or22 4 3.( 2 )24=+ −wEI y l x x lx
or42 2 3(2 )
24=+ −wyx l x l x
EIFig. 23.10.  Shaft fixed at both ends
carrying a uniformly distributed load.

926       /G6C               Theory of Machines
Integrating the above equation within limits from 0 to l,
 42 2 3
00(2 )24l lwydx x l x l x dxEI=+ −∫∫
          52 3 4 55 5
022
24 5 3 4 24 5 3 4lwxl x l x wll l
EI EI  =+ − =+ −  
  
          55
24 30 720wl w l
EI EI=× =
Now integrating y2 within the limits from 0 to l,
           2
24 2 2 3 2
00(2 )24l lwyd x x l x l x d xEI=+ −∫∫
          2
84 4 2 6 2 6 7 3 5
0(4 2 4 2 )24lwx l xl xl x l xl x d xEI=+ + + − −∫

2
84 4 2 6 7 3 5
0(6 4 2 )24lwx l xl xl xl x d xEI=+ + + −∫
          294 5 2 7 8 3 6
064 2
2 4 9 5786lw x lx lx l x lx
EI =+ + − −   
          229 999 9 9642
24 9 5 7 8 6 24 630wl l l l l wl
EI EI  =+ + − − =     
    520 2
29 4
2
0(24 ) 630 504
720l
lgy d x
wl EI EIggEI wl w l
yd x×ω= =× × =∫
∫
∴4504EIg
wlω=
and natural frequency,
     4415 0 43.57322nEIg E I gf
wl wlω== =
ππ

4415 0 43.57322nEIg E I gf
wl wlω== =
ππ
Since the static deflection of a shaft fixed at both ends and carrying a uniformly distributed
load is
     4
S384wl
EIδ=       or    4S1
384EI
wl=δ
∴                    
S S0.5713.573384ngf==

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          927
23.11.23.11.23.11.23.11.23.11. Natural Frequency of Free Transverse Vibrations For a ShaftNatural Frequency of Free Transverse Vibrations For a ShaftNatural Frequency of Free Transverse Vibrations For a ShaftNatural Frequency of Free Transverse Vibrations For a ShaftNatural Frequency of Free Transverse Vibrations For a Shaft

Subjected to a Number of Point LoadsSubjected to a Number of Point LoadsSubjected to a Number of Point LoadsSubjected to a Number of Point LoadsSubjected to a Number of Point Loads
Consider a shaft AB of negligible mass loaded with
point loads W1 , W2, W3 and W4 etc. in newtons, as shown
in Fig. 23.11. Let m1, m2, m3 and m4 etc. be the corre-
sponding masses in kg. The natural frequency of such a
shaft may be found out by the following two methods :
1. Energy (or Rayleigh’s) method

Let y1, y2, y3, y4 etc. be total deflection under loads
W1, W2, W3 and W4 etc. as shown in Fig. 23.11.
We know that maximum potential energy
      11 2 2 3 3 4 411 1 1.. .. .. .. . . . . .22 2 2mg y mg y mg y m g y=× +× + +× +
      1..2mgy=Σ
and maximum kinetic energy
      22 22
11 2 2 33 4 411 1 1( . ) ( . ) ( . ) ( . ) ......22 2 2my my my my=× ω +× ω +× ω +× ω +
      22 2 2 2
1 1 22 33 441() () () () . . . .2m y my my my=× ω + + + +

22 1.2my =× ω Σ . . . ( where ω = Circular frequency of vibration)
Equating the maximum kinetic energy to the maximum potential energy, we have
22 11.. .22my mgy×ω Σ = Σ
∴2
22.. .
..mgy g my
my myΣΣω= =
ΣΣ          or      2.
.gm y
myΣω=
Σ
∴   Natural frequency of transverse vibration,
 21.
22 .ngm yf
myωΣ==ππΣ
2. Dunkerley’s method
The natural frequency of transverse vibration for a shaft carrying a number of point loads

and uniformly distributed load is obtained from Dunkerley’s empirical formula. According to this
         2222 2
1231111 1....
() ( ) ( ) ( ) ( )nn n n n sff f f f=+++ +Fig. 23.11.  Shaft carrying a
number of point loads.

928       /G6C               Theory of Machines
where                       fn = Natural frequency of transverse vibration of the shaft
  carrying point loads and uniformly distributed load.
        123,,nn nff f , etc.  =   Natural frequency of transverse vibration of each point load.
   nsf =  Natural frequency of transverse vibration of the uniformly
  distributed load (or due to the mass of the shaft).
Now, consider a shaft AB loaded as shown in Fig. 23.12.

Fig. 23.12.  Shaft carrying a number of point loads and a uniformly distributed load.
Let123,,,δδδ  etc. =  Static deflection due to the load W1, W2, W3 etc. when
  considered separately.
     Sδ =   Static deflection due to the uniformly distributed load or due
   to the mass of the
   shaft.
We know that natural frequency of transverse
vibration due to load W1,
    1
10.4985
nf=
δ Hz
Similarly, natural frequency of transverse vibra-
tion due to load W2,
   2
20.4985
nf=
δ Hz

   2
20.4985
nf=
δ Hz
and, natural frequency of transverse vibration due to load
W3,
   
3
30.4985
nf=
δ Hz
Also natural frequency of transverse vibration
due to uniformly distributed load or weight of the shaft,
    
S0.5615
nsf=
δ Hz
Therefore, according to Dunkerley’s empirical
formula, the natural frequency of the whole system,
2223 2
() ( ) ( ) ( ) ( )nn n n n sfff f f=+++ +
         3S 12
222 2....
(0.4985) (0.4985) (0.4985) (0.5615)δδ δδ=+++ +

         S
123 21....1.27 (0.4985)δ = δ +δ +δ + +
Suspension spring of an automobile.

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          929
or     
S
1230.4985
....1.27nf=
δδ+ δ+ δ+ + Hz
Notes : 1. When there is no uniformly distributed load or mass of the shaft is negligible, then S0δ= .
∴
1230.4985
....nf=
δ+ δ+ δ+ Hz
2. The value of 123,,δδδ  etc. for a simply supported shaft may be obtained from the relation
       22
3Wa b
EIlδ=
where        δ = Static deflection due to load W,
             a  and b = Distances of the load from the ends,

E  = Young’s modulus for the material of the shaft,
        I  = Moment of inertia of the shaft, and
       l  = Total length of the shaft.
Example 23.4. A shaft 50 mm diameter and 3 metres long is simply supported at the ends
and carries three loads of 1000 N, 1500 N and 750 N at 1 m, 2 m and 2.5 m from the left support.
The Young's modulus for shaft material is 200 GN/m2. Find the frequency of transverse vibration.

Solution.  Given : d = 50 mm = 0.05 m ; l = 3 m, W1 = 1000 N ; W2 = 1500 N ;
W3 = 750 N; E = 200 GN/m2 = 200 × 109 N/m2
The shaft carrying the loads is shown in Fig. 23.13
       44 6 4(0.05) 0.307 10 m
64 64Id− ππ=×= = ×
and the static deflection due to a point load W,
      22
3Wa b
EIlδ=
Fig. 23.13
∴     Static deflection due to a load of 1000 N,
     22
3
1 961000 1 27.24 10
3 200 10 0.307 10 3−
−××δ= = ×
××× × × m

−××δ= = ×
××× × × m
. . . (Here a = 1 m, and b = 2 m)

930       /G6C               Theory of Machines
Similarly, static deflection due to a load of 1500 N,
      22
3
2 961500 2 110.86 10
−××δ= = ×
××× × × m
. . . (Here a = 2 m, and b = 1 m)
and static deflection due to a load of 750 N,
     22
3
3 96750(2.5) (0.5)2.12 10
−δ= = ×
××× × × m
. . . (Here a = 2.5 m, and b = 0.5 m)
We know that frequency of transverse vibration,
                     33 31230.4985 0.4985
7.24 10 10.86 10 2.12 10nf
−− −==

−− −==
δ+ δ+ δ ×+ ×+×
          0.4985
0.1422= = 3.5 Hz  Ans.
23.12.23.12.23.12.23.12.23.12. Critical or Whirling Speed of a ShaftCritical or Whirling Speed of a ShaftCritical or Whirling Speed of a ShaftCritical or Whirling Speed of a ShaftCritical or Whirling Speed of a Shaft
In actual practice, a rotating shaft carries different mountings and accessories in the form
of gears, pulleys, etc. When the gears or pulleys are put on the shaft, the centre of gravity of the

pulley or gear does not coincide with the centre line of the bearings or with the axis of the shaft,when the shaft is stationary. This means that the centre of gravity of the pulley or gear is at acertain distance from the axis of rotation and due to this, the shaft is subjected to centrifugal force.This force will bent the shaft which will further increase the distance of centre of gravity of thepulley or gear from the axis of rotation. This correspondingly increases the value of

This correspondingly increases the value of centrifugalforce, which further increases the distance of centre of gravity from the axis of rotation. This effectis cumulative and ultimately the shaft fails. The bending of shaft not only depends upon the valueof eccentricity (distance between centre of gravity of the pulley and the axis of rotation) but alsodepends upon the speed at which the shaft rotates.

The speed at which the shaft runs so that the additional deflection of the shaft from
the axis of rotation becomes infinite, is known as critical  or whirling speed .
(a) When shaft is stationary. ( b) When shaft is rotating.
Fig. 23.14.  Critical or whirling speed of a shaft.
Consider a shaft of negligible mass carrying a rotor, as shown in Fig.23.14 ( a). The point
O is on the shaft axis and G is the centre of gravity of the rotor. When the shaft is stationary, the

centre line of the bearing and the axis of the shaft coincides. Fig. 23.14 ( b) shows the shaft when
rotating about the axis of rotation at a uniform speed of ω rad/s.
Let        m  = Mass of the rotor,
         e  = Initial distance of centre of gravity of the rotor from the centre
             line of the bearing or shaft axis, when the shaft is stationary,

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          931
        y  = Additional deflection of centre of gravity of the rotor when
  the shaft starts rotating at ω rad/s, and
         s  = Stiffness of the shaft i.e. the load required per unit deflection
  of the shaft.
Since the shaft is rotating at ω rad/s, therefore centrifugal force acting radially outwards
through G causing the shaft to deflect is given by
     2
C .( ) Fm y e=ω +

2
The shaft behaves like a spring. Therefore the force resisting the deflection y,
         = s.y
For the equilibrium position,
    2.( ) .my e s yω+ =
or            22.. .. .my me s yω+ ω=     or   22(. ) . .ys m m e−ω=ω
∴        22
22.. .
./me ey
sm s mωω==
−ω − ω. . . (i)
We know that circular frequency,
     ns
mω=     or    2
22.
()neyω=
ω− ω. . . [ From equation (i) ]
A little consideration will show that when nω>ω , the value of y will be negative and the

shaft deflects is the opposite direction as shown dotted in Fig 23.14 ( b).
In order to have the value of y always positive, both plus and minus signs are taken.
∴              2
22 2 2()11n nceeeyω± ±=± = =
ω− ω ωω −− ωω 
... (Substituting ncω= ω )
We see from the above expression that when ncω= ω , the value of y becomes infinite.
Therefore cω is the  critical or whirling speed.
∴     Critical or whirling speed,
    cnsg
mω= ω= =δ Hz . . . .mg
sδ=∵

cnsg
If Nc is the critical or whirling speed in r.p.s., then
 2cgNπ=δ     or   1 0.4985r.p.s.
2cgN==
πδ δ
where      δ = Static deflection of the shaft in metres.
Hence the critical or whirling speed is the same as the natural frequency of transverse
vibration but its unit will be revolutions per second.

932       /G6C               Theory of Machines
Notes : 1.  When the centre of gravity of the
rotor lies between the centre line of the shaft
and the centre line of the bearing, e is taken
negative. On the other hand, if the centre ofgravity of the rotor does not lie between thecentre line of the shaft and the centre line ofthe bearing (as in the above article) the valueof e is taken positive.
2. To determine the critical speed of a

shaft which may be subjected to point loads,uniformly distributed load or combination ofboth, find the frequency of transverse vibrationwhich is equal to critical speed of a shaft inr.p.s. The Dunkerley’s method may be used forcalculating the frequency.
3. A shaft supported is short bearings
(or ball bearings) is assumed to be a simply sup-ported shaft while the shaft supported in longbearings (or journal bearings) is assumed to
have both ends fixed.

Example 23.5. Calculate the
whirling speed of a shaft 20 mm diameter
and 0.6 m long carrying a mass of 1 kg atits mid-point. The density of the shaft ma-terial is 40 Mg/m
3, and Young’s modulus is 200 GN/m2. Assume the shaft to be freely supported.
Solution.  Given : d = 20 mm = 0.02 m ; l = 0.6 m ; m1 = 1 kg ; ρ = 40 Mg/m3
= 40 × 106 g/m3 = 40 × 103 kg/m3 ; E = 200 GN/m2 = 200 × 109 N/m2
The shaft is shown in Fig. 23.15.

       44 4(0.02) m64 64Idππ=×=
          = 7.855 × 10–9 m4
Since the density of shaft material is 40 × 103 kg/m3,
therefore mass of the shaft per metre length,
     23
SArea length density (0.02) 1 40 104mπ=× × = × × ×  = 12.6 kg/m
We know that static deflection due to 1 kg of mass at the centre,
        33
6
991 9.81(0.6)28 1048 48 200 10 7.855 10Wl
EI−
−×δ= = = ×
××× × m
and static deflection due to mass of the shaft,
       44
3

44
3
S 995 5 12.6 9.81(0.6)0.133 10384 384 200 10 7.855 10wl
EI−
−××δ= = = ×
××× × mFig. 23.15
Diesel engines have several advantages over petrol
engines. They do not need an electrical ignition system;they use cheaper fuel; and they do not need acarburettor. Diesel engines also have a greater abilityto convert the stored energy in the fuel into mechanicalenergy, or work.

not a direct example of the current chapter.ExhaustvalveIntakevalveCompression Exhaust
Fuel injector PowerInduction
Burnedgases
PistonAir intake
CrankshaftCompressed airand fuel mixtureFuel injectionand combustion

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          933
    3S 60.4985 0.4985
0.133 1028 101.27 1.27nf
−
−=+
δ ×δ+ ×+
         30.4985
11.52 10−=
× = 43.3 Hz
Let       Nc= Whirling speed of a shaft.
We know that whirling speed of a shaft in r.p.s. is equal to the frequency of transverse
vibration in Hz , therefore
    Nc = 43.3 r.p.s. = 43.3 × 60 = 2598 r.p.m. Ans.

Example 23.6. A shaft 1.5 m long, supported in flexible bearings at the ends carries two
wheels each of 50 kg mass. One wheel is situated at the centre of the shaft and the other at a
distance of 375 mm from the centre towards left. The shaft is hollow of external diameter 75 mmand internal diameter 40 mm. The density of the shaft material is 7700 kg/m
3 and its modulus of
elasticity is 200 GN/m2. Find the lowest whirling speed of the shaft, taking into account the mass
of the shaft.

Solution.  l = 1.5 m ; m1 = m2 = 50 kg ;
d1= 75 mm = 0.075 m ; d2 = 40 mm = 0.04 m ;
ρ= 7700 kg/m3 ; E = 200 GN/m2 = 200 × 109
N/m2
The shaft is shown in Fig. 23.16.
      44 4 4 6 4
12( ) ( ) (0.075) (0.04) 1.4 10 m
64 64Id d− ππ  =− = − = ×  
Since the density of shaft material is 7700 kg/m3, therefore mass of the shaft per metre
length,
     mS = Area × length × density
         22(0.075) (0.04) 1 77004π=− × = 24.34 kg/m

We know that the static deflection due to a load W
         22 22.
33Wa b m ga b
EIl EIl==
∴    Static deflection due to a mass of 50 kg at C,
     22 22
1
19650 9.81(0.375) (1.125)
3 3 200 10 1.4 10 1.5mg a b
EIl−×δ= =
××× × × = 70 × 10–6 m
. . . (Here a = 0.375 m, and b = 1.125 m)
Similarly, static deflection due to a mass of 50 kg at D
    22 22
1
29650 9.81(0.75) (0.75)
EIl−×δ= =
××× × × = 123 × 10–6 m
. . . (Here a = b = 0.75 m)Fig. 23.16

934       /G6C               Theory of Machines
We know that static deflection due to uniformly distributed load or mass of the shaft,
 44
S 965 5 24.34 9.81(1.5)
384 384 200 10 1.4 10wl
EI−×δ= × = ×
××× = 56 × 10–6 m
. . . (Substituting, w = mS × g)
6S 66120.4985 0.4985
56 1070 10 123 101.27 1.27nf
−
−−==
δ ×δ+ δ+ ×+ ×+ Hz
   = 32.4 Hz
Since the whirling speed of shaft ( Nc) in r.p.s. is equal to the frequency of transverse

vibration in Hz, therefore
Nc = 32.4 r.p.s. = 32.4 × 60 = 1944 r.p.m.  Ans.
Example 23.7. A vertical shaft of 5 mm diameter is 200 mm long and is supported in long
bearings at its ends. A disc of mass 50 kg is attached to the centre of the shaft. Neglecting any

increase in stiffness due to the attachment of the disc to the shaft, find the critical speed of rotationand the maximum bending stress when the shaft is rotating at 75% of the critical speed. The centreof the disc is 0.25 mm from the geometric axis of the shaft. E = 200 GN/m
2.
Solution.  Given : d = 5 mm = 0.005 m ; l = 200 mm = 0.2 m ; m = 50 kg ; e = 0.25 mm
= 0.25 × 10–3 m ; E = 200 GN/m2 = 200 × 109 N/m2
Critical speed of rotation

  44 1 2 4(0.005) 30.7 10 m
Since the shaft is supported in long bearings, it is assumed to be fixed at both ends. We
know that the static deflection at the centre of the shaft due to a mass of 50 kg,
  33
3
91 250 9.81(0.2)3.33 10 m192 192 200 10 30.7 10Wl
EI−
−×δ= = = ×
××××
 . . .  (∵ W = m.g)
We know that critical speed of rotation (or natural frequency of transverse vibrations),
30.4985
3.33 10cN
−=
× = 8.64 r.p.s.  Ans.

30.4985
3.33 10cN
−=
Maximum bending stress
Let σ = Maximum bending stress in N/m2, and
  N = Speed of the shaft = 75% of critical speed = 0.75 Nc . . . (Given)
When the shaft starts rotating, the additional dynamic load ( W1) to which the shaft is
subjected, may be obtained by using the bending equation,
1M
Iyσ=       or      
1.IMyσ=

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          935
We know that for a shaft fixed at both ends and carrying a point load ( W1) at the centre, the
maximum bending moment
                   1.
8WlM=
∴      1.
8/ 2Wl I
dσ⋅=  . . .(∵ y1 = d/2 )
and        12
6
1. 8 30.7 10 80.49 10 N/ 2 0.005/ 2 0.2IWdl−
− σσ × ×=× = × =× σ
∴   Additional deflection due to load W1,
         631 2 1 0.49 103.33 10 3.327 1050 9.81WyW−−− ×σ=× δ = × × = × σ×

                    2
11cceeyN
N±±==
 ω −− ω           . . . (Substituting ccNω= and Nω= )
   312 3
20.25 103 . 3 2 71 0 0 . 3 21 0
10.75c
cN
N−−−±××σ = = ± ×
− 

       31 2 9 20.32 10 / 3.327 10 0.0962 10 N / m−−σ= × × = ×  …( Taking + ve sign )
          = 96.2 × 106 N/m2 = 96.2 MN/m2 Ans.
Example 23.8. A vertical steel shaft 15 mm diameter is held in long bearings 1 metre

apart and carries at its middle a disc of mass 15 kg. The eccentricity of the centre of gravity of the
disc from the centre of the rotor is 0.30 mm.
The modulus of elasticity for the shaft material is 200 GN/m2 and the permissible stress is
70 MN/m2. Determine : 1. The critical speed of the shaft and 2. The range of speed over which it
is unsafe to run the shaft. Neglect the mass of the shaft.
[For a shaft with fixed end carrying a concentrated load (W) at the centre assume 3
192Wl
EIδ= ,

192Wl
EIδ= ,
and M = .
8Wl, where δ and M are maximum deflection and bending moment respectively ].
Solution.  Given : d = 15 mm = 0.015 m ; l = 1 m ; m = 15 kg ; e = 0.3 mm
= 0.3 × 10–3 m ; E= 200 GN/m2 = 200 × 109 N/m2 ; σ = 70 MN/m2 = 70 × 106 N/m2
44 9 4(0.015) 2.5 10 m64 64Id− ππ=×= =×
1. Critical speed of the shaft
Since the shaft is held in long bearings, therefore it is assumed to be fixed at both ends. We

know that the static deflection at the centre of shaft,
                     33
3
9915 9.81 11.5 10 m192 192 200 10 2.5 10Wl
EI−
−××δ= = = ×
××× ×    …(∵ W = m.g)

936       /G6C               Theory of Machines
∴    Natural frequency of transverse vibrations,
     30.4985 0.498512.88Hz
1.5 10nf
−== =
δ×
We know that the critical speed of the shaft in r.p.s. is equal to the natural frequency of
transverse vibrations in Hz.
∴     Critical speed of the shaft,
     Nc = 12.88 r.p.s. = 12.88 × 60 = 772.8 r.p.m.  Ans.
Let       N1 and N2 = Minimum and maximum speed respectively.

When the shaft starts rotating, the additional dynamic load ( W1 = m1.g) to which the shaft
is subjected may be obtained from the relation
    
1M
Iyσ= or
1.IMyσ=
Since       11.. .
88Wl mg lM==  ,    and    12dy=, therefore
1.. .
8/ 2mg l I
dσ=
or      69
18 2 8 2 70 10 2.5 1019 kg. . 0.015 9.81 1Imdgl−×× σ × ×× × × ×== =××
∴   Additional deflection due to load W1 = m1g,
   33 11 191.5 10 1.9 10 m
15Wmy
Wm−−=× δ =× δ =×× =×
      
2
1cey±=
ω−ω     or    21
1cy
e N
N±=

2
1cey±=
1cy
e N
N±=
−
. . . (Substituting, ccNω=  , and Nω= )
∴     3
321.9 10 1
0.3 101cN
N−
−×±=
×−      or    20.310 . 1 61.9cN
N−= ± = ±
         2
10 . 1 61 . 1 6cN
N=± = or     0.84
 . . . (Taking first plus sign and then negative sign)
or      
1.16cNN=           or      
0.84cN

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          937
∴ 1772.8718r.p.m.
1.16 1.16cNN===
and 2772.8843 r.p.m.
0.84 0.84cNN===
Hence the range of speed is from 718 r.p.m. to 843 r.p.m. Ans.

23.13. Frequency of Free Damped Vibrations (Viscous Damping)23.13. Frequency of Free Damped Vibrations (Viscous Damping)23.13. Frequency of Free Damped Vibrations (Viscous Damping)23.13. Frequency of Free Damped Vibrations (Viscous Damping)23.13. Frequency of Free Damped Vibrations (Viscous Damping)
We have already discussed that the motion of a body is resisted by frictional forces. In
vibrating systems, the effect of friction is referred to as damping. The damping provided by fluid

resistance is known as viscous damping .
We have also discussed that in damped
vibrations, the amplitude of the resulting vibrationgradually diminishes. This is due to the reason thata certain amount of energy is always dissipated toovercome the frictional resistance. The resistanceto the motion of the body is provided partly by themedium in which the vibration takes place andpartly by the internal friction, and in some casespartly by a dash pot or other external dampingdevice.

Consider a vibrating system, as shown in
Fig. 23.17, in which a mass is suspended from oneend of the spiral spring and the other end of whichis fixed. A damper is provided between the massand the rigid support.
Let   m= Mass suspended from the spring,
 s= Stiffness of the spring,
  x= Displacement of the mass from
the mean position at time t,
δ= Static deflection of  the spring
=m.g/s , and
 c= Damping coefficient or the damping
    force per unit velocity.

force per unit velocity.
Since in viscous damping, it is assumed that the frictional
resistance to the motion of the body is directly proportional tothe speed of the movement, therefore
Damping force or frictional force on the mass acting in
opposite  direction to the motion of the mass
dxc
dt=×
Accelerating force on the mass, acting along  the
motion of the mass
2
2dxm
dt=×Riveting Machine

direct example of the current chapter.Fig. 23.17.  Frequency of free damped
vibrations.

938       /G6C               Theory of Machines
and spring force on the mass, acting in opposite  direction to the motion of the mass,
          = s.x
Therefore the equation of motion becomes
         2
2.dx d xmc s x
dt dt×= − × + 
 …(Negative sign indicates that the force opposes the motion)
or      2
2.0dx d xmc s xdt dt×+ × + =
or    * 2
20dx c d x sxmd t m dt+× +× =
This is a differential equation of the second order. Assuming a solution of the form

x = ek t where k is a constant to be determined. Now the above differential equation reduces to
2.. 0kt kt kt cske k e emm+× +× = …22
2, and .kt kt dx d xke k edt dt
==
∵
or20cskk
mm+× +=  . . . (i)
and       2
4
2cc s
mm mk−± − ×=
          2
22cc s
mm m=− ± − 
∴     The two roots of the equation are
     2
122cc skmm m=− + − 
and       2
222cc skmm m=− − − 
The most general solution of the differential equation (i) with its right hand side equal to

zero has only complementary function and it is given by
       1212kt k txC e C e=+  . . . (ii)
where C1 and C2 are two arbitrary constants which are to be determined from the initial conditions
of the motion of the mass.
It may be noted that the roots k1 and k2 may be real, complex conjugate (imaginary) or
equal. We shall now discuss these three cases as below :
* A system described by this equation is said to be a single degree of freedom harmonic oscillator with
viscous damping.

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          939
1. When the roots are real (overdamping)
If  2
2cs
mm>, then the roots k1 and k2 are real but negative. This is a case of overdamping
or large damping and the mass moves slowly to the equilibrium position. This motion is known as
aperiodic . When the roots are real, the most general solution of the differential equation is
       1212kt k txC e C e=+
         22
22 2 2
12cc s c c sttmm m m m m

22
22 2 2
Ce C e   −+ − −− −      =+
Note :  In actual practice, the overdamped vibrations are avoided.
2. When the roots are complex conjugate (underdamping)
If 2
2sc
mm>, then the radical ( i.e. the term under the square root) becomes negative.
The two roots k1 and k2 are then known as complex conjugate. This is a most practical case of
damping and it is known as underdamping  or small damping . The two roots are
      2

2
122cs ckimmm=− + − 
and      2
222cs ckimmm=− − − 
where i is a Greek letter known as iota and its value is 1−. For the sake of mathematical calcu-
lations, let
    2;( ) ;2ncsamm== ω  and 2
22()2dnscamm−= ω = ω −
Therefore the two roots may be written as
     1 d ka i=− + ω  ;     and     2 d ka i=− − ω
We know that the general solution of a differential equation is
       12 () ()
12 1 2dd ai t ai t kt ktxC e C e C e C e−+ ω −− ω=+ = +
         .

.
12()ddit i t ateC e C eω− ω −=+    …(Using em + n = em × en) …(iii)
Now according to Euler’s theorem
   cos siniei+θ=θ +θ  ; and cos siniei−θ=θ −θ
Therefore the equation (iii) may be written as
       []12(cos . sin . ) (cos . sin . )at
dd dd x e C t it C t it−=ω + ω + ω − ω
          []12 12() c o s . () s i n . )at
dd eC C t i C C t−=+ ω + − ω
Let            12 , CC A+=  and 12()iC C B−=

940       /G6C               Theory of Machines
∴        (c o s . s i n . )at
dd xe A tB t−=ω + ω  . . . (iv)
Again, let  cos AC=θ , and sin BC=θ , therefore
       22CA B=+  , and    tanB
Aθ=
Now the equation (iv) becomes
       (c o sc o s . s i ns i n . )at
dd xe C tC t−=θ ω + θ ω
          cos ( . )at
d Ce t−=ω − θ . . . (v)
If t is measured from the instant at which the mass m is released after an initial displace-
ment A, then

       cos AC=θ . . . [Substituting x = A and t = 0 in equation (v)]
and            when 0θ= , then A = C
∴    The equation  (v) may be written as
      cos .at
d xA e t−=ω  . . . (vi)
where      2
22–( )2dnscammω= = ω −  ; and 2ca
m=
We see from equation (vi), that the motion of the mass is simple harmonic whose circular
damped frequency is dω and the amplitude is atAe− which diminishes exponentially with time as

shown in Fig. 23.18. Though the mass eventually returns to its equilibrium position because of its
inertia, yet it overshoots and the oscillations may take some considerable time to die away.
Fig. 23.18. Underdamping or small damping.
We know that the periodic time of vibration,
 22 222 2
()
2p
dnt
a sc
mmππ π== =
ωω− −
and frequency of damped vibration,
    2
22 11 1()22 2 2ddn
pscf atm mω == = ω−= − ππ π  . . . (vii)

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          941
Note :  When no damper is provided in the system, then c = 0. Therefore the frequency of the undamped
vibration,
      1
2nsfm=π
. . . [Substituting c = 0, in equation (vii)]
It is the same as discussed under free vibra-
tions.
3. When the roots are equal (critical damping)
If 2
2cs
mm=, then the radical becomes
zero and the two roots k1 and k2 are equal. This is a

case of critical damping.  In other words, the critical
damping is said to occur when frequency of damped
vibration ( fd) is zero ( i.e. motion is aperiodic). This
type of damping is also avoided because the massmoves back rapidly to its equilibrium position, inthe shortest possible time.
For critical damping, equation (ii) may be
written as
      
−−ω=+ =+212 12() ()nctt m xCC e CC e... == ω
∵
2ncs
mm

∵
2ncs
mm
Thus the motion is again aperiodic. The critical damping coefficient ( cc) may be obtained
by substituting cc for c in the condition for critical damping, i.e.
          2
2cc s
mm=      or     22cnscm mm== × ω
The critical damping coefficient is the amount of damping required for a system to be
critically damped.

23.14.23.14.23.14.23.14.23.14. Damping Factor or Damping RatioDamping Factor or Damping RatioDamping Factor or Damping RatioDamping Factor or Damping RatioDamping Factor or Damping Ratio
The ratio of the actual damping coefficient ( c) to the critical damping coefficient ( cc) is
known as damping factor  or damping ratio . Mathematically,
Damping factor           2.cncc
cm==ω. . . (2 . ) cnc=π ω∵

The damping factor is the measure of the relative amount of damping in the existing system
with that necessary for the critical damped system.
23.15.23.15.23.15.23.15.23.15. Logarithmic DecrementLogarithmic DecrementLogarithmic DecrementLogarithmic DecrementLogarithmic Decrement
It is defined as the natural logarithm of the amplitude reduction factor. The amplitude
reduction factor is the ratio of any two successive amplitudes on the same side of the mean position.

If x1 and x2 are successive values of the amplitude on the same side of the mean position,
and is not a direct example of the current chapter.In a disc brake, hydraulic pressure forces
friction pads to squeeze a metal disc that
rotates on the same axle as the wheel.
Here a disc brake is being tested.

942       /G6C               Theory of Machines
as shown in Fig. 23.18, then amplitude reduction factor,
    1
()2p
patat
at txA eexAe−
−+==  = constant
where tp is the period of forced oscillation or the time difference between two consecutive amplitudes.
As per definition, logarithmic decrement,
       1
2log logpat xexδ= =

or                   1
22222log .
()ep
dnxaat axa π× πδ= = = × =ω  ω−
. . . 22()dn aω= ω −∵
         2
222
()2nc
m
c
m×π
=
ω−. . . 2ca
m=∵

222
()2nc
m
c
m×π
=
m=∵
          
2222 2
112.nc
ncc
c m
cccmc×π×π==
  ω− −  ω           …(2 . )cncm=ω∵
         222
()cc
ccπ×=
−
In general, amplitude reduction factor,
     3 12
234 1....pat n
nxx xxexxx x +=== = =  = constant
∴     Logarithmic decrement,
      2212log .
()n
ep
ncx catxcc + π×δ= = =
 −
Example 23.9. A vibrating system consists of a mass of 200 kg, a spring of stiffness

80 N/mm and a damper with damping coefficient of 800 N/m/s. Determine the frequency of vibration
of the system.
Solution.  Given : m = 200 kg ; s = 80 N/mm = 80 × 103 N/m ; c = 800 N/m/s
We know that circular frequency of undamped vibrations,
     380 10
200ns
m×ω= =  = 20 rad/s

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          943
and circular frequency of damped vibrations,
   22 2 2() () ( / 2 ) dn n ac m ω= ω − = ω − …(/ 2 )acm=∵
= 22(20) (800 / 2 200) 19.9 rad/s−× =
∴    Frequency of vibration of the system,
/ 2 19.9/ 2 ddf=ω π= π  = 3.17 Hz   Ans.
Example 23.10. The following data are given for a vibratory system with viscous damp-
ing:
Mass = 2.5 kg ; spring constant = 3 N/mm and the amplitude decreases to 0.25 of the

initial value after five consecutive cycles.
Determine the damping coefficient of the damper in the system.
Solution.  Given : m = 2.5 kg ; s = 3 N/mm = 3000 N/m ; x6 = 0.25 x1
We know that natural circular frequency of vibration,
                  3000
2.5ω= =ns
m = 34.64 rad/s
Let             c  = Damping coefficient of the damper in N/m/s,
x1 = Initial amplitude, and
        x6 = Final amplitude after five consecutive cycles = 0.25 x1    …(Given)
35 12 4
23456xx xx x
xxxxx====

35 12 4
xxxxx====
or5
35 11 2 4 1
6 23456 2xx xx x x x
x xxxxx x=××××= 

∴1/5 1/5
1/5 11 1
26 1(4) 1.320.25xx x
xx x == = =  
 
1
2222log
()e
nxaxa π=×
 ω−
222log (1.32)
(34.64)e a
aπ=×
−     or        
220.2776
1200a
a×π=
−
2
239.50.077
1200a
a=
−                  or      2292.4 0.077 39.5 aa−=
∴         22.335 a=      or     a = 1.53

We know that     a = c/2m      or     c = a × 2m =  1.53 × 2 × 2.5 = 7.65 N/m/s   Ans.

944       /G6C               Theory of Machines
Example 23.11.  An instrument vibrates with a frequency of 1 Hz when there is no damping.
When the damping is provided, the frequency of damped vibrations was observed to be 0.9 Hz.
Find 1. the damping factor, and  2. logarithmic decrement.
Solution.  Given : fn = 1 Hz ; fd = 0.9 Hz
1.  Damping factor
Let      m = Mass of the instrument in kg,
     c = Damping coefficient
or damping force per unit
velocity in N/m/s, and

     cc= Critical damping coefficient in
N/m/s.
We know that natural circular frequency of undamped vibrations,
     2 2 1 6.284ω=π × =π ×=nn f  rad/s
     2 2 0.9 5.66ω=π × =π × =dd f  rad/s
We also know that circular frequency of damped vibrations ( ωd),
   22 225.66 ( ) (6.284)=ω − = −n aa
(5.66)2 = (6.284)2 – a2 or 32 = 39.5 – a2

∴                  a2 = 7.5           or        a  = 2.74
We know that,        a  = c/2m        or        c  = a × 2m = 2.74 × 2 m = 5.48 m N/m/s
and      2 . 2 6.284=ω =×cncm m  = 12.568 m N/m/s
∴    Damping factor,
 / 5.48 /12.568=ccc m m  = 0.436 Ans.
2.  Logarithmic decrement
We know that logarithmic decrement,
       22 2 22 2 5.48 34.4
11.3( ) (12.568 ) (5.48 )ππ ×δ= = =
−− ccm
cc m m = 3.04  Ans.

−− ccm
Example 23.12. The measurements on a mechanical vibrating system show that it has a
mass of 8 kg and that the springs can be combined to give an equivalent spring of stiffness5.4 N/mm. If the vibrating system have a dashpot attached which exerts a force of 40 N when themass has a velocity of 1 m/s, find : 1. critical damping coefficient, 2. damping factor,  3. logarithmic
decrement, and 4. ratio of two consecutive amplitudes.

Solution.  Given : m = 8 kg ; s = 5.4 N/mm = 5400 N/m
Since the force exerted by dashpot is 40 N, and the mass has a velocity of 1 m/s , therefore
Damping coefficient (actual),
       c  = 40 N/m/s
Guitar

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          945
1. Critical damping coefficient
We know that critical damping coefficient,
     54002. 2 288cnscm mm=ω =× = ×  = 416 N/m/s  Ans.
2. Damping factor
We know that damping factor
         40
416==
cc
c = 0.096  Ans.
3. Logarithmic decrement
     22 2 222 4 0
( ) (416) (40)ππ ×δ= =
−− cc
cc = 0.6  Ans.
4. Ratio of two consecutive amplitudes

Let     xn and xn+1 = Magnitude of two consecutive amplitudes,
      
1log
+δ=n
e
nx
x or 0.6
1(2.7)δ
+==n
nxex = 1.82  Ans.
Example 23.13. A mass suspended from a helical
spring vibrates in a viscous fluid medium whose resistance

varies directly with the speed. It is observed that the frequencyof damped vibration is 90 per minute and that the amplitudedecreases to 20 % of its initial value in one complete vibration.Find the frequency of the free undamped vibration of thesystem.
Solution.  Given : f
d = 90/min = 90/60 = 1.5 Hz
We know that time period,
     tp = 1/ fd =  1/1.5 = 0.67 s
Let       x1 = Initial amplitude, and
      x2 = Final amplitude after one
  complete vibration
= 20% x1 = 0.2 x1
 . . . (Given)

       1
2log .=
epxatx    or   1
1log 0.670.2=×exax
∴                      loge 5 = 0.67 a  or   1.61 = 0.67 a or  a = 2.4 . . . (∵ loge 5 = 1.61)
Helical spring suspension of a
two-wheeler.

946       /G6C               Theory of Machines
We also know that frequency of free damped vibration,
     22 1()
2=ω −
πdnfa
or                         22 2() ( 2 )ω= π × +nd fa . . . (By squaring and arranging)
          22(2 1.5) (2.4) 94.6=π × + =
∴                   9.726ω=n  rad/s
We know that frequency of undamped vibration,
     9.726
22ω==ππn
nf  = 1.55 Hz  Ans.
Example 23.14. A coil of spring stiffness 4 N/mm supports vertically a mass of 20 kg at

the free end. The motion is resisted by the oil dashpot. It is found that the amplitude at the beginning
of the fourth cycle is 0.8 times the amplitude of the previous vibration. Determine the dampingforce per unit velocity. Also find the ratio of the frequency of damped and undamped vibrations.
Solution.  Given : s = 4 N/mm = 4000 N/m ; m = 20 kg
Damping force per unit velocity
Let         c  = Damping force in newtons per unit velocity i.e. in N/m/s
       x

x
n = Amplitude at the beginning of the third cycle,
    xn+1 = Amplitude at the beginning of the fourth cycle = 0.8 xn
We know that natural circular frequency of motion,
      400014.1420ω= = =ns
m rad/s
and     
222log1() π=×+ ω−n
e
nnxaxa
or
222log0.8(14.14) π=×
 −n
e
nxaxa
         
22log 1.25
200π=×
−e a
a    or    
220.223
200π=×
−a
a
Squaring both sides
   22 2
2243 9 . 50.05
200 200×π==
−−aa
aa

22 2
−−aa
aa
      0.05 × 200 – 0.05 a2 = 39.5 a2               or      39.55 a2 = 10
∴ a2 = 10 / 39.55 = 0.25   or      a  = 0.5
We know that        a  = c/2m
∴ c = a × 2m = 0.5 ×2 × 20 = 20 N/m/s  Ans.
Ratio of the frequencies
Let      1nf = Frequency of damped vibrations = 2dω
π

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          947
    2nf = Frequency of undamped vibrations = 2nω
π
∴
                                      22 22
1
2() 2 (14.14) (0.5)
21 4 . 1 4nd d n
nn n nfa
fωω ω −π−=×== =πω ω ω
. . . ()22()dn a ω= ω −∵
          = 0.999 Ans.
Example 23.15. A machine of mass 75 kg is mounted on springs and is fitted with a
dashpot to damp out vibrations. There are three springs each of stiffness 10 N/mm and it is found

that the amplitude of vibration diminishes from 38.4 mm to 6.4 mm in two complete oscillations.Assuming that the damping force varies as the velocity, determine : 1. the resistance of the dash-
pot at unit velocity ; 2. the ratio of the frequency of the damped vibration to the frequency of the
undamped vibration ; and 3. the periodic time of the damped vibration.
Solution.  Given : m = 75 kg ; s = 10 N/mm = 10 ×10
3 N/m ; x1 = 38.4 mm = 0.0384 m ;
x3 = 6.4 mm = 0.0064 m

Since the stiffness of each spring is 10 × 103 N/m and there are 3 springs, therefore total
stiffness,
       3331 01 0 3 01 0s=× × = ×  N/m
     330 102075ns
m×ω= = =  rad/s
1. Resistance of the dashpot at unit velocity
Let         c  =  Resistance of the dashpot in newtons at unit velocity i.e. in
  N/m/s,
       x2 = Amplitude after one complete oscillation in metres, and

x3 = Amplitude after two complete oscillations in metres.
We know that      12
23xx
xx=
∴2
11
23xx
xx=
. . . 2
11 21 1 1
32 32 2 2xx xx x x
xxxxx x=×=×=  ∵
or      1/2 1/2
11
230.03842.450.0064xx
xx == =  
       1
2222log
()e
 ω−

948       /G6C               Theory of Machines
       222log 2.45
(20)e a
aπ=×
−
220.8951
400a
a×π=
−  or   2
239.50.8
400a
a×=
− . . . (Squaring both sides)
∴     a2 = 7.94      or    a = 2.8
We know that      a = c/2m
∴ c = a × 2m = 2.8 × 2 × 75 = 420 N/m/s  Ans.
2. Ratio of the frequency of the damped vibration to the frequency of undamped vibration
Let    1nf = Frequency of damped vibration = 2dω
π
  2nf = Frequency of undamped vibration = 2nω
π
∴    22 22
1
2() (20) (2.8) 2
22 0n nd d

π
∴    22 22
1
22 0n nd d
nn n na f
fω− − ωωπ=×== =πω ω ω = 0.99  Ans.
3. Periodic time of damped vibration
We know that periodic time of damped vibration
        22 2 222 2
( ) (20) (2.8) dn aππ π== =ωω− − = 0.32 s  Ans.
Example 23.16. The mass of a single degree damped vibrating system is 7.5 kg and makes
24 free oscillations in 14 seconds when disturbed from its equilibrium position. The amplitude of

vibration reduces to 0.25 of its initial value after five oscillations. Determine : 1. stiffness of the
spring, 2. logarithmic decrement, and 3. damping factor, i.e. the ratio of the system damping to
critical damping.
Solution.  Given : m = 7.5 kg
Since 24 oscillations are made in 14 seconds, therefore frequency of free vibrations,
   fn = 24/14 = 1.7
and   2 2 1.7 10.7 nn f ω=π × =π × =  rad/s
1. Stiffness of the spring
Let       s  =  Stiffness of the spring in N/m.

We know that      2() /n smω=  or 22( ) (10.7) 7.5n sm=ω =  = 860 N/m  Ans.
2. Logarithmic decrement
Let      x1 =  Initial amplitude,
x6 = Final amplitude after five oscillations = 0.25 x1... (Given)
∴          5
35 11 2 4 1
xx x x x x x=××××= 
12345
23456xxxxx
xxxxx====
…∵

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          949
or      1/5 1/5
1/5 11 1
 
    1
2log log 1.32eex
xδ= =
 = 0.28  Ans.
3. Damping factor
Let         c  = Damping coefficient for the actual system, and
       cc = Damping coefficient for the critical damped system.
We know that logarithmic decrement ( δ),
 22 22220.28
( ) (10.7)naa
aa×π ×π==
ω− −
2
239.50.0784
114.5a
a×=

aa×π ×π==
ω− −
2
114.5a
a×=
         8.977 – 0.0784 a2 = 39.5 a2     or       a2 = 0.227     or    a = 0.476
We know that             a = c/2m         or    c = a × 2m = 0.476 × 2 × 7.5 = 7.2 N/m/s Ans.
and       2. 27 . 51 0 . 7cncm=ω = ××  = 160.5 N/m/s  Ans.
∴              Damping factor = c/cc = 7.2 / 160.5 = 0.045  Ans.

23.16.23.16.23.16.23.16.23.16. Frequency of Under Damped Forced VibrationsFrequency of Under Damped Forced VibrationsFrequency of Under Damped Forced VibrationsFrequency of Under Damped Forced VibrationsFrequency of Under Damped Forced Vibrations
Consider a system consisting of spring, mass and
damper as shown in Fig. 23.19. Let the system is acted
upon by an external periodic ( i.e. simple harmonic)
disturbing force,
      cos . xFF t=ω
where                     F  = Static force, and

                   ω= Angular velocity of
     the periodic disturbing
   force.
When the system is constrained to move in vertical
guides, it has only one degree of freedom. Let at sometime
t, the mass is displaced downwards through a distance x
from its mean position.
Using the symbols as discussed in the previous article, the equation of motion may be
written as
2
2.c o s .dx d xmc s x F t
dt dt×= − × − + ω
or    2

or    2
dt dt×+ × + = ω . . . (i)Fig. 23.19.  Frequency of under
damped forced vibrations.

950       /G6C               Theory of Machines
This equation of motion may be solved either by differential equation method or by graphi-
cal method as discussed below :
1. Differential equation method
The equation (i) is a differential equation of the second degree whose right hand side is
some function in t. The solution of such type of differential equation consists of two parts ; one
part is the complementary function and the second is particular integral. Therefore the solution

may be written as
        x  = x1 + x2
where         x1 = Complementary function, and
      x2 = Particular integral.
The complementary function is same as discussed in the previous article, i.e.
     1 cos ( )at
d xC e t−=ω − θ  . . . (ii)
where C and θare constants. Let us now find the value of particular integral as discussed below :
Let the particular integral of equation (i) is given by
      21 2 sin . cos . xB t B t=ω +ω   . . . (where B1 and B2 are constants)
∴12.c o s. .s i n.dxBt Bt

dt=ωω −ωω
and   222
122.s i n . .c o s .dxBt B t
dt=− ω ω − ω ω
Substituting these values in the given differential equation (i), we get
    22
12 1 2 1 2( . s i n . . c o s .) ( . c o s . . s i n .) ( s i n . c o s .)mB t B t c B tB t s B tB t−ω ω − ω ω+ ω ω − ωω+ ω + ω
          cos .Ft=ω
or22
12 1 2 1 2 (. . . . . ) s i n . (. . . . . ) c o s .mB c B sB t m B c B sB t−ω − ω + ω + − ω+ ω + ω
          cos .Ft=ω
or22

cos .Ft=ω
or22
12 1 2 (. ) . . s i n .. . (. ) c o s .sm B cB t cB sm B t −ω − ω ω +ω+−ω ω 
          cos . 0sin .Ft t=ω + ω
Comparing the coefficients of sin ωt and cos ωt on the left hand side and right hand side
separately, we get
    2
12 (. ) . .0sm B c B−ω − ω = . . . (iii)
and    2
12.. ( . )cB sm B Fω+−ω =  . . . (iv)
Now from equation (iii)
    2
12 (. ) . .sm B cB−ω =ω
∴2
21.
.smBBc−ω=×ω. . . (v)
Substituting the value of B2 in equation (iv)
22

22
11(. ) (. )...sm smcB B Fc−ω −ωω+ ×=ω

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          951
22 2 2
11 .. ( .) . .cB s m B c Fω+ − ω = ω
22 2 2
1 .(. ) . . Bc sm c Fω+ − ω =ω
∴       122 2 2..
.(. )cFB
cs mω=
ω+ − ω
and                2
222 2 2.. .
. .(. )sm cFBc cs m−ω ω=×ω ω+ − ω . . . [From equation (v)]
          2
22 2 2(. )
.(. )Fs m
cs m−ω=
ω+ − ω
∴     The particular integral of the differential equation (i) is
      21 2 sin . cos . xB t B t=ω +ω
          2
2 2 22 2 2 22.. ( . )sin . cos .

2
.(. ) .(. )cF F smtt
cs m cs mω− ω=× ω +× ω
ω+ − ω ω+ − ω
          2
22 2 2.s i n. ( . ) c o s.
.(. )Fct s m t
cs m =ω ω + − ω ωω+ − ω. . . (vi)
Let      .s i n ;cXω= φ  and 2.c o s sm X−ω = φ
∴22 2 2.(. ) Xc s m=ω + − ω . . . (By squaring and adding)
Note : This picture is given as additional information and is not a direct example of the current chapter.This machine performs pressing operation, welding operation and material handling.

952       /G6C               Theory of Machines
and    2.tan
.c
smωφ=
−ω       or       1
2.tan
.c
sm−ωφ= −ω
Now the equation (vi) may be written as
       [] 222 2 2sin .sin cos cos
.(. )FxX t X t
cs m=φ ω + φ ω
ω+ − ω
           22 2 2.cos ( . )
.(. )FXt
cs m=× ω − φ
ω+ − ω
                                  22 2 2
22 2 2.(. )cos ( . )
.(. )Fc s mt
cs mω+ − ω=× ω − φ
ω+ − ω
               22 2 2cos ( . )
.(. )Ft
cs m=× ω − φ
ω+ − ω

.(. )Ft
cs m=× ω − φ
ω+ − ω
∴     The complete solution of the differential equation  (i) becomes
                                x  = x1 + x2
                     22 2 2. cos ( . ) cos( . )
.(. )at
dFCe t t
cs m−=ω − θ + × ω − φ
ω+ − ω
In actual practice, the value of the complementary function x1 at any time t is much smaller
as compared to particular integral x2. Therefore, the displacement x, at any time t, is given by the
particular integral x2 only.
∴                     
22 2 2cos ( . )

∴                     
.(. )Fxt
cs m=× ω − φ
ω+ − ω... (vii)
This equation shows that motion is simple harmonic whose circular frequency is ω and the
amplitude is 22 2 2.(. )F
cs mω+ − ω.
A little consideration will show that the frequency of forced vibration is equal to the angular
velocity of the periodic force and the amplitude of the forced vibration is equal to the maximum
displacement of vibration.
∴     Maximum displacement or the amplitude of forced vibration,

22 2 2.(. )maxFx
cs m=
ω+ − ω . . . (viii)
Notes : 1. The equations (vii) and (viii) hold good when steady vibrations of constant amplitude takes
place.
           2. The equation (viii) may be written as
                          22 2 2
22/
.(. )maxFsx
cs m
ss=
ω− ω+
. . . (Dividing the numerator and denominator by s)

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          953
                          222 2
2..1ox
cm
s s=
ωω+− . . . (Substituting F/s = xo)
where xo is the deflection of the system under the static force F. We know that the natural frequency of free
vibrations is given by
 2() /n smω=
∴      222 2
22.1
()omax
nxx
c
s=
ωω+−ω. . . (ix)
3. When damping is negligible, then c = 0.
∴                     2
22 2 2 2
2() /
() ()1
()oo n o
max
nn
nxx x s mxω×== =

2() /
() ()1
()oo n o
max
nn
ωω − ω ω − ω−
ω         . . . 2() /n sm ω=∵
∴  22()nF
m=ω− ω . . . () .o Fx s=∵  . . . (x)
4. At resonance nω=ω . Therefore the angular speed at which the resonance occurs is
 ns
mω=ω =  rad/s
and         .max o
nsxxc=×ω . . . [From equation (ix)]
The solution of the equation of motion for a forced and damped vibration may be easily
obtained by graphical method as discussed below :

Let us assume that the displacement of the mass ( m) in the system, as shown in Fig. 23.19,
under the action of the applied simple harmonic force F cos .tω is itself simple harmonic, so that
it can be represented by the equation,
  cos ( ) xA t=ω − φ
where A is the amplitude of vibration.
Now differentiating the above equation,
[] .s i n (. ) .c o s 9 0 (. )dxAt A t
dt=−ω ω −φ =ω °+ ω −φ
and []222
2. c o s (. ) . c o s 1 8 0 (. )dxAt A t

954       /G6C               Theory of Machines
∴     Elastic force i.e. the force required to extend the spring
          .. c o s ( . )sx sA t== ω − φ
Disturbing force i.e. the force required to overcome the resistance of dashpot
          [] .. c o s 9 0 (. )dxcc A tdt=× =ω ° + ω− φ
and inertia force i.e. the force required to accelerate the mass m
          []22
2. . cos 180 ( . )dxmm A t
dt=× =ω ° + ω − φ
Fig. 23.20. Graphical method.

The algebraic sum of these three forces at any given instant must be equal to the applied
force cosFtω. These forces are represented graphically in Fig. 23.20 ( a). The vector OP repre-
sents, to some suitable scale, the elastic force (of maximum value s.A), at an inclination (. )tω− φ
to the vertical. The vector OQ (of maximum value .cAω) and vector OR (of maximum value

2.mAω ) represents, to the same scale, the disturbing force and inertia force respectively. The vec-
tors OP, OQ and OR are at successive intervals of 90°.
The projected lengths Op, Oq and Or represent the instantaneous values of these forces at
time t and Os (the algebraic sum of Op, Oq and Or) must represent the value F cos .tωof the
applied force at the same instant. Thus the force vector OS must be the vector sum of OP, OQ and

OR or force F must be the vector sum of s.A, ..cAω and 2..mAω , as shown in Fig. 23.20 ( b). From
      22 22() () ( ) () Fo c o d c d o a a d c d== + = − +
          22 2 2 2 2 2(. . . ) (. . ) ( . ) .sA m A c A A s m c=− ω + ω =− ω + ω
∴                
22 2 2(or )
(. ) .maxFAx
sm c=
−ω + ω. . . (Same as before)
and    2
22.. .tan
.. . .cd c A c
od sA m A s mωωφ= = =
−ω −ω. . . (Same as before)(a) (b)

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          955
23.17. Magnification Factor or Dynamic Magnifier23.17. Magnification Factor or Dynamic Magnifier23.17. Magnification Factor or Dynamic Magnifier23.17. Magnification Factor or Dynamic Magnifier23.17. Magnification Factor or Dynamic Magnifier
It is the ratio of maximum displacement of the forced vibration (xmax ) to the deflection

due to the static force  F(xo). We have proved in the previous article that the maximum displace-
ment or the amplitude of forced vibration,
   222 2
22.1
()o
max
nxx
c
s=
ωω+−ω
    Fig. 23.21.  Relationship between magnification factor and phase angle for different values of /nωω .
∴    Magnification factor or dynamic magnifier,
       
222 2
221
.1
()max
o
nxDx
c
s==
ωω+−ω... (i)
          2 22
21
2.1. () cn nc
c=
 ωω+−  ω ω  
. . . 2.2 . 2 . 2 .

. 2( ) 2 cn ncc c c
s sc m mm
ωω ω ω== =ω ω  ×∵

956       /G6C               Theory of Machines
The magnification factor or dynamic magnifier gives the factor by which the static deflection
produced by a force F (i.e. xo) must be multiplied in order to obtain the maximum amplitude of the
forced vibration ( i.e. xmax) by the harmonic force F cos .ωt
∴ max oxx D=×
Fig. 23.21 shows the relationship between the magnification factor ( D) and phase angle φ
for different value of /ωωn and for values of damping factor c/cc = 0.1, 0.2 and 0.5.

Notes: 1. If there is no damping ( i.e. if the vibration is undamped), then c = 0. In that case, magnification
factor,
      2
22 22
21( )
()
1
()max n
o n
nxDxω== =
ω− ωω−ω
2. At resonance, nω=ω . Therefore magnification factor,
      .max
onxsDxc==ω
Example 23.17.  A single cylinder vertical petrol engine of total mass 300 kg is mounted
upon a steel chassis frame and causes a vertical static deflection of 2 mm. The reciprocating parts

of the engine has a mass of 20 kg and move through a vertical stroke of 150 mm with simpleharmonic motion. A dashpot is provided whose damping resistance is directly proportional to thevelocity and amounts to 1.5 kN per metre per second .
Considering that the steady state of vibration is reached ; determine :  1. the amplitude of
forced vibrations, when the driving shaft of the engine rotates at 480 r.p.m., and 2. the speed of the
driving shaft at which resonance will occur.

Solution : Given. m = 300 kg; 
δ = 2 mm = 2 × 10–3 m ; m1 = 20 kg ; l = 150 mm
= 0.15 m ; c= 1.5 kN/m/s = 1500 N/m/s ; N = 480 r.p.m. or 2 480/ 60ω= π×  = 50.3 rad/s
Depending upon the case bridges can be treated as beams subjected to
uniformly distributed leads and point loads.

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          957
1. Amplitude of the forced vibrations
We know that stiffness of the frame,
       s  = m.g/δ = 300 × 9.81/2 × 10–3 = 1.47 × 106 N/m
Since the length of stroke ( l) = 150 mm = 0.15 m, therefore radius of crank,
        r  = l/ 2 = 0.15 / 2 = 0.075 m
We know that the centrifugal force due to the reciprocating parts or the static force,
     2
1..=ωFm r  = 20 (50.3)2 0.075 = 3795 N

2
∴    Amplitude of the forced vibration (maximum),
   22 2 2.(. )maxFx
cs m=
ω+ − ω
          22 6 2 23795
(1500) (50.3) [1.47 10 300(50.3) ]=
+× −
          3
3 993795 37955.3 10
710 10 5.7 10 500 10−== = ×
× ×+× m
           = 5.3 mm  Ans.
2. Speed of the driving shaft at which the resonance occurs
Let         N  = Speed of the driving shaft at which the resonance occurs in
 r.p.m.
We know that the angular speed at which the resonance occurs,

61.47 1070300×ω=ω = = =ns
m rad/s
∴                  60/ 2 70 60/ 2=ω× π= × πN  = 668.4 r.p.m.  Ans.
Example 23.18.  A mass of 10 kg is suspended from one end of a helical spring, the other
end being fixed. The stiffness of the spring is 10 N/mm. The viscous damping causes the amplitude

to decrease to one-tenth of the initial value in four complete oscillations. If a periodic force of150 cos 50 t N is applied at the mass in the vertical direction, find the amplitude of the forcedvibrations. What is its value of resonance ?
Solution.  Given :  m = 10 kg ; s = 10 N/mm = 10 × 10
3 N/m ; 1
510xx=
Since the periodic force, cos . 150cos50xFF t t=ω = , therefore
Static force,        F  = 150 N
and angular velocity of the periodic disturbing force,
       50ω= rad/s

50ω= rad/s
We know that angular speed or natural circular frequency of free vibrations,
    310 10
10×ω= =ns
m = 31.6 rad/s

958       /G6C               Theory of Machines
Amplitude of the forced vibrations
Since the amplitude decreases to 1/10th of the initial value in four complete oscillations,
therefore, the ratio of initial  amplitude ( x1) to the final amplitude after four complete oscillations
(x5) is given by
                             4
3 11 2 4 1
5 2345 2=×××= 
x xx x x x
xx x x x x        . . . 1234
2345xxxx
xxxx===
∵
∴                   1/4 1/4
1/4 11 1
25 1(10) 1.78/10xx x

1/4 11 1
xx x  == = = 
 . . . 1
510xx=
        1
2222log
() π=×
 ω−e
nxaxa
          222log 1.78
(31.6)π=×
−e a
a or 220.576
1000×π=
−a
a
Squaring both sides and rearranging,
         39.832 a2 = 332     or    a2 = 8.335    or     a = 2.887
We know that           a  = c/2m   or     c = a × 2m = 2.887 × 2 × 10 = 57.74 N/m/s
and deflection of the system produced by the static force F,
       xo = F/s = 150/10 × 103 = 0.015 m

xo = F/s = 150/10 × 103 = 0.015 m
We know that amplitude of the forced vibrations,
  222 2
22.1
()o
max
nxx
c
s=
ωω+−ω
           22 22
320.015 0.015
0.083 2.25(57.74) (50) 50131.6 (10 10 )==
++− × 
           3 0.0159.8 10
1.53−== ×  m = 9.8 mm  Ans.
Amplitude of forced vibrations at resonance
We know that amplitude of forced vibrations at resonance,
    3
010 100.015 0.0822. 57.54 31.6max
nsxxc×=× = × =ω× m = 82.2 mm Ans.

Example 23.19. A body of mass 20 kg is suspended from a spring which deflects 15 mm
under this load. Calculate the frequency of free vibrations and verify that a viscous damping force
amounting to approximately 1000 N at a speed of 1 m/s is just-sufficient to make the motionaperiodic .

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          959
If when damped to this extent, the body is subjected to a disturbing force with a maximum
value of 125 N making 8 cycles/s, find the amplitude of the ultimate motion .
Solution .  Given : m = 20 kg ; δ  = 15 mm = 0.015 m ; c = 1000 N/m/s ; F = 125 N ;
f = 8 cycles/s
Frequency of free vibrations
We know that frequency of free vibrations,
      11 9 . 8 1
2 2 0.015==πδ πngf  = 4.07 Hz Ans.

The critical damping to make the motion aperiodic is such that damped frequency is zero,
i.e.
 2
2=cs
mm
∴          2 .44 . 4=× = = ××δsm gcm s m mm. . . .mgs=δ∵
          20 9.814 20 10230.015×=× ×=  N/m/s
This means that the viscous damping force is 1023 N at a speed of 1 m/s. Therefore a
viscous damping force amounting to approximately 1000 N at a speed of 1 m/s is just sufficient to
make the motion aperiodic. Ans.
Amplitude of ultimate motion

We know that angular speed of forced vibration,
       2 2 8 50.3ω= π× = π× = f  rad/s
and stiffness of the spring,      s  = m.g/δ =  20 × 9.81 / 0.015 = 13.1 × 103 N/m
∴     Amplitude of ultimate motion i.e. maximum amplitude of forced vibration,
  22 2 2.(. )maxFx
cs m=
ω+ − ω
          22 3 2 2125
(1023) (50.3) [13.1 10 20(50.3) ]=
+× −
          366125 125
63.7 10 2600 10 1406 10==
× ×+ × = 1.96 × 10–3 m
           = 1.96 mm  Ans.

Example 23.20. A machine part of mass 2 kg vibrates in a viscous medium. Determine the
damping coefficient when a harmonic exciting force of 25 N results in a resonant amplitude of12.5 mm with a period of 0.2 second. If the system is excited by a harmonic force of frequency4 Hz what will be the percentage increase in the amplitude of vibration when damper is removedas compared with that with damping .

Solution .  Given : m = 2 kg ; F = 25 N ; Resonant x
max = 12.5 mm = 0.0125 m ;
tp = 0.2 s ; f =4H z

960       /G6C               Theory of Machines
Damping coefficient
Let         c  = Damping coefficient in N/m/s.
We know that natural circular frequency of the exicting force,
    2/ 2/ 0 . 2ω=π =πnp t  = 31.42 rad/s
We also know that the maximum amplitude of vibration at resonance ( xmax ),
25 0.7960.0125. 31.42== =ω×nF
cc c or c = 63.7 N/m/s  Ans.
Percentage increase in amplitude
Since the system is excited by a harmonic force of frequency (  f ) = 4 Hz, therefore corre-

sponding circular frequency
       2 2 4 25.14ω= π× = π× = f  rad/s
We know that maximum amplitude of vibration with damping,
 22 2 2.(. )maxFx
cs m=
ω+ − ω
         22 2 2 225
(63.7) (25.14) [2(31.42) 2(25.14) ]=
+−
. . . 22() / o r ()nn sm s m ω= =ω∵
          
6625 250.014317492.56 10 0.5 10== =
×+× m = 14.3 mm
and the maximum amplitude of vibration when damper is removed,
 22 2225 25
710 2[(31.42) (25.14) ] ()max
nFx
m== =
 − ω− ω = 0.0352 m
          = 35.2 mm

m== =
          = 35.2 mm
∴    Percentage increase in amplitude
          35.2 14.3
14.3−=  = 1.46    or    146%  Ans.
Example 23.21. The time of free vibration of a mass hung from the end of a helical spring
is 0.8 second. When the mass is stationary, the upper end is made to move upwards with a
displacement y metre such that y = 0.018 sin 2 πt, where t is the time in seconds measured from

the beginning of the motion. Neglecting the mass of the spring and any damping effects, determinethe vertical distance through which the mass is moved in the first 0.3 second .
Solution.  Given : t
p = 0.8 s ; y = 0.018 sin 2 πt
Let        m  =  Mass hung to the spring in kg, and
        s  = Stiffness of the spring in N/m.
We know that time period of free vibrations ( tp),
     0.8 2m
s=π         or      20.80.01622==πm
s

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          961
If x metres is the upward displacement of mass m from its equilibrium position after time t
seconds, the equation of motion is given by
          2
2() ×= −dxms y x
dt      or      2
20.018sin 2 ×+ = = πmd xxy t
sdt
       20.018sin 2sin cos
21
/π=× +× +
π−ss txA t B tmm
sm
 . . . (where A and B are constants)
          20.018sin 2sin cos

20.018sin 2sin cos
0.0162 0.0162 1 4 0.0162tt tABπ=++
−π×
          sin 7.85 cos7.85 0.05sin 2=++ πAt B t t  . . . (i)
Now when     t = 0, x = 0, then from equation (i), B = 0.
Again when   t = 0, dx/dt = 0.
Therefore differentiating equation (i) and equating to zero, we have
 / 7.85 cos7.85 0.05 2 cos 2 0=+ × π π = dx dt A t t . . . (∵   B = 0 )
or             7.85 cos 7.85 0.05 2 cos 2 At t =− × π π
∴                   0.05 2 / 7.85 0.04 A=− × π =− . . . (∵   t = 0 )

Now the equation (i) becomes
      0.04sin 7.85 0.05sin 2 xt t=− + π . . . (∵   B = 0)  . . . (ii)
∴  Vertical distance through which the mass is moved in the first 0.3 second ( i.e.
when t = 0.3 s),
         0.04sin (7.85 0.3) 0.05sin (2 0.3)=− × + π×
 . . .  [ Substituting t = 0.3 in equation (ii)]
          0.04 0.708 0.05 0.951 0.0283 0.0476 0.0193= − ×+ ×= − + =  m
         =  19.3 mm  Ans.

=  19.3 mm  Ans.
23.18.23.18.23.18.23.18.23.18.  Vibration Isolation and Transmissibility Vibration Isolation and Transmissibility Vibration Isolation and Transmissibility Vibration Isolation and Transmissibility Vibration Isolation and Transmissibility
A little consideration will show that when an
unbalanced machine is installed on the foundation, it produces

vibration in the foundation. In order to prevent these vibrationsor to minimise the transmission of forces to the foundation,the machines are mounted on springs and dampers or on somevibration isolating material, as shown in Fig. 23.22. Thearrangement is assumed to have one degree of freedom, i.e. it
can move up and down only.
It may be noted that when a periodic ( i.e. simple
harmonic) disturbing force F cos 
ωt is applied to a machineFig. 23.22.  Vibration isolation.

962       /G6C               Theory of Machines
of mass m supported by a spring of stiffness s, then the force is transmitted by means of the spring
and the damper or dashpot to the fixed support or foundation.
The ratio of the force transmitted ( FT) to the force applied ( F) is known as the isolation
factor  or transmissibility ratio of the spring support.
We have discussed above that the force transmitted to the foundation consists of the fol-
lowing two forces :

1. Spring force or elastic force which is equal to s. xmax, and
2. Damping force which is equal to c.ω.xmax.
Since these two forces are perpendicular to one another, as shown in Fig.23.23, therefore
the force transmitted,
     22
T (. ) (. . )max max Fs x c x=+ ω
          22 2. maxxs c=+ ω
∴    Transmissibility ratio,
        22 2
T .maxxs c F
FF+ωε= =
  max oFxx D D
s=×=× . . . oFxs=∵
∴2222 2
2..1ωε= + ω = +Dcsc Ds s
          221
cncD

          221
cncD
cω=+ ×ω. . . .2
cncc
scωω=×ω ∵
We have seen in Art. 23.17 that the magnification factor,
      2 22
21
2.1. ()=
 ωω+− ω ω   cn nD
c
c
∴2
2 22
22.1.
2.1. ()ω+ωε=
 ωω+− ω ω  cn
cn nc
c
c
c. . . (i)
When the damper is not provided, then c = 0, and
        21
1(/ )ε=
−ωωn . . . (ii)Fig. 23.23

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          963
From above, we see that when /1 ,ωω> ε n  is negative. This means that there is a phase
difference of 180° between the transmitted force and the disturbing force (c o s . )ω Ft . The value of
/ωωn must be greater than 2 if ε is to be less than 1 and it is the numerical value of ε,
independent of any phase difference between the forces that may exist which is important. It is

therefore more convenient to use equation (ii) in the following form, i.e.
        21
(/ ) 1ε=
ωω −n . . . (iii)
Fig. 23.24 is the graph for different values of damping factor c/cc to show the variation of
transmissibility ratio ( ε) against the ratio /ωω n.
1. When /2ωω=n , then all the curves pass through the point ε = 1 for all values of
damping factor c/cc .
Fig. 23.24. Graph showing the variation of transmissibility ratio.

2. When /2ωω< n , then ε > 1 for all values of damping factor c/cc. This means that the
force transmitted to the foundation through elastic support is greater than the force applied.
3. When /2ωω > n  , then ε < 1 for all values of damping factor c/cc. This shows that
the force transmitted through elastic support is less than the applied force. Thus vibration isolation
is possible only in the range of /2ωω> n .

964       /G6C               Theory of Machines
We also see from the curves in Fig. 23.24 that the damping is detrimental beyond
/2ωω>n and advantageous only in the region /2ωω<n. It is thus concluded that for the
vibration isolation, dampers need not to be provided but in order to limit resonance amplitude,
stops may be provided.
Example 23.22.  The mass of an electric motor is 120 kg and it runs at 1500 r.p.m. The

armature mass is 35 kg and its C.G. lies 0.5 mm from the axis of rotation.  The motor is mounted
on five springs of negligible damping so that the force transmitted is one-eleventh of the impressedforce. Assume that the mass of the motor is equally distributed among the five springs.
Determine : 1. stiffness of each spring; 2. dynamic force transmitted to the base at the
operating speed; and 3. natural frequency of the system.
Solution.  Given m

1 = 120 kg ;  m2 = 35 kg;   r = 0.5 mm = 5 × 10–4 m;  ε = 1 / 11;
N = 1500 r.p.m.  or  ω = 2π × 1500 / 60 = 157.1 rad/s ;
1. Stiffness of each spring
Let                   s  = Combined stiffness of the spring in N-m, and
            ωn = Natural circular frequency of vibration of the machine in
  rad/s.
We know that transmissibility ratio ( ε),
     22
22 2 2 2() () 11
11 () ( 1 5 7 . 1 ) ()
1nn
nn
nωω== =
ω−ω −ω ω−ω

1nn
nn
nωω== =
or                    22 2(157.1) ( ) 11( )nn−ω = ω    or   2( ) 2057ω=n  or   45.35ω=n  rad/s
We know that     1/n smω=
                 2
1( ) 120 2057 246 840 N / mn sm=ω=× =
Since these are five springs, therefore stiffness of each spring
        = 246 840 / 5 = 49 368 N/m  Ans.
2.  Dynamic force transmitted to the base at the operating speed (i.e. 1500 r.p.m.  or  157.1 rad/s)
We know that maximum unbalanced force on the motor due to armature mass,
     22 4

22 4
2 35(157.1) 5 10 432 N Fm r−=ω ⋅ = × =
∴ Dynamic force transmitted to the base,
     T1.4 3 2 3 9 . 2 7 N11FF=ε = × =  Ans.
3. Natural frequency of the system
We have calculated above that the natural frequency of the system,
     45.35ω=n  rad/s   Ans.

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          965
Example 23.23. A machine has a mass of 100 kg and unbalanced reciprocating parts of
mass 2 kg which move through a vertical stroke of 80 mm with simple harmonic motion. The
machine is mounted on four springs, symmetrically arranged with respect to centre of mass, in
such a way that the machine has one degree of freedom and can undergo vertical displacements
only.

only.
Neglecting damping, calculate the combined stiffness of the spring in order that the force
transmitted to the foundation is 1 / 25 th of the applied force, when the speed of rotation of ma-chine crank shaft is 1000 r.p.m.
When the machine is actually supported on the springs, it is found that the damping reduces
the amplitude of successive free vibrations by 25%. Find : 1. the force transmitted to foundation at

1000 r.p.m., 2. the force transmitted to the foundation at resonance, and 3. the amplitude of the
forced vibration of the machine at resonance.
1 = 100 kg ; m2 = 2 kg ; l = 80 mm = 0.08 m ; ε = 1 / 25 ;
N = 1000 r.p.m. or 2 1000 / 60ω= π×  = 104.7 rad/s
Combined stiffness of springs
Let         s  = Combined stiffness of springs in N/m, and
      ωn = Natural circular frequency of vibration of the machine in rad/s.
     22

     22
25 ( ) (104.7) ( )
1ωω== =
ω−ω −ω ω−ωnn
nn
n
or            22 2(104.7) ( ) 25( )−ω = ωnn      or    2( ) 421.6ω=n or 20.5ω=n  rad/s
We know that      1/ω=n sm
∴        2
1( ) 100 421.6=ω=×n sm  = 42 160 N/m  Ans.
1. Force transmitted to the foundation at 1000 r.p.m.
Let      FT = Force transmitted, and
       x1 = Initial amplitude of vibration.

x1 = Initial amplitude of vibration.
Since the damping reduces the amplitude of successive free vibrations by 25%, therefore
final amplitude of vibration,
     210.75=xx
       1
2222log
() ×π=
 ω−e
nxa
xa      or      1
212log0.75421.6 ×π=  − exa
xa
        222
24(0.2877)
421.6×π=
−a
a         or       2
239.50.083
421.6=
−a
a
1... log log 1.333 0.28770.75ee==∵

a
    2235 0.083 39.5−= aa               or    20.884=a     or    a = 0.94

966       /G6C               Theory of Machines
We know that damping coefficient or damping force per unit velocity,
       12 0.94 2 100=× = ××ca m  = 188 N/m/s
and critical damping coefficient,
      2 . 2 100 20.5=ω = ××cncm  = 4100 N/m/s
∴ Actual value of transmissibility ratio,
        2
2 22
22.1.
cn nc
c
c
c
           2
2222 188 104.714100 20.5
2 188 104.7 104.714100 20.5 20.5××+×=
××  +−   ×    10 . 2 2
0.22 629+=

0.22 629+=
+
           1.1040.044
25.08==
We know that the maximum unbalanced force on the machine due to reciprocating parts,
      22
2. . 2(104.7) (0.08/ 2) 877=ω = =Fm r  N . . . (∵   r = l/2 )
∴    Force transmitted to the foundation,
     T . 0.044 877=ε = ×FF  = 38.6 N  Ans. . . . (∵    T/FFε= )
2. Force transmitted to the foundation at resonance
Since at resonance, ω=ω n, therefore transmissibility ratio,
        22
222 2 1881 14100 10 . 0 0 8 410.920.0922 188 2

4100 ×+ +  +   ε= = = =
×   
  c
cc
c
c
c
and maximum unbalanced force on the machine due to reciprocating parts at resonance speed ωn,
      22
2( ) 2(20.5) (0.08/ 2) 33.6=ω= =n Fm r  N . . . (∵  r = l/2 )
∴    Force transmitted to the foundation at resonance,
     T . 10.92 33.6=ε = ×FF  = 367 N Ans.
3. Amplitude of the forced vibration of the machine at resonance
We know that amplitude of the forced vibration at resonance

3 Force transmitted at resonance 3678.7 10Combined stiffness 42 160−== = ×  m
          = 8.7 mm  Ans.

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          967
Example 23.24. A single-cylinder engine of total mass 200 kg is to be mounted on an
elastic support which permits vibratory movement in vertical direction only. The mass of the piston

is 3.5 kg and has a vertical reciprocating motion which may be assumed simple harmonic with astroke of 150 mm. It is desired that the maximum vibratory force transmitted through the elasticsupport to the foundation shall be 600 N when the engine speed is 800 r.p.m. and less than this atall higher speeds.
1. Find the necessary stiffness of the elastic support, and the amplitude of vibration at 800
r.p.m., and

2. If the engine speed is reduced below 800 r.p.m. at what speed will the transmitted force
again becomes 600 N?
1 = 200 kg ; m2 = 3.5 kg ; l = 150 mm = 0.15 mm or r = l/2 = 0.075 m ;
FT = 600 N ; N = 800 r.p.m. or 2 800/ 60ω= π×  = 83.8 rad/s
We know that the disturbing force at 800 r.p.m.,
      F  = Centrifugal force on the piston
2
2..=ωmr  = 3.5 (83.8)2 0.075 = 1843 N
1. Stiffness of elastic support and amplitude of vibration

Let        s  = Stiffness of elastic support in N/m, and
    xmax = Max. amplitude of vibration in metres.
Since the max. vibratory force transmitted to the foundation is equal to the force on the
elastic support (neglecting damping), therefore
Max. vibratory force transmitted to the foundation,
     FT = Force on the elastic support
= Stiffness of elastic support × Max. amplitude of vibration
          = s × xmax 22()nFs
m=×
ω−ω
          22.
.=× =
 ω−ω−FF sss msmm. . . 2()ns

.=× =
mω=∵
∴261843 1843600
200(83.8) 1.4 10×==
−× −ss
ss. . . (Substituting m = m1)
* The equation (x) of Art. 23.16 is
 22()max
nFx
m=ω− ω
Since the max. vibratory force transmitted to the foundation through the elastic support decreases at all
higher speeds ( i.e. above N = 800 r.p.m. or ω = 83.8 rad/s), therefore we shall use
                22()max
nFx
m=ω−ω

968       /G6C               Theory of Machines
or         840 × 106 – 600 s = 1843 s
∴ s = 0.344 × 106 = 344 × 103 N/m  Ans.
and maximum amplitude of vibration,
  22 3 31843 1843
. 200(83.8) 344 10 1056 10maxFx
ms== =
ω− − × ×m
           = 1.745 × 10–3 m = 1.745 mm  Ans.
2. Speed at the which the transmitted force again becomes 600 N
The transmitted force will rise as the speed of the engine falls and passes through reso-

nance. There will be a speed below resonance at which the transmitted force will again equal to
600 N. Let this speed be 1ω rad/s (or  N1 r.p.m.).
∴     Disturbing force, 22 2
21 1 1( ) 3.5( ) 0.075 0.2625( )=ω=ω = ωFm r N
Since the engine speed is reduced below N1 = 800 r.p.m., therefore in this case, max,
amplitude of vibration,
  2 2221 11() () ( )()max
nFF Fxs sm mmm== =
   −ω ω− ω−ω  
and           Force transmitted = 2
1()×
−ωFs
sm
∴                          23 23 11
32 22

−ωFs
sm
∴                          23 23 11
32 22
110.2625( ) 90.3 10 ( )600 344 10
344 10 200( ) 344 10 200( )ω× ω=×× =
×− ω ×− ω
. . . (Substituting m = m1)
63 2 3 2
11 206.4 10 120 10 ( ) 90.3 10 ( )×−×ω= × ω    or   2
1( ) 981ω=
∴                   131.32ω=  rad/s  or  131.32 60 / 2=× πN   = 299 r.p.m.   Ans.
1. A shaft of 100 mm diameter and 1 metre long is fixed at one end and other end carries a flywheel

of mass 1 tonne. Taking Y oung’s modulus for the shaft material as 200 GN/m2, find the natural
frequency of longitudinal and transverse vibrations. [Ans. 200 Hz ; 8.6 Hz]
2. A beam of length 10 m carries two loads of mass 200 kg at distances of 3 m from each end togetherwith a central load of mass 1000 kg. Calculate the frequency of transverse vibrations. Neglect themass of the beam and take I = 10
9 mm4 and E = 205×103 N/mm2. [Ans. 13.8 Hz]

3. A steel bar 25 mm wide and 50 mm deep is freely supported at two points 1 m apart and carries amass of 200 kg in the middle of the bar. Neglecting the mass of the bar, find the frequency oftransverse vibration.
If an additional mass of 200 kg is distributed uniformly over the length of the shaft, what will be
the frequency of vibration ? Take E = 200 GN/m
2. [Ans. 17.8 Hz ; 14.6 Hz]

4. A shaft 1.5 m long is supported in flexible bearings at the ends and carries two wheels each of 50kg mass. One wheel is situated at the centre of the shaft and the other at a distance of 0.4 m fromthe centre towards right. The shaft is hollow of external diameter 75 mm and inner diameter 37.5mm. The density of the shaft material is 8000 kg/m
3. The Young’s modulus for the shaft material is
200 GN/m2. Find the frequency of transverse vibration. [Ans. 33.2 Hz]

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          969
5. A shaft of diameter 10 mm carries at its centre a mass of 12 kg. It is supported by two short
bearings, the centre distance of which is 400 mm. Find the whirling speed : 1. neglecting the massof the shaft, and 2. taking the mass of the shaft also into consideration. The density of shaft materialis 7500 kg/m
3. [Ans. 748 r.p.m.; 744 r.p.m.]

6. A shaft 180 mm diameter is supported in two bearings 2.5 metres apart. It carries three discs ofmass 250 kg, 500 kg and 200 kg at 0.6 m, 1.5 m and 2 m from the left hand. Assuming the mass ofthe shaft 190 kg/m, determine the critical speed of the shaft. Young’s modulus for the material ofthe shaft is 211 GN/m
2. [Ans. 18.8 r.p.m.]

7. A shaft 12.5 mm diameter rotates in long bearings and a disc of mass 16 kg is secured to a shaft atthe middle of its length. The span of the shaft between the bearing is 0.5 m. The mass centre of thedisc is 0.5 mm from the axis of the shaft. Neglecting the mass of the shaft and taking E = 200
GN/m
2, find : 1 critical speed of rotation in r.p.m., and 2. the range of speed over which the stress

in the shaft due to bending will not exceed 120 MN/m2. Take the static deflection of the shaft for a
beam fixed at both ends, i.e. 3
192Wl
EIδ= .                           [Ans. 1450 r.p.m. ; 1184 to 2050 r.p.m.]

8. A vertical shaft 25 mm diameter and 0.75 m long is mounted in long bearings and carries a pulleyof mass 10 kg midway between the bearings. The centre of pulley is 0.5 mm from the axis of theshaft. Find ( a) the whirling speed, and ( b) the bending stress in the shaft, when it is rotating at 1700
r.p.m. Neglect the mass of the shaft and E = 200 GN/m
2. [Ans. 3996 r.p.m ; 12.1 MN/m2]

9. A shaft 12 mm in diameter and 600 mm long between long bearings carries a central mass of 4 kg.If the centre of gravity of the mass is 0.2 mm from the axis of the shaft, compute the maximumflexural stress in the shaft when it is running at 90 per cent of its critical speed. The value ofYoung’s modulus of the material of the shaft is 200 GN/m
2. [Ans. 14.8 kN/m2]

10. A vibrating system consists of a mass of 8 kg, spring of stiffness 5.6 N/mm and a dashpot ofdamping coefficient of 40 N/m/s. Find ( a) damping factor, ( b) logarithmic decrement, and ( c) ratio
of the two consecutive amplitudes. [Ans. 0.094 ; 0.6 ; 1.82]

11. A body of mass of 50 kg is supported by an elastic structure of stiffness 10 kN/m. The motion ofthe body is controlled by a dashpot such that the amplitude of vibration decreases to one-tenth of itsoriginal value after two complete vibrations. Determine : 1. the damping force at 1 m/s ; 2. thedamping ratio, and 3. the natural frequency of vibration. [Ans. 252 N/m/s ; 0.178 ; 2.214 Hz]

12. A mass of 85 kg is supported on springs which deflect 18 mm under the weight of the mass. Thevibrations of the mass are constrained to be linear and vertical and are damped by a dashpot whichreduces the amplitude to one quarter of its initial value in two complete oscillations. Find : 1. themagnitude of the damping force at unit speed, and 2. the periodic time of damped vibration.
[Ans. 435 N/m/s ; 0.27 s]

13. The mass of a machine is 100 kg. Its vibrations are damped by a viscous dash pot which diminishesamplitude of vibrations from 40 mm to 10 mm in three complete oscillations. If the machine ismounted on four springs each of stiffness 25 kN/m, find ( a) the resistance of the dash pot at unit
velocity, and ( b) the periodic time of the damped vibration. [Ans. 6.92 N/m/s ; 0.2 s]

14. A mass of 7.5 kg hangs from a spring and makes damped oscillations. The time for 60 oscillationsis 35 seconds and the ratio of the first and seventh displacement is 2.5. Find ( a) the stiffness of the
spring, and ( b) the damping resistance in N/m/s. If the oscillations are critically damped, what is the
damping resistance required in N/m/s ? [Ans. 870 N/m ; 3.9 N/m/s ; 162 N/m/s]
15. A mass of 5 kg is supported by a spring of stiffness 5 kN/m. In addition, the motion of mass is

controlled by a damper whose resistance is proportional to velocity. The amplitude of vibrationreduces to 1/ 15th of the initial amplitude in four complete cycles. Determine the damping force per
unit velocity and the ratio of the frequencies of the damped and undamped vibrations.
[Ans. 34 N/m/s : 0.994]

16. A mass of 50 kg suspended from a spring produces a statical deflection of 17 mm and when inmotion it experiences a viscous damping force of value 250 N at a velocity of 0.3 m/s. Calculate theperiodic time of damped vibration. If the mass is then subjected to a periodic disturbing forcehaving a maximum value of 200 N and making 2 cycles/s, find the amplitude of ultimate motion.
[Ans. 0.262 s ; 8.53 mm]

970       /G6C               Theory of Machines
Fig. 23.2517. A mass of 50 kg is supported by an elastic structure of total stiffness 20 kN/m. The damping ratio
of the system is 0.2. A simple harmonic disturbing force acts on the mass and at any time t seconds,
the force is 60 cos 10 t newtons. Find the amplitude of the vibrations and the phase angle caused by
the damping. [Ans. 3.865 mm ; 14.93°]

18. A machine of mass 100 kg is supported on openings of total stiffness 800 kN/m and has a rotatingunbalanced element which results in a disturbing force of 400 N at a speed of 3000 r.p.m. Assum-ing the damping ratio as 0.25, determine : 1. the amplitude of vibrations due to unbalance ; and 2.the transmitted force. [Ans. 0.04 mm ; 35.2 N]

19. A mass of 500 kg is mounted on supports having a total stiffness of 100 kN/m and which providesviscous damping, the damping ratio being 0.4. The mass is constrained to move vertically and is
subjected to a vertical disturbing force of the type F cos 
ωt. Determine the frequency at which
resonance will occur and the maximum allowable value of F if the amplitude at resonance is to be
restricted to 5 mm. [Ans. 2.25 Hz ; 400 N]

20. A machine of mass 75 kg is mounted on springs of stiffness 1200 kN/m and with an assumed

damping factor of 0.2. A piston within the machine of mass 2 kg has a reciprocating motion with astroke of 80 mm and a speed of 3000 cycles/min. Assuming the motion to be simple harmonic,find : 1. the amplitude of motion of the machine, 2. its phase angle with respect to the excitingforce, 3. the force transmitted to the foundation, and 4. the phase angle of transmitted force withrespect to the exciting force. [Ans. 1.254 mm ; 169.05° ; 2132 N ; 44.8°]

1. What are the causes and effects of vibrations ?
2. Define, in short, free vibrations, forced vibrations and damped vibrations.
3. Discuss briefly with neat sketches the longitudinal, transverse and torsional free vibrations.
4. Derive an expression for the natural frequency of free transverse and longitudinal vibrations byequilibrium method.

5. Discuss the effect of inertia of the shaft in longitudinal and transverse vibrations.
6. Deduce an expression for the natural frequency of free transverse vibrations for a simply supportedshaft carrying uniformly distributed mass of m kg per unit length.
7. Deduce an expression for the natural frequency of free transverse vibrations for a beam fixed atboth ends and carrying a uniformly distributed mass of m kg per unit length.

8. Establish an expression for the natural frequency of free transverse vibrations for a simply sup-ported beam carrying a number of point loads, by ( a) Energy method ; and ( b) Dunkerley’s method.
9. Explain the term ‘whirling speed’ or ‘critical speed’ of a shaft. Prove that the whirling speed for arotating shaft is the same as the frequency of natural transverse vibration.

10. Derive the differential equation characterising the motion of an oscillation system subject to vis-cous damping and no periodic external force. Assuming the solution to theequation, find the frequency of oscillation of the system.
11. Explain the terms ‘under damping, critical damping’ and ‘over damping’
12. A thin plate of area A and mass m is attached to the end of a spring and is
allowed to oscillate in a viscous fluid, as shown in Fig. 23.25. Show that
22() µ= ω − ω dm
A

A
where the damping force on the plate is equal to ..µAv; v being the velocity.
The symbols ω and ωd indicate the undamped and damped natural circular frequencies of
oscillations.
13. Explain the term 'Logarithmic decrement' as applied to damped vibrations.
14. Establish an expression for the amplitude of forced vibrations.
15. Explain the term ‘dynamic magnifier’.
16. What do you understand by transmissibility ?

Chapter 23 : Longitudinal and Transverse Vibrations            /G6C          971
1. When there is a reduction in amplitude over every cycle of vibration, then the body is said to have
(a) free vibration ( b) forced vibration ( c) damped vibration
2. Longitudinal vibrations are said to occur when the particles of a body moves

(a) perpendicular to its axis ( b) parallel to its axis
(c) in a circle about its axis
3. When a body is subjected to transverse vibrations, the stress induced in a body will be(a) shear stress ( b) tensile stress ( c) compressive stress
4. The natural frequency (in Hz) of free longitudinal vibrations is equal to
(a)1
2πs
m(b) 1
2πδg(c ) 0.4985
δ
(d) any one of these
where    m = Mass of the body in kg,
s = Stiffness of the body in N/m, and
δ = Static deflection of the body in metres.

5. The factor which affects the critical speed of a shaft is
(a) diameter of the disc ( b) span of the shaft
(c) eccentricity ( d) all of these
6. The equation of motion for a vibrating system with viscous damping is
2
20dx c d x sxmd t m dt+×+× =
If the roots of this equation are real, then the system will be(a) over damped ( b) under damped ( c) critically damped
7.  In under damped vibrating system, if x

1 and x2 are the successive values of the amplitude on the
same side of the mean position, then the logarithmic decrement is equal to(a)  x
1/x2(b) log ( x1/x2)( c) loge (x1/x2)( d) log ( x1.x2)
8. The ratio of the maximum displacement of the forced vibration to the deflection due to the static
force, is known as
(a)  damping factor ( b) damping coefficient
(c)  logarithmic decrement ( d) magnification factor

9.  In vibration isolation system, if /nωω  is less than 2, then for all values of the damping factor, the
transmissibility will be(a) less than unity   ( b)  equal to unity ( c) greater than unity ( d) zero
   where   ω = Circular frequency of the system in rad/s, and
      ω
n = Natural circular frequency of vibration of the system in rad/s.
10. In vibration isolation system, if ω/ωn > 1, then the phase difference between the transmitted force

and the disturbing force is(a) 0° ( b) 90° ( c) 180° ( d) 270°
1.  (c) 2.  (b) 3.  (b) 4.  (d) 5.  (d)
6.  (a) 7.  (b) 8.  (d) 9.  (c) 10.  (c)

972       /G6C               Theory of Machines
972TTTTTorororororsionalsionalsionalsionalsional
VVVVVibraibraibraibraibrationstionstionstionstions24
2. Natural Frequency of Free
Torsional Vibrations.
3. Effect of Inertia of the
Constraint on Torsional
Vibrations.
4. Free Torsional Vibrations of
a Single Rotor System.
5. Free Torsional Vibrations of
a Two Rotor System.
6. Free Torsional Vibrations of
a Three Rotor System.

7. Torsionally Equivalent
Shaft.
8. Free Torsional Vibrations of
a Geared System .24.1.24.1.24.1.24.1.24.1. IntroductionIntroductionIntroductionIntroductionIntroduction
We have already discussed in the previous chapter
that when the particles of a shaft or disc move in a circleabout the axis of a shaft, then the vibrations are known astorsional vibrations . In this case, the shaft is twisted and

untwisted alternately and torsional shear stresses are inducedin the shaft. In this chapter, we shall now discuss the fre-quency of torsional vibrations of various systems.
24.2.24.2.24.2.24.2.24.2. Natural Frequency of Free TorsionalNatural Frequency of Free TorsionalNatural Frequency of Free TorsionalNatural Frequency of Free TorsionalNatural Frequency of Free Torsional
Consider a shaft of negligible mass whose one end

is fixed and the other end carrying a disc as shown in Fig.24.1.
Let        
θ= Angular displacement of the shaft
from mean position after time t
in radians,
m= Mass of disc in kg,
             I = Mass moment of inertia of disc
in kg-m2  = m.k2,
k= Radius of gyration in metres,
q= Torsional stiffness of the shaft in
N-m.

Chapter 24 : Torsional Vibrations            /G6C          973
Fig 24.1.  Natural frequency of
free torsional vibrations.∴    Restoring force        = .θq  ... (i)
and accelerating force          2
2θ=×dI
dt... (ii)
Equating equations (i) and (ii), the equation of
motion is
2
2.dIq
dtθ×= − θ
or2
2.0dIq
dtθ×+ θ =
∴2
20θ+× θ =dq
I dt. . . (iii)
The fundamental equation of the simple harmonic motion is
       22
2.0θ+ω =dx
dt. . . (iv)
Comparing equations (iii) and (iv),
       ω=q
I

       ω=q
I
∴    Time period,        22π== πωpItq
and natural frequency ,        11
2==
πn
pqf
tI
A modern lathe can create an artificial hip joint from information fed into it by a computer. Accurate
drawings of the joint are first made on a computer and the information about the dimensions fed is
directly into the lathe.Note : This picture is given as additional information and is not a direct example of the current chapter.

974       /G6C               Theory of Machines
Note : The value of the torsional stiffness q may be obtained from the torsion equation,
       . TC
Jlθ=       or     . TC J
l=θ
∴.CJq
l= . . . Tq=θ∵
where       C = Modulus of rigidity for the shaft material,
        J  = Polar moment of inertia of the shaft cross-section,
          4
32dπ=  ; d is the diameter of the shaft, and
         l  = Length of the shaft.

l  = Length of the shaft.
Example 24.1.  A shaft of 100 mm diameter and 1 metre long has one of its end fixed and
the other end carries a disc of mass 500 kg at a radius of gyration of 450 mm. The modulus of
rigidity for the shaft material is 80 GN/m2. Determine the frequency of torsional vibrations.
Solution. Given : d = 100 mm = 0.1 m ; l = 1 m ; m = 500 kg ; k = 450 mm = 0.45 m ;
C = 80 GN/m2 =  80 × 109 N/m2

44 6 4(0.1) 9.82 10 m32 32− ππ=×= = ×Jd
∴    Torsional stiffness of the shaft,
     963 . 8 01 0 9 . 8 21 0785.6 101−×× ×== = ×CJqlN-m
We know that mass moment of inertia of the shaft,
       22. 500(0.45) 101.25== =Im k  kg-m2
∴    Frequency of torsional vibrations,
     31 1 785.6 10 88.1
2 2 101.25 2×== =ππ πnqfI = 14 Hz Ans.
Example 24.2.  A flywheel is mounted on a vertical shaft as shown in Fig 24.2. The both

ends of a shaft are fixed and its diameter is 50 mm. The flywheel has a mass of 500 kg and itsradius of gyration is 0.5 m. Find the natural frequency of torsional vibrations, if the modulus ofrigidity for the shaft material is 80 GN/m
2.
Solution.  Given : d = 50 mm = 0.05 m ; m = 500 kg ; k = 0.5m; G = 80 GN/m2
= 84 × 109 N/m2
       44 4(0.05) m32 32Jdππ=×=
         640.6 10 m−=×
∴    Torsional stiffness of the shaft for length l1,
     96
1

96
1
1.8 4 1 0 0 . 6 1 0
0.9−×××==CJql
         = 56 × 103 N-mFig 24.2

Chapter 24 : Torsional Vibrations            /G6C          975
Similarly torsional stiffness of the shaft for length l2,
  96
2
2.8 4 1 0 0 . 6 1 0
0.6−×××==CJql = 84 × 103 N-m
∴   Total torsional stiffness of the shaft,
      33 3
12 56 10 84 10 140 10 = + = ×+ ×=×qq q  N-m
       22. 500(0.5) 125== =Im k  kg-m2
∴    Natural frequency of torsional vibration,
      31 1 140 10 33.5
22 1 2 5 2×== =ππ πnqfI = 5.32 Hz  Ans.

24.3.24.3.24.3.24.3.24.3. Effect of Inertia of the Constraint on Torsional VibrationsEffect of Inertia of the Constraint on Torsional VibrationsEffect of Inertia of the Constraint on Torsional VibrationsEffect of Inertia of the Constraint on Torsional VibrationsEffect of Inertia of the Constraint on Torsional Vibrations
Consider a constraint i.e. shaft whose one end is fixed and the
other end free, as shown in Fig.24.3.

Let      ω= Angular velocity of free end,
 m= Mass of constraint for unit length,
          l = Length of constraint,
mC= Total mass of constraint = m.l,
  k= Radius of gyration of constraint,
 IC= Total mass moment of inertia of
        constraint
                       = mC.k2 = m.l.k2.
Consider a small element at a distance x from the fixed end and
of length δx. Therefore,
Mass moment of inertia of the element
22(. ) . .δ=δ =×xmx k m kl
l

l
... (Dividing and multiplying by l)
    Cδ=×xIl... (Substituting m.k2.l = IC)
and angular velocity of the element
          ω=× xl
Kinetic energy possessed by the element
          2 22
C
C3.. 1
2 2Ix xIx x
ll lω δω  =× × = × δ    
∴  Total kinetic energy of the constraint
22 322 CC C
33
0 0.. 1
32 3 22l lII I xxd x
ll ωω =× = = ω   ∫   . . . (i)Fig 24.3.  Effect of inertia
of the constraint on
torsional vibrations.

976       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G49/G66/G20/G61/G20/G6D/G61/G73/G73/G20/G77/G68/G6F/G73/G65/G20/G6D/G61/G73/G73/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G69/G73/G20/G65/G71/G75/G61/G6C/G20/G74/G6F/G20 /G49/G43/G20/G2F/G20/G33/G20/G69/G73/G20/G70/G6C/G61/G63/G65/G64/G20/G61/G74/G20/G74/G68/G65/G20/G66/G72/G65/G65/G20/G65/G6E/G64/G20/G61/G6E/G64/G20/G74/G68/G65

/G43/G31
/G32/G2F /G33/G6E/G71/G66/G49/G49
24.4.24.4.24.4.24.4.24.4. Free Torsional Vibrations of a Single Rotor SystemFree Torsional Vibrations of a Single Rotor SystemFree Torsional Vibrations of a Single Rotor SystemFree Torsional Vibrations of a Single Rotor SystemFree Torsional Vibrations of a Single Rotor System

/G6F/G66/G20/G61/G20/G73/G69/G6E/G67/G6C/G65/G20/G72/G6F/G74/G6F/G72/G20/G73/G79/G73/G74/G65/G6D/G2EWhen loads are applied on the above two pulleys, the shaft is subject to torsional vibration

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          977
/G6B/G20/G3D/G20/G52/G61/G64/G69/G75/G73/G20/G6F/G66/G20/G67/G79/G72/G61/G74/G69/G6F/G6E/G20/G6F/G66/G20/G72/G6F/G74/G6F/G72/G2C/G20/G61/G6E/G64
/G49/G20/G3D/G20/G4D/G61/G73/G73/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G6F/G66/G20/G72/G6F/G74/G6F/G72/G20/G3D/G20 /G6D/G2E/G6B/G32

/G72/G65/G6D/G61/G69/G6E/G73/G20/G75/G6E/G61/G66/G66/G65/G63/G74/G65/G64/G20/G62/G79/G20/G74/G68/G65/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G2E
24.5.24.5.24.5.24.5.24.5. Free Torsional Vibrations of a Two Rotor SystemFree Torsional Vibrations of a Two Rotor SystemFree Torsional Vibrations of a Two Rotor SystemFree Torsional Vibrations of a Two Rotor SystemFree Torsional Vibrations of a Two Rotor System

/G53/G69/G6E/G63/G65 /G66/G6E/G41/G20/G3D/G20/G66/G6E/G42/G2C/G20/G74/G68/G65/G72/G65/G66/G6F/G72/G65
/G41/G41 /G42/G42/G31/G2E/G31/G2E
/G32/G2E/G32/G2E/G43/G4A /G43/G4A
/G6C/G49 /G6C/G49/G20/G20/G20/G20/G20/G6F/G72/G20/G20/G20/G20 /G6C/G41/G20/G2E/G20/G49/G41/G20/G3D/G20/G6C/G42/G20/G2E/G20/G49/G42/G2E/G20/G2E/G20/G2E/G20/G28/G69/G69/G69/G29
/G42/G42
/G41
/G41/G2E/G6C/G49/G6C/G49
/G57/G65/G20/G61/G6C/G73/G6F/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74

978       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G46/G72/G6F/G6D/G20/G65/G71/G75/G61/G74/G69/G6F/G6E/G73 /G20/G28/G69/G69/G69/G29/G20/G61/G6E/G64/G20/G28/G69/G76/G29/G2C/G20/G77/G65/G20/G6D/G61/G79/G20/G66/G69/G6E/G64/G20/G74/G68/G65/G20/G76/G61/G6C/G75/G65/G20/G6F/G66/G20 /G6C/G41/G20/G61/G6E/G64/G20 /G6C/G42/G20/G61/G6E/G64/G20/G68/G65/G6E/G63/G65/G20/G74/G68/G65/G20/G70/G6F/G73/G69/G74/G69/G6F/G6E/G20/G6F/G66

24.6.24.6.24.6.24.6.24.6. Free Torsional Vibrations of a Three Rotor SystemFree Torsional Vibrations of a Three Rotor SystemFree Torsional Vibrations of a Three Rotor SystemFree Torsional Vibrations of a Three Rotor SystemFree Torsional Vibrations of a Three Rotor System

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          979
/G4E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G73/G20/G66/G6F/G72/G20/G72/G6F/G74/G6F/G72/G20 /G42/G2C
/G20/G20/G20 /G20/G20/G20/G20/G20/G20/G20/G20 /G20/G20 /G2A/G42
/G42/G31/G41/G32/G43/G31/G2E/G31 /G31

/G53/G69/G6E/G63/G65/G20 /G66/G6E/G41/G20/G3D/G20/G66/G6E/G42/G20/G3D/G20/G66/G6E/G43/G2C/G20/G74/G68/G65/G72/G65/G66/G6F/G72/G65/G20/G65/G71/G75/G61/G74/G69/G6E/G67/G20/G65/G71/G75/G61/G74/G69/G6F/G6E/G73/G20 /G28/G69/G29/G20/G61/G6E/G64/G20/G28/G69/G49/G69/G29
/G20/G20 /G20/G20/G20
/G41/G41 /G43/G43/G31/G2E/G31/G2E
/G6C/G49 /G6C/G49/G6F/G72/G20/G20/G20/G20/G20/G20/G20/G20/G6C/G41/G2E/G49/G41/G20/G3D/G20/G6C/G43/G2E/G49/G43
/G43/G43
/G41

/G43/G43
/G41
/G41/G2E/G6C/G49/G6C/G49/G2E/G20/G2E/G20/G2E/G20/G28/G69/G76/G29
/G4E/G6F/G77/G20/G65/G71/G75/G61/G74/G69/G6E/G67/G20/G65/G71/G75/G61/G74/G69/G6F/G6E/G73/G20 /G28/G69/G69/G29/G20/G61/G6E/G64/G20/G28/G69/G69/G69/G29/G2C
/G42/G31/G41/G32/G43 /G43 /G43/G31/G2E/G31 /G31 /G31/G2E
/G32/G32 /G2E/G43/G4A /G43/G4A
/G49/G6C /G6C/G6C /G6C /G6C /G49
/G6F/G72 /G2E/G20/G2E/G20/G2E/G20/G28/G76/G29

/G6C/G6C
/G61/G6E/G64/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G73/G74/G69/G66/G66/G6E/G65/G73/G73/G20/G64/G75/G65/G20/G74/G6F/G20/G28 /G6C/G32/G20/G96/G20/G6C/G43/G29 /G20/G20/G20/G20/G20/G20 /G3D/G20/G20/G32/G43/G2E/G43/G4A
/G6C/G6C
/G20/G20/G20/G20/G20/G20/G20/G54/G6F/G74/G61/G6C/G20/G73/G74/G69/G66/G66/G6E/G65/G73/G73/G20/G6F/G66/G20/G74/G68/G65/G20/G72/G6F/G74/G6F/G72/G20 /G42 /G20/G20/G20/G20 /G20
/G31/G41/G32/G43/G31/G31/G2E/G43/G4A
/G6C/G6C /G6C /G6C
Inside view of a workshop.

980       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G49/G74/G20/G6D/G61/G79/G20/G62/G65/G20/G6E/G6F/G74/G65/G64/G20/G74/G68/G61/G74

/G32/G31
/G41/G6C/G6C/G61/G61/G6C
/G61/G6E/G64/G20/G61/G6D/G70/G6C/G69/G74/G75/G64/G65/G20/G6F/G66/G20/G72/G6F/G74/G6F/G72/G20 /G43/G2C /G20/G20/G20/G20/G20/G20/G43
/G33/G32
/G43/G32/G6C/G61/G61/G6C/G6C

/G6E/G6F/G64/G65/G20/G61/G6E/G64/G20/G74/G77/G6F/G20/G6E/G6F/G64/G65/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G73/G2E
24.7.24.7.24.7.24.7.24.7. Torsionally Equivalent ShaftTorsionally Equivalent ShaftTorsionally Equivalent ShaftTorsionally Equivalent ShaftTorsionally Equivalent Shaft

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          981
/G4C/G65/G74 /G20/G20/G20/G20/G64/G31/G2C/G20/G64/G32/G20/G61/G6E/G64/G20 /G64/G33/G20/G3D/G20/G44/G69/G61/G6D/G65/G74/G65/G72/G73/G20/G66/G6F/G72/G20/G74/G68/G65/G20/G6C/G65/G6E/G67/G74/G68/G73/G20 /G6C/G31/G2C/G20/G6C/G32/G20/G61/G6E/G64/G20 /G6C/G33/G20/G72/G65/G73/G70/G65/G63/G74/G69/G76/G65/G6C/G79/G2C

/G64/G69/G66/G66/G65/G72/G65/G6E/G74/G20/G6C/G65/G6E/G67/G74/G68/G73/G2C/G20/G74/G68/G65/G72/G65/G66/G6F/G72/G65
/G20/G20/G20/G20/G20/G20/G31/G32/G33
/G6F/G72 /G20/G20/G33 /G31/G32
/G31/G32/G33/G2E /G2E/G2E/G2E
/G2E/G2E/G2E /G2E/G54/G6C /G54/G6C /G54/G6C /G54/G6C
/G43/G4A /G43/G4A /G43/G4A /G43/G4A
/G20/G20/G20/G20/G20/G33 /G31/G32
/G31/G32/G33/G6C /G6C/G6C/G6C
/G4A/G4A /G4A /G4A
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G33 /G31/G32
/G34/G34/G34/G34

/G31/G32/G33/G28/G29 /G28/G29 /G28/G29/G33/G32 /G33/G32 /G33/G32 /G33/G32/G6C /G6C/G6C /G6C
/G64/G64 /G64 /G64
/G31/G32/G33/G28/G29 /G28/G29 /G28/G29/G6C /G6C/G6C /G6C

/G6F/G6E/G65/G20/G6F/G66/G20/G74/G68/G65/G20/G64/G69/G61/G6D/G65/G74/G65/G72/G20/G6F/G66/G20/G74/G68/G65/G20/G61/G63/G74/G75/G61/G6C/G20/G73/G68/G61/G66/G74/G2E/G20/G4C/G65/G74/G20/G75/G73/G20/G61/G73/G73/G75/G6D/G65/G20/G74/G68/G61/G74/G20 /G64/G20/G3D/G20/G64/G31/G2E
/G20/G33 /G31/G32
/G31/G31/G32/G33/G28/G29 /G28/G29 /G28/G29 /G28/G29/G6C /G6C/G6C /G6C
/G64/G64 /G64/G64
/G6F/G72 /G20/G20/G20/G20/G20/G20/G34 /G34
/G31/G31
/G31/G32/G33

/G45/G78/G61/G6D/G70/G6C/G65/G20/G32/G34/G2E/G33/G2E/G20 /G41/G20 /G73/G74/G65/G65/G6C/G20 /G73/G68/G61/G66/G74/G20 /G31/G2E/G35/G20 /G6D/G20 /G6C/G6F/G6E/G67/G20 /G69/G73/G20 /G39/G35/G20 /G6D/G6D/G20 /G69/G6E/G20 /G64/G69/G61/G6D/G65/G74/G65/G72/G20 /G66/G6F/G72/G20 /G74/G68/G65/G20 /G66/G69/G72/G73/G74/G20 /G30/G2E/G36/G20 /G6D/G20 /G6F/G66/G20 /G69/G74/G73

982       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G53/G6F/G6C/G75/G74/G69/G6F/G6E/G2E/G20 /G47/G69/G76/G65/G6E/G20/G3A/G20 /G4C/G20/G3D/G20/G31/G2E/G35/G20/G6D/G20/G3B/G20 /G64/G31/G20/G3D/G20/G39/G35/G20/G6D/G6D/G20/G3D/G20/G30/G2E/G30/G39/G35/G20/G6D/G20/G3B/G20 /G6C/G31/G20/G3D/G20/G30/G2E/G36/G20/G6D/G20/G3B/G20 /G64/G32/G20/G3D/G20/G36/G30/G20/G6D/G6D/G20/G3D/G20/G30/G2E/G30/G36/G20/G6D/G20/G3B

/G20/G20/G20/G20/G20/G20/G20/G20/G34 /G34 /G34/G34
/G31/G31
/G32/G33/G30/G2E/G30/G39/G35 /G30/G2E/G30/G39/G35/G30/G2E/G36 /G30/G2E/G35 /G30/G2E/G34/G30/G2E/G30/G36 /G30/G2E/G30/G35/G64/G64/G6C/G6C /G6C /G6C/G64/G64
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G3D/G20/G30/G2E/G36/G20/G2B/G20/G33/G2E/G31/G34/G20/G2B/G20/G35/G2E/G32/G31/G20/G3D/G20/G38/G2E/G39/G35/G20/G6D
/G4C/G6F/G63/G61/G74/G69/G6F/G6E/G20/G6F/G66/G20/G74/G68/G65/G20/G6E/G6F/G64/G65

/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74/G20/G6D/G61/G73/G73/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G6F/G66/G20/G66/G6C/G79/G77/G68/G65/G65/G6C/G20 /G41/G2C
/G20/G20/G20/G32/G32
/G41/G41 /G41/G28 /G29 /G39/G30/G30/G28/G30/G2E/G38/G35/G29 /G36/G35/G30 /G49/G6D /G6B /G20/G6B/G67/G2D/G6D/G32

/G61/G6E/G64/G20/G6D/G61/G73/G73/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G6F/G66/G20/G66/G6C/G79/G77/G68/G65/G65/G6C/G20 /G42/G2C
/G20/G20/G20/G20/G32/G32
/G42/G42 /G42/G28 /G29 /G37/G30/G30/G28/G30/G2E/G35/G35/G29 /G32/G31/G32 /G49/G6D /G6B /G20/G6B/G67/G2D/G6D/G32
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74 /G20/G41/G41 /G42/G42/G2E/G2E/G6C/G49 /G6C/G49 /G20/G20/G20/G20/G20/G20/G20/G6F/G72/G20/G20/G20/G20/G42/G42 /G42
/G41/G42

/G41/G42
/G41/G2E/G32 /G31/G32/G30/G2E/G33/G32/G36/G36/G35/G30/G6C/G49 /G6C/G6C/G6C/G49
/G41/G6C/G73/G6F/G2C/G41/G42 /G38/G2E/G39/G35 /G6C/G6C/G6C /G6D/G20/G20/G6F/G72/G20/G20/G20/G42/G42/G30/G2E/G33/G32/G36 /G38/G2E/G39/G35 /G6C/G6C /G20/G20/G20/G6F/G72/G20/G20 /G20/G20/G6C/G42/G20/G3D/G20/G36/G2E/G37/G35/G20/G6D
/G61/G6E/G64 /G20/G20/G20/G20/G20 /G20/G6C/G41/G20/G3D/G20/G38/G2E/G39/G35/G20/G96/G20/G36/G2E/G37/G35/G20/G3D/G20/G32/G2E/G32/G20/G6D

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          983

/G73/G68/G61/G66/G74/G2E
∴ /G20/G50/G6F/G73/G69/G74/G69/G6F/G6E/G20/G6F/G66/G20/G6E/G6F/G64/G65/G20/G6F/G6E/G20/G74/G68/G65/G20/G6F/G72/G69/G67/G69/G6E/G61/G6C/G20/G73/G68/G61/G66/G74/G20/G66/G72/G6F/G6D/G20/G66/G6C/G79/G77/G68/G65/G65/G6C/G20 /G41
/G20/G20/G34 /G34
/G32
/G31/G41/G31
/G31/G30/G2E/G30/G36/G28/G29 /G30 /G2E /G36 /G28 /G32 /G2E /G32 /G30 /G2E /G36 /G29/G30/G2E/G30/G39/G35/G64/G6C/G6C/G6C/G64/G20 /G3D/G20/G30/G2E/G38/G35/G35/G20/G6D/G20 /G41/G6E/G73/G2E

/G34/G34 /G36 /G34
/G31/G28/G29 /G28 /G30 /G2E /G30 /G39 /G35 /G29 /G38 /G31/G30/G6D
/G33/G32 /G33/G32/G4A/G64
/G20/G4E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G66/G72/G65/G65/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G73/G2C
/G41 /G6E/G6E/G66/G66/G20/G20/G6F/G72/G20/G20/G42/G6E/G66
/G39/G36
/G41/G41/G31 /G2E/G31 /G38 /G30 /G31/G30 /G38 /G31/G30
/G32/G2E /G32 /G32 /G2E /G32 /G36 /G35 /G30/G43/G4A

/G31/G2E/G20 /G74/G68/G65/G20 /G64/G69/G61/G6D/G65/G74/G65/G72/G20 /G91/G64/G92/G20 /G6F/G66/G20 /G74/G68/G65/G20 /G70/G6F/G72/G74/G69/G6F/G6E/G20 /G43/G44/G20 /G73/G6F/G20 /G74/G68/G61/G74/G20 /G74/G68/G65/G20 /G6E/G6F/G64/G65/G20 /G6F/G66/G20 /G74/G68/G65/G20 /G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20 /G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G20 /G6F/G66/G20 /G74/G68/G65

Note : This picture is given as additional information and is not a direct example of the current chapter.The above machine tools include grinding machine, drill, router, milling machines, lathe and a
circular saw.

984       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G54/G68/G65/G20/G61/G63/G74/G75/G61/G6C/G20/G73/G68/G61/G66/G74/G20/G69/G73/G20/G73/G68/G6F/G77/G6E/G20/G69/G73/G20/G46/G69/G67/G2E/G20/G32/G34/G2E/G31/G30/G20/G28 /G61/G29/G2E/G20/G46/G69/G72/G73/G74/G20/G6F/G66/G20/G61/G6C/G6C/G2C/G20/G6C/G65/G74/G20/G75/G73/G20/G66/G69/G6E/G64/G20/G74/G68/G65/G20/G6C/G65/G6E/G67/G74/G68/G20/G6F/G66/G20/G74/G68/G65/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74

/G46/G69/G67/G2E/G20/G32/G34/G2E/G31/G30
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74/G20/G6C/G65/G6E/G67/G74/G68/G20/G6F/G66/G20/G74/G68/G65/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G73/G68/G61/G66/G74/G2C
/G31/G31
/G32/G33/G30/G2E/G30/G35 /G30/G2E/G30/G35/G30/G2E/G34 /G30/G2E/G35 /G30/G2E/G36/G30/G2E/G30/G36/G64/G64/G6C/G6C /G6C /G6C/G64/G64 /G64

/G66/G6C/G79/G77/G68/G65/G65/G6C/G20 /G41/G2C
/G41/G41 /G41/G28 /G29 /G36/G30/G30/G28/G30/G2E/G36/G29 /G49/G6D /G6B /G20/G3D/G20/G32/G31/G36/G20/G6B/G67/G2D/G6D/G32
/G61/G6E/G64/G20/G6D/G61/G73/G73/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G6F/G66/G20/G66/G6C/G79/G77/G68/G65/G65/G6C/G20 /G44/G2C
/G20/G20/G20/G20/G20/G32/G32

/G44/G44 /G44/G28 /G29 /G38/G30/G30/G28/G30/G2E/G39/G29 /G49/G6D /G6B /G20/G3D/G20/G36/G34/G38/G20/G6B/G67/G2D/G6D/G32
A pulley is being processed at a turning centre.

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          985
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G41/G41 /G44/G44/G2E/G2E/G6C/G49 /G6C/G49
/G6F/G72/G41/G41 /G41 /G41
/G44/G2E/G32 /G31/G36
/G36/G34/G38 /G33/G44/G6C/G49 /G6C /G6C/G6C/G49

/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G6C/G65/G6E/G67/G74/G68/G20/G66/G72/G6F/G6D/G20/G72/G6F/G74/G6F/G72/G20 /G41/G2C
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G34 /G34
/G32/G31
/G41/G31
/G32/G30/G2E/G35 /G30/G2E/G30/G35/G30/G2E/G34 /G30/G2E/G35/G32/G6D/G32/G32 /G30 /G2E /G30 /G36/G6C/G64/G6C/G6C/G64
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G41
/G44/G30/G2E/G35/G32
/G33/G33/G6C/G6C /G20/G3D/G20/G30/G2E/G31/G37/G33/G20/G6D

/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74/G20/G20 /G20/G20/G20/G20/G20/G20/G20 /G6C/G20/G3D/G20/G6C/G41/G20/G2B/G20/G6C/G44
/G6F/G72 /G20/G20/G20/G36
/G34/G33/G2E/G37/G35 /G31/G30/G30/G2E/G36/G34 /G30/G2E/G35/G32 /G30/G2E/G31/G37/G33
/G64/G2E/G20/G2E/G20/G2E/G20/G5B/G46/G72/G6F/G6D/G20/G65/G71/G75/G61/G74/G69/G6F/G6E/G20 /G28/G69/G29/G5D
/G36
/G34/G33/G2E/G37/G35 /G31/G30/G30/G2E/G35/G32 /G30/G2E/G31/G37/G33 /G30/G2E/G36/G34 /G30/G2E/G30/G35/G33

/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74/G20/G70/G6F/G6C/G61/G72/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G6F/G66/G20/G74/G68/G65/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G73/G68/G61/G66/G74/G2C
/G20/G34/G34 /G36 /G34
/G31/G28/G29 /G28 /G30 /G2E /G30 /G35 /G29 /G30 /G2E /G36 /G31/G34/G31/G30/G6D/G32/G33 /G32/G4A/G64

/G20/G20/G4E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G2C
/G41 /G6E/G6E/G66/G66/G20/G6F/G72/G20/G20/G44/G6E/G66
/G20/G20/G20/G20/G39/G36
/G41/G41/G31 /G2E/G31 /G38 /G30 /G31/G30 /G30 /G2E /G36 /G31/G34 /G31/G30
/G32/G2E/G32 /G30 /G2E /G35 /G32 /G32 /G31/G36/G43/G4A
/G6C/G49/G20/G48/G7A/G20/G3D/G20/G33/G2E/G33/G33/G20/G48/G7A /G20/G41/G6E/G73/G2E

986       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G43/G43
/G41/G43 /G43
/G41/G30/G2E/G30/G39/G30/G2E/G36/G30/G2E/G31/G35/G6C/G49/G6C/G6C /G6C/G49
/G41/G6C/G73/G6F /G20
/G43/G43 /G42 /G31/G41 /G32 /G43/G31/G31/G31 /G31
/G2E/G6C/G49 /G49 /G6C /G6C /G6C /G6C
/G20/G20/G20/G20/G20/G20/G20/G20/G20
/G43/G41 /G43/G31/G31 /G31/G31
/G30/G2E/G30/G39 /G30/G2E/G33 /G31/G2E/G35 /G31/G6C/G6C /G6C

/G43/G43 /G43/G30/G2E/G33 /G31/G31
/G30/G2E/G30/G39 /G31/G2E/G35 /G30/G2E/G36 /G31/G6C/G6C /G6C/G2E/G20/G2E/G20/G2E/G20/G28/G53/G75/G62/G73/G74/G69/G74/G75/G74/G69/G6E/G67/G20 /G6C/G41/G20/G3D/G20/G30/G2E/G36/G20 /G6C/G43/G29
/G20/G20 /G20/G43/G43 /G43
/G32
/G43/G43 /G43 /G43/G43/G31 /G31/G2E/G35 /G30/G2E/G36 /G32/G2E/G35 /G31/G2E/G36/G31/G30

/G33/G28 /G31/G2E /G35 /G30 /G2E /G36 /G29 /G28 /G31 /G29 /G31/G2E/G35 /G32/G2E/G31 /G30/G2E/G36/G28 /G29/G6C/G6C /G6C
/G6C/G6C /G6C /G6C/G6C
/G4F/G6E/G20/G63/G72/G6F/G73/G73/G2D/G6D/G75/G6C/G74/G69/G70/G6C/G79/G69/G6E/G67/G20/G61/G6E/G64/G20/G72/G65/G2D/G61/G72/G72/G61/G6E/G67/G69/G6E/G67/G2C
/G20/G31/G30/G2E/G38/G20/G28 /G6C/G43/G29/G32/G20/G96/G20/G32/G38/G2E/G35/G20 /G6C/G43/G20/G2B/G20/G31/G35/G20/G3D/G20/G30
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G32

/G43/G32/G38/G2E/G35 /G28/G32/G38/G2E/G35/G29 /G34 /G31/G30/G2E/G38 /G31/G35 /G32/G38/G2E/G35 /G31/G32/G2E/G38
/G32/G31/G30 /G2E /G38 /G32 /G31/G2E /G36/G6C
/G20/G20/G20/G20/G20/G20/G20 /G3D/G20/G31/G2E/G39/G31/G20/G6D/G20/G20/G20/G20/G20/G20/G6F/G72/G20/G20/G20/G20/G20/G20/G30/G2E/G37/G32/G36/G20/G6D
/G61/G6E/G64 /G20/G20/G20/G20/G20 /G20/G6C/G41/G20/G3D/G20/G30/G2E/G36/G20 /G6C/G43/G20/G3D/G20/G31/G2E/G31/G34/G36/G20/G6D/G20/G20/G6F/G72/G20/G20/G30/G2E/G34/G33/G35/G36/G20/G6D

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          987
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74/G20/G70/G6F/G6C/G61/G72/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G6F/G66/G20/G74/G68/G65/G20/G73/G68/G61/G66/G74/G2C
/G20/G20/G20/G20/G20/G20/G20/G34/G34/G36 /G34/G28/G30/G2E/G30/G37/G29 /G32/G2E/G33/G36 /G31/G30 /G6D/G33/G32 /G33/G32/G4A/G64

/G20/G20/G20/G20/G4E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G20/G66/G6F/G72/G20/G61/G20/G73/G69/G6E/G67/G6C/G65/G20 /G6E/G6F/G64/G65/G20/G73/G79/G73/G74/G65/G6D/G2C
/G31
/G41/G41/G31 /G2E/G31 /G38 /G34 /G31/G30 /G32 /G2E /G33 /G36 /G31/G30
/G32/G2E/G32 /G31/G2E /G31/G34 /G36 /G30 /G2E /G31/G35/G6E/G43/G4A/G66/G6C/G49/G20/G48/G7A

/G53/G69/G6D/G69/G6C/G61/G72/G6C/G79/G2C/G20/G6E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G20/G66/G6F/G72/G20/G61/G20/G74/G77/G6F/G20/G6E/G6F/G64/G65/G20/G73/G79/G73/G74/G65/G6D/G2C
/G20/G20/G20/G39/G36
/G32
/G32 /G2E /G32 /G30/G2E/G34/G33/G35/G36 /G30/G2E/G31/G35/G6E/G43/G4A/G66/G6C/G49/G20/G48/G7A

/G69/G73/G20 /G38/G34/G20 /G47/G4E/G2F/G6D/G32/G2E/G20 /G54/G68/G65/G20 /G73/G79/G73/G74/G65/G6D/G20 /G63/G61/G6E/G20 /G76/G69/G62/G72/G61/G74/G65/G20 /G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G6C/G79/G20 /G77/G69/G74/G68/G20 /G6F/G6E/G65/G20 /G6E/G6F/G64/G65/G20 /G61/G74/G20 /G39/G35/G20 /G6D/G6D/G20 /G66/G72/G6F/G6D/G20 /G41/G2C/G20 /G74/G68/G65/G20 /G66/G6C/G79/G77/G68/G65/G65/G6C

/G61/G6E/G64/G20/G33/G2E/G20/G74/G68/G65/G20/G72/G61/G64/G69/G75/G73/G20/G6F/G66/G20/G67/G79/G72/G61/G74/G69/G6F/G6E/G20/G6F/G66/G20/G74/G68/G65/G20/G61/G72/G6D/G61/G74/G75/G72/G65/G20/G43/G2E
/G54/G68/G65/G20/G6F/G74/G68/G65/G72/G20/G64/G61/G74/G61/G20/G61/G72/G65/G20/G67/G69/G76/G65/G6E/G20/G62/G65/G6C/G6F/G77/G20/G3A
/G2C/G96
/G2C/G50 /G61 /G72 /G74 /G69 /G63 /G75 /G6C /G61 /G72 /G73 /G41/G42/G43

/G52/G61/G64/G69/G75/G73 /G6F/G66 /G67/G79/G72/G61/G74/G69/G6F/G6E /G6D/G6D /G33/G30/G30 /G33/G37/G35
/G4D/G61/G73/G73 /G6B/G67 /G34/G30/G30 /G35/G30/G30 /G33/G30/G30
/G46/G69/G67/G2E/G20/G32/G34/G2E/G31/G33

988       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G31/G2E/G20/G50/G6F/G73/G69/G74/G69/G6F/G6E/G20/G6F/G66/G20/G61/G6E/G6F/G74/G68/G65/G72/G20/G6E/G6F/G64/G65

/G46/G69/G67/G2E/G20/G32/G34/G2E/G31/G34
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20
/G42/G31/G41 /G33/G43 /G41 /G41/G31/G31 /G31 /G31
/G2E /G49/G6C /G6C /G6C /G6C /G6C /G49
/G43/G31/G31 /G31 /G31
/G37/G30/G2E/G33 /G30/G2E/G33 /G30/G2E/G30/G39/G35 /G30/G2E/G33/G30/G35 /G30/G2E/G30/G39/G35 /G33/G36 /G6C
/G20
/G43/G31/G31 /G37/G30/G2E/G33/G32/G30/G2E/G35/G36/G30/G2E/G32/G30/G35 /G30/G2E/G33/G30/G35 /G30/G2E/G30/G39/G35 /G33/G36 /G6C

/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20
/G43/G31/G31/G32/G30/G2E/G35/G36 /G31/G35/G2E/G36/G38/G30/G2E/G33/G30/G35 /G30/G2E/G32/G30/G35 /G6C
/G20/G30/G2E/G33/G30/G35/G20/G96/G20 /G6C/G43/G20/G3D/G20/G31/G2F/G31 /G35/G2E/G36/G38/G20/G3D/G20/G30/G2E/G30/G36/G34/G20/G20/G6F/G72/G20/G20 /G6C/G43/G20/G3D/G20/G30/G2E/G32/G31/G20/G6D/G28/G61/G29
/G28/G62/G29
/G28/G63/G29

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          989
/G20/G20/G20/G20/G43/G6F/G72/G72/G65/G73 /G70/G6F/G6E/G64/G69/G6E/G67/G20/G76/G61/G6C/G75/G65/G20/G6F/G66/G20 /G6C/G43/G20/G69/G6E/G20/G61/G6E/G20/G6F/G72/G69/G67/G69/G6E/G61/G6C/G20/G73/G79/G73/G74/G65/G6D/G20/G66/G72/G6F/G6D/G20/G72/G6F/G74/G6F/G72/G20 /G43
/G20/G20/G20/G34 /G34
/G32

/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G34/G34 /G36 /G34
/G31/G28/G29 /G28 /G30 /G2E /G31/G29 /G39 /G2E /G38 /G32 /G31/G30/G6D/G33/G32 /G33/G32/G4A/G64

/G20/G20/G20/G20/G4E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G73/G2C
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G39/G36
/G41/G41/G31 /G2E/G31 /G38 /G34 /G31/G30 /G39 /G2E /G38 /G32 /G31/G30
/G32/G2E/G32 /G30 /G2E /G30 /G39 /G35 /G33 /G36/G6E/G43/G4A/G66/G6C/G49/G20/G20/G48/G7A
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G3D/G20/G37/G38/G2E/G31/G20/G48/G7A /G20/G41/G6E/G73/G2E

/G20/G20/G20/G20/G20/G20/G20/G49/G43/G20/G3D/G20/G4D/G61/G73/G73/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G6F/G66/G20/G61/G72/G6D/G61/G74/G75/G72/G65/G20 /G43/G20/G3D/G20/G6D/G43/G20/G28/G6B/G43/G29/G32/G20/G69/G6E/G20/G6B/G67/G2D/G6D/G32/G2E
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74/G32
/G41/G41 /G43/G43 /G43 /G43 /G43/G2E/G2E/G2E /G28 /G29/G6C/G49 /G6C/G49 /G6C/G6D/G6B
/G6F/G72 /G20/G32/G32/G41/G41
/G43

/G43
/G43/G43/G2E /G30/G2E/G30/G39/G35 /G33/G36/G28 /G29 /G30/G2E/G30/G35/G34/G33/G6D/G2E/G30 /G2E /G32 /G31/G33 /G30 /G30/G6C/G49/G6B/G6C/G6D
/G20/G20/G20/G20/G20/G20/G6B/G43/G20/G3D/G20/G30/G2E/G32/G33/G33/G20/G6D/G20 /G41/G6E/G73/G2E

/G4E/G65/G67/G6C/G65/G63/G74/G69/G6E/G67/G20 /G74/G68/G65/G20 /G69/G6E/G65/G72/G74/G69/G61/G20 /G6F/G66/G20 /G74/G68/G65/G20 /G73/G68/G61/G66/G74/G20 /G61/G6E/G64/G20 /G74/G61/G6B/G69/G6E/G67/G20 /G69/G74/G73/G20 /G6D/G6F/G64/G75/G6C/G75/G73/G20 /G6F/G66/G20 /G72/G69/G67/G69/G64/G69/G74/G79/G20 /G61/G73/G20 /G38/G30/G20 /G47/G4E/G2F/G6D/G32/G2C

/G3D/G20/G30/G2E/G30/G35/G6D/G2E
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74/G20/G6C/G65/G6E/G67/G74/G68/G20/G6F/G66/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G73/G68/G61/G66/G74/G2C
/G20/G20/G20/G20/G20/G20/G20/G20/G34 /G34
/G31
/G31/G32
/G32/G30/G2E/G30/G35/G32/G32 /G33 /G34 /G6D/G30/G2E/G30/G32/G35/G64/G6C/G6C /G6C/G64

/G20/G20/G20/G20/G20/G20/G20/G6C/G43/G20/G3D/G20/G44/G69/G73/G74/G61/G6E/G63/G65/G20/G6F/G66/G20/G6E/G6F/G64/G65/G20 /G4E/G32/G20/G66/G72/G6F/G6D/G20/G72/G6F/G74/G6F/G72/G20 /G43/G2E
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74 /G20/G20/G20/G20/G20/G20/G20/G20/G41/G41 /G43/G43/G2E/G2E/G6C/G49 /G6C/G49
/G43/G43 /G43
/G41/G43
/G41/G2E/G32 /G30/G30/G2E/G30/G32/G35/G38/G30/G30/G6C/G49 /G6C/G6C/G6C/G49

990       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G41/G6C/G73/G6F/G2C /G20/G20
/G43/G43 /G42 /G31/G41 /G33 /G43/G31/G31 /G31 /G31
/G43/G43 /G43/G31/G31 /G31 /G31
/G32/G30 /G33/G32/G30 /G32 /G30/G2E/G30/G32/G35 /G33/G32/G6C/G6C /G6C/G2E/G20/G2E/G20/G2E/G20/G28 /G20/G20/G20/G20 /G6C/G33/G20/G3D/G20/G6C/G20/G96/G20/G6C/G31/G29
/G43/G43 /G43/G33/G32/G30 /G31/G31

/G32/G30 /G32 /G30/G2E/G30/G32/G35 /G33/G32/G6C/G6C /G6C
/G20/G20/G20/G20/G20/G20/G43/G43 /G43
/G32
/G43/G43 /G43 /G43/G43/G33/G32 /G32 /G30/G2E/G30/G32/G35 /G33/G34 /G31/G2E/G30/G32/G35/G31/G36
/G28/G32 /G30/G2E/G30/G32/G35 /G29/G28/G33/G32 /G29 /G36/G34 /G32/G2E/G38 /G30/G2E/G30/G32/G35/G28 /G29/G6C/G6C /G6C

/G6F/G72/G20/G20/G20/G20/G31/G2E/G34/G32/G35/G20/G28 /G6C/G43/G29/G32/G20/G96/G20/G37/G38/G2E/G38/G20 /G6C/G43/G20/G2B/G20/G31/G30/G32/G34/G20/G3D/G20/G30
/G46/G69/G67/G2E/G20/G32/G34/G2E/G31/G35
/G20/G32
/G43/G37/G38/G2E/G38 /G28/G37/G38/G2E/G38/G29 /G34 /G31/G2E/G34/G32/G35 /G31/G30/G32/G34 /G37/G38/G2E/G38 /G31/G39/G2E/G33
/G32 /G31/G2E/G34/G32/G35 /G32/G2E/G38/G35/G6C

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          991
/G20/G20/G20/G20/G20/G20/G34/G34 /G36 /G34

/G31/G28/G29 /G28 /G30 /G2E /G30 /G35 /G29 /G30 /G2E /G36 /G31/G34/G31/G30/G6D
/G20/G20/G20/G20/G4E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G73/G20/G66/G6F/G72/G20/G61/G20/G73/G69/G6E/G67/G6C/G65/G20 /G6E/G6F/G64/G65/G20/G73/G79/G73/G74/G65/G6D/G2C
/G20/G20/G20/G20/G20/G39/G36
/G31

/G31
/G32/G2E/G32 /G30 /G2E /G38 /G36 /G38 /G30 /G30/G6E/G43/G4A/G66/G6C/G49/G20/G48/G7A
/G2E/G20/G2E/G20/G2E/G20/G28/G53/G75/G62/G73/G74/G69/G74/G75/G74/G69/G6E/G67/G20 /G6C/G41/G20/G3D/G20/G30/G2E/G38/G36/G20/G6D/G29
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G3D/G20/G31/G2E/G33/G34/G35/G20/G48/G7A/G20 /G41/G6E/G73/G2E

/G53/G69/G6D/G69/G6C/G61/G72/G6C/G79/G20/G6E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G73/G20/G66/G6F/G72/G20/G61/G20/G74/G77/G6F/G20/G6E/G6F/G64/G65/G20/G73/G79/G73/G74/G65/G6D/G2C
/G32
/G32/G2E/G32 /G30 /G2E /G35 /G32 /G38 /G30 /G30/G6E/G43/G4A/G66/G6C/G49/G20/G48/G7A

/G61/G74/G20/G61/G20/G64/G69/G73/G74/G61/G6E/G63/G65/G20/G6F/G66/G20 /G6C/G43/G20/G3D/G20/G31/G2E/G33/G20/G6D/G20/G66/G72/G6F/G6D/G20/G74/G68/G65/G20/G70/G72/G6F/G70/G65/G6C/G6C/G65/G72/G2E/G20 /G41/G6E/G73/G2E
24.8.24.8.24.8.24.8.24.8. Free Torsional Vibrations of a Geared SystemFree Torsional Vibrations of a Geared SystemFree Torsional Vibrations of a Geared SystemFree Torsional Vibrations of a Geared SystemFree Torsional Vibrations of a Geared System

992       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G46/G69/G67/G2E/G20/G32/G34/G2E/G31/G36
/G54/G68/G65/G20/G66/G6F/G6C/G6C/G6F/G77/G69/G6E/G67/G20/G74/G77/G6F/G20/G63/G6F/G6E/G64/G69/G74/G69/G6F/G6E/G73/G20/G6D/G75/G73/G74/G20/G62/G65/G20/G73/G61/G74/G69/G73/G66/G69/G65/G64/G20/G62/G79/G20/G61/G6E/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G73/G79/G73/G74/G65/G6D/G20/G3A

/G42/G42 /G42 /G42/G31/G31/G28/G29 /G28/G29/G32/G32/G49/G49 /G20/G6F/G72/G20/G20/G20/G20/G32/G32
/G42/G41 /G42 /G42/G28/G29 /G28/G29/G49/G49 /G20/G20/G2E/G20/G2E/G20/G2E/G20/G28/G42/G41/G29
/G20/G20/G32
/G42/G42
/G42/G42/G32/G41/G49/G49/G49
/G47/G41
/G42/G47 /G20/G20/G2E/G20/G2E/G20/G2E/G20/G28/G69/G29

/G20/G20/G20/G20/G20/G53/G74/G72/G61/G69/G6E/G20/G65/G6E/G65/G72/G67/G79/G20/G6F/G66/G20 /G6C/G33/G20/G3D/G20/G53/G74/G72/G61/G69/G6E/G20/G65/G6E/G65/G72/G67/G79/G20/G6F/G66/G20 /G6C/G32
/G6F/G72 /G20/G20/G20/G20/G20/G20/G20/G20 /G33/G33 /G32/G32/G31/G31
/G32/G32/G54/G54 /G20/G20/G20/G6F/G72/G20/G20/G20/G33/G32
/G32/G33/G54
/G54/G2E/G20/G2E/G20/G2E/G20/G28/G69/G69/G29/G28/G61/G29

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          993
/G77/G68/G65/G72/G65 /G20/G20/G20/G20/G20/G20/G20 /G54/G32/G20/G61/G6E/G64/G20 /G54/G33/G20/G3D/G20/G54/G6F/G72/G71/G75/G65/G20/G6F/G6E/G20/G74/G68/G65/G20/G73/G65/G63/G74/G69/G6F/G6E/G73/G20 /G6C/G32/G20/G61/G6E/G64/G20 /G6C/G33/G2C/G20/G61/G6E/G64

/G20/G20/G20/G20/G20/G2E /G54/G43 /G4A/G71
/G6C
/G77/G68/G65/G72/G65 /G20/G20/G20/G20 /G20 /G4A/G20/G3D/G20/G50/G6F/G6C/G61/G72/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G6F/G66/G20/G74/G68/G65/G20/G73/G68/G61/G66/G74/G2E
/G46/G6F/G72/G20/G73/G65/G63/G74/G69/G6F/G6E/G20 /G20/G6C/G33/G2C/G20/G33/G33
/G33/G33/G2E /G54/G43 /G4A
/G6C/G2E/G20/G2E/G20/G2E/G20/G28/G76/G29

/G61/G6E/G64 /G20/G20/G20/G20/G20/G20/G46/G6F/G72/G20/G73/G65/G63/G74/G69/G6F/G6E/G20 /G6C/G32/G20/G2C/G20/G32/G32
/G32/G32/G2E /G54/G43 /G4A
/G6C/G2E/G20/G2E/G20/G2E/G20/G28/G76/G69/G29
/G44/G69/G76/G69/G64/G69/G6E/G67/G20/G65/G71/G75/G61/G74/G69/G6F/G6E /G20/G28/G76/G29/G20/G62/G79/G20/G65/G71/G75/G61/G74/G69/G6F/G6E/G20 /G28/G76/G69/G29/G2C
/G20/G20/G20/G20/G20/G20/G20/G20 /G33/G33/G32/G32
/G32/G33/G33/G32/G54/G4A /G6C

/G54/G6C /G4A/G20/G20/G20/G20/G6F/G72/G33/G33 /G33 /G32/G32/G32 /G32 /G33/G2E
/G2E/G2E/G54/G4A /G6C
/G54/G4A/G6C
/G6F/G72 /G20/G20/G20/G20/G33/G32/G32/G33/G31/G4A /G6C/G47/G47/G4A /G6C/G5B/G46/G72/G6F/G6D/G20/G65/G71/G75/G61/G74/G69/G6F/G6E/G20 /G28/G69/G76/G29/G5D
/G32/G33
/G33/G32
/G32/G4A/G6C/G47 /G6C/G4A/G2E/G20/G2E/G20/G2E/G20/G28/G76/G69/G69/G29

/G32/G32/G28/G29/G33/G32/G4A/G64
/G20/G20/G20/G34
/G33/G31
/G32/G32/G4A /G64
/G4A/G64
/G4E/G6F/G77/G20/G74/G68/G65/G20/G65/G71/G75/G61/G74/G69/G6F/G6E /G20/G28/G76/G69/G69/G29/G20/G6D/G61/G79/G20/G62/G65/G20/G77/G72/G69/G74/G74/G65/G6E/G20/G61/G73
/G20/G20/G20/G20/G20/G34
/G32 /G31
/G33/G32
/G32/G64/G6C/G47 /G6C/G64/G2E/G20/G2E/G20/G2E/G20/G28/G76/G69/G69/G69/G29

994       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G4C/G65/G74/G20/G74/G68/G65/G20/G6E/G6F/G64/G65/G20/G6F/G66/G20/G74/G68/G65/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G73/G79/G73/G74/G65/G6D/G20/G6C/G69/G65/G73/G20/G61/G74/G20 /G4E/G20/G61/G73/G20/G73/G68/G6F/G77/G6E/G20/G69/G6E/G20/G46/G69/G67/G2E/G20/G32/G34/G2E/G31/G36/G20/G28 /G63/G29/G2C/G20/G74/G68/G65/G6E/G20/G74/G68/G65/G20/G6E/G61/G74/G75/G72/G61/G6C

/G42
/G42/G42/G31/G2E
/G32/G2E/G6E/G43/G4A/G66
/G6C/G49
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74/G20/G41/G42/G6E/G6E/G66/G66
/G20/G20/G20/G20 /G20
/G41/G41/G42/G42/G31/G2E/G31/G2E
/G32/G2E/G32 /G2E/G43/G4A /G43/G4A
/G6C/G49 /G6C/G49
/G6F/G72/G41/G41 /G42/G42/G2E/G2E/G6C/G49 /G6C/G49 /G2E
/G2E/G20/G2E/G20/G28/G78/G29
/G41/G42/G6C/G6C /G6C /G2E/G20/G2E/G20/G2E/G20/G28/G78/G69/G29

/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74/G20/G6D/G61/G73/G73/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G74/G68/G65/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G72/G6F/G74/G6F/G72/G20 /G42/G2C
/G20/G20/G20/G20/G32/G32 /G32
/G42/G42/G2F /G31/G35/G30/G30/G2F/G33 /G31/G36/G36/G2E/G37 /G6B/G67/G96/G6D /G49/G49 /G47

/G61/G6E/G64/G20/G61/G64/G64/G69/G74/G69/G6F/G6E/G61/G6C/G20/G6C/G65/G6E/G67/G74/G68/G20/G6F/G66/G20/G74/G68/G65/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G73/G68/G61/G66/G74/G2C
/G20/G20/G20/G20/G20/G20/G20/G34 /G34
/G32/G32/G31
/G33/G32
/G32/G30/G2E/G30/G36/G2E /G33 /G30/G2E/G36 /G30/G2E/G37/G6D /G37/G30/G30/G6D/G6D/G30/G2E/G31/G64/G6C/G47 /G6C/G64
CNC lathe ata turning centre.
is not a direct example of the current chapter.

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          995
/G20/G54/G6F/G74/G61/G6C/G20/G6C/G65/G6E/G67/G74/G68/G20/G6F/G66/G20/G74/G68/G65/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G73/G68/G61/G66/G74/G2C
/G20/G20/G31/G33/G33/G30/G30 /G37/G30/G30 /G31/G30/G30/G30 /G6D/G6D /G31/G6D /G6C/G6C /G6C

/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G41/G41 /G42/G42/G2E/G2E/G6C/G49 /G6C/G49 /G20/G20/G6F/G72/G20/G20/G20/G41/G41/G34/G30/G30 /G28/G31/G29/G31 /G36/G36/G2E/G37/G6C/G6C /G20/G2E/G20/G2E/G20/G2E/G20/G42/G41/G28/G29/G6C/G6C /G6C
/G41/G30/G2E/G32/G39/G34/G6D /G32/G39/G34 /G6D/G6D /G6C

/G46/G69/G67/G2E/G20/G32/G34/G2E/G31/G37
/G20/G46/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G2C
/G41/G41/G31 /G2E/G31 /G38 /G30 /G31/G30 /G31/G2E /G32 /G37 /G31/G30/G48/G7A/G32/G2E/G32 /G30 /G2E /G32 /G39 /G34 /G34 /G30 /G30/G6E/G43/G4A/G66/G6C/G49
/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20 /G20/G20/G20/G20 /G3D/G20/G34/G2E/G37/G20/G48/G7A /G20/G41/G6E/G73/G2E

996       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G4E/G65/G67/G6C/G65/G63/G74/G69/G6E/G67/G20/G74/G68/G65/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G65/G66/G66/G65/G63/G74/G20/G6F/G66/G20/G74/G68/G65/G20/G67/G65/G61/G72/G73/G2C/G20/G66/G69/G6E/G64/G20/G74/G68/G65/G20/G76/G61/G6C/G75/G65/G20/G6F/G66/G20/G4C/G20/G69/G66/G20/G74/G68/G65/G20/G67/G65/G61/G72/G73/G20/G61/G72/G65/G20/G74/G6F/G20/G62/G65/G20/G61/G74/G20/G74/G68/G65

/G6E/G6F/G64/G65/G20 /G66/G6F/G72/G20 /G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20 /G6F/G73/G63/G69/G6C/G6C/G61/G74/G69/G6F/G6E/G20 /G6F/G66/G20 /G74/G68/G65/G20 /G73/G79/G73/G74/G65/G6D/G20 /G61/G6E/G64/G20 /G68/G65/G6E/G63/G65/G20 /G64/G65/G74/G65/G72/G6D/G69/G6E/G65/G20 /G74/G68/G65/G20 /G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20 /G6F/G66/G20 /G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C

/G20/G20/G20/G20/G20/G32/G32 /G32
/G41/G41 /G41/G28 /G29 /G33/G37/G2E/G35/G28/G30/G2E/G31/G29 /G30/G2E/G33/G37/G35/G6B/G67/G2D/G6D /G49/G6D /G6B
/G61/G6E/G64/G20/G6D/G61/G73/G73/G20/G6D/G6F/G6D/G65/G6E/G74/G20/G6F/G66/G20/G69/G6E/G65/G72/G74/G69/G61/G20/G6F/G66/G20/G74/G68/G65/G20/G63/G65/G6E/G74/G72/G69/G66/G75/G67/G65/G2C
/G42/G42 /G42/G28 /G29 /G33/G30/G28/G30/G2E/G31/G34/G29 /G30/G2E/G35/G38/G38/G6B/G67/G2D/G6D /G49/G6D /G6B

/G61/G6E/G64/G20/G61/G64/G64/G69/G74/G69/G6F/G6E/G61/G6C/G20/G6C/G65/G6E/G67/G74/G68/G20/G6F/G66/G20/G65/G71/G75/G69/G76/G61/G6C/G65/G6E/G74/G20/G73/G68/G61/G66/G74/G2C
/G20/G20/G20/G20/G20/G34 /G34
/G33/G32
/G32/G30/G2E/G30/G35/G34/G2E /G28/G30/G2E/G32/G35/G29 /G30/G2E/G34 /G30/G2E/G30/G35/G31/G38/G6D/G30/G2E/G30/G34/G35/G64/G6C/G47 /G6C/G64

/G20/G20/G20/G20/G20/G41/G2C /G6C/G4C/G20/G20/G20/G20/G61/G6E/G64/G20/G20/G20/G20/G42/G33 /G30/G2E/G30/G35/G31/G38/G6D /G6C/G6C
Note : This picture is given as additional information and is not a direct example of the current chapter.Grinding is a commonly used method for removing the excess material from the cestings,
forgings and weldments

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          997
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74 /G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G41/G41 /G42/G42/G2E/G2E/G6C/G49 /G6C/G49

/G20/G20/G20/G20/G20/G20/G20/G34/G34/G36 /G34
/G31/G28 /G29 /G28/G30/G2E/G30/G35/G34/G29 /G30/G2E/G38/G33/G35 /G31/G30 /G6D

/G20/G46/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G6F/G73/G63/G69/G6C/G6C/G61/G74/G69/G6F/G6E/G73/G2C
/G20/G20/G20/G20/G39/G36/G31 /G2E/G31 /G38 /G34 /G31/G30 /G30 /G2E /G38 /G33 /G35 /G31/G30
/G32/G2E/G32 /G31/G2E /G33 /G30 /G2E /G33 /G37 /G35/G6E
/G41/G41/G43/G4A/G66/G6C/G49/G20/G20/G3D/G20/G36/G30/G2E/G34/G20/G48/G7A /G20/G41/G6E/G73/G2E

998       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73
/G53/G6F/G6C/G75/G74/G69/G6F/G6E/G2E/G20 /G47/G69/G76/G65/G6E/G20/G3A/G20 /G64/G31/G20/G3D/G20/G36/G30/G20/G6D/G6D/G20/G3D/G20/G30/G2E/G30/G36/G20/G6D/G20/G3B/G20 /G6C/G31/G20/G3D/G20/G39/G35/G30/G20/G6D/G6D/G20/G3D/G20/G30/G2E/G39/G35/G20/G6D/G20/G3B/G20 /G64/G32/G20/G3D/G20/G34/G30/G20/G6D/G6D/G20/G3D/G20/G30/G2E/G30/G34/G20/G6D/G20/G3B

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          999
/G57/G65/G20/G6B/G6E/G6F/G77/G20/G74/G68/G61/G74 /G20/G20/G20/G20/G41/G41 /G42/G42/G2E/G2E/G6C/G49 /G6C/G49 /G20/G20/G20/G20/G20/G6F/G72/G20/G20/G20/G42
/G41/G42 /G42 /G42
/G41/G32/G37/G32/G30/G2E/G33/G34/G38/G30/G30/G49/G6C/G6C /G6C /G6C/G49
/G41/G6C/G73/G6F /G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20

/G31/G41/G42/G42 /G45 /G33 /G42/G31/G31 /G31 /G31
/G2E /G6C/G6C /G6C/G49 /G49 /G6C /G6C
/G42/G42 /G42/G31/G31 /G31 /G31
/G37/G39/G32/G37/G32 /G30/G2E/G39/G35 /G30/G2E/G33/G34 /G30/G2E/G30/G39/G35 /G6C/G6C /G6C
/G20/G20/G20/G20/G20/G20/G20/G20/G42/G42
/G42/G42 /G42/G28/G30/G2E/G30/G39/G35 /G30/G2E/G39/G35 /G30/G2E/G33/G34 /G29/G37/G39
/G32/G37/G32 /G28/G30/G2E/G39/G35 /G30/G2E/G33/G34 /G29/G28/G30/G2E/G30/G39/G35 /G29/G6C/G6C
/G6C/G6C /G6C
/G20/G20/G42
/G32

/G32
/G42/G42 /G42/G31/G2E/G30/G34/G35 /G31/G2E/G33/G34 /G30/G2E/G32/G39
/G30/G2E/G30/G39 /G30/G2E/G39/G38 /G30/G2E/G33/G34/G28 /G29/G6C
/G32/G32
/G42/G42 /G42 /G42 /G30/G2E/G30/G32/G36 /G30/G2E/G32/G38 /G30/G2E/G31/G28 /G29 /G31/G2E/G30/G34/G35 /G31/G2E/G33/G34/G28 /G29 /G6C/G6C /G6C /G6C
/G20/G32
/G42/G42/G31/G2E/G34/G34/G28 /G29 /G31/G2E/G33/G32/G35 /G30/G2E/G30/G32/G36 /G30 /G6C/G6C
/G32

/G32
/G42/G31/G2E/G33/G32/G35 /G28/G31/G2E/G33/G32/G35/G29 /G34 /G31/G2E/G34/G34 /G30/G2E/G30/G32/G36 /G31/G2E/G33/G32/G35 /G31/G2E/G32/G36/G37
/G32 /G31/G2E/G34/G34 /G32/G2E/G38/G38/G6C
/G20/G20/G20/G20/G20/G20/G20 /G3D/G20/G30/G2E/G39/G20/G6D/G20/G20/G6F/G72/G20/G20/G30/G2E/G30/G32/G20/G6D
/G61/G6E/G64 /G20/G20/G20/G20/G41/G42/G30/G2E/G33/G34 /G30/G2E/G33/G30/G36/G6D /G6C/G6C /G20/G20/G6F/G72/G20/G20/G20/G30/G2E/G30/G30/G36/G38/G6D

/G20/G20/G20/G20/G20/G20/G20/G20/G34/G34/G36 /G34
/G31/G28/G29 /G28 /G30 /G2E /G30 /G36 /G29 /G31/G2E /G32 /G37/G31/G30 /G6D/G33/G32 /G33/G32/G4A/G64

/G20/G4E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74 /G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G6F/G73/G63/G69/G6C/G6C/G61/G74/G69/G6F/G6E/G73/G20/G66/G6F/G72 /G20/G61/G20/G73/G69/G6E/G67/G6C/G65/G20/G6E/G6F/G64/G65/G20/G73/G79/G73/G74/G65/G6D/G2C
/G31
/G41/G41/G31 /G2E/G31 /G38 /G34 /G31/G30 /G31/G2E /G32 /G37 /G31/G30/G48/G7A/G32/G2E/G32 /G30 /G2E /G33 /G30 /G36 /G38 /G30 /G30/G6E/G43/G4A/G66/G6C/G49

/G53/G69/G6D/G69/G6C/G61/G72/G6C/G79/G20/G6E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G6F/G73/G63/G69/G6C/G6C/G61/G74/G69/G6F/G6E/G73/G20/G66/G6F/G72/G20/G61/G20/G74/G77/G6F/G20/G6E/G6F/G64/G65/G20/G73/G79/G73/G74/G65/G6D/G2C
/G32
/G41/G41/G31 /G2E/G31 /G38 /G34 /G31/G30 /G31/G2E /G32 /G37 /G31/G30/G48/G7A/G32/G2E/G32 /G30 /G2E /G30 /G30 /G36 /G38 /G38 /G30 /G30/G6E/G43/G4A/G66/G49/G49

1000       /G6C               /G54/G68/G65/G6F/G72/G79/G20/G6F/G66/G20/G4D/G61/G63/G68/G69/G6E/G65/G73

/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G20/G44/G61/G74/G61/G20/G6F/G66/G20/G72/G6F/G74/G6F/G72/G73
/G20/G20/G20/G4D/G61/G73/G73/G2C /G6B/G67 /G34/G35/G30 /G35/G34/G30 /G33/G36/G30
/G52/G61/G64/G69/G75/G73/G20/G6F/G66 /G32/G35/G30 /G33/G30/G30
/G67/G79/G72/G61/G74/G69/G6F/G6E/G2C/G20/G6D/G6D/G52/G6F/G74/G6F/G72 /G50 /G51 /G52

/G43/G68/G61/G70/G74/G65/G72/G20/G32/G34/G20 /G3A/G20/G54/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G56/G69/G62/G72/G61/G74/G69/G6F/G6E/G73            /G6C          1001

/G31/G2E/G54/G68/G65/G20/G6E/G61/G74/G75/G72/G61/G6C/G20/G66/G72/G65/G71/G75/G65/G6E/G63/G79/G20/G6F/G66/G20/G66/G72/G65/G65/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G73/G20/G6F/G66/G20/G61/G20/G73/G68/G61/G66/G74/G20/G69/G73
/G20/G28 /G61/G29/G20/G32/G71
/G49/G28/G62/G29/G32/G2E/G71/G49 /G28/G63/G29/G31
/G32/G71
/G49/G28/G64/G29/G31/G2E

/G32/G2E/G41/G74/G20/G61/G20/G6E/G6F/G64/G61/G6C/G20/G70/G6F/G69/G6E/G74/G20/G69/G6E/G20/G61/G20/G73/G68/G61/G66/G74/G2C/G20/G74/G68/G65/G20/G61/G6D/G70/G6C/G69/G74/G75/G64/G65/G20/G6F/G66/G20/G74/G6F/G72/G73/G69/G6F/G6E/G61/G6C/G20/G76/G69/G62/G72/G61/G74/G69/G6F/G6E/G20/G69/G73/G28/G61/G29/G20/G20/G7A/G65/G72/G6F /G28 /G62/G29 /G6D/G69/G6E/G69/G6D/G75/G6D /G28 /G63/G29/G6D /G61 /G78 /G69 /G6D /G75 /G6D

/G32/G31
/G33/G64/G6C/G6C
/G64
/G28/G63/G29/G20/G20/G34
/G32/G31/G32
/G32/G64/G6C/G6C /G6C/G64/G28/G64/G29/G20/G34 /G34
/G31/G31/G31/G32/G33
/G32/G33/G64/G64/G6C/G6C /G6C /G6C
/G64/G64
/G46/G69/G67/G2E/G20/G32/G34/G2E/G32/G31

/G64/G64
/G34/G2E/G41/G20/G73/G68/G61/G66/G74/G20/G63/G61/G72/G72/G79/G69/G6E/G67/G20/G74/G77/G6F/G20/G72/G6F/G74/G6F/G72/G73/G20/G61/G73/G20/G69/G74/G73/G20/G65/G6E/G64/G73/G20/G77/G69/G6C/G6C/G20/G68/G61/G76/G65/G28/G61/G29 /G6E/G6F/G20/G6E/G6F/G64/G65 /G28 /G62/G29 /G6F/G6E/G65/G20/G6E/G6F/G64/G65 /G28 /G63/G29 /G74/G77/G6F/G20/G6E/G6F/G64/G65/G73 /G28 /G64/G29 /G74/G68/G72/G65/G65/G20/G6E/G6F/G64/G65/G73

/G35/G2E/G41/G20/G73/G68/G61/G66/G74/G20/G63/G61/G72/G72/G79/G69/G6E/G67/G20/G74/G68/G72/G65/G65/G20/G72/G6F/G74/G6F/G72/G73/G20/G77/G69/G6C/G6C/G20/G68/G61/G76/G65/G28/G61/G29 /G6E/G6F/G20/G6E/G6F/G64/G65 /G28 /G62/G29 /G6F/G6E/G65/G20/G6E/G6F/G64/G65 /G28 /G63/G29 /G74/G77/G6F/G20/G6E/G6F/G64/G65/G73 /G28 /G64/G29 /G74/G68/G72/G65/G65/G20/G6E/G6F/G64/G65/G73

/G31/G2E/G20/G28/G63/G29 /G32/G2E/G20/G28/G61/G29 /G33/G2E/G20/G28/G64/G29 /G34/G2E/G20/G28/G62/G29 /G35/G2E/G20/G28/G63/G29

1002       /G6C               Theory of Machines1002       /G6C               Theory of Machines
1002ComputerComputerComputerComputerComputer
AidedAidedAidedAidedAided
Analysis andAnalysis andAnalysis andAnalysis andAnalysis and
Synthesis ofSynthesis ofSynthesis ofSynthesis ofSynthesis of
MechanismsMechanismsMechanismsMechanismsMechanisms25
2. Computer Aided Analysis for
Four Bar Mechanism

(Freudenstein’ s Equation).
3. Programme for Four Bar
4. Computer Aided Analysis for
Slider Crank Mechanism.
6. Coupler Curves.
7. Synthesis of Mechanisms.8. Classifications of Synthesis
Problem.
9. Precision Points for Function
Generation.
10. Angle Relationship for
function Generation.
11. Graphical Synthesis of Four
Bar Mechanism.
12. Graphical Synthesis of Slider
13. Computer Aided (Analytical)

Synthesis of Four Bar
15. Least square Technique.
16. Programme using Least
Square Technique.
17. Computer Aided Synthesis of
Four Bar Mechanism With
Coupler Point.
18. Synthesis of Four Bar
Mechanism for BodyGuidance
19. Analytical Synthesis for
slider Crank Mechanism25.1. Introduction
We have already discussed in chapters 7 and 8, the

graphical methods to determine velocity and accelerationanalysis of a mechanism. It may be noted that graphicalmethod is only suitable for determining the velocity andacceleration of the links in a mechanism for a single positionof the crank. In order to determine the velocity andacceleration of the links in a mechanism for differentpositions of the crank, we have to draw the velocity andacceleration diagrams for each position of the crank whichis inconvenient. In this chapter, we shall discuss

inconvenient. In this chapter, we shall discuss the analyticalexpressions for the displacement, velocity and accelerationin terms of general parameters of a mechanism andcalculations may be performed either by a desk calculatoror digital computer.

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1003
25.2. Computer Aided Analysis for Four Bar Mechanism (Freudenstein’s
Equation)
Consider a four bar mechanism ABCD , as shown in Fig. 25.1 ( a), in which AB = a, BC = b,
CD = c, and DA = d. The link AD is fixed and lies along X-axis. Let the links AB (input link), BC
(coupler)  and DC (output link)  make angles ,a n dθβ φ  respectively along the X-axis or fixed link
AD.

AD.
(a) Four bar mechanism.         ( b) Components along X-axis and Y-axis.
Fig. 25.1
The relation between the angles and link lengths may be developed by considering the
links as vectors. The expressions for displacement, velocity and acceleration analysis are derived
1. Displacement analysis
For equilibrium of the mechanism, the sum of the components along X-axis and along
Y-axis must be equal to zero. First of all, taking the sum of the components along X-axis as shown

in Fig. 25.1 ( b), we have
cos cos cos 0abcdθ+ β− φ− = . . . (i)
or cos cos cosβ= φ+ − θ bcd a
22 2cos ( cos cos )β= φ+ − θ bc d a
        = 22 2 22cos 2 cos coscd c d aφ+ + φ+ θ
               2 cos cos 2 cosac ad−φ θ −θ . . . (ii)
Now taking the sum of the components along Y-axis, we have
  sin sin sin 0abcθ+ β− φ= . . . (iii)
or sin sin sinbc aβ= φ− θ
       22 2sin ( sin sin )bc aβ= φ− θ
22 22sin sin 2 sin sinca a c=φ +θ −φ θ . . . (iv)

Adding equations  (ii) and (iv),
       2 22 2 22 2 2 22(cos sin ) (cos sin ) 2 cos (cos sin )β+ β = φ+ φ + + φ+ θ+ θ bcd c d a
2( c o s c o ss i n s i n ) 2 c o sac ad−φ θ + φ θ −θ

1004       /G6C               Theory of Machines
or      222 22 cos 2 (cos cos sin sin ) 2 cos bc d c d a a c a d=++ φ +− φθ +φθ − θ
or222 22 (cos cos sin sin ) 2 cos 2 cosac a b c d cd adφθ +φθ =−+++ φ − θ
     222 2
cos cos sin sin cos cos2abcd d d
ac a c−++φ θ + φ θ = +φ −θ . . . (v)
Let222 2
12 3;; a n d2dd a b c dkk kac a c−++== = . . . (vi)
Equation (v) may be written as
         123 cos cos sin sin cos cos kk k φθ +φθ = φ − θ + . . . (vii)

or   cos ( φ – θ) or     123 cos( ) cos cosθ−φ = φ− θ+ kk k
The equation (vii) is known as  Freudenstein’s equation.
Since it is very difficult to determine the value of φ for the given value of θ, from
equation (vii), therefore it is necessary to simplify this equation.
From trigonometrical ratios, we know that
22t a n ( /2 )sin
1t a n ( / 2 )φφ=
+φ   and   2
21t a n ( / 2 )cos
1t a n ( / 2 )−φφ=
+φ
Substituting these values of sinφand cosφ in equation (vii),
   2

2
221 tan ( / 2) 2 tan( / 2)cos sin
1t a n ( / 2 ) 1t a n ( / 2 )−φ φ×θ + ×θ
+φ +φ
           2
12 321t a n ( / 2 )cos
1t a n ( / 2 )kk k−φ=× − θ +
+φ
2cos [1 tan ( / 2)] 2sin tan ( / 2)θ− φ + θ φ
          22 2
12 3[1 tan ( / 2)] cos [1 tan ( / 2)] [1 tan ( / 2)]kk k=− φ− θ + φ++ φ
 2cos cos tan ( / 2) 2sin tan ( / 2)θ− θ φ + θ φ
                                   22 2
11 2 2 33 tan ( / 2) cos cos tan ( / 2) tan ( / 2) kk k k kk=− φ − θ − θ φ ++ φ
Rearranging this equation,
22 22

22 22
12 3 cos tan ( / 2) tan ( / 2) cos tan ( / 2) tan ( / 2) 2sin tan ( / 2) kk k −θ φ+ φ+ θ φ− φ+ θ φ
           12 3 cos cos kk k =− θ+ − θ+
2
12 3 13 2 tan ( / 2) [cos cos ] 2sin tan ( / 2) cos (1 ) 0−φ θ − − θ + +θ φ − − + θ + = kk k kk k
2
23 1 1 3 2 [ ( 1) c o s ] t a n / 2 ( 2 s i n ) t a n / 2 [ ( 1) c o s ] 0 kk k k k k− θ +− φ+ − θ φ++− + θ =
(By changing the sign)
or2tan ( / 2) tan( / 2) 0φ+ φ+ = AB C . . . (viii)

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1005
where 23 1 ,
13 2(1 ) cos
2sin ,and
(1 ) cosAk k k
B
Ck k k=− θ ++ 
=− θ =+− + θ. . . (ix)
The equation (viii) is a quadratic equation in tan( / 2).φ  Its two roots are
          24tan ( / 2)2BB A C
A−± −φ=
or          2
1 42t a n2BB A C
A−−± − φ=
. . . (x)
From this equation (x), we can find the position of output link CD (i.e. angle φ) if the

length of the links ( i.e. a, b, c and d) and position of the input link AB (i.e. angle θ) is known.
If the relation between the position of input link AB (i.e. angle θ) and the position of
coupler link BC (i.e. angle β) is required, then eliminate angle φ from the equations (i) and (iii).
The equation (i) may be written as
   cos cos cosφ= θ+ β− ca b d ... (xi)
           22 22 22cos cos cos 2 cos cosφ= θ+ β+ θ β ca b a b
   + 22 cos 2 cos−θ −βda d b d . . . (xii)

+ 22 cos 2 cos−θ −βda d b d . . . (xii)
Now equation (iii) may be written as
   sin sin sinφ= θ+ β ca b . . . (xiii)
Inner view of an aircraft engine.

1006       /G6C               Theory of Machines
           22 22 22sin sin sin 2 sin sinφ= θ+ β+ θ β ca b a b ... (xiv)
Adding equations  (xii) and (xiv),
 22 2 22 2 22 2(cos sin ) (cos sin ) (cos sin )φ+ θ = θ+ θ + β+ β ca b
    22 (cos cos sin sin ) 2 cos 2 cosab d a d b d+θ β + θ β + − θ − β
or      22 22 (cos cos sin sin ) =++ θβ +θβca b a b
                                       22 cos 2 cos+− θ − βda d b d

or22222 (cos cos sin sin ) 2 cos 2 cosθβ +θβ =−−−+ θ + β a b c a b d ad bd
      2222
cos cos sin sin cos cos2−−−θβ +θβ = + θ + βcabd d d
ab b a. . . (xv)
Let                14;;==ddkkab   and   2222
52−−−=cabdkab. . . (xvi)
∴ Equation (xvi) may be written as
145 cos cos sin sin cos cosθβ +θβ = β + θ + kk k . . . (xvii)
     22t a n ( /2 )sin ,
1t a n ( / 2 )ββ=
+β   and   2
1t a n ( / 2 )−ββ=
+β

+β
Substituting these values of sinβ and cosβ in equation (xvii),
   2
1t a n ( / 2 ) 1t a n ( / 2 )  −β βθ+ θ 
+β +β    
2
14 521t a n ( / 2 )cos
1t a n ( / 2 )−β=+ θ +
+βkk k
2cos [1 tan ( / 2)] 2sin tan( / 2)θ− β + θ β
                      22
141 tan ( / 2) cos 1 tan ( / 2) =−β+ θ +β kk  2
51t a n ( / 2 )++ βk
2cos cos tan ( / 2) 2sin tan( / 2)θ− θ β + θ β
            = 22

= 22
11 4 4 tan ( / 2) cos cos tan ( / 2)− β +θ +θ βkk k k
                                   2
55 tan ( / 2)++ βkk
22 2 2
14 5 cos tan ( / 2) tan ( / 2) cos tan ( / 2) tan ( / 2)−θ β+ β− θ β− β kk k
14 5 2sin tan( / 2) cos cos 0 kk k +θ β − − θ − + θ =
2
45 1 45 1 tan ( / 2)[( 1)cos ] 2sin tan( / 2) [( 1)cos ] 0− β + θ +−+ θ β − − θ ++ = kk k kk k

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1007
or2
45 1 45 1[( 1) cos ] tan ( / 2) ( 2sin ) tan( / 2) [( 1) cos ] 0+θ + − β + − θ β + −θ + + =kk k kk k
or2tan ( / 2) tan( / 2) 0β+ β+ = DE F . . . (xviii)
where45 1
45 1(1 ) c o s ,
2sin , and
[( 1)cos ]=+ θ + − 
=− θ −θ + + Dk kk
E
F= k k k... (xix)
The equation (xviii) is a quadratic equation in tan( / 2)β . Its two roots are
              24tan( / 2)2−± −β=EE D F
D

24tan( / 2)2−± −β=EE D F
D
or            21 42t a n2−−± − β=
EE D F
D. . . (xx)
From this equation (xx), we can find the position of coupler link BC (i.e. angle β).
Note:   The angle αmay be obtained directly from equation (i) or (iii) after determining the angle φ.
2. Velocity analysis
Let    1ω = Angular velocity of the link /=θAB d dt ,
  2ω = Angular velocity of the link / BC d dt=β , and
 3ω = Angular velocity of the link / CD d dt=φ .

3ω = Angular velocity of the link / CD d dt=φ .
Differentiating equation (i) with respect to time,
    sin sin sin 0dddabcdt dt dtθβφ−θ × −β × +φ × =
or     123sin sin sin 0abc−ω θ −ω β +ω φ = ... (xxi)
Again, differentiating equation (iii) with respect to time,
    cos cos cos 0dddabc
dt dt dtθβφθ× + β× − φ× =
or    123cos cos cos 0abcωθ + ωβ − ωφ = ... (xxii)
Multiplying the equation  (xxi) by cosβ and equation (xxii) by sinβ,
123sin cos sin cos sin cos 0abc− ωθβ − ω ββ + ω φβ = ... (xxiii)

and123cos sin cos sin cos sin 0abcωθ β + ωβ β − ωφ β = ... (xxiv)
Adding equations  (xxiii) and (xxiv),
   13sin( ) sin ( ) 0acωβ − θ + ωφ − β =
∴     1
3sin( )
sin( )a
c−ω β − θω=φ−β... (xxv)

1008       /G6C               Theory of Machines
Again, multiplying the equation (xxi) by cosφ and equation (xxii) by sinφ,
 123sin cos sin cos sin cos 0abc−ω θ φ −ω β φ +ω φ φ = ... (xxvi)
and    123cos sin cos sin cos sin 0abcω θ φ+ ω β φ− ω φ φ= ... (xxvii )
Adding equations (xxvi) and (xxvii ),
   12sin( ) sin( ) 0abωφ − θ + ωφ − β =
∴1
2sin( )
sin( )a
b−ω φ − θω=φ−β... (xxviii )
From equations (xxv) and (xxviii ), we can find ω3 and 2ω, if θφβ ,,, ,,abc  and 1ω are
known.

known.
3. Acceleration analysis
Let1α = Angular acceleration of the link 1/ AB d dt=ω ,
2α = Angular acceleration of the link 2/ BC d dt=ω , and
3α = Angular acceleration of the link 3/ CD d dt=ω .
Differentiating equation (xxi) with respect to time,
12
12cos sin cos sindd ddab
dt dt dt dtωω θβ  −ω θ × +θ × −ω β × +β ×  
3
3cos sin 0d dc
dt dtω φ +ω φ × + φ × =
. . . =× +∵ ()dduu
dx dxvd uvv ×
dx
or22
11 22cos sin cos sin aa bb−ω θ − θ α−ω β − β α
           2

2
33cos sin 0cc+ω φ + φ α= ... (xxix)
Again, differentiating equation  (xxii) with respect to time,
   12
12 sin cos sin cosdd ddab
dt dt dt dtωω θβ  ω× − θ × + θ × + ω× − β × + β ×  
  3
3 sin cos 0d dc
dt dtω φ −ω × −φ × + φ × =
or22
11 22sin cos sin cosaabb−ω θ + θ α−ω β + β α
 2
33sin cos 0+ω φ − φ α=cc ... (xxx)
Multiplying equation  (xxix) by cosφ, and equation (xxx) by sinφ,
22
11 2cos cos sin cos cos cos− ω θφ − αθφ − ω βφaa b
          22

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1009
and22
112sin sin cos sin sin sin− ω θφ + α θφ − ω βφaab
22
23 3cos sin sin cos sin 0+α β φ +ω φ −α φ φ =bc c ... (xxxii )
Adding equations  (xxxi) and (xxxii ),
2
11(cos cos sin sin ) (sin cos cos sin )−ω φ θ + φ θ+α φ θ − φ θaa
2
22(cos cos sin sin ) (sin cos cos sin )bb−ω φ β + φ β+α φ β − φ β
22 2
3(cos sin ) 0+ω φ + φ=c
222

22 2
222
11 22 3cos( ) sin ( ) cos( ) sin ( ) 0− ω φ−θ + α φ−θ − ω φ−β + α φ−β + ω =aa bbc
∴    22 2
11 23
2sin ( ) cos( ) cos( )
sin ( )−α φ − θ +ω φ − θ +ω φ − β −ωα=φ−βaa bc
b ... (xxxiii )
Again multiplying equation (xxix) by cosβ and equation (xxx) by sinβ,
22 2
11 2 2cos cos sin cos cos sin cos− ω θβ − α θβ − ω β − α ββaa b b
        2
33cos cos sin cos 0+ω φ β +α φ β =cc ... (xxxiv )
and22 2
112 2sin sin cos sin sin cos sin−ω θ β +α θ β −ω β +α β βaab b

2
33sin sin cos sin 0+ ω φβ − α φβ =cc ... (xxxv )
Adding equations (xxxiv ) and (xxxv ),
    22 2 2
112(cos cos sin sin ) (sin cos cos sin ) (cos sin )−ω β θ + β θ+α β θ − β θ−ω β + βaab
2
33(cos cos sin sin ) (sin cos cos sin ) 0cc+ω φ β + φ β +α φ β − φ β =
22 2
11 2 33cos ( ) sin ( ) cos ( ) sin ( ) 0−ω β − θ+α β − θ−ω+ω φ − β+α φ − β=aa b cc
∴    22 2
11 2 3
3sin ( ) cos ( ) cos( )
sin ( )−α β − θ+ω β − θ+ω−ω φ − β=φ−βaa b cac... (xxxvi )

From equations (xxxiii ) and (xxxvi ), the angular acceleration of the links BC and CD (i.e.
2α and 3α) may be determined.
25.3. Programme for Four Bar Mechanism
The following is a programme in Fortran for determining the velocity and acceleration of
the links in a four bar mechanism for different position of the crank.
C PROGRAM TO FIND THE VELOCITY AND ACCELERATION IN A FOUR-BAR
C MECHANISM

DIMENSION PH (2), PHI (2), PP (2), BET (2), BT (2), VELC (2), VELB (2), ACCC (2),ACCB (2), C1 (2), C2 (2), C3 (2), C4 (2), B1 (2), B2 (2), B3 (2), B4 (2)
READ (*, *) A, B, C, D, VELA, ACCA, THETA
PI = 4.0 * A TAN (1.0)THET = 0IHT = 180/THETADTHET = PI/IHT

1010       /G6C               Theory of Machines
DO 10 J = 1, 2 * IHT
THET = (J – 1) * DTHETAK = (A * A – B * B + C * C + D * D) * 0.5)TH = THET * 180/PIAA = AK – A * (D – C) * COS (THET) – (C * D)
BB = – 2.0 * A* C * SIN (THET)

CC = AK – A * (D + C) * COS (THET) + (C * D)AB = BB * * 2 – 4 * AA * CCIF     (AB . LT . 0) GO TO 10PHH = SQRT (AB)PH (1) = – BB + PHHPH (2) = – BB – PHHDO 9 I =   1, 2PHI (I) = ATAN (PH (I) * 0.5/AA) * 2PP (I) = PHI (I) * 180/PIBET (I) = ASIN ((C * SIN (PHI (I)) – A * SIN (THET)) / B)BT (I) = BET (I) * 180/PI
VELC (I) = A * VELA * SIN (BET (I) – THET) / (C * SIN (BET (I) – PHI (I)))

VELB (I) = (A * VELA * SIN (PHI (I) – THET) ) / (B * SIN (BET (I) – PHI (I))))C1 (I) = A * ACCA * SIN (BET (I) – THET)C2 (I) = A * VELA * * 2 * COS (BET (I) – THET) + B * VELB (I) * * 2C3 (I) = C * VELC (I) * * 2 * COS (PHI (I) – BET (I) )C4 (I) = C * SIN (BET (I) – PHI (I))ACCC (I) = (C1 (I) – C2 (I) + C3 (I) ) / C4 (I)B1 (I) = A* ACCA* SIN (PHI (I) – THET )B2 (I) = A * VELA * * 2 * COS (PHI (I) – THET )B3 (I) = B * VELB (I) * * 2 * COS (PHI (I) – BET (I) ) – C * VELC (I) * * 2B4 (I) = B *

(I) – BET (I) ) – C * VELC (I) * * 2B4 (I) = B * (SIN (BET (I) – PHI (I))))

9 ACCB (I) = (B1 (I) – B2 (I) – B3 (I)) / B4 (I)
IF (J . NE . 1) GO TO 8WRITE (*, 7)
7 FORMAT (4X,’ THET’, 4X,’ PHI’, 4X,’ BETA’, 4X,’ VELC’, 4X,’ VELB’, 4X,’ ACCC’, 4X,’
ACCB’)
8 WRITE (*, 6) TH, PP (1), BT (1), VELC (1), VELB (1), ACCC (1), ACCB (1)6 FORMAT (8F8 . 2)
WRITE (*, 5) PP (2), BT (2), VELC (2), VELB (2), ACCC (2), ACCB (2)
5 FORMAT (8X, 8F8 . 2)10 CONTINUE
STOP
END
The various input variables are
   A, B, C, D  = Lengths of the links AB, BC, CD,  and DA  respectively in mm,

THETA  = Interval of the input angle in degrees,
        VELA  = Angular Velocity of the input link AB in rad/s, and
        ACCA  = Angular acceleration of the input link in rad/s2.
The output variables are :
         THET  = Angular displacement of the input link AB in degrees,
           PHI  = Angular displacement of the output link DC in degrees,
         BET A  = Angular displacement of the coupler link BC in degrees,
        VELC  = Angular velocity of the output link DC in rad/s,

VELB  = Angular velocity of the coupler link  BC in rad/s,
        ACCC  = Angular acceleration of the output link DC in rad/s2,
        ACCB  = Angular acceleration of the coupler link BC in rad/s2.

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1011
Example 25.1.  ABCD  is a four bar mechanism, with link AD fixed. The lengths of the links
are
AB = 300 mm; BC = 360 mm; CD  = 360 mm and AD  = 600 mm.
The crank AB has an angular velocity of 10 rad/s and an angular retardation of 30 rad/s2,
both anticlockwise. Find the angular displacements, velocities and accelerations of the links BC
and CD, for an interval of 30° of the crank AB .
Solution.

Solution.
Given input :
A = 300, B = 360,   C = 360,   D = 600, VA = 10, ACCA = –30, THETA = 30OUTPUT :
THET PHI BETA VELC VELB ACCC ACCB
.00 – 114.62 – 65.38 – 10.00 – 10.00 – 61.67 121.67
114.62 65.38 – 10.00 – 10.00 121.67 – 61.67
30.00 – 144.88 – 82.70 – 8.69 – .84 101.52 181.43
97.30 35.12 – .84 – 8.69 181.43 101.52
60.00 – 166.19 – 73.81 – 6.02 6.02 38.02 77.45
106.19 13.81 6.02 – 6.02 77.45 38.02
90.00 174.73 – 47.86 – 8.26 12.26 – 180.18 216.18
132.14 – 5.27 12.26 – 8.26 216.18 – 180.18

270.00 – 132.14 5.27 12.26 – 8.26 – 289.73 229.73
– 174.73 47.86 – 8.26 12.26 229.73 – 289.73
300.00 – 106.19 – 13.81 6.02 – 6.02 – 113.57 – 1.90
166.19 73.81 – 6.02 6.02 – 1.90 – 113.57
330.00 – 97.30 – 35.12 – .84 – 8.69 – 170.39 – 49.36
144.88 82.70 – 8.69 – .84 – 49.36 – 176.39
25.4. Computer Aided Analysis For Slider Crank Mechanism
A slider crank mechanism is shown in Fig. 25.2 ( a). The slider is attached to the connecting

rod BC of length b. Let the crank AB of radius a rotates in anticlockwise direction with uniform
angular velocity 1ω rad/s and an angular acceleration 1α rad/s2. Let the crank makes an angle θ
with the X-axis and the slider reciprocates along a path parallel to the X-axis, i.e. at an eccentricity
CD =  e, as shown in Fig. 25.2 ( a).(a)                  ( b)
Fig. 25.2 Slider crank mechanism.

1012       /G6C               Theory of Machines
The expressions for displacement, velocity and acceleration analysis are derived as
For equilibrium of the mechanism, the sum of the components along X-axis and along Y-
axis must be equal to zero. First of all, taking the sum of the components along X-axis, as shown in
Fig. 25.2 ( b), we have
cos cos( ) 0ab xθ+ −β − = ... (β in clockwise direction  from X-axis is taken – ve)
or cos cosbx aβ= − θ ... (i)

22 222cos cos 2 cosbx a x aβ= + θ− θ ... (ii)
Now taking the sum of components along Y-axis, we have
sin( ) sin 0be a−β + + θ=
or sin sinbe a−β + =θ
∴ sin sinbe aβ= − θ ... (iii)
      22 2 22sin sin 2 sinbe a e aβ= + θ− θ ... (iv)
Adding equations (ii) and (iv),
    22 2 2 2 22 2(cos sin ) (cos sin ) 2 cos 2 sinb x e a xa eaβ+ β = + + θ+ θ − θ− θ
  22 2 22c o s2s i n bx e a x a e a=++− θ − θ

22 2 22c o s2s i n bx e a x a e a=++− θ − θ
or           22 2 2(2c o s) 2 s i n 0 xa x a b e e a+− θ + − + − θ =
or           2
12 0 xk x k++ = ... (v)
where 1 2c o s ka=− θ , and 222
2 2s i n ka b e e a=−+− θ ... (vi)
The equation (v) is a quadratic equation in x. Its two roots are
               2
11 2 4
2kk kx−± −= ... (vii)
From this expression, the output displacement x may be determined if the values of a, b, e

and  θ are known. The position of the connecting rod BC (i.e. angle β) is given by
                  sinsin ( )ae
bθ−−β =
or                       sinsinea
b−θβ=
∴                        1 sinsinea
b−−θβ=... (viii)

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1013
Note :  When the slider lies on the X-axis, i.e. the line of stroke of the slider passes through the axis of
rotation of the crank, then eccentricity, e = 0. In such a case, equations (vi) and (viii) may be written as
     1 2c o s ka=− θ ,   and   22
2kab=−
and       1 sinsina
b−−θβ= 
Let     1ω = Angular velocity of the crank /, AB d dt=θ

2ω = Angular velocity of the connecting rod / BC d dt=β , and
    Sv = Linear velocity of the slider = /.dx dt
sin sindd x dbadt dt dtβθ×− β× = − ×− θ×
or12sin sin 0dxab
dt−ω θ −ω β − = ... (ix)
Again, differentiating equation  (iii) with respect to time,
    cos cosddba
dt dtβθβ× =− θ×
or         12cos cos 0abωθ + ωβ = ... (x)
Multiplying equation  (ix) by cosβ and equation (x) by sinβ,
    12sin cos sin cos cos 0dxab

12sin cos sin cos cos 0dxab
dt−ω θ β −ω β β − × β =  ... (xi)
and12cos sin cos sin 0abωθ β + ωβ β = ... (xii)
Adding equations  (xi) and (xii),
1(sin cos cos sin ) cos 0dxa
dtωβ θ − β θ − × β =
1sin( ) cosdxa
dtωβ − θ = × β
∴1sin( )
cosa dx
dtωβ − θ=β... (xiii)
From this equation, the linear velocity of the slider (Sv) may be determined.
The angular velocity of the connecting rod BC (i.e. 2ω) may be determined from equa-
tion (x) and it is given by
    1
2cos
cosa
b−ω θω=β... (xiv)

1014       /G6C               Theory of Machines
Let    1α = Angular acceleration of the crank 1/ AB d dt=ω ,
   2α = Angular acceleration of the connecting rod = 2/,dd tω and
   Sa = Linear acceleration of the slider = 22/dxd t
Differentiating equation (ix) with respect to time,
 2
12
122cos sin cos sin 0dd dd d xab
dt dt dt dt dtωω θβ   −ω θ × + θ × −ω β × +β × − =    
 222
11 2 22sin cos sin cos 0dxab
dt −α θ + ω θ −α β + ω β − = ... (xv)

Differentiating equation  (x) with respect to time,
12
12 sin cos sin cos 0dd ddabdt dt dt dtωω θβ   ω× − θ × + θ × + ω× − β × + β × =    
22
11 2 2cos sin cos sin 0 ab α θ−ω θ + α β−ω β = ... (xvi)
Multiplying equation  (xv) by cosβ and equation (xvi) by sinβ,
   22 2
11 2 2sin cos cos cos sin cos cos ab  −α θ β + ω θ β −α β β + ω β  
2
2cos 0dx
dt−× β = ... (xvii)
and                  22 2

and                  22 2
11 2 2cos sin sin sin cos sin sin 0 ab  α θ β−ω θ β + α β β−ω β =  ... (xviii)
The chain-belt at the bottom of a bulldozer provides powerful grip, spreads weight
and force on the ground, and allows to exert high force on the objects to be moved.

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1015
Adding equations (xvii) and (xviii),
2
11(sin cos cos sin ) (cos cos sin sin ) aα βθ −βθ − ω βθ +βθ
222 2
22(cos sin ) cos 0dxb
dt−ω β + β− × β =
 222
11 22sin ( ) cos ( ) cos 0dxaa b
dtα β−θ − ω β−θ − ω − × β=
∴22 2
11 2
2sin ( ) cos ( )
cosaa b dx
dtα β−θ − ω β−θ − ω=β... (xix)
From this equation, the linear acceleration of the slider ( aS) may be determined.

The angular acceleration of the connecting rod BC (i.e. 2α) may be determined from
equation  (xvi) and it is given by,
22
11 22( cos sin ) sin
cosab
bαθ − ωθ − ωβα=β... (xx)
25.5. Programme for a Slider Crank Mechanism
The following is a programme in Fortran to find the velocity and acceleration in a slider
crank mechanism.
c PROGRAM TO FIND THE VELOCITY AND ACCELERATION IN A SLIDER
c CRANK MECHANISM

READ (*, *) A, B, E, V A, ACC, THAPI = 4 * ATAN (1.)TH = 0IH = 180/THADTH = PI / IHDO 10 I = 1, 2 * I HTH = (I – 1) * DTHBET = ASIN (E – A * SIN (TH) ) / B)VS = – A * VA * SIN (TH – BET) / (COS (BET) * 1000)VB = – A * V A * COS (TH) / B * COS (BET)AC1 = A * ACC * SIN (BET – TH) – B * VB * * 2AC2 = A * VA * * 2 * COS (BET – TH)ACS = (AC1 – AC2) / (COS (BET) * 1000)AC3 = A * ACC * COS (TH) – A * V A * * 2 * SIN (TH)AC4 = B * VB * * 2 * SIN (BET)ACB = – (AC3 – AC4) / (B * COS (BET) )I F (i . EQ .

9 FORMAT (3X,’ TH’, 5X,’ BET’, 4X,’ VS,’ 4X,’ VB,’ 4X,’ ACS’, 4X,’ ACB’)10 WRITE (*, 8) TH * 180 / P I , BET * 180 / P I, VS, VB, ACS, ACB8 FORMAT (6 F 8 . 2)
STOP
END

1016       /G6C               Theory of Machines
The input variables are :
A, B, E  = Length of crank AB (a), connecting rod BC (b) and offset ( e) in mm,
     V A  =  Angular velocity of crank AB (input link) in rad/s,
   ACC  = Angular acceleration of the crank AB (input link) in rad/s2, and
  THA  = Interval of the input angle in degrees.
   THA  = Angular displacement of the crank or input link AB in degrees,

BET  = Angular displacement of the connecting rod BC in degrees,
     VS  = Linear velocity of the slider in m/s,
     VB  = Angular velocity of the crank or input link AB in rad/s,
   ACS  = Linear acceleration of the slider in m/s2, and
   ACB  = Angular acceleration of the crank or input link AB in rad/s2.
Example 25.2.  In a slider crank mechanism, the crank AB = 200 mm and the connecting
rod BC = 750 mm. The line of stroke of the slider is offset by a perpendicular distance of 50 mm.

If the crank rotates at an angular speed of 20 rad/s and angular acceleration of 10 rad/s2, find at
an interval of 30° of the crank,  1. the linear velocity and acceleration of the slider, and 2. the
angular velocity and acceleration of the connecting rod .
Solution.
A = 200, B = 750, E = 50, VA = 20, ACC = 10, THA = 30
OUTPUT :
T H B E T V S V B ACS AC B
.00 3.82 .27 – 5.32 – 101.15 – .78
30.00 – 3.82 – 2.23 – 4.61 – 83.69 49.72
60.00 – 9.46 – 3.80 – 2.63 – 35.62 91.14

90.00 – 11.54 – 4.00 .00 14.33 108.87
120.00 – 9.46 – 3.13 2.63 44.71 93.85
150.00 – 3.82 – 1.77 4.61 55.11 54.35180.00 3.82 – .27 5.32 58.58 4.56210.00 11.54 1.29 4.53 62.42 – 47.90
240.00 17.31 2.84 2.55 57.93 – 93.34
270.00 19.47 4.00 .00 30.28 – 113.14
300.00 17.31 4.09 – 2.55 – 21.45 – 96.14330.00 11.54 2.71 – 4.53 – 75.44 – 52.61
25.6. Coupler Curves
It is often desired to have a mechanism to guide a point along a specified path. The path

generated by a point on the coupler link is known as a coupler curve  and the generating point is
called a coupler point  (also known as  tracer point ). The straight line mechanisms as discussed in
chapter 9 (Art. 9.3) are the examples of the use of coupler curves. In this article, we shall discuss

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1017
the method of determining the co-ordinates of the cou-
pler point in case of a four bar mechanism and a slidercrank mechanism.
1. Four bar mechanism
Consider a four bar mechanism ABCD  with an
offset coupler point E on the coupler link BC, as shown
in Fig. 25.3. Let the point E makes an angle 
α with BC
in the anticlockwise direction and its co-ordinates are E
(xE, yE).

(xE, yE).
First of all, let us find the value of BD, γ and β.
From right angled triangle BB1 D,
    11
11sintancosBB BB a
B D AD AB d aθγ= = =−− θ
or        1 sintan
cosa
da−θγ= −θ
and2222 2
11 1 1 ( )() ()() ( )BD BB B D BB AD AB=+=+ −
         22(s i n) ( c o s)=θ + − θad a
         22 2 2 2sin cos 2 cos=θ + +θ − θad a a d
         22 2 2(sin cos ) 2 cosad a d=θ + θ + −θ
         222c o s ad a d=+− θ
Now in triangle DBC ,
222()()()cos( )2BD BC CD

BC BD+−γ+β =×... (cosine law of triangle)
          222
2fb c
bf+−=
or   2221cos2fb c
bf−+−γ+β= 
∴     2221cos2fb c
bf−+−β= −γ  ... (i)
Let us now find the co-ordinates xE and yE. From Fig. 25.3, we find that
     E2 1 1 2 1 1xA EA B B EA B B E== += + ...  (∵ 12 1BE B E= )
        cos cos ( )ae=θ +α + β ... (ii)
and      E2 2 1 1 1 1y E EE E E EB B E E==+= + ... (∵ 21 1EE B B= )
         sin sin ( )ae=θ +α + β ... (iii)

sin sin ( )ae=θ +α + β ... (iii)
From the above equations, the co-ordinates of the point E may be determined if a, e, θ, α
and β are known.Fig. 25.3. Four bar mechainsm
with a coupler point.

1018       /G6C               Theory of Machines
2. Slider crank mechanism
Consider a slider crank mechanism with an offset coupler point E, as shown in Fig. 25.4.
Let the point E makes an angle α with BC in the anticlockwise direction and its co-ordinates are
E (xE, yE).
First of all, let us find the angle β. From right angled triangle BC1C,
  11 1 1 1 sinsinBC BB B C a e
BC BC b−θ −β= = =
∴     1 1 sinsinae
b−θ−β=      ... (iv)
Now          E11 1 1 1 2xA E A B B E A B B B==+=+

         cos cos ( )ae=θ +α − β       ... (v)
and      E1 1 2 2 1 2y E EE B B EB B B E== +=+
         sin sin ( )ae=θ +α − β       ... (vi)
From the above equations, the co-ordinates of the
point E may be determined, if a, b, e, e1, θ, α and β
are known.

rotation of the crank, then eccentricity e1 = 0. In such a case equation (iv) may be written as
1sinsina
b−θβ=
25.7. Synthesis of Mechanisms
In the previous articles, we have discussed the computer-aided analysis of mechanisms, i.e.
the determination of displacement, velocity

and acceleration for the given proportionsof the mechanism. The synthesis is theopposite of analysis. The synthesis ofmechanism is the design or creation of amechanism to produce a desired outputmotion for a given input motion. In otherwords, the synthesis of mechanism dealswith the determination of proportions of amechanism for the given input and outputmotion. We have already discussed theapplication of synthesis in designing a cam(Chapter 20) to give follower  a knownmotion from the

20) to give follower  a knownmotion from the displacement diagram andin the determination of number of teeth onthe members in a gear train (Chapter 13)to produce a desired velocity ratio.

In the application of synthesis, to
the design of a mechanism, the problemdivides itself into the following three parts:Fig. 25.4 Slider crank mechanism with
coupler point.
Roller conveyor.

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1019
1.  Type synthesis , i.e. the type of mechanism to be used,
2.  Number synthesis , i.e. the number of links and the number of joints needed to produce
the required motion, and
3. Dimensional synthesis , i.e. the proportions or lengths of the links necessary to satisfy
the required motion characteristics.
In designing a mechanism, one factor that must be kept in mind is that of the accuracy

required of the mechanism. Sometimes, it is possible to design a mechanism that will theoretically
generate a given motion. The difference between the desired motion and the actual motion producedis known as structural error . In addition to this, there are errors due to manufacture. The error
resulting from tolerances in the length of links and bearing clearances is known as  mechanical
error .
25.8. Classifications of Synthesis Problem

The problems in synthesis can be placed in one of the following three categories :
1. Function generation ; 2. Path generation ; and 3. Body guidance.
These are discussed as follows :
1. Function generation . The major classification of the synthesis problems that arises in
the design of links in a mechanism is a function generation. In designing a mechanism, the frequent

requirement is that the output link should either rotate, oscillate or reciprocate according to aspecified function of time or function of the motion of input link. This is known as function genera-tion. A simple example is that of designing a four bar mechanism to generate the function y = f(x).
In this case, x represents the motion of the input link and the mechanism is to be designed so that
the motion of the output link approximates the function y.

Note :  The common mechanism used for function generation is that of a cam and a follower in which the
angular displacement of the follower is specified as a function of the angle of rotation of the cam. The
synthesis problem is to find the shape of the cam surface for the given follower displacements.
2. Path generation . In a path generation, the mechanism is required to guide a point (called
a tracer point or coupler point) along a path having a prescribed shape. The common requirements

are that a portion of the path be a circular arc, elliptical or a straight line.
3. Body guidance . In body guidance, both the position of a point within a moving body
and the angular displacement of the body are specified. The problem may be a simple translationor a combination of translation and rotation.
25.9. Precision Points for Function Generation
In designing a mechanism to generate a particular function, it is usually impossible to

accurately produce the function at more than a few points. The points at which the generated anddesired functions agree are known as  precision points  or accuracy points  and must be located so
as to minimise the error generated between these points.
The best spacing of the precision points, for the first trial, is called Chebychev  spacing .
According to Freudenstein and Sandor, the Chebychev spacing for n points in the range 
SFxx x≤≤
(i.e. when x varies between xS and xF) is given by

SF FS11 ( 2 1 )() () c o s22 2jjxx x x xnπ−=+ −−... (i)
    SF11 ( 2 1 )() c o s22 2jxx xnπ−=+ − × ∆ ×

1020       /G6C               Theory of Machines
where    xj = Precision points
           x∆ = Range in FS , xx x=−  and
    j = 1, 2, ... n
The subscripts S and F indicate start and finish positions respectively.
The precision or accuracy points may be easily obtained by using the graphical method as
discussed below.
1. Draw a circle of diameter equal to the range FS xx x∆= − .
2. Inscribe a regular polygon having the number of sides equal to twice the number of

precision points required, i.e. for three precision points, draw a regular hexagon inside the circle,
as shown in Fig. 25.5.
3. Draw perpendiculars from each corner which intersect the diagonal of a circle at preci-
sion points x1, x2, x3.
Now for the range SF 13 , 1 ; 3xx x≤≤ = = , and
∴   FS 31 2 xx x∆= − =−=
or radius of circle,               /2 2/2 1 rx=∆ = =
∴               2S S21222xxxr x∆=+ =+ = + =
              12 2 cos30 cos302xxxr x∆=− ° =− °
      22 cos30 1.1342=− ° =

22 cos30 1.1342=− ° =
and               32 2 cos30 cos302xxx r x∆=+ ° =+ °
       22 cos30 2.8662=+ ° =
25.10. Angle Relationships for Function Generation
(a) Four bar mechanism. ( b) Linear relationship between x and θ.
Fig. 25.6Fig. 25.5. Graphical method for
determining three precision
points.

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1021
Consider a four bar mechanism, as shown in Fig. 25.6 ( a) arranged to generate a function
y = f(x) over a limited range. Let the range in x is (xF – xS) and the corresponding range in θ is
FS()θ− θ . Similarly, let the range in y is (FS()yy−  and the corresponding range in φ is FS()φ− φ .
The linear relationship between x and θ is shown in Fig. 25.6 ( b). From the figure, we find
that
      FS
SS

that
      FS
SS
FS()xxxxθ− θθ=θ + −−... (i)
Similarly, the linear relationship between y and φ may be written as
      FS
SS
FS()yyyyφ− φφ=φ + −−... (ii)
For n points in the range, the equation (i) and (ii) may be written as
      FS
SS
FS()jj xxxxθ− θθ= θ+ −− SS ()jxxx∆θ=θ + −∆
and     FS
SS
FS()jj yyyyφ− φφ= φ+ −− SS ()jyyy∆φ=φ + −∆
where       j  = 1, 2, ... n,
     FS xx x∆= − ; FS∆θ=θ −θ ,
     FS yy y∆= − ;     and  FS∆φ=φ −φ
An automatic filling and sealing machine.

1022       /G6C               Theory of Machines
Example 25.3.  A four bar mechanism is to be designed, by using three precision points, to
generate the function
      1.5yx= , for the range 1x4≤≤ .
Assuming 30° starting position and 120° finishing position for the input link and 90°
starting position and 180° finishing position for the output link, find the values of x, y, θ and φ
corresponding to the three precision points.

Solution :  Given : xS = 1 ; xF = 4 ; S30θ= °  ; F120θ= °  ; S90φ= °  ; F180φ= °
Values of x
The three values of x corresponding to three precision points ( i.e. for n = 3) according to
Chebychev’s spacing are given by
SF FS11 ( 2 1 )() () c o s22 2jjxx x x xnπ−=+ −−,      where j = 1, 2 and 3
∴ 111 ( 2 1 1 )(1 4) (4 1) cos22 2 3xπ× −=+ −−× = 1.2  Ans. ... (∵ j = 1)

211 ( 2 2 1 )(1 4) (4 1)cos22 2 3xπ×−=+ −−× = 2.5  Ans.  ... (∵ j = 2)
and 31 1 ( 231 )(1 4) (4 1) cos22 2 3xπ× −=+ −−× = 3.8 Ans. ... (∵ j = 3)
Note :  The three precision points x1, x2 and x3 may be determined graphically as discussed in Art. 25.9.
Values of y
Since y = x1.5 , therefore the corresponding values of y are
1.5 1.5
11() ( 1 . 2 )yx==  = 1.316  Ans.
1.5 1.5
22() ( 2 . 5 ) yx==  = 3.952  Ans.
1.5 1.5
33() ( 3 . 8 ) yx==  = 7.41  Ans.
Note :      1.5 1.5

SS() ( 1 ) 1== =yx   and  1.5 1.5
FF() ( 4 ) 8== =yx
Values of θθθθθ
The three values of θ corresponding to three precision points are given by
FS
SS
FS()jj xxxxθ− θθ= θ+ −−, where j = 1, 2 and 3
∴ 1120 3030 (1.2 1)41−θ= + −− = 36° Ans.
2120 3030 (2.5 1)41−θ= + −− = 75°  Ans.
and 3120 3030 (3.8 1)41−θ= + −− = 114°  Ans.

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1023
Values of  φ
The three values of φ corresponding to three precision points are given by
             FS
SS
FS()jj yyyyφ− φφ= φ+ −−
∴   1180 9090 (1.316 1)
81−φ= + −
− = 94.06°  Ans.
            2180 9090 (3.952 1)
81−φ= + −
− = 127.95°  Ans.
and             3180 9090 (7.41 1)
81−φ= + −
− = 172.41°  Ans.
25.11. Graphical Synthesis of Four Bar Mechanism

The synthesis of four bar mechanism consists of determining the dimensions of the links in
which the output link is to occupy three specified positions corresponding to the three given positions
of the input link. Fig. 25.7 shows the layout of a four bar mechanism in which the starting angle of
the input link AB1 (link 2) of known length is θ. Let 12θ, 23θ and 13θ be the angles between the

positions B1B2, B2B3 and B1B3 measured anticlockwise. Let the output link DC1 (link 4) passes
through the desired positions C1, C2 and C3 and 12 23,φφ  and 13φ are the corresponding angles
between the positions 12 23, CC C C  and 13CC . The length of the fixed link (link 1) is also known.
Now we are required to determine the lengths of links B1C1 and DC1 (i.e. links 3 and 4) and the
starting position of link 4 ( φ).

The easiest way to solve the problem is based on inverting the mechanism on link 4. The
procedure is discused as follows :
1.Draw AD equal to the known length of fixed link, as shown in Fig. 25.8.
2.At A, draw the input link 1 in its three specified angular positions AB1, AB2 and AB3.
3.Since we have to invert the mechanism on link 4, therefore draw a line B2D and and rotate

it clockwise (in a direction opposite to the direction in which link 1 rotates) through an
angle 12φ (i.e. the angle of the output link 4 between the first and second position) in
order to locate the point 2B′.
Fig. 25.7.  Layout of four bar mechanism.
4.Similarly, draw another line B3D and rotate it clockwise through an angle 13φ (i.e. angle
of the output link between the first and third position) in order to locate point 3B′.

1024       /G6C               Theory of Machines
Fig. 25.8.  Design of four bar mechanism (Three point synthesis).
5.Since the mechanism is to be inverted on the first design position, therefore B1 and 1B′ are
coincident.
6.Draw the perpendicular bisectors of the lines 12BB′′ and 23BB′′. These bisectors intersect
at point C1.
7.Join 11BC′ and 1CD . The figure 11AB C D′  is the required four bar mechanism. Now the

length of the link 3 and length of the link 4 and its starting position ( φ) are determined.
25.12. Graphical Synthesis of Slider Crank Mechanism
Consider a slider crank mechanism for which the three positions of the crank AB (i.e.
12,θθ  and 3θ) and corresponding three positions of the slider C (i.e. s1, s2 and s3) are known, as
shown in Fig. 25.9.
In order to synthesis such a mechanism, the following procedure is adopted.

1.First of all, draw the crank AB1 in its initial position. If the length of crank is not speci-
fied, it may be assumed.
2.Now find the *relative poles P12 and P13 as shown in Fig. 25.10. The relative poles are
obtained by fixing the link A and observing the motion of the crank AB1 in the reverse
direction. Thus, to find P12, draw angle YAP12 equal to half of the angle between the first
and second position (12θ) in the reverse direction and from AY draw IP12 equal to half of

the slider displacement between the first and second position ( i.e. s12). Similarly P13 may
be obtained.
3.From P12 and P13, draw two lines P12 Q12 and P13 Q13 such that 12 1 12 12AP I B P Q∠= ∠
and 13 1 13 13AP I B P Q∠= ∠ . The lines P12 Q12 and P13 Q13 intersect at C1, which is the
location of the slider at its first position. Now the length of the connecting rod B1C1 and
the offset ( e) may be determined.

* The relative pole is the centre of rotation of the connecting rod relative to the crank rotation and
the corresponding slider displacement .

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1025
(a) Three positions of the crank. ( b) Three positions of the slider.
Fig. 25.9
Fig. 25.10
25.13. Computer Aided (Analytical) Synthesis of Four Bar Mechanism
(a) Four bar mechanism.      ( b) Three positions of input and output link.
Fig. 25.11
Consider a four bar mechanism as shown in Fig. 25.11.
The synthesis of a four bar mechanism, when input and output angles are specified, is

Let the three positions i.e. angular displacements (12,θθ  and 3θ) of the input link AB and

1026       /G6C               Theory of Machines
the three positions (12,φφ  and 3φ) of the output link, as shown in Fig. 25.11 ( b), are known and
we have to determine the dimensions a, b, c  and d of the four bar mechanism.
We have discussed in Art. 25.2 that the Freudenstein’s equation is
123cos cos cos ( )kk kφ− θ+ = θ−φ ... (i)
where       1dk
a= ; 2dk
c= ; and 222 2
32abcdkac−++=  ... (ii)
For the three different positions of the mechanism, the equation  (i) may be written as

11 21 3 1 1cos cos cos ( )φ− θ+ = θ− φ kk k ... (iii)
12 22 3 2 2cos cos cos ( )φ− θ+ = θ− φ kk k ... (iv)
and13 23 3 3 3cos cos cos ( )φ− θ+ = θ− φ kk k ... (v)
The equations (iii), (iv) and (v) are three simultaneous equations and may be solved for k1,
k2 and k3 either by elimination method (See Examples 25.4 and 25.5) or by using Cramer’s rule of
determinants as discussed below :
              11
22
33cos cos 1
cos cos 1
cos cos 1φθ
∆= φ θ
φθ
                         11 1
12 2 2

∆= φ θ
φθ
                         11 1
12 2 2
33 3cos ( ) cos 1
cos ( ) cos 1
cos ( ) cos 1θ− φ θ
∆= θ− φ θ
θ− φ θ
An off-shore oil well.

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1027
          11 1
22 2 2
33 3cos cos ( ) 1
cos cos ( ) 1
cos cos ( ) 1φθ − φ
∆= φ θ− φ
φθ − φ
           11 1 1
32 22 2
33 3 3cos cos cos( )
cos cos cos( )
cos cos cos ( )φθ θ − φ
∆= φ θ θ− φ
φθ θ − φ
Now the values of k1, k2 and k3 are given by
12
12 , kk∆∆==∆∆ and 3
3k∆=∆
Once the values of k1, k2 and k3 are known, then the link lengths a, b, c  and d are determined

by using equation  (ii). In actual practice, either the value of a or d is assumed to be unity to get the
proportionate values of other links.
Note :  The designed mechanism may not satisfy the input and output angle co-ordination at positions other
than these three positions. It is observed that a four bar mechanism can be designed precisely for five positions
of the input and output links provided θ and φ are measured from some arbitrary reference rather than from

the reference fixed link AD. In such cases, the synthesis equations become non-linear and some other means
are required to solve such synthesis equations.
25.14. Programme to Co-ordinate the Angular Displacement of the Input
and Output Links
The following is the programme in Fortran to co-ordinate the angular displacements of the
input and output links.
C PROGRAM TO COORDINATE ANGULAR DISPLACEMENTS OF
C THE INPUT AND OUTPUT LINKS IN THREE POSITIONS

READ (*, *) Q1, Q2, Q3, P1, P2, P3RAD = 4 * ATAN (1.0) / 180
QA = COS (Q1 * RAD)QB = COS (Q2 * RAD)
QC = COS (Q3 * RAD)
PA = COS (P1 * RAD)
PB = COS (P2 * RAD)
PC = COS (P3 * RAD)
AA = COS ( (Q1 – P1) * RAD )
BB = COS ( (Q2 – P2) * RAD )CC = COS ( (Q3 – P3) * RAD )
D = PA * (QB – QC) + QA * (PC – PB) + (PB * QC – PC * QB)
D1 = AA * (QB – QC) + QA * (CC – BB) + (BB * QC – CC * QB)
D2 = PA * (BB – CC) + AA * (PC – PB) + (PB * CC – PC * BB)

D3 = PA * (QB * CC – QC * BB) + QA * (BB * PC – CC * PB) + AA * (PB * QC – PC * QB)A1 = D/D1
A2 = SQRT (A1 * A1 + A3 * A3 + 1.0 – 2 * A1 * A3 * D3 / D)
A3 = – D/D2
WRITE (*, 1) A1, A2, A3, 1
1 FORMAT (6X, A1’, 7X,’ A2’, 7X,’ A3’ 7X,’ A4,’ / 4F8 . 2)
STOP
END

1028       /G6C               Theory of Machines
Q1, Q2, Q3 = Angular displacement of the
       input link AB in degrees,
 P1, P2, P3 = Angular displacement of the
        output link DC in degrees.
  A, B, C, D  =  Ratio of length of the links AB,
        BC, CD  and AD respectively
Example 25.4.  Design a four bar mechanism
to co-ordinate the input and output angles as follows :
Input angles = 15°, 30° and 45° ; Output angles

= 30°, 40° and 55° .
Solution.  Given : 115θ= °  ; 230θ= °  ;
345θ= ° ; 130φ= °  ; 240φ= °  ; 355φ= °
The Freudenstein’s equation for the first position
of the input and output link ( i.e. when 115θ= °  and
130φ= ° ) may be written as
123cos30 cos15 cos (15 30 )kkk°− °+ = °− °
or  0.866 k1 – 0.966 k2 + k3  = 0.966 ... (i)
Similarly, for the second position ( i.e. when 230θ= °  and 240φ= ° ),
123cos 40 cos30 cos (30 40 )kkk°− °+ = °− °

or           0.766 k1 – 0.866 k2 + k3 = 0.985 ... (ii)
and for the third position ( i.e. when 345θ= °  and 355φ= ° ),
 123cos55 cos 45 cos (45 55 )kk k°− °+ = °− °
or           0.574 k1 – 0.707 k2 + k3 = 0.985 ... (iii)
Solving the three simultaneous equations (i), (ii) and (iii), we get
k1 = 0.905 ; k2 = 1.01 and k3 = 1.158
Fig. 25.12
Assuming the length of one of the links, say a as one unit, we get the length of the other
links as follows :

We know that k1 = d/a        or      d  = k1 a = 0.905 units Ans.
Grinding machine.

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1029
    k2 = d/c or c = d/k2 = 0.905 / 1.01 = 0.896 units  Ans.
and     222 2
32abcdkac−++=
or              22 2 2
32( ) bk a c ac d−=× − ++
        = 1.158 × 2 × 1 × 0.896 – [12 + (0.896)2 + (0.905)2]
        = 2.075 – 2.622 = – 0.547     or     b = 0.74 units  Ans.
The designed mechanism with AB = a = 1 unit, BC = b = 0.74 units ; CD = c = 0.896 units
and AD = d = 0.905 units, is shown in Fig. 25.12.

Example 25.5.  Determine the proportions of four bar mechanism, by using three precision
points, to generate y = x1.5, where x varies between 1 and 4. Assume S30θ= ° ; 90∆θ= °  ;
S90φ= ° ; and 90∆φ= ° . Take length of the fixed link AD as 25 mm.
Solution.  Given : xS = 1 ; xF = 4 ; S30θ= °  ; FS 90 ∆θ=θ −θ = ° ; S90φ= ° ;
FS 90 ∆φ=φ −φ = °  ; d = 25 mm
We have already calculated the three values of x and y for the above given data in Example

25.3. These values are :
    x1 = 1.2 ;        x2 = 2.5 ;        and      x3 = 3.8
    y1 = 1.316 ;     y2 = 3.952 ;     and      y3 = 7.41
The corresponding values of θ and φ are
    136θ= ° ; 275θ= °  ; and 3114θ= °
    194.06φ= °  ; 2127.95φ= °  ; and 3172.41φ= °
We know that the Freudenstein’s equation is
   123cos cos cos ( )kk kφ− θ+ = θ−φ ... (i)
where     1dka= ; 2dk
c= ;     and     222 2
32abcdkac−++= ... (ii)

Now for the three different positions of the mechanism, the equation (i) may be written
three times as follows :
    12 3cos94.06 cos36 cos (36 94.06 )kk k °− °+ = °− °
or       – 0.0708 k1 – 0.809 k2 + k3 = 0.529 ... (iii)
Similarly    12 3cos127.95 cos 75 cos(75 127.95 )kk k °− °+ = °− °
or        – 0.615 k1 – 0.259 k2 + k3 = 0.6025 ... (iv)
and            12 3cos172.41 cos114 cos (114 172.41 )kk k °− °+ = °− °

– 0.9912 k1 + 0.4067 k2 + k3 = 0.5238 ... (v)
Solving three simultaneous equations (iii), (iv) and (v), we get
k1 = 0.6 ; k2 = 0.453 ; and k3 = 0.12
12541.7mm0.6dak== = Ans.

1030       /G6C               Theory of Machines
         
22555.2mm0.453dck== =  Ans.
and          22 2 1 / 2
3 (2 )bacd k a c=+ + − ×
1/222 2(41.7) (55.2) (25) 0.12 2 41.7 55.2=+ + − × × × = 69.7 mm Ans.
The designed four bar mechanism AB2 C2D in one position ( i.e. for 22,xθ  and 22,yφ ) is
shown by thick lines in Fig. 25.13.
Fig. 25.13
The other two positions of the four bar mechanism may be drawn by joining B1C2 (i.e.
11,xθ  and 11,yφ ) and B3C3 (i.e. 33,xθ  and 33,yφ ).

Note :  In the above example, the motion of input link and output link is taken clockwise.
Example 25.6.  Synthesize a four bar linkage, as shown in Fig. 25.14, using Freudenstein’s
equation to satisfy in one of  its positions. The specification
of position θ, velocity ω and acceleration α are as follows :
θ = 60°,   ω2 = 5 rad/s; α2 = 2 rad/s2;
φ = 90°;   ω4 = 2 rad/s; α4 = 7 rad/s2.
Solution :  Given : θ = 60° ; ω2 = 5 rad/s ; α2 = 2 rad/s2;
φ = 90°, ω4 = 2 rad/s; α4 = 7 rad/s2

The four bar linkages is shown in Fig. 25.15. Let
AB = Input link = a,
BC = Coupler = b,
CD = Output link = c, and
AD = Fixed link = d.
The Freudenstein’s equation is given by
123cos cos cos( )φ− θ+ = θ−φ kk k   ... (i)Fig. 25.14
Fig. 25.15

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1031
where 1=dka;   2 ;=dkc   and   222 2
32−++=abcdkac
Substituting the value of θ and φ in equation (i),
123cos90 cos 60 cos(60 90 )°− °+ = °− ° kk k
        12 30 0.5 0.866×− × + =kk k
      230.5 0.866kk−+ = ... (ii)
12()sin sin sin( )φθ θ − φ×− φ× − ×− θ× =− θ−φ ×dd dkkdt dt dt
   14 22 2 4sin sin sin( )( )kk−φ ω + θ ω = − θ − φ ω − ω ... (iii)

42;a n ddd
dt dtφθ=ω =ω…∵
  12sin 90 2 sin 60 5 sin(60 90 ) (5 2)kk−× ° ×+ ° ×= − ° −° −
  1253 3222−+ =kk
or122.165 0.75=−kk .... (iv)
Now differentiating equation (iii) with respect to time,
42
14 22s i nc o ss i nc o sdd ddkkdt dt dt dtωω φθ  − φ× +ω φ× + θ× +ω θ×  
          24
24() ()sin( ) ( ) cos( )ω− ω θ−φ=− θ−φ + ω −ω θ−φ×d d
dt dt
22
14 4 22 2sin cos sin cos −φ × α + ω φ + θ × α + ω θ kk
          2
24 24 sin( ) ( ) ( ) cos( )=− θ−φ α −α + ω −ω θ−φ

22
12sin 90 7 2 cos90 sin 60 2 5 cos60kk −° × +° + ° × +° 
          2sin(60 90 ) (2 7) (5 2) cos(60 90 )=− °− ° − + − °− °
12(7 0) (1.732 12.5) (2.5 7.794)kk−+ + + = −+
127 14.232 10.294kk−+ = −
or        122.033 1.47kk=+ ... (v)
  22 2 2.165 0.75 2.033 1.47 or 16.8 kk k−= + =

1032       /G6C               Theory of Machines
From equation (v)
            122.165 0.75 2.165 16.8 0.75 35.6=− = × − =kk
and from equation (ii),
k3 = 0.5 k2 + 0.866 = 0.5 × 16.8 + 0.866 = 9.266
Assuming the length of one of links say a as one unit, we get the length of the links as
follows :
We know that k1 = d/a  or   d = k1 . a = 35.6 units  Ans.
k2 = d/c   or  c = d/k2 = 35.6 / 16.8 = 2.12 units  Ans.
and2 2 222 2 2 2
31 (2.12) (35.6)
22 1 2 . 1 2−++ −+ +==××abcd bkac

       221 4.494 1267.36 1272.8549.2664.24 4.24−+ + −==bb
b2 = 1272.854 – 9.266 × 4.24 = 1233.566
∴ b = 35.12 units  Ans.
Example 25.7.  Synthesize a four-bar mechanism to generate a function y = sin x for
. 0x9 0≤≤ °  The range of the output crank may be chosen as 60° while that of inut crank be 120°.
Assume three precision points which are to be abtained from Chebyshev spacing. Assume fixed
link to be 52.5 mm long and 1105θ=° and 1 .66φ= °

Solution.  Given : S0;x= F90 ; x=°  60 ;∆φ = °  120 ;∆θ= °  52.5 mm;d=
11.05 ;θ= °166φ= °
The three values of x corresponding to three precision points ( i.e. for n = 3), according to
Chebyshev spacing are given by
SF S F1( 2 1 )() () c o s ,22jjxx x x xnπ−=+− − where j = 1, 2, 3
∴ 111 ( 2 1 1 )(0 90) (90 0) cos22 2 3xπ× −=+ − −×
    = 45 – 45 cos 30° = 6° ... ( ∵   j = 1)

= 45 – 45 cos 30° = 6° ... ( ∵   j = 1)
           21 1 ( 221 )(0 90) (90 0) cos22 2 3xπ×−=+ − −×
               45 45cos90 45=− ° =° ... (∵  j = 2)
and                       31 1 ( 231 )( 0 9 0 ) ( 9 0–0 )c o s22 2 3xπ×−=+ −×
    45 45cos150 84=− ° =°
Since y = sin x, therefore corresponding values of y are
11sin sin 6 0.1045== ° =yx

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1033
22sin sin 45 0.707 yx==° =
and33sin sin84 0.9945yx==° =
Alsosssin sin 0 0 yx== ° =
andFFsin sin 90 1 yx== ° =
The relation between the input angle ( θ) and x is given by
FS
SS
FS() ,jj xxxxθ− θθ= θ+ −− where j = 1, 2 and 3.
The above expression may be written as
SS ()jj xxx∆θθ= θ+ −∆
1S 1 x
x∆θθ= θ+ ×
∆... () S0x=∵ ... (i)

1S 1 x
x∆θθ= θ+ ×
2S 2 x
x∆θθ= θ+ ×
∆... (ii)
and 3S 3 x
x∆θθ= θ+ ×
∆.... (iii)
From equations (i), (ii) and (iii),
21 21120()( 4 5 6 ) 5 2
90∆θθ− θ= − = − = °
∆xx
x... (iv)
... FS (9 0 0 9 0 ) xx x∆= − = −=∵
32 32120()( 8 4 4 5 ) 5 2
90xx
x∆θθ− θ= − = − = °
∆... (v)
and 31 31120( ) (84 6) 104
90xx
∆... (iv)
Since1105θ= °   (Given), therefore
21 52 105 52 157 θ =θ + °= °+ °= °
32 52 157 52 209 θ= θ+ ° = ° + = °

The relation between the output angle ( φ) and y is given by
FS
SS
FS() ,jj yyyyφ− φφ= φ+ −− when j = 1, 2 and 3
SS ()jj yyy∆φφ= φ+ −∆

1034       /G6C               Theory of Machines
1S 1 yy∆φφ= φ+ ×∆.... (∵ yS = 0) ... (vii)
2S 2 yy∆φφ= φ+ ×∆... (viii)
and 3S 3 yy∆φφ= φ+ ×∆... (ix)
From equations (vii), (viii) and (ix),
21 2160( ) (0.707 0.1045) 36.151yyy∆φφ− φ= − = − = °∆... (x)
... (∵ ∆y = yF – yS = 1 – 0 = 1)
32 3260( ) (0.9945 0.707) 17.251∆φφ− φ= − = − = °∆yyy... (xi)
31 3160( ) (0.9945 0.1045) 53.41∆φφ− φ= − = − = °∆yyy... (xii)

Since 166φ= °  (Given), therefore
21 36.15 66 36.15 102.15 φ= φ+ ° = ° + ° = °
32 17.25 102.15 17.25 119.40 φ =φ + °= °+ °= °
We have calculated above the three positions i.e. the angular  displacements ( θ1, θ2 and θ3)
of the input crank and the three positions ( φ1, φ2 and φ3) of the output crank. Now let us find the
dimensions of the four bar mechanism.
Let  a = Length of the input crank,
 b = Length of the coupler,
 c = Length of the output crank, and

c = Length of the output crank, and
 d = Length of the fixed crank = 52.5 mm (Given)
We know that the Freudenstein displacement equation is
123cos cos cos( )kk kφ− θ+ = θ−φ ... (xiii)
where            1 ;=dk
a2=dk
c     and      222 2
The equation (xiii) for the first position of input and output crank ( i.e. when θ1 = 45° and
φ = 66°) may be written as
    11 21 3 1 1cos cos cos( )φ− θ+ = θ− φ kk k
12 3cos 66 cos105 cos(105 66 )kk k°− °+ = °− °

  12 3 0.4067 0.2588 0.7771 ++ =kk k ... (xiv)

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1035
Similarly, for the second position ( i.e. when θ2 = 157° and φ2 = 102.15°),
12 22 3 2 2cos cos cos( )φ− θ+ = θ− φ kk k
12 3cos102.15 cos157 cos(157 102.15 )kk k °− °+ = °− °
12 3 0.2105 0.9205 0.5757 kk k−++ = ...(xv)
   and for the third position ( i.e. when θ3 = 209° and φ3 = 119.4°),
13 23 3 3 3cos cos cos( )φ− θ+ = θ− φ kk k
    12 3cos119.4 cos 209 cos(209 119.4 )kk k °− °+ = °− °

12 3 0.4909 0.8746 0.007 kk k−++ = ... (xvi)
Solving the three simultaneous equations (xiv), (xv) and (xvi), we get
k1 = 1.8; k2 = 1.375  and k3 = – 0.311
Since the length of the fixed link ( i.e. d = 52.5 mm) is known, therefore we get the length
of other links as follows:
k1 = d/a or a = d/k1 = 52.5 / 1.8 = 29.17 mm  Ans.
k2 = d/c or c = d/k2 = 52.5 / 1.375 = 38.18 mm    Ans.
and222 2
or b2 = 22 2
32 ++−×acdk a c

or b2 = 22 2
  = (29.17)2 + (38.18)2 + (52.5)2 – (– 0.311)× 2 × 29.17 × 38.18 = 5758
∴ b = 75.88 mm  Ans.
25.15.Least Square Technique
Most of the mechanisms are not possible to design even for five positions of the input and
output links. However, it is possible to design a mechanism to give least deviation from the specified
positions. This is done by using least square technique as discussed below :
We have already discussed that the Freudenstein’s equation is

123cos cos cos ( ) 0kk kφ− θ+ − θ−φ =
The angles θ and φ are specified for a position. If iθ and iφ are the angles for ith
position, then Freudenstein’s equation may be written as
123cos cos cos ( ) 0ii i i kkkφ− θ+ − θ− φ =
Let e be the error which is defined as
 2
123
1[c o s c o s c o s ( ) ]n
ii i i
iek k k
==φ − θ + − θ − φ∑
For e to be minimum, the partial derivatives of e with respect to k1, k2, k3 separately must
be equal to zero, i.e.
10e
k∂=∂ ; 
20e
k∂=∂, and 
30e
k∂=∂

1036       /G6C               Theory of Machines
∴ 123
112 [ cos cos cos ( )]cos 0n
ii i i i
iekk kk=∂=φ − θ + − θ − φ φ =∂∑
or2
12 3
11 1 1cos cos cos cos cos ( ) cos
== = =φ− θ φ+ φ= θ− φ φ∑∑ ∑ ∑nn n n
ii ii i i i
ii i ikk k ... (i)
Similarly,   123
212 [ cos cos cos ( )]cos 0n
ii i i i
iekk kk=∂= − φ− θ+ − θ− φ θ=∂∑
or           2
12 3
11 1 1cos cos cos cos cos ( ) cosnn n n
ii i i i i i
ii i ikk k
== = =φθ + θ + θ = θ − φ θ∑∑ ∑ ∑... (ii)
Now 123
312[ c o s c o s c o s ( ) ] 0n
ii i i

Now 123
ii i i
iekk kk=∂=φ − θ + − θ − φ =∂∑
or 123
111 1cos cos 1 cos ( )nnn n
ii i i
iii ikk k
=== =φ+ θ+ = θ− φ∑∑∑ ∑ ... (iii)
The equations  (i), (ii) and (iii) are simultaneous, linear, non-homogeneous equations in
three unknowns k1, k2 and k3. These equations can be solved by using Cramer’s rule.
25.16. Programme Using Least Square Technique
The following is a programme in Fortrans to find the ratio of lengths for different links by

using the least square technique.
    J = Number of specified positions
          TH  (I) = Angular displacements of the input link AB for I = 1 to J (degrees), and
          PH  (I) = Angular displacements of the output link DC for I = 1 to J (degrees).
    A, B, C, D  = Ratio of the lengths of the links AB, BC , CD and AD respectively.
C PROGRAM TO COORDINATE ANGULAR DISPLACEMENT OF THE

C INPUT AND OUTPUT LINKS IN MORE THAN THREE POSITIONS TOC FIND RATIO OF DIFFERENT LINKS USING LEAST SQUARE TECHNIQUE
DIMENSIONREAD (*, *) JREAD (*, *) (TH (I), I = 1, J), PH (I), I = 1, J)RAD = 4 * ATAN (1.0) / 180DO 10 K = 1 . JA1 = A1 + (COS (PH (K) * RAD ) ) * * 2A2 = A2 + (COS (TH (K) RAD ) ) * (COS (PH (K) * RAD ) )A3 = A3 + (COS (PH (K) * RAD ) )
B1 = A2
B2 = B2 + (COS (TH (K) * RAD ) ) * * 2

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1037
B3 = B3 + (COS (TH (K) * RAD ) )
P1 = A3P2 = B3P3 = JTT = COS ( ( TH (K) – PH (K) * RAD )
Q1 = Q1 + TT * COS (PH (K) * RAD )
Q2 = Q2 + TT * COS (TH (K) * RAD
10 Q3 = Q3 + TT

D = A1 * (B2 * P3 – B3 * P2) + B1 * (P2 * A3 – P3 * A2) + P1 * (A2 * B3 – A3 * B2)D1 = Q1 * (B2 * P3 – B3 * P2) + B1 * (P2 * Q3 – P3 * Q2) + P1 * (Q2 * B3 – Q3 * B2)D2 = A1 * (Q2 * P3 – Q3 * P2) + Q1 *  (P2 * A3 – P3 * A2) + P1 * (A2 * Q3 – A3 * Q2)D3 = A1 * (B2 * Q3 – B3 * Q2) + B1 * (Q2 * A3 – Q3 * A2) + Q1 * (A2 * B3 – A3 * B2)Q = D / D1R = – D / D2P = SQRT (Q * Q + r * r + 1. – 2. * r * r * 03 / D)WRITE (* , 9) Q, P, r, 1.

9 FORMAT (6X, ’ Q’, 7X,’ P’, 7X,’ r’, 7X,’ D’ / 4F8 . 2)
STOP
END
25.17. Computer Aided Synthesis of Four Bar Mechanism with Coupler
Point
Consider a four bar mechanism ABCD  with a couple point E, as shown in Fig. 25.16,
which is specified by r and γ.
Fig. 25.16.  Four bar mechanism with a couple point.
Let  12,θθ  and 3θ = Three positions of the input link AB,
     r1, r2 and r3 = Three positions of the coupler point E from point O, and

12,γγ  and 3γ = Three angular positions of the coupler point E from OX.
The dimensions a, c, e, f  and the location of points A and D specified by ( q, β) and ( p, α)
respectively, may be determined as discussed below :

1038       /G6C               Theory of Machines
Considering the loop OABE , the horizontal and vertical components of vectors q, a, e  and
r are
cos cos cos cosβ+ θ+ δ= γ qaer ... (i)
and sin sin sin sinβ+ θ+ δ= γ qaer ... (ii)
Squaring equations  (i) and (ii) and adding in order to eliminate angle δ, we have
      2222[2 cos( )] [2 cos( )] [2cos( )]qr ar e q a r q a γ − β + θ − γ +−−=+ θ − β ... (iii)
Let  q = k1 ; a = k2 ; 222
3 −−=eqa k  ; and q a = k4 = k1 k2... (iv)

Now the equation (iii) may be written as
2
123 4[2 cos( )] [2 cos( )] [2cos( )]γ−β + θ−γ + = + θ−β kr k r k r k ... (v)
Since 41 2=kk k , therefore the equation  (v) is difficult to solve for 123,,kk k  and k4. Such
type of non-linear equations can be solved easily by making them linear by some substitutions as
Let11 1=+ λkl m  ; 22 2=+ λkl m  ; and 33 3=+ λkl m ... (vi)
where  41 2 1 1 2 2 () ( ) λ= = = +λ +λkk k l m l m
  2

2
12 1 2 12 1 2=+ λ + λ + λll l m ml mm
or                2
12 12 21 1 2 (1 ) 0λ+ + −λ + = mm lm l m ll
or20 λ+ λ + =AB C
∴24
2−± −λ=BB A C
A... (vii)
where 12=Am m  ; 12 21(1 )=+ −Bl ml m  ; and C = l1 l2... (viii)
Substituting the values of k1, k2, k3 and k4 in equation  (v),
11 2 2 33( )[2 cos( )] ( )[2 cos( )] ( )+λ γ−β + +λ θ−γ + +λlm r l mr lm
   2([2cos( )]=+ λ θ − βr
Equating the terms with λ and without λ separately equal to zero, we get the components

into two groups, one with λ and the other without λ. These components are
2
123[2 cos( )] [2 cos( )]γ−β + θ−γ + = lr lr l r ... (ix)
and        123[2 cos( )] [2 cos( )] 2cos( )γ−β + θ−γ + = θ−β mr m r m ... (x)
The equation  (ix) for the three positions of θ, r and γ may be written three times as
follows :
  2
11 1 21 1 1 3 1[2 cos( )] [2 cos( )]lr l r l r γ− β+ θ− γ + = ... (xi)
2
12 2 22 2 2 3 2[2 cos( )] [2 cos( )]lr lr l r γ− β+ θ− γ + = ... (xii)
 2

2
13 3 23 3 3 3 3[2 cos( )] [2 cos( )]lr l r l r γ− β+ θ− γ + = ... (xiii)
Similarly, equation (x) for the three positions of θ, r and γ may be written three  times as
follows :

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1039
1121 1 3 1[2 cos( )] [2 cos( )] 2cos( )γ− β+ θ− γ + = θ− β mr m r m ... (xiv)
1222 2 3 2[2 cos( )] [2 cos( )] 2cos( )γ− β+ θ− γ + = θ− β mr m r m ... (xv)
132 3 3 3 3[2 cos( )] [2 cos( )] 2cos( )γ− β+ θ− γ + = θ− β mr m r m ... (xvi)
The equations (xi), (xii) and (xiii) are three linear equations in l1, l2, l3. Similarly, equations

(xiv), (xv) and (xvi) are three linear equations in m1, m2 and m3. Assuming a suitable value of β,
the values of l1, l2, l3 and m1, m2, m3 may be determined by using elimination method or Cramer’s
rule.
Knowing the values of l1, l2, l3 and m1, m2, m3, we can find the value of λ from equation
(vii). Now the values of k1, k2 and k3 are determined from equation  (vi) and hence q, a and e are
known from equation (iv). Using equation (i) or (ii), we can find the three valves of δ i.e. 12,δδ

and 3δ. From equation (i), we have
cos cos cos cosδ= γ− β− θ erq a
∴              111 11cos cos coscos− γ− β − θδ=rq a
e... (xvii)
Similarly,     122 22cos cos coscos− γ− β − θδ=rq a
e... (xviii)
and      133 3
3cos cos coscos− γ− β − θδ=rq a
e... (xix)
Thus by considering the loop OABE , we can find the values of q, a, e,  β and δ.
Now considering the loop ODCE  in order to find  p, c, f, α and ψ. The horizontal and
vertical components of vectors p, c, f  and r are

cos cos cos cosα+ φ+ ψ= γ pcf r ... (xx)
and sin sin sin sinα+ φ+ ψ= γ pcf r ... (xxi)
Since these equations are similar to equations  (i) and (ii), therefore we shall proceed in the
similar way as discussed for loop OABE .
Squaring equations (xx) and (xxi) and adding in order to eliminate angle φ, we have
2222[2 cos( )] [2 cos( )] [2cos( )]pr fr c p f r p f γ−α + ψ−γ + − − = + ψ−α ... (xxii)

Let p = k5 ; f = k6 ; c2 – p2 – f2 = k7 and p f = k8 = k5 k6... (xxiii)
Now equations (xxii) may be written as
2
567 8[2 cos( )] [2 cos( )] [2cos( )]kr kr k r k γ−α + ψ−γ + = + ψ−α ... (xxiv)
The equation (xxiv) is a non-linear equation and can be solved easily by making it linear
by some substitutions as given below :
Let55 1 566 1 6 ; kl m kl m=+ λ =+ λ  ;  and  77 1 7=+ λkl m ... (xxv)
where185 6 51 5 61 6 () () kk k l m l mλ= = = + λ + λ
    2
56 5 1 6 1 56 1 5 6ll l m ml mm=+ λ+ λ + λ

2
or                           2
56 1 56 65 1 5 6 (1 ) 0 mm lm lm llλ+ + − λ+ =

1040       /G6C               Theory of Machines
or   2
11 0 DEFλ+ λ+ =
∴      2
14
2EE D F
D−± −λ= ... (xxvi)
where     56 56 65;( 1 ) Dm mE l m l m== + −  ; and F = l5 l6... (xxvii )
Substituting the values of k5, k6, k7 and k8 in equation  (xxiv),
51 5 61 6 71 7( )[2 cos ( )] ( )[2 cos( )]lm r lm r lm+λ γ−α + +λ ψ−γ + +λ
 2
1[2cos( )]=+ λ ψ − αr

2
567[2 cos( )] [2 cos( )]γ−α + ψ−γ + = lr lr l r ... (xxviii )
and567[2 cos( )] [2 cos( )] 2cos( )γ−α + ψ−γ + = ψ−α mr mr m ... (xxix)
The equation  (xxviii ) for the three positions of r, γ and ψ may be written three times as
follows :
2
51 1 61 1 1 7 1[2 cos( )] [2 cos( )]γ− α+ ψ− γ + = lr lr l r ... (xxx)
2
52 2 62 2 2 7 2[2 cos( )] [2 cos( )]γ− α+ ψ− γ + = lr lr l r ... (xxxi)
2

2
53 3 63 3 3 7 3[2 cos( )] [2 cos( )]γ− α+ ψ− γ + = lr lr l r ... (xxxii )
Similarly, equation  (xxix) for the three positions of r, γ and ψ may be written three times
            51 1 61 1 1 7 1[2 cos( )] [2 cos( )] 2cos( )γ− α+ ψ− γ + = ψ− α mr mr m ... (xxxiii )
52 2 62 2 2 7 2[2 cos( )] [2 cos( )] 2cos( )γ− α+ ψ− γ + = ψ− α mr mr m  ..  (xxxiv )
            53 3 63 3 3 7 3[2 cos( )] [2 cos( )] 2cos( )γ− α+ ψ− γ + = ψ− α mr mr m                    ... (xxxv )

The equations (xxx), (xxxi) and (xxxii ) are three linear equations in l5, l6 and l7. Similarly,
equations (xxxiii ), (xxxiv ) and (xxxv ) are linear equations in m5, m6 and m7. Assuming a suitable
value of α, the values of l5, l6, l7 and m5, m6, m7 may be determined by using elimination method
or Cramer’s rule.
Knowing the values of l5, l6, l7 and m5, m6, m7, we can find the value of 1λ from equation

(xxvi). Now the values of k5, k6 and k7 are determined from equation  (xxv) and hence p, f and c are
known from equation (xxiii).
Assuming the value of 1ψ, the corresponding values of 2ψ and 3ψ may be calculated as
follows :
Since the angular displacements of the coupler link BCE is same at the points B and C,
therefore
           212 1ψ− ψ= δ− δ
or       212 1 () ψ= ψ+ δ− δ ... (xxxvi )
Similarly,                  313 1 () ψ= ψ+δ− δ ... (xxxvii )

If the mechanism is to be designed for more than three positions of the input link AB and

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1041
the same number of positions of the couple point E, then the least square technique is used. The
error function from equations (ix) and (x) are defined as
22
11 2 3 [{ 2c o s ( ) } { 2c o s ( ) } ]=γ − β + θ − γ + −∑e l rl rl r ... (xxxviii )
and2
21 2 3 [ { 2c o s ( ) } { 2c o s ( ) } 2 c o s ( ) ]=γ − β + θ − γ + − θ − β∑e m rm rm ... (xxxix )

For e1 and e2 to be minimum, the partial derivatives of e1 with respect to l1, l2, l3 and
partial derivatives of e2 with respect to m1, m2, m3 separately must be equal to zero, i.e.
and11 1
12 3
222
1230; 0 ; 0
0; 0; 0ee e
ll l
eee
mm m∂∂ ∂ == =∂∂ ∂
 ∂∂∂=== ∂∂∂ ... (xxxx )
First consider when 1
10∂=∂e
l,
2
12 3
12 2 cos ( ) 2 cos ( ) 2 cos ( ) 0n
lr l r l r r γ−β + θ−γ + − γ−β =∑
or  2
12
11[2 cos( )] [2 cos( )][2 cos( )]γ−β + θ−γ γ−β∑∑nn
lr l r r
                2
3

lr l r r
                2
3
11[2 cos( )] [2 cos( )]+ γ−β = γ−β∑∑nn
lr r r ... (xxxxi )
Similarly, for 1
20∂=∂e
l,
2
12
11[2 cos( )][2 cos( )] [2 cos( )]γ−β θ−γ + θ−γ∑∑nn
lr r l r
             2
3
11[2 cos( )] [2 cos( )]+ θ−γ = θ−γ∑∑nn
lr r r ... (xxxxii )
Note : This picture is given as additional information and is not a direct example of the current chapter.An aircraft assembling plant.

1042       /G6C               Theory of Machines
and for   1
30∂=∂e
l, 2
123
111[2 cos( )] [2 cos( )] 1γ−β + θ−γ + =∑∑∑ ∑nnn
lr l r l r... (xxxxiii )
The above three equations can be solved by using Cramer’s rule to find l1, l2 and l3.
In the similar way as discussed above, for 2
10∂=∂e
m
         2
12
mr m r r
        3
11[2 cos ( )] [2 cos ( )][2 cos ( )]+γ − β = θ − β γ − β∑∑nn
mr r r  ... (xxxxiv )
Similarly, for 2
20,e
m∂=∂
2
12

20,e
m∂=∂
2
12
11[2 cos( )] [2 cos( )] [2 cos( )]γ−β + θ−γ + θ−γ∑∑nn
mr r m r
        3
11[2 cos( )] [2 cos( )][2 cos( )]+θ − γ = θ − β θ − γ∑∑nn
mr r r  ... (xxxxv )
and for 2
30∂=∂e
m,
         123
111[2 cos( )] [2 cos( )] 1 [2 cos( )]γ−β + θ−γ + = θ−γ∑∑∑ ∑nnn
mr m r m r  ... (xxxxvi )
The above three equations can be solved by using Cramer’s rule to find m1, m2 and m3.

(vii) and k1, k2, k3 from equation (vi). Thus q, a and e are determined. Now 123,,δδδ  may be
determined by using equation (i) or (ii).
The values of p, c and f are obtained by solving equation (xxiv) in the similar way as
discussed earlier.
25.18. Synthesis of Four Bar Mechanism For Body Guidance
Consider the three positions of a rigid planer body containing the points A and B as shown
in Fig. 25.17 ( a). The four bar mechanism for body guidance, considering the three positions of the

body, may be designed graphically as discussed below.
1. Consider the three positions of the points A and B such as A1, A2, A3 and B1, B2, B3 as
shown in Fig. 25.17 ( a).
2. Find the centre of a circle which passes through three points A1, A2, A3. This is obtained
by drawing the perpendicular bisectors of the line segments A1 A2 and A2 A3. Let these bisectors
intersect at OA. It is evident that a rigid link A OA pinned to the body at point A and pinned to the

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1043
ground at point OA will guide point A through its three positions A1, A2 and A3.
(a)( b)
Fig. 25.17.  Four bar mechanism for body guidance.
3. Similarly, find the centre OB of a circle which passes through three points B1, B2, B3. It
is evident that a rigid link B OB pinned to the body at point B and pinned to the ground at point OB
will guide point B through its three positions B1, B2 and B3.

4. The above construction forms the four bar mechanism OA ABOB which guides the body
through three specified positions. Fig. 25.17 ( b) shows a four bar mechanism in these three positions.
The points OA and OB may be determined analytically as discussed below :
Consider the three positions of the point A such as A1, A2, A3. Let the co-ordinates of these
points are A1 (x1, y1) ; A2 (x2 , y2) and A3 (x3, y3). Let the co-ordinates of the point OA are ( x, y).
Now we know that

Distance between points A1 and OA,
22 1 / 2
1A 1 1 [( ) ( ) ]=−+− AO x x y y ... (i)
Similarly, distance between points A2 and OA,
22 1 / 2
2A 2 2 [( ) ( ) ]=− + − AO x x y y ... (ii)
and distance between points A3 and OA,
           22 1 / 2
3A 3 3 [( ) ( ) ]=−+ − AO x x y y ... (iii)
For the point OA to be the centre of a circle passing through the points A1, A2 and A3, the
distances A1OA, A2OA and A3OA must be equal. In other words,
1A 2A 3A== AO A O AO

Now considering  1A 2A= AO A O , we have
1/2 1/222 2 2
11 2 2() () ( ) ( )  −+− = −+−  xx yy x x y y ... (iv)
Similarly, considering 2A 3A= AO AO , we have
             1/2 1/222 22
22 33() ( ) () () −+− = −+− xx yy xx yy ... (v)
Squaring both sides of the equations  (iv) and (v) and simplifying, we get the following two
equations in the unknowns x and y.
22 22
21 21 1 2 1 2 2( ) 2( ) ( ) ( ) 0−+ −+−+ − = xx x yy y x x y y ... (vi)
and22 22

and22 22
32 32 23 23 2( ) 2( ) ( ) ( ) 0−+ −+−+−= xx x yy y x x y y ... (vii)

1044       /G6C               Theory of Machines
The equations  (vi) and (vii) are simultaneous equations and may be solved to find the
co-ordinates x, y of the point OA. This point OA becomes the location of the fixed pivot guiding the
point A. The length of the guiding link OAA may be determined by any of the equations (i), (ii) or
(iii).
In the similar way, as discussed above, we can find the location of the fixed pivot point OB
and the length of the link OBB.

Example 25.8.  Synthesize a four bar mechanism to guide a rod AB through three consecu-
tive positions A1B1, A2B2 and A3B3 as shown in Fig. 25.18.
Fig. 25.18
Solution :  In order to synthesize a four bar mechanism, we shall use the graphical method
1. Join points A1, A2 and A2, A3. Draw the perpendicular bisectors of line segments A1A2
and A2A3 to intersect at OA, as shown in Fig. 25.19. It is evident that a rigid link OAA1 pinned to the

body at point A1 and pinned to the ground at point OA will guide point A1 through its three posi-
tions.
Fig. 25.19

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1045
2. Similarly, join points B1, B2 and B2, B3. Draw the perpendicular bisectors of line seg-
ments B1B2 and B2B3 to intersect at OB as shown in Fig. 25.19. It is evident that a rigid link OBB1
pinned to the body at point B1 and pinned to the ground at point OB will guide point B, through its
three positions.
3. From above we see that the points OA and OB are the required fixed points and

OA A1 B1 OB is one position of the four bar mechanism. The other two positions of the mechanism
will be A22BO ABO  and A33BOA B O .
25.19. Analytical Synthesis for Slider Crank Mechanism
A slider crank mechanism is shown in Fig. 25.20. In the sysnthesis problem of the slider
crank mechanism, the displacement ( s) of the slider C has to co-ordinate with the crank angle ( θ)
in a specified manner. For example, consider that the displacement of the slider is proportional to

crank angle over a given interval, i.e.
SS () ss C−=θ − θ , for SFθ≤ θ ≤ θ ... (i)
where      C  = Constant of proportionality, and
      s  = Displacement of the slider when crank angle is θ.
The subscripts S and F denote
starting and finishing positions.
The synthesis of a slider crank
mechanism for three precision pointsis obtained as discussed below.
The three positions of the crank
(
12,θθ  and 3θ) may be obtained in

(
the similar way as discussed in Art.25.10 and the corresponding threepositions of the slider ( s
1, s2 and s3) are
obtained from the given condition as inequation (i). Now the dimensions a, b
and c may be determined as discussed
below :
In a right angled triangle 
′BC C ,
;s i n BC b BC a c ′== θ −  , and cos′=− θCC s a
∴  222(s i n ) ( c o s)=θ − + − θba c s a
     22 2 2 2 2sin 2 sin cos 2 cos=θ + −θ + +θ −θac a c s a s a

22 22s i n 2c o s =+− θ −− θac a c s a s
or              22222c o s 2s i nθ+ θ+ − − = as ac b a c s
           2
12 3cos sinθ+ θ− = ks k k s ... (ii)
where             12=ka  ; 22=ka c  and 222
3=−+kabc ... (iii)Fig. 25.20. Slider crank mechanism.

1046       /G6C               Theory of Machines
For the three different positions of the mechanism i.e. for 12 3(, )θθθ  and ( s1, s2, s3), the
equation (ii) may be written as
 2
11 1 2 1 3 1cos sinks k k sθ+ θ− = ... (iv)
           2
12 2 2 2 3 2cos sinks k k sθ+ θ− = ... (v)
2
13 3 2 3 3 3cos sinks k k sθ+ θ− = ... (vi)
The equations (iv), (v) and (vi) are three simultaneous equations and may be solved for

three unknowns k1, k2 and k3. Knowing the values of k1, k2 and k3, the lengths a, b and c may be
obtained from equations (iii).
Example 25.9.  Synthesize a slider crank mechanism so that the displacement of the slider
is proportional to the square of the crank rotation in the interval 45 135°≤θ≤ ° . Use three preci-
sion points with Chebyshev’s spacing .
Solution :  Given. S45θ= °  ; F135θ= °
First of all, let us find the three precision points ( i.e. x1, x2 and x3). We know that

SF FS11 ( 2 1 )() () c o s22 2jjxx x x xnπ−=+ −− ; where  j = 1, 2 and 3
Assuming the starting displacement of the slider ( sS) = 100 mm and final displacement of
the slider ( sF) = 30 mm. It may be noted that for the crank rotating in anticlockwise direction, the
final displacement will be less than the starting displacement.
∴   111 ( 2 1 1 )(100 30) (30 100)cos 95.322 2 3xπ× −=+ − − =× mm.
. . . (∵    SSxs= ; FFxs= and n = 3)

211 ( 2 2 1 )(100 30) (30 100)cos22 2 3xπ×−=+ − −× = 65 mm
and 311 ( 2 3 1 )(100 30) (30 100)cos22 2 3xπ× −=+ − −× = 34.7 mm
The corresponding three values of θ are given by
FS
SS
FS()jj xxxxθ− θθ= θ+ −− ; j = 1, 2, and 3
∴ 1135 4545 (95.3 100) 51.04
30 100−θ= + − = °
−
2135 4545 (65 100) 9030 100−θ= + − = °−
and             3135 4545 (34.7 100) 128.96
−
Since it is given that the displacement of the slider ( s)

is proportional to the square of the crank rotation ( θ), therefore,
for the displacement from initial position ( sS) to s when crank
rotates from initial position (Sθ) to θ, we haveA belt-conveyor that can trans-
port small components.

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1047
   2
SS () ss C−=θ − θ ... (θ is expressed in degrees)
∴    S
22
S30 100 7
810 ( ) (135 45)ssC− −−== =
θ−θ −... (Taking s = sF ; and Fθ=θ )
Now the three positions for the slider displacement ( s) corresponding to the three positions
of the crank angle ( θ) are given by
22
1S 1S7( ) 100 (51.4 45) 99.7
810ssC=+θ − θ = − − =  mm
22
2S 2S7( ) 100 (90 45) 82.5
810ss C=+θ − θ = − − =  mm
    22

    22
3S 3S7( ) 100 (128.96 45) 39.08
Now the three equations relating the (11,sθ ), (22,sθ ) and (θ3, s3) are written as
2
12 399.7cos51.04 sin 51.04 (99.7)kk k×° + ° − =
or             12 3 62.7 0.7776 9940 kk k+− = ... (i)
Similarly, 2
12 382.5cos90 sin 90 (82.5)kk k×° + ° − =
or   23 6806−=kk ... (ii)
and2
12 339.08cos128.96 sin128.96 (39.08)kk k×° + ° − =
or            12 3 24.57 0.776 1527 kk k−+− = ... (iii)

The equations (i), (ii) and (iii) are three simultaneous equations in three unknowns k1, k2and k3. On solving, we get
       k1 = 96.4 ;  k2 = 13 084 ; and k3 = 6278
We know that   k1 = 2a, or  a = k1 / 2 = 96.4 / 2 = 48.2 mm  Ans.
        k2 = 2a.c or  c = k2 / 2a = 13 084 / 2 × 48.2 = 135.7 mm  Ans.
and       222
3=−+kabc
       22 2 2 2
3(48.2) (135.7) 6278 14 460 ba c k=+−= + − =
or         b  = 120.2 mm  Ans.

1. In a four bar mechanism PQRS , the link PS is fixed. The length of the links are : PQ = 62.5 mm ;
QR = 175 mm ; RS = 112.5 mm and PS = 200 mm. The crank PQ rotates at 10 rad/s clockwise.
Find the angular velocity and angular acceleration of the links QR and RS for the values of angle
QPS at an interval of 60°.
2. In a slider crank mechanism, the crank AB = 100 mm and the connecting rod BC = 300 mm. When

the crank is at 120° from the inner dead centre, the crank shaft has a speed of 75 rad/s and an
angular acceleration of 1200 rad/s2 both clockwise. Find at an interval of 60° 1. the linear velocity
and acceleration of the slider, and 2. the angular velocity and angular acceleration of the rod, when

1048       /G6C               Theory of Machines
(a) the line of stroke of the slider is offset by 30 mm, and
(b) the line of stroke of the slider is along the axis of rotation of the crank.
3. A mechanism is to be designed to generate the function
y = x0.8
for the range 13x≤≤ , using three precision points. Find the three values of x and y.
[Ans. 1.134, 2, 2.866 ; 1.106, 1.741, 2.322]
4. Determine the three precision positions of input and output angles for a mechanism to generate  a
function

function
y = x1.8
when x varies from 1 to 5, using Chebyshev’s spacing. Assume that the initial values for the input
and output crank are 30° and 90° respectively and the difference between  the final and initialvalues for the input and output cranks are each equal to 90°.
[Ans. 36°, 75°, 94.48°; 91.22°, 144.57°, 181.22°]
5. Synthesize a four bar linkage using Freudenstein’s equation to generate the function y = x
1.8 for

1.8 for
the interval 15 .x≤≤  The input crank is to start from S30θ= °  and is to have a range of 90°. The
output follower is to start at S0φ=°  and is to have a range of 90°. Take three accuracy points at x
= 1, 3 and 5.
6. A four bar function generator is used to generate the function y = 1/ x for 13x≤≤  between the

input angle of a crank and the angle the follower makes with the frame. Find the three precisionpoints from Chebyshev’s spacing if the initial values of input angle ( i.e. crank angle) and output
angle ( i.e. follower angle) are 30° and 200° respectively. The difference between the final and
initial values of the crank and follower angles are each equal to 90°.
7. Synthesize a four bar linkage that will generate a function y = x
1.2 for the range 15 .x≤≤  Take

three precision points : θ= °S30 ; φ= °S60 and 90 , ∆θ= ∆φ= °  where θS and φS represent respec-
tively the initial angular positions of the input and output crank; ∆θ and ∆φ are respectively the
ranges of the angular movements of the input and output crank.
8. Synthesize a four bar mechanism to generate the function y = log x, where x varies between 1 and
10. Use three accuracy points with Chebyshev’s spacing. Assume θS = 45°; θF = 105°; φS = 135°

and φF = 225°. Take the length of the smallest link equal to 50 mm.
9. Synthesize a four bar mechanism to move the rod AB as shown in Fig. 25.21, through the positions
1, 2 and 3. The end points A and B are used as moving pivot points.
Fig. 25.21 Fig. 25.22

Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms            /G6C          1049
Fig. 25.2310. Design a four bar mechanism to guide the door in and out with little rotation until it clears the
surrounding structure, after which it swings fully open to one side. The three positions of such adoor under going  this type of motion is shown in Fig. 25.22. The points A and B are used as
moving pivots that guides the body through the three positions.

1. Explain Freudenstein’s method of three point synthesis of mechanisms.
2. Derive the expressions for displacement, velocity and acceleration of a four bar mechanism.
3. What do you understand by coupler curves ? Describe the method of obtaining the co-ordinates ofa coupler point in a slider crank mechanism.
4. Explain synthesis of mechanism with examples. What do you understand by

(a) Type synthesis ;     ( b) Number synthesis ; and ( c) dimensional synthesis.
5. Describe the classifications of synthesis problem.
6. Write an expression for determining the precision points.
7. Discuss the method of determining the angles for input and output link in a four bar mechanism for
function generation.
8. Describe the method of designing a four bar mechanism as a function generation.

1. The analysis of mechanism deals with
(a)  the determination of input and output angles of a mechanism
(b)  the determination of dimensions of the links in a mechanism
(c)  the determination of displacement, velocity and acceleration of the links in a mechanism
(d)  none of the above

2.  The synthesis of mechanism deals with(a)  the determination of input and output angles of a mechanism
(c)   the determination of displacement, velocity and acceleration of the links in a mechanism
3. The three precision points in the range 
13x≤≤  are
 (a) 1.1, 2, 2.6 ( b) 1.6, 2.5, 2.95
 (c) 1.134, 2, 2.866 ( d) 1.341, 2 , 2.686

(c) 1.134, 2, 2.866 ( d) 1.341, 2 , 2.686
4. For a four bar mechanism, as shown in Fig. 25.23 the Freudenstein’s equation is
(a)  123cos cos cos ( )kk kθ+ φ+ = θ−φ
(b)  123cos cos cos( )kk kθ− φ+ = θ−φ
(c)  123cos cos cos ( )kk kφ+ θ+ = θ−φ
(d)  123cos cos cos( )kk kφ− θ+ = θ−φ
where 1dka= ; 2dkc= ; 222 2
1. (c) 2.  (b) 3. (c) 4. (d)

1050       /G6C               Theory of Machines
1050AutomaticAutomaticAutomaticAutomaticAutomatic
ContrContrContrContrContr ololololol26
2. Terms Used in Automatic
control of Systems.
3. Types of Automatic Control
System.
4. Block Diagrams.5. Lag in Response.6. Transfer Function.7. Overall Transfer Function.8. Transfer Function for a
System with Viscous
Damped Output.
9. Transfer Function of a
Hartnell Governor.
10. Open-Loop Transfer

Function.
11. Closed-Loop Transfer
Function.26.1.26.1.26.1.26.1.26.1. IntrIntrIntrIntrIntroductionoductionoductionoductionoduction
The automatic control of system (or machine) is a very
accurate and effective means to perform desired function by

the system in which the human operator is replaced by adevice thereby relieving the human operator from the job thussaving physical strength. The automatic control systems arealso called as self-activated systems.  The  centrifugally
actuated ball governor which controls the throttle valve tomaintain the constant speed of an engine is an example of anautomatically controlled system.
The automatic control systems are very fast, produces

uniform and quality products. It reduces the requirement ofhuman operators thus minimising wage bills.
26.2.26.2.26.2.26.2.26.2. TTTTTerererererms used in ms used in ms used in ms used in ms used in AAAAAutomautomautomautomautoma tic Contrtic Contrtic Contrtic Contrtic Contr ol ofol ofol ofol ofol of
SystemsSystemsSystemsSystemsSystems
The following terms are generally used in automatic
control of systems :
1. Command.  The result of the act of adjustment, i.e.

closing a valve, moving a lever, pressing a button etc., isknown as command.
2. Response.  The subsequent result of the system to
the command is known as response.
3. Process control.  The automatic control of variables
i.e. change in pressure, temperature or speed etc. in machine
is termed as process control.

Chapter 26 : Automatic Control            /G6C          1051
4. Process controller.  The device
which controls a process is called a
process controller.
5. Regulator.  The device used to
keep the variables at a constant desired
value is called as regulator.
6. Kinetic control.  The automatic
control of the displacement or velocity or
acceleration of a member of a machine iscalled as kinetic control.
7. Feed back.  It is defined as

measuring the output of the machine forcomparison with the input to the machine.
8. Error detector.  A differential
device used to measure the actualcontrolled quantity and to compare itcontinuously with the desired value is
called an error detector. It is also known
as deviation sensor.
9. Transducer.  It is a device to

change a signal which is in one physical form to a corresponding signal in another physical form.A Bourdon tube is an example of transducer because it converts a pressure signal into a displacement,thereby facilitating the indication of the pressure on a calibrated scale. The other examples oftransducer are a loud speaker (because it converts electrical signal into a sound) and a photo-electric cell (because it converts a light signal into an electric signal). Similarly, the primary elementsof

signal). Similarly, the primary elementsof all the many different forms of thermometers are transducers.

10.  Amplification.  It is defined as increasing the amplitude of the signal without affecting its
waveform. For example, an error detector itself has insufficient power output to actuate the correcting
mechanism and hence the error signal has to be amplified. This is generally done by employingmechanical or hydraulic or pneumatic amplifying elements like levers, gears and venturimeters etc.

26.3.26.3.26.3.26.3.26.3. TTTTTypes of ypes of ypes of ypes of ypes of AAAAAutomautomautomautomautoma tic Contrtic Contrtic Contrtic Contrtic Contr ol Systemol Systemol Systemol Systemol System
The automatic control systems are of the following two types :1. Open-loop or unmonitored system. When the input to a system is independent of the

output from the system, then the system is called an open-loop or unmonitored system. It is alsocalled as a calibrated system. Most measuring instruments are open-loop control systems, as for
the same input  signal, the readings will depend upon things like ambient temperature and pressure.Following are the examples of open-loop system :
(a) A simple Bourdon tube pressure gauge commonly used for measuring pressure.

(b) A simple carburettor in which the air-fuel ratio adjusted through venturi remains same
irrespective of load conditions.
(c) In traffic lights system, the timing of lights is preset irrespective of intensity of traffic.
2. Closed-loop or monitored system.  When output of a system is measured and is continu-
ously compared with the required value, then it is known as closed-loop or monitored system.  In

this system, the output is measured and through a feedback transducer, it is sent to an error detectorwhich detects any error in the output from the required value thus adjusting the input in a way toget the required output. Following are the examples of  a closed-loop system :
(a) In a traffic control system, if the flow of traffic is measured either by counting the number

of vehicles by a person or by counting the impulses due to the vehicles passing over a pressure padand then setting the time of signal lights.
A rail-track maintenance machine.

1052       /G6C               Theory of Machines
(b) In a thermostatically controlled water heater, whenever the temperature of water heater
rises above the required point, the thermostate senses it and switches the water heater off so as to
bring the temperature down to the required point. Similarly, when the temperature falls below therequired point, the thermostate switches on the water heater to raise the temperature of water to therequired point.

26.4.26.4.26.4.26.4.26.4. Block DiagramsBlock DiagramsBlock DiagramsBlock DiagramsBlock Diagrams
Fig. 26.1.  Block diagram of a single carburettor.
The block diagrams are used to study the automatic control systems in a simplified way. In
this, the functioning of a system is explained by the interconnected blocks where each block represents

a labelled rectangle and is thought of as a block box with a definite function. These blocks areconnected to other blocks by lines with arrow marks in order to indicate the sequence of events thatare taking place. Fig. 26.1 shows the diagram of a simple carburettor. The reduction of a controlsystem to a block diagram greatly facilitates the analysis of the system performance or response.
26.5.26.5.26.5.26.5.26.5. Lag in ResponseLag in ResponseLag in ResponseLag in ResponseLag in Response

We know that response is the subsequent result of the system to the command. In any control
system, there is a delay in response (output) due to some inherent cause and it becomes difficult tomeasure the input and output simultaneously. This delay in response is termed as lag in response .

For example, in steam turbines, with the sudden decrease in load, the hydraulic relay moves in thedirection to close the valve. But unless the piston valve ports are made with literally zero overlap,there would be some lag in operation, since the first movement of the piston valve would not besufficient to open the ports. This lag increases the probability of unstable operation.
26.6 26.6 26.6 26.6 26.6 TTTTTransfer Functionransfer Functionransfer Functionransfer Functionransfer Function

The transfer function is an expression showing the relation between output and the input to
each unit or block of a control system. Mathematically,
Transfer function = /oiθθ
whereoθ = Output signal of the block of a system, and
iθ =  Input signal to the block of a system.

Chapter 26 : Automatic Control            /G6C          1053
Thus, the output from an element may be obtained by multiplying the input signal with the
transfer function.
Note :  From the transfer function of the individual blocks, the equation of motion of system can be formu-
lated.
26.7 Ov26.7 Ov26.7 Ov26.7 Ov26.7 Ov erall erall erall erall erall TTTTTransfer Functionransfer Functionransfer Functionransfer Functionransfer Function

In the previous article, we have discussed the transfer function of a block. A control system
actually consists of several such blocks which are connected in series. The overall transfer function
of the series is the product of the individual transfer function. Consider a block diagram of anycontrol system represented by the three blocks as shown in Fig. 26.2.
Fig. 26.2. Overall transfer function.
Thus, if F1 (D), F2 (D), F3 (D) are individual transfer functions of three blocks in series, then

the overall transfer function of the system is given as
12
123
12() () () ()oo
iiFD F D FD K G Dθθθθ=××= × × =θθθθ
where K = Constant representing the overall amplification or gain, and
        G(D) = Some function of the operator D.
Note:  The above equation is only true if there is no interaction between the blocks, that is the output from
one block is not affected by its connection to the subsequent blocks.

26.8.26.8.26.8.26.8.26.8.      TTTTTransfer Function fransfer Function fransfer Function fransfer Function fransfer Function f or a System with viscous Damped Outputor a System with viscous Damped Outputor a System with viscous Damped Outputor a System with viscous Damped Outputor a System with viscous Damped Output
Consider a shaft, which is used to position a load (which may be pulley or gear) as shown in
Fig. 26.3. The movement of the load is resisted by a viscous damping torque.

Fig. 26.3. Transfer function for a system with viscous damped output.
Letiθ = Input signal to the shaft,
           oθ = Output signal of the shaft,
  q = Stiffness of the shaft,
  I = Moment of Inertia of the load, and
 Td = Viscous damping torque per unit angular velocity.

1054       /G6C               Theory of Machines
After some time t,
Twist in the shaft io=θ −θ
∴ Torque transmitted to the load ()ioq=θ − θ
We also know that damping torque = 0o
dddT= T
dtθω... 0 (/ )odd tω=θ∵
According to Newton’s Second law, the equation of motion of the system is given by
          2
2()oo
io dddIqTdt dtθθ =θ − θ − ... (i)
         2
2oo
io dddIq q Tdt dtθθ =θ −θ − 
Replacing d / dt by D in above equation, we get

2() ( )oi o d o ID q q T Dθ=θ −θ − θ
or2() ( )odo o i ID T D q qθ+ θ+θ =θ
           2() ( ) ( )d
oo o iT qqDDII Iθ+ θ + θ = θ
        22 2( ) () ()d
oo n o nTDDIθ+ θ +ω θ=ω θ ... (ii)
where nω = Natural frequency of the shaft = q
I
Also we know that viscous damping torque per unit angular velocity,
 =2dnTIξω   or   /= 2dnTI ξω
where   ξ = Damping factor or damping ratio.
Material being moved via-belt conveyor.

or

Chapter 26 : Automatic Control            /G6C          1055
The equation (ii) may now be written as
    22 22() ( ) ( )no n o n i DDθ+ ξω θ + ω θ = ω θ
or22 2[2 ( ) ]( )nn o n i DD+ξ ω + ω θ=ω θ
∴                   Transfer function 2
22()
2( )on
i nn DDθω==θ+ξ ω +ω
                   221
21 TD T D=
+ξ +
where                                           T = Time constant = 1/nω
Note:  The time constant ( T) may also be obtained by dividing the periodic time ( td) of the undamped natural

oscillations of the system by 2π. Mathematically,
                            211
22d
nntTπ==×=πω πω... 2
d
nt π=ω∵
Example 26.1. The motion of  a pointer over a scale is resisted by a viscous damping torque
of magnitude 0.6 N-m at an angular velocity of 1 rad / s. The pointer, of negligible inertia, is
mounted on the end of  a relatively flexible shaft of stiffness 1.2 N-m / rad, and this shaft is driventhrough a 4 to 1 reduction gear box. Determine its overall transfer function.

If the input shaft to the gear box is suddenly rotated through 1 completed revolution, determine
the time taken by the pointer to reach a position within 1 percent of its final value.
Solution.  Given:
T
d = 0.6/1 = 0.6 N-ms/rad;
 q = 1.2 N-m/rad
The control system along with its
block diagram is shown in Fig 26.4 ( a)
1. Overall transfer function
Since the inertia of the pointer is

negligible, therefore the torque generatedby the twisting of the shaft has only toovercome the damping torque.
Therefore
1() ( / )od o qT d d tθ− θ = θ
where           θ1 = Output from the gear box.
∴            1 ()od o qq T Dθ− θ= θ . . . (/ )dd t D= ∵
or          1 ()do qT D q+θ = θ
∴                     
111
1( /) 1o
ddq
qT D T q D T Dθ== =θ+ + +...(i)
where  T = Time constant = Td/q = 0.6/1.2 = 0.5sFig. 26.4

1056       /G6C               Theory of Machines
Substituting this value in equation (i), we get
θ=θ+0
11
10 . 5 D
We know that overall transfer function for the control system is
12
111×4( 1 + 0 . 5 )o
ii Dθθθ=×=θθ θ Ans. ... 11/ (Given)
4iθθ =∵
2. Time taken by the pointer
Let    t = Time taken by the pointer.
Since the input shaft to the gear box is rotated through 1 complete revolution, therefore
2,iθ=π  a constant.
We know that transfer function for the control system is

11or (1 0.5 )4( 10 . 5 ) 4oi
o
iDDθθ=× + θ=θ+
∴ 0.5
4oi
od
dtθθ+θ =... (∵ D ≡ d/dt)
Substituting 2iθ=π  in the above equation, we get
20.542o
od
dtθ ππ +θ = =
or        0.5
2o
od
dtθπ =− θ
Separating the variables, we get
2
oddtθ=π−θ
Aircraft engine is being assembled.

Chapter 26 : Automatic Control            /G6C          1057
Integrating the above equation, we get
log 2 constant2eo tπ−− θ = +... (ii)
Applying initial conditions to the above equation i.e. when t = 0, 0 oθ= , we get
         constant = log2eπ−
Substituting the value of constant in equation (i),
log 2 log22eo e tππ  −− θ = −    
or  log 2 log22eo e tππ  −θ =− +    
∴            2
22t
oe− ππ−θ = ×
or         2 /2
/2t oe− π− θ=
π

22t
oe− ππ−θ = ×
or         2 /2
π
i.e                  −πθ= −2(1 )2t
o e ... (iii)
The curve depicted by above equation is shown in Fig. 26.5 and is known as simple expo-
nential time delay curve .
Fig. 26.5
The output oθ will be within 1 percent of its final value when 00.99( / 2).θ= π  Substituting
this value in equation (iii), we get
        ()20.99 122te− ππ=−
                             220.99 1 or 0.01ttee−−=− =
∴                  2 log 100 4.6 or 2.3s e tt== =  Ans.

1058       /G6C               Theory of Machines
26.9.26.9.26.9.26.9.26.9.      TTTTTransfer Function of a ransfer Function of a ransfer Function of a ransfer Function of a ransfer Function of a HHHHHararararartnell Gotnell Gotnell Gotnell Gotnell Go vvvvvererererernornornornornor
Consider a Hartnell governor * as shown in Fig. 26.6 ( a). The various forces acting on the
governor are shown in Fig. 26.6 ( b).
Let m = Mass of the ball
           M  = mass of the sleeve,

M  = mass of the sleeve,
r  = Radius of rotation of the governor in mid position,
         r∆ = Change in radius of rotation,
           ω = Angular speed of rotation in mid position,
            ∆ω = Change in angular speed of rotation,
(a) Hartnell governor.  ( b) Forces acting on a Hartnell governor.
Fig. 26.6
 x = Length of the vertical or ball
      arm of the lever,
 y =Length of the horizontal or
      sleeve arm of the lever,
h = compression of spring with

      balls in vertical position,
          h′ = Displacement of the sleeve,
 s = Stiffness of the spring,
c = Damping coefficient i.e.
   damping force per unit
                 velocity, and
           ξ = Damping factor.
* For details on Hartnell governor, refer chapter 18, Art. 18.8 .
Bucket conveyor

Chapter 26 : Automatic Control            /G6C          1059
The various forces acting on the governor at the given position are as follows :
1.  Centrifugal force due to ball mass,
2() ( )cFm r r=+ ∆ ω + ∆ ω
                 = 2() (+ )xmr h
y′+ω ∆ ω
2. Inertia force of the balls,            2
2imxd hFmy dt′ =
3. Inertia force of the sleeve mass,    2
2′=iMdhFM
dt
4. Damping force, ddxFcdt=
5. Spring force,              () sFs h h ′=+

It is assumed that the load on the sleeve, weight of the balls and the friction force are negligible
as compared to the inertia forces. Now, taking moments about the fulcrum O, considering only one
half of the governor,
22
2
221()2x x dh dhmr h x m x M yyy dt dt ′′ ′+ω + ∆ ω = × + ×      
               11()22dhcy s h h ydt′′ +× +× +
Neglecting the product of small terms, we get
222( )′ ω+××ω+ ω ∆ ωxmr x m h x mr xy
= 22 2

= 22 2
2211
22m x dh dh d hMy c yyd t dt dt ′′ ′ +× +×      1()2syh h′ +× +
...(i)
Also, we know that at equilibrium position,
2 1
2mr x s h yω=×
222 2 11 12( ) () () ( )22 2xm xs hy m h x m r x Dh M yDh cyD hyy′′ ′ ′ ×+ × × ω +ω ∆ ω = + +
           1()2sy h h′ ++ ...(/ )dd t D= ∵

1060       /G6C               Theory of Machines
or22
22 11 12( )22 2mx mxMy D h cy Dh sy D h mr xyy   ′′ ′ ++ × + − × = ω ∆ ω        
Multiplying the above equation throughout by 2 y, we get
22 2 2 2 2 2(2 ) ( ) ( 2 ) 4 ( )mx My D h c y dh sy mx h mr x y ′′ ′ ++ + − ω = ω ∆ ω
22 2 222 2
22 222(2 ) 4 ( )
22cy sy mxmx My D h mr xy
mx My mx My −ω′ ++ + = ω ∆ ω++ 
or22 2 22
22 22 2224 ( )
222cy sy mx mr xyDDh
mx My mx My mx My −ω ω ∆ ω′ +× + =+++ 
or22

or22
224( )2( )
2nnmr xyDD h
mx Myω∆ ω′ +ξ ω +ω =
+
∴22 2 24( ) 1
22 ( )nnmr x yh
mx M y DDω∆ ω′=×
++ ξ ω + ω
where       2
222
2ncy
mx Myξω =
+
            ξ = Damping factor, and
          nω = Natural frequency  = 22 2
222
2sy mx
mx My−ω
+
Thus, transfer function for the Hartnell governor,
   Output signal Displacement of sleeve ( )=Input signal Change in speed ( )h′=∆ω
              22 2 241
22 ( ) nnmr x y
mx M y DDω=×
++ ξ ω + ω

mx M y DDω=×
++ ξ ω + ω
26.10.26.10.26.10.26.10.26.10. Open-Loop Open-Loop Open-Loop Open-Loop Open-Loop TTTTTransfer Functionransfer Functionransfer Functionransfer Functionransfer Function
Fig. 26.7. Open loop control system.
Fig. 26.8 Simplified open loop control system.
The open loop transfer function is defined as the overall transfer function of the forward path
elements. Consider an open loop control system consisting of several elements having individual

transfer function such F1(D), F2(D), F3 (D) as shown in Fig. 26. 7. Thus

Chapter 26 : Automatic Control            /G6C          1061
Open loop transfer function = 12
12oo
iiθθθθ=××θθ θθ
            = 123() () () ()FD F D FD K G D××=
The simplified block diagram of open loop transfer function is shown in Fig. 26.8.
26.11.26.11.26.11.26.11.26.11.  Closed - Loop  Closed - Loop  Closed - Loop  Closed - Loop  Closed - Loop TTTTTransfer Functionransfer Functionransfer Functionransfer Functionransfer Function

The closed loop transfer function is defined as the overall transfer function of the entire
control system. Consider a closed loop transfer function consisting of several elements as shown in
Fig. 26.9.
Fig 26.9 Closed-loop transfer function.
Now, for the forward path element, we know that
1()oo
ioKGDθθ==θθ − θ
where      123 () () () ()K G D F DF DF D=××
On rearranging, we get
           () ()oi oKG D KG Dθ= θ− θ
or     [1 ( )] ( )oi KG D KG D+θ = θ
∴            () O p e n l o o p

1 ( ) 1+ Open loopo
iKGD T F
KGD T Fθ==θ+
The above expression shows the transfer function for the closed-loop control system.
Thus the block diagram may be further simplified as shown in Fig. 26.10, where the entire
system is represented by a single block.
Fig. 26.10.  Simplified closed-loop system.
1. Define the following terms:
(a) Response ( b) Process control
(c) Regulator ( d) Transducer

2. What do you understand by open-loop and closed loop control system? Explain with an example.
3. Discuss the importance of block diagrams in control systems.

1062       /G6C               Theory of Machines
4. Draw the block diagrams for the following control systems:
(a) A simple carburettor,
(b) A thermostatically controlled electric furnace.
5. What is a transfer function ?
1. The device used to keep the variables at a constant desired value is called a
(a) process controlled ( b) regulator
(c) deviation sensor ( d) amplifier

2. The transfer function of a 4 to 1 reduction gear box is
(a)4 ( b)2
(c) 1/4 ( d) 1/2
3. A simple Bourdon tube pressure gauge is a
(a) closed-loop control system
(b) open-loop control system
(c) manually operated system
4. The overall transfer function of three blocks connected in series is
(a)12
3() ()
()× FD F D
FD                          ( b) 13
2() ()
()× FD F D
FD
(c) 123() () ()×× FD F D F D                         ( d) 1231

() () ()×× FD F D FD
where F1 (D), F2 (D) and F3 (D) are the individual transfer functions of the three blocks.
5. The transfer function for a closed-loop control system is
(a)()
1( )+KG D
KG D(b) () [ 1 () ]+ KGD K GD
(c)1( )
()+KGD
KG D(d)()
1( )−KG D
KG D
1. (b) 2. (c) 3. (b) 4. (c) 5. (a)