gyroscope - engineering colleges in rajkot of... · 1. gyroscope theory of machines (2151902)...

Department of Mechanical Engineering Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot .1

1 Gyroscope

Course Contents

1.1 Principle of Gyroscope

1.2 Angular Velocity

1.3 Angular Acceleration

1.4 Gyroscopic Torque (Couple)

1.5 Gyroscopic Effect on Aero-

planes

1.6 Gyroscopic Effect on Naval

Ships

1.7 Stability of an Automobile

1.8 Stability of two-wheel vehicle

1.9 Rigid Disc at an Angle Fixed To

a Rotating Shaft

1.10 Summary

1. Gyroscope Theory of Machines (2151902)

Department of Mechanical Engineering Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.2

1.1 Principle of Gyroscope

If the axis of spinning or rotating body is given an angular motion about an axis perpendicular to

the axis of spin, an angular acceleration acts on the body about the third perpendicular axis. The

torque required to produce this acceleration is known as active gyroscopic torque. A reactive

gyroscopic torque or couple also acts similar to the concept of centripetal and centrifugal forces

on a reacting body. The effect produced by the reactive gyroscopic couple is known as gyro-

scopic effect. Thus aeroplanes, ships, automobiles, etc., that have rotating parts in the form of

wheels or rotors of engines experiences this effect while taking turn, i.e., when the axes of spin

is subjected to some angular motion.

1.2 Angular Velocity

The angular velocity of a rotating body is specified by

the magnitude of velocity

the direction of the axis of rotor

the sense of rotation of the rotor, i.e., clockwise or counter-clockwise

Angular velocity is represented by a vector in the following manner:

(i) Magnitude of the velocity is represented by the length of the vector.

(ii) Direction of axis of the rotor is represented by drawing the vector parallel to the axis of the

rotor or normal to the plane of the angular velocity.

(iii) Sense of rotation of the rotor is denoted by taking the direction of the vector in a set rule.

The general rule is that of a right-handed screw, i.e., if a screw is rotated in the clockwise direc-

tion, it goes away from the viewer and vice-versa.

For example, Fig. (a) shows a rotor which rotates in the clockwise direction when viewed from

the end l. Its angular motion has been shown vectorially in Fig. (b). The vector has been taken to



a scale parallel to the axis of the rotor. The sense of direction of the vector is from a to b accord-

ing to the screw rule. However, if the direction of rotation of the rotor is reversed, it would be

from b to a [Fig.(c)].

1.3 Angular Acceleration

Let a rotor spin (rotate) about the horizontal axis Ox at a speed of ω rad/s in the direction as

shown in Fig.(a). Let oa represent its angular velocity Fig. (b).

Now, if the magnitude of the angular velocity changes to (ω+δω) and the direction of the axis of

spin to Ox’ (in time δt). The vector ob would represent its angular velocity in the new position.

Join ab which represents the change in the angular velocity of the rotor. The vector ab can be

resolved into two components:

(i) ac representing angular velocity change in a plane normal to ac or x-axis, and

(ii) cb representing angular velocity change in a plane normal to cb or y-axis.

Change of angular velocity,

𝑎𝑐 = (𝜔 + 𝛿𝜔) 𝑐𝑜𝑠 𝛿𝜃 − 𝜔

Rate of change of angular velocity

=(𝜔 + 𝛿𝜔) cos 𝛿𝜃 − 𝜔

𝛿𝑡

Angular acceleration

lim𝛿𝑡→0

(𝜔 + 𝛿𝜔) cos 𝛿𝜃 − 𝜔

𝛿𝑡



𝐴𝑠 𝛿𝑡 → 0, 𝛿𝜃 → 0 𝑎𝑛𝑑 cos 𝛿𝜃 → 1


= lim𝛿𝑡→0

𝜔 + 𝛿𝜔 − 𝜔

𝛿𝑡=

𝑑𝜔

𝑑𝑡

Change of angular velocity,

𝑐𝑑 = (𝜔 + 𝛿𝜔) sin 𝛿𝜃

Rate of change of angular velocity

=(𝜔 + 𝛿𝜔) sin 𝛿𝜃

𝛿𝑡


= lim𝛿𝑡→0

(𝜔 + 𝛿𝜔) sin 𝛿𝜃

𝛿𝑡

𝐴𝑠 𝛿𝑡 → 0, 𝛿𝜃 → 0 𝑎𝑛𝑑 cos 𝛿𝜃 → 𝛿𝜃


= lim𝛿𝑡→0

(𝜔 + 𝛿𝜔)𝛿𝜃

𝛿𝑡= 𝜔

𝑑𝜃

𝑑𝑡

Total angular acceleration,

𝛼 =𝑑𝜔

𝑑𝑡+ 𝜔

𝑑𝜃

𝑑𝑡

This shows that the total angular acceleration of the rotor is the sum of

1. dω/dt, representing change in the magnitude of the angular velocity of the rotor

2. ω dθ/dt, Representing change in the direction of the axis of spin, the direction of cb is from

c to b in the vector diagram (being a component of ab), the acceleration acts clockwise in the

vertical plane xy. (When viewed from front along they-axis)



1.4 Gyroscopic Torque (Couple)

Let I be the moment of inertia of a rotor and 𝜔 its angular velocity about a horizontal axis of spin

Ox in the direction as shown in Fig. (a). Let this axis of spin turn through a small angle 𝛿𝜃 in the

horizontal plane (xy) to the position Ox' in time δt,

Figure (b) shows the vector diagram. oa represents the angular velocity vector when the axis is

ox and ob when the axis is changed to ox'. Then ab represent the change in the angular velocity

due to change in direction of the axis of spin of the rotor. This change in the angular velocity is

clockwise when viewed from a towards b and is in the vertical plane xz. This change results in

angular acceleration, the sense and direction of which are the same as that of the change in the

angular velocity.

Change in the angular velocity,

𝒂𝒃 = 𝜔 × 𝛿𝜃

Angular acceleration,

𝛼 = 𝜔𝑑𝜃

𝑑𝑡

In the limit,

𝑤ℎ𝑛 𝛿𝑡 → 0, 𝛼 = 𝜔𝑑𝜃

𝑑𝑡

Usually, dθ/dt the angular velocity of the axis of spin is called the angular velocity of precession

and is denoted by ωp




𝛼 = 𝜔 ∙ 𝜔𝑝

The torque required to produce this acceleration is known as the gyroscopic torque and is a

couple which must be applied to the axis of spin to cause it to rotate with angular velocity ωp

about the axis of precession Oz.

Acceleration torque,

𝑇 = 𝐼 ∙ 𝛼 = 𝐼 ∙ 𝜔 ∙ 𝜔𝑝

For the configuration of Fig.(a)

Ox is known as the axis of spin

Oz is known as the axis of precession

Oy is known as the axis of gyroscopic couple 00

yz is plane of spin

xy is plane of precession

yz is plane of gyroscopic couple

The torque obtained above is that which is required to cause the axis of spin to precess in

the horizontal plane and is known as the active gyroscopic torque or the applied torque. A re-

active gyroscopic torque or reaction torque is also applied to the axis which tends to rotate

the axis of spin in the opposite direction i.e., in the counter-clockwise direction in the above

case. Just as the centrifugal force on a rotating body tends move the body tends to move

the body outwards, while a centripetal acceleration (and thus centripetal force) acts on it in-

wards, in the same way, the effects of active and reactive gyroscopic torques can be un-

derstood.

The effect of the gyroscopic couple on a rotating body is known as the gyroscope effect on the

body. A gyroscope is a spinning body which is free to move in other directions under the action

of external forces.



1.5 Gyroscopic Effect on Aeroplanes

Figure (a) shows an aeroplane in space. Let the propeller be rotating in the clockwise direction

when viewed from the rear end. The angular momentum vector oa due to the angular velocity is

shown in Fig. (b).

If the plane takes a left turn, the angular momentum vector is shifted and may be represented

by the vector ob. The change is shown by the vector ab and is the active gyroscopic couple. This

vector is in the horizontal plane and is perpendicular to the vector oa in the limit. The reactive

vector is given by b'a' which is equal and opposite to the vector ab. The interpretation of this

vector shows that the couple acts in the vertical plane and is counter-clockwise when viewed

from the right-hand side of the plane. This indicates that it tends to raise the nose and depress

the tail of the aeroplane.

Figure (c) shows the gyroscopic effect, when the aeroplane takes the right turn. The change is

shown by the vector cd and is the active gyroscopic couple. It is perpendicular to the vector oc

in the limit in the horizontal plane. The reactive couple is given by d'c'. The couple acts in the

vertical plane and is clockwise when viewed from the right-hand side of the plane. Thus, it tends

to dip the nose and raise the tail of the aeroplane.

If the rotation of the engine is reversed, i.e., it rotates counter-clockwise when viewing from the

rear end, the angular momentum vector is oe as shown in Fig. (d). On taking a left turn, it

changes to of. The active gyroscopic vector is ef and the reactive fe'. Viewing from the right-hand

side of the plane, it indicates that the nose is dipped and the tail is raised. Similarly, when the



plane takes a right tum, the effect is indicated in Fig. (e). The nose is raised and the tail is de-

pressed.

It can be concluded from the above cases that if the direction of either the spin of the rotor or of

the precession is changed, the gyroscopic effect is reversed, but if both are changed, the effect

remains same.

1.6 Gyroscopic Effect on Naval Ships

Some of the terms used in connection with the motion of naval ships or sea vessels are given be-

low.

Bow is the fore or the front end.

Stern or aft is the rear end.

Starboard is the right hand side when looking from stern.

Port is the left-hand side when looking form the stern.

Steering is turning on the side when viewing from the top.

Pitching is limited angular motion of the ship about the transverse axis.

Rolling is limited angular motion of the ship about the longitudinal axis.

Let the plane of spin the rotor and other rotating masses be horizontal and across the breath of

the ship. Assume 𝜔 to the angular velocity of the rotor in the clockwise direction when viewed

from stern.



Gyroscopic Effect during Turning

When the ship turns left the angular momentum vector changes from oa to ob. The reaction cou-

ple is found to be b ̍ a ̍ which tends to raise the bow and lower the stern. On the turning right the

reaction couple is revered. So that bow is lower and stern is raised.

Gyroscopic Effect on Pitching

Pitching of the ship is usually considered to take place with simple harmonic motion. A simple

harmonic motion is represented by x = X sin ω0t

Such a motion is obtain by the projection of a rotating vector X on a diameter while rotating

around a circle with a constant angular velocity ω0 and where x is a displacement from the time

mean position in time t. in such the way angular displacement θ of the axis of the spin from its

mean position is given by

𝜃 = 𝜑 sin 𝜔0𝑡

Where φ = amplitude (angular) of swing or the maximum angle turned from the mean position

in radius

𝜔0 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑆𝐻𝑀 =2𝜋

𝑇𝑖𝑚𝑒 𝑃𝑒𝑟𝑖𝑜𝑑

Angular velocity of precession,

𝑑𝜃

𝑑𝑡= 𝜑 ∙ 𝜔0 ∙ cos 𝜔0𝑡

This is maximum when

𝑐𝑜𝑠 𝜔0𝑡 = 1

Therefore, maximum angular velocity of precession

𝜔𝑝 = 𝜑 ∙ 𝜔0

Gyroscopic Couple,

𝐼 ∙ 𝜔 ∙ 𝜔𝑝 = 𝐼 ∙ 𝜔 ∙ (𝜑 ×2𝜋

𝑇𝑖𝑚𝑒 𝑃𝑒𝑟𝑖𝑜𝑑)



When the bow is rising, the reaction couple is clockwise on viewing from top and thus the ship

would move towards right or starboard side. Similarly, when the bow is lowered, the ship turns

toward left or port side.


= 𝜑 ∙ 𝜔02 sin 𝜔0𝑡

Maximum angular acceleration

= 𝜑 ∙ 𝜔02

Gyroscopic Effect on Rolling

As the axes of the rolling of the ship of the rotor are parallel, there is no precession of the axis of

spin and thus there is no gyroscopic effect.

In the same way, the effect on steering, pitching or rolling can be observed when the plane of the

spin of the rotating masses is horizontal but along the longitudinal axis of the vessel or the axis

is vertical.

1.7 Stability of an Automobile



In case of a four wheeled vehicle, it is essential that that no wheel is lifted off the ground while

the vehicle takes a turn. The condition is fulfilled as long as the vertical reaction of the ground

on any the wheels is positive

Fig. shows a four wheeled vehicle having a mass m. assuming that the weight is equally divided

among the four wheels,

Weight on each wheel =𝑤

4=

𝑚𝑔

4 (downwards)

Reaction of ground on each wheel, 𝑅𝑤 =𝑤

4=

𝑚𝑔

4 (upwards)

Effect of Gyroscopic Couple

Gyroscopic couple due to four wheels,

𝐶𝑤 = 4 ∙ 𝐼𝑤 ∙ 𝜔𝑤 ∙ 𝜔𝑝 = 4 ∙ 𝐼𝑤 ∙𝑣2

𝑟𝑅

Where

𝐼𝑤 = mass moment of inertia of each wheel

𝜔𝑤 = angular velocity of wheel =𝑣

𝑟

𝜔𝑝 = angular velocity of precession = 𝑣

𝑅

𝑣 = linear velocity of the vehicle

𝑅 = radius of curvature

Gyroscopic couple due to engine rotating parts,

𝐶𝑒 = 𝐼𝑒 ∙ 𝜔𝑒 ∙ 𝜔𝑝 = 𝐼𝑒 ∙ 𝐺 ∙ 𝜔𝑤 ∙ 𝜔𝑝

Where G is the gear ratio,

𝐺 =𝜔𝑒

𝜔𝑤



Total gyroscopic couple,

𝐶𝐺 = 𝑐𝑤 ± 𝑐𝑒

Positive sign is used when the engine parts rotate in same direction as the wheels and the nega-

tive sign when they rotate in the opposite.

Assuming that CG is positive and the vehicle takes a left turn, the reaction gyroscopic couple on it

is clockwise when viewed from the rear of the vehicle. The reaction couple is provided by equal

and opposite forces on the outer and the inner wheels of the vehicle.

Forces on two outer wheels =𝑐𝐺

𝑤 (downwards)

Forces on the two inner wheels =𝑐𝐺

𝑤 (upwards)

Forces on each of the outer wheels =𝑐𝐺

𝑤 (downwards)

Forces on each of the inner wheels =𝑐𝐺

𝑤 (upwards)

Thus the forces on each of the outer wheels is similar to the weight. On the inner wheels it is in

the opposite direction. Thus,

Reaction of ground on each outer wheel, 𝑅𝐺 =𝑐𝐺

2𝑤(upwards)

Reaction of ground on each inner wheel, 𝑅𝐺 =𝑐𝐺

2𝑤(downward)

Effect of Centrifugal Couple

As the vehicle moves on a curved path, a centrifugal force also acts on the vehicle in the outward

direction at the centre of mass of the vehicle.

Centrifugal force = 𝑚 ∙ 𝑅 ∙ ω𝑝2 = 𝑚 ∙ 𝑅 ∙ (

𝑣

𝑅)

2

= 𝑚 ∙𝑣2

𝑅

This force would tend to overturn the vehicle outwards and the overturning couple will be

𝐶𝑐 = 𝑚 ∙ 𝑅 ∙ 𝜔𝑝2 = 𝑚 ∙

𝑣2

𝑅∙ ℎ



This is equivalent to a couple to equal and opposite forces on outer and inner wheels.

Forces on each outer wheel =𝑐𝑐

2𝑤(𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑𝑠)

Forces on each inner wheel =𝑐𝑐

2𝑤(upwards)

Again, the force on each of the outer wheels is similar to the weight on each of the inner wheels,

it is opposite.

Reaction of ground on each outer wheel, 𝑅𝑐 =𝑐𝑐

2𝑤(upwards)

Reaction of ground on each inner wheel, 𝑅𝑐 =𝑐𝑐

2𝑤(𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑𝑠)

Vertical reaction on each outer wheel =𝑤

4+

𝑐𝐺

2𝑤+

𝑐𝑐

2𝑤(upwards)

Vertical reaction on each inner wheel =𝑤

4+

𝑐𝐺

2𝑤+

𝑐𝑐

2𝑤(𝑢𝑝𝑤𝑎𝑟𝑑𝑠)

It can be observed that there are chances that the reaction of the ground on the inner wheels

may not be upwards and thus the wheels are lifted from the ground. For positive reaction, the

Conditions will be

𝑤

4−

𝑐𝐺

2𝑤−

𝑐𝑐

2𝑤≥ 0 𝑜𝑟

𝑤

4≥

𝑐𝐺 + 𝑐𝑐

2𝑤 𝑜𝑟 𝑅𝑤 ≥ 𝑅𝐺 + 𝑅𝑐

1.8 Stability of two-wheel vehicle



The case of two-wheel vehicle can be taken in the same way as that of an auto mobile. However,

it easier to tilt such a vehicle inwards to neutralise the overturning effect and the vehicle can

stay in equilibrium while taking a turn.

Let a vehicle take a left turn as shown in fig. (a). The vehicle is inclined to the vertical (inwards)

for equilibrium. The angle of inclination of the vehicle to the vertical is known as the angle of

heel.

Let

𝑣 = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑣𝑒ℎ𝑖𝑐𝑙𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑡𝑟𝑎𝑐𝑘

𝑟 = radius of the wheel

𝑅 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑇𝑟𝑎𝑐𝑘

𝐼𝑥 = moment of inertia of each wheel

𝐼𝑒 = moment of inertia of rotating parts of the engine

𝑚 = total mass of the vehicle and the rider

𝜔𝑤 = angular velocity of the wheel

𝜔𝑒 = angular velocity of rotating parts of engine

𝐺 = 𝑔𝑒𝑎𝑟 𝑟𝑎𝑡𝑖𝑜

ℎ = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒ℎ𝑖𝑐𝑙𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑟𝑖𝑑𝑒𝑟

𝜃 = 𝑖𝑛𝑐𝑙𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑣𝑒ℎ𝑖𝑐𝑙𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 (𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 ℎ𝑒𝑒𝑙)

As the axis of spin is not horizontal but inclined to the vertical at an angle θ and the axis of pre-

cession is vertical, it is necessary to take the horizontal component of the spin vector.

Spin vector (horizontal) = 𝐼𝑤 cos θ = (2 ∙ 𝐼𝑤 ∙ 𝜔𝑤 + 𝐼𝑒 ∙ 𝜔𝑒) cos θ

𝑎𝑛𝑑 𝐺𝑦𝑟𝑜𝑠𝑐𝑜𝑝𝑖𝑐 𝐶𝑜𝑢𝑝𝑙𝑒 = (2 ∙ 𝐼𝑤 ∙ 𝜔𝑤 + 𝐺 ∙ 𝐼𝑒 ∙ 𝜔𝑤) cos 𝜔𝑝

= (2 ∙ 𝐼𝑤 + 𝐺 ∙ 𝐼𝑒 )𝑣

𝑟 𝑣

𝑅 cos θ = (2 ∙ 𝐼𝑤 + 𝐺 ∙ 𝐼𝑒 )

𝑣2

𝑟𝑅 cos θ

The reaction couple b’ a’ is clockwise when viewed for the rear (back) of the vehicle and tends

to overturn it in the outward direction.



𝑂𝑣𝑒𝑟𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑐𝑜𝑢𝑝𝑙𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑐𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 = (𝑚𝑣2

𝑅) ℎ ∙ 𝑐𝑜𝑠 𝜃

Total overturning = (2 ∙ 𝐼𝑤 + 𝐺 ∙ 𝐼𝑒)𝑣2

𝑟𝑅 cos θ + 𝑚 ∙

𝑣2

𝑅∙ ℎ cos 𝜃

=𝑣2

𝑅(

2 ∙ 𝐼𝑤 + 𝐺 ∙ 𝐼𝑒

𝑟+ 𝑚ℎ) cos 𝜃

𝑅𝑖𝑔ℎ𝑡𝑒𝑛𝑖𝑛𝑔 (𝑏𝑎𝑙𝑎𝑛𝑐𝑖𝑛𝑔)𝑐𝑜𝑢𝑝𝑙𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑔ℎ 𝑠𝑖𝑛 𝜃

For Equilibrium,𝑣2

𝑅(

2 ∙ 𝐼𝑤 + 𝐺 ∙ 𝐼𝑒

𝑟+ 𝑚ℎ) cos θ = 𝑚𝑔ℎ sin θ

From This Relation, The angle of heel θ can be determined to avoid skidding of the vehicle.

1.9 Rigid Disc at an Angle Fixed To a Rotating Shaft

Consider a circular disc fixed rigidly to a rotating shaft in such a way that the polar axis of the

shaft makes angel θ with the axis of the shaft (Fig.). Assume that the shaft rotates clockwise with

angular velocity ω when viewed along the left end of the shaft.

Let

OX be the axis of the shaft

OP be the polar axis of the disc and

OD the horizontal diametral axis of the disc

Also, let m, r and t be the mass, radius and the thickness of the disc. Then

𝐼𝑝 = moment of inertia of disc about polar Axis OP =𝑚𝑟2

2



𝐼𝑑 = moment of inertia of disc about diametral axis = 𝑚 (𝑡2

12+

𝑟2

4) = (

𝑚𝑟2

4)

(If the disc is thin and t is neglected)

First consider the spinning about the polar axis

Angular velocity of spin = Angular velocity of Disc about the polar axis OP = ω cos θ

Angular velocity of precession = Angular velocity of disc about the diametral axis OD = ω sin θ

𝐺𝑦𝑟𝑜𝑠𝑐𝑜𝑝𝑖𝑐 𝑐𝑜𝑢𝑝𝑙𝑒 = 𝐼𝑝 × 𝜔 ∙ 𝑐𝑜𝑠 𝜃 × 𝜔 ∙ 𝑠𝑖𝑛 𝜃 =1

2∙ 𝐼𝑝 ∙ 𝜔2 ∙ 𝑠𝑖𝑛 2𝜃

Its effect is to rotate the disc counter-clockwise when viewing from the top.

Now consider the spinning about the diametral axis

Angular velocity of spin = Angular velocity of disc about the diametral axis OD = ω sin θ

Angular velocity of precession = = Angular velocity o disc about the polar axis OP = ω cos θ

𝐺𝑦𝑟𝑜𝑠𝑐𝑜𝑝𝑖𝑐 𝑐𝑜𝑢𝑝𝑙𝑒 = 𝐼𝑑 × 𝜔 ∙ 𝑠𝑖𝑛 𝜃 × 𝜔 ∙ 𝑐𝑜𝑠 𝜃 =1

2∙ 𝐼𝑑 ∙ 𝜔2 ∙ 𝑠𝑖𝑛 2𝜃

Its effect is to rotate the disc clockwise when viewing from the top. (Angular velocity of preces-

sion is counter-clockwise when viewing from the right end along OP.)

Resultant gyroscopic couple on the disc,

𝐶 =1

2(𝐼𝑝−𝐼𝑑) 𝜔2 sin 2𝜃 =

1

2(

𝑚𝑟2

2−

𝑚𝑟2

4) 𝜔2 sin 2𝜃 =

𝑚𝑟2

8𝜔2 sin 2𝜃

1.10 Summary

1. The angular velocity is represented by a vector by drawing the parallel to the axis of the ro-

tor and representing the magnitude by the length of the vector to some scale. Sense of rota-

tion of the rotor is denoted by the rule of a right- handed screw, i.e., if a screw is rotated in

the clockwise direction, it goes away from the viewer and vice-versa.

2. The axis of spin, the axis of precession and the axis of gyroscopic couple are in three perpen-

dicular planes.



3. The torque required to cause the axis of spin to process in a plane is known as the active gy-

roscopic torque or the applied torque.

4. A reactive gyroscopic torque or reaction torque tends to rotate the axis of spin in the oppo-

site direction.

5. The effect of the gyroscopic couple on a rotating body is known as the gyroscopic effect on

the body.

6. A gyroscopic is a spinning body which is free to move in other directions under the action of

external forces.

7. A four-wheel vehicle tends to turn outwards when taking a turn due to the effect of gyro-

scopic couple and the centrifugal force.

8. A two-wheel vehicle stabilises itself by tilting towards inside while taking a turn to nullify

the effects of gyroscopic couple and the centrifugal force.

Department of Mechanical Engineering Prepared By: Vimal Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page 2.1

2 FRICTION DEVICES: CLUTCHES, BRAKES AND DYNAMOMETERS

Course Contents

2.1 Introduction to Clutch

2.2 Classification of Clutches

Positive Contact Clutch

Friction Clutch

2.3 Brakes

2.4 Classification of Brakes

2.5 Energy Equations

Block or Shoe Brake

Band Brake

Band and Block Brake

Internal Expanding Brake

2.6 Braking of Vehicle

2.7 Dynamometer

2.8 Types of Dynamometer

Prony Brake Dynamometer

Rope Brake Dynamometer

Belt Transmission Dynamometer

Epicyclic Train Dynamometer

Bevis Gibson Torsion

Dynamometer

2. Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (2151902)

Prepared By: Vimal G. Limbasiya Department of Mechanical Engineering Page 2.2 Darshan Institute of Engineering & Technology, Rajkot

2.1 Introduction to clutch

The clutch is a mechanical device, which is used to connect or disconnect the source of

power from remaining parts of power transmission system at the will of the operator.

An automotive clutch can permit the engine to run without driving the car. This is

desirable when the engine is to be started or stopped, or when the gears are to be

shifted.

Very often, three terms are used together, namely, couplings, clutches and brakes.

There is a basic difference between the coupling and the clutch.

A coupling, such as a flange coupling, is a permanent connection. The driving and

driven shafts are permanently attached by means of coupling and it is not possible to

disconnect the shafts, unless the coupling is dismantled.

On the other hand, the clutch can connect or disconnect the driving and driven shafts,

as and when required by the operator. Similarly, there is a basic difference between

initial and final conditions in clutch and brake operations.

In the operation of clutch, the conditions are as follows:

(i) Initial Condition The driving member is rotating and the driven member is at rest.

(ii) Final Condition Both members rotate at the same speed and have no relative motion.

In the operation of brake, the conditions are as follows:

(i) Initial Condition One member such as the brake drum is rotating and the braking

member such as the brake shoe is at rest.

(ii) Final Condition Both members are at rest and have no relative motion.

Functions of clutches

(i) It maintains constant load on shaft

(ii) It transmits or starts high inertia load of the machine with a small motor.

(iii) It acts like emergency device to disengage the shaft of the machine from the shaft of

the motor in case of condition of machine jam.

2.2 Classification of clutches

Clutches are classified into the following four groups:

(i) Positive contact Clutches: They include square jaw clutches; spiral jaw clutches and

toothed clutches. In these clutches, power transmission, is achieved by means of

interlocking of jaws or teeth. Their main advantage is positive engagement and once

coupled, they can transmit large torque with no slip.

(ii) Friction Clutches: They include single and multi-plate clutches, cone clutches and

centrifugal clutches. In these clutches, power transmission is achieved by means of

friction between contacting surfaces.

(iii) Fluid Clutches and Couplings: In these clutches, power transmission is achieved by

means of hydraulic pressure. A fluid coupling provides extremely smooth starts and

absorbs shock.

Theory of Machines (2151902) 2. Friction Devices: Clutches, Brakes and Dynamometers

Department of Mechanical Engineering Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page 2.3

(iv) Electromagnetic Clutches: They include magnetic particle clutches, magnetic

hysteresis clutches and eddy current clutches. In these clutches, power transmission

is achieved by means of the magnetic field. These clutches have many advantages,

such as rapid response time, ease of control, and smooth starts and stops.

Positive Contact Clutches

The simplest form of positive contact clutches is the square jaw clutch as shown in Fig.

2.1. It consists of two halves carrying projections or jaws. One clutch half is fixed and

the other can move along the axis of the shaft over either a feather key or splines by

means of shift lever.

During the engagement, the jaws of the moving half enter into the sockets of the

mating half. The engaging surfaces of jaw and socket form a rigid mechanical junction.

Jaw clutches can be used to connect shafts, when the driving shaft is stationary or

rotating at very low velocity. There are two types of jaws, namely, square and spiral.

The spiral jaws can be engaged at slightly higher speed without clashing. Frequent

engagement results in wear of jaws.

The jaw clutches have the following advantages:

(a) They do not slip and engagement is positive.

(b) No heat is generated during engagement or disengagement.

Fig. 2.1 Square Jaw Clutch

The jaw clutches have the following drawbacks:

(a) Jaw clutches can be engaged only when both shafts are stationary or rotate with very

small speed difference.

(b) They cannot be engaged at high speeds because engagement of jaws and sockets

results in shock.

In general, positive contact clutches are rarely used as compared with friction

clutches. However, they have some important applications where synchronous

operation is required like power presses and rolling mills.



Friction Clutches

Single Plate Clutch

A single plate friction clutch consisting of two flanges is shown in Fig. 2.2. One flange is

rigidly keyed to the driving shaft, while the other is connected to the driven shaft by

means of splines.

The splines permit free axial movement of the driven flange with respect to the driven

shaft. This axial movement is essential for engagement and disengagement of clutch.

The actuating force is provided by a helical compression spring, which forces the

driven flange to move towards the driving flange.

Power is transmitted from the driving shaft to the driving flange by means of the key.

Power is then transmitted from the driving flange to the driven flange by means of

frictional force.

Finally, power is transmitted from the driven flange to the driven shaft by means of

the splines. Since the power is transmitted by means of frictional force between the

driving and driven flanges, the clutch is called 'friction' clutch.

In order to disengage the clutch, a fork is inserted in the collar on the driven flange to

shift it axially to the right side. This relieves the pressure between the driving and the

driven flanges and no torque can be transmitted.

In the working condition, the clutch is in an engaged position under the action of

spring force. Levers or forks are operated to disengage the clutch.

Fig. 2.2 Single Plate Clutch

Advantages of Friction Clutch

The main advantages of friction clutch are as follows:

(i) The engagement is smooth.

(ii) Slip occurs only during engaging operation and once the clutch is engaged, there is

no slip between the contacting surfaces. Therefore, power loss and consequent heat

generation do not create problems, unless the operation requires frequent starts

and stops.



(iii) In certain cases, the friction clutch serves as a safety device. It slips when the torque

transmitted through it exceeds a safe value. This prevents the breakage of parts in

the transmission chain.

Depending upon the number of friction surfaces, the friction clutches are classified as

single-plate or multi-plate clutches.

Depending upon the shape of the friction material, the clutches are classified as disk

clutches, cone clutches or expanding shoe clutches.

The following factors should be considered while designing friction clutches:

(i) Selection of a proper type of clutch that is suitable for the given application

(ii) Selection of suitable friction material at the contacting surfaces

(iii) Designing the clutch for sufficient torque capacity

(iv) Engagement and disengagement should be without shock or jerk

(v) Provision for holding the contacting surfaces together by the clutch itself and

without any external assistance

(vi) Low weight for rotating parts to reduce inertia forces, particularly in high-speed

applications

(vii) Provision for taking or compensating wear of rubbing surfaces

(viii) Provision for carrying away the heat generated at the rubbing surfaces

Torque Transmitting Capacity

A friction disk of a single plate clutch is shown in Fig 2.3. The following notations are used in

the derivation:

rO = outer diameter of friction disk (mm)

ri = inner diameter of friction disk (mm)

p = intensity of pressure at radius r (N/mm2)

W = total operating force (N)

Ft = frictional force applied on plate

T = torque transmitted by the clutch (N-mm)

Fig. 2.3 Friction Disk

Assume elemental ring at radius r whose thickness is dr.

dW = axial force on elemental ring

dW = p x 2r dr



Elemental area = 2r dr

Elemental axial force = p (2r dr)

O

i

r

r

Total axial force, W dW

O

i

r

r

Total axial force, p 2 r) drW (

O

i

r

r

Total axial force 2 p r dr, W

Two theories are used to obtain the torque capacity of the clutch. They are called uniform

pressure theory and uniform wear theory.

(i) Uniform Pressure Theory: In case of new clutches employing a number of springs, the

pressure remains constant over the entire surface area of the friction disk. With this

assumption, p is assumed to be constant. This constant pressure distribution is illustrated in

Fig. 2.4.

Fig. 2.4 Pressure Distribution

Axial force on elemental ring O

i

r

r

2 p r drW

O

i

r2

r

Wr

2 p2

2 2O ip r rW

2 2

O ir r

Wp

……………(i)

Total torque transmitted,

O O

i i

r r

t t

r r

M T dT dF r O

i

r

r

dW r

O

i

r

2

r

p 2 r dr



O

i

r

2

r

2 p r dr

O

i

r3

r

r2 p

3

3 3O ir

3p r

2

Substituting value of p from equation (i)

3 3

O i2 2O i

Wr r

r r

2T

3

3 3O i

2 2O i

r r2T

3W

r r

…………….(ii)

Mt = T = μ.W.rm

3 3O i

2O i

m 2

2Where,r

3

r r

r r

The above equations have been derived for a single pair of contacting surfaces. When there

are a number of friction surfaces in contact, as in the case of the multi-disk clutch, Eq. (ii)

should be multiplied by the number of pairs of contacting surfaces to obtain the resultant

torque transmitting capacity.

(ii) Uniform Wear Theory: According to the second theory it is assumed that the wear is

uniformly distributed over the entire surface area of the friction disk. This assumption is used

for worn-out clutches. The axial wear of the friction disk is proportional to the frictional work.

The work done by the friction force at radius r is proportional to the frictional force (μp) and

rubbing velocity (2rn) where n is speed in rev/min.

Fig. 2.5 Pressure Distribution

Wear ∝ (μp) (2r n)

Assuming speed n and the coefficient of friction μ as constant for a given configuration,

Wear ∝ pr

When the wear is uniform,

pr = constant



The pressure distribution according to uniform wear theory is illustrated in Fig. 2.5. In this

case, p is inversely proportional to r. Therefore, pressure is maximum at the inner radius and

minimum at the outer periphery. The maximum pressure intensity at the inner radius ri is

denoted by pa. It is also the permissible intensity of pressure. O

i

r

r

Total axial force, p 2 r) drW ( O

i

r

r

2 pr dr

O

i

r

r2 c r

O ic r rW 2

O i

cr r

W

2 …………..(iii)

Total torque transmitted,

O

i

r

t

r

M T dT

O

i

r

t

r

dF r

O

i

r

r

dW r

O

i

r

2

r

p 2 r dr

O

i

r

r

(2 r) p r dr

O

i

r2

r

r2 c

2

2 2

O ir rc

22

2 2O ic r r

Substituting value of c from equation (iii)

2 2

O i

O i

Wr r

2 r r

O i(r r)W

2

Mt = T = μ.W.rm mO iW

(rhere,r

r)

2

Conclusions

(i) The uniform-pressure theory is applicable only when the friction lining is new.

(ii) The uniform-wear theory is applicable when the friction lining gets worn out.

(iii) The friction radius for new clutches is slightly greater than that of worn-out clutches.

(iv) The torque transmitting capacity of new clutches is slightly more than that of worn-

out clutches.



(v) A major portion of the life of friction lining comes under the uniform wear criterion.

(vi) It is more logical and safer to use uniform-wear theory in the design of clutches.

Therefore, the torque transmitting capacity can be increased by three methods:

(a) Use friction material with a higher coefficient of friction (μ)

(b) Increase the plate pressure (p)

(c) Increase the mean radius of the friction disk (rm)

In design of clutches, the following factors should be considered:

(a) Service Factor In order to start the machine from rest and accelerate it to the rated

speed, the clutch should have torque capacity substantially higher than the nominal

torque rating.

In most of the cases, the accelerating or starting torque is much more than the

running torque. If the clutch is not designed for this increased torque, it will slip under

the load and no power can be transmitted.

There is another factor to account for additional torque. In many applications, the

torque developed by the prime mover fluctuates and also, the torque requirement by

driven machinery fluctuates as in the case of presses. These two factors are accounted

by means of service factor.

(b) Location of clutch Let us consider a mill driven by a diesel engine. The optimum

operating speed of the engine is too high for direct connection to the mill shaft.

Therefore, a gearbox is provided to reduce the speed.

In this set-up, a clutch is also required so that the engine can be started and brought

up to the full speed before connecting to the mill shaft.

In such applications, the question arises about the location of clutch-whether the

clutch should be located between the engine and the gearbox or between the gearbox

and the mill?

The clutch is required to transmit a given power. The power transmitted by the clutch

is the product of torque and speed. Therefore, greater the speed, lower is the torque

to be transmitted. It is, therefore, logical to place the clutch at the high-speed side,

that is, between the engine and the gearbox.

Since the torque capacity is low, the cost of the clutch is also low. On the other hand,

the speed is low between the gearbox and the mill and, the clutch will have to

transmit high torque, increasing the cost.

(c) The coefficient of friction for automotive clutches, which use asbestos lining in contact

with a cast iron surface, is taken from 0.3 to 0.4.The allowable pressure on the friction

lining varies from 0.1 N/mm2 for large heavy-duty double-plate clutches to 0.25

N/mm2 for an average passenger car clutch. The allowable pressure for clutches with

metal plates is from 0.7 to 1.05 N/mm2.



Multi-Disk Clutches

A multi-disk clutch, as shown in Fig. 2.6, consists of two sets of disks – A and B. Disks of

Set A are usually made of hardened steel, while those of set B are made of bronze.

Disks of Set A are connected to the driven shaft by means of splines. Because of

splines, they are free to move in an axial direction on the splined sleeve. There are

four through bolts, which pass through the holes in the disks of Set B.

A clearance fit between the bolt and the holes in the plates allows disks of Set B to

move in an axial direction. The bolts are rigidly fixed to a rotating drum, which is keyed

to the driving shaft.

The axial force P (or W), which is required to hold the disks together, is provided by

means of levers or springs. When the driving shaft rotates, the drum, along with bolts

and disks of Set B, also rotate.

Power is transmitted from the disks of Set B to those of A by means of friction. When

the disks of Set A rotate, they transmit the power to the driven shaft through splined

sleeve.

Fig. 2.6 Multi-Disk Clutch

Equations derived for torque transmitting capacity of the single plate clutch are modified to

account for the number of pairs of contacting surfaces in the following way:

For the uniform pressure criterion,

3 3O i

2 2O i

r r2T Wn

3 r r

For the uniform wear criterion,

O i(r r)T Wn

2

Where n is the number of pairs of contacting surfaces.



Fig. 2.7 Number of Disks

In the design of multi-disk clutches, very often it is required to determine the number

of disks rather than the number of pairs of contacting surfaces.

In multi-disk clutch, illustrated in Fig. 2.6, there are two types of disks called disks of

Set A and disks of Set B.

We can use steel disks for Set A and bronze disks for Set B. Or, we can use plane steel

disks for Set A and Set B consisting of steel disks with asbestos lining.

Let us consider 5 disks – 3 disks of Set A and 2 disks of Set B. As shown in Fig 2.7, the

number of pairs of contacting surfaces is 4. Therefore,

Number of disks = Number of pairs of contacting surfaces +1 = n + 1 …….(a)

Suppose,

n1= number of disks on driving shaft

n2= number of disks on driven shaft

Substituting in (a),

n1+ n2 = Number of pairs of contacting surfaces + 1

or number of pairs of contacting surfaces = n1+ n2 – 1

It should be noted that the two outer disks have contacting surface on one side only.

Difference between Single and Multi-Plate Clutches

The difference between single and multi-plate clutches is as follows:

(i) The number of pairs of contacting surfaces in the single plate clutch is one or at the

most two. There are more number of contacting surfaces in the multi-disk clutch.

(ii) As the number of contacting surfaces is increased, the torque transmitting capacity is

also increased, other conditions being equal. In other words, for a given torque

capacity, the size of the multi-plate clutch is smaller than that of the single plate

clutch, resulting incompact construction.

(iii) The work done by friction force during engagement is converted into heat is

generated in the multi-plate clutch due to increased number of contacting surfaces.

Heat dissipation is a serious problem in the multi-plate clutch. Therefore, multi-plate

clutches are wet clutches, while single plate clutches are dry.



(iv) The coefficient of friction decreases due to cooling oil, thereby reducing the torque

transmitting capacity of the multi-plate clutch. The coefficient of friction is high in

dry single plate clutches.

(v) Single plate clutches are used in applications where large radial space is available,

such as trucks and cars. Multi-disk clutches are used in applications where compact

construction is desirable, e.g., scooter and motorcycle.

Difference between Dry and Wet Clutches

The difference between dry and wet clutches is as follows:

(i) A dry clutch has higher coefficient of friction. In wet clutches, the coefficient of

friction is reduced due to oil. The coefficient of friction for dry operation is 0.3 or

more, while it is 0.1 or less for wet operation.

(ii) The torque capacity of dry clutch is high compared with the torque capacity of wet

clutch of the same dimensions.

(iii) For dry clutch, it is necessary to prevent contamination due to moisture or near by

lubricated machinery, by providing seals. Such a problem is not serious in wet

clutches.

(iv) Heat dissipation is more difficult in dry clutches. In wet clutches, the lubricating oil

carries away the frictional heat.

(v) Rate of wear is far less in wet clutches compared to dry clutches. The wear rate in

wet clutches is about 1% of the rate expected in dry clutches.

(vi) The engagement in wet clutch is smoother than in the case of dry clutch.

(vii) In wet clutches, the clutch facings are grooved to provide for passage of lubricant.

This reduces the net face area for transmitting torque.

Friction Material

For light loads and low speeds, wood, cork and leather are used as friction materials. The

present trend for high speeds and heavy loads has given a stimulus to the development of

new friction materials, which are capable of withstanding severe service conditions.

The desirable properties of a good friction material are as follows:

(i) It should have high coefficient of friction.

(ii) The coefficient of friction should remain constant over the entire range of,

temperatures encountered in applications.

(iii) It should have good thermal conductivity.

(iv) It should remain unaffected by environmental conditions like moisture, or dirt

particles.

(v) It should have high resistance to abrasive and adhesive wear.

(vi) It should have good resilience to provide good distribution of pressure at the

contacting surfaces.



The coefficient of friction depends upon a number of factors. They include materials of

contacting surfaces, surface finish, surface temperature, rubbing speed, foreign

particles on rubbing surfaces and atmospheric conditions like moisture.

There are two types of friction materials in common use – asbestos-base and sintered

metals.

There are two types of asbestos friction materials-woven and moulded. A woven

asbestos friction disk consists of asbestos fiber woven around brass, copper or zinc

wires and impregnated with rubber or asphalt.

Asbestos material whether woven or moulded is anorganic material and is subject to

destruction by heat at comparatively low temperature. Sintered-metal friction materials solve

this difficulty.

There are two varieties of friction disks made from sintered metals – bronze-base and iron-

base, depending upon the major constituents. The advantages of sintered-metal friction disks

are as follows:

(i) They have higher wear resistance.

(ii) They can be used at high temperatures.

(iii) The coefficient of friction is constant over a wide range of temperature and pressure.

(iv) They are unaffected by environmental conditions, such as dampness, salt water or

fungi.

Sintered-metal friction materials offer an excellent design with lighter, cheaper and

compact construction. The maximum permissible intensity of pressure for woven and

moulded asbestos materials is 0.3 N/mm2 and 1 N/mm2 respectively, while for sintered

metals it can be taken between 1 and 2 N/mm2.

It has been found that if asbestos dust is inhaled, it may lead to cancer. The body cells

which come in contact with asbestos particles are agitated and develop into cancer

cells. Lung cancer is common among operators working in atmospheres of asbestos.

There are federal regulations in a number of countries, which prohibit the use of

asbestos in clutch or brake linings. Nowadays, metallic or semi-metallic fibres or

powder is used in place of asbestos fibres.

Modern friction lining consists of four basic ingredients, namely, fibres, filler, binder

and friction modifiers. Fibres provide rigidity and strength for the friction lining

Nowadays, steel wool or aramid is used as fibre material instead of asbestos.

A filler fills the space between the fibres and extend the lining life. Filler materials are

barytes, clay and calcium carbonate. In case of metallic lining, fine metal power is used

as filler material.

Binder is a glue that holds the lining ingredients together. Phenolformaldehyde is

extensively used as binder material. Friction modifier improves frictional and wear

properties. Brass and zinc particles are added as friction modifiers to control the

abrasive properties of the lining.



Centrifugal Clutches

Whenever it is required to engage the load after the driving member has attained a

particular speed, a centrifugal clutch is used. The centrifugal clutches permit the drive-

motor or engine to start, warm up and accelerate to the operating speed without

load.

Then the clutch is automatically engaged and the driven equipment is smoothly

brought up to the operating speed. These clutches are particularly useful with internal

combustion engines, which can not start under load.

The centrifugal clutch works on the principle of centrifugal force. The centrifugal force

increases with speed. Construction of centrifugal clutch is shown in Fig. 2.8.

It consists of a spider, which is mounted on the input shaft, and which is provided with

four equally spaced radial guides. A sliding shoe is retained in each guide by means of

a spring.

The outer surface of the sliding shoe is provided with a lining of friction material like

asbestos. The complete assembly of spider, shoes and springs is enclosed in a coaxial

drum, which is mounted on the output shaft.

As the angular speed of the input shaft increases, the centrifugal force acting on the

sliding shoes increases, causing the shoes to move in a radially outward direction. The

shoes continue to move with increasing speed until they contact the inner surface of

the drum.

Power is transmitted due to frictional force between the shoe lining and the inner

surface of the drum. The clutch is disengaged automatically. When the angular

velocity of the shoes decreases, the centrifugal force decreases.

This reduces the normal farce between the friction lining and the drum. The friction

force, which is proportional to normal force, also reduces. The lining slips with respect

to the drum and no torque can be transmitted.

Fig. 2.8 Centrifugal Clutch



Fig. 2.9 Forces on Shoe

The forces acting on the shoe are shown in Fig. 2.9, where following notations are used:

m = mass of each shoe in kg

r = distance of centre of gravity of shoe from centre of spider at the time of

engagement in m

R = inner radius of rim in m

z = No. of shoes

N = driving speed in rpm

= angular speed of driving shaft (rad/s)

g = angular speed at time of engagement (rad/s)

x = g /

= angle subtended by each shoe at centre of spider

FC = centrifugal force in N

FS = spring force in N

F = outward force on each shoe in N

FF = Friction force in N

T = Mt = torque transmitted

l = contact length

b = face width

Centrifugal force, FC = mr2

Spring force, FS = mrg2

Total or Net outward force on shoe, F = FC – FS

= mr2 – mrg2

= mr (2 – g2)

2

g2F mr 1

2 2F mr 1 x

Tangential frictional force, FF = μ x F

= μ mr2 (1– x2)

Torque transmitted, T = FF x R x z

= μ mr2 z R (1– x2) (If not given, assume r ≤ R and = 60)



Example – 1:- A single plate clutch with both sides effective is to transmit 75 kW at 900

rpm. The axial pressure is limited to 0.07 MPa. The coefficient of friction may be taken as

0.25. The ratio of face width to mean radius is 0.25. Determine the outer and inner radii of

clutch plate.

Solution: P = 75 KW N = 900 rpm

p = 0.07 MPa μ = 0.25

b/rm = 0.25

Power, 2 NT

P60

3 2 900 T75 10

60

T = 795.77 N.m

Assuming uniform wear theory,

mT W r n

mT (A p) r n

m mT (2 r b p) r n

mm m

rT 2 r p r n

4

3mT r p n

2

3 3m795.77 10 0.25 r 0.07 2

2

rm = 243.706 mm

O im

r rr

2

O ir r243.706

2

rO + ri = 487.412 mm …………….(i)

face width of plate is equal to difference of outer and inner radii

b = rO – ri

m

b0.25

r

b = 60.926 mm

rO – ri = 60.926 mm …………….(ii)

From equation (i) and (ii)

ri = 213.24 mm

rO = 274.17 mm



Example – 2:- Design a single plate clutch considering uniform wear criterion and effective

two pair of contacting surfaces from the following specification:

Power to be transmitted = 40 kW

Speed = 1560 r.p.m.

Service factor = 1.25

Permissible pressure for the lining = 0.4 MPa

Coefficient of friction = 0.30

Outer diameter = 300 mm

Permissible stress for shaft material = 45 MPa

Solution: P = 40 KW N = 1560 rpm

p = 0.4 MPa μ = 0.03

dO= 300 mm τ = 45 MPa

Power, 2 NT

P60

3 2 1560 T40 10

60

T = 244.85 N.m


mT W r n

mT (A p) r n

m mT (2 r b p) r n

O i O iO i

r r r rT 2 r r p n

2 2

2 2

3 O iO i

r r244.85 10 2 r r p n

4

3 2 2

i i244.85 10 0.3 150 r 150 r 0.4 22

ri = 142.4 mm

Example – 3:- A car engine has a maximum torque of 100 N.m. The clutch is a single plate

with two acting surfaces; the axial pressure is not to exceed 0.85 bar. The ratio of outer

diameter to inner diameter can be considered as 1.2. Find the dimensions of friction plate &

axial force required by springs. Assume µ = 0.3.

Solution: T = 100 N.m

n = 2

p = 0.85 bar = 0.85 x 105 N/m2

dO/di = 1.2

μ = 0.3




mT W r n

O ir r100 0.3 W 2

2

O iW r r 333.33

Oi

i

rWr 1 333.33

r

iWr 2.2 333.33

iWr 151.51 ……………(i)

O iW 2 c r r

c = p x ri

= 85000 x ri

i iW 2 (85000 r)r 1.2 1

2iW 106814.15 r

……………(ii)

From equation (i) and (ii)

3

i106814.15 r 151.51

ri = 0.1123 m

rO = 0.1348 m

W = 1347.06 N

Example – 4:- A multiple disc clutch is to transmit 4 kW at 750 rpm. Available steel and

bronze discs are 40 mm inner radius and 70 mm outer radius are to be assembled alternately

in appropriate numbers. The clutch is to operate in oil with an expected coefficient of friction

of 0.1 and maximum allowable pressure is not to exceed 350 KPa. Assume uniform wear

condition to prevail and specify the number of steel (driving) and bronze (driven) discs

required. Also determine what axial force is to be applied to develop the full torque.

Solution: P = 4 kW N = 750 rpm

ri = 40 mm μ = 0.1

rO = 70 mm p = 350 KPa

Power, 2 NT

P60

3 2 750 T4 10

60

T = 50.93 N.m


mT W r n



3 70 4050.93 10 0.1 W n

2

Wn = 9260 ……………(i)

O iW 2 c r r

max i O iW 2 (p r) r r

W 2 0.35 40 70 40

W = 2638.94 N

Wn = 9260

n = 3.5

n ≈ 4

No. of plates required = n + 1

= 5

No. of steel (driving) discs = 3

No. of bronze (driven) discs = 2

Example – 5:- A multiplate clutch is used to transmit 5 kw power at 1440 rpm. The inner &

outer diameters of contacting surfaces are 50 mm and 80 mm respectively. The coefficient of

friction and the average allowable pressure intensity for the lining may be assumed as 0.1 and

350 kPa respectively. Determine

(i) Number of friction plates & pressure plates (ii) Axial force required to transmit power (iii) The actual average pressure (iv) Actual maximum pressure intensity after wear.


di= 50 mm μ = 0.1

dO= 80 mm p = 350 KPa

Power, 2 NT

P60

3 2 1440 T5 10

60

T = 33.157 N.m


mT W r n

3 40 2533.157 10 0.1 W n

2

Wn = 10202.15 ……………(i)

O iW 2 c r r



avg O iW 2 (p r ) r r

2 23 0.04 0.025

W 2 350 102

W = 1072.068 N

Wn = 10202.15

n ≈ 10

No. of plates required = n + 1= 11

Plates on drivingshafts = 6

Plates on driven shafts = 5

Actual number of active surfaces = 10

act mT W r n

act33.157 0.1 W 0.0325 10 Wact = 1020.215 N

act O iW 2 c r r

1020.215 2 c 0.04 0.025

c = 10824.817

c = pavg x ravg

10824.817 = pavg x 0.0325

pavg = 333071.3 N/m2

c = pmax x ri

10824.817 = pmax x 0.025

pmax = 432992.68 N/m2

Example – 6:- A Centrifugal clutch is to be designed to transmit 15 kW at 900 rpm. The

shoes are four in number. The speed at which the engagement begins is 3/4th of the running

speed. The inside radius of the pulley rim is 150 mm. The shoes are lined with Ferrodo for

which the coefficient of friction may be taken as 0.25.

Determine: (i) Mass of the shoes and

(ii) Size of the shoes.


z = 4 g = ¾

R = 150 mm μ = 0.25

Assume, r =120 mm

Power, 2 NT

P60

3 2 900 T15 10

60

T = 159.15N.m



2 N 2 90094.26

60 6 rad c

0/ se

2 2T mr zR(1 x )

2

2 3159.15 0.25 m 120 (94.26) 4 150 1

4

m = 2.27 kg

Assuming that the arc of contact of the shoes subtend an angle of = 60 at the centre of the

spider, therefore,

Contact length, l R

l 60 150

180

= 157 mm

Assuming intensity of pressure (pn) exerted on shoes is 0.1 N/mm2

2 2

n

F mr (1 x )p

l b l b

2

2

6

32.27 0.12 (94.26) 1

40.1 10

l b

l x b = 0.01058

b = 0.06744 m

Example – 7:- A centrifugal clutch consists of four shoes, each having a mass of 2 Kg. Inner

radius of the drum is 140 mm in the engaged position. The distance of C.G. of the shoe from

the axis of rotation of the spider is 115 mm. µ = 0.2. The spring force at the beginning of the

engagement is 1400 N.

Calculate (i) The speed at which the engagement begins

(ii)The power transmitted by the clutch at 1200 rpm.

Solution: z = 4 m = 2 kg

R = 140 mm r = 115 mm

μ = 0.2 FS = 1400 N

N = 1200 rpm

Engagement occurs only whenever the spring force is overcomes by centrifugal force.

FS = mrg2

1400 = 2 x 0.115 x g2

g = 78.019 rad/sec

2 N 2 1200125.66

60 6 rad / sec

0

g 78.019x 0.6208

125.66



2 2T mr zR(1 x )

22T 0.2 2 0.115 (125.66) 4 0.14 1 0.62

= 250.4 N.m

Power, 2 NT

P60

2 900 250.4P

60

P = 31.46kW

Example – 8:- Find out mass of a shoe, volume of a shoe and maximum load of a spring of a

centrifugal clutch for following data:

Power to be transmitted = 15 kW

No. of shoes = 4

Engagement begins at 75 % of the Running Speed

Inside diameter of pulley rim = 32.5 cm

Distance of C.G. of shoe from the centre of the spider = 120 mm

Coefficient of friction the shoe and drum = 0.25

Running Speed = 720 r.p.m.

Shoe is made up of C.I. for which the density is 7260 Kg/m3

Solution: P = 15 kW z = 4

x = 0.75 D = 32.5 cm

r = 120 mm μ = 0.25

N = 720 rpm ρ = 7260 Kg/m3

Power, 2 NT

P60

3 2 720 T15 10

60

T = 198.94 N.m

2 N 2 72075.398

60 60 rad / sec

2 2T mr zR(1 x )

22198.94 0.25 m 0.12 (75.398) 4 0.1625 1 0.75

m = 4.1 kg

m 4.1

Volume, V7260

= 5.65 x 10–4 m3

g = 0.75 = 56.548 rad/sec



FSmax = mrg2

FSmax = 4.1 x 0.12 x 56.548 = 1573.25 N

Example – 9:- The following specification refers to a centrifugal clutch: Power to be transmitted = 30 kW

No. of shoes = 4

Running Speed = 950 r.p.m.

The speed at which the engagement starts is 80 % of the Running Speed

Inner radius of drum = 120 mm

The radial distance of centre of gravity from the axis of spider is 100 mm

The normal intensity of pressure between friction lining and the drum = 0.2 MPa

The arc of contact subtended by friction lining of shoe at centre of spider is 600

The material of friction lining is Ferrodo with coefficient of friction = 0.25

Find (i) capacity of the clutch Mass of each shoe and (ii) size of each shoe.

Solution: P = 30 kW z = 4

N = 950 rpm x = 0.8

R = 0.12 m r = 0.1 m

p = 0.2 MPa = 60

μ = 0.25

Power, 2 NT

P60

3 2 950 T30 10

60

T = 301.5 N.m

2 N 2 95099.48

60 6 rad c

0/ se

2 2T mr zR(1 x )

22301.5 0.25 m 0.1 (99.48) 4 0.12 1 0.8

m = 7.05 kg

Contact length, l R

l 60 0.12

180

= 0.125 m

2 2

n

F mr (1 x )p

l b l b

22

67.05 0.1 (99.48) 1 0.8

0.2 10l b

l x b = 0.0125 Face width, b = 0.099 m



2.3 Brakes

A brake is defined as a mechanical device which is used to absorb the energy

possessed by a moving system or mechanism by the friction.

The primary purpose of the brake is to slow down or completely stop the motion of

moving system such as rotating drum, machine or vehicle. It is also used to hold the

parts of the system in position at rest.

An automobile brake is used either to reduce the speed of the car or bring it to rest. It

also used to keep the car stationary on the downhill road.

The energy absorbed by the brake can be either kinetic or potential or both. In

automobile application, the brake absorbs kinetic energy of moving vehicle.

In hoists and elevators, the brake absorbs the potential energy released by the object

during the braking period. The energy absorbs by the brake converts into heat energy

and dissipated to surrounding. Heat dissipation is a serious problem in brake

application.

2.4 Classification of brakes

Brakes are classified into the following three groups:

(a) Mechanical brakes which is operated by mechanical means such as levers, springs and

pedals. Depending upon the shape of the friction material, the mechanical brakes are

classified as Block brakes, Band Brakes, Block and Band Brake and internal or

external shoes brakes.

(b) Hydraulic brakes and pneumatic brakes which are operated by fluid pressure such as

oil pressure or air pressure.

(c) Electrical brakes which are operated by magnetic force and which include magnetic

particle brakes, hysteresis brakes and eddy current brakes.

Brake capacity depends upon the following factor.

(i) The unit pressure between braking surface.

(ii) The contacting area of braking surface.

(iii) The radius of the brake drum

(iv) The coefficient of friction

(v) The ability of the brakes to dissipate heat that is equivalent to the energy being

absorbed.

2.5 Energy Equations

The first step in the design of a mechanical brake is to determine the braking-torque capacity

for the given application. The braking-torque depends upon the amount of energy absorbed

by the brake. When a mechanical system of mass m moving with a velocity v1 is slowed down

to the velocity v2during the period of braking, the kinetic energy absorbed by the brake is

given by



2 21 2

1KE m v v

2 ……………..(i)

Where, KE = kinetic energy absorbed by the brake (J)

m = mass of the system (kg)

v1 and v2 = initial and final velocities of the system (m/s)

Similarly, the kinetic energy of the rotating body is given by

2 21 2

1KE I

2 ……………..(ii)

2 2 21 2

1KE mk

2 …………….(iii)

where, I =mass moment of inertia of the rotating body (kg-m2)

k = radius of gyration of the body (m)

ω1, ω2 = initial and final angular velocities of the body (rad/s)

In certain applications like hoists, the brake absorbs the potential energy released by the

moving weight during the braking period. When a body of mass m falls through a distance h,

the potential energy absorbed by the brake during the braking period is given by

PE =mgh …………….(iv)

where, g =gravitational constant (9.81 m/s2)

Depending upon the type of application, the total energy absorbed by the brake is

determined by adding the respective quantities of Eqs (i) to (iv). This energy is equated to the

work done by the brake.

Therefore,

E = Tb θ

where, E = total energy absorbed by the brake (J)

Tb = braking torque (N-m)

θ = angle through which the brake drum rotates during the braking period (rad)

Block or Shoe Brake:



Fig. 2.10 Block or Shoe Brake

A block or shoe brake consists of a block or shoe which is pressed against a rotating drum.

The force on the drum is increased by using a lever [Fig. 2.10(a)]. If only one block is used

for the purpose, a side thrust on the bearing of the shaft supporting the drum will act.

This can be prevented by using two blocks on the two sides of the drum [Fig.2.10(b)]. This

also doubles the braking torque.

A material softer than that of the drum or the rim of the wheel is used to make the blocks

so that these can be replaced easily on wearing. Wood and rubber are used for light and

slow vehicles and cast steel for heavy and fast ones.

Let

r

n

f n

r Radius of drum

Co efficient of friction

F Radial force applied on the drum

R Normalreaction on the block

F Force applied at the lever end

F Frictional force R

Assuming that normal reaction nR and frictional force

fF act at the mid point of the block.

n

Breaking Torque Frictional force Radius

R r

The direction of the frictional force on the drum is to be opposite to that of its rotation

while on the block it is in the same direction. Taking moments about the pivot O [Fig.

2.10(a)],

n n

n

n

F a R b R c 0

F aR

b c

b cF R _________(1)

a

When b c , F = 0 which implies that the force needed to apply the brake is virtually

zero, or that once contact is made between the block and the drum, the brake is applied

itself. Such a brake is known as a self – locking brake.



As the moment of the force fF about O is in the same direction as that of the applied force

F, fF aids in applying the brake. Such a brake is known as a self – enegised brake.

If the rotation of the drum is reversed, i.e., it is made clockwise,

nF R b c / a

which shows that the required force F will be far greater than what it would be when the

drum rotates counter – clockwise.

If the pivot lies on the line of action of Ef , i.e., at O’, c = 0 and n

aF R ,

b

which is valid for clockwise as well as for counter – clockwise rotation.

If c is made negative, i.e., if the pivot is at O’’,

n

b cF R for counter clockwiserotation

a

and

n

b cF R for clockwiserotation

a

In case the pivot is provided on the same side of the applied force and the block as shown

in Fig. 2.10 (c), the equilibrium condition can be considered accordingly.

In the above treatment, it is assumed that the normal reaction and the frictional force act

at the mid – point of the block. However, this is true only for small angles of contact.

When the angles of contact is more than 40°, the normal pressure is less at the ends than

at the centre. In that case, µ has to be replaced by an equivalent coefficient of friction µ’

given by

4 sin2

'sin

Band Brake:

Fig. 2.11 Band Brake



It consists of a rope, belt or flexible steel band (lined with friction material) which is

pressed against the external surface of a cylindrical drum when the brake is applied. The

force is applied at the free end of a lever (Fig.2.11).

Brake torque on the drum = ( T1 – T2) r

where r is the effective radius of the drum.

The ratio of the tight and slack side tensions is given by 1 2T / T e on the assumption

that the band is on the point of slipping on the drum.

The effectiveness of the force F depends upon the

direction of rotation of the drum

ratio of length a and b

direction of the applied force F.

To apply the brke to the rotating drum, the band has to be tightened on the drum. This is

possible if

1. F is applied in the downward direction when a > b

2. F is applied in the upward direction when a < b

If the force applied is not as above, the band is further loosened on the drum which

means no braking effect is possible.

1. a > b, F Downwards

(a) Rotation Counter – Clockwise

For counter – clockwise rotation of the drum, the tight and the slack sides of the band will be

as shown in Fig. 2.11.

Considering the forces acting on the lever and taking moments about the pivot,

1 2

1 2

F l T a T b 0

or

T a T bF _________(1)

l

As T1> T2 and a > b under all conditions, the effectiveness of the brake will depend upon the

force F.

(b) Rotation Clockwise

Fig. 2.12



In this case, the tight and the slack sides are reversed as shown in Fig. 2.12.

Now,

2 1

2 1

F l T a T b 0

or

T a T bF _________(2)

l

As T2< T1 and a > b, the brake will be effective as long as T2. a is greater than T1. b

Or

22 1

1

T bT a T b or

T a

i.e., as long as the ratio of T2 to T1 is greater than the ratio b/a.

When

2

1

T a

T b

, F is zero or negative, i.e., the brake becomes self – locking as no force is

needed to apply the brake. Once the brake has been engaged, no further force is required to

stop the rotation of the drum.

2. a < b, F upwards

Fig. 2.13

(a) Rotation Counter – Clockwise

The tight and the slack sides will be as shown in Fig.2.13(a).

Therefore,

1 2

2 1

F l T a T b 0

or

T b T aF

l

As T2< T1 and b > a, the brake is operative only as long as

22 1

1

T aT b T a or

T b

Once T2 / T1 becomes equal to a/b, F required is zero and the brake becomes self – locking.



(b) Rotation Clockwise

The tight and the slack sides are shown in Fig. 2.13(a).

From Fig. 2.13(b),1 2

1 2

T b T aF l T b T a 0 or F

l

As T1> T2 and b > a, under all conditions, effectiveness of the brake will depend upon the

force F.

When a = b, the band cannot be tightened and thus, the brake cannot be applied.

Fig. 2.14

The band brake just discussed is known as a differential band brake. However, if either a

or b is made zero, a simple band brake is obtained. If b = 0 (Fig. 2.14) and F downwards,

1

1

F l T a 0

or

aF T

l

Similarly, the force can be found for the other cases.

Note that such a brake can neither have self – energising properties nor it can be self –

locked.

The brake is said to be more effecctive when maximum braking force is applied with the

least effort F.

For case (i), when a > b and F is downwards, the force (effort) F required is less when the

rotation is clockwise assuming that the brake is effective.

For case (ii), when a < b and F is upwards, F required is less when the rotation is counter –

clockwise assuming that the brake is effective.

Thus, for the given arrangement of the differential brake, it is more effective when

(a) a > b, F downwards, rotation clockwise

(b) a > b, F upwards, rotation counter – clockwise.

The advantages of self – locking is taken in hoists and conveyers where motion is

permissible in only one direction. If the motion gets reversed somehow, the self – locking

is engaged which can be released only by reversing the applied force.



It is seen that a differential band brake is more effective only in one direction of rotation

of the drum.

Fig. 2.15

However, a two-way band brake can also be deigned which is equally effective for both

the directions of rotation of the drum (Fig. 2.15). In such a brake, the two lever arms are

made equal.

For both directions of rotation of the drum,

1 2

1 2

F l T a T a 0

or

aF T T

l

Band brake offers the following advantages:

(i) Band brake has simple construction. It has small number of parts. These features

reduce the cost of band brake.

(ii) Most equipment manufacturers can easily produce band brake without requiring

specialized facilities like foundry or forging shop. The friction lining is the only part

which must be purchased from outside agencies.

(iii) Band brake is more reliable due to small number of parts.

(iv) Band brake requires little maintenance.

The disadvantages of band brake are as follows:

(a) The heat dissipation capacity of a band brake is poor.

(b) The wear of friction lining is uneven from one end to the other.

Band brakes are used in applications like bucket conveyors, hoists and chain saws. They are

more popular as back-stop devices.



Band and Block Brake:

Fig. 2.16 Band and Block brake

A band and block brake consists of a number of wooden blocks secured inside a flexible

steel band. When the brake is applied, the blocks are pressed against the drum. Two sides

of the band becomes tight and slack as usual. Wooden blocks have a higher coefficient of

friction. Thus, increasing the effectiveness of the brake. Also, such blocks can be easily

replaced on being worn out [ Fig.2.16(a)].

Each block subtends a small angle of 2θ at the centre of the drum. The frictional force on

the blocks acts in the direction of rotation of the drum. For n blocks on the brake,

Let

0

1

2

n

n

T Tension on the slack side.

T Tension on the tight side after one block.

T Tension on the tight side after two blocks.

T Tension on the tight side after nblocks.

Coefficient of friction.

R Normalreaction on the block.

Resolving the forces horizontally and vertically….

1 0 n

1 0 n

1 0 n

1 0 n

T T cos R ________(1)

T T sin R ________(2)

T T cos R

T T sin R

1 0

1 0

1 0 1 0

1 0 1 0

T T tan

T T 1

T T T T tan 1

tan 1T T T T



1

0

1

0

2T 1 tan

2T 1 tan

T 1 tan

T 1 tan

Similarly…

2

1

n

n 1

T 1 tanand so on

T 1 tan

T 1 tan

T 1 tan

n n n 1 2 1

0 n 1 n 2 1 0

n

T T T T T

T T T T T

1 tan

1 tan

Internal Expanding Brake

The construction of an internal expanding brake is shown in Fig. 2.17. It consists of a

shoe, which is pivoted at one end and subjected to an actuating force P at the other

end. A friction lining is fixed on the shoe and the complete assembly of shoe lining and

pivot is placed inside the brake drum.

Fig.2.17 Internal Expanding Brake

Internal shoe brakes, with two symmetrical shoes, are used on all automobile vehicles.

The actuating force is usually provided by means of a hydraulic cylinder or a cam

mechanism. The analysis of the internal shoe brake is based on the following

assumptions:

(a) The intensity of normal pressure between the friction lining and the brake drum at any

point is proportional to its vertical distance from the pivot.

(b) The brake drum and the shoe are rigid.

(c) The centrifugal force acting on the shoe is negligible.

(d) The coefficient of friction is constant.



2.6 Braking of 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 ofbraking, it is required to determine theretardationof

the vehicle when brakes are applied. Since thevehicle retards, therefore it is a problem of

dynamics. But it may be reduced to an equivalentproblem of statics byincluding the inertia

force inthe system of forces actually applied to the vehicle.The inertia force is equal and

opposite to thebraking force causing retardation.

Fig. 2.18

Consider a vehicle moving up an inclined plane shown in Fig.2.18.

Let

Angle of inclination of the plane to the horizontal.

m mass of vehicle (Such that its weight mg inNewton).

h Height of C.G. of the vehicle above the road surface (mtr).

x Perpendicular dis tance of C.G. from the rear axle (mtr).

L Distan

A B

2

ce between centre of rear and front wheel (wheelBase)(mtr).

R ,R Reactions of the ground on the front & rear wheel (N).

Coefficient of frictionbetween tyre androad surface.

a Retardation of the vehicle (m / s ).

a. Brakes applied to Rear wheel only:

It is a common way of braking a vehicle in which the braking froce acts at the rear wheel.

Resolving the forces……

B

A B

F m g sin m a _____(1)

R R m g cos _____(2)



B

B

F Totalbreaking force acting at

the rear wheel due to application of brakes

R

Taking moment about G…..

B B A

B B B

F h R x R L x

R h R x (m g cos R ) L x

B B

A B

F R

R m g cos R

B BR h x mg cos L x R L x

BR h x L x mg cos L x

B

m g cos L xR

L h

and

A B

A

R m g cos Rm g cos L x

m g cosL h

m g cos x hR

L h

From equation (1)

BF m g sin m

B

B

BB

F m g sina

mF

g sinm

Ra g sin Putting value of R in equation

m

m g cos (L x)

L ha g sin

m

g cos (L x)a g sin

L h

Note : (1) On Road level α = 0 so…..

g (L x)

aL h

(2) When vehicle moves down a plane

equation (1) becomes BF mg sin ma



B

B

Fa g sin

mR

g sinmg cos (L x)

a g sinL h

b. Brake applied to Front wheels only:

It is a very rare way of braking the vehicle, in which the braking froce acts at the front

wheels only.

Let

A

A

F Braking force acting at the front wheel.

R

Resolving the forces horizontally and vertically…..

A

A B

F m g sin m a _______(1)

R R m g cos _______(2)

Taking moments about G……

A B AF h R x R (L x )

A A

B A

F R

R m g cos R

Putting values....

A A AR h m g cos R x R L x

A AR h mg cos x R L

AR L h m g cos x

A

m g cos xR

L h

From equation (2)

A BR R m g cos

B A

B

R m g cos R

m g cos xm g cos

L h

xm g cos 1

L h

L h xR m g cos

L h

From equation (1)

AF mg sin ma



A

AA

F m g sina

mR m g sin

Putting Value of Rm

m g cos xm g sin

L h

m

g cos xa g sin

L h

Note : (1) On a Road level α = 0

A

m g cos x m g xR cos 0 1

L h L h

B

m g L h xL h xR m g cos

L h L h

g cos x cos 0 1a g sin

sin 0 0L h

g x

L h

(2) When vehicle moves down the plane…..

AF mg sin ma

A

A

m g sinFa

m mR

g sinmg cos x

g sinL h

c. Brakes applied to all 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.

A A

B B

F Braking force for front wheels R

F Braking force for rear wheels R

Note : A little consideration will show that whenthe brakes are applied to all the four wheels,

thebraking distance (i.e. the distance in which thevehicle 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.

Resolving the forces…….

A B

A B

F F m g sin m a ________(1)

R R m g cos ________(2)



Taking moment about G…..

A B B AF F h R x R (L x ) _______(3)

A A

B B

B A

F R

F R

R m g cos R

Putting values in eq.(3)....

A B A AR R h mg cos R x R (L x )

A A A AR mg cos R h mg cos R x R (L x )

A

m g cos h xR _______(4)

L

B AR m g cos R

h xm g cos m g cos

L

L h xm g cos _______(5)

L

From equation (1)

A BF F mg sin ma

A B

A B

R R m g sin m a

m g cos m g sin m a R R m g cos

a g cos sin ______(6)

Note : (1) On a Road level α = 0

A

m g cos h x mg h xR

L L

B

L h xR m g

L

a g

(2) If vehicle moves down the plane, then equation (1) may be written as…..

A BF F mg sin ma

A B

A B

R R m g sin m a

m g cos m g sin m a R R m g cos

a g cos sin



2.7 Dynamometer A Dynamometer is a brake but in addition it has a device to measure the frictional resistance.

Knowing the frictional resistance, may obtain the torque transmitted and hence the power of

the engine.

2.8 Types of Dynamometers: There are mainly two types of dynamometers:

1. Absorption Dynamometers: In this type, the work done is converted into heat by friction

while being measured. They can be used for the measurement of moderate powers only.

Examples are prony brake dynamometer and rope brake dynamometer.

2. Transmission Dynamometers: In this type, the work is not absorbed in the process, but is

utilised after the measurement. Examples are the belt – transmission dynamometer and

the trosion dynamometer.

Prony Brake Dynamometer:

Fig. 2.19 Prony Brake Dynamometer

A prony brake dynamometer consists of two wooden blocks clamped together on a

revolving pulley carrying a lever (Fig. 2.19).

The friction between the blocks and the pulley tends to rotate the blocks in the

direction of rotation of the shaft. However, the weight due to suspended mass at the

end of the lever prevents this tendency. The grip of the blocks over the pulley is

adjusted using the bolts of the clamp until the engine runs at the required speed.

The mass added to the scale pan is such that the arm remains horizontal in the

equilibrium position; the power of the engine is thus absorbed by the friction.

Frictional torque Wl Mg l

2 NPower of themachine under test T Mg l

60MNk

where k is a constant for a particular brake.

Note that the expression fro power is independent of the size of the pulley and the coefficient

of friction.



Rope Brake Dynamometer:

In a rope brake dynamometer (Fig. 2.20), a rope is wrapped over the rim of a pulley

keyed to the shaft of the engine. The diameter of the rope depends upon the power of

the machine.

The spacing of the ropes on the pulley is done by 3 to 4 U-shaped wooden blocks

which also prevent the rope from slipping off the pulley. The upper end of the rope is

attached to a spring balance where as the lower end supports the weight of

suspended mass.

t

Power of the machine TF r

2 NM g s r

60

Fig. 2.20 Rope Brake Dynamometer

If the power produced is high, so will be the heat produced due to friction between

the rope and the wheel, and a cooling arrangement is necessary.

For this, the channel of the flywheeel usually has flanges turned inside in which water

from a rope is supplied. An outlet pipe with a flattenend end takes the water out.

A rope brake dynamometer is frequently used to test the power of engines. It is easy

to manufacture, inexpensive, and requires no lubrication.

If the rope is wrapped several times over the wheel, the tension on the slack side of

the rope, i.e., the spring balance reading can be reduced to a negligible value as

compared to the tension of the tight side (as T1 / T2 = eµθ and θ is increaased). Thus,

one can even do away with the spring balance.



Belt Transmission Dynamometer:

Fig. 2.21 Belt Transmission Dynamometer

The belt transmission dynamometer occupies a prominent position among

transmission dynamometers. When a belt transmits power from one pulley to

another, there exists a difference in tensions between the tight and slack sides. A

dynamometer measures directly the difference in tensions (T1 – T2) while the belt is

running.

Fig. 2.21 shows a Tatham dynamometer. A continuous belt runs over the driving and

the driven pulleys through two intermediate pulleys. The intermediate pulleys have

their pins fixed to a lever with its fulcrum at the midpoint of the two pulley centres.

As the lever is not pivoted at its midpoint, a mass at the left end is used for its initial

equilibrium. When the belt transmits power, the lever tends to rotate in the counter –

clockwise direction due to the difference of tensions on the tight and slack sides.

To maintain its horizontal position, a weight of the required amount is provided at the

right end of the lever. Two stops, one on each side of the lever arm, are used to limit

the motion of the lever.

Taking moments about the fulcrum,

1 2

1 2

1 2

M g l 2 T a 2 T a 0

M g l 2a T T 0

M g lT T

2a

1 2Power,P T T v

where v =belt speed in metres per second.



Epicyclic - Train Dynamometer:

An epicyclic – train dynamometer is another transmission type of dynamometer. As

shown in Fig. 2.22, it consists of a simple epicyclic train of gears. A spur gear A is the

driving wheel which drives an annular driven wheel B through an intermediate pinion

C.

Fig. 2.22 Epicyclic - Train Dynamometer

The intermediate gear C is mounted on a horizontal lever, the weight of which is

balanced by a counterweight at the left end when the system is at rest.

When the wheel A rotates counter – clockwise, the wheel B as well as the wheel C

rotates clockwise.

Two tangential forces, each equal to F, act at the ends of the pinion C, one due to the

driving force by the wheel A and the other due to reactive force of the driven wheel B.

Both forces are equal if friction is ignored.

This tends to rotate the lever in the counter – clockwise direction and it no longer

remains horizontal. To maintain it in the same position as earlier, a balancing weight

W is provided at the right end of the lever.

Two stops, one on each side of the lever arm, are used to limit the motion of the lever.

For the equilibrium of the lever,

W l2F a W l or F

2 a

and torque transmitted = F . r where r is the radius of the driving wheel

Thus power,

2 NP T F r

60



Bevis – Gibson Torsion Dynamometer:

Fig. 2.23 Bevis – Gibson Torsion Dynamometer

A Bevis – Gibson torsion dynamometer consists of two discs A and B, a lamp and a

movable torque finder arranged as shown in Fig. 2.23(a). The two discs are similar and

are fixed to the shaft at a fixed distance from each other.

Thus, the two discs revolve with the shaft. The lamp is masked and fixed on the

bearing of the shaft. The torque finder has an eyepiece capable of moving

circumferentially.

Each disc has a small radial slot near its periphery. Similar slots are also made at the

same radius on the mask of the lamp and on the torque finder.

When the shaft rotates freely and does not transmit any torque, all the four slots are

in a line and a ray of light from the lamp can be seen through the eyepiece after every

revolution.

When a torque is transmitted, the shaft twists and the slot in the disc B shifts its

position. The ray of light can no longer pass through the four slots. However, if the

eyepiece is moved circumferentially by an amount equal to the displacement, the

flash will again be visible once in each revolution of the shaft.

The eyepeice is moved by a micrometer spindle. The angle of twist may be measured

up to one hundredth of the degree.

In case the torque is varied during each revolution of the shaft as in reciprocating

engines and it is required to measure the angle of twist at different angular position,

then each disc can be perforated with several slots arranged in the form of a spiral at

varying radii [Fig. 2.23(b)].

The lamp and the torque finder have to be moved radiallly to and from the shaft so

that they come opposite each pair of slots in the discs.



Example – 10:- A brake as shown in Fig. 2.24 is fitted with a C.I. brake shoe. The braking torsional moment = 360N.m The drum diameter = 300mm The coefficient of friction = 0.3

Find: (i) force P for counter clockwise rotation

(ii) force P for clockwise rotation

(iii) where must pivot be placed to make brake self locking with clockwise rotation?

Fig. 2.24

Solution:

Tb = 360 N.m

d = 300 mm

μ = 0.3

(a) (b)

Fig. 2.25

for the clockwise rotation of the brake drum, the frictional force or the tangential force (Ft)

acting at the contact surfaces is shown in Fig. 2.25 (a)

Braking torque, Tb = Ft x r

= μ RN x r

360 = 0.3 x RN x 0.15

Normal force, RN = 8000 N

Now taking moments about the fulcrum O

P (600 + 200) + Ft x 50 = RN x 200

P x 800 + (0.3 x 8000) 50 = 8000 x 200

P x 800 = 1480 x 103

P = 1850 N



for the counter clockwise rotation of the brake drum, the frictional force or the tangential

force (Ft) acting at the contact surfaces is shown in Fig. 2.25 (b)

Now taking moments about the fulcrum O,

P (600 + 200) = Ft x 50 + RN x 200

P x 800 = (0.3 x 8000) 50 + 8000 x 200

P x 800 = 1720 x 103

P = 2150 N

Location of the pivot or fulcrum to make the brake self-locking

The clockwise rotation of the brake drum is shown in Fig. 2.25 (a). Let x be the distance of the

pivot or fulcrum O from the line of action of the tangential force (Ft). Taking moments about

the fulcrum O,

P (600 + 200) + Ft x x – RN x 200 = 0

In order to make the brake self-locking, Ft x x must be equal to RN x 200, so that the force P is

zero.

Ft x x = RN x 200

2400 x x = 8000 x 200

x = 667 mm

Example – 11:- A simple band brake is shown in Fig. 2.26 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

rpm. The co-efficient of friction is 0.2 and the brake drum diameter is 240 mm. Take b = 120

mm, l = 420 mm, = 210, then find out the followings:

(i) The torque applied due to hand load of 100 N

(ii) The number of turns of the flywheel before it is brought to rest

(iii) The time required to bring it to rest from the moment of the application of the brake.

Fig. 2.26

Solution: m = 400 kg k = 450 mm

N = 300 rpm μ = 0.2

d = 240 mm = 210 = 3.665 rad



Taking moments about the fulcrum O,

T1 x b = P x l

T1 x 0.12 = P x 0.42

T1 x 0.12 = 100 x 0.42

T1 = 350 N

Now, 0.2 3.6651

2

Te e

T

1

2

T2.081

T

12

T 350T

2.081 2.081

T2 = 168.188 N


= (T1– T2) x r

= (350 – 168.188) x 0.12

= 21.817 N.m

Work done against friction due to absorption of K.E.

2t

1F S mv

2

2t

1F S m(r )

2

2

t

m dNS

F 2 60

2400 0.24 300

S181.81 2 60

= 15.63 m

S = D x n

SNo. of turns, n

D

15.63

0.24

= 20.73

Tb = I α

= (mk2) α

21.817 = (400 x 0.4502) α

α = 0.269

= o + αt

2 N

0 t60



2 3000 0.269 t

60

t = 116.78 sec

Example – 12:- A band brake shown in Fig. 2.27 is used to balance a torque of 980 N-m at

the drum shaft. The drum diameter is 400 mm (rotating in clockwise direction) and the

allowable pressure between lining and drum is 0.5 MPa. The coefficient of friction is 0.25.

Design the steel band, shaft, brake lever and fulcrum pin, if all these elements are made from

steel having permissible tensile stress 70 MPa and shear stress 50 MPa.

Fig. 2.27

Solution:

Tb = 980 N.m d = 400 mm

p = 0.5 MPa μ = 0.25

σt = 70 MPa τ = 50 MPa

= 120 = 2.094 rad


= (T1– T2) x r

980 x 103 = (T1– T2) x 200

T1– T2 = 4.9 x 103 ………………(i)

Now, 0.25 2.0941

2

Te e

T

1

2

T1.688

T

………………(ii)

1.688 T2– T2 = 4.9 x 103

T2 = 7122.09 N

T1 = 12022.09 N

Taking moments about the fulcrum O,

T2 x 80 = P x 600

7122.09 x 80 = P x 600

P = 949.612 N



Design of Shaft

Since the shaft has to transmit torque equal to braking torque

3

b ST d16

3 3S980 10 d 50

16

dS = 46.388 mm

Shaft diameter, dS ≈ 50 mm

Design of Lever

t1 = thickness of the lever

B = width of the lever

Maximum bending moment at fulcrum O due to force P

M = P x l = 949.612 x 600

= 569767.2 N.mm

Section modulus,

2 2

1 1 1

1 1Z t B t (2t )

6 6 (Assuming B = 2t1)

= 0.67 t13 mm3

MBending stress

Z

31

569767.270

0.67 t

t1 = 22.988 mm

≈ 23 mm

B = 2t1 = 46 mm

Design of Pins

d1 = diameter of pins

l1 = length of pins = 1.25 d1

The pins are designed for maximum tension in the band (i.e. T1 = 12022.09 N)

Considering bearing of the pins, maximum tension (T1),

12022.09 = d1.l1.p = d1 x 1.25 d1 x 0.5

d1 = 138.69 mm

d1 ≈ 140 mm

l1 = 1.25 d1 = 1.25 x 140

= 175 mm

Check the pin for induced shearing stress. Since the pin is in double shear, therefore

maximum tension (T1),

2 2

112022.09 2 (d ) 2 (140)4 4

30787.6



12022.09

30787.6

= 0.39 MPa

This induced stress is quite within permissible limits.

The pin may be checked for induced bending stress.

1

5 5Maximum bending moment, M W.l 12022.09 175

24 24 (W= T1)

= 438305.36 N-mm

3 31section modulus, Z (d ) (140)

32 32

3269391.57mm

M 438305.36Bending stress

Z 269391.57

= 1.627 MPa

The induced bending stress is within safe limits of 70 MPa.

The lever has an eye hole for the pin and connectors at band have forked end.

12Thickness of each eye

l 175t 87.5mm

2 2,

Outer diameter of the eye, D= 2d1 = 2 x 140 = 280 mm

A clearance of 1.5 mm is provided on either side of the lever in the fork.

A brass bush of 3 mm thickness may be provided in the eye of the lever.

Diameter of hole in the lever = d1 + 2 x 3 = 140 + 2 x 3 = 146 mm

The boss is made at pin joints whose outer diameter is taken equal to twice the diameter of

the pin and length equal to length of the pin.

The inner diameter of the boss is equal to diameter of the hole in the lever.

Outer diameter of the boss = 2d1 = 2(140) = 280 mm

Length of the boss = l1 = 175 mm

Check the bending stress induced in the lever at the

fulcrum.

Maximum bending moment at the fulcrum,

M = P x l = 949.612 x 600

= 569767.2 N-mm

3 31175 280 146

12Section modulus, Z280 / 2

= 1962485.83 mm3

M 569767.2Bending stress 0.29 MPa

Z 1962485.83

The induced bending stress is within safe limits of 70 MPa. Fig. 6.28



Example – 13:- A differential band brake has a drum with a diameter of 800 mm. The two

ends of the band are fixed to the pins on the opposite sides of the fulcrum of the lever at

distances of 40 mm and 200 mm from the fulcrum. The angle of contact is 270° and the

coefficient of friction is 0.2. Determine the brake torque when a force of 600 N is applied to

the lever at a distance of 800 mm from the fulcrum.

Solution:

d = 800 mm θ = 270°

μ = 0.2 F = 600 N

l = 800 mm

Fig.2.29

Assuming a = 200 mm, b = 40 mm, i.e., a > b, F must act downward direction to apply the

brake (Fig. 2.29).

2701 180

2

1

2

Te e

T

T2.57

T

(a) Anticlockwise Rotation

Taking moment about fulcrum O….

1 2

2 2

2 2

2

2

1

F l T a T b 0 (Fig.2.29)

600 800 (2.57 T ) 200 T 40 0

480000 514 T 40 T 0

474 T 480000

T 1012.658 N and

T 2602.53N

B 1 2T T T r

2602.5 101

Braking T

2.7 0.4

orque

63 .9 m

,

5 2N

(b) Clockwise Rotation

Taking moment about fulcrum O…. (Fig.2.30)



1 2

2 2

2

1 2

F l T b T a 0

600 800 2.57 T 40 T 200 0

T 4938 N and

T 2.57 T

12691N

Fig.2.30

B 1 2T T T r

12691 4938 0.4

Braking Torque

3101N m

,

Assuming a = 40 mm, b = 200 mm, i.e., a < b, F must act upward to apply the brake.

(a) Anticlockwise Rotation

Fig.2.31


1 2

2 2

2

1

F l T a T b 0

600 800 2.57 T (40) T 200 0

T 4938 N and

T 12691N

B 1 2T T T r

12691 4938 0.4

Braking Torque

3101N m

,



(b) Clockwise Rotation

Fig.2.32


2 1

2 2

2

1

F l T a T b 0

600 800 T 40 2.57 T 200 0

T 1012.7N and

T 2602.5N

B 1 2T T T r

1012.7 2602.5 0.4

Braking Torque,

636 N m

Note: The above results show that the effectiveness of the brake in one direction of rotation

is equal to the effectiveness in the other direction if the distance of the pins on the opposite

sides of the fulcrum are changed and the force is applied in the proper direction so that the

band is tightened.

Example – 14:- A simple band brake (Fig.2.33) is applied to a shaft carrying a flywheel of

250 kg mass and of radius of gyration of 300 mm. the shaft speed is 200 rpm. The drum

diameter is 200 mm and the coefficient of friction is 0.25. Determine the

1. brake torque when a force of 120 N is applied at the lever end

2. number of turns of the flywheel before it comes to rest

3. time taken by the flywheel to come to rest.

Solution:

Fig.2.33



m 250 kg k 300 mm

N 200 rpm d 200 mm

0.25 360 135 225 3.93rad180

1. Taking moment about O….

1

1

1

12

F l T a

120 280 T 100

T 336 N

TT 125.8 N

2.67

1 0.25 3.93

2

1

2

Te e

T

T2.67

T

B 1 2T T T r

336 125.8 0.1

Braking Torque,

21 N m

2. K.E of flywheel 2

2 2

2

2

1 1 2 NI mk

2 2 60

2 2001250 0.3

2 60

4935 N m

This K.E is used to overcome the work done by the braking torque in n revolutions.

BK.E of flywheel T Angular displacement

4935 21 2 n

n 37.4 Revolution

3. For uniform retardation,

200Average speed 100 rpm

2n 37.4

Time taken min.N 100

37.4 =

100 / 60

22.44 sec

Example – 15:- A simple band brake is applied to a drum of 560 mm diameter which

rotates at 240 rpm. The angle of contact of the band is 270°. One end of the band is fastened

to a fixed pin and the other end to the brake lever, 140 mm from the fixed pin. The brake

lever is 800 mm long and is placed perpendicular to the diameter that bisects the angle of

contact. Assuming the coefficient of friction as 0.3, determine the necessary pull at the end of

the lever to stop the drum if 40 kW of power is being absorbed. Also, find the width of the

band if its thickness is 3 mm and the maximum tensile stress is limited to 40 N/mm2.



Solution:

,

l 800 mm

d 560 mm N 240 rpm

270 a 140 mm

0.3

P 40 kW

2thickness t 3mm 40N/mm

Fig.2.34

Note: It can be observed from the figure that to tighten the band, the force is to be applied

upwards. If the drum rotates counter – clockwise, the tight and slack sides will be as shown.

B

1 2

31 2

1 2

P T

2 NT T r

602 240

40 10 T T 0.2860

T T 5684 ________(1)

and

0.3 2701 180

2

Te e 4.11 _______(2)

T

From equation (1) & (2)

2

1

T 1828 N

T 7514 N

Taking the moment about O……

2F l T 140 cos 45

F 800 1828 140 cos 45

F 226.2N

Let maximum tension……

1T b t

7514 40 b 3

b 62.6 mm



Note: If drum rotates clockwise, the brake is less effective as in that case tight and slack sides

are interchange and the force requied to apply the same braking torque is more.

1F l T 140 cos 40

F 800 7514 140 cos 40

F 930 N

Example – 16:- A crane is required to support a load of 1.2 tonnes on the rope round its

barrel of 400 mm diameter (Fig.2.35). The brake drum which is keyed to the same shaft as the

barrel has a diameter of 600 mm. The angle of contact of the band brake is 275° and the

coefficient of friction is 0.22. Determine the force required at the end of the lever to support

the load. Take a = 150 mm and l = 750 mm.

Fig.2.35

Solution: W 1.2 1000 9.81N

R 300 mm r 200 mm

0.22 275

For equilibrium position…..

1 2

2 2

2

1

T T R W r

2.87 T T 300 1.2 1000 9.81 200

T 4197N m

and T 12045N m

0.22 2751 180

2

Te e 2.87

T

Taking moment about O…..

1F l T a

F 750 12045 150

F 2409 N



Example – 17:- A band and block brake has 14 blocks. Each block subtends an angle of 14°

at the centre of the rotating drum. The diameter of the drum is 750 mm and the thickness of

the blocks is 65 mm. The two ends of the band are fixed to the pins on the lever at distances

of 50 mm and 210 mm from the fulcrum on the opposite sides. Determine the least force

required to be applied at the lever at a distance of 600 mm from the fulcrum if the power

absorbed by the blocks is 180 kW at 175 rpm. Coefficient of friction between the blocks and

drum is 0.35.

Solution:

, b 50 mm

n 14 blocks 2 14

d 750 mm t 65mm

a 210 mm

l 600 mm P 180 kW

N 175rpm 0.35

F ?

Let

14 0

14 0

314 0

14 0

P T T v

DNT T

60

0.75 2 0.065 175180 10 T T

60

T T 22323N _____(1)

Let n

14

0

14

14

0

T 1 tan

T 1 tan

T 1 0.35 tan73.334 _______(2)

T 1 0.35 tan7

From equation (1) and (2)

0

14

T 9564 NT 31887N

Assume a > b, F must be downward and clockwise rotation for maximum braking torque.

Moment about O…. (Fig. in theory)

0 14F l T a T b 0

F 600 9564 210 31887 50 0

F 690 N



Example – 18:- A band and block brake having 12 blocks, each of which subtends an angle

of 16° at the centre, is applied to a rotating drum with a diameter of 600 mm. The blocks are

75 mm thick. The drum and the flywheel mounted on the same shaft have a mass of 1800 kg

and have a combined radius of gyration of 600 mm. The two ends of the band are attached to

pins on the opposite sides of the brake fulcrum at distances of 40 mm and 150 mm from it. If

a force of 250 N is applied on the lever at a distance of 900 mm from the fulcrum, find the

1. Maximum braking torque

2. Angular retardation of the drum

3. Time taken by the system to be stationary from the rated speed of 300 rpm.

Solution:

, b 40 mm

16 n 12 blocks

d 600 mm t 75mm

m 1800 kg k 600 mm

a 150 mm

F 250 N l 900 mm

N 300 rpm 0.3

Refer the Fig.1.7

1. n

12

0

12

14

0

T 1 tan

T 1 tan

T 1 0.3 tan8

T 1 0.3 tan8

2.752

Assume a = 150 mm, b = 40 mm as a > b, F must be downwardsand rotation is clockwise.

Taking moment about fulcrum O….

0 12

0 12

0

0

12

F l T a T b 0

250 900 T (150) T 40 0

T 150 2.752 40 250 900

T 5636 N

T 15511N

Maximum Braking torque….

B 12 0T T T r

0.6 0.075 215511 5636

23703N m

d 2 tdr

2 2



2. Let Braking torque

2B

2

2

T I mk

3703 1800 0.6

5.71 rad / sec

4. Initial Angular Speed

0

Final Angular speed 02 N 2 300

31.4 rad / sec60 60

0 t

0 31.4 5.71 t

t 5.5 sec

Example – 19:- A car moving on a level road at a speed 50 km/h has a wheel base

2.8metres, distance of C.G. from ground level 600 mm, and the distance of C.G. from rear

wheels 1.2metres. 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: u 50 km / hr 13.89 m / sec

L 2.8 m h 600 mm

x 1.2m 0.6

1. Rear Wheels

Here vehicle moves on a level road, so retardation of car is…..

20.6 9.81 2.8 1.2g (L x)a 2.98 m / sec

L h 2.8 0.6 0.6

If retardation is uniform 2 2

2

2 2

v u 2 s a

0 u 2 s a

u 13.89s 32.4 m

2 a 2 2.98

2. Front Wheels

Here vehicle moves on a level road, so retardation of car is….

2g x 0.6 9.81 1.2a 2.9 m / sec

L h 2.8 0.6 0.6



For uniform retardation….

2

2

us

2 a13.89

s2 2.9

33.26 m

3. All the four wheels

Here vehicle moves on a level road, so retardation of car is…..

a = gμ = 9.81 x 0.6

= 5.886 m/sec2


2 2u 13.89s

2 a 2 5.886 16.4 m

Example – 20:- 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 vehicle before 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:

10 u 36 km / h 10 m / sec

L 1.8 m x 0.8 m

h 0.9 m 0.5

1. The vehicle moves up (All Four wheels)

2

a g cos sin

9.81 0.5 cos 10 sin10

6.53 m / sec


2 2u 10s

2 a 2 6.53 7.657m

Final velocity of vehicle….

v u a t ve sign due to retardation

0 10 6.53 t

t 1.53 sec

2. The vehicle moves down the plane (All Four wheels)



2

a g cos sin

9.81 0.5 cos 10 sin10

3.13 m / sec


2 2u 10s 16 m

2 a 2 3.13

Final velocity of vehicle….

v u a t

0 10 3.13 t

t 3.2 sec

Example – 21:- The following data refer to a laboratory experiment with a rope brake:

Diameter of the flywheel = 800 mm Diameter of the rope = 8 mm Dead weight on the brake = 40 kg Speed of the engine = 150 rpm Spring balance reading = 100 N Find the power of the engine.

Solution:

2 NP M g s r

602 150

40 9.81 100 0.4 0.00460

1855.6 W

Example – 22:- In a belt transmission dynamometer, the driving pulley rotates at 300 rpm.

The distance between the centre of the driving pulley and the dead mass is 800 mm. The

diameter of each of the driving as well as the intermediate pulleys is equal to 360 mm. Find

the value of the dead mass required to maintain the lever in a horizontal position when the

power transmitted is 3 kW. Also, find its value when the belt just begins to slip on the driving

pulley, µ being 0.25 and the maximum tension in the belt 1200 N.

Solution: N 300 rpm a 0.36 m

l 0.8 m P 3000 W

1 2(i) P T T vM g l

r2a

M g l 2 Nr

2a 60

M 9.81 0.8 2 3003000 0.18

2 0.36 60

M 48.7kg



1

0.251

2

2

1 2

(ii) T 1200 N, 0.25, rad

T e e 2.19

T

1200 T 548 N

2.19

M g l T T

2a

M 9.81 0.81200 548

2 0.36

M 59.8 kg

Example – 23:- A torsion dynamometer is fitted to a propeller shaft of a marine engine. It is

found that the shaft twists 2o in a length of 20 meters at 120 r.p.m. If the shaft is hollow with

400 mm external diameter and 300 mm internal diameter, find the power of the engine. Take

modulus of rigidity for the shaft material as 80 GPa.

Solution: l 20 m N 120 rpmD 400 mm d 300 mm

G 80 GPa 2 2 0.035rad180

Polar moment of Inertia of the shaft,

4 4J (D d )32

4 4J [(0.4) (0.3) ]32

J = 0.0017 m4

T G

J l

GT J

l

980 10 0.035T 0.0017

20

Torque applied to the shaft, T = 238 x 103 N.m

2 NTP

60

32 120 238 10P

60

Power of the engine, P = 2990 kW


3 Flywheels

Course Contents

3.1 Turning-Moment Diagram

3.2 Single-Cylinder Double-Acting

Steam Engine

3.3 Single cylinder Four Stroke En-

gine

3.4 Multi Cylinder Engines

3.5 Fluctuation of energy

3.6 Flywheel

3.7 Dimensions of Flywheel Rim

3.8 Punching Press

3.9 Summary

3. Flywheels Theory of Machines (2151902)


3.1 Turning-Moment Diagrams

During revolution of the crankshaft of a steam engine or IC engine, the torque on it varies and is

given by,

𝑇 = 𝐹𝑡 × 𝑟 = 𝐹𝑟 (sin 𝜃 +sin 2𝜃

2√𝑛2 − sin 𝜃2)

Where F is the net piston effort.

A plot of T vs θ is known as the turning-moment diagram. The inertia effect of the connecting

rod is, usually ignored while drawing these diagrams, but can be taken into account if desired.

As T=Ft x r, a plot of Ft vs θ (known as crank effort diagram) is identical to a turning-moment

diagram. The turning-moment diagram for different types of engines are being given below:

3.2 Single-Cylinder Double-Acting Steam Engine

Figure shows a turning-moment diagram for a single-cylinder double-acting steam engine. The

crank angle θ is represented along x-axis & the turning-moment along y-axis. It can be observed

that during the outstroke (ogp) the turning moment is maximum when the crank angle is a little

less than 90. & zero with the crank angle is zero & 180. A somewhat similar turning moment di-

agram is obtained and the angle turned.

Note that the area of the turning-moment diagram is proportional to the work done per revolu-

tion as the work is the product of the turning-moment diagram and the angle turned.

The mean torque against which the engine works is given by

𝑜𝑒 =𝐴𝑟𝑒𝑎 𝑜𝑔𝑝𝑘𝑝

2𝜋



Where oe is the mean torque & is the mean height of the turning-moment diagram.

When the crank turns from the angle oa to ob, the work done bt the engine is represented by the

area afghb. But the work done against the resisting torque is represented by afhb. Thus, the en-

gine has work done more work than what has been taken from it. The excess work is repre-

sented by the area fgh. This excess work increases the speed of the engine & is stored in the fly-

wheel.

During the crank travel from the ob or oc, the work needed for the external resistance is propor-

tional to bhjc whereas the work produced by the engine is represented by the area under hpj.

Thus, during this period, more work has been taken from the engine that is produced. The loss

is made up by the flywheel which gives up some of its energy& the speed decreases during this

period.

Similarly, during the period of crank travel from oc to od, excess work is again developed and is

stored in the flywheel & the speed of the engine increases. During the crank travel from od to oa,

the loss of work is made up by the flywheel and the speed again decreases.

The areas fgh, hpj, jkl & lqf represent fluctuations of energy of the flywheel. When the crank is at

b, the flywheel has absorbed energy while the crank has moved from a to b & thereby, the speed

of the engine is maximum. At c, the flywheel has given out energy while the crank has moved

from b to c & thus the engine has a maximum speed. Similarly, the engine speed is again maxi-

mum at d & minimum at a. Thus, there are two maximum & two minimum speeds for the turn-

ing-moment diagram.

The greatest speed is the greater of the two maximum speeds & the least speed is the lesser of

the two minimum speeds.

The difference between the greatest & the least speeds of the engine over one revolution is

known as fluctuation of speed.

3.3 Single cylinder Four Stroke Engine



3.4 Multi Cylinder Engines

3.5 Fluctuation of energy

Let a1, a2, and a5 be the areas in work units of the portions above the mean torque ab of the

turning moment diagram (fig.). These areas represent quantities of energies added to flywheel.

Similarly, areas a2, a4 and a6 below ab represent quantities of energies taken from the flywheel.

The energies of the flywheel corresponding to positions of the crank are as follows:

Crank position Flywheel energy

𝒄

𝒅

𝒆

𝒇

𝒈

𝒉

𝒄

𝐸

𝐸 + 𝑎1

𝐸 + 𝑎1 − 𝑎2

𝐸 + 𝑎1 − 𝑎2 + 𝑎3

𝐸 + 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4

𝐸 + 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4 + 𝑎5

𝐸 + 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4 + 𝑎5 − 𝑎6

From the two values of the energies of the flywheel corresponding to the position c, it is con-

cluded that

𝐸 + 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4 + 𝑎5 − 𝑎6 = 0

The greatest of these energies is the maximum kinetic energy of the flywheel and for the corre-

sponding crank position, the speed is maximum.

The least of these energies is the least kinetic energy of the flywheel and for the corresponding

crank position, the speed is minimum.



The different between the maximum and minimum kinetic energies of the flywheel is known as

the maximum fluctuation of energy whereas the ratio of this maximum fluctuation of energy to

the work done per cycle is define as the coefficient fluctuation of energy.

The different between the greatest speed and the least speed is known as the maximum fluctua-

tion of speed and the ratio of maximum fluctuation of speed to the mean speed is the coefficient

of fluctuation of speed.

3.6 Flywheel

A flywheel is used to control the variations in speed during each cycle of an engine. A flywheel of

suitable dimensions attached to the crankshaft, makes the moment of the rotating part quite

large and thus, acts as a reservoir of energy. During the periods when the supply of energy is

more than required, it stores energy and during the periods the requirements is more than the

supply, it release energy.

𝐼 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑦𝑤ℎ𝑒𝑒𝑙

𝜔1 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑝𝑒𝑒𝑑

𝜔2 = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑠𝑝𝑒𝑒𝑑

𝜔 = 𝑚𝑒𝑎𝑛 𝑠𝑝𝑒𝑒𝑑

E = 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑦𝑤ℎ𝑒𝑒𝑙 𝑎𝑡 𝑚𝑒𝑎𝑛 𝑠𝑝𝑒𝑒𝑑

𝑒 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑐𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦

𝐾 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓𝑙𝑢𝑐𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑝𝑒𝑒𝑑 =𝜔1 − 𝜔2

𝜔

Maximum fluctuation of energy,

𝑒 =1

2𝐼𝜔1

2 −1

2𝐼𝜔2

2 = 1

2𝐼(𝜔1

2 − 𝜔22) = 𝐼 (

𝜔1 + 𝜔2

2) (𝜔1 − 𝜔2)

= 𝐼 ∙ 𝜔 ∙ (𝜔1 − 𝜔2) = 𝐼 ∙ 𝜔2 ∙ (𝜔1 − 𝜔2

𝜔) = 𝐼 ∙ 𝜔2 ∙ 𝐾

∴ 𝐾 =𝑒

𝐼 ∙ 𝑤2=

𝑒

2 ×12 𝐼𝜔2

=𝑒

2𝐸



3.7 Dimensions of Flywheel Rims

The inertia of a flywheel is provided by the hub, spokes and the Rim. However, as the inertia due

to the hub and the spokes is very small, usually it is ignored. In case it is known, it can be taken

into account.

Consider a rim of flywheel as shown in Fig

Let

𝜔 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝑟 = 𝑚𝑒𝑎𝑛 𝑟𝑎𝑑𝑖𝑢𝑠

𝑡 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑖𝑚

𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑖𝑚

Consider an element of the rim,

Centrifugal force on the element/unit length =

= [𝜌 ∙ (𝑟 ∙ 𝑑𝜃) ∙ 𝑡] ∙ 𝑟 ∙ 𝜔2

Total vertical force /unit length

= ∫ 𝜌 ∙ 𝑟2 ∙𝜋

0

𝑑𝜃 ∙ 𝑡 ∙ 𝜔2 ∙ sin 𝜃 = 𝜌. 𝑟2. 𝑡. 𝜔2 ∙ ∫ sin 𝜃 ∙ 𝑑𝜃𝜋

0

= 𝜌 ∙ 𝑟2 ∙ 𝑡 ∙ 𝜔2 ∙ (− cos 𝜃)𝜋

0= 2𝜌 ∙ 𝑟2 ∙ 𝑡 ∙ 𝜔2



Let

𝜎 = 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛𝑑𝑢𝑐𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑖𝑚

(Circumferential stress is also known as hoop stress.)

Then for equilibrium,

𝜎(2𝑡) ∙ 𝑙 = 2𝜌 ∙ 𝑟2 ∙ 𝑡 ∙ 𝜔2

𝜎 = 𝜌 ∙ 𝑟2 ∙ 𝜔2 = 𝜌 ∙ 𝑣2

The above relation provides the limiting tangential velocity at the mean radius of rim of the fly-

wheel. Then the diameter can be calculated from the relation,

𝑣 = 𝜋𝑑𝑛/60

Also, mass = density ×volume = density × circumference× cross-sectional area

𝑚 = 𝜌 × 𝜋𝑑 ∙ 𝑏 ∙ 𝑡

The relation can be used to find the width and the thickness of the rim.

3.8 Punching Press

From the previous discussion, it can be observed that when the load on the crankshaft is con-

stant or varies and the input torque varies continuously during a cycle, a flywheel is used to re-

duce the fluctuation of speed. A flywheel can perform the same purpose in a punching press or a

riveting machine in which the torque available is constant but the load varies during the cycle.



Figure shows the sketch of a punching press. It is a slider crank mechanism in which a punch re-

places the slider. A motor provides a constant torque to the crankshaft through a flywheel. It

may be observed that the actual punching process is performed only during the downward

stroke of the punch and that also for a limiting period when the punch travels through the thick-

ness of the plate. Thus, the load is applied during the actual punching process only and during

the rest of the downward stroke and the return stroke, there is no load on the crankshaft. In the

absence of a flywheel, the decrease in the speed of the crankshaft will be very large during the

actual punching period whereas it will increase to a much higher value during the no-load pe-

riod as the motor will continue to supply the energy all the time.

3.9 Summary

1. Dynamic forces are associated with accelerating masses. As all machines have some

accelerating parts, dynamic forces are always present when the machines operate.

2. D' Alembert's principle states that the inertia forces and couples, and the external forces

and torques on a body together give statically equilibrium.

3. In graphical solutions, it is possible to replace inertia force and inertia couple by an

equivalent offset inertia force which can account for both. This is done by displacing the

line of action of the inertia force from the centre of mass.

4. The sense of angular acceleration of the connecting rod is such that it tends to reduce

the angle of the connecting rod with the line of stroke.

5. The piston effort is the net or effective force applied on the piston.

6. Inertia force on the piston,

𝐹𝑏 = 𝑚 ∙ 𝑓 = 𝑚 ∙ 𝑟 ∙ 𝜔2 (cos 𝜃 +cos 2𝜃

𝑛)

7. Crank effort is the net effort (force) applied at the crankpin perpendicular to the crank

which gives the required turning moment on the crankshaft.

8. Turning moment due to force F on the piston

𝑇 = 𝐹𝑟 (sin 𝜃 +sin 2𝜃

2√𝑛2 − sin 𝜃2)

9. A dynamically equivalent system means that the rigid link is replaced by a link with two

point masses in such a way that it has the same motion as the rigid link when subjected

to the same force, i.e., the centre of mass of the equivalent link has the same linear

acceleration and the link has the same angular acceleration.

10. The distributed mass of a rod can be replaced by two point masses to have the same



dynamical properties if the sum of the two masses is equal to the total mass, the

combined centre of mass coincides with that of the rod and the moment of inertia of two

point masses about the perpendicular axis through their combined centre of mass is

equal to that of the rod.

11. In the analysis of the connecting rod, the two point masses are considered to be located

at the center of the two end bearings and then a correction is applied for the error

involved.

12. A plot of T vs. 𝜃 is known as the turning- moment diagram.

13. The difference between the maximum and minimum kinetic energies of the flywheel is

known as the maximum fluctuation of energy.

14. The difference between the greatest speed and the least speed is known as the

maximum fluctuation of speed.

15. A flywheel is used to control the variations in speed during each cycle of an engine.

16. Coefficient of fluctuation of speed is given by

∴ 𝐾 =𝑒

𝐼𝜔2=

𝑒

2𝐸

.

Department of Mechanical Engineering Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.1

4 Governors

Course Contents

4.1 Introduction.

4.2 Types of Governors.

4.3 Terms used in Governor.

4.4 Watt Governor.

4.5 Porter Governor.

4.6 Proell Governor.

4.7 Terms related with

Governor.

4.8 Controlling Force.

4.9 Problems.

4. Governors Theory of Machines (2151902)

Prepared By: Prof. A. J. Makadia Department of Mechanical Engineering Page 4.2 Darshan Institute of Engineering & Technology, Rajkot

4.1 Introduction

Function of Governor:

The function of a governor is to regulate the meanspeed 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 workingfluid. 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 loadincreases, 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 increasesand the governor

decreases the supply of workingfluid.

4.2Types of Governors 1. Centrifugal Governor.

2. Inertia Governor.

Centrifugal Governor

Pendulum type Loaded Type

Watt Governor

Dead Weight Spring Controlled

Governor Governor

Porter Governor Proell Governor

Hartnell Hartung Wilson – Hartnell Pickering

Governor Governor Governor Governor

Theory of Machines (2151902) 4. Governors


Centrifugal Governor

Fig. 4.1 Centrifugal Governor

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 figure. 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 upper ends of the arms are pivoted to the spindle, so that

the balls may rise up or fall down as they revolve about the vertical axis. The arms are

connected by the links to a sleeve, which is keyed to the spindle. This sleeve revolves with

the spindle; but can slide up and down. The balls and the sleeve rise when the spindle speed

increases, and falls when the speed decreases. In order to limit the travel of the sleeve in

upward and downward directions, two stops S, S are provided on the spindle. The sleeve is

connected by a bell crank lever to a throttle valve. The supply of the working fluid decreases

when the sleeve rises and increases when it falls.

When the load on the engine increases, the engine and the governor speed decreases. This

results in the decrease of centrifugal force on the balls. Hence the balls move inwards and

the sleeve moves downwards. The downward movement of the sleeve operates a throttle

valve at the other end of the bell crank lever to increase the supply of working fluid and thus

the engine speed is increased. In this case, the extra power output is provided to balance

the increased load. When the load on the engine decreases, the engine and the governor

speed increases, which results in the increase of centrifugal force on the balls. Thus the balls

move outwards and the sleeve rises upwards. This upward movement of the sleeve reduces

the supply of the working fluid and hence the speed is decreased. In this case, the power

output is reduced.



4.3Terms used in Governor 1. Height of governor (h): It is the verticaldistance from the centre of the ball to a

pointwherethe axes of the arms (or arms produced) intersecton the spindle axis. It is usually

denoted by h.

2. Equilibrium speed: It is the speed atwhich the governor balls, arms etc., are in complete

equilibrium and the sleeve does not tend to moveupwards or downwards.

3. Mean equilibrium speed: It is the speedat the mean position of the balls or the sleeve.

4. Maximum and Minimum equilibrium speed: The speeds at the maximum and minimum

radius of rotation of the balls, without tending tomove either way are known as maximum

and minimumequilibrium speeds respectively.

5. Sleeve lift: It is the vertical distance which the sleeve travels due to change in equilibrium

speed.

4.4Watt Governor

Fig. 4.2 Watt Governor

The simplest form of a centrifugal governor is a Watt governor, as shown in Figure. 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 figure 4.2 (a).

2. The pivot (P), may be offset from the spindle axis and the arms when produced intersect

at O, as shwon in figure 4.2 (b).

3. The pivot (P), may be offset, but the arms cross the axis at O, as shown in figure 4.2 (c).

Let

m Mass of ballkg,

w Weight of ball innewtons mg,

T Tensionin the arminnewtons,



C

Angular velocity of the arm & ball about the spindle axis inrad / s,

r Radius of the path of rotation of the ball i.e.Horizontal dist. from centre of the

ball to the spindle axis inmetres,

F Centrifugal force acting on the ball innewtons

2m r, and

h Height of governor inmetres.

It is assumed that the weight of arms, link & sleeve are negligible as comparedto the

weight of the balls.

Governor ball is equilibrium under the following

1. Centrifugal force FC

2. Tension (T)

3. Weight of ball (w)

Taking moment about O….

C2

2

2

2 2

F h w r(m r )h m g r

where g gravitational force 9.81m / sg2 Nh

60

9.81 895h metres

N2 N

60

4.5Porter Governor

Fig. 4.3 Porter Governor

The Porter governor is a modification of a Watt’s governor, with central load attached to

the sleeve as shown in Fig. 4.3 (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 predetermined level.

Consider the forces acting on one – half of the governor as shown in Fig. 4.3 (b).

Let

m Mass of eachball inkg,

w Weight of eachball innewtons mg,

W Weight of Central load innewtons mg,

r radius of rotation inmetres,

h Height of governor inmetres,

N Speed of the balls inr.p.m.,

Angular speed of the balls inrad / s 2

2C

1

2

N/ 60,

F Centrifugal force acting on the ball m r,

T Force in the arm innewtons,

T Force in the link innewtons,

Angle of inclination of the arm (or upper link) to the vertical, and

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,

2

2

M gWT cos

2 2

M gor T ................(i)

2 cos

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 figure 4.3(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,

1 2

M gT cos T cos w m g .........(ii)

2



2

M gT cos

2w m g

Resolving the forces horizontally,

1 2 CT sin T sin F

1 C 2

Mg MgT sin sin F T

2 cos 2 cos

1 C

MgT sin tan F

2

1 C

MgT sin F tan ..............(iii)

2

Dividing equation (3) by equation (2)

C

1

1

M g tanF

T sin 2

M gT cos m g2

C

Mg Mg tanm g tan F

2 2

CMg F Mg tanm g

2 tan 2 tan

2C

tan rTake q,tanhtan

F m r

2Mg m r Mgm g q

r2 2h

2 Mgm h m g 1 q

2

2

M gm g 1 q

2h .................(iv)

m

2

M g 1 qm g

2

h m

2 1 qgm M

h m 2



2

2 N g Mm 1 q

60 h m 2

2

2 60 g MN m 1 q

2 h m 2

2Take g 9.81m / s

2 895 MN m 1 q ............(v)

h m 2

Note: 1. When the length of arms are equal to the length of links and the points P & D lie on

the same vertical line, then

tan tan

tan

then q 1tan

Therefore, the equation (5) becomes

2 895N (m M) ..............(vi)

h m

2.When the loaded sleeve moves up & 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 (5) and (6) may be

written as

2 Mg F895N mg 1 q

h m g 2

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.

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. 4.4.

Taking moments about the point I,

C

WF BM w IM ID

2

Fig. 4.4 Instantaneous centre method

C

MgF BM mg IM IM MD

2



C

IM Mg IM MDF mg

BM 2 BM BM

C

MgF mg tan tan tan

2

IM MDtan , tan

BM BM

Dividing the equation by tan…

CF Mg tanmg 1

tan 2 tan

2m r Mg

mg 1 qr 2h

2C

tan rq, tan ,F m rhtan

2 Mgm h mg 1 q

2

2

Mgmg 1 q

2hm

2

g Mh m 1 q

m 2

2

m Mh g

m

When tan tantan

then q 1tan

4.6Proell 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. 4.5 (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. 4.5

(b).

The instantaneous centre (I) lies on the intersection of the line PF produced and the line

from D drawn perpendicular to the spindle axis. The perpendicular BM is drawn on ID.



Fig. 4.5 Proell Governor

Taking moment about I……

C

WF BM w IM ID

2

C

MgF BM mg IM IM MD .........(i)

2

C

IM Mg IM MDF mg

BM 2 BM BM

FM IM Mg IM MD

mg Multiplying and dividing by FMBM FM 2 FM FM

FM Mg IM MD

mg tan tan tan tan , tanBM 2 FM FM

FM Mg tantan mg 1

BM 2 tan

2C

tan rPutting q , tan ,F m rhtan

2 FM r Mgm r mg 1 q

BM h 2

2 FM g Mm 1 q ...........(ii)

BM hm 2

22 N, g 9.81m / s

60

2 FM 895 MN m 1 q ...........(iii)

BM mh 2



4.7 Terms related with Governor

1. 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 sleeveof 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 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

andminimum equilibrium speeds to the mean equilibrium speed.

Let

1

2

1 2

N Minimum equilibrium speed,

N Maximum equilibrium speed,

N Mean equilibrium speed

N N.

2

2 12 1

2 1

2 N NN NSensitiveness

N N N

2 1

1 2

2......(In terms of angular speeds)

2. 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.

A governor is said to be unstable, if the radius of rotation decreases as the speed

increases.



3. 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. The isochronism is the stage of infinite sensitivity.

Let us consider the case of a Porter governor running at speeds N1 and N2 r.p.m.

21

22

895 MN m 1 q ........(i)

h m 2

895 MN m 1 q .........(ii)

h m 2

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.

1C1

21

1 C1 1

M g Sy F x

22 Nx x

M g S 2 F 2 m ry 60 y

.......(iii)

2C2

22

2 C2 2

M g Sy F x

22 Nx x

M g S 2 F 2 m ry 60 y

.......(iv)

For isochronism, N2 = N.Therefore from equation (iii) and (iv),

1 1

2 2

Mg S r(conditionof Isochronous governor)

Mg S r

4. Hunting

A governor is said to be hunt if the speedof 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 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

effectof this will be to cause wide fluctuations in the engine speed or in other words, the

engine will hunt.

4.8 Controlling Force We have seen earlier that when a body rotates in a

circularpath, 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 equaland opposite to the centrifugal reaction.

∴ Controlling force, FC = m.ω2.r Fig. 4.6 Controlling force diagram

The controlling force is provided by the weight of the sleeve and balls as in Porter governor

and by the spring and weight as in Hartnell governor (or spring controlled governor).

When the graph between the controlling force (FC) 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.

1. Controlling Force diagram for Porter Governor

The controlling force diagram for a porter governor is a curve as shown in Fig. 4.12.

2

2C

2 NF m r m r

60

Or

2 2

2 C C

12

F F1 60 1 60N tan tan

m 2 r m 2 r

60 tanN

2 m

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.

2. Controlling Force diagram for Spring - controlled Governors

The controlling force diagram for the spring controlled governors is a straight line, as

shownin Fig. 4.7. We know that controlling force,



2 2C CF m r or F lr m

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. 4.7.

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

CF a r b .......(i)

Where a and b are constants. Fig. 4.7

2. The value of b in equation (i) may be made either zero or positive by increasing the initial

tension of the spring. If b is zero, the controlling force line CD passes through the origin and

the governor becomes isochronous because FC /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,

CF 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.

Such a governor is said to be unstable and the relation between the controlling force and

the radius of rotation is, therefore

CF a r b .......(iii)



4.9 Problems

Problem – 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:

1

2

N 60 rpm

N 61rpm

Let

1 2 21

895895h 0.248 m

N 60

2 2 22

895895h 0.240 m

N 61

Changeinheight 0.248 0.240 0.008m

Problem – 2:-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:

1 2

BP BD 250 mm

m 5kg

M 30 kg

r 150 mm,r 200 mm

F 20 N(Friction)

For Minimum Speed



21

1

1

895 MN m 1 q

h m 2

895 305 1 1

0.2 5 2

31325

N 177 rpm

2 2

1

Arms are same length andmountedon the same axis

tanq 1

tan

h 250 150 200mm 0.2m

For Maximum speed

22

2

2

895 MN m 1 q

h m 2

895 305 1 1

0.15 5 2

41766.67

N 204.4 rpm

2 22

tanHere q 1

tan

h 250 200 150mm 0.15m

Speed range of governor

2 1N N 204.4 177 27.4 rpm

(a) Considering friction F(N) for minimum speed.

21

1

1

1

Mg F895 tanN mg 1 q q 1

h m g 2 tan

895mg Mg F

h m g

8955 9.81 30 9.81 20

0.2 5 9.81

29500

N 172rpm



(b) Considering friction F = 20N for maximum speed.

22

2

2

895N mg Mg F

h m g

8955 9.81 30 9.81 20

0.15 5 9.81

44200

N 210 rpm

Speed range of governor

2 1N N 210 172 38rpm

Problem – 3:-In an engine governor of the Porter type, the upper and lower arms are

200mmand 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 24 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:

PB 200

BD 250

M 15kg

m 2 kg

F 24 N

For Minimum Condition Figure (a)



21 1

1

1

Mg F895N mg 1 q

h m g 2

15 9.81 248952 9.81 1 0.753

0.1732 2 9.81 2

N 183.3rpm

11

2 21

1

1 1

1

11

1

From fig.

hcos30 h 0.1732m

200

r 200 173.20 100.00mm 0.1m

tan tan30 0.5774

100sin 23.578

250tan tan23.578 0.4364

tanq 0.753

tan

For Maximum Condition Figure (b)

22 2

2

2

Mg F895N mg 1 q

m g h 2

15 9.81 248952 9.81 1 0.703

2 9.81 0.1532 2

49236

N 222 rpm

21

22

2

2

22

2

2

22

2

From fig.

hcos40 h 0.1532m

200

tanq

tan

Here tan tan40 0.8390

rsin40 r 0.1286 m

200

0.1286sin 30.957

0.250

tan tan30.957 0.590

tan 0.59q 0.703

tan 0.8390

2 1N N 222Ran 183ge o .3f speed 38.7rpm



Problem – 4:-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:

2 1

BP BD 300 mm

DH 40 mm

M 70 kg

m 10 kg

r 200 mm N ?

F 24 N N N ?

(1) Equilibrium speed at r = 200 mm

2 895 MN m 1 q

h m 2

895 7010 1 0.705

0.224 10 2

27840

N 167 rpm

2 2

2 2

From figure....

h 300 200 0.224 m

tan 200q ,tan 0.893

tan 224

DF 300 160 254 mm 0.254 m

160 0.690tan 0.630, q 0.705

254 0.254

(2) Range of speed when F = 20N.



(a) For minimum condition……

21 1

1

1

Mg F895N mg 1 q

h m g 2

70 9.81 2089510 9.81 1 0.705

0.224 10 9.81 2

27144

N 164.8 rpm

1

1

Here

h 0.224 h

tanq 0.705

tanF 20 N

(b) For maximum condition……

22 2

2

2

Mg F895N mg 1 q

h m g 2

70 9.81 2089510 9.81 1 0.705

0.224 10 9.81 2

28533

N 169 rpm

2

2

Here

h 0.224 h

q q 0.705

F 20 N

2 1N N

169

Ran

16

ge of

4.8

speed

4.2rpm

Problem – 5:-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 150mm at minimum speed and at a radius of 200 mm at maximum

speed. Determine the range of speed.



Solution:

1

2

2 1

PB BD 250

m 3kg

M 20 kg

DH 40 mm

r 150 mm

r 200 mm

N N ?

Let

21

1

1

895 MN m 1 q

h m 2

895 203 2

0.306 3 2

22424

N 150 rpm

1

1

1

tanHere q 1

tan110

and sin 26.10250150

tan26.10 h 0.306 mh

Then

22

2

2

895 MN m 1 q

h m 2

895 203 2 28590

0.240 3 2

N 169 rpm



2 2

2

2

tanHere q 1

tan160

and sin 39.8250

200tan39.8 h 0.240 m

h

2 1Rang N N

16

e of s

9 150 1

peed

9 rpm

Problem – 6:-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 the speed? The mass of each ball is 9 kg.

Solution:

2 C2

1 C1

When r 200 mm F 1200 N

r 100 mm F 450 N

C

C

C

F 4504500

r 0.1

12006000

0.2

Fincrease as r increase so far stability

rconditionF ar b

(1) For the stability of spring controlled governor

CF ar b

1When r r 100 mm

0.1m

450 a 0.1 b .......(1)

2and r r 200 mm

0.2m

1200 a 0.2 b .......(2)

By solving equation (1) & (2)

a 7500

b 300

Now we have



C

250F ar b 7500 0.125 300 r 125 given

2637.5N

2C

2

F m r

637.5 9 0.125

23.80 rad / sec

2 NN 227.3rpm

60

(2) Initial tension on spring for Isochronism

An Isochronism 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 spring to 300 N.

Initial tension on the spring for Isochronism = 300 N

(3) Isochronism speed

C

2

2

For Isochronism......F ' ar

m ' r ar

2 N'9 7500

60

N' 275 rpm


5 Introduction to Dynamics

Course Contents

5.1 Force Convention

5.2 Free-Body Diagrams

5.3 Equilibrium of Two-And-Three

–Force Members

5.4 Member with Two Forces and

a Torque

5.5 Equilibrium of Four-Force

Members

5.6 Superposition

5.7 Mass Moments and Products

of Inertia

5.8 Inertia Forces and D’alem-

bert’s Principle

5.9 The Principle of Superposi-

tion

5.10 Measuring Mass Moment of

Inertia

5. Introduction to Dynamics Theory of Machines (2151902)


5.1 Force Convention

The force exerted by the member i on the member i is represented by Fij.

(a) Points of application of forces F1 and F2 to a rigid body may or may not be important. (b) The

rectangular components of a force vector.

5.2 Free-Body Diagrams

A free-body diagram is a sketch or diagram of a part isolated from the mechanism in order to

determine the nature of forces acting on it.



Fig. (a) Shows a four-link mechanism. The free-body diagrams of its members 2, 3 and 4 are shown

in Figs (b) (c) and (d) respectively. Various forces acting on each member are also shown. As the

mechanism is in static equilibrium, each of its members must be in equilibrium individually.

Member 4 is acted upon by three forces F, F34 and F14.

Member 3 is acted upon by two forces F23 and F43.

Member 2 is acted upon by two forces F32 and F12 and a torque T.

Initially, the direction and the sense of some of the forces may not be known.

Assume that the force F on the member 4 is known completely. To know the other two forces

acting on this member completely, the direction of one more force must be known.

Link 3 is a two-force member and for its equilibrium, F23 and F43 must act along BC. Thus. F34,

being equal and opposite to F43, also acts along BC. For the member 4 to be in equilibrium, F14

passes through the intersection of F and F34, by drawing a force triangle (F is completely known),

magnitudes of F14 and F34 can be known [(e)].

Now, F34 = F43 = F23 = F32

Member 2 will be in equilibrium if F12 is equal, parallel and opposite to F32 and

T = F12 × 𝒉 = F32× 𝒉

5.3 Equilibrium of Two-And-Three –Force Members

A member under the action of two forces will be in equilibrium if

The forces are of the same magnitude,

The force acting along the same line, and

The forces are in opposite directions.

Figure shows such a member. A member under the action of three forces will be in equilibrium if

The resultant of the forces is zero, and The lines of action of the forces intersect at a point (known

as point of concurrency).



(a) Two-force member not in equilibrium; (b) two-force member in equilibrium if FA and FB are

equal, opposite, and share the same line of action; (c) three-force member not in equilibrium; and

(d) three-force member in equilibrium if FA, FB, and FC are coplanar, if their lines of action intersect

at a common point O, and if their vector sum is zero.

Figure (a) shows a member acted upon by three forces F1, F2, and F3 and is in equilibrium as the

lines of action of forces intersect at one point O and the resultant is zero. This is verified by adding

the forces vector ally [(b)]. As the head of the last vector F3 meets the tail of the first vector F1,

the resultant is zero. It is not necessary to add the three vectors in order to obtain the resultant

as is shown in Fig. (c) in which F2 is added to F3 and then F1 is taken.

Figure shows a case where the magnitudes and directions of the forces are the same as before, but

the lines of action of the forces do not intersect at one point. Thus, the member is not in

equilibrium.

Consider a member in equilibrium in which the force F1 is completely known, F2 is known in

direction only and F3 is completely unknown. The point of application of F1, F2, and F3 are A, B

and C respectively. To solve such a problem, first find the point of concurrency O from the two

forces with known directions, i.e., from F1 and F2. Joining O with C gives the line of action of the

third force F3. To know the magnitudes of the forces F2 and F3, take a vector of proper magnitude

and direction to represent the force F1. From its two ends, draw lines parallel to the lines of action

of the forces F2 and F3 forming of force triangle. Mark arrowheads on F2 and F3 so that F1, F2



and F3 are in the same order.

If the lines of the action of the two forces are parallel then the point of concurrency lies at infinity

and, therefore, the third force is also parallel to the first two.

5.4 Member with Two Forces and a Torque

A member under the action of two forces and an applied torque will be in equilibrium if The forces

are equal in magnitude, parallel in direction and opposite in sense, and The forces form a couple

which is equal and opposite to the applied torque.

Figure shows a member acted upon by two equal forces F1 and F2 and an applied torque T. For

equilibrium, T = F1 * h = F2 * h

Where T, F1 and F2 are the magnitudes of T, F1 and F2 respectively is clockwise whereas coupled

formed by F1 and F2 is counter clockwise.

5.5 Equilibrium of Four-Force Members

Normally, in most of the cases the above conditions for equilibrium of a member are found to be

sufficient. However, in some problems, it may be found that the number of forces on a member is

four or even more than that. In such cases, first look for the forces completely known and combine

them into a single force repenting the sum of the known forces. This may reduce the number of

forces acting on a body to two or three. However, in planer mechanism, a four-force system is also

solvable if one force is known completely along with lines of action of the others. The following

example illustrates the procedure.

5.6 Superposition

In linear systems, if a number of loads act on a system of forces, the net effect is equal to the

superposition of the effects of the individual loads taken one at a time. A linear system is one in

which the output force is directly proportional to the input force, i.e., in mechanisms where

coulomb or dry friction is neglected.



5.7 Mass Moments and Products of Inertia

Another problem that often arises when forces are distributed over an area or volume is that of

calculating their moment about a specified point or axis of rotation. Sometimes the force intensity

varies according to its distance from the point or axis of rotation. Although we will save a more

through derivation of this equation until previous Section and later, we will point out here that

such problems always give rise to integrals of the form ʃ(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2𝑑𝑚

In three-dimensional problems, three such integrals are defined as follows:

(𝐼)𝑥𝑥 = ∫(Î ∗ 𝑅)(Î ∗ 𝑅)𝑑𝑚 = ʃ [(Ry)2 + (Rz)2]𝑑𝑚

(𝐼)𝑦𝑦 = ∫(𝑗̂ ∗ 𝑅)(𝑗̂ ∗ 𝑅)𝑑𝑚 = ʃ [(Rz)2 + (Rx)2]𝑑𝑚

(𝐼)𝑧𝑧 = ∫(�̂� ∗ 𝑅)(�̂� ∗ 𝑅)𝑑𝑚 = ʃ [(Rx)2 + (Ry)2]𝑑𝑚

These three integrals are called the mass moment for inertia of the body. Another three similar

integrals are

(𝐼)𝑥𝑦 = (𝐼)𝑦𝑥 = ∫(Î ∗ 𝑅)(𝑗̂ ∗ 𝑅)𝑑𝑚 = −ʃ (𝑅𝑥𝑅𝑦)𝑑𝑚

(𝐼)𝑦𝑧 = (𝐼)𝑧𝑦 = ∫(𝑗̂ ∗ 𝑅)(�̂� ∗ 𝑅)𝑑𝑚 = −ʃ (𝑅𝑦𝑅𝑧)𝑑𝑚

(𝐼)𝑧𝑥 = (𝐼)𝑥𝑧 = ∫(�̂� ∗ 𝑅)(Î ∗ 𝑅)𝑑𝑚 = −ʃ (𝑅𝑧𝑅𝑥)𝑑𝑚

And these three integrals are called the mass products of inertia of body. Sometimes it is

convenient to arrange these mass moments of inertia and the mass product of inertia into a

symmetric square array or matrix format called the inertia tensor of the body:

𝐼 = [𝐼𝑥𝑥 −𝐼𝑥𝑦 −𝐼𝑥𝑧

𝐼𝑦𝑥 𝐼𝑦𝑦 −𝐼𝑦𝑧

−𝐼𝑧𝑥 −𝐼𝑧𝑦 𝐼𝑧𝑧]

A careful look at the above integrals will indicate that they represents the mass distribution of the

body with respect to the coordinate system about which they are determine, but that they change

if evaluated in a different coordinate system.to keep their meaning direct and simple ,we assume

that the coordinate system chosen for each body is attached to that body in a convenient location

and orientation .therefore, for rigid bodies ,the mass moments and products of inertia are



constant properties of body and it’s mass distribution and they do not changed when the body is

moves; they do ,however ,depend on the coordinate system chosen.

An interesting property of these integrals is that it is always possible to choose the coordinate

system so that it is origin it located at the center of mass of the body and oriented such that all of

the mass product of inertia become zero such choice of coordinate axes of the body is called is its

principal axes, and the corresponding values of eqs. (14.7) are then called the principal mass

moment of inertia. A variety of simple geometric solid, the orientation of their principal axes, and

formulas for their principal mass moments of inertia are included in Appendix A.

If we note that mass moments of inertia have units of mass times distance squared, it seems

natural to define a radius value of the body as

𝐼𝐺 = 𝑘2𝑚 𝑜𝑟 𝑘 =√𝐼

𝑚

This distance k is called the radius of gyration of the bodies, and it is always calculated or means

med from the center of mass of the part about one of the principal axes. For three-dimensional

motions of parts there are three radii of gyration kx, ky and kz associated with the three principal

axes, Ixx, Iyy and Izz.

It is often necessary to determine the moments and products of inertia of bodies, which are

composed of several simpler sub shapes for which formula are known, such as those given in table

5 in Appendix A. The easiest method of finding these is to compute the mass moment about the

principal axes of each sub shape, then to shift the origins of each to the mass center of the

composite body, and then to sum the results .This require that we develop methods of redefining

mass moments and products of inertia when the axes are translated to a new position. The form

of the transfer, or parallel-axis theorem for mass moment of inertia, is written

𝐼 = 𝐼𝐺 + 𝑚𝑑2

Where IG is one of the principal mass moments of inertia about some known principal axis and I

is mass moment of inertia about a parallel axis at distance d from that principal axis. Equation

must be used for translation of inertia axes starting from a principal axis. Also, the rotation of

these axes results in the introduction of product of inertia terms. More will be said in general

transformation of inertia in section.

Only one mass moment of inertia Izz was requested or terminated. This does not mean that Ixx and



Iyy are zero , but rather that will likely not be needed for further analysis .In problems with only

planer motion, only Izz is needed because Ixx and Iyy are used only rotation out of the xy plan. These

other mass moment and products of inertia are used in a section and later, where we treat

problems with spatial motion, and they are determine in identical fashion.

5.8 Inertia Forces and D’alembert’s Principle

Next, let us consider a moving rigid body of mass m acted upon by any system of forces, say F1,

F2, and F3, as illustrated in fig. We designate the center of mass of the body as point G, and we find

the resultant of the system of forces from the equation

∑F = 𝐹1 + 𝐹2 + 𝐹3

(a) An unbalanced set of forces on a rigid body. (b) The accelerations that result from the

unbalanced forces.

In the general case, the line of action of this resultant will not be through the mass center but will

be displaced by some distance, illustrated in fig. (a) as distance h. We demonstrated in Eq. that the

effect of this unbalanced force system is to produce an acceleration of the center of mass of the

body:

∑𝐹𝑖𝑗 = 𝑚𝑖𝐴𝐺𝑗

In a very similar way, it has been proven that the unbalanced moment effect of this resultant force

about the center of mass causes angular acceleration of the body that obeys the equation:

∑𝑀𝐺 = 𝐼𝐺𝑗𝛼𝑗



However, this equation is restricted to use in taking moments about the center of mass G. It cannot

be used for taking moments about an arbitrary point.

The quantity ∑F is the resultant of all external forces acting upon the body, and ∑MG is the sum

of all applied external moments and the moments of all externally applied forces about point G.

The mass moment of inertia is designated as IG, Signifying that it must be taken with respect to

the mass center G.

Equations demonstrate that when an unbalanced system of forces acts upon a rigid body, the body

experiences a rectilinear acceleration AG of its mass center in the same direction as the resultant

force ∑F. The body also experiences an angular acceleration α in the same direction as the

resultant moment ∑MG, caused by the moments of the forces and the torques about the mass

center. This situation is illustrated in fig. (b). If the forces and moments are known, Eqs. may be

used to determine the resulting acceleration pattern that is, the resulting motion of the body.

During engineering design, however, the motions of the machine members are often specified in

advance by other machine requirements. The problem then is this: given the motions of the

machine elements, what forces are required to produce these motions? The problem requires (1)

a kinematic analysis to determine the translational and rotational accelerations of the various

members and (2) definitions of the actual shapes, dimensions, and material specifications to

determine the centroids and mass moments of inertia of the members. In the examples presented

here, only the results of the kinematic analysis are included because methods of finding these have

been presented in previous Chapter. The selection of the materials, shapes and many of the

dimensions of machine members form the subject of machine design and are not further

discussed here.

In the dynamic analysis of machines, the acceleration vectors are usually known; therefore, an

alternative form of Equations. Are often convenient in determining the forces required to produce

these known accelerations. Thus, we can write

∑𝐹 + (−𝑚𝐴𝐺) = 0

And

∑𝑀𝐺 + (−𝐼𝐺𝛼) = 0

Both of these are vector equations applying to the planar motion of a rigid body. Equation states

that vector sum of all external forces acting upon the body plus the fictitious force -mAG sum to



zero. This new fictitious force -mAG is called an inertia force. It has the same line of action as the

absolute acceleration AG, but is opposite in sense. Equation states that the sum of all external

moments and the moments of all external forces acting upon the body about an axis through G

and perpendicular to the plane of motion plus the fictitious torque -IGα sum to zero. This new

fictitious torque -IGα is called an inertia torque. The inertia torque is opposite in sense to the

angular acceleration vector α. We recall that Newton’s first law states that a body perseveres in

its state of uniform motion except when compelled to change by impressed; in other words, bodies

resist any change in motion. In a sense, we can picture the fictitious inertia force and inertia torque

vectors as resistances of the body to the change of motion required by the net unbalanced forces.

(a) Unbalanced forces and resulting accelerations. (b) Inertia force and inertia couple. (c) Inertia

force offset from center of mass.

Equations are known as d’Alembert’s principle, because d’Alembert was the first to call attention

to the fact that addition of the inertia force and inertia torque to the real system of forces and

torques enables a solution from the equations of static equilibrium. We note that the equations

can also be written

∑𝐹 = 0 𝑎𝑛𝑑 ∑M = 0

Where it is understood that both the external and the inertia forces and torques are to be included

in the summations. Equations are useful because they permit us to take the summation of

moments about any axis perpendicular to the plane of motion.

D’Alembert’s principle is summarized as follows: The vector sum of all external forces and inertia

forces acting upon a system of rigid bodies is zero. The vector sum of all external moments and

inertia torques acting upon a system of rigid bodies is also separately zero.

When a graphical solution by a force polygon is desired, Eqs. can be combined. In Fig. (a), link 3 is

acted upon by the external forces F23 and F43. The resultant F23 + F43 produces an acceleration



of the center of mass AG and an angular acceleration of link α3 because the line of action of the

resultant does not pass through the center of mass. Representing the inertia torque -IGα3 as a

couple, as illustrated in Fig. (b), we intentionally choose the two forces of this couple to +be m3AG.

For the moment of the couple to be of magnitude -IGα, the distance between the forces of the

couple must be

ℎ =𝐼𝐺α3

m3AG

Because of this particular choice for the couple, one force of the couple exactly balances the inertia

forces itself and leaves only a single forces, as illustrated in Fig. (c). This force includes the

combined effects of the inertia force and the inertia torque, yet appears as only a single inertia

force offset by the distance h to give the effect of the inertia torque.

5.9 The Principle of Superposition

Linear systems are those in which effect is proportional to cause. This means that the response or

output of a linear system is directly proportional to the drive or input to the system. An example

of a linear system is a spring, where the deflection (output) is directly proportional to the force

(input) exerted on the spring.

The principle of superposition may be used to solve problems involving linear systems by

considering each of the inputs to the systems separately. If the system is linear, the responses to

each of these inputs can be summed or superposed on each other to determine the total response

of the system.

Thus, the principle of superposition states that for a linear system the individual responses to

several disturbances, or driving functions, can be superposed on each other to obtain the total

response of the system.

The principle of superposition does not apply to non-linear systems. Some examples of non-linear

systems, where superposition may not be used, are systems with static or Coulomb friction,

systems with clearances or backlash, or systems with springs that change stiffness as they are

deflected.

We have now reviewed all of the principles necessary for making a complete dynamic-force

analysis of a planar motion mechanism. The steps in using the principle of superposition for



making such an analysis are summarized as follows:

1. Perform a kinematic analysis of the mechanism. Locate the centre of mass of each link and

find its acceleration; also find the angular acceleration of each link.

2. If the inertia forces are attached to all links simultaneously, along with other applied forces

and moments, then there are often no two-or-three-force members and it may become diffi-

cult to find the lines of action of unknown constraint forces. Instead of doing this, it is some-

times more convenient to ignore the masses and applied forces and moments on all but one

or two links and to leave other links as two-or-three-force members. By choosing in this man-

ner, a solution may become possible for the constraint forces caused by the masses or applied

forces and moments being considered, but without those caused by the masses and applied

forces and moments being ignored.

3. Those masses and applied forces and moments considered in Step 2 can now be ignored while

a solution is obtained for additional constraint force components caused by some of the pre-

viously ignored masses or applied forces and moments. This process can be continued until

constraint force components caused by all masses and all applied forces and moments are

found.

4. The results of Steps 2 and 3 can now be vectorially added to obtain the resultant forces and

torques on each link caused by the combined effects of all masses and all applied forces and

moments.

5.10 Measuring Mass Moment of Inertia

(a) Simple pendulum. (b) Torsional pendulum.

Sometimes the shapes of machine parts are so complicated that it is extremely tedious and time



consuming to calculate the mass moment(s) of inertia. Consider, for example, the problem of

finding the mass moment of inertia of an automobile body about a vertical axis through its center

of mass. For such problems it is usually possible to determine the mass moment of inertia of a

body by observing its dynamic behavior caused by a known rotational disturbance.

Many bodies, connecting rods and cranks for example, are shaped so that their masses can be

assumed to lie in a single plane. Once such bodies have been weighed and their mass centers

located, they can be suspended like a pendulum and caused to oscillate. The mass moment of

inertia of such a body can then be computed from an observation of its period or frequency of

oscillation. For best experimental results, the part should be suspended with the pivot located

close to, but not coincident with, the center of mass. If is not usually necessary to drill a hole to

suspend the body; for example, a spoked wheel or a gear can be suspended on a knife edge at its

rim.

When the body of Fig. a is displaced through an angle θ, a gravity force mg acts at G. Summing

moments about the pivot O gives

∑ 𝑀0 = −𝑚𝑔(𝑟𝐺 sin 𝜃) − 𝐼0�̈�

We intend that the pendulum be displaced only through a small angle, so that sinθ can be

approximated by θ. Equation (a) can then be written as

θ̈ +𝑚𝑔𝑟𝐺

𝐼0𝜃 = 0

This second- order, linear differential equation has the well- known solution

𝜃 = 𝑐1 sin √𝑚𝑔𝑟𝐺

𝐼0𝑡 + 𝑐2 cos √

𝑚𝑔𝑟𝐺

𝐼0𝑡

Where c1 and c2 are constants of integration.

We shall start pendulum motion by displacing it through a small angle θ0 and releasing it with no

initial velocity from this position. Thus, at time t = 0, we obtain θ = θ0 and θ̇ = 0. Substituting these

conditions into Eq. (c) and its first time derivative enables the two constants. They are c1 = 0 and

c2 = θ0. Therefore,



𝜃 = 𝜃0 cos √𝑚𝑔𝑟𝐺

𝐼0𝑡

Because a cosine function repeats itself every 360𝑜 or 2π radians, the period of the motion is

𝜏 = 2𝜋√𝐼0

𝑚𝑔𝑟𝐺

Therefore, the mass moment of inertia of the body about the pivot O is

𝐼0 = 𝑚𝑔𝑟𝐺 (𝜏

2𝜋)

2

This equation demonstrates that the body must be weighted to get mg, the distance rG must be

measured, and then pendulum must be suspended and oscillated so that the period τ can be

observed; Eq. can then be solved to give I0 about O. If the moment of inertia about the mass center

is desired, it can be obtained by using the parallel axis theorem, Eq.

Figure b illustrates how the mass moment of inertia can be determined without actually weighing

the body. The body is connected to a slender rod or wire at the mass center. The torsional stiffness

kt of the rod or wire is defined as the torque necessary to twist the rod through a unit angle. If the

body of Fig. b is turned through a small angle θ and released, the equation of motion becomes

�̈� +𝑘𝑡

𝐼𝐺𝜃 = 0

This is similar to Eq. (b) and, with the same starting conditions, has the solution

𝜃 = 𝜃0 cos √𝑘𝑡

𝐼𝐺𝑡

The period of oscillation is then

𝜏 = 2𝜋√𝐼𝐺

𝑘𝑡

and



𝐼𝐺 = 𝑘𝑡 (𝜏

2𝜋)

2

The torsional stiffness is often known or can be computed from knowledge of the length and

diameter of the rod or wire and its material. Then the oscillation of the body can be observed and

Eq. can be used to compute the mass moment of inertia IG. Alternatively, when the torsional

stiffness kt is unknown, a body with known mass moment of inertia IG can be mounted and Eq.

can be used to determine kt.

A trifilar pendulum, also called a three-string torsional pendulum, illustrated in Fig., can provide

a very accurate method of measuring mass moment of inertia. Three strings of equal length

support a lightweight platform and are equally spaced about its center. A round platform serves

just as well as the triangular one shown. The part whose mass moment of inertia is to be

determined is carefully placed on the platform so that the center of mass of the object coincides

with the platform center. The platform is then made to oscillate, and the number of oscillations is

counted over a specified period of time.

The notation for the three-string torsional pendulum analysis is as follows.

𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑟𝑡

𝑚𝑝 = mass of the platform

𝐼𝐺 = mass moment of inertia of the part

𝐼𝑝 = mass moment of inertia of the platform

𝑟 = 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚 𝑟𝑎𝑑𝑖𝑢𝑠

𝜃 = 𝑠𝑡𝑟𝑖𝑛𝑔 𝑙𝑒𝑛𝑔𝑡ℎ

∅ = string angular displacement

𝑧 = vertical axis through the center of the platform

We begin by writing the summation of moments about the z axis. This gives

∑𝑀𝑧 = −𝑟(𝑚 + 𝑚𝑝)𝑔 sin ∅ − (𝐼𝐺 + 𝐼𝑝)�̈� = 0

Because we are assuming small displacements, the sine of an angle can be approximated by the

angle itself. Therefore,



∅ ≈𝑟

𝑙𝜃

And Eq. becomes

𝜃 +(𝑚 + 𝑚𝑝)𝑔𝑟2

(𝐼𝐺 + 𝐼𝑝)𝑙= 0

This equation can be solved in the same manner as Eq. (b). The result is

𝐼𝐺 + 𝐼𝑝 =(𝑚 + 𝑚𝑝)𝑔𝑟2

𝑙(

𝜏

2𝜋)

This equation should be used first with an empty platform to find Ip. With mp and Ip known, the

equation can then be used to find IG of the part being measured.

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