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Module 12

Design of brakes

Version 2 ME , IIT Kharagpur

Lesson 1

Design of shoe brakes Version 2 ME , IIT Kharagpur

Instructional Objectives: After reading the lesson the students should learn:

• Different types of shoe brakes and their operating principles

• Design procedure of different shoe brakes

1. Types of brakes

Brakes are devices that dissipate kinetic energy of the moving parts of a

machine. In mechanical brakes the dissipation is achieved through sliding

friction between a stationary object and a rotating part. Depending upon the

direction of application of braking force, the mechanical brakes are primarily of

three types

• Shoe or block brakes – braking force applied radially

• Band brakes – braking force applied tangentially.

• Disc brake – braking force applied axially.

2. Shoe or block brake In a shoe brake the rotating drum is brought in contact with the shoe by suitable

force. The contacting surface of the shoe is coated with friction material.

Different types of shoe brakes are used, viz., single shoe brake, double shoe

brake, internal expanding brake, external expanding brake. These are

sketched in figure 12.1.1.

drum

shoe

lever

Figure 1(a) Single shoe brake


drum

shoe

lever

Figure 1(b) Double shoe brake

drum

shoe

Figure 1(c): Internal expanding shoe brake


drum

shoe

Figure 1(d): External expanding shoe brake

Figure 12.1.1: Different shoe brakes

Single Shoe brake The force needed to secure contact is supplied by a lever. When a force F is

applied to the shoe (see figure 12.1.2a ) frictional force proportional to the

applied force 'frF Fμ= develops, where 'μ depends of friction material and the

geometry of the shoe. A simplified analysis is done as discussed below.

P F

O

Ffr

Figure 12.1.2a: Free body diagram of a brake shoe

Though the exact nature of the contact pressure distribution is unknown, an

approximation (based on wear considerations) is made as


0( ) cosp pθ θ=

Where the angle is measured from the centerline of the shoe. If Coulomb’s law

of friction is assumed to hold good, then

θμθ cos)( 0pf fr =

Since the net normal force of the drum is F, one has

0

0

( )cos ,Rb p d Fθ

θ

θ θ θ−

=∫

Where R and b are the radius of the brake drum and width of the shoe

respectively.

The total frictional torque is

∫−

=0

0

2)(θ

θ

θθ dRfbT fr

If the total frictional force is assumed to be a concentrated one, then the

equivalent force becomes frTFR

= . A simple calculation yields,

0

0 0

4 sin2 sin 2

μ θμθ θ

′ =+

2θ

Figure 12.1.2(b): Pressure distribution on brake


It may be seen that for very small value of 0 , '.θ μ μ= Even when , 00 30θ =

' 1.0453 .μ μ= Usually if the contact angle is below , the two values of friction

coefficient are taken to be equal.

060

Consider, now single shoe brakes as shown in figures 12.1.3(a) and 3(b).

Suppose a force P is applied at the end of a lever arm with length l. The shoe

placed at a distance x from the hinge experiences a normal force N and a

friction force F, whose direction depends upon the sense of rotation of the

drum. Drawing free body diagram of the lever and taking moment about the

hinge one gets

(a) for clockwise rotation of the brake wheel,

Nx + Fa = Pl

(b) for anticlockwise rotation of the brake wheel,

Nx – Fa = Pl.

Where a is the distance between the hinge and the line of action of F and is

measured positive when F acts below point O as shown in the figure. Using

Coulomb’s law of friction the following results are obtained,

(a) for clockwise rotation PlFx aμμ

=+

,

(b) for anticlockwise rotation PlFx aμμ

=−

,

It may be noted that for anticlockwise rotating brake, if xa

μ > , then the force P

has negative value implying that a force is to applied in the opposite direction to

bring the lever to equilibrium. Without any force the shoe will, in this case, draw

the lever closer to the drum by itself. This kind of brake is known as ‘self-

locking, brake. Two points deserve attention.

(1) If a < 0, the drum brake with clockwise rotation becomes self-energizing and

if friction is large, may be self locking.

(2) If the brake is self locking for one direction, it is never self locking for the

opposite direction. This makes the self locking brakes useful for ‘back stop’s

of the rotors.


Double shoe brake

P

N

lx

Fa

Figure 12.1.3(a): FBD of shoe (CW drum rotation)

P

N

lx

Fa

Figure 12.1.3(b): FBD of shoe (CCW drum rotation)

Since in a single shoe brake normal force introduces transverse loading on the

shaft on which the brake drum is mounted two shoes are often used to provide

braking torque. The opposite forces on two shoes minimize the transverse

loading. The analysis of the double shoe brake is very similar to the single shoe

brake.

External expanding shoe brake An external expanding shoe brake consists of two symmetrically placed shoes

having inner surfaces coated with frictional lining. Each shoe can rotate about

respective fulcrum (say, and ). A schematic diagram with only one shoe is 1O 2O


presented (figure 12.1.4) When the shoes are engaged, non-uniform pressure

develops between the friction lining and the drum. The pressure is assumed to

be proportional to wear which is in turn proportional to the perpendicular

distance from pivoting point (O1N in figure 12.1.4). A simple geometrical

consideration reveals that this distance is proportional to sine of the angle

between the line joining the pivot and the center of the drum and the line joining

the center and the chosen point. This means

0( ) sin , p pθ θ=

where the angle is measured from line OO1 and is limited as 1 2.θ θ θ≤ ≤

Drawing the free body diagram of one of the shoes (left shoe, for example) and

writing the moment equilibrium equation about (say) the following equation is

resulted for clockwise rotation of the drum :

1O

1 ,p fF l M M= −

Where is the force applied at the end of the shoe, and 1F

( )0 2 1 1 21 1( ) sin 2 sin 2 ,2 2pM p bRδ θ θ θ θ⎡ ⎤= − + −⎢ ⎥⎣ ⎦

( )0 1 2 11 (cos ) cos 2 cos 2 ,2 4fM bR R δμ δ θ θ θ θ2

⎡ ⎤= − − −⎢ ⎥⎣ ⎦

where δ is the distance between the center and the pivot (OO1 in figure 12.1.4)

and is the distance from the pivot to the line of action of the force (O1F 1C in

the figure). In a similar manner the force to be applied at the other shoe can be

obtained from the equation

2 .p fF l M M= +

The net braking torque in this case is 2

0 1(cos cos ).T p bR 2μ θ θ= −


C

B

A

F

θN O

O1

Figure 12.1.4: Force distribution in externally expanding brake.

Internal expanding shoe brake Here the brake shoes are engaged with the internal surface of the drum.

The analysis runs in the similar fashion as that of an external shoe brake.

The forces required are

1 ( )p fF M M= + l

and 2 ( )p fF M M l= − ,

respectively.

One of the important member of the expanding shoe brakes is the anchor

pin. The size of the pin is to be properly selected depending upon the face

acting on it during brake engagement.


Module 12

Design of Brakes Version 2 ME , IIT Kharagpur

Lesson 2

Design of Band and Disc Brakes


Instructional Objectives: After reading this lesson the students should learn:

• Different types of band brakes

• Design of band brakes

• Design of disc brakes

• Properties of friction materials

1. Band brakes: The operating principle of this type of brake is the following. A flexible band of

leather or rope or steel with friction lining is wound round a drum. Frictional

torque is generated when tension is applied to the band. It is known (see any

text book on engineering mechanics) that the tensions in the two ends of the

band are unequal because of friction and bear the following relationship:

1

2

,T eT

μ β=

where = tension in the taut side, 1T

= tension in the slack side, 2T

μ = coefficient of kinetic friction and

β = angle of wrap.

If the band is wound around a drum of radius R, then the braking torque is

( ) ( )1 2 1 1brT T T R T e μ β−= − = − R

Depending upon the connection of the band to the lever arm, the member

responsible for application of the tensions, the band brakes are of two types,

(a) Simple band brake: In simple band brake one end of the band is attached to the fulcrum of the

lever arm (see figures 12.2.1(a) and 1(b) ). The required force to be applied to

the lever is

1bP Tl

= for clockwise rotation of the brake drum and


2bP Tl

= for anticlockwise rotation of the brake drum,

where l = length of the lever arm and

b = perpendicular distance from the fulcrum to the point of attachment of

other end of the band.

Figure 12.2.1: Band brakes

(b) Differential band brake: In this type of band brake, two ends of the band are attached to two points on

the lever arm other than fulcrum (see figures 12.2.2(a) and 12.2.2(b)). Drawing

the free body diagram of the lever arm and taking moment about the fulcrum it

is found that

2 1aP T Tl l

= −b , for clockwise rotation of the brake drum and

1 2aP T Tl l

= −b , for anticlockwise rotation of the brake drum.

Hence, P is negative if

b l

T1 T2

P

1(a): Band brake with CW rotating drum

b l

T2 T1

P

1(b): Band brake with CCW rotating drum


1

2

T aeT b

μ β = > for clockwise rotation of the brake drum

and 1

2

T aeT b

μ β = < for counterclockwise rotation of the brake drum. In

these cases the force is to be applied on the lever arm in opposite direction to

maintain equilibrium. The brakes are then self locking.

The important design variables of a band brake are the thickness

and width of the band. Since the band is likely to fail in tension, the following

relationship is to be satisfied for safe operation.

1 TT wts=

where w = width of the band,

t = thickness of the band and

= allowable tensile stress of the band material. The steel bands of the

following dimensions are normally used

Ts

w 25-40 mm 40-60 mm 80 mm 100 mm 140-200

mm

t 3 mm 3-4 mm 4-6 mm 4-7 mm 6-10 mm

Fig.12.2.2(a): Differential Band brake with CW rotation

b l

T1

T2

P

a


b l

T2

T1

P

a

Fig 12.2.2(b): Differential Band brake with CCW rotation

2. Band and block brakes: Sometimes instead of applying continuous friction lining along the band, blocks

of wood or other frictional materials are inserted between the band and the

drum. In this case the tensions within the band at both sides of a block bear

the relation

1

1

1 tan1 tan

TT

μ θμ θ

+=

′ −,

where = tension at the taut side of any block 1T

= tension at the slack side of the same block 1T ′

2θ = angle subtended by each block at center.

If n number of blocks are used then the ratio between the tensions at taut side to slack side becomes

1

2

1 tan1 tan

nTT

μ θμ θ

⎛ ⎞+= ⎜ ⎟−⎝ ⎠

.

The braking torque is ( )1 2brT T T= − R 3. Disc brake:


In this type of brake two friction pads are pressed axially against a rotating

disc to dissipate kinetic energy. The working principle is very similar to friction

clutch. When the pads are new the pressure distribution at pad-disc interface

is uniform, i.e.

constantp = .

If F is the total axial force applied then FpA

= , where A is the area of the pad.

The frictional torque is given by

brakingA

FT rA

dAμ= ∫

where μ = coefficient of kinetic friction and r is the radial distance of an

infinitesimal element of pad. After some time the pad gradually wears

away. The wear becomes uniforms after sufficiently long time, when

constant = (say)pr c=

where dAF p dA cr

= =∫ ∫ . The braking torque is

'brakingAFT pr dA Ac dAr

μμ μ= = =∫∫

It is clear that the total braking torque depends on the geometry of the pad. If

the annular pad is used then 3 3

1 22 2

1 2

23br

R RT FR R

μ⎛ ⎞−

= ⎜ ⎟−⎝ ⎠

1 2

2brR RT Fμ +⎛ ⎞′ = ⎜ ⎟

⎝ ⎠

where 1 and 2R R are the inner and outer radius of the pad.


4. Friction materials and their properties. The most important member in a mechanical brake is the friction material. A

good friction material is required to possess the following properties:

• High and reproducible coefficient of friction.

• Imperviousness to environmental conditions.

• Ability to withstand high temperature (thermal stability)

• High wear resistance.

• Flexibility and conformability to any surface.

Some common friction materials are woven cotton lining, woven asbestos

lining, molded asbestos lining, molded asbestos pad, Sintered metal pads etc.

Review questions and answers: Q.1. A double shoe brake has diameter of brake drum 300mm and contact

angle of each shoe 90 degrees, as shown in figure below. If the coefficient of

friction for the brake lining and the drum is 0.4, find the spring force necessary

to transmit a torque of 30 Nm. Also determine the width of the brake shoe if the

braking pressure on the lining material is not to exceed 0.28 MPa.

S S

250

225

Figure 12.2.3

Ans. The friction force required to produce the given torque is


( )1 230 200

0.150F F N+ = =

The normal forces on the shoes are 11 2, N ,

' 'FN 2Fμ μ

= = where

00

0 0

4 sin' ( ) 0.44.2 sin 2 4

μ θ πμ θθ θ

= =+

= Writing the moment equilibrium equations about

the pivot points of individual shoes (draw correct FBDs and verify)

1 1 10 F 0.718 ,

'

SlSl N x Fa Sxaμ

− + + = ⇒ = =+

and

2 2 20 F 1.1314

'

SlSl N x F a Sx aμ

− + = ⇒ = =−

This yields S = 98.4(N).

Width of the friction lining :

According to the pressure distribution assumed for a shoe brake, the maximum

bearing pressure occurs at the centerline of the shoe. The width of the brake

lining must be selected from the higher values of the normal forces, in this

case . Noting that 2N

/ 42

2 max/ 4

cos ,N Rbp dπ

π

θ θ−

= ∫

Where R = 0.150, the value of b is

calculated to be 5.4 mm or 6 mm (approx.).

6max 20.28 10 , N 1.314 98.4 / 0.44,p X= = ×

Q2. A differential band brake has brake drum of diameter 500mm and the

maximum torque on the drum is 1000 N-m. The brake embraces 2/3rd of the

circumference. If the band brake is lined with asbestos fabric having a

coefficient of friction 0.3, then design the steel band. The permissible stress is

70 MPa in tesnion. The bearing pressure for the brake lining should not

exceed 0.2 MPa.


Ans. The design of belt is to be carried out when the braking torque is

maximum i.e. Tbr = 1000 N-m. According to the principle of band brake

25.01)1( 343.0

11 ×⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

×−−π

μβ eTReTTbr

Which yield In order to find the pressure on

the band, consider an infinitesimal element. The force balance along the radial

direction yields

1 2 15587 , T T 1587 .T N e μ β−= = = N

N T θ= Δ

Since N p b R θ= Δ so p = TbR

.

The maximum pressure is 1max

TpbR

= .

Hence 6

55870.25 0.2 10

b = 0.112 m (approx.)

Δθ

T

T+ΔT

F

N

=× ×

The thickness t of the band is calculated from the relation

1tS bt T=

Which yields 6

558770 10 0.1117

t =× ×

= 0.0007145 m or 1 mm (approx.).


Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

MODULE III

Brakes, Clutches and Flywheel,

Brakes

• A brake is a device by means of which artificial resistance is applied on

to a moving machine member in order to retard or stop the motion of the

member or machine

Types of Brakes

• Different types of brakes are used in different applications

• Based on the working principle used brakes can be classified as

mechanical brakes, hydraulic brakes, electrical (eddy current) magnetic

and electro-magnetic types.

Mechanical Brakes

• Mechanical brakes are invariably based on the frictional resistance principles•

In mechanical brakes artificial resistances created using frictional contact

between the moving member and a stationary member, to retard or stop the

motion of the moving member.

Basic mechanism of braking

The illustration below explains the working of mechanical brakes. An element

dA of the stationary member is shown with the braked body moving past at

velocity v. When the brake is actuated contact is established between the

stationary and moving member and a normal pressure is developed in the

contact region. The elemental normal force dN is equal to the product of contact

pressure p and area of contact dA. As one member is stationary and the other is

in relative motion, a frictional force dF is developed between the members. The



magnitude of the frictional force is equal to the co-efficient of friction times the

normal force dN

dA

v dFf =µ.dN=µ.p.dA

dN=p.dA

Figue 3.1.1

The moment of the frictional force relative to the point of motion contributes to the

retardation of motion and braking. The basic mechanism of braking is illustrated

above.

Design and Analysis

To design, select or analyze the performance of these devices knowledge on the

following are required.

• The braking torque

• The actuating force needed

• The energy loss and temperature rise

At this beginning stage attention will be focused mainly on some preliminary

analysis related to these aspects, namely torque, actuating force, energy

absorbed and temperature rise. Torque induced is related to the actuating force,

the geometry of the member and other contact conditions. Most mechanical

brakes that work on the frictional contact basis are classified based on the

geometry.

There are two major classes of brakes, namely drum brakes and disc brakes.

Design and analysis of drum brakes will be considered in detail in following



sections, the discussion that follow on disc or plate clutches will from the basis

for design of disc type of brakes.

Drum brakes basically consists of a rotating body called drum whose motion is

braked together with a shoe mounted on a lever which can swing freely about a

fixed hinge H. A lining is attached to the shoe and contacts the braked body. The

actuation force P applied to the shoe gives rise to a normal contact pressure

distributed over the contact area between the lining and the braked body. A

corresponding friction force is developed between the stationary shoe and the

rotating body which manifest as retarding torque about the axis of the braked

body.

Brakes Classification

O

Lining

O

G

O

ωωωRotating body(drum)

ShoeStationary member

Rigid Pivoted

Short Shoe Long shoe

Figure 3.1.2

Various geometric configurations of drum brakes are illustrated above.

Drum Brakes are classified based on the shoe geometry. Shoes are classified as

being either short or long. A short shoe is one whose lining dimension in the

direction of motion is so small that contact pressure variation is negligible, i.e. the

pressure is uniform everywhere.



When the area of contact becomes larger, the contact may no longer be with a

uniform pressure, in which case the shoe is termed as long shoe. The shoes are

either rigid or pivoted, pivoted shoes are also some times known as hinged

shoes. The shoe is termed rigid because the shoes with attached linings are

rigidly connected to the pivoted posts. In a hinged shoe brake - the shoes are

not rigidly fixed but hinged or pivoted to the posts. The hinged shoe is connected

to the actuating post by the hinge, G, which introduces another degree of

freedom

Preliminary Analysis

The figure shows a brake shoe mounted on a lever, hinged at O, having an

actuating force Fa, applied at the end of the lever. On the application of an

actuating force, a normal force Fn is created when the shoe contacts the rotating

drum. And a frictional force Ff of magnitude f.Fn, f being the coefficient of friction,

develops between the shoe and the drum. Moment of this frictional force about

the drum center constitutes the braking torque.



Fa

ro

(a) Brake assembly

Fa a

RyFf cRx o

b

(b) Free-body diagram

Fn

θ

ω

shoe

drum

Ry

Rx

T

Figure 3.1.3

Short Shoe Analysis

For a short shoe we assume that the pressure is uniformly distributed over the

contact area. Consequently the equivalent normal force Fn = p .A, where = p is

the contact pressure and .A is the surface area of the shoe. Consequently the

friction force Ff = f.Fn where f is the co-efficient of friction between the shoe

lining material and the drum material.

The torque on the brake drum is then,

T = f Fn. r = f.p.A.r

A quasi static analysis is used to determine the other parameters of braking.

Applying the equilibrium condition by taking moment about the pivot ‘O’ we can

write

O a n nM F a F b f F c 0= − + =∑

Substituting for Fn and solving for the actuating force, we get,



Fa = Fn(b+-fc)/a

The reaction forces on the hinge pin (pivot) are found from a summation of

forces,

i.e.

F 0,R fp Ax x a= =

F 0,R p A Fy y a a= = −

Self- energizing

The principle of self energizing and leading and trailing shoes

With the shown direction of the drum rotation (CCW), the moment of the frictional

force f. Fn c adds to the moment of the actuating force, Fa

As a consequence, the required actuation force needed to create a known

contact pressure p is much smaller than that if this effect is not present. This

phenomenon of frictional force aiding the brake actuation is referred to as self-

energization.

Leading and trailing shoe

• For a given direction of rotation the shoe in which self energization is present is

known as the leading shoe

• When the direction of rotation is changed, the moment of frictional force now

will be opposing the actuation force and hence greater magnitude of force is

needed to create the same contact pressure. The shoe on which this is prevailing

is known as a trailing shoe



Self Locking

At certain critical value of f.c the term (b-fc) becomes zero. i.e no actuation force

need to be applied for braking. This is the condition for self-locking. Self-locking

will not occur unless it is specifically desired.

Short and Long Shoe Analysis

• Foregoing analysis is based on a constant contact pressure p.

• In reality constant or uniform constant pressure may not prevail at all points of

contact on the shoe.

• In such case the following general procedure of analysis can be adopted

General Procedure of Analysis

• Estimate or determine the distribution of pressure on the frictional surfaces

• Find the relation between the maximum pressure and the pressure at any point

• For the given geometry, apply the condition of static equilibrium to find the

actuating force, torque and reactions on support pins etc.



Drum Brakes

Among the various types of devices to be studied, based on their practical use,

the discussion will be limited to “Drum brakes” of the following types which are

mainly used in automotive vehicles and cranes and elevators.

Drum Brake Types:

• Rim types with internal expanding shoes

• Rim types with external contracting shoes

Internal expanding Shoe

The rim type internal expanding shoe is widely used for braking systems in

automotive applications and is generally referred as internal shoe drum brake.

The basic approach applied for its analysis is known as long-rigid shoe brake

analysis. Long –rigid Shoe Analysis

• A schematic sketch of a single shoe located inside a rotating drum with relevant

notations, is shown in the figure below. In this analysis, the pressure at any point

is assumed to be proportional to the vertical distance from the hinge pin, the

vertical distance from the hinge pin, which in this case is proportional to sine of

the angle and thus,

p d sin sin∝ θ ∝ θ

Since the distance d is constant, the normal pressure at any point is just

proportional to sinΘ. Call this constant of proportionality as K



>

><

X

Y

RX

dNcos

d dN

N Ndf

sindNcos

df NsinF

FX

FY

RY

d

θ θ

θθf

θ

dθ

Figure 3.1.4

Thus p K sin= θ

It the maximum allowable pressure for the lining material is pmax then the constant

K can be defined as

max

max

ppKsin sin

= =θ θ

max

max

pp s

sinin= θ

θ



• The normal force dN is computed as the product of pressure and area and the

frictional force as the product of normal force and frictional coefficient i.e. f dN.•

By integrating these over the shoe length in terms of its angle the braking torque

T, and other brake parameters are computed.

To determine the actuating force F, the moment equilibrium about the pivot point

is applied. For this we need to determine the moment of the normal force MN and

moment of the frictional force about the pivot point. Moment of the normal force

is equal to the normal force times its moment arm about the pivot point. From

the figure it is clear that the moment arm in this case is equal to d sin Θ where d

is the distance between the drum center and pivot center

1 1

2 2

1

2

2max

M p.b.r.d .d sin b.p.r.d.sin .dN

p = b.d.r. sin d

sin

θ θ

θ θ

θ

θ

= θ θ = θ

θ θθ

∫ ∫

∫

θ

( )p b.d.r 1 1maxM (sin 2 sin 2 )N 2 1 2sin 2 4a1

⎡ ⎤= θ − θ − θ − θ⎢ ⎥θ ⎣ ⎦

On similar lines the moment of friction force is computed

( )

( )

1

2

1

2

max

max

M f.p.b.r.d r d sinF

p = f .b.r. sin r d sin d

sin

θ

θ

θ

θ

= θ − θ

θ − θθ

∫

∫ θ

( ) ( )max.

a

f .p b.r d 2 2M r cos cos sin sif 2 1 2sin 2⎡ ⎤= − θ − θ − θ − θ⎢ ⎥θ ⎣ ⎦

n 1



The actuating force F is determined by the summation of the moments of normal

and frictional forces about the hinge pin and equating it to zero.

Summing the moment about point O gives

M MN fFc

±=

where,

• MN and Mf are the moment of the normal and frictional forces respectively,

about the shoe pivot point.

The sign depends upon the direction of drum rotation,

(- sign for self energizing and + sign for non self energizing shoe)Where the

lower sigh is for a self energizing shoe and the upper one for a self deenergizing

shoe.

The reaction forces are determined by applying force summation and equilibrium

( ) ( )

2 2

1 1

2 2

1 1

2max max

max max

max.b

max

R dN.cos dF.sinx

b.r.p cos d f b.r.psin d

p p b.r. sin cos d f b.r. sin d

sin sin

p r 1 1 1 1 2 2( ) sin 2 sin 2 f sin sin2 1 2 1 2 1sin 2 2 4 2

θ θ

θ θ

θ θ

θ θ

= θ + θ

= θ θ + θ θ

= θ θ θ + θ θθ θ

⎛ ⎞⎡ ⎤= θ − θ − θ − θ ± θ −⎜ ⎟⎢ ⎥θ ⎣ ⎦⎝ ⎠

∫ ∫

∫ ∫

∫ ∫

θ

The equations can be simplified and put as

p braR (Ax sin a

=θ

∓ fB)

p braR (B fA)y xsin aF= ± −

θ

Where

( )1 2 2A sin sin2 12= θ − θ



1 1 1B ( ) sin 2 sin 22 2 42 1 2 1⎡ ⎤⎛ ⎞= θ − θ − θ − θ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

The braking torque T on the drum by the shoe is of the frictional forces f.dN times

the radius of the drum and resulting equation is,

1

1

1

1

2

max

T f .b.p.r.d .r

pmax = fbr sin dsin

θ

θ

θ

θ

= θ

θ θθ

∫

∫

2fdp r (cos cos )a 1Tsin a

2θ − θ=

θ

T

Double Shoe Brakes Twin Shoe Brakes

Behavior of a single shoe has been discussed at length. Two

such shoes are combined into a complete practical brake unit,

two being used to cover maximum area and to minimize the

unbalanced forces on the drum, shaft and bearings.

• If both the shoes are arranged such that both are leading shoes in which

self energizing are prevailing, then all the other parameters will remain

same and the total braking torque on the drum will be twice the value

obtained in the analysis.

• However in most practical applications the shoes are arranged such that

one will be leading and the other will be trailing for a given direction of

drum rotation



• If the direction of drum rotation changes then the leading shoe will

become trailing and vice versa.

• Thus this type of arrangement will be equally effective for either direction

of drum rotation. Further the shoes can be operated upon using a single

cam or hydraulic cylinder thus provide for ease of operation

One leading shoe & one trailing

Two Leading shoe

Figure 3.1.5

However the total braking torque will not be the twice the value of a single

shoe, if the same normal force is applied or created at the point of force

application on both the brake shoes which is the normal practice as they

are actuated using a common cam or hydraulic cylinder.

• This is because the effective contact pressure (force) on the trailing shoe

will not be the same, as the moment of the friction force opposes the

normal force, there by reducing its actual value as in most applications



the same normal force is applied or created at the point of force

application on the brake shoe as noted above

• Consequently we may write the actual or effective pressure prevailing on

a trailing shoe

F.a'p p .a a (M M )n f

⎡ ⎤= ⎢ ⎥

+⎢ ⎥⎣ ⎦

Resulting equation for the braking torque

p2 aT f .w.r . (cos - cos )(p pB 1 2sin a= θ θ

θ')a a+

Some pictorial illustrations of the automotive drum brakes are presented

below

Figure 3.1.6



Figure 3.1.7

Oblong Cam Actuator

Leading shoe

Rotating Drum

Pivot point(Fixed axis)

The anatomy of the single leading shoe drum Brake

Trailing shoe

Animation



Figure 3.1.9

Figure 3.1.10



External Contracting Shoe

• The same analysis can be extended to a drum brake with external contracting

type of shoes, typically used in elevators and cranes.

A schematic sketch of as single shoe located external to the rotating drum is with

all relevant notations is shown in the figure below.

Figure 3.1.11

• Corresponding contact geometry is shown in the figure

• The resulting equations for moment of normal and frictional force as well as the

actuating force and braking torque are same as seen earlier.

• For convenience they are reproduced here again

( )2fbp r cos cosa 1Tsin a

2θ − θ=

θ



M MN fFc

±=

( )p bra 1 1aM (sin 2 sin 2 )N 2 1 2sin 2 4a1

⎡ ⎤= θ − θ − θ − θ⎢ ⎥θ ⎣ ⎦

( )a

a

fp br d 2 2M r cos cos (sin sinf 1 2 2sin 2⎡ ⎤= θ − θ − θ −⎢ ⎥θ ⎣ ⎦

)1θ

TWIN SHOE BRAKES

As noted earlier for the internal expanding shoes, for the double shoe brake the

braking torque for one leading and one trailing shoe acted upon a common cam

or actuating force the torque equation developed earlier can be applied.

i.e p2 aT f .w.r . (cos - cos )(p pB 1 2sin a

= θ θθ

')a a+



Brake with a pivoted long shoe

When the shoe is rigidly fixed to the lever, the tendency of the

frictional force (f.Fn) is to unseat the block with respect to the

lever. This is eliminated in the case of pivoted or hinged shoe

brake and it also provides some additional advantages.

Long Hinged Shoe

In a hinged shoe brake - the shoes are not rigidly fixed but

hinged or pivoted to the posts. The hinged shoe is connected to

the actuating post by the hinge, G, which introduces another

degree of freedom - so the shoe tends to assume an optimum

position in which the pressure distribution over it is less peaked

than in a rigid shoe.

Bo

H

FF

G

OH=a

OG=b

X

BG

φ

y

x

Figure 3.1.12

As wear proceeds the extra degree of freedom allows the linings

to conform more closely to the drum than would be the case to



rigid shoes. This permits the linings to act more effectively and

also reduces the need for wear adjustment.

The extra expense of providing another hinge is thus justified on

the grounds of more uniform lining wear and consequently a

longer life. This is the main advantage of the pivoted shoe brake

This is possible only if the shoe is in equilibrium.

For equilibrium of the shoe moments of the forces

about the hinge pin should balance -

i.e ΣMG=T+Fxby-Fybx=0 where bx = b.cos θG

by = b.sinθG

This needs that the resultant moment due to the frictional force

(and due to the normal force) about the pivot point should be

zero, so that no rotation of the shoe will occur about the pivot

point. To facilitate this location of the pivot is to be selected

carefully.

The actuating force P is applied to the post HG so the shoe itself

is subject to actual and ideal contacts only - the (ideal) at pin G

and the actual as distributed contact with the drum.

The location is in such a way that the moment of frictional force

(and the normal force) about the pivot is zero. i.e the actual

distributed contact leads to the ideal (concentrated)contact at the

hinge or pivot.

i.e the actual distributed contact leads to the ideal contact at the

hinge or pivot Further it is desirable to minimize the effect of pin

reaction for which the shoe pivot and post pivot points are made

con current.



Let us now look how this can be met, satisfying the conditions set above and

consequently the derive the equations relating the location of the pivot from the

center of the drum

A schematic sketch of a single shoe is shown in the figure

dN cosθrcosθ

h

θ

fdN

cosfdN θ

θ

(hcosθ−r)

Rx

yR

fdN sin θ

dNr

Force acting on shoe

An element of friction lining located at an angle Θ and

subtending to a small angle dθ is shown in figure. The area if

the element is ( )r.d .bθ , where b is the width of the friction lining

parallel to the axis of the brake drum. If the intensity of pressure

at the element is p, the normal reaction dN on the element is

given by

dN (rd b)p= θ (a)



Distribution of pressure

If the shoe is long then the pressure will not be uniform

We need to determine the distribution of pressure along the

lining; the pressure distribution should be conducive for

maintaining a uniform wear

Since the brake drum is made of a hard material like cast iron or

steel, the wear occurs on the friction lining, which is attached to

the shoe. As shown in fig the lining need to retain the cylindrical

shape of the brake drum when wear occurs. After the radial

wear has take place, a point such as X’ moves to X in order to

maintain contact on the lining with the brake drum. In figure δx is

the wear in the X direction and δr is the wear in the radial

direction. If it is assumed that the shoe is constrained to move

towards the brake drum to compensate to wear, δx should be

constant because it need to be same for all points. Therefore,

rxcosδ

δ =θ

= constant (b)

θ

Y

X

x

δx

x'θ

δx

wear

wear of friction lining



The radial wear δr is proportional to the work done by the

frictional force. The work done by the frictional force depends

upon the frictional force ( fdN ) and the rubbing velocity. Since the

rubbing velocity is constant for all points on friction lining,

r fdNδ ∝

Or ( )r frd bpδ ∝ θ

Therefore (c) r pδ ∝

From the expression (b) and (c)

1p cons tan t or p C cos

cos= =

θθ

C

(d)

Where C1 is the constant of proportionality. The pressure is

maximum when . 0θ =

Substituting,

max 1p = (e)

From Eqs (d) and (e),

maxp p cos= θ

Substituting this value in Eq. (a

maxdN (rd b)p cos= θ θ (f)

The forces acting on the element of the friction lining are shown

in figure. The distance h of the pivot is selected in such a manner

that the moment of frictional force about it is zero.

Therefore, fM 0==

dMf=f.dN moment arm

moment arm in this case = (h cos r)θ −



fM fdN(h cos r)θ

−θ

0= θ − =∫

f maxM fp rd cos (h r cos )θ

−θ

= θ θ −∫ θ

Substituting dN from Eq. (f),

( )

( )

2

0

0 0

0

0

h cos r cos d 0

1 cos 2or h d r cos d 02

1 sin 22or h R sin 0

2

4R sin h2 sin 2

θ

θ θ

θ

θ

θ − θ θ =

+ θ⎡ ⎤ θ − θ θ =⎢ ⎥⎣ ⎦

⎡ ⎤φ + φ⎢ ⎥− φ =⎢ ⎥

⎢ ⎥⎣ ⎦

θ=

θ + θ

∫

∫ ∫

The elemental torque of frictional force fdN about the axis of

brake drum is fdNR . Therefore

BT 2 fdNθ

−θ

= r∫

Substituting the value of dN from Eq.(f)

2B max

2B max

T 2fr bp cos d

T 2fr bp sin

θ

−θ

= θ θ

= θ

∫

The reaction can be determined by considering two components XR

( ) ( )dN cos and fdN sinθ θ .



Due to symmetry, the other two vertical components of the force

balances

i.e

fdN sin 0

dN sin 0

θ =

θ =

∫∫

Therefore,

x

2max

max

x max

R dN cos

rbp cos d

2 sin 2 2rbp4

1or R rbp (2 sin 2 )2

θ

−θθ

−θ

= θ

= θ θ

θ + θ⎡ ⎤= ⎢ ⎥⎣ ⎦

= θ + θ

∫

∫

Note that is also = xR nF

The reaction yR can be determined by considering two

components ( ) ( )dN sin and fdN cosθ θ

Due to symmetry,

=0 dN sin∫ θ

Therefore,



y

2max

y max

R fdN cos

frbp cos d

1or R frbp (2 sin 2 )2

θ

−θθ

−θ

= θ

= θ θ

= θ + θ

∫

∫

As noted earlier,

2B maT 2fr bp si= θx n

Rewriting it,

B max

n

2 sin 4r sinT frbp .2 2 sin

= fF h

θ + θ θ=

θ + θ

DOUBLE BRAKE SHOE

A double block brake with two symmetrical and pivoted shoes is show in

figure.

If the same magnitude of actuating forces are acted upon the posts, then

B n1 n2 nT f .(F F ).h 2f .F .h= + =

Θ

Q

P P

Pivoted double block brake



Pivoted shoe brakes are mainly used in hoists and cranes. Their

applications are limited because of the physical problem in

locating pivot so close to the drum surface.



Energy Consideration

It has been noted that the most common brakes employ friction to transform

the braked system's mechanical energy, irreversibly into heat which is then

transferred to the surrounding environment -

• Kinetic energy is absorbed during slippage of either a clutch or brake, and this

energy appears as heat.

• If the heat generated is faster than it is dissipated, then the temperature rises.

Thorough design of a brake therefore requires a detailed transient thermal

analysis of the interplay between heat generated by friction, heat transferred

through the lining and the surrounding metalwork to the environment, and the

instantaneous temperature of the surface of the drum as well as the lining. For

a given size of brake there is a limit to the mechanical power that can be

transformed into heat and dissipated without lthe temperatures reaching

damaging levels. Temperature of the lining is more critical and the brake size is

characterized by lining contact area, A.

The capacity of a clutch or brake is therefore limited by two factors:

1. The characteristics of the material and,

2. The ability of the brake to dissipate heat.

Heat Generated In Braking

During deceleration, the system is subjected to an essentially constant torque

T exerted by the brake, and in the usual situation this constancy implies constant

deceleration too.

Application of the work or energy principle to the system enables the torque

exerted by the brake and the work done by the brake, U, to be calculated from:-



U = ∆E = T ∆θ (2)

Where ∆E is the loss of system total energy which is absorbed by the brake

during deceleration, transformed into heat, and eventually dissipated.

The elementary equations of constant rotational deceleration apply, thus when

the brake drum is brought to rest from an initial speed ωo :-

Deceleration = ωo2/ 2 (1) ∆θ = ωm.∆t ; ωm = ωo/2 =

∆θ/∆t where ωm is the mean drum speed over the deceleration period.

The mean rate of power transformation by the brake over the braking period is

:-

Pm = U / ∆t = T ωm ( 3 ) which forms a basis for the selection or

the design of the necessary brake dimensions.

The rise in temperature in the lining material is also important as rate of wear is

also a function of the temperature. Further for any lining material, the maximum

allowable temperature is also another performance criteria.

Temperature Rise

The temperature rise of the brake assembly can be approximated by the classic

expression,,

ETC.m

∆ =

Where is temperature ∆T is rise in temperature in oC,’C’ is the specific heat of

the brake drum material – (500J/Kg for steel or Cast Iron) and m is the mass

(kg) of the brake parts dissipating the heat into the surroundings.

Though the equation appears to be simple, there are so many variables involved

that it would be most unlikely that such an analysis would even approximate

experimental results.



On the other hand the temperature-rise equations can be used to explain what

happens when a clutch or brake is operated frequently. For this reason such

analysis are most useful, for repetitive cycling, in pin pointing those design

parameters that have the greatest effect on performance.

An object heated to a temperature T1 cools to an ambient temperature Ta

according to the exponential relation

Time-temperature relation

(AU / WC)tT T (T T )ei a 1 a−− = −

Where T1 = instantaneous temperature at time t, ˚C

A= heat transfer area, m2

U= Heat Transfer coefficient, W/(m2.s. ˚C)

T1 = Initial temperature, ˚C

Ta = Ambient temperature, ˚C

C - Specific heat

t - time of operation, s



T

T

t t t

C

B

T

A

T

A

a

B C

1

T2

Time t

Figure 3.1.16

Figure shows an application of Eq. (a). At time tA a clutching or braking

operation causes the temperature to rise to T1 at A. Though the rise occurs in a

finite time interval, it is assumed to occur instantaneously. The temperature then

drops along the decay line ABC unless interrupted by another braking operation.

If a second operation occurs at time tB, the temperature will rise along the dashed

line to T2 and then begin an exponential drop as before. About 5 -10 % of the

heat generated at the sliding interface of a friction brake must be transferred

through the lining to the surrounding environment without allowing the lining to

reach excessive temperatures, since high temperatures lead to hot spots and

distortion, to fade (the fall-off in friction coefficient) or, worse, to degradation and

charring of the lining which often incorporates organic constituents



In order to determine the brake dimensions the energy need to be absorbed

during critical braking conditions is to be estimated.

Energy to be Absorbed

If t is the time of brake application and ωm the mean or average angular velocity

then the energy to be absorbed in braking E

E = T. ωm .t = Ek+ Ep+ Ei

where Ek is the kinetic energy of the rotating system

Ep is the potential energy of the moving system

Ei is the inertial energy of the system

Energy to be absorbed

( )

k

2

2 22 1

1E2 g1 mv21 v v2 g

ω=

=

ω= −

( )

p

2i

2 22 1

E mgh h

1E I21 I2

= = ω

= ω

= ω − ω

Frictional Material

A brake or clutch friction material should have the following characteristics to a

degree, which is dependent upon the severity of the service.

• A high and uniform coefficient of friction.

• Imperviousness to environmental conditions, such as moisture.

• The ability to withstand high temperatures together with good thermal

conductivity.

• Good resiliency.

• High resistance to wear, scoring, and galling.



Linings

The choice of lining material for a given application is based upon criteria such as

the expected coefficient of friction; fade resistance, wear resistance, ease of

attachment, rigidity or formability, cost, abrasive tendencies on drum, etc. The

lining is sacrificial - it is worn away. The necessary thickness of the lining is

therefore dictated by the volume of material lost - this in turn is the product of the

total energy dissipated by the lining throughout its life, and the specific wear rate

Rw (volume sacrificed per unit energy dissipated) which is a material property

and strongly temperature dependent. The characteristics of a typical moulded

asbestos lining material is illustrated in the figure below. The coefficient of

friction, which may be taken as 0.39 for design purposes, is not much affected by

pressure or by velocity - which should not exceed 18 m/s. The maximum

allowable temperature is 400°C. However at this temperature the wear is very

high. From a lower wear or higher life point, the maximum temperature should

not exceed about 200 oC

Temperature ( C ) 4000

0

0.5

o

Figure 3.1.17



Linings traditionally were made from asbestos fibers bound in an organic matrix,

however the health risks posed by asbestos have led to the decline of its use.

Non-asbestos linings generally consist of three components - metal fibers for

strength, modifiers to improve heat conduction, and a phenolic matrix to bind

everything together.

Brake Design Section

The braked system is first examined to find out the required brake capacity that is

the torque and average power developed over the braking period. - The brake

is then either selected from a commercially available range or designed from

scratch ff a drum brake has to be designed for a particular system (rather than

chosen from an available range) then the salient brake dimensions may be

estimated from the necessary lining area, A, together with a drum diameter- to-

lining width ratio somewhere between 3:1 and 10:1, and an angular extent of 100

°C say for each of the two shoes.

Worked out Example 1

An improved lining material is being tried on an existing passenger car drum

brake shown in Figure. Quality tests on the material indicated permissible

pressure of 1.0 MPa and friction co-efficient of 0.32. Determine what maximum

actuating force can be applied for a lining width of 40 mm and the corresponding

braking torque that could be developed. While cruising on level road at 100

kmph, if it is to decelerated at 0.5g and brought to rest, how much energy is

absorbed and what is the expected stopping distance?



While cruising on level road at 100 kmph, if it is to decelerated at 0.5g and

brought to rest, how much energy is absorbed and what is the expected stopping

distance?

300

300

0

R=125

0

100.9

50 50300 0

5

AUTOMOTIVE DOUBLE SHOE BRAKE

05

30

F F

120 120Pin Pin

0

300

86.6

Figure 3.1.18

Analysis based on leading shoe

a

2 2

P 1 MPa f 0.32 a 187.5 mm 0b 40 mm 90 d = 100 +86.1max

= = =

= θ =

= 99.99 100mm0 05 120 r =125 mm1 2

≈

θ = θ =



( )

( )

a 2 1n 2 1

a6 -3 3

n

p rd 1M (sin 2 sin 2 )

sin 2 4

10 *40*10 *125 *10 * 0.1 115 1 = * sin 240 sin10

1 2 180 41

=40*125*0.1 1.003- 1.034

M 631.459 N.m

b

−

θ − θ= − θ − θ

θ

π− ° − °

−

=

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎣ ⎦

( )f .b.rpm d 2 2M r(cos - cos ) - sin - sinf 1 2 2 1sin 2a-3 -3 6 2 20.32*40*10 *125*10 *10 0.125(cos5 - cos120 ) - 0.04(sin 120 - sin 5 )

M 224.85 N - mf

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎣ ⎦

= θ θ θ θθ

= ° °

=

° °

F*a= N fM M−

n fM M 631.459 224.85F 2174.3Na 0.187 Max. actutating force

− −= = =

( )2 a a

B 1 2

a n f

3 2 6

B

B

p FT fbr cos cos ) 1

sin M M

406.609T 0.32 * 40 *10 * (0.125) *10 (Cos5 Cos120) 1856.36

T 441.329 N-m

−

⎡ ⎤= θ − θ +⎢ ⎥α +⎣ ⎦

⎡ ⎤= − +⎢ ⎥⎣ ⎦=

Running at 100 kmph =100*5/8

= 27.7 m/s



U= 27.7 m/s

Deceleration =0.5*9.8=4.9

2 2

2

2

V U 2aS

0 (27.7) 2* ( 4.9) *S

27.7S 78.29 m2* 4.9

− =

− = −

= =

E T. .tav1 27.7 78.24 441.329 .2 0.125 27.7

138206 138.2KJ

= ω

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

==

Worked out example 2

A spring set, hydraulically released double shoe drum brake, schematically

shown at Fig 2 is to be designed to have a torque capacity of 600 N.m under

almost continuous duty when the brake drum is rotating at 400 rpm in either

direction. Assume that the brake lining is to be molded asbestos having a

friction coefficient of 0.3 and permissible pressure of 0.8 MPa. The width of the

brake shoe is to be third of drum diameter and the remaining proportion's are as

shown in figure.

0

1 413

0 6 0 69330

=

=

= =

a . D

b D

. Dd .cos

D



Double block brake

90

30ο

ο1.4 D

0.6 D

Figure 3.1.19

Determine the required brake drum diameter, width of the lining and the spring

force required to be set.

0

a 1.4D1b D30.6Dd 0.693D

cos30

=

=

= =



( )

( ) ( )( ) ( )

1

1

6

3N

aJ

2 2a 1 2 2

6

3

1 D sin(210 ) sin 300.8*10 * D * * 0.6028D.3 2 4 4

M 92373.3D *1.035

P brfM r a cos sin d

1

d P brf r cos cos sin sin2

D D D 0.6928 0.8*10 * * * 0.3 1.225 0.8663 2 2 2

12500D

θ

θ

1

π ° − °⎡ ⎤= −⎢ ⎥⎣ ⎦

=

= − θ θ θ

⎡ ⎤= θ − θ − θ −⎢ ⎥⎣ ⎦θ

⎡ ⎤= −⎢ ⎥⎣ ⎦

=

∫

a

' N Fa a

N F

a

a

P 'for the trailing shoe

M MP P

M M

95606.36 12500 P95606.36 12500

0.7687P

⎛ ⎞−= ⎜ ⎟+⎝ ⎠

−⎛ ⎞= ⎜ ⎟+⎝ ⎠=

Torque due to trailing and leading shoe=

( ) ( )

2 2

1 1

a

a

2

a aa

2

a a 1 2a

fdN.r

P sin f .r .rd .b

sin

fbr P sin d P ' sin dsin

fbr . P P ' . cos cossin

θ θ

θ θ

τ =

θ= θ

θ

⎡ ⎤⎢ ⎥= θ θ +

θθ θ

⎢ ⎥⎣ ⎦

= + θ −θ

∫

∫

∫ ∫

θ

( )

total

2

6

3

D D0.3* *3 4 1.7687 *0.8*10 cos15 cos1.51

43324 DT 600 Nm

Therefore D 240.14mm

⎡ ⎤= °⎣ ⎦

==

=

− °



Actuating force due to spring

2N FM M 83106 * 0.240F 3423.22N

1.4 * D 1.4−

= = =

Actuating force =3423.22 N

DLiving width3

80.04 mm

=

=



CLUTCH

Clutch Introduction

A Clutch is ia machine member used to connect the driving shaft to a driven

shaft, so that the driven shaft may be started or stopped at will, without stopping

the driving shaft. A clutch thus provides an interruptible connection between two

rotating shafts

Clutches allow a high inertia load to be stated with a small power.

A popularly known application of clutch is in automotive vehicles where it is used

to connect the engine and the gear box. Here the clutch enables to crank and

start the engine disengaging the transmission Disengage the transmission and

change the gear to alter the torque on the wheels. Clutches are also used

extensively in production machinery of all types

Mechanical Model

Two inertia’s and traveling at the respective angular velocities ωI and I1 2 1

and ω2, and one of which may be zero, are to be brought to the same speed

by engaging. Slippage occurs because the two elements are running at different

speeds and energy is dissipated during actuation, resulting in temperature rise.

ω1

Ι1 Ι1

ω2

Clutch or brake

Dynamic Representation of Clutch or Brake

Figure 3.2.1



Animated

Figure 3.2.2

To design analyze the performance of these devices, a knowledge on the

following are required.

1. The torque transmitted

2. The actuating force.

3. The energy loss

4. The temperature rise

FRICTION CLUTCHES

As in brakes a wide range of clutches are in use wherein they vary in their are in

use their working principle as well the method of actuation and application of

normal forces. The discussion here will be limited to mechanical type friction



clutches or more specifically to the plate or disc clutches also known as axial

clutches

Frictional Contact axial or Disc Clutches

An axial clutch is one in which the mating frictional members are moved in a

direction parallel to the shaft. A typical clutch is illustrated in the figure below. It

consist of a driving disc connected to the drive shaft and a driven disc

co9nnected to the driven shaft. A friction plate is attached to one of the

members. Actuating spring keeps both the members in contact and

power/motion is transmitted from one member to the other. When the power of

motion is to be interrupted the driven disc is moved axially creating a gap

between the members as shown in the figure.

Figure 3.2.3



Flywheel

Clutch plate

Pressure plate

Clutch cover

Diaphragmspring

to transmission

Throw outBearing

Animated

Figure 3.2.4

METHOD OF ANALYSIS

The torque that can be transmitted by a clutch is a function of its geometry and

the magnitude of the actuating force applied as well the condition of contact

prevailing between the members. The applied force can keep the members

together with a uniform pressure all over its contact area and the consequent

analysis is based on uniform pressure condition



Uniform Pressure and wear

However as the time progresses some wear takes place between the contacting

members and this may alter or vary the contact pressure appropriately and

uniform pressure condition may no longer prevail. Hence the analysis here is

based on uniform wear condition

Elementary Analysis

Assuming uniform pressure and considering an elemental area dA

dA = 2Π.r dr

The normal force on this elemental area is

dN 2 .r.dr.p= π

The frictional force dF on this area is therefore

dF f .2 .r.dr.p= π



<

< >do

Fr

di

dr

lining

A single-Surface Axial Disk Clutch

Figure 3.2.5

Now the torque that can be transmitted by this elemental are is equal to the

frictional force times the moment arm about the axis that is the radius ‘r’

i.e. T = dF. r = f.dN. r = f.p.A.r

= f.p.2.π.r. dr .r

The total torque that could be transmitted is obtained by integrating this equation

between the limits of inner radius ri to the outer radius ro

ro 22 3T 2 pfr dr pf (r r )o i3ri

= π = π −∫ 3

Integrating the normal force between the same limits we get the actuating force

that need to be applied to transmit this torque.



( )

a

2 2a o i

roF 2 prdr

riF r r

= π∫

= π − .p

Equation 1 and 2 can be combined together to give equation for the torque

3 3o i

a 2 2o i

(r r )2T fF .3 (r r )

−=

−

Uniform Wear Condition

According to some established theories the wear in a mechanical system is

proportional to the ‘PV’ factor where P refers the contact pressure and V the

sliding velocity. Based on this for the case of a plate clutch we can state

The constant-wear rate Rw is assumed to be proportional to the

product of pressure p and velocity V.

Rw= pV= constant

And the velocity at any point on the face of the clutch is V r.= ω

Combining these equation, assuming a constant angular velocity ω

pr = constant = K

The largest pressure pmax must then occur at the smallest radius ri ,

max iK p r=

Hence pressure at any point in the contact region

imax

rp p

r=



In the previous equations substituting this value for the pressure term p and

integrating between the limits as done earlier we get the equation for the torque

transmitted and the actuating force to be applied.

I.e The axial force Fa is found by substituting imax

rp p

r= for p.

and integrating equation dN 2 prdr= π

r ro o riF 2 prdr 2 p rdr 2 p r (r r )max max i o irr ri i

⎛ ⎞= π = π = π −⎜ ⎟∫ ∫ ⎜ ⎟

⎝ ⎠

Similarly the Torque

ro 2 2T f 2 p r rdr f p r (r rmax i max i o iri

= π = π −∫ )

Substituting the values of actuating force Fa

The equation can be given as

(r r )o iT fF .a 2

+=

Single plate dry Clutch – Automotive application

The clutch used in automotive applications is generally a single plate dry clutch.

In this type the clutch plate is interposed between the flywheel surface of the

engine and pressure plate.



Flywheel

Frictionplanes

Clutch plate(driven disk)

Pressure platePressure spring

HousingReleasebearing

Enginecrankshaft

To release

Totransmission

Figure 3.2.6

Single Clutch and Multiple Disk Clutch

Basically, the clutch needs three parts. These are the engine flywheel, a friction

disc called the clutch plate and a pressure plate. When the engine is running

and the flywheel is rotating, the pressure plate also rotates as the pressure plate

is attached to the flywheel. The friction disc is located between the two. When



the driver has pushed down the clutch pedal the clutch is released. This action

forces the pressure plate to move away from the friction disc. There are now air

gaps between the flywheel and the friction disc, and between the friction disc and

the pressure plate. No power can be transmitted through the clutch.

Operation Of Clutch

When the driver releases the clutch pedal, power can flow through the clutch.

Springs in the clutch force the pressure plate against the friction disc. This action

clamps the friction disk tightly between the flywheel and the pressure plate. Now,

the pressure plate and friction disc rotate with the flywheel.

As both side surfaces of the clutch plate is used for transmitting the torque, a

term ‘N’ is added to include the number of surfaces used for transmitting the

torque

By rearranging the terms the equations can be modified and a more general form

of the equation can be written as

T N.f .F .Ra m=

T is the torque (Nm).

N is the number of frictional discs in contact.

f is the coefficient of friction

Fa is the actuating force (N).

Rm is the mean or equivalent radius (m).

Note that N = n1 + n2 -1

Where n1= number of driving discs

n2 = number of driven discs

Values of the actuating force F and the mean radius for the two conditions of

analysis are summarized and shown in the table

mr



Clutch Construction

Two basic types of clutch are the coil-spring clutch and the diaphragm-spring

clutch. The difference between them is in the type of spring used. The coil

spring clutch shown in left Fig 3.2.6 uses coil springs as pressure springs (only

two pressure spring is shown). The clutch shown in right figure 3.2.6 uses a

diaphragm spring.

Figure 3.2.6

The coil-spring clutch has a series of coil springs set in a circle.

At high rotational speeds, problems can arise with multi coil spring clutches

owing to the effects of centrifugal forces both on the spring themselves and the

lever of the release mechanism.

These problems are obviated when diaphragm type springs are used, and a

number of other advantages are also experienced



Clutch or Driven Plate

More complex arrangements are used on the driven or clutch plate to facilitate

smooth function of the clutch

The friction disc, more generally known as the clutch plate, is shown partly cut

away in Fig. It consists of a hub and a plate, with facings attached to the plate.

Figure 3.2.7

First to ensure that the drive is taken up progressively, the centre plate, on which

the friction facings are mounted, consists of a series of cushion springs which is

crimped radially so that as the clamping force is applied to the facings the

crimping is progressively squeezed flat, enabling gradual transfer of the force

On the release of the clamping force, the plate springs back to its original

position crimped (wavy) state



This plate is also slotted so that the heat generated does not cause distortion that

would be liable to occur if it were a plain plate. This plate is of course thin to keep

rotational inertia to a minimum.

Plate to hub Connection

Secondly the plate and its hub are entirely separate components, the drive being

transmitted from one to the other through coil springs interposed between them.

These springs are carried within rectangular holes or slots in the hub and plate

and arranged with their axes aligned appropriately for transmitting the drive.

These dampening springs are heavy coil springs set in a circle around the hub.

The hub is driven through these springs. They help to smooth out the torsional

vibration (the power pulses from the engine) so that the power flow to the

transmission is smooth.

In a simple design all the springs may be identical, but in more sophisticated

designs the are arranged in pairs located diametrically opposite, each pair having

a different rate and different end clearances so that their role is progressive

providing increasing spring rate to cater to wider torsional damping

The clutch plate is assembled on a splined shaft that carries the rotary motion to

the transmission. This shaft is called the clutch shaft, or transmission input shaft.

This shaft is connected to the gear box or forms a part of the gear box.

Friction Facings or Pads

It is the friction pads or facings which actually transmit the power from the fly

wheel to hub in the clutch plate and from there to the out put shaft. There are



grooves in both sides of the friction-disc facings. These grooves prevent the

facings from sticking to the flywheel face and pressure plate when the clutch is

disengaged. The grooves break any vacuum that might form and cause the

facings to stick to the flywheel or pressure plate. The facings on many friction

discs are made of cotton and asbestos fibers woven or molded together and

impregnated with resins or other binding agents. In many friction discs, copper

wires are woven or pressed into the facings to give them added strength.

However, asbestos is being replaced with other materials in many clutches.

Some friction discs have ceramic-metallic facings.

Such discs are widely used in multiple plate clutches

The minimize the wear problems, all the plates will be enclosed in a covered

chamber and immersed in an oil medium

Such clutches are called wet clutches

Multiple Plate Clutches

Figure 3.2.8

The properties of the frictional lining are important factors in the design of the

clutches



Typical characteristics of some widely used friction linings are given in the table

Table Properties of common clutch/ Brake lining materials

Friction MaterialAgainst Steel or Cl

Dynamic Coefficient of Friction

Maximum Pressure Maximum Temprerature

Molded

Woven

Sintered metal

Cast iron of hard steel

0.25-0.45

0.25-0.45

0.15-0.45

0.15-0.25

0.06-0.09

0.08-0.10

0.05-0.08

0.03-0.06

1030-2070

345-690

1030-2070

690-720

204-260

204-260

232-677

260

dry in oilKPa oC

Table 3.2.1



Energy considerations

Kinetic energy is absorbed during slippage of a clutch and this energy appears

as heat.

The clutch or brake operation is completed at the instance in which the two

angular velocities θ1 and θ2 become equal. Let the time required for the entire

operation be t1, then,

( )( )

I I1 2 1 2t1 T I I1 2

ω −ω=

+

This is derived by writing the equations of motion involving inertia

i.e

1 11 11

2 22 22

TI T tI

TI T tI

• • •θ = − θ = − + ω

•• •θ = − θ = − + ω

1 21 2

1 21 2

1 2

T Tt t1 2 I I

I IT t

I I

• • • ⎛ ⎞θ = θ − θ = − + ω − + ω⎜ ⎟

⎝ ⎠⎛ ⎞+

= ω −ω − ⎜ ⎟⎝ ⎠

from which ( )( )

1 2 1 2

1 2

I It

T I Iω −ω

=+

as at the instance of completion of clutching

operation ω1-ω2 = 0

Assuming the torque to be constant, the rate of energy dissipation during the

operation is then,

1 21 2

1 2

I IU T T T .t

I I⎡ ⎤⎛ ⎞+•= θ = ω − ω −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦



The total energy dissipated during the clutching operation or braking cycle is

obtained by integrating the above equation from t=0 to t = t1 . The result can be

summed up as,

( )( )

( )( )

1 2 1 2

1 2

1 21 2

1 2

1 1 1 21 2

1 2

2I IE

2 I I

I IU T T T .t

I I

t t I IE udt T T t dt

I I0 02

I I1 2 1 2E2 I I1 2

ω −ω=

+

⎡ ⎤⎛ ⎞+•= θ = ω −ω −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞+= = ω −ω −∫ ∫ ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

ω − ω=

+

Thus the energy absorbed during clutch slip is a function of the magnitude of the

inertia and the angular velocities only. This energy compared to the brake

energy may be negligible. Heat dissipation and temperature rise are governed

by the same equations presented during brakes. To contain the temperature rise

when very frequent clutching operations, wet clutches rather than dry clutches

are often use

WORKED OUT EXAMPLE 1

Design an automotive plate clutch to transmit a torque of 550 N-m. The

coefficient of friction is 0.25 and the permissible intensity of pressure is 0.5

N/mm2. Due to space limitations, the outer diameter of the friction disc is fixed as

250 mm.

Using the uniform wear theory, calculate:

The inner diameter of the friction disc

the spring force required to keep the clutch in engaged position



Solution: As noted the friction disc of the automotive clutch is fixed between the

flywheel on one side and the pressure plate on the other. The friction lining is

provided on both sides of the friction disc.

Therefore two pairs of contacting surfaces-one between the fly wheel and the

friction disc and the other between the friction disc and the pressure plate.

Therefore, the torque transmitted by one pair of contacting surfaces is (550/2) or

275 N-m

( ) ( )( ) ( )( ) ( )3 2

i i

2 2M p r r rt a i o if

275 10 0.25 0.5 r 125 r

= πµ −

× = π − 2

From the Eqr ( )2 28r 125 r 5602254i i− =

Rearranging the terms, we have

The above equation is solved by the trial and error method. It is a cubic equation,

with following three roots:

(i) ri = 87.08 mm

(ii) ri = 56.145 mm

(iii) ri =-143.23 mm

Mathematically, all the three answer are correct. The inner radius cannot be

negative. As a design engineer, one should select the inner radius as 87.08 mm,

which results in a minimum area of friction lining compared with 56.145. For

minimum cost of friction lining.

ri=87 mm

Actuating force needed can be determined using the equation

( ) ( )aF 2 p r r r 2 (0.5)(87) 125 87 10390.28Na i o i= π − = π − =




A multiple-disc wet clutch is to be designed for transmitting a torque of 85 N.m.

Space restriction limit the outside disk diameter to 100 mm. Design values for the

molded friction material and steel disks to be used are f=0.06(wet) and pmax

=1400 kPa. Determine appropriate values for the disc inside diameter, the total

number of discs, and the clamping force.

Solution

Known: A multiple – disc with outside disc diameter, d0≤ 100 mm,

dynamic friction coefficient, f=0.06(wet)

and maximum disc allowable pressure, pmax =1400 kPa,

To transmits a torque, T= 85 N.m

Find: Determine the disc inside diameter di, the total number of disks N, and the

clamping force Fa.

Decisions and Assumptions

Use the largest allowable outside disc diameter, do=100 mm (ro =50 mm).

Select ri =29 mm (based on the optimum d/D ratio of 0.577)

The coefficient of friction f is a constant.

The wear rate is uniform at the interface.

The torque load is shared equally among the disc.

Design Analysis:

Using design equation for torque under constant wear gives

( )2 2N T / p r f r r 6.69max i o i⎡ ⎤= π − =⎢ ⎥⎣ ⎦

Since N must be an even integer, use N= 8. It is evident that this requires a total

of 4+5, or nine discs, remembering that the outer disks have friction surfaces on

one side only. 3. With no other changes, this will give a clutch that is over

designed by a factor of 8/6.69= 1.19. Possible alternatives include (a) accepting



the 19 percent over design, (b) increasing ri, (c) decreasing ro, and (d) leaving

both radii unchanged and reducing both pmax and F by a factor of 1.19

4. With the choice of alternative d, the clamping force is computed to be just

sufficient to produce the desired torque:

r r 0.050 0.029o iT Ff N 85 N.m F(0.06) m 8,2 2

F 4483 N

+⎛ ⎞ +⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠=

Rounding up the calculated value of F, we

Find that the final proposed answers are (a) inside diameter= 58 mm, (b)

clamping force= 4500 N and (C) a total of nine discs.



Flywheel

A flywheel is an inertial energy-storage device. It absorbs mechanical

energy and serves as a reservoir, storing energy during the period

when the supply of energy is more than the requirement and releases

it during the period when the requirement of energy is more than the

supply.

Flywheels-Function need and Operation

The main function of a fly wheel is to smoothen out variations in the

speed of a shaft caused by torque fluctuations. If the source of the

driving torque or load torque is fluctuating in nature, then a flywheel is

usually called for. Many machines have load patterns that cause the

torque time function to vary over the cycle. Internal combustion

engines with one or two cylinders are a typical example. Piston

compressors, punch presses, rock crushers etc. are the other systems

that have fly wheel.

Flywheel absorbs mechanical energy by increasing its angular

velocity and delivers the stored energy by decreasing its velocity

T

T

T A B C D θ

ω

2

1

m

A B C D θ

max

ωmin

1 CYCLE

Figure 3.3.1



Design Approach

There are two stages to the design of a flywheel.

First, the amount of energy required for the desired degree of

smoothening must be found and the (mass) moment of inertia needed

to absorb that energy determined.

Then flywheel geometry must be defined that caters the required

moment of inertia in a reasonably sized package and is safe against

failure at the designed speeds of operation.

Design Parameters

Flywheel inertia (size) needed directly depends upon the acceptable

changes in the speed.

Speed fluctuation

The change in the shaft speed during a cycle is called the speed

fluctuation and is equal to ωmax- ωmin

Fl max min= ω − ω

We can normalize this to a dimensionless ratio by dividing it by the

average or nominal shaft speed (ωave) .

max minCfω − ω

=ω

Where ωavg is nominal angular velocity

Co-efficient of speed fluctuation

The above ratio is termed as coefficient of speed fluctuation Cf and it is defined as

max minCfω − ω

=ω



Where ω is nominal angular velocity, and ωave the average or mean

shaft speed desired. This coefficient is a design parameter to be

chosen by the designer.

The smaller this chosen value, the larger the flywheel have to be and

more the cost and weight to be added to the system. However the

smaller this value more smoother the operation of the device

It is typically set to a value between 0.01 to 0.05 for precision

machinery and as high as 0.20 for applications like crusher

hammering machinery.

Design Equation

The kinetic energy Ek in a rotating system

= ( )1 2I2

ω

Hence the change in kinetic energy of a system can be given as,

1 2 2E IK m max min2⎛ ⎞= ω − ω⎜ ⎟⎝ ⎠

K 2E E E1= −

( )max minavg 2

ω + ωω =

( )( )1E I 2 CK s avg f avg22E E C I2 1 f

EkIs 2Cf avg

= ω ω

− = ω

=ω

Thus the mass moment of inertia Im needed in the entire rotating

system in order to obtain selected coefficient of speed fluctuation is

determined using the relation



( )( )1E I 2 CK s avg f avg2EkIs 2Cf avg

= ω ω

=ω

The above equation can be used to obtain appropriate flywheel inertia

Im corresponding to the known energy change Ek for a specific value

coefficient of speed fluctuation Cf,

Torque Variation and Energy

The required change in kinetic energy Ek is obtained from the known

torque time relation or curve by integrating it for one cycle.

( )@ max

T T d El avg @ min

K

θ ω− θ =∫

θ ω

Computing the kinetic energy Ek needed is illustrated in the following example

Torque Time Relation without Flywheel

A typical torque time relation for example of a mechanical punching

press without a fly wheel in shown in the figure.

In the absence of fly wheel surplus or positive enregy is avalible

initially and intermedialty and enery absorbtion or negative energy

during punching and stripping operations. A large magitidue of speed

fluctuation can be noted. To smoothen out the speed fluctuation fly

wheel is to be added and the fly wheel energy needed is computed as

illustrated below



Torque

Area+20 073

Area+15 388

BA

Area-26 105

Area-9 202

7 020

0

34 200

C

D A

rms

Average

Shaft angle time t

3600

-34 200

θ

ωmaxωmin

Figure 3.3.2

Accumulation of Energy pulses under a Torque- Time curve

From Area= Accumulated sum =E Min & maxE

A to B

B to C

C to D

D to A

+20 073

-26 105

+15 388

-9 202

+20 073

-6 032

+9 356

+154

ωmin@B

ωmax@C

Total Energy= E E@ωmin@ωmin-=(-6 032)-(+20 073)= 26 105 Nmm2

Figure 3.3.3



Torque Time Relation with Flywheel

7020

8730

0

Average

Time t

C =0.05f

Torque

360Shaft angleθ

Figure 3.3.4

Geometry of Flywheel

The geometry of a flywheel may be as simple as a cylindrical disc of

solid material, or may be of spoked construction like conventional

wheels with a hub and rim connected by spokes or arms Small fly

wheels are solid discs of hollow circular cross section. As the energy

requirements and size of the flywheel increases the geometry

changes to disc of central hub and peripheral rim connected by webs

and to hollow wheels with multiple arms.



d

D

b

d

b

D0 Do

Figure 3.3.5

D

b

d

a

D

0

Arm Type Flywheel

Figure 3.3.6

The latter arrangement is a more efficient of material especially for

large flywheels, as it concentrates the bulk of its mass in the rim which

is at the largest radius. Mass at largest radius contributes much more

since the mass moment of inertia is proportional to mr2



For a solid disc geometry with inside radius ri and out side radius ro,

the mass moment of inertia I is

m2 2I mk (r rm o2= = + 2 )i

The mass of a hollow circular disc of constant thickness t is

( )W 2 2m ro ig gγ

= = π − r t

Combing the two equations we can write

( )4 4I r rm o i2 gtπ γ

= −

Where is material’s weight density γ

The equation is better solved by geometric proportions i.e by

assuming inside to out side radius ratio and radius to thickness ratio.

Stresses in Flywheel

Flywheel being a rotating disc, centrifugal stresses acts upon its

distributed mass and attempts to pull it apart. Its effect is similar to

those caused by an internally pressurized cylinder

2 2 2t i o

2 22 2 2 2i o

r i o 2

3 v 1 3vr r rg 8 3 v

r r3 v r r rg 8 r

γ + +⎛ ⎞⎛σ = ω + −⎜ ⎟⎜ +⎝ ⎠⎝⎛ ⎞γ +⎛ ⎞σ = ω + − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

2 ⎞⎟⎠

γ = material weight density, ω= angular velocity in rad/sec. ν= Poisson’s ratio, is the

radius to a point of interest, ri and ro are inside and outside radii of the solid disc

flywheel.

Analogous to a thick cylinder under internal pressure the tangential

and radial stress in a solid disc flywheel as a function of its radius r is

given by:



σ

σ

t

r

Tang. stress

Radial stress

Radius

Radius

The point of most interest is the inside radius where the stress is a

maximum. What causes failure in a flywheel is typically the tangential

stress at that point from where fracture originated and upon fracture

fragments can explode resulting extremely dangerous consequences,

Since the forces causing the stresses are a function of the rotational

speed also, instead of checking for stresses, the maximum speed at

which the stresses reach the critical value can be determined and safe

operating speed can be calculated or specified based on a safety

factor. Generally some means to preclude its operation beyond this

speed is desirable, for example like a governor.

Consequently

F.O.S (N) = Nos yield

ω=

ω




A 2.2 kw, 960 rpm motor powers the cam driven ram of a press through a gearing of 6:1

ratio. The rated capacity of the press is 20 kN and has a stroke of 200 mm. Assuming

that the cam driven ram is capable of delivering the rated load at a constant velocity

during the last 15% of a constant velocity stroke. Design a suitable flywheel that can

maintain a coefficient of Speed fluctuation of 0.02. Assume that the maximum diameter

of the flywheel is not to exceed 0.6m.

Work done by the press=

3U 20*10 *0.2*0.15 600Nm

==

Energy absorbed= work done= 600 Nm

Mean torque on the shaft:

32.2*10 21.88Nm9602* *60

=π

Energy supplied= work don per cycle

2 * 21.88* 6825 Nm

Thus the mechanical efficiency of the system is =600 0.727 72%825

= π=

η = = =

There fore the fluctuation in energy is =

E Energy absorbed - Energy suppliedk =



( )

( )

600 825*0.075 21.88*6* *0.15538.125Nm

E 538.125kI2 2960C 0.02 2 *f avg 60

2 2.6622 kg m

− π

= =⎛ ⎞ω π⎜ ⎟⎝ ⎠

=

( )

( )

r 2 2I . r r .to i2 griAssuming 0.8ro

78500 4 42.6622 * 0.30 0.24 t2 9.86

59.805t

π= −

=

π= −

=

2 .6622 t 0.044559.805

or 45 mm

∴ = =

r 3 1 32 2 2 2r r rt i og 8 378500 3 0.3 1.92 2 2. 0.24 0.3 *0.24t 9.81 8 3.3

29600.543* 2 *t 602 55667N / m

0.556MPa

⎛ ⎞+ γ + γ⎛ ⎞σ = ω + −⎜ ⎟⎜ ⎟+ γ⎝ ⎠⎝ ⎠+⎛ ⎞⎛σ = ω + −⎜ ⎟⎜

⎝ ⎠⎝

⎛ ⎞σ = π⎜ ⎟⎝ ⎠

==

2 ⎞⎟⎠

)

( )( )( )(

or if 150 MPat6 2150 *10 7961.4 0.4125 0.0376 0.090 0.0331

2 0.548216544 rad / sec

σ =

= ω

= ω

ω =

16544yieldNOS 32 164.65

ω= =

ω π=

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