Department of Mchanical Engineering 3

Clutches
Clutch is a mechanical device, which is used to connect/
disconnect the source of power from the remaining parts of
the power transmission system at the will of the operator

Dynamic representation of a clutch/brake (Mechanical Engineering Design:

Richard G Budynas and J. Keith)

Department of Mechanical Engineering 4

 Clutches
Initial Condition
• Driving member rotating and driven member at rest
Final Condition
• Both members rotate at same speed with no relative motion

Brakes
Initial Condition
• One member (brake drum) rotating and braking member (brake
shoe) is stationary
Final Condition
• Both members are at rest with no relative motion
Department of Mechanical Engineering 5
 Classifications
Clutches
Clutches
Clutches

Positive
Positive
Positive Electro
Electro
Electro Fluid
Fluid
Fluid
Friction
Friction
Friction
contact
contact
contact magnetic
magnetic
magnetic clutches
clutches
clutches

Square
Square
Squarejaw
jaw
jaw Spiral
Spiral
Spiraljaw
jaw
jaw

Singe/multi
Singe/multi
Singe/multi
Cone
Cone
Cone Centrifugal
Centrifugal
Centrifugal
plate
plate
plate

Department of Mechanical Engineering 6

 Positive Contact Clutches

• Two halves carrying projections or halves

• One half is fixed and the other can move along the axis
• Jaws of moving half enter into socket of mating half
• Eg. Square jaw, spiral jaw
Advantages
• No slip and positive engagement
• No heat during engagement/disengagement
Drawbacks
• Engagement only when stationary or rotate at very low speed
• High speed engagement results in shocks
Department of Mechanical Engineering 7
Positive Contact Clutches

Department of Mechanical Engineering 8

 Friction Clutch: Single plate

• Consists of two flanges, one rigidly keyed to driving shaft and one
connected to driven shaft with splines
• Actuating force is given by spring
• Power transmission between driving and driven flange is through
friction
• Fork inserted on the collar of the driven flange for axial
movement of driven flange
Department of Mechanical Engineering 9
Friction Clutch: Single plate

Department of Mechanical Engineering 10

 Friction Clutch: Single plate
Advantages
• Smooth engagement
• Slip only during engagement
• Acts as safety device
Design Considerations
• Selection of type of clutch
• Selection of friction material of contact surfaces
• Designing for sufficient torque capacity
• Smooth engagement/disengagement
• Provision for holding contact surface without any external
assistance
• Low weight of rotating parts to reduce inertia forces
• Provision for taking or compensating wear
• Provision for carrying away the heat generated
Department of Mechanical Engineering 11
Single plate clutch: Torque transmitting capacity

D : Outer diameter of friction disc (mm) : Do

d : Inner diameter of friction disc (mm) : Di
p : Intensity of pressure at radius r (N/mm2) : p
P : Operating axial force (N) : Fap
Mt : Torque transmitted (N-mm) : T
μ : Friction coefficient : μ
r : radius of the element dr from the centre of the disc : r
Department of Mechanical Engineering 12
Single plate clutch: Torque transmitting capacity

𝐷𝑜 /2
• Total axial load, 𝐹𝑎 = 2𝜋 ‫ 𝐷׬‬/2 𝑝𝑟𝑑𝑟
𝑖
𝐷 /2
• Total torque, 𝑇 = 2𝜋𝜇 ‫𝑜 𝐷׬‬/2 𝑝𝑟 2 𝑑𝑟
𝑖

Using uniform pressure theory (New clutch)

(𝐷𝑜2 −𝐷𝑖2 )
• Total axial load, 𝐹𝑎𝑝 = 𝜋𝑝 eqn 13.9a
4
1 2(𝐷𝑜3 −𝐷𝑖3 )
• Total torque, 𝑇 = 𝜇𝐹𝑎𝑝 𝐷𝑚𝑝 : 𝐷𝑚𝑝 = eqn 13.9b
2 3(𝐷𝑜2 −𝐷𝑖2 )
Using uniform wear theory (Worn out clutch)
(𝐷𝑜 −𝐷𝑖 )
• Total axial load,𝐹𝑎𝑤 = 𝜋𝑝𝑚𝑎𝑥 𝐷𝑖 eqn 13.9d
2
1 1
• Total torque, 𝑇 = 𝜇𝐹𝑎𝑤 𝐷𝑚𝑤 : 𝐷𝑚𝑤 = (𝐷𝑜 + 𝐷𝑖 ) eqn 13.9e
2 2
Department of Mechanical Engineering 13
Comparison: Uniform Pressure & Wear Theory

1
Total torque, T = 𝜇𝐹𝑎 𝐷𝑚 :
2
3 3
2(𝐷𝑜 −𝐷𝑖 )
• 𝐷𝑚𝑝 = Using uniform pressure theory
3(𝐷𝑜2 −𝐷𝑖2 )
1
• 𝐷𝑚𝑤 = (𝐷𝑜 + 𝐷𝑖 ) Using uniform wear theory
2

2(𝐷𝒐3 − 𝐷𝒊3 ) 1
𝐷𝒐 𝐷𝒊 (𝐷𝒐 + 𝐷𝒊 ) % difference
3(𝐷𝒐2 − 𝐷𝒊2 ) 2

140 80 112.72 110 2.4

200 100 155.56 150 3.5

200 180 190.18 190 0.08

Department of Mechanical Engineering 14

 Single plate clutch
Conclusions
• Uniform pressure theory applicable for new friction lining
• Uniform wear theory applicable for worn-out friction lining
• Friction radius for new clutch is slightly greater than worn-out
clutches
• Torque transmitting capacity of new clutch slightly higher
• Major life of friction lining comes under uniform wear criterion
• Logical and safer to use uniform wear theory in design of clutches

Factors Considered for design

• Service factor: To compensate for the starting torque
• Location of clutch: Low torque capacity requirement
• Coefficient of friction:

Department of Mechanical Engineering 15

 Problem 1.1
A plate clutch consists of one pair of contacting surfaces. The inner
and outer diameters of the friction disk are 100 and 200 mm
respectively. The coefficient of friction is 0.1 and the permissible
intensity of pressure is 1 N/mm2. Assuming uniform wear theory,
calculate the power transmitting capacity of the clutch at 750 rpm.
Calculate the power transmitting capacity of the clutch using
uniform pressure theory.

Department of Mechanical Engineering 16

 Problem 1.2
An automotive clutch consists of two pairs of contacting surfaces
with an asbestos friction lining. The torque transmitting capacity of
the clutch is 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 disk is fixed as 250
mm. Using uniform wear theory, calculate
• The inner diameter of the friction disk: and
• The spring force required to keep the clutch in an engaged
position

Department of Mechanical Engineering 17

 Problem 1.3
An automotive plate clutch consists of two pairs of contacting surfaces
with asbestos friction lining. The maximum engine torque is 250 N-m. The
coefficient of friction is 0.35. The inner and outer diameters of friction
lining are 175 and 250 mm respectively. The clamping force is provided by
nine springs, each compressed by 5 mm to give a force of 800 N, when the
clutch is new.
• What is the factor of safety with respect to slippage when the clutch is
brand new?
• What is the factor of safety with respect to slippage after initial wear
has occurred?
• How much wear of friction lining can take place before the clutch will
slip?

Department of Mechanical Engineering 18

 Problem 1.4
A single plate clutch consists of one pair of contacting surfaces. Because of
space limitations, the outer diameter of the friction disk is fixed as D2. The
permissible intensity of pressure is p and the coefficient of friction, μ.
Assuming uniform wear theory, plot the variation of the torque
transmitting capacity against the ratio of diameters (D1/D2). Show that the
torque transmitting capacity of the clutch is maximum when (D1/D2) is
equal to 0.577

Department of Mechanical Engineering 19

 Multi-Disk Clutches
Disc B:
• Made of Bronze
• Connected by four bolts through
holes
• Bolts are rigidly fixed to a rotating
drum keyed to driving shaft
• Axial force required are provided
by means of springs

Disc A:
• Hardened steel
• Connected to driven shaft
using splines
• Free to move in axial
direction
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Multi-Disk Clutches

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 Governing Equations

1
• Total torque, 𝑀𝑡 = 𝑖𝜇𝐹𝑎 𝐷𝑚 :
2
2(𝐷23 −𝐷13 )
𝐷𝑚 = (New clutch)
3(𝐷22 −𝐷12 )
1
𝐷𝑚 = (𝐷1 + 𝐷2 ) (worn-out clutch)
2

• Number of friction surface, 𝑖 = 𝑖1 + 𝑖2 − 1 (eqn 19-93)

𝑖
• Number of driving disks, 𝑖1 = (eqn 19-94)
2
𝑖
• Number of driven disks, 𝑖2 = +1 (eqn 19-95)
2

Department of Mechanical Engineering 22

 Difference Single Plate & Multi Plate
Parameters Single plate Multi plate

Can be more than

Contacting surface Maximum of two
two

Torque transmitting More (decreases

Less
capacity with cooling oil)

Heat generation Less More

Space Requirement More Less

Cooling oil Not required Required

Department of Mechanical Engineering 23
 Difference Wet and Dry clutch
Parameters Wet clutch Dry clutch
Coefficient of
Less More
friction
Torque capacity Less High
Moisture Proper sealing
Not a serious issue
contamination required
Heat dissipation Lubricating oil Difficult
Wear rate Less More
Clutch engagement Smooth Rough
Surface area Reduced (grooving's) More
Department of Mechanical Engineering 24
 Problem 1.5
An oil immersed multi-disk clutch with cork sheet as the
friction material is used on a scooter engine. The torque
transmitted by the clutch is 10 N-m. The coefficient of
friction between the cork sheet and the steel plate in the
wet condition is 0.2. The permissible pressure on the cork
sheet is 0.1 N/mm2. The inner and outer diameters of the
friction lining are 65 and 95 mm, respectively. There are
radial slots, on the friction surface for the circulation of the
coolant, which reduces the effective friction area. To account
for these slots, the number of contacting surfaces can be
increased by 5%. Assuming uniform wear theory, calculate
the required number of contacting surfaces.

Department of Mechanical Engineering 25

 Problem 1.6
An oil immersed multi-disk clutch with moulded asbestos on
one side and steel disks on the other, is used in an
application. The torque transmitted by the clutch is 75 N-m.
The coefficient of friction between the asbestos lining and
the steel plate in the wet condition is 0.1. The permissible
intensity of pressure on the asbestos lining is 500 kPa. The
outer diameter of the friction lining is kept as 100 mm due to
the space limitation. Assuming uniform wear theory,
calculate the inside diameter of the disks, the required
number of disks and the clamping force.

Department of Mechanical Engineering 26

 Cone Clutches

• Consists of inner and outer conical surfaces

• Outer cone keyed to driving shaft and inner cone connected to driven
shaft using splines
• Wedge action causes considerable friction force with a small engaging
force
• Recommended semi cone angle – 12.50
• Strict co-axiality required
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Cone Clutches

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 Cone Clutches: Governing Equations

(𝐷22 −𝐷12 )
Total axial load, 𝐹𝑎 = 𝜋𝑝 (Eqn. 19-86b)
4
(𝐷 −𝐷 )
𝐹𝑎 = 𝜋𝑝𝐷1 2 1 (Eqn. 19-83)
2
𝜇𝐹𝑎 𝐷𝑚
• Total torque, 𝑀𝑡 = (Eqn. 19-78)
2 sin 𝛼
3 3
2(𝐷 −𝐷 )
𝐷𝑚 = 22 12 (New clutch)
3(𝐷2 −𝐷1 )
1
𝐷𝑚 = (𝐷1 + 𝐷2 ) (worn-out clutch)
2

Department of Mechanical Engineering 29

 Observations: Torque Equation

• Torque capacity is inversely proportional to sin 𝛼

• 𝛼 < ∅, can cause self engagement

• For 𝑀𝑡 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝐹𝑎 ∞ sin 𝛼 , small axial force

for decrease in semi cone angle

𝑀𝑡 𝑐𝑜𝑛𝑒 1 1
• = = = 4.62
𝑀𝑡 𝑝𝑙𝑎𝑡𝑒 sin 𝛼 sin 12.5

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Friction materials: Desirable Properties

• High coefficient of friction

• Constant coefficient of friction for a wide range of
temperature
• Good thermal conductivity
• Unaffected by environmental conditions
• Resistance to abrasive and adhesive wear
• Good resilience to provide good distribution of
pressure at contacting surface

Department of Mechanical Engineering 31

 Friction materials: Types
Woven Moulded

Asbestos fibre woven around Wet mixture of brass chips and

brass, copper or zinc wires and asbestos poured into mould.
impregnated with rubber or Heated and pressed for curing
asphalt time
Flexible Rigid
Higher friction coefficient Less
Cheap cost Higher cost
Higher wear rate Less
Department of Mechanical Engineering 32
 Friction materials: Four Basic Ingredients
• Fibers: Rigidity and strength (Steel wool or aramid
replacing asbestos)
• Filler: Fills space between fibers and extend lining life
(barytes, clay and calcium carbonate
For metallic lining: fine powder is used as filler
• Binder: Glue that holds lining ingredients together
(phenylformaldehyde)
• Friction modifiers: Control abrasive properties of lining
(Brass and zinc)
Department of Mechanical Engineering 33
 Problem 1.7

A cone clutch with asbestos friction lining transmits 30 kW

power at 500 rpm. The coefficient of friction is 0.2 and the
permissible intensity of pressure is 0.35 N/mm2. The semi-
cone angle is 12.50. The outer diameter is fixed as 300 mm
from space limitations. Assuming uniform wear theory,
Calculate:
• The inner diameter
• The face width of the friction lining
• The force required to engage the clutch
Department of Mechanical Engineering 34
 Problem 1.8
A cone clutch is used to connect an electric motor running at
1440 rpm with a machine which is stationary. The machine is
equivalent to a rotor of 150 kg mass and radius of gyration as 250
mm. The machine has to be brought to the full speed of 1440
rpm from stationary condition in 40s. The semi-cone angle is
12.50. The mean radius of the clutch is twice the face width. The
coefficient of friction is 0.2 and the normal intensity of pressure
between contacting surfaces should not exceed 0.1N/mm2.
Assuming uniform wear criterion, calculate:
• The inner and outer diameter
• The face width of friction lining
• The force required to engage the clutch
• The amount of heat generated during each engagement of
clutch

Department of Mechanical Engineering 35

 Centrifugal clutches

• Engages the load after a particular speed

• Helps in warm-up and acceleration without load
• Centrifugal force increases with speed which causes engagement of
clutch
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Centrifugal clutches: Animation

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 Centrifugal clutches

Notations used:
𝑟 ′ = 𝑖𝑛𝑛𝑒𝑟 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑢𝑚 (𝑚𝑚)
𝑟 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑜𝑓 𝑠ℎ𝑜𝑒 𝑖𝑛 𝑒𝑛𝑔𝑎𝑔𝑒𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 (𝑚𝑚)
𝑤 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑠ℎ𝑜𝑒(𝑁)
𝐹𝑐 = 𝐶𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 (𝑁)
𝐹𝑠 = 𝑆𝑝𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 (𝑁)
𝑖 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠ℎ𝑜𝑒𝑠
𝜔2 = 𝑟𝑢𝑛𝑛𝑖𝑛𝑔 𝑠𝑝𝑒𝑒𝑑(𝑟𝑎𝑑/𝑠)
𝜔1 = 𝑠𝑝𝑒𝑒𝑑 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝑒𝑛𝑔𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝑠𝑡𝑎𝑟𝑡𝑠 (𝑟𝑎𝑑/𝑠)
Department of Mechanical Engineering 38
 Centrifugal clutches: Governing Equations

𝑤
𝐹𝑜𝑟𝑐𝑒 𝑑𝑢𝑟𝑖𝑛𝑔 𝑐𝑙𝑢𝑡𝑐ℎ 𝑒𝑛𝑔𝑎𝑔𝑒𝑚𝑒𝑛𝑡, 𝐹𝑐1 = 𝜔12 𝑟 Eqn. 19-114
1000 𝑔
𝑤
𝑂𝑢𝑡𝑠𝑖𝑑𝑒 𝑟𝑎𝑑𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒, 𝐹𝑐2 = 𝜔22 𝑟 Eqn. 19-115
1000 𝑔
𝑤
𝑆𝑝𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒, 𝐹𝑠 = 𝜔12 𝑟
1000 𝑔
𝑤
𝑁𝑒𝑡 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑑𝑟𝑢𝑚, 𝐹𝑐 = (𝜔22 −𝜔12 )𝑟 Eqn.19-116a,b
1000 𝑔
𝑤
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒, 𝐹𝑓 = 𝜇 (𝜔22 −𝜔12 )𝑟
1000 𝑔
𝑤
𝐹𝑟𝑖𝑡𝑐𝑡𝑖𝑜𝑛 𝑡𝑜𝑟𝑞𝑢𝑒, 𝑀𝑡 = 𝜇 (𝜔22 −𝜔12 )𝑟𝑟 ′ Eqn.19-118
1000 𝑔
Department of Mechanical Engineering 39
 Problem 1.9
A centrifugal clutch, transmitting 20kW at 750 rpm consists of four
shoes. The clutch is to be engaged at 500 rpm. The inner radius of
the drum is 165 mm. The radius of the centre of gravity of the
shoes is 140 mm, when the clutch is engaged. The coefficient of
friction is 0.3, while the permissible pressure on friction lining is 0.1
N/mm2. Calculate:
• The mass of each shoe
• The dimensions of friction lining

Department of Mechanical Engineering 40

 Problem 1.10
A centrifugal clutch consists of four shoes, each having a mass of
1.5 kg. In the engaged position, the radius to the centre of gravity
of each shoes is 110 mm, while the inner radius of the drum is 140
mm. The coefficient of friction is 0.3. The pre-load in the spring is
adjusted in such a way that the spring force at the beginning of
engagement is 700 N. The running speed is 1440 rpm. Calculate:
• The speed at which the engagement begins
• The power transmitted by the clutch at 1440 rpm

Department of Mechanical Engineering 41

 Problem 1.11
Design a single plate automobile clutch to transmit a maximum
torque of 250 N-m at 2000 rpm. The outside diameter of the clutch is
250 mm and the clutch is engaged at 55 km/h. Find
• The inner diameter
• The number of revolutions of the clutch slip during engagement and
• Heat to be dissipated by the clutch for each engagement
Following additional data available are
• Engine torque during engagement= 100 N-m
• Mass of automobile= 1500 kg
• Diameter of the automobile wheel = 0.7
• Moment of inertia of combined engine parts, flywheel and input side of
the clutch = 1 kg-m2
• Gear reduction ratio at differential = 5
• Torque at rear wheels available for accelerating automobile = 175 Nm
• Coefficient of friction of clutch material = 0.3
• Permissible pressure = 0.13 N/mm2
Department of Mechanical Engineering 42
 Problem 1.12
A single dry plate clutch is to be designed to transmit 7.5 KW at 900
rpm. Find
• Diameter of the shaft, Allowable Shear stress of shaft, 𝜏1 = 40 𝑁/𝑚𝑚2
• Mean radius and face width of friction lining assuming ratio of mean
radius to the face width as 4
• Outer and inner radii of the clutch plate
• Dimensions of the spring assuming no of springs are 6 and spring index=
6. The allowable shear stress of the spring wire may be taken as 420 MPa.
• Assume missing data

Department of Mechanical Engineering 43

 Brakes
A mechanical device, which is used to absorb the energy
possessed by a moving system or mechanism by means of friction.
Energy absorbed can be:
• Potential energy- Hoists and elevators
• Kinetic energy-Moving vehicle

Based on actuation

Department of Mechanical Engineering 44

 Brakes: Types

Mechanical Hydraulic

Pneumatic Electrical
Department of Mechanical Engineering 45
 Brakes: Energy Equations
For a mechanical system moving with a mass, m and velocity v1
slowing down to v2 during the period of braking
1
𝐾. 𝐸 = 𝑚 𝑣12 − 𝑣22 (Eqn 19:135a)
2
For a rotating body,

1
𝐾. 𝐸 𝑟 = 𝐼 𝜔12 − 𝜔22 (Eqn 19:136)
2

In hoists
𝑃. 𝐸 = 𝑚𝑔ℎ
Total energy, E
𝐸 = 𝐾. 𝐸 + 𝐾. 𝐸 𝑟 + 𝑃. 𝐸

Total energy = work done

𝐸 = 𝑀𝑡 𝜃
Department of Mechanical Engineering 46
 Problem 1.13
A four wheeled automobile car has a total mass of 1000 kg. The moment
of inertia of each wheel about a transverse axis through its centre of
gravity is 0.5 kg-m2. The rolling radius of the wheel is 0.35 m. The
rotating and reciprocating parts of the engine and the transmission
system are equivalent to a moment of inertia of 2.5 kg-m2, which rotates
at 5 times the road-wheel speed. The car is travelling at a speed of 100
km/h on a plane road. When the brakes are applied, the car decelerates
at 0.5 g. The are brakes on all four wheels. Calculate:
• The energy absorbed by each brake
• The torque capacity of each brake
Department of Mechanical Engineering 47
 Problem 1.14
A mass of 2500 kg is lowered at a velocity of 1.5 m/s from the drum as
shown in Figure. The mass of the drum is 50 kg and its radius of
gyration can be taken as 0.7 m. On applying the brake, the mass is
brought to rest in a distance of 0.5 m. Calculate:
• The energy absorbed by the brake
• The torque capacity of the brake

Department of Mechanical Engineering 48

Block Brake: Short Shoe
• Block brake: A simple block
pressed against rotating drum
• Friction between block and brake
drum causes retardation
• Commonly employed in railway
wagons
• Angle of contact between the
block and brake drum is usually
small
• Intensity of pressure is uniform
Department of Mechanical Engineering 49
Block Brakes: Short Shoe

Department of Mechanical Engineering 50

 Brake Capacity
• Pressure between braking
surfaces
• Contacting area of braking
surface
• Radius of brake drum
• Coefficient of friction
• Ability of brake to dissipate
heat

Department of Mechanical Engineering 51

 Block Brake: Short Shoe
Assumptions:
• Block is rigidly attached to the lever
• Angle of contact small, resulting in uniform
pressure distribution
• Brake drum rotating in clockwise direction
Notations:
• 𝑀𝑡 = 𝑏𝑟𝑎𝑘𝑖𝑛𝑔 𝑡𝑜𝑟𝑞𝑢𝑒
• 𝑅 = 𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑢𝑚
• 𝜇𝑁 = 𝐹𝜃 = 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
• 𝑝 = 𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦
• 𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑙𝑜𝑐𝑘
• 𝑤 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑙𝑜𝑐𝑘

𝑴𝒕 = 𝝁𝑵𝑹
𝐹𝜃 𝑎 1 𝑐 𝑵 = 𝒑𝒍𝒘
𝐹= − Optimum width
𝑎+𝑏 𝜇 𝑎
1 1
(Eqn:19-145) 𝑑𝑟𝑢𝑚 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 < 𝑤 < 𝑑𝑟𝑢𝑚 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
4 2
Department of Mechanical Engineering 52
 Problem 1.15
A single block brake with a torque capacity of 250 N-m is shown in
Figure. The brake drum rotates at 100 rpm and the coefficient of
friction is 0.35. Calculate:
• The actuating force and the hinge-pin reaction for clockwise
rotation of the drum:
• The actuating force and hinge-pin reaction for anti clockwise
rotation of the drum
• The rate of heat generated during the braking action
• The dimensions of the block, if the intensity of pressure between
the block and brake drum is 1 N/mm2. The length of the block is
twice its width.
State whether the brake is self-locking

Department of Mechanical Engineering 53

Problem 1.15

Department of Mechanical Engineering 54

 Problem 1.16
A double brake is shown in
Figure. The brake drum
rotates in clockwise
direction and the actuating
force is 500 N. The
coefficient of friction
between the blocks and the
drum is 0.35. Calculate the
torque absorbing capacity of
the brake.

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Problem 1.16

Department of Mechanical Engineering 56

 Problem 1.17
A block brake with a short shoe is shown in Figure. It is to be designed so that the
product pv is limited to 2.
Where,
p = normal pressure between friction lining and the brake drum (N/mm2)
v= peripheral velocity of brake drum (m/s)
The coefficient of friction between the brake drum and the friction lining is 0.2. The
cable drum is connected to the brake by means of a pair of spur gears. The brake
drum rotates four times as fast as the cable drum. The permissible intensity of
pressure on friction lining is 1N/mm2. Calculate:
(i) The magnitude of the brake shoe force
(ii) The area of friction lining
(iii) The uniform velocity at which the
mass can be lowered

Department of Mechanical Engineering 57

 Block Brake: Long Shoe
• Short Shoe: Angle of contact less than 450
• Assumed normal reaction (N) and frictional force (μN) are
concentrated at the midpoint of the shoe
• Not applicable for long shoe

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 Block Brake: Long Shoe
Notations
• θ- Semi block angle
• dφ- Element of friction lining
• φ –Location of friction lining element
• Rdφw- Area of the element
• w- width of the friction lining parallel to
the axis
• dN-Normal reaction on the element

Department of Mechanical Engineering 59

 Block Brake: Long Shoe

4 sin 𝜃
Assumptions: 𝑀𝑡 = 𝜇𝑁𝑅
2𝜃 + sin 2𝜃
• Wear occurs at the brake shoe 𝐷 4 sin 𝜃
• δy=wear in y direction (constant) 𝑀𝑡 = 𝜇𝐹
2 2𝜃 + sin 2𝜃
• δr= wear in radial direction
(Eqn 19-141)
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Block Brake: Pivoted Long Shoe

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Pivoted Long Shoe: Force Diagram
4𝑅 sin 𝜃
ℎ=
2𝜃 + sin 2𝜃

𝑀𝑡 = 2𝜇𝑅2 𝑤𝑝𝑚𝑎𝑥 sin 𝜃

1
𝑅𝑥 = 𝑅𝑤𝑝𝑚𝑎𝑥 (2θ + sin 2𝜃)
2
1
𝑅𝑦 = 𝜇𝑅𝑤𝑝𝑚𝑎𝑥 (2θ + sin 2𝜃)
2

Department of Mechanical Engineering 62

 Problem 1.18
A pivoted double-block brake, similar to that in Figure has two
shoes, which subtend an angle of 1000. The diameter of the brake
drum is 500 mm and the width of the friction lining is 100 mm. The
coefficient of friction is 0.2 and the maximum intensity of pressure
between the lining and the brake drum is 0.5 N/mm2. The pivot of
each shoe is located in such a manner that the moment of the
frictional force on the shoe is zero. Calculate:
• The distance of the pivot from the axis of the brake drum
• The torque capacity of each shoe
• The reactions at the pivot

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Internal Expanding Brake: Animation

Department of Mechanical Engineering 64

 Internal Expanding Brake
• A shoe pivoted at one end and subject
to an actuating force
• Friction lining fixed on the shoe
• Complete assembly of shoe, lining and
pivot is placed inside the brake drum
• Internal shoe brakes with two
symmetrical shoes are used on all
automobiles

Assumptions
• 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
• The brake drum and shoes are rigid
• The centrifugal force acting on the shoe is negligible
• Coefficient of friction is constant
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Internal Expanding Brake: Force Diagram

Department of Mechanical Engineering 66

 Internal Expanding Brake: Final Equations
Moment due to friction force about pivot
Eqn 19-174a
𝜇𝑝𝑎 𝑟𝑏 𝜃2 𝜇𝑝𝑎 𝑟𝑏 4𝑟 cos 𝜃1 − cos 𝜃2 − 𝑎 cos 2𝜃1 − cos 2𝜃2
𝑀𝑡𝜇 = න sin 𝜃 𝑟 − 𝑎 cos 𝜃 𝑑𝜃 =
sin 𝜃𝑎 𝜃1 4 sin 𝜃𝑎
Moment due to normal force about pivot
Eqn 19-175
𝜃2
𝑝𝑎 𝑟𝑏𝑎 𝑝𝑎 𝑟𝑏𝑎 2 𝜃2 − 𝜃1 − sin 2𝜃2 − sin 2𝜃1
𝑀𝑡𝑛 = න sin2 𝜃 𝑑𝜃 =
sin 𝜃𝑎 𝜃1 4 sin 𝜃𝑎
Frictional torque
𝜇𝑟 2 𝑝𝑎 𝑏 cos 𝜃1 − cos 𝜃2 Eqn 19-177
𝑀𝑡 =
sin 𝜃𝑎
For the clockwise rotation of the brake drum
𝑀𝑡𝑛 − 𝑀𝑡𝜇 Eqn 19-176
𝑃/𝐹 =
𝑐
For the anti- clockwise rotation of the brake drum
𝑀𝑡𝑛 + 𝑀𝑡𝜇 Eqn 19-180
𝑃/𝐹 =
𝑐
Department of Mechanical Engineering 67
 Internal Expanding Brake: Force Diagram
For forward motion of the vehicle For backward motion of the vehicle
𝑀𝑡𝑛 − 𝑀𝑡𝜇 𝑀𝑡𝑛 + 𝑀𝑡𝜇
𝑃/𝐹 = 𝑃/𝐹 =
𝑐 𝑐

Advantages
• Simple construction
• More reliable
• Little maintenance
• Small braking force can produce a large braking torque
• Protection against foreign particles

Disadvantages
• Relatively poor heat dissipating capacity
• Can become self locking due to wear if improperly
designed

Department of Mechanical Engineering 68

 Problem 1.19
An internal expanding double shoe
brake is shown in Figure. The face
width of friction lining is 40 mm
and the maximum intensity of
normal pressure is limited to 1
N/mm2. The coefficient of friction
is 0.32. The angle θ1 can be
assumed to be zero. Calculate:
• The actuating force: F/P
• The torque absorbing capacity
Department of Mechanical Engineering 69
 Problem 1.20
An internal expanding brake with four
identical shoes is shown in Figure.
The actuating hinge pin support a pair
of shoes. The actuating mechanism is
designed in such a way that it
produces the same force P/F on each
of the four shoes. The face width of
the friction lining is 50 mm and the
maximum intensity of normal
pressure is limited to 1N/mm2. The
coefficient of friction is 0.30.
Calculate:
• The actuating force
• The torque absorbing capacity of
the brake
Department of Mechanical Engineering 70
Simple Band Brakes
𝐹1 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑛 𝑡𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒
𝐹2 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑛 𝑙𝑜𝑜𝑠𝑒 𝑠𝑖𝑑𝑒
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝜃 = 𝑎𝑛𝑔𝑒 𝑜𝑓 𝑤𝑟𝑎𝑝
The torque absorbing capacity
𝑀𝑡 = 𝐹1 − 𝐹2 𝑅

𝐹1
= 𝑒 𝜇𝜃
𝐹2

Department of Mechanical Engineering 71

Differential Band Brakes
𝐹1 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑛 𝑡𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒
𝐹2 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑛 𝑙𝑜𝑜𝑠𝑒 𝑠𝑖𝑑𝑒
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝜃 = 𝑎𝑛𝑔𝑒 𝑜𝑓 𝑤𝑟𝑎𝑝

Department of Mechanical Engineering 72

 Problem 1.21
A differential band brake is shown in Figure. The width and the
thickness of the steel band are 100 mm and 3 mm respectively
and the maximum tensile stress in the band is 50 N/mm2. The
coefficient of friction between the friction lining and the brake
drum is 0.25. Calculate:
• The tensions in the band
• The actuating force
• The torque capacity of the brake
• Find out whether the brake is self locking

Department of Mechanical Engineering 73