Chapter10 clutches and_brakes

Mechanical DesignPRN Childs, University of Sussex

Clutches and brakes

Mechanical Design


Aims

• A clutch is a device that permits the smooth, gradual connection of two shafts rotating at different speeds.

• A brake enables the controlled dissipation of energy to slow down, stop or control the speed of a system.

• This section describes the basic principles of frictional clutches and brakes and outlines design and selection procedures for disc clutches, disc and drum brakes.


Learning objectives

At the end of this section you should be able to determine:

• the primary dimensions for a single disc clutch, • the principal dimensions and number of discs for

a multiple disc clutch,• torque capacity for short or long, internal or

external brakes,• the configuration of a brake to be self-

energising.


Introduction

• When a rotating machine is started it must be accelerated from rest to the desired speed.

• A clutch is a device used to connect or disconnect a driven component from a prime mover such as an engine or motor.


Automotive clutches• A familiar application is the use of a clutch between a car

engine’s crankshaft and the gearbox. • The need for the clutch arises from the relatively high

torque requirement to get a vehicle moving and the low torque output from an internal combustion engine at low levels of rotational speed.

• The disconnection of the engine from the drive enables to engine to speed up unloaded to about 1000 rpm where it is generating sufficient torque to drive the transmission.

• The clutch can then be engaged, allowing power to be transmitted to the gearbox, transmission shafts and wheels.


Brakes

• A brake is a device used to reduce or control the speed of a system or bring it to rest.


Typical applications for a clutch and a brake

MOTOR ORENGINE

COUPLING

CLUTCH GEAR BOX

COUPLING

DRIVENMACHINE

MOTOR ORENGINE BRAKE

CONNECTIONTO DRIVENMACHINE


Classification

• Clutches and brakes are similar devices providing frictional, magnetic or mechanical connection between two components.

• If one component rotates and the other is fixed to a non-rotating plane of reference the device will function as a brake and if both rotate then as a clutch.


Force, torque and energy

• Whenever the speed or direction of motion of a body is changed there is force exerted on the body.

• If the body is rotating, a torque must be applied to the system to speed it up or slow it down.

• If the speed changes, so does the energy, either by addition or absorption.


Acceleration

• The acceleration, α, of a rotating machine is given by

• where • T is the torque (N m) and • I is the mass moment of inertia (kg m2).

IT=α


Mass moment of inertia

• The mass moment of inertia can often be approximated by considering an assembly to be made up of a series of cylinders and discs and summing the individual values for the disc and cylinder mass moments of inertia.


Mass moments of inertia

• The mass moments of inertia for a cylinder and a disc are given by

( )4i

4ocylinder rrL

21

I −ρπ=

4odisc Lr

21

I ρπ=


Clutch or brake location

• Torque is equal to the ratio of power and angular velocity.

• In other words torque is inversely proportional to angular velocity.

• This implies that it is usually advisable to locate the clutch or brake on the highest speed shaft in the system so that the required torque is a minimum.

• Size, cost and response time are lower when the torque is lower.


Friction clutches• Friction type clutches and

brakes are the most common. • Two or more surfaces are

pushed together with a normal force to generate a friction torque.

• Normally, at least one of the surfaces is metal and the other a high friction material referred to as the lining.

• The frictional contact can occur radially, as for a cylindrical arrangement, or axially as in a disc arrangement.

ACTUATORPUSHES DISCS

TOGETHER

INPUT

OUTPUT

DRIVINGDISC

DRIVENDISC

FRICTIONMATERIAL


Function

• The function of a frictional clutch or brake surface material is to develop a substantial friction force when a normal force is applied.

• Ideally a material with a high coefficient of friction, constant properties, good resistance to wear and chemical compatibility is required.

• Clutches and brakes transfer or dissipate significant quantities of energy and their design must enable the absorption and transfer of this heat without damage to the component parts of the surroundings.


Supply

• With the exception of high volume automotive clutches and brakes, engineers rarely need to design a clutch or a brake from scratch.

• Clutch and brake assemblies can be purchased from specialist suppliers and the engineer’s task is to specify the torque and speed requirements, the loading characteristics and the system inertias and to select an appropriately sized clutch or brake and the lining materials.


Clutches

• The function of a clutch is to permit the connection and disconnection of two shafts, either when both are stationary or when there is a difference in the relative rotational speeds of the shafts.

• Clutch connection can be achieved by a number of techniques from direct mechanical friction, electromagnetic coupling, hydraulic or pneumatic means or by some combination.


Types of clutch

METHOD OFENGAGEMENT

MECHANICAL

PNEUMATICAND HYDRAULIC

ELECTRICAL

MAGNETIC

POSITIVECONTACT

FRICTION

OVERRUNNING

MAGNETIC

FLUIDCOUPLING

SQUARE JAW

SPIRAL JAW

TOOTHED

DISC

DRUM

CONE

ROLLER

SPRAG

SPRING WOUND

MAGNETIC PARTICLE

HYSTERESIS

EDDY CURRENT

DRY FLUID

HYDRAULIC

ACTUATIONMETHOD OF


Requirements

Clutches must be designed principally to satisfy four requirements:

• the necessary actuation force should not be excessive,

• the coefficient of friction should be constant,• the energy converted to heat must be dissipated,• wear must be limited to provide reasonable

clutch life.


Positive contact clutches

• Positive contact clutches have teeth or serrations, which provide mechanical interference between mating components.


Over-running clutches• Over-running clutches operate automatically based on

the relative velocity of the mating components. • They allow relative motion in one direction only. • If the rotation attempts to reverse the constituent

components of the clutch grab the shaft and lock up. • Applications include backstops, indexing and

freewheeling. • The range of overrunning clutches the simple ratchet

and pawl, roller, sprag and spring wound clutches.


Ratchet and pawl clutch


Roller clutch


Sprag clutch


Centrifugal clutches• Centrifugal clutches engage automatically when the shaft speed

exceeds some critical value. • Friction elements are forced radially outwards and engage against

the inner radius of a mating cylindrical drum. • Common applications of centrifugal clutches include chainsaws,

overload-releases and go-karts.

RETAINING SPRING

DRIVING ELEMENTDRIVEN ELEMENT

SHOE


Clutch selection criteria (Neale(1994))

Machine tool gearboxes. Numerical control machine tools.

Compact. Low wear.Magnetic

Electric motor drives. Industrial diesel drives.

Automatic engagement at a critical speed.Centrifugal clutch

Machine tool head stocks. Motorcycles.

The power transmitted can be increased by using more plates allowing a reduction in diameter.

Multiple disc clutch

Automobile drives.Used when diameter is not restricted. Simple construction

Single disc clutch

Contractor’s plant. Feed drives for machine tools.

Embodies the mechanical principle of the wedge which reduces the axial force required to transmit a given torque.

Cone clutch

One way operation.One way clutch. Rollers ride up ramps and drive by wedging into place.

Roller

One way operation. e.g. backstop for hoists.

One way clutch. Profiled elements jam against the outer edge to provide drive. High torque capacity.

Sprag

TYPICAL APPLICATIONSCHARACTERISTICSTYPE OF CLUTCH


Clutch design

• Clutches are rarely designed from scratch. • Either an existing design is available and is

being modified for a new application or a clutch can be bought in from a specialist manufacturer.

• In the latter case the type, size and the materials for the clutch lining must be specified.

• This requires determination of the system characteristics such as speed, torque, loading characteristic and operating temperatures.

• Many of these factors have been lumped into a multiplier called a service factor.


Service factor

• A lining material is typically tested under steady conditions using an electric motor drive.

• The torque capacity obtained from this test is then de-rated by the service factor according to the particular application to take account of vibrations and loading conditions.


Disc clutch PRESSURE

PLATE

HELICALSPRINGS

OR CASTINGPRESSING

THRUSTBEARING

MOVE AXIALLYTO DISENGAGE

SPLINE ORKEYED DRIVE

CLUTCHPLATE DRIVE

SPIGOTBEARING

FLYWHEELCASTING

FRICTIONPLATE

CRANKSHAFT ORMOTOR SHAFT

DRIVEN ORGEARBOX END

COVER


Disc clutchEngaged Disengaged

PRESSUREPLATE

HELICALSPRINGS

OR CASTINGPRESSING

THRUSTBEARING

MOVE AXIALLYTO DISENGAGE

SPLINE ORKEYED DRIVE

CLUTCHPLATE DRIVE

SPIGOTBEARING

FLYWHEELCASTING

FRICTIONPLATE

CRANKSHAFT ORMOTOR SHAFT

DRIVEN ORGEARBOX END

COVER

PLATE DRIVE

MOTOR SHAFTCRANKSHAFT OR

MOVE AXIALLY

GEARBOX ENDDRIVEN OR

KEYED DRIVESPLINE OR

TO ENGAGE

CLUTCH

PLATEPRESSURE


Multiple disc clutch

BUSH

OIL SUPPLY FORHYDRAULIC CYLINDER

BEARING

DRIVENEND

DRIVINGEND

DRIVENDISCS

DRIVINGDISCS

HYDRAULIC CYLINDER.PRESSURISE TOENGAGE CLUTCH

SEALS

SHAFTSPLINED


Disc clutches

• Disc clutches can consist of single or multiple discs.

• Generally multiple disc clutches enable greater torque capacity but are harder to cool.

• Frictional clutches can be run dry or wet using oil.

• Typical coefficients of friction are 0.07 for a wet clutch and 0.45 for a dry clutch.


Running a clutch wet

• While running a clutch wet in oil reduces the coefficient of friction it enhances heat transfer and the potential for cooling of the components.

• The expedient solution to the reduction of the friction coefficient is to use more discs and hence the use of multiple disc clutches.


Assumptions

• Two basic assumptions are used in the development of procedures for disc clutch design based upon a uniform rate of wear at the mating surfaces or a uniform pressure distribution between the mating surfaces.

• The equations for both of these methods are outlined in this section.


Elemental annular ring

• The area of an elemental annular ring on a disc clutch is δA=2πrδr.

• Now F=pA, where p is the assumed uniform interface pressure, so δF=2πrpδr.

r rδ


Normal force

• For the disc the normal force acting on the entire face is

• Note that F is also the necessary force required to clamp the clutch discs together.

o

i

o

i

r

r

2r

r 2r

p2rpdr2F ��

��

�π=π= �

( )2i

2o rrpF −π=


Friction torque

• The friction torque δT that can be developed on an elemental ring is the product of the elemental normal force, given by µδF and the radius:

δT=rµδF=2µπr2pδr• where µ is the coefficient of friction which

models the less than ideal frictional contact which occurs between two surfaces.


Total torque

• The total torque is given by integration between the limits of the annular ring, riand ro:

• This equation represents the torque capacity of a clutch with a single frictional interface. In practice clutches use an even number of frictional surfaces.

( )3i

3o

r

r

r

r

32 rrp

32

3r

p2pdrr2To

i

o

i

−µπ=��

��

�µπ=µπ= �


Torque capacity with N faces

• For a clutch with N faces the torque capacity is given by:

( )3i

3o rrNp

32

T −µπ=


Torque capacity

• Substituting for the pressure, p, gives an equation for the torque capacity as a function of the axial clamping force.

��

��

−−

µ=2i

2o

3i

3o

rrrr

FN32

T


Uniform wear

• The equations assuming uniform wear are developed below.

• The wear rate is assumed to be proportional to the product of the pressure and velocity. So

prω = constant


Maximum pressure

• For a constant angular velocity the maximum pressure will occur at the smallest radius.

pmaxriω = constant• Eliminating the angular velocity and

constant gives a relationship for the pressure as a function of the radius:

rr

pp imax=


Dynamic friction coefficients, permissible contact pressures and temperature limits

0.10 - 0.14-Graphite/resin

0.10 - 0.17-Paper based

2600.69- 1.7250.03 - 0.060.15 - 0.25Cast iron

900.345 - 0.620.12 - 0.160.20 - 0.45Wood

800.055 - 0.10.15 - 0.250.30 - 0.50Cork

230 - 6801.035 -2.070.05 - 0.080.15 - 0.45Sintered metal

200 - 2600.345 - 0.690.08 - 0.100.25 - 0.45Woven materials

200 - 2601.035 - 2.070.06 - 0.100.25 - 0.45Moulded compounds

T (oC)pmax (MN/m)µoilµdryMATERIAL


Axial force

• The elemental axial force on an elemental annular ring is given by

• Integrating to give the total axial force: rpr2F δπ=δ

( )ioimax

r

r

imax

r

rrrrp2rdr

rr

p2prdr2F o

i

o

i

−π=π=π= ��


Torque

• The elemental torque is given by

• Rearranging gives

FrT δµ=δ

( )2i

2oimax

r

r imax rrrprdrrp2T o

i

−µπ=µπ= �

( )ioimax rrr2

Fp

−π=


Torque

• Substituting gives

• For N frictional surfaces

( )ioio

2i

2o rr

2F

rrrr

2F

T +µ=��

��

−−µ=

( )2i

2oimax

r

r imax rrNrprdrrNp2T o

i

−µπ=µπ= �

( )io rr2NF

T +µ=


maximum torque for any outer radius

• By differentiating with respect to ri and equating the result to zero, the maximum torque for any outer radius ro is found to occur when

• This useful formula can be used to set the inner radius if the outer radius is constrained to a particular value.

oi r1/3r =


Design

• Clutches are usually designed based on uniform wear.

• The uniform wear assumption gives a lower torque capacity clutch than the uniform pressure assumption.

• The preliminary design procedure for disc clutch design requires the determination of the torque and speed, specification of space limitations, selection of materials, and the selection of principal radii, ro and ri.

• Common practice is to set the value of ribetween 0.45ro and 0.8ro.


Procedure

The procedure for determining the initial geometry is itemised below.

1) Determine the service factor.2) Determine the required torque capacity,

T=power/ω.3) Determine the coefficient of friction µ.4) Determine the outer radius ro.5) Find the inner radius ri.6) Find the axial actuation force required.


Materials

• The material used for clutch plates is typically grey cast iron or steel.

• The friction surface will consist of a lined material which may be moulded, woven, sintered or solid.

• Moulded linings consist of a polymeric resin used to bind powdered fibrous material and brass and zinc chips.


Example

• A clutch is required for transmission of power between a four cylinder internal combustion engine and a small machine.

• Determine the radial dimensions for a single face dry disc clutch with a moulded lining which should transmit 5 kW at 1800 rpm.

• Base the design on the uniform wear assumption.


Example


Solution

• A service factor of two should be used.

• The design will therefore be undertaken using a power of 2×5 kW=10 kW.

3.73.43.23.0Stone crushers, roll mills, heavy mixers, single cylinder compressors

overloads, cycling, high inertia starts, high power, pulsating power source

3.22.92.72.5Presses, Punches, piston pumps, Cranes, hoists

Frequent start-stops,

2.72.42.22.0Larger conveyor belts, larger machines, reciprocating pumps

with some irregularity of load up to 1.5 times nominal power

2.72.42.01.8Light machinery for wood, metal and textiles, conveyor belts

Steady power source

2.21.91.71.5Belt drive, small generators, centrifugal pumps, fans, machine tools

Steady power source, steady load, no shock or overload

SINGLE CYLINDER ENGINE

IC ENGINES (2 OR 3 CYLINDERS)

IC ENGINES (4 TO 6 CYLINDERS). MEDIUM TO LARGE ELECTRIC MOTORS

SMALL ELECTRIC MOTORS, TURBINE

TYPICAL DRIVEN SYSTEM

DESCRIPTION OF GENERAL SYSTEM

TYPE OF DRIVER


Solution cont.

• The torque is given by

• From Table 10.3 taking mid range values for the coefficient friction and the maximum permissible pressure for moulded linings gives µ=0.35 and pmax=1.55 MN/m2.

( ) mN5360/21800

10000PowerT =

π×=

ω=


Solution cont.

0.10 - 0.14-Graphite/resin

0.10 - 0.17-Paper based

2600.69- 1.7250.03 - 0.060.15 - 0.25Cast iron

900.345 - 0.620.12 - 0.160.20 - 0.45Wood

800.055 - 0.10.15 - 0.250.30 - 0.50Cork

230 - 6801.035 -2.070.05 - 0.080.15 - 0.45Sintered metal

200 - 2600.345 - 0.690.08 - 0.100.25 - 0.45Woven materials

200 - 2601.035 - 2.070.06 - 0.100.25 - 0.45Moulded compounds

T (oC)pmax (MN/m)µoilµdryMATERIAL


Solution cont.

• Taking oi r3/1r =

( )3omax

3o

3omax

2

o2omaxo

rp4/27��r31

rp1/3��

r1/3rpr1/3��T

=�

��

−

=��

� −=


Solution cont.

m0.043244/27101.55�0.35

53.05

4/27��pT

r

1/3

6

1/3

maxo

=�

��

××

=��

��

=

m02497.0r3/1r oi ==


Solution cont.

• So the clutch consists of a disc of inner and outer radius 25 mm and 43 mm respectively, with a moulded lining having a coefficient of friction value of 0.35 and a maximum permissible contact pressure of 1.55 MPa and an actuating force of 4.4 kN.

( )( )N4443

0.024970.04324101.550.024972�

rrpr2F6

iomaxi

=−×××

=−= �


Example

• A disc clutch, running in oil, is required for a vehicle with a four-cylinder engine. The design power for initial estimation of the clutch specification is 90 kW at 4500 rpm. Determine the radial dimensions and actuating force required. Base the design on the uniform wear assumption.


Solution

• From Table 10.2 a service factor of 2.7 should be used to account for starts and stops and the four cylinder engine. The design will therefore be undertaken using a power of 2.7×90 = 243 kW.

3.73.43.23.0Stone crushers, roll mills, heavy mixers, single cylinder compressors

overloads, cycling, high inertia starts, high power, pulsating power source

3.22.92.72.5Presses, Punches, piston pumps, Cranes, hoists

Frequent start-stops,

2.72.42.22.0Larger conveyor belts, larger machines, reciprocating pumps

with some irregularity of load up to 1.5 times nominal power

2.72.42.01.8Light machinery for wood, metal and textiles, conveyor belts

Steady power source

2.21.91.71.5Belt drive, small generators, centrifugal pumps, fans, machine tools

Steady power source, steady load, no shock or overload

SINGLE CYLINDER ENGINE

IC ENGINES (2 OR 3 CYLINDERS)

IC ENGINES (4 TO 6 CYLINDERS). MEDIUM TO LARGE ELECTRIC MOTORS

SMALL ELECTRIC MOTORS, TURBINE

TYPICAL DRIVEN SYSTEM

DESCRIPTION OF GENERAL SYSTEM

TYPE OF DRIVER


Solution cont.


• From Table 10.3 taking mid range values for the coefficient friction and the maximum permissible pressure for moulded linings gives µ=0.35 and pmax=1.55 MN/m2.

( ) mN7.51560/24500

243000PowerT =

π×=

ω=


Solution cont.

• Takingoi r3/1r =

m0.073254/27101.552�0.35

515.7

4/27��NpT

r

1/3

6

1/3

maxo

=�

��

×××

=��

��

=

m04229.0r3/1r oi ==


Solution cont.

• So the clutch consists of a disc of inner and outer radius 42.3 mm and 73.3 mm respectively, with a moulded lining having a coefficient of friction value of 0.35 and a maximum permissible contact pressure of 1.55 MPa and an actuating force of 25.5 kN.

( )( ) N255000.042290.07325101.55

0.0422922�rrprN2�F6

iomaxi

=−×

×××=−=


Example

• A multiple disc clutch, running in oil, is required for a motorcycle with a three-cylinder engine.

• The power demand is 75 kW at 8500 rpm. • The preliminary design layout indicates

that the maximum diameter of the clutch discs should not exceed 100 mm.


Example cont.

• In addition previous designs have indicated that a moulded lining with a coefficient of friction of 0.068 in oil and a maximum permissible pressure of 1.2 MPais reliable.

• Within these specifications determine the radii for the discs, the number of discs required and the clamping force.


Solution


( ) mN286.5/60�28500

750003.4�

PowerfactorServiceT

=××

=×=


Solution cont.

• Select the outer radius to be the largest possible, i.e. ro=50 mm.

• Using • ri=28.87 mm.

oi r3/1r =


Solution cont.

• The number of frictional surfaces, N

• This must be an even number, so the number of frictional surfaces is taken as N=24. This requires thirteen driving discs and twelve driven discs to implement.

( )

( ) 23.230.028870.050.0680.02887101.2�

286.5rr�rp�

TN

226

2i

2oimax

=−×××

=−

=


Solution cont.

• The clamping force can be calculated:

( ) ( ) N445202887.005.024068.0

5.2862rrN

T2F

io

=+×

×=+µ

=


Brakes

• The basic function of a brake is to absorb kinetic energy and dissipate it in the form of heat.


Example

• An idea of the magnitude of energy that must be dissipated can be obtained from considering the familiar example of a car undergoing an emergency stop in seven seconds from 60 mph (96 km/h).


Solution

• If the car’s mass is 1400 kg and assuming that 65% of the car’s weight is loaded onto the front axles during rapid braking then the load on the front axle is

N892765.081.91400 =××


Solution cont.

• This will be shared between two brakes so the energy that must be absorbed by one brake is

( )2f

2i VVm

21

E −=

kJ161.803600

10960.5

9.818927

21

E 2

23

=��

��

�

−��

��

××�

��

××=


Solution cont.

• If the car brakes uniformly in seven seconds then the heat that must be dissipated is 161.8×103/7=23.1 kW.

• From your experience of heat transfer from say 1 kW domestic heaters you will recognise that this is a significant quantity of heat to transfer away from the relatively compact components that make up brake assemblies.


Heat transfer• Convective heat transfer can be modelled by Fourier’s

equation:

• This equation indicates that the ability of a brake to dissipate the heat generated increases as the surface area increases or as the heat transfer coefficient rises.

• For air, the heat transfer coefficient is usually dependent on the local flow velocity and on the geometry.

• A method often used for disc brakes to increase both the surface area and the local flow is to machine multiple axial or radials holes in the disc.

( )fs TThAThAQ −=∆=


Types of brake

METHOD OF

AUTOMATIC

MAGNETIC

AND HYDRAULIC

MECHANICAL

ELECTRICAL

PNEUMATIC

ELECTRICALLY ON

FULL DISC

DRUM

DISC

ELECTRICALLY OFF

CALIPER DISC

LONG SHOE

SHORT SHOE

BAND

FRICTION

METHOD OFACTUATION

ENGAGEMENT


Comparative table of brake performance

Vehicles (rear axles on passenger cars)

Good if sealedUnstable if humid, ineffective if wet

LowHighHigher than external brake

Internal drum brake (two leading shoes)

Vehicles (rear axles on passenger cars)

Good if sealedUnstable if humid, ineffective if wet

MediumMediumHigher than external brake

Internal drum brake (leading trailing edge)

Mills, elevators, winders

GoodUnstable if humid, poor if wet

MediumMediumLowExternal drum brake (leading trailing edge)

Winches, hoist, excavators, tractors

GoodUnstable but still effective

LowHighLowDifferential band brake

TYPICAL APPLICATIONS

DUST AND DIRT

DRYNESSSTABILITYBRAKE FACTOR

MAXIMUM OPERATING TEMPERATURE

TYPE OF BRAKE


Self-energising brakes

• Brakes can be designed so that, once engaged the actuating force applied is assisted by the braking torque.

• This kind of brake is called a self-energising brake and is useful for braking large loads.


Self-locking

• Great care must be exercised in brake design.

• It is possible and sometimes desirable to design a brake, which once engaged, will grab and lock up (called self-locking action).


Disc brakes• Disc brakes are familiar

from automotive applications where they are used extensively for vehicle wheels.

• These typically consist of a cast iron disc, bolted to the wheel hub.

• This is sandwiched between two pads actuated by pistons supported in a calliper mounted on the stub shaft.

DISC

WHEELHUB

STUB-AXLE

CALIPER

PADS

HYDRAULICCYLINDER

SEAL

SEAL


Torque capacity

• With reference to the figure, the torque capacity per pad is given by

• where re is an effective radius.

r iro

err

ANNULAR PAD CIRCULAR PAD

θ

R

FF

eFrT µ=


Actuating force

• The actuating force assuming constant pressure is given by

• or assuming uniform wear by2

rrpF

2i

2o

av

−θ=

( )ioimax rrrpF −θ=


• The relationship between the average and the maximum pressure for the uniform wear assumption is given by

oi

oi

max

av

r/r1r/r2

pp

+=


Effective radius

• For an annular disc brake the effective radius is given, assuming constant pressure, by

• and assuming uniform wear by

( )( )2

i2o

3i

3o

e rr3rr2

r−−

=

2rr

r oie

+=


Circular pad disk brake design values (Fazekas, (1972))

• For circular pads the effective radius is given by re=rδ, where values for δ are given in the table as a function of the ratio of the pad radius and the radial location, R/r

1.8750.9380.5

1.5780.9470.4

1.3670.9570.3

1.2120.9690.2

1.0930.9830.1

1.0001.0000

pmax/pavδ=re/rR/r


Actuating force

• The actuating force for circular pads can be calculated using:

av2pRF π=


Example

• A calliper brake is required for the front wheels of a sport’s car with a braking capacity of 820 N m for each brake.

• Preliminary design estimates have set the brake geometry as ri=100 mm, ro=160 mm and θ=45o.

• A pad with a coefficient of friction of 0.35 has been selected.

• Determine the required actuating force and the average and maximum contact pressures.


Solution

• The torque capacity per pad = 820/2 = 410 N m.

• The effective radius is

m13.02

16.01.0re =+=


Solution cont.

• The actuating force is given by

• The maximum contact pressure is

kN011.913.035.0

410rT

Fe

=×

=µ

=

( )

( ) ( )2

3

ioimax

MN/m1.9120.10.160.12��/3645

109.011

rr�rF

p

=−×××

×

=−

=


Solution cont.

• The average pressure is given by

2

oi

oimaxav m/MN471.1

r/r1r/r2

pp =+

=


Drum brakes• Drum brakes apply friction to the external or internal

circumference of a cylinder. • A drum brake consists of the brake shoe, which has the

friction material bonded to it, and the brake drum. • For braking, the shoe is forced against the drum

developing the friction torque. • Drum brakes can be divided into two groups depending

on whether the brake shoe is external or internal to the drum.

• A further classification can be made in terms of the length of the brake shoe: short, long or complete band.


Short and long shoe internal drum brakes

• Short shoe internal brakes are used for centrifugal brakes that engage at a particular critical speed.

• Long shoe internal drum brakes are used principally in automotive applications.

• Drum brakes can be designed to be self-energising.

• Once engaged the friction force increases the normal force non-linearly, increasing the friction torque as in a positive feedback loop.


Stability• One problem associated some drum brakes is stability. • If the brake has been designed so that the braking

torque is not sensitive to small changes in the coefficient of friction, which would occur if the brake is worn or wet, then the brake is said to be stable.

• If a small change in the coefficient of friction causes a significant change to the braking torque the brake is unstable and will tend to grab if the friction coefficient rises or the braking torque will drop noticeably if the friction coefficient reduces.


Short shoe external drum brakes

a

b

c

r

SHOE

DRUM

PIVOT

BRAKELEVER Fa

nF

fFy

xθ

ω



• If the included angle of contact between the brake shoe and the brake drum is less than 45o, the force between the shoe and the drum is relatively uniform and can be modelled by a single concentrated load Fnat the centre of the contact area.

• If the maximum permissible pressure is pmax the force Fn can be estimated by

Fn = pmaxrθw



• The frictional force, Ff, is given byFf = µFn

• where µ is the coefficient of friction.• The torque on the brake drum is

T = Ffr = µFnr



• Summing moments, for the shoe arm, about the pivot gives:

� =+−= 0cFbFaFM fnapivot

acb

Fa

cFbFF n

fna

µ−=−

=



• Resolving forces gives the reactions at the pivot:

Rx=-Ff

Ry=Fa-Fn

a

b

c

r

SHOE

DRUM

PIVOT

BRAKELEVER Fa

nF

fFy

xθ

ω



• Note that for the configuration and direction of rotation shown, the friction moment µFnc adds or combines with the actuating moment aFa.

• Once the actuating force is applied the friction generated at the shoe acts to increase the braking torque.

• This kind of braking action is called self-energising.



• If the brake direction is reversed the friction moment term µFnc becomes negative and the applied load Fa must be maintained to generate braking torque.

• This combination is called self de-energising.



• From

• note that if the brake is self-energising and if µc>b then the force required to actuate the brake is zero or negative and the brake action is called self-locking.

• If the shoe touches the drum it will grab and lock.

• This is usually undesirable with exceptions being hoist stops or over-running clutch type applications.

acb

Fa

cFbFF n

fna

µ−=−

=


Long shoe external drum brakes

• If the included angle of contact between the brake shoe and the drum is greater than 45o then the pressure between the shoe and the brake lining cannot be regarded as uniform and the approximations made for the short shoe brake analysis are inadequate.

• Most drum brakes use contact angles greater than 90o.



• For a single block brake the force exerted on the drum by the brake shoe must be supported by the bearings.

• To balance this load and provide a compact braking arrangement two opposing brake shoes are usually used in a calliper arrangement.

PIVOT

DRUM

r

LEVER

SHOE

Fa

1 2

a

b

fF

nFθθ

θ

ω

SHOE

DRUM

1 2

r

Fa

aF

ω

θθθ



• The braking torque T is given by

• This is based on the assumption that the local pressure p at an angular location θ is related to the maximum pressure, pmax, by

( ) ( )21max

max2 coscossin

pwrT θ−θ

θµ=

( )max

max

sinsinp

pθ

θ=



• With the direction of rotation shown (i.e. the brake is self-energising), the magnitude of the actuation force is given by

aMM

F fna

−=

PIVOT

DRUM

r

LEVER

SHOE

Fa

1 2

a

b

fF

nFθθ

θ

ω



• The normal and frictional moments can be determined using

( ) ( ) ( )��

��

� θ−θ−θ−θθ

= 1212max

maxn 2sin2sin

41

21

sinwrbp

M

( ) ( ) ( )��

��

� θ−θ+θ−θθ

µ= 1221

max

maxf 2cos2cos

4b

coscosrsinwrp

M



• If the direction of rotation for the drum shown is reversed, the brake becomes self de-energising and the actuation force is given by

PIVOT

DRUM

r

LEVER

SHOE

Fa

1 2

a

b

fF

nFθθ

θ

ω

aMM

F fna

+=


Example

• Design a long shoe drum brake to produce a friction torque of 75 N m to stop a drum rotating at 140 rpm.

• Initial design calculations have indicated that a shoe lining with µ=0.25 and using a value of pmax=0.5×106 N/m2 in the design will give suitable life.


Solution

• First propose trial values for the brake geometry, say r=0.1 m, b=0.2 m, a=0.3 m, θ1=30o, θ2=150o.


Solution cont.

• Solving for the width of the shoe,

• Select the width to be 35 mm as this is a standard size.

( )( )

( ) m0.0346cos150cos30100.50.10.25

75sin90cos�cos�p�r

sin�Tw

62

21max2

max

=−×××

=−

=


Solution cont.

• The actual maximum pressure experienced, will be:

26max m/N494900

035.00346.0

105.0p =×=


Solution cont.

• The moment of the normal force with respect to the shoe pivot is:

( ) mN512.8sin60sin30041

3602�

12021

sin90100.49490.20.10.035

M6

n

=��

��

� −−�

��

×

×××××=


Solution cont.

• The moment of the frictional forces with respect to the shoe pivot is:

=75 N m

( ) ( )��

��

� −+−

×××××=

cos60cos3004

0.2cos150cos300.1

sin90100.49490.10.0350.25

M6

f


Solution cont.

• The actuation force is

N14593.0

758.512a

MMF fn

a =−=−

=


Double long shoe external drum brake

• For the double long shoe external drum brake illustrated, the left hand shoe is self-energising and the frictional moment reduces the actuation load.

• The right hand shoe, however, is self de-energising and its frictional moment acts to reduce the maximum pressure which occurs on the right hand brake shoe.

SHOE

DRUM

1 2

r

Fa

aF

ω

θθθ


Normal and frictional moments

• The normal and frictional moments for a self-energising and self de-energising brake are related by

max

'maxn

n ppM

'M =

max

'maxf

f ppM

'M =


Example

• For the double long shoe external drum brake illustrated in following figure determine the limiting force on the lever such that the maximum pressure on the brake lining does not exceed 1.4 MPa and determine the torque capacity of the brake.

130o

20o

Fa

R100

115

120

79.37 200

50

20


Example cont.

• The face width of the shoes is 30 mm and the coefficient of friction between the shoes and the drum can be taken as 0.28.


Solution

• First it is necessary to calculate values for θ1 and θ2 as these are not indicated directly on the diagram.

o1o1 54.10

12020

tan20 =�

��

−=θ −

o1oo2 5.140

12020

tan13020 =�

��

−+=θ −


Solution cont.

• The maximum value of sinθ would be sin90=1.

• The distance between the pivot and the drum centre,

m1217.012.002.0b 22 =+=


Solution cont.

• The normal moment is given by

( ) ( ) ( )��

��

� θ−θ−θ−θθ

= 1212max

maxn 2sin2sin

41

21

sinwrbp

M

( ) ( )

mN751.1

sin21.08sin28141

3602�

10.54140.521

sin90101.40.12170.10.03 6

=��

��

� −−�

��

×−

×××××=


Solution cont.

( ) ( ) ( )��

��

� θ−θ+θ−θθ

µ= 1221

max

maxf 2cos2cos

4b

coscosrsinwrp

M

( ) ( )

mN179.8

cos21.08cos2814

0.1217cos140.5cos10.540.1

sin90101.40.10.030.28 6

=��

��

� −+−

×××××=


Solution cont.

• The orthogonal distance between the actuation force and the pivot,

a=0.12+0.115+0.05=0.285 m.• The actuation load on the left hand shoe is

given by

N2004285.0

8.1791.751a

MMF fn

shoelefta =−=−

=


Solution cont.

• The torque contribution from the left hand shoe is given by

=206.4 N m.

( ) ( )21max

max2shoeleft coscos

sinp

wrT θ−θθ

µ=

( )5.140cos54.10cos104.11.003.028.0 62 −××××=


Solution cont.

• The actuation force on the right hand shoe can be determined by considering each member of the lever mechanism as a free body.

CV V

C

HC H

CV

B

HB

VBH

B

HA

VA F

79.37

50

200

F=501 N

Fa left shoe

Fa right shoe

14.04 o

200

50

=2004 N

=2065 N


Solution cont.

F-AV+BV=0.AH=BH.BH=CH

AH=CH.0.2F=0.05BH,

F=BH/4. BH=2004 N.

F=2004/4=501 N.

CV V

C

HC H

CV

B

HB

VBH

B

HA

VA F

79.37

50

200

F=501 N

Fa left shoe

Fa right shoe

14.04 o

200

50

=2004 N

=2065 N


Solution cont.

• So the limiting lever force is F=501 N.CV=0, BV=0.

• The actuating force for the right hand lever is the resultant of F and BH.

• The resultant angle is given by tan-1(0.05/0.2)=14.04o.

N206504.14cos

2004F shoerighta ==


Solution cont.

• The orthogonal distance between the actuation force vector and the pivot, is given by

• a=(0.235-0.01969tan14.04)cos14.04 = 0.2232 m

14.04o

14.04 o

223.2

235

19.69


Solution cont.

• The normal and frictional moments for the right hand shoe can be determined using

6

'max

max

'maxn

n 104.1p1.751

ppM

'M×

==

6

'max

max

'maxf

f 104.1p8.179

ppM

'M×

==


Solution cont.

• For the right hand shoe the maximum pressure can be determined from

0.2232101.4179.8p751.1p

2065a

'M'MF

6

'max

'max

fnshoerighta

××−

==+=

26'max m/N10130.1p ×=


Solution cont.

• The torque contribution from the right hand shoe is

=166.6 N m

( ) ( )21max

'max2

shoeright coscossin

pwrT θ−θ

θµ=

( )= × × × × −0 28 0 03 01 113 10 1054 14052 6. . . . cos . cos .


Solution cont.

• The total torque is given byTtotal=Tleft shoe+Tright shoe=206.4+166.6=373 N m


Long Shoe Internal Drum Brakes

• Most drum brakes use internal shoes that expand against the inner radius of the drum. Long shoe internal drum brakes are principally used in automotive applications.

• An automotive drum brake typically comprises two brake shoes and linings supported on a back plate bolted to the axle casing.

• The shoes are pivoted at one end on anchor pins or abutments fixed onto the back plate.



c

r1

r2

DRUM

ROTATION

RETRACTIONSPRING

ANCHORPINS OR

BRAKELINING

HYDRAULICCYLINDER

p

SHOELEADING

TRAILINGSHOE

ABUTMENTS HEELSHOE

TOESHOE

θ

θ



• The brake can be actuated by a double hydraulic piston expander, which forces the free ends of the brake apart so that the non-rotating shoes come into frictional contact with the rotating brake drum.

• A leading and trailing shoe layout consists of a pair of shoes pivoted at a common anchor point.


Leading shoe• The leading shoe is

identified as the shoe whose expander piston moves in the direction of rotation of the drum.

• The frictional drag between the shoe and the drum will tend to assist the expander piston in forcing the shoe against the drum and this action is referred to as self-energising or the self-servo action of the shoe.

c

r1

r2

DRUM

ROTATION

RETRACTIONSPRING

ANCHORPINS OR

BRAKELINING

HYDRAULICCYLINDER

p

SHOELEADING

TRAILINGSHOE

ABUTMENTS HEELSHOE

TOESHOE

θ

θ


Trailing shoe• The trailing shoe is the

one whose expander piston moves in the direction opposed the rotation of the drum.

• The frictional force opposes the expander and hence a trailing brake shoe provides less braking torque than an equivalent leading shoe actuated by the same force.

c

r1

r2

DRUM

ROTATION

RETRACTIONSPRING

ANCHORPINS OR

BRAKELINING

HYDRAULICCYLINDER

p

SHOELEADING

TRAILINGSHOE

ABUTMENTS HEELSHOE

TOESHOE

θ

θ



• The equations developed for external long shoe drum brakes are also valid for internal long shoe drum brakes.


Example• Determine the actuating

force and the braking capacity for the double internal long shoe brake illustrated.

• The lining is sintered metal with a coefficient of friction of 0.32 and the maximum lining pressure is 1.2 MPa.

• The drum radius is 68 mm and the shoe width is 25 mm.

15

55

48

16

FaaF

120

o

o

DRUM

SHOE

LINING

R68

20


Solution

• As the brake lining angles relative to the pivot, brake axis line, are not explicitly shown on the diagram, they must be calculated.

• As θ2>90o, the maximum value of sinθ is sin90=1=(sinθ)max.

m05701.0055.0015.0b 22 =+=

o1 745.4=θ o

2 7.124=θ


Solution cont.

• For this brake with the direction of rotation as shown the right hand shoe is self-energising.

15

55

48

16

FaaF

120

o

o

DRUM

SHOE

LINING

R68

20


Solution cont.

• For the right hand shoe:

( ) ( )��

��

� −−�

��

×−

×××××=

sin9.49sin249.441

3602�

4.745124.721

1101.20.057010.0680.025

M6

n

mN8.153=


Solution cont.

( ) ( )��

��

� −+−

×××××=

cos9.49cos249.44

0.05701cos124.7cos4.7450.068

1101.20.0680.0250.32

M6

f

mN57.1=


Solution cont.

• The actuating force is 938.9 N.

m103.0048.0055.0a =+=

N9.938103.0

1.578.153a

MMF fn

a =−=−

=


Solution cont.

• The torque applied by the right hand shoe is given by

( ) ( )21max

max2

shoeright coscossin

pwrT θ−θ

θµ

=

( ) mN69.54cos124.7cos4.7451

101.20.0680.0250.32 62

=−

×××××=


Solution cont.

• The torque applied by the left hand shoe cannot be determined until the maximum operating pressure pmax’ for the left hand shoe has been calculated.


Solution cont.

• As the left hand shoe is self de-energising the normal and frictional moments can be determined.

6

'max

max

'maxn'

n 102.1p8.153

ppM

M×

==

6

'max

max

'maxf'

f 102.1p1.57

ppM

M×

==


Solution cont.

• The left hand shoe is self de-energising, so,

• Fa=938.9 N as calculated earlier.a

MMF fn

a+=

103.0102.1p1.57p8.153

9.9386

'max

'max

××+

=

26'max m/N105502.0p ×=


Solution cont.

• The torque applied by the left hand shoe is given by

( ) ( )21max

'max

2

shoeleft coscossin

pwrT θ−θ

θµ

=

( ) mN31.89cos124.7cos4.7451

100.55020.0680.0250.32 62

=−

×××××=


Solution cont.

• The total torque applied by both shoes is:

mN101.431.8969.54

TTT shoeleftshoerighttotal

=+

=+=


Comment

• From this example the advantage in torque capacity of using self-energising brakes is apparent.

• Both the left hand and the right hand shoes could be made self-energising by inverting the left hand shoe, having the pivot at the top. 15

55

48

16

FaaF

120

o

o

DRUM

SHOE

LINING

R68

20


Comment cont.

• This would be advantageous if rotation occurred in just one direction.

• If, however, drum rotation is possible in either direction, it may be more suitable to have one brake self-energising for forward motion and one self-energising for reverse motion.


Band brakes

• One of the simplest types of braking device is the band brake.

• This consists of a flexible metal band lined with a frictional material wrapped partly around a drum.

• The brake is actuated by pulling the band against the drum

FF1 2F

r

a

ac

θ


Band brakes

• For the clockwise rotation shown the friction forces increase F1 relative to F2.

FF1 2F

r

a

ac

θ


Band brakes

• The relationship between the tight and slack sides of the band is given by

• F1 = tension in the tight side of the band (N),• F2 = tension in the slack side of the band (N),• µ = coefficient of friction,• θ = angle of wrap (rad).

µθ= eFF

2

1


Point of maximum contact pressure

• The point of maximum contact pressure for the friction material occurs at the tight end and is given by:

• where w is the width of the band (m).

rwF

p 1max =


Torque braking capacity

• The torque braking capacity is given by

( )rFFT 21 −=


Moments

• The relationship, for the band brake shown between the applied lever force Fa and F2can be found by taking moments.

0aFcF 2a =−

ca

FF 2a =

FF1 2F

r

a

ac

θ


Self-energising band brakes• The brake configuration

shown in the top figure opposite is self-energising for clockwise rotation.

• The level of self-energisation can be enhanced by using the differential band brake configuration shown in the bottom figure opposite.

FF1 2F

r

a

ac

θ

ab

c

F12F Fa

θ


Summation of the moments

• Summation of the moments about the pivot gives

ab

c

F12F Fa

θ

0bFaFcF 12a =+−


Force relationship

• So the relationship between the applied load Fa and the band brake tensions is given by:

cbFaF

F 12a

−=


Value of b

• Note that the value of b must be less than a so that applying the lever tightens F2more than it loosens F1.

• Substituting:( )

cbeaF

F 2a

µθ−=


Self-locking

• The brake can be made self-locking if a<beµθ and the slightest touch on the lever would cause the brake to grab or lock abruptly.

• This principle can be used to permit rotation in one direction only as in hoist and conveyor applications.


Example

• Design a band brake to exert a braking torque of 85 N m.

• Assume the coefficient of friction for the lining material is 0.25 and the maximum permissible pressure is 0.345 MPa.


Solution

• Propose a trial geometry, say r=150 mm, θ=225o and w=50 mm.

N258705.015.010345.0rwpF 6max1 =×××==

( ) N969e

5.2587eF

F 360/222525.01

2 === π×µθ

( ) ( ) mN7.24215.09695.2587rFFT 21 =−=−=


Solution cont.

• This torque is much greater than the 80 N m desired, so try a different combination of r, θ and w until a satisfactory design is achieved.


Solution cont.

• Try r=0.1 m, θ=225o and w=50 mm.

N172505.01.010345.0rwpF 6max1 =×××==

( ) N3.646e

1725eF

F360/222525.0

12 === π×µθ

( ) ( ) mN9.1071.03.6461725rFFT 21 =−=−=


Solution cont.

• Try r=0.09 m, θ=225o and w=50 mm.

N5.155205.009.010345.0rwpF 6max1 =×××==

( ) N7.581e

5.1552eF

F360/222525.0

12 === π×µθ

( ) ( ) mN4.8709.07.5815.1552rFFT 21 =−=−=


Solution cont.

• The actuating force is given by Fa=F2a/c. • If a=0.08 m and c=0.15 m then,

N2.31015.008.0

7.581Fa =×=


Conclusions

• Clutches are designed to permit the smooth, gradual engagement or disengagement of a prime mover from a driven load.

• Brakes are designed to decelerate a system. • Clutches and brakes are similar devices

providing frictional, magnetic or direct positive connection between two components.

• This section has concentrated on rotating clutches and brakes and specifically on the design of friction based devices.


Conclusions cont.

• The detailed design of a clutch or braking system involves integration of a wide range of skills such as bearings, shafts, splines, teeth, flywheels, casings, frictional surfaces, hydraulics, sensors and control algorithms.

• Both brakes and clutches can be purchased from specialist suppliers or alternatively key components such as brake pads or clutch discs can be specified and bought in from specialist suppliers and integrated into a fit for purpose machine design.