Chapter10 clutches and_brakes

1. Mechanical Design
PRN Childs, University of Sussex
Clutches and brakes
Mechanical Design

2. Mechanical Design
PRN Childs, University of Sussex
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.

3. Mechanical Design
PRN Childs, University of Sussex
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.

4. Mechanical Design
PRN Childs, University of Sussex
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.

5. Mechanical Design
PRN Childs, University of Sussex
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.

6. Mechanical Design
PRN Childs, University of Sussex
Brakes
• A brake is a device used to reduce or
control the speed of a system or bring it to
rest.

7. Mechanical Design
PRN Childs, University of Sussex
Typical applications for a clutch
and a brake
MOTOR OR
ENGINE
COUPLING
CLUTCH GEAR BOX
COUPLING
DRIVEN
MACHINE
MOTOR OR
ENGINE BRAKE
CONNECTION
TO DRIVEN
MACHINE

8. Mechanical Design
PRN Childs, University of Sussex
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.

9. Mechanical Design
PRN Childs, University of Sussex
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.

10. Mechanical Design
PRN Childs, University of Sussex
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).
I
T
=α

11. Mechanical Design
PRN Childs, University of Sussex
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.

12. Mechanical Design
PRN Childs, University of Sussex
Mass moments of inertia
• The mass moments of inertia for a cylinder
and a disc are given by
( )4
i
4
ocylinder rrL
2
1
I −ρπ=
4
odisc Lr
2
1
I ρπ=

13. Mechanical Design
PRN Childs, University of Sussex
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.

14. Mechanical Design
PRN Childs, University of Sussex
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.
ACTUATOR
PUSHES DISCS
TOGETHER
INPUT
OUTPUT
DRIVING
DISC
DRIVEN
DISC
FRICTION
MATERIAL

15. Mechanical Design
PRN Childs, University of Sussex
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.

16. Mechanical Design
PRN Childs, University of Sussex
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.

17. Mechanical Design
PRN Childs, University of Sussex
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.

18. Mechanical Design
PRN Childs, University of Sussex
Types of clutch
METHOD OF
ENGAGEMENT
MECHANICAL
PNEUMATIC
AND HYDRAULIC
ELECTRICAL
MAGNETIC
POSITIVE
CONTACT
FRICTION
OVERRUNNING
MAGNETIC
FLUID
COUPLING
SQUARE JAW
SPIRAL JAW
TOOTHED
DISC
DRUM
CONE
ROLLER
SPRAG
SPRING WOUND
MAGNETIC PARTICLE
HYSTERESIS
EDDY CURRENT
DRY FLUID
HYDRAULIC
ACTUATION
METHOD OF

19. Mechanical Design
PRN Childs, University of Sussex
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.

20. Mechanical Design
PRN Childs, University of Sussex
Positive contact clutches
• Positive contact
clutches have teeth or
serrations, which
provide mechanical
interference between
mating components.

21. Mechanical Design
PRN Childs, University of Sussex
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.

22. Mechanical Design
PRN Childs, University of Sussex
Ratchet and pawl clutch

23. Mechanical Design
PRN Childs, University of Sussex
Roller clutch

24. Mechanical Design
PRN Childs, University of Sussex
Sprag clutch

25. Mechanical Design
PRN Childs, University of Sussex
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

26. Mechanical Design
PRN Childs, University of Sussex
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

27. Mechanical Design
PRN Childs, University of Sussex
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.

28. Mechanical Design
PRN Childs, University of Sussex
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.

29. Mechanical Design
PRN Childs, University of Sussex
Disc clutch
PRESSURE
PLATE
HELICAL
SPRINGS
OR CASTING
PRESSING
THRUST
BEARING
MOVE AXIALLY
TO DISENGAGE
SPLINE OR
KEYED DRIVE
CLUTCH
PLATE DRIVE
SPIGOT
BEARING
FLYWHEEL
CASTING
FRICTION
PLATE
CRANKSHAFT OR
MOTOR SHAFT
DRIVEN OR
GEARBOX END
COVER

30. Mechanical Design
PRN Childs, University of Sussex
Disc clutch
Engaged Disengaged
PRESSURE
PLATE
HELICAL
SPRINGS
OR CASTING
PRESSING
THRUST
BEARING
MOVE AXIALLY
TO DISENGAGE
SPLINE OR
KEYED DRIVE
CLUTCH
PLATE DRIVE
SPIGOT
BEARING
FLYWHEEL
CASTING
FRICTION
PLATE
CRANKSHAFT OR
MOTOR SHAFT
DRIVEN OR
GEARBOX END
COVER
PLATE DRIVE
MOTOR SHAFT
CRANKSHAFT OR
MOVE AXIALLY
GEARBOX END
DRIVEN OR
KEYED DRIVE
SPLINE OR
TO ENGAGE
CLUTCH
PLATE
PRESSURE

31. Mechanical Design
PRN Childs, University of Sussex
Multiple disc clutch
BUSH
OIL SUPPLY FOR
HYDRAULIC CYLINDER
BEARING
DRIVEN
END
DRIVING
END
DRIVEN
DISCS
DRIVING
DISCS
HYDRAULIC CYLINDER.
PRESSURISE TO
ENGAGE CLUTCH
SEALS
SHAFT
SPLINED

32. Mechanical Design
PRN Childs, University of Sussex
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.

33. Mechanical Design
PRN Childs, University of Sussex
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.

34. Mechanical Design
PRN Childs, University of Sussex
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.

35. Mechanical Design
PRN Childs, University of Sussex
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

36. Mechanical Design
PRN Childs, University of Sussex
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
2
r
r 2
r
p2rpdr2F π=π=
( )2
i
2
o rrpF −π=

37. Mechanical Design
PRN Childs, University of Sussex
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.

38. Mechanical Design
PRN Childs, University of Sussex
Total torque
• The total torque is given by integration
between the limits of the annular ring, ri
and 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.
( )3
i
3
o
r
r
r
r
3
2
rrp
3
2
3
r
p2pdrr2T
o
i
o
i
−µπ=µπ=µπ=

39. Mechanical Design
PRN Childs, University of Sussex
Torque capacity with N faces
• For a clutch with N faces the torque
capacity is given by:
( )3
i
3
o rrNp
3
2
T −µπ=

40. Mechanical Design
PRN Childs, University of Sussex
Torque capacity
• Substituting for the pressure, p, gives an
equation for the torque capacity as a
function of the axial clamping force.
−
−
µ= 2
i
2
o
3
i
3
o
rr
rr
FN
3
2
T

41. Mechanical Design
PRN Childs, University of Sussex
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

42. Mechanical Design
PRN Childs, University of Sussex
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:
r
r
pp i
max=

43. Mechanical Design
PRN Childs, University of Sussex
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

44. Mechanical Design
PRN Childs, University of Sussex
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
i
max
r
r
rrrp2rdr
r
r
p2prdr2F
o
i
o
i
−π=π=π=

45. Mechanical Design
PRN Childs, University of Sussex
Torque
• The elemental torque is given by
• Rearranging gives
FrT δµ=δ
( )2
i
2
oimax
r
r
imax rrrprdrrp2T
o
i
−µπ=µπ=
( )ioi
max
rrr2
F
p
−π
=

46. Mechanical Design
PRN Childs, University of Sussex
Torque
• Substituting gives
• For N frictional surfaces
( )io
io
2
i
2
o
rr
2
F
rr
rr
2
F
T +
µ
=
−
−µ
=
( )2
i
2
oimax
r
r
imax rrNrprdrrNp2T
o
i
−µπ=µπ=
( )io rr
2
NF
T +
µ
=

47. Mechanical Design
PRN Childs, University of Sussex
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 =

48. Mechanical Design
PRN Childs, University of Sussex
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 ri
between 0.45ro and 0.8ro.

49. Mechanical Design
PRN Childs, University of Sussex
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.

50. Mechanical Design
PRN Childs, University of Sussex
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.

51. Mechanical Design
PRN Childs, University of Sussex
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.

52. Mechanical Design
PRN Childs, University of Sussex
Example

53. Mechanical Design
PRN Childs, University of Sussex
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
CYLINDER
S)
IC
ENGINES
(4 TO 6
CYLINDER
S).
MEDIUM
TO LARGE
ELECTRIC
MOTORS
SMALL
ELECTRIC
MOTORS,
TURBINE
TYPICAL
DRIVEN
SYSTEM
DESCRIPTION OF
GENERAL
SYSTEM
TYPE OF
DRIVER

54. Mechanical Design
PRN Childs, University of Sussex
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.
( )
mN53
60/21800
10000Power
T =
π×
=
ω
=

55. Mechanical Design
PRN Childs, University of Sussex
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

56. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• Taking oi r3/1r =
( )
3
omax
3
o
3
omax
2
o
2
omaxo
rp4/27r
3
1
rp1/3
r1/3rpr1/3T
=−
=−=

57. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
m0.04324
4/27101.550.35
53.05
4/27p
T
r
1/3
6
1/3
max
o
=
××
==
m02497.0r3/1r oi ==

58. Mechanical Design
PRN Childs, University of Sussex
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
rrpr2F
6
iomaxi
=−×××
=−=

59. Mechanical Design
PRN Childs, University of Sussex
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.

60. Mechanical Design
PRN Childs, University of Sussex
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
CYLINDER
S)
IC
ENGINES
(4 TO 6
CYLINDER
S).
MEDIUM
TO LARGE
ELECTRIC
MOTORS
SMALL
ELECTRIC
MOTORS,
TURBINE
TYPICAL
DRIVEN
SYSTEM
DESCRIPTION OF
GENERAL
SYSTEM
TYPE OF
DRIVER

61. Mechanical Design
PRN Childs, University of Sussex
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.
( )
mN7.515
60/24500
243000Power
T =
π×
=
ω
=

62. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• Taking oi r3/1r =
m0.07325
4/27101.5520.35
515.7
4/27Np
T
r
1/3
6
1/3
max
o
=
×××
==
m04229.0r3/1r oi ==

63. Mechanical Design
PRN Childs, University of Sussex
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
0.0422922rrprN2F
6
iomaxi
=−×
×××=−=

64. Mechanical Design
PRN Childs, University of Sussex
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.

65. Mechanical Design
PRN Childs, University of Sussex
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 MPa
is reliable.
• Within these specifications determine the
radii for the discs, the number of discs
required and the clamping force.

66. Mechanical Design
PRN Childs, University of Sussex
Solution
• The torque is given by
( )
mN286.5
/6028500
750003.4
PowerfactorService
T
=
×
×
=
×
=

67. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• Select the outer radius to be the largest
possible, i.e. ro=50 mm.
• Using
• ri=28.87 mm.
oi r3/1r =

68. Mechanical Design
PRN Childs, University of Sussex
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.
( )
( ) 23.23
0.028870.050.0680.02887101.2
286.5
rrrp
T
N
226
2
i
2
oimax
=
−×××
=
−
=

69. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The clamping force can be calculated:
( ) ( )
N4452
02887.005.024068.0
5.2862
rrN
T2
F
io
=
+×
×
=
+µ
=

70. Mechanical Design
PRN Childs, University of Sussex
Brakes
• The basic function of a brake is to absorb
kinetic energy and dissipate it in the form
of heat.

71. Mechanical Design
PRN Childs, University of Sussex
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).

72. Mechanical Design
PRN Childs, University of Sussex
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 =××

73. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• This will be shared between two brakes so
the energy that must be absorbed by one
brake is
( )2
f
2
i VVm
2
1
E −=
kJ161.80
3600
1096
0.5
9.81
8927
2
1
E 2
23
=−
×
×××=

74. Mechanical Design
PRN Childs, University of Sussex
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.

75. Mechanical Design
PRN Childs, University of Sussex
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 −=∆=

76. Mechanical Design
PRN Childs, University of Sussex
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 OF
ACTUATION
ENGAGEMENT

77. Mechanical Design
PRN Childs, University of Sussex
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
APPLICATIO
NS
DUST AND
DIRT
DRYNESSSTABILITYBRAKE
FACTOR
MAXIMUM
OPERAT
ING
TEMPER
ATURE
TYPE OF BRAKE

78. Mechanical Design
PRN Childs, University of Sussex
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.

79. Mechanical Design
PRN Childs, University of Sussex
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).

80. Mechanical Design
PRN Childs, University of Sussex
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
WHEEL
HUB
STUB-AXLE
CALIPER
PADS
HYDRAULIC
CYLINDER
SEAL
SEAL

81. Mechanical Design
PRN Childs, University of Sussex
Torque capacity
• With reference to the figure, the torque capacity
per pad is given by
• where re is an effective radius.
ri
ro
er
r
ANNULAR PAD CIRCULAR PAD
θ
R
FF
eFrT µ=

82. Mechanical Design
PRN Childs, University of Sussex
Actuating force
• The actuating force assuming constant
pressure is given by
• or assuming uniform wear by
2
rr
pF
2
i
2
o
av
−
θ=
( )ioimax rrrpF −θ=

83. Mechanical Design
PRN Childs, University of Sussex
• The relationship between the average and
the maximum pressure for the uniform
wear assumption is given by
oi
oi
max
av
r/r1
r/r2
p
p
+
=

84. Mechanical Design
PRN Childs, University of Sussex
Effective radius
• For an annular disc brake the effective
radius is given, assuming constant
pressure, by
• and assuming uniform wear by
( )
( )2
i
2
o
3
i
3
o
e
rr3
rr2
r
−
−
=
2
rr
r oi
e
+
=

85. Mechanical Design
PRN Childs, University of Sussex
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

86. Mechanical Design
PRN Childs, University of Sussex
Actuating force
• The actuating force for circular pads can
be calculated using:
av
2
pRF π=

87. Mechanical Design
PRN Childs, University of Sussex
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.

88. Mechanical Design
PRN Childs, University of Sussex
Solution
• The torque capacity per pad =
820/2 = 410 N m.
• The effective radius is
m13.0
2
16.01.0
re =
+
=

89. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The actuating force is given by
• The maximum contact pressure is
kN011.9
13.035.0
410
r
T
F
e
=
×
=
µ
=
( )
( ) ( )
2
3
ioi
max
MN/m1.912
0.10.160.12 /3645
109.011
rrr
F
p
=
−×××
×
=
−
=

90. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The average pressure is given by
2
oi
oi
maxav m/MN471.1
r/r1
r/r2
pp =
+
=

91. Mechanical Design
PRN Childs, University of Sussex
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.

92. Mechanical Design
PRN Childs, University of Sussex
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.

93. Mechanical Design
PRN Childs, University of Sussex
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.

94. Mechanical Design
PRN Childs, University of Sussex
Short shoe external drum brakes
a
b
c
r
SHOE
DRUM
PIVOT
BRAKE
LEVER
Fa
nF
fFy
x
θ
ω

95. Mechanical Design
PRN Childs, University of Sussex
Short shoe external drum brakes
• 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 Fn
at the centre of the contact area.
• If the maximum permissible pressure is
pmax the force Fn can be estimated by
Fn = pmaxrθw

96. Mechanical Design
PRN Childs, University of Sussex
Short shoe external drum brakes
• The frictional force, Ff, is given by
Ff = µFn
• where µ is the coefficient of friction.
• The torque on the brake drum is
T = Ffr = µFnr

97. Mechanical Design
PRN Childs, University of Sussex
Short shoe external drum brakes
• Summing moments, for the shoe arm,
about the pivot gives:
=+−= 0cFbFaFM fnapivot
a
cb
F
a
cFbF
F n
fn
a
µ−
=
−
=

98. Mechanical Design
PRN Childs, University of Sussex
Short shoe external drum brakes
• Resolving forces
gives the reactions at
the pivot:
Rx=-Ff
Ry=Fa-Fn
a
b
c
r
SHOE
DRUM
PIVOT
BRAKE
LEVER
Fa
nF
fFy
x
θ
ω

99. Mechanical Design
PRN Childs, University of Sussex
Short shoe external drum brakes
• 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.

100. Mechanical Design
PRN Childs, University of Sussex
Short shoe external drum brakes
• 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.

101. Mechanical Design
PRN Childs, University of Sussex
Short shoe external drum brakes
• 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.
a
cb
F
a
cFbF
F n
fn
a
µ−
=
−
=

102. Mechanical Design
PRN Childs, University of Sussex
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.

103. Mechanical Design
PRN Childs, University of Sussex
Long shoe external drum brakes
• 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
nF
θθ
θ
ω
SHOE
DRUM
1 2
r
Fa
aF
ω
θ
θθ

104. Mechanical Design
PRN Childs, University of Sussex
Long shoe external drum brakes
• 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
( )
( )21
max
max2
coscos
sin
p
wrT θ−θ
θ
µ=
( )max
max
sin
sinp
p
θ
θ
=

105. Mechanical Design
PRN Childs, University of Sussex
Long shoe external drum brakes
• With the direction of
rotation shown (i.e.
the brake is self-
energising), the
magnitude of the
actuation force is
given by
a
MM
F fn
a
−
=
PIVOT
DRUM
r
LEVER
SHOE
Fa
1 2
a
b
fF
nF
θθ
θ
ω

106. Mechanical Design
PRN Childs, University of Sussex
Long shoe external drum brakes
• The normal and frictional moments can be
determined using
( )
( ) ( )θ−θ−θ−θ
θ
= 1212
max
max
n 2sin2sin
4
1
2
1
sin
wrbp
M
( )
( ) ( )θ−θ+θ−θ
θ
µ
= 1221
max
max
f 2cos2cos
4
b
coscosr
sin
wrp
M

107. Mechanical Design
PRN Childs, University of Sussex
Long shoe external drum brakes
• 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
nF
θθ
θ
ω
a
MM
F fn
a
+
=

108. Mechanical Design
PRN Childs, University of Sussex
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.

109. Mechanical Design
PRN Childs, University of Sussex
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.

110. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• Solving for the width of the shoe,
• Select the width to be 35 mm as this is a
standard size.
( )
( )
( )
m0.0346
cos150cos30100.50.10.25
75sin90
coscospr
sinT
w
62
21max
2
max
=
−×××
=
−
=

111. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The actual maximum pressure
experienced, will be:
26
max m/N494900
035.0
0346.0
105.0p =×=

112. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The moment of the normal force with
respect to the shoe pivot is:
( ) mN512.8sin60sin300
4
1
360
2
120
2
1
sin90
100.49490.20.10.035
M
6
n
=−−×
×
××××
=

113. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The moment of the frictional forces with
respect to the shoe pivot is:
=75 N m
( ) ( )−+−
×
××××
=
cos60cos300
4
0.2
cos150cos300.1
sin90
100.49490.10.0350.25
M
6
f

114. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The actuation force is
N1459
3.0
758.512
a
MM
F fn
a =
−
=
−
=

115. Mechanical Design
PRN Childs, University of Sussex
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
θ
θθ

116. Mechanical Design
PRN Childs, University of Sussex
Normal and frictional moments
• The normal and frictional moments for a
self-energising and self de-energising
brake are related by
max
'
maxn
n
p
pM
'M =
max
'
maxf
f
p
pM
'M =

117. Mechanical Design
PRN Childs, University of Sussex
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.
130
o
20
o
F
a
R100
115120
79.37 200
50
20

118. Mechanical Design
PRN Childs, University of Sussex
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.

119. Mechanical Design
PRN Childs, University of Sussex
Solution
• First it is necessary to calculate values for
θ1 and θ2 as these are not indicated
directly on the diagram.
o1o
1 54.10
120
20
tan20 =−=θ −
o1oo
2 5.140
120
20
tan13020 =−+=θ −

120. Mechanical Design
PRN Childs, University of Sussex
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
=+=

121. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The normal moment is given by
( )
( ) ( )θ−θ−θ−θ
θ
= 1212
max
max
n 2sin2sin
4
1
2
1
sin
wrbp
M
( ) ( )
mN751.1
sin21.08sin281
4
1
360
2
10.54140.5
2
1
sin90
101.40.12170.10.03 6
=−−×−
×
××××
=

122. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
( )
( ) ( )θ−θ+θ−θ
θ
µ
= 1221
max
max
f 2cos2cos
4
b
coscosr
sin
wrp
M
( ) ( )
mN179.8
cos21.08cos281
4
0.1217
cos140.5cos10.540.1
sin90
101.40.10.030.28 6
=−+−
×
××××
=

123. Mechanical Design
PRN Childs, University of Sussex
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
N2004
285.0
8.1791.751
a
MM
F fn
shoelefta =
−
=
−
=

124. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The torque contribution from the left hand
shoe is given by
=206.4 N m.
( )
( )21
max
max2
shoeleft coscos
sin
p
wrT θ−θ
θ
µ=
( )5.140cos54.10cos104.11.003.028.0 62
−××××=

125. Mechanical Design
PRN Childs, University of Sussex
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.
C
V V
C
H
C H
C
V
B
H
B
V
BH
B
H
A
V
A F
79.37
50
200
F=501 N
F
a left shoe
F
a right shoe
14.04
o
200
50
=2004 N
=2065 N

126. Mechanical Design
PRN Childs, University of Sussex
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.
C
V V
C
H
C H
C
V
B
H
B
V
BH
B
H
A
V
A F
79.37
50
200
F=501 N
F
a left shoe
F
a right shoe
14.04
o
200
50
=2004 N
=2065 N

127. Mechanical Design
PRN Childs, University of Sussex
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.
N2065
04.14cos
2004
F shoerighta ==

128. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The orthogonal
distance between the
actuation force vector
and the pivot, is given
by
• a=(0.235-
0.01969tan14.04)cos
14.04 = 0.2232 m
14.04
o
14.04
o
223.2
235
19.69

129. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The normal and frictional moments for the
right hand shoe can be determined using
6
'
max
max
'
maxn
n
104.1
p1.751
p
pM
'M
×
==
6
'
max
max
'
maxf
f
104.1
p8.179
p
pM
'M
×
==

130. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• For the right hand shoe the maximum
pressure can be determined from
0.2232101.4
179.8p751.1p
2065
a
'M'M
F
6
'
max
'
max
fn
shoerighta
××
−
==
+
=
26'
max m/N10130.1p ×=

131. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The torque contribution from the right hand
shoe is
=166.6 N m
( )
( )21
max
'
max2
shoeright coscos
sin
p
wrT θ−θ
θ
µ=
( )= × × × × −028 003 01 113 10 1054 14052 6
. . . . cos . cos .

132. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The total torque is given by
Ttotal=Tleft shoe+Tright shoe=
206.4+166.6=373 N m

133. Mechanical Design
PRN Childs, University of Sussex
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.

134. Mechanical Design
PRN Childs, University of Sussex
Long Shoe Internal Drum Brakes
c
r
1
r
2
DRUM
ROTATION
RETRACTION
SPRING
ANCHOR
PINS OR
BRAKE
LINING
HYDRAULIC
CYLINDER
p
SHOE
LEADING
TRAILING
SHOE
ABUTMENTS HEEL
SHOE
TOE
SHOE
θ
θ

135. Mechanical Design
PRN Childs, University of Sussex
Long Shoe Internal Drum Brakes
• 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.

136. Mechanical Design
PRN Childs, University of Sussex
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
r
1
r
2
DRUM
ROTATION
RETRACTION
SPRING
ANCHOR
PINS OR
BRAKE
LINING
HYDRAULIC
CYLINDER
p
SHOE
LEADING
TRAILING
SHOE
ABUTMENTS HEEL
SHOE
TOE
SHOE
θ
θ

137. Mechanical Design
PRN Childs, University of Sussex
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
r
1
r
2
DRUM
ROTATION
RETRACTION
SPRING
ANCHOR
PINS OR
BRAKE
LINING
HYDRAULIC
CYLINDER
p
SHOE
LEADING
TRAILING
SHOE
ABUTMENTS HEEL
SHOE
TOE
SHOE
θ
θ

138. Mechanical Design
PRN Childs, University of Sussex
Long Shoe Internal Drum Brakes
• The equations developed for external long
shoe drum brakes are also valid for
internal long shoe drum brakes.

139. Mechanical Design
PRN Childs, University of Sussex
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

140. Mechanical Design
PRN Childs, University of Sussex
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
=+=
o
1 745.4=θ
o
2 7.124=θ

141. Mechanical Design
PRN Childs, University of Sussex
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

142. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• For the right hand shoe:
( ) ( )−−×−
×
××××
=
sin9.49sin249.4
4
1
360
2
4.745124.7
2
1
1
101.20.057010.0680.025
M
6
n
mN8.153=

143. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
( ) ( )−+−
×
××××
=
cos9.49cos249.4
4
0.05701
cos124.7cos4.7450.068
1
101.20.0680.0250.32
M
6
f
mN57.1=

144. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The actuating force is 938.9 N.
m103.0048.0055.0a =+=
N9.938
103.0
1.578.153
a
MM
F fn
a =
−
=
−
=

145. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The torque applied by the right hand shoe
is given by
( )
( )21
max
max
2
shoeright coscos
sin
pwr
T θ−θ
θ
µ
=
( ) mN69.54cos124.7cos4.745
1
101.20.0680.0250.32 62
=−
×
××××
=

146. Mechanical Design
PRN Childs, University of Sussex
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.

147. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• As the left hand shoe is self de-energising
the normal and frictional moments can be
determined.
6
'
max
max
'
maxn'
n
102.1
p8.153
p
pM
M
×
==
6
'
max
max
'
maxf'
f
102.1
p1.57
p
pM
M
×
==

148. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The left hand shoe is self de-energising,
so,
• Fa=938.9 N as calculated earlier.
a
MM
F fn
a
+
=
103.0102.1
p1.57p8.153
9.938 6
'
max
'
max
××
+
=
26'
max m/N105502.0p ×=

149. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The torque applied by the left hand shoe is
given by
( )
( )21
max
'
max
2
shoeleft coscos
sin
pwr
T θ−θ
θ
µ
=
( ) mN31.89cos124.7cos4.745
1
100.55020.0680.0250.32 62
=−
×
××××
=

150. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The total torque applied by both shoes is:
mN101.431.8969.54
TTT shoeleftshoerighttotal
=+
=+=

151. Mechanical Design
PRN Childs, University of Sussex
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

152. Mechanical Design
PRN Childs, University of Sussex
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.

153. Mechanical Design
PRN Childs, University of Sussex
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
2
F
r
a
a
c
θ

154. Mechanical Design
PRN Childs, University of Sussex
Band brakes
• For the clockwise
rotation shown the
friction forces
increase F1 relative to
F2.
FF1
2
F
r
a
a
c
θ

155. Mechanical Design
PRN Childs, University of Sussex
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
= e
F
F
2
1

156. Mechanical Design
PRN Childs, University of Sussex
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).
rw
F
p 1
max =

157. Mechanical Design
PRN Childs, University of Sussex
Torque braking capacity
• The torque braking capacity is given by
( )rFFT 21 −=

158. Mechanical Design
PRN Childs, University of Sussex
Moments
• The relationship, for
the band brake shown
between the applied
lever force Fa and F2
can be found by
taking moments.
0aFcF 2a =−
c
a
FF 2a =
FF1
2
F
r
a
a
c
θ

159. Mechanical Design
PRN Childs, University of Sussex
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
2
F
r
a
a
c
θ
ab
c
F1
2F Fa
θ

160. Mechanical Design
PRN Childs, University of Sussex
Summation of the moments
• Summation of the
moments about the
pivot gives
ab
c
F1
2F Fa
θ
0bFaFcF 12a =+−

161. Mechanical Design
PRN Childs, University of Sussex
Force relationship
• So the relationship between the applied
load Fa and the band brake tensions is
given by:
c
bFaF
F 12
a
−
=

162. Mechanical Design
PRN Childs, University of Sussex
Value of b
• Note that the value of b must be less than
a so that applying the lever tightens F2
more than it loosens F1.
• Substituting:
( )
c
beaF
F 2
a
µθ
−
=

163. Mechanical Design
PRN Childs, University of Sussex
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.

164. Mechanical Design
PRN Childs, University of Sussex
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.

165. Mechanical Design
PRN Childs, University of Sussex
Solution
• Propose a trial geometry, say r=150 mm,
θ=225o and w=50 mm.
N258705.015.010345.0rwpF 6
max1 =×××==
( ) N969
e
5.2587
e
F
F 360/222525.0
1
2 === π×µθ
( ) ( ) mN7.24215.09695.2587rFFT 21 =−=−=

166. Mechanical Design
PRN Childs, University of Sussex
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.

167. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• Try r=0.1 m, θ=225o and w=50 mm.
N172505.01.010345.0rwpF 6
max1 =×××==
( ) N3.646
e
1725
e
F
F 360/222525.0
1
2 === π×µθ
( ) ( ) mN9.1071.03.6461725rFFT 21 =−=−=

168. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• Try r=0.09 m, θ=225o and w=50 mm.
N5.155205.009.010345.0rwpF 6
max1 =×××==
( ) N7.581
e
5.1552
e
F
F 360/222525.0
1
2 === π×µθ
( ) ( ) mN4.8709.07.5815.1552rFFT 21 =−=−=

169. Mechanical Design
PRN Childs, University of Sussex
Solution cont.
• The actuating force is given by Fa=F2a/c.
• If a=0.08 m and c=0.15 m then,
N2.310
15.0
08.0
7.581Fa =×=

170. Mechanical Design
PRN Childs, University of Sussex
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.

171. Mechanical Design
PRN Childs, University of Sussex
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.