Mechanical Design II

Brakes and clutches

Emerson Escobar Nunez

Department of Energy and Mechanics
Universidad Autónoma de Occidente

1
 Introduction
The purpose of clutches and brakes is to interrupt the
connection between two components. There are some
differences between clutches and brakes:

Brake: Provides an interruptible connection between one

rotating element and a nonrotating ground

Clutch: Provides an interruptible connection between one

rotating element and a nonrotating ground

2
 Automobile disk brakes
Clutch: Permit smooth,
gradual connection and
disconnection of two
rotating members

Brake: As a clutch with

one member being fixed

AA Book of the car

3
 Clutch types

Taken from: Machine Design; An Integrated Approach, Four Edition. Robert

Norton

4
 Clutch types (cont..)
So far we have discussed components for power
transmission, which have in common:

Minimize friction (and wear)

Exactly the opposite is true for Clutches and Brakes

Want high (and stable) friction (and LOW WEAR)

Clutch/Brake main types:

• Friction
• Hydraulic/Fluid
• Magnetic

5
 Other clutch types
Spring clutch

Overrunning clutch

https://wikis.engrade.com/planetarygearsetsoperati/planetarycontrols2

6
 Automobile drum brakes

AA Book of the car

7
 Automobile clutch

AA Book of the car

http://videos.howstuffworks.com/howstuffworks/115-how-clutches-work-definition-video.htm
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Automobile clutch (cont..)
Flywheel, clutch cover and
pressure plate are rotating
with the crankshaft.
Springs push pressure plate
against rotating flywheel
clutch

Press the pedal and the

release lever is actuated to
move the pressure plate
away from the rotating
flywheel plate No motion
transmitted
Operates under dry conditions
9
 Magnetic and electromagnetic

Used to maintain constant torque

Transfer load without contact
Magnetic levitation (maglev) is a method by which an object is
suspended above another object with no support other than
magnetic fields (maximum train speed, 361 mph in Japan in 2003)
10
 Fluid drive/coupling
It acts as an automatic clutch
between the engine and gearbox
It allows the engine to idle when
stationary but takes up the drive
smoothly and progressively when
RPM increases (accelerating)
Fluid Drive is the trademark name
that Chrysler Corporation
assigned to a transmission
driveline combination offered from
1939 through 1953 in Chryslers
and Desotos.
The fluid drive element was a
hydraulic coupling inserted in
place of the flywheel
AA Book of the car

11
 Fluid torque converter

Similar to the
fluid drive but
in addition at
low RPM
increase the
torque output
(uses a
reactor)

AA Book of the car

https://www.youtube.com/watch?v=11Q4g-oOLr8
https://www.youtube.com/watch?v=leCEmJA0WsI
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Inertial coupling

13
 Friction clutches / brakes
Hamrock et al
Expanding rim Thrust disk

Contacting rim

Cone disk

Band

1. Kinetic Energy is converted to frictional heating needs to be dissipated!

2. To what extend is possible to obtain low wear with high friction?
14
 Friction and wear relationships
E. Rabinowicz, 1995

Metals: wear goes as 4-to-1 with friction

Nonmetals: wear goes as 2-to-1 with friction
15
Multiple-disk clutch, hydraulically operated

Disks a rotate with

the input shaft
Disks b rotate with
the output shaft
Using hydraulic
pressure can
disengage clutch
(separate disks a
and disks b

4 disks, 6 friction
surfaces
Operates under Wet (Oil) conditions (as in an A/C
automotive compressor)
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 Analysis of friction clutch
Analysis depends if
you assume:

1. Either uniform
pressure
assumption, OR
2. Uniform wear rate
ro at interface
ri

F Always assume
constant friction
dr µ p dA coefficient
throughout, f
dθ

Pressure p is independent of r and θ (uniform)

17
 Analysis of friction clutch (uniform p)
2π ro
Total force on plate, F F= ∫ ∫ p r dr dθ
0 ri

(
= π p ro − ri
2 2
)
2π
T = ∫0 ∫ µ pr dr dθ
ro 2
Friction torque, T ri

= π µ p (ro − ri )
2 3 3

3
2 (ro − ri )
3 3

T = µF
3 (ro − ri )
Eliminate P 2 2

Torque capacity as a function of axial clamping force

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Analysis of friction clutch (uniform WR)

Recall Archard’s wear equation:

Wear rate (WR) = (K/H) pV [p = pressure; V = velocity]
I.e., WR is proportional to p V
But V=rω (radius times angular velocity)
I.e., WR ∝ r ω p = constant (Assume)
pmax occurs at the inside radius, ri
ri
And a friction material has a certain pmax p = pmax
Clutch design is then based on: r
p r = Const. = pmax ri
19
 Analysis of friction clutch (uniform WR)
2π ri
Total Force F = ∫0 ∫r pmax r dr dθ
ro

i
r N=number of
= 2π ri pmax (ro − ri ) friction interfaces
2π ri 2
Torque T = ∫0 ∫r µ pmax r dr dθ
ro
Uniform p
r
(ro − ri )
i
3 3

Eliminate pmax = π µ ri pmax (ro − ri )

2
T = µF 2
(ro − ri )
2 2
2
3
T= µF
(r +r )
N o i What should ri /ro be
2 for max
ri = 1
3
T
ro = 0 . 58 ro
Example
For f=0.1; F=2000 N, ri=30 mm, and ro=60 mm, N=1, find T
TWR=9 Nm; Tp=9.33 Nm
i.e. assumption of uniform WR gives lower clutch capacity!
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Friction coef. and pmax of friction materials

21
Friction coefficient values in oil

22
 Disk brakes
So just fix the one plate of the clutch and you have a brake?

Why it will never work?

Same equations (F,T) apply as before

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 Brakes: Energy absorption and cooling
Main function of brakes (and clutches) is to convert and
absorb energy:
KE (0.5MV2) or (0.5Iω2) and PE (Wd) changes to frictional
heating Needs to dissipate, else high temperatures will
damage (distort) and/or melt materials
Time rate of heat dissipation, H in Watts or hp:
H = CA(t s − t a )
C=Overall heat transfer coefficient (W/(m2 oC)
A=exposed heat dissipating surface area (m2)
ts=average temperature of heat dissipating surfaces (oC)
ta=air temperature in the vicinity (oC)

Want large A (fins), increased air flow, increased mass and

C
24
 Brakes: Energy absorption and cooling
Rate at which heat is generated on a unit area of a
frictional interface is equal to the normal clamping pressure
p times friction coef. µ times velocity V (pVµ)

Below typical empirical values of pV

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 Cone clutches and brakes
Disk clutch is a special case of a cone clutch with cone angle
α=90o. Also, here N=1 (practical). Advantage? is that the shaft
thrust is much less

As before
(divide by
sinα)

T =
(
2 f F ro − ri
3 3
) Smaller
Uniform pressure (
3 sin α ro 2 − ri 2 ) α less
clamping
(ro + ri ) force needed
Uniform wear T = fF
2 sin α (8o<α<15o)
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 In class-Problem
The wheels of a standard adult bicycle have a rolling
radius of 13.5 in and a radius to the center of the caliper
disk brake pad of 12.5 in. The combined weight of bike
plus rider is 225 lb, equally distributed between the two
wheels. If friction coefficient between tires and road
surface is twice that between the brake pads and the metal
wheel rim, what clamping force must be exerted at the
caliper in order to slide the wheels?

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 Problem (Solution)
µ F on each side of wheel
112.5
12.5 µ= coef of friction of brake
o pads
F=caliper clamping force
13.5

112.5 ( 2 µ )

112.5

(ΣM)o = 0 112.5 ( 2 µ) ( 13.5 ) = 2 µ F ( 12.5 )

F = 121.5 lbs
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 Short-shoe drum brakes
1. External shoes against the outer drum surface (bicycles)
2. Internal shoes that expand to contact the inner drum surface
(trucks) F

F=force applying the brake

N=normal force
T=inertial and load torque
equaling the friction torque

Summing up moments about A and O

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Short-shoe drum brakes (self-energizing)
(ΣM)A = 0
Fc + fNa − Nb = 0
Fc
N =
b − fa
Friction increases the
(ΣM)O = 0 normal force

T = fNr fF cr
T =
b − fa
Self-energizing brake - friction moment (fNa) assists
applied force F (Is this good or bad?)
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Short-shoe drum brakes (self-deenergizing)

Fc
N =
b + fa
Opposite direction
fF cr
F T =
b + fa

Self-deenergizing brake -
friction opposes the normal
force
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 Self-locking brake (self-energizing)

(ΣM)A = 0
Fc + fNa − Nb = 0
Fc
N =
b − fa
if b ≤ fa

Then what happens?

No force is needed to activate the system, which is very
unstable
Also, use of one shoe means all force has to be taken by shaft
bearings. Better to use two opposing shoes (as in bikes)
32
 External long-shoe drum brakes
The same analysis is true for the automotive drum brakes
If the shoe over the drum has an arc > 45o then new analysis

θ1=to start of shoe

θ2=to end of shoe
Maximum pressure (and
maximum wear) occur at
θ=90o. The pressure, p
p = pmax sin θ

Considering ΣMo2=0
F c + Mn + Mf = 0
Mn moment of normal force
Mf moment of friction force

33
 External long-shoe drum brakes

Mn = −
pmax brd
[2(θ 2 − θ1 ) − sin 2θ 2 + sin 2θ1 ]
4(sin θ ) max

f pmax br  d 
Mf =
(sin θ ) max r (cosθ1 − cos θ 2 ) + 4 (cos 2θ 2 − cos 2θ1 )
 

For equilibrium considerations of the moments

acting on the drum, the friction (braking) torque:

r 2 fbpmax
T =− [cosθ1 − cosθ 2 ]
(sin θ ) max
34
 External long-shoe drum brakes

F c + Mn + Mf = 0

A self energizing brake (want it) is also self locking (do not
want it) If Mf ≥ Mn

Want to make shoe strongly self-energizing while staying away

from a self locking condition:
Design consideration: Design Mf based on f that is 25 -50%
greater than its true value equal to Mn
35
 External long-shoe drum brakes

Mn = −
pmax brd
[2(θ 2 − θ1 ) − sin 2θ 2 + sin 2θ1 ]
4(sin θ ) max

f pmax br  d 
Mf =
(sin θ ) max r (cosθ1 − cos θ 2 ) + 4 (cos 2θ 2 − cos 2θ1 )
 

For equilibrium considerations of the moments

acting on the drum, the friction (braking) torque:

r 2 fbpmax
T =− [cosθ1 − cosθ 2 ]
(sin θ ) max
36
 External long-shoe drum brakes

F c + Mn + Mf = 0

A self energizing brake (want it) is also self locking (do not
want it) If Mf ≥ Mn

Want to make shoe strongly self-energizing while staying

away from a self locking condition:
Design consideration: Design Mf based on f that is 25 -
50% greater than its true value equal to Mn
37
 Internal long-shoe drum brakes
Typical in conventional automobiles (rear brakes)
Exactly same equations for external brakes apply

r 2 fbpmax
T =− [cosθ1 − cosθ 2 ]
(sin θ ) max

Left shoe: Self-energizing

(friction tends to close
shoe)
Right shoe: Self-de-
energizing (friction tends
to open shoe)

Want: Self energizing to reduce pedal forces but not self

locking! Are the shoes self-energizing or self-deenergizing?
38
 Internal long-shoe drum brakes
Both shoes are self-energizing for forward car motion
Used on front brakes with previous design used on rear brakes
Car going forward: 6 self-energizing brakes
Car going backwards: 2 self-energizing brakes
New CARS: Front brakes are disk brakes (better cooling, less fade …)

39
 Band brakes
Simplest brake?
Band is usually steel
CW rotation, friction
increases P1 and
decreases P2
Braking torque, T
F= lever force
pmax occurs at θ=φ

P1
= e fφ
T = ( P1 − P2 ) r F=
P2 a P2

Self-energizing
c P1 = pmax rb
40
 Differential band brake
Greater self-energizing

P2 a − P1s
F=
c
T = ( P1 − P2 ) r
41
 Braking Torque (Summary)

Cone brake T = fF
(ro + ri )
2 sin α

Short shoe fF cr
T =
(self-energizing)
b − fa

r 2 fbpmax
Long shoe T =− [cosθ1 − cosθ 2 ]
(self-energizing) (sin θ ) max

Band brake T = ( P1 − P2 ) r

42
 In-class problem
The fig below shows a brake with only one shoe, being applied by a 1.5 kN force.
(The complete brake would normally have a second shoe in order to balance
the forces, but only one shoe is considered here to keep the problem short.)
Four seconds after force F is applied, the drum comes to a stop. During this
time the drum makes 110 revolutions. Use the short shoe approximation and
an estimated coefficient of friction of 0.35.
(a) Draw the brake shoe and arm assembly as a free body in equilibrium.
(b) Is the brake self-energizing or de energizing for the direction of drum rotation
involved?
(c) What is the magnitude of the torque developed by the brake?
(d) How much work does the brake
do in bringing the drum to a stop?
(e) What is the average braking power
during the 4-second interval?
(f) How far below the drum center would
the arm pivot need to be to make the
brake self-locking for f = 0.35?

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