15 cutches & brakes s

7/23/2019 15 Cutches & Brakes S

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CLUTCHES & BRAKES

Clutches, Brakes & FlywheelsDesign calculations:

1. Actuating force

2. Torque transmitted

3. Energy loss4. Temperature rise

Types:

1. Rim type with internalshoes

2. Rim type with externalshoes

3. Band

4. Disk or axial

5. Others


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Multi-disc Clutch/Brake

D

Coefficient of frictionf= constant

For a new (not worn) clutch

with rigid discs, pressure is

uniformly distributed over

interface area

d

r

dr

Required: torque capacity

Take a ring element of disc

Normal force: 2adF p rdr 2

2

2

D

ad

F p rdr

2 2

4

apF D d

This is the axial force clamping the driving & driven discs

ap p


3/17

D

d

r

drFriction Torque

2adT frdF fr p rdr

2

2 3 3

2

22

24

D

aa

d

fpT fp r dr D d

This is the total torque that can be developed over

the entire surface

3 3

2 23

fF D dT

D d

or

With successive engagement and

disengagement, the discs will start wearing

Wear rate friction force x rubbing velocity

Butfis constant

Wear rate pressure x sliding velocity

Wearing a piece of wood with sandpaper But on the clutch surface, velocity radius

Therefore wear rate pressure X radius

Consequently, the disc will initially wear more on

the outer radius

After this initial run-in wear, the friction lining will

tend to wear at a uniform wear rate Pressure x radius = constant pr const


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Uniform Wear

2 2a ad d

pr p p p r

/ 2 / 2

/ 2 / 2

22

D D

aa

d d

p dF prdr p d dr D d

Normal force:

/ 2 / 2

2 2 2

/ 2 / 2

28

D D

aa

d d

fp dT fpr dr fp d rdr D d

Torque:

4

FfT D d Or:

D

d

r

drpr constMaximum pressure will now occur

at minimum radius

2dF p rdr

2dT frdF fr p rdr

Assumption of uniform wear rate gives a lower

calculated clutch capacity than the assumption

of uniform pressure

Clutches are designed on the basis of uniform

wear & have a little extra torque capacity whennew

3 32

24

afpT D d

Uniform pressure:

2 2

8

afp dT D d

Uniform wear rate:

Conclusion


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Conclusion

F T

r

D

d

Number of disks are nwith2n frictionsurfaces.

For2n friction surfacesthe axial force doesnot change.

For2n friction surfacesthe torque ismultiplied by 2n

Multiple Disk Clutches

Conclusion

Part of diskangle (rad) isat both sides ofrotor.

Axial force is afunction of .

Torque is a

function of .

Dd

Caliper Disk Brakes


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Automobile Disk Clutch

Automobile Caliper Brake

Caliper

Rotor


7/17

Cone Clutch

Friction Materials


8/17

Slip Time of Clutches

The figure above shows Je and e as the collective mass

moment of inertia of the driving parts, and Ja and a as

the collective mass moment of inertia of the driven parts.

The torque Tc is the torque needed by the clutch to drive

the driven parts, while Te is the engine (or driving) torque

and Ta is the torque of the driven parts.

Te

Ta

e

a

TcJe

Ja


0

The slip time ts is the time needed for clutch to let a = e. The time te is the engagement time needed by the clutch

to achieve the value Tc

Ta

Tc

B

C

e

a

te

ts

T T

Ta

Tc

e

a

ts

A

Actual torque/omega vs time Simplified torque/omega vs time


9/17


cae TdtdJT aac Tdt

dJT

The differential equation of motion of masses of the

system is:


system is:

tJ

TT

dJdtTT

e

cee

e

t

ce

e

0

0 0


system is:

tJ

TT

dJdtTT

a

aca

a

t

ac

a

0

0 0


The slipping ceases when e and a become equal to

each other, thus we can find the slip time ts :

ceaaceea

sTTJTTJ

JJt

0


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The rate of energy dissipation (i.e.

power) is:

Energy Dissipation in Clutches

i.e. max at t=0 or

e ac e a c

e a

J JE T T T t

J J

max ( )c e aE T

The total energy dissipated from t=0tot=tt is then:


which is not a function of Tc.

t tt t

c e a c

e c0 0

1 1E E dt T T t dtJ J

Or ( )2e a

e a

E1 1

2

J J


11/17

The temperature rise To can beobtained approximately as:


where:

To = temperature rise in oCC= specific heat

m = clutch or brake mass in kg

mC

ETo

A multi-disk clutch transmits 25 kW at 1575 rpm. It

has 3 disks on driving shaft and 2 on driven shaft.

Disks outside and inside diameters are 240 mm

and 120 mm. Coefficient of friction is 0.3. Find

maximum axial pressure

Sample Application

Data: H=25 kW, N=1575 rpm, D=0.25 m, d=0.125 m,

f=0.3

Number of contact surfaces:

n = 2(3+2-1)= 8

Torque T= 60H/2 N =151.6 N.m


12/17

For uniform wear,

OrF=2T/n f(D+d)= 1404 N.

Since ,

p=1404/0.0072=0.06 MN/m2

Sample Application

Solution:

( )1 2T n f F D d

)()2( dDdpF

(i) Relations:

Uniform pressure:

Design Process

Preliminary Design:

Fa

T

Rm

D

dmT fN R 2 n

aFN

max( )2 2

4 Np

D d

max ( )3 3T n f p D d

8

Usuallyor

8060Dd ../ 33 D5020d )..(

max( . . ) 3T 0 3 0 2 n f p D


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(ii) Procedure:

Design Process

Preliminary Design:

1. UsuallyHand are given.CalculateTas

rpm

kW

N

H00010T ,

Multiply by 2.55 for safety.

2. Select materials (Tables andFigs.) which gives fandpmax.

(ii) Procedure (cont.):

Design Process

Preliminary Design:

3. Findn andD from

Taken = 38 disks (usually 5) andfindD. Then d=(0.60.8)D

4. If large D or large n are obtained,

change materials or lining.

max)../( p2030TDn 3


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Design ProcessPreliminary Design:

Ranges of Material Properties

Friction Coefficientf:Dry 0.15-0.6,Wet (Oil immersed) 0.04-0.12

Working Pressure ~ pmax(MN/m2):

Dry 0.175 2.8,

Wet (Oil immersed) 0.7 - 4.2 Pressure * Velocity (pV) (MPa*m/s):

1.05-3.0

Design Process

PreliminaryDesign Chart

Clutch Diameter (m)

0.001

0.01

0.1

1

10

100

Log [Power(kW)/(number of disks*N(rpm))]

Diameter(m)

1

0.1

0.01

0.001

0.0001

Base

-5 -4 -3 -2 -1 0 1 2 3 4 5

1.pmax= 0.05 MN/m2

0.1.pmax= 0.5 MN/m2

0.01.pmax= 5 MN/m2

0.001.pmax= 50 MN/m2

0.0001.pmax= 500 MN/m2

Legend:


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Sample Application

Design a multi-disk clutch that

can transmit 25 kW at 1575 rpm.Assume that it has 3 disks, thecoefficient of friction is 0.3, andthe maximum axial pressure is0.06 MN/m2.

Find the disks outside and insidediameters.

Sample Application

SOLUTION

Data:H=25 kW, N=1575 rpm, n =3,f= 0.3, and pmax= 0.06 MN/m

2.

TorqueT= 60H/2N=151.6 N.m (T

10,000HkW/N=158.7 N.m)max/( . . )

/[( . . ) . ( . )]

/( ) . .

3

6

n D T 0 3 0 2 f p

152 0 3 0 2 0 3 0 06 10

152 5400 3600 0 042 0 028

Taken = 3 gives:

m....21002400D009300140D

3

and m..)..( 19001250D8060d


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Sample Application

(i) Relations:

Detailed Design:

Uniform Wear:

)(max dD2

dpFa

max ( )2 2T n f p d D d

4

Uniform Pressure:

)(max22

a dDp2

F


12

Fa

T

RmD

d

Sample Application

(ii) Procedure:Detailed Design:

1. Usually Hand are given.CalculateTas

N.m,rpm

kW

N

H9549T

2. Find design torque Td.

TKKT servicestartd

1.2-2 1-2.5


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Sample Application


Detailed Design:

3. Select uniform wear or uniformpressure and select materials ofparts which gives pmax and f.

4. With preliminary n,D andd, check if

dTT

Tfrom max ( )2 2T n f p d D d

4


12

Uniform wear

Uniform pressureOr

Sample Application


Detailed Design:

5. Find requiredFa from eqs.

6. Find temperature rise T, checkmaximum temperature (Table),checkpV(Table,V= velocity), and findwear (Table). Check power rate(Area/Power) (Table)

7. Iterate if required.

)(max dD

2

dpFa

)(max22

a dDp2

F

Uniform wear

Uniform pressure

15 cutches & brakes s

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