Md-16 Clutches and brakes .pdf

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16. Clutches and Brakes16. Clutches and Brakes

ObjectivesObjectives•• Recognize the basic geometries of clutch and brakeRecognize the basic geometries of clutch and brake

systems.systems.

•• Calculate the frictional forces and torque capabilitiesCalculate the frictional forces and torque capabilities

in brake systems.in brake systems.

•• Understand the principles of heat generation andUnderstand the principles of heat generation and

heat removal from brake systems.heat removal from brake systems.

•• Calculate frictional brake horsepower and recognizeCalculate frictional brake horsepower and recognize

how to use it.how to use it.

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IntroductionIntroduction

Clutch is a device that connects and disconnectsClutch is a device that connects and disconnects

two collinear shafts.two collinear shafts. Similar to couplingsSimilar to couplings

Friction and hence heat dissipationFriction and hence heat dissipation

Purpose of a brake is to stop the rotation of aPurpose of a brake is to stop the rotation of ashaft.shaft.

Braking action is produced by friction as aBraking action is produced by friction as astationary part bears on a moving part.stationary part bears on a moving part.

Heat dissipation is a problemHeat dissipation is a problem

Brake fade during continuous application of brakingBrake fade during continuous application of brakingdue to heat generateddue to heat generated

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Plate clutchPlate clutch

Uses Spring loaded flat surfaces

Transmit power in either direction

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Cone clutchCone clutch

Uses tapered friction surfaces

Easy to engage

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Caliper disc brakeCaliper disc brake

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Drum brakeDrum brake



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Drum brakeDrum brake

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Block brakeBlock brake –– Wagon brakeWagon brake

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Band brakeBand brake

Belt wrapped around the wheel

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Fig. 16Fig. 16--7 Disc brake7 Disc brake

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Friction MaterialsFriction Materials

Asbestos fibers embedded in an epoxyAsbestos fibers embedded in an epoxy--type materialtype material

Good thermal propertiesGood thermal properties

High friction coefficient (0.35 to 0.50)High friction coefficient (0.35 to 0.50)

Environmental concernsEnvironmental concerns

Polymer compounds with impregnated materialPolymer compounds with impregnated material

Metal shavingsMetal shavings

GraphiteGraphite

Sintered ironSintered iron

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Torque and ForcesTorque and Forces

Sliding frictionSliding friction

Friction force, FFriction force, Ff f = f N= f N

f = coefficient of frictionf = coefficient of friction

N = normal force N = normal force

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Plate type clutchPlate type clutch

Rotating torque,Rotating torque, TTf f ==

R R oo = outside radius= outside radius

R R ii = inside radius= inside radius

⎟ ⎠

⎞⎜⎝

⎛ +

2

R Nf io R

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Plate type clutchPlate type clutch

Friction power can be calculated as,Friction power can be calculated as,

PPf f ==63,000

nTf

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• A plate-type clutch has the following properties:

– Ro

= 12 in

– Ri= 9 in

– engagement force of 120 lb (normal force)

– turns at 2000 rpm

• Friction disc has coefficient of friction of .3.

• Determine torque and power that can be transmitted by this system.

Example Problem 16-1: Torques and Forces onClutches and Brakes

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• Torque capacity:(16-2)

T f = f N( r o + r i )

2

T f = .3 (120 lb)

⎝

⎜⎛

⎠

⎟ ⎞

12 in + 9 in

2

T f = 378 in-lb

• Power:(16-3)

P f =T f n

63,000

P f =378 (2000)

63,000

P f = 12 hp

Example Problem 16-1: Torques and Forces on

Clutches and Brakes (cont’d.)

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Clutches and Brakes

• For the short-shoe drum brake

shown, determine the braking

torque for the following dimensions:

– a = 4 in

– L = 20 in

– D = 12 in

– f = .4

– W = 100 lb:



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• Find moments to determine normal force:

Σ M p = WL – a N

Σ M p = 100 lb 20 in – 4 in N

N = 500 lb

• Torque friction:

T f = f N D

2

T f = .4 (500 lb)12 in

2

T f = 1200 in-lb

• This analysis assumes the lever arms stay approximately horizontal.



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This becomes a bit complicated because ofThis becomes a bit complicated because of

the cone anglethe cone angle

In this case the frictional force is given byIn this case the frictional force is given by

TTf f = F= Ff f r r mm= f N= f N r r mm

Normal force, N = Normal force, N =

FFaa = axial force= axial force

ΑΑ = cone angle= cone angle

αcosf αsin

Fa

+

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Fig. 16.8 Cone clutch geometryFig. 16.8 Cone clutch geometry

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Combining the above two equations we getCombining the above two equations we get

αcosf αsin

Fr f T am

f +

=

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Clutches and Brakes

• For the cone clutch shown,

determine the torque-

transmitting capacity based on

the following parameters:

– Dmean

= 12 inches

– Fa

= 75 lb

– f = .35

– α = 20°

• Also solve if α = 10° and

compare the results.

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(16-5)

T f = f r m F a

sin α + f cos α

T f =

.35⎝ ⎜⎛

⎠⎟ ⎞

12 in

2 75 lb

sin 20° + .35 cos 20°

T f = 235 in-lb

For α = 10°:

T f =

.35⎝ ⎜⎛

⎠⎟ ⎞

12 in

2 75 lb

sin 10° + .35 cos 10°

T f = 304 in-lb

• The smaller angle creates a greater wedging force and, correspondingly,larger torque capacity.





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Clutches and Brakes

• A truck has total weight of 40,000 lb and is

traveling 60 mph.

• The brake design calls for it to be able to stop in

400 feet.

• Determine stopping force required.

• Determine stopping torque required if wheels are

36 inches in diameter.

• Determine torque per brake, assuming there are

10 sets of brakes.

• Assuming each brake is a disc brake with mean

radius of 10 inches, determine normal brake force if

f = .4.August 15, 2007August 15, 2007 2626

60 mph 5280 ft mile

hr 3600 sec

= 88 ft/sec


Clutches and Brakes

•First, find the rate of deceleration. Converting 60 mph to ft/sec:

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• Find the stopping rate:

D = V a t

t = D

V a

t =400 ft

88

2 ft/sec

t = 9 sec

V = a t

a =V

t

a =88 ft/sec

9 sec

a = 9.8 ft/sec2

• Find the stopping force:

F =W

g a

F = 40,000 lb32.2 ft/sec2 9.8 ft/sec2

F = 12,100 lb



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• Find the torque, if the wheels are 36 inches in diameter :

T = F r

T = 12,100 lb36 in

2

T = 217,800 in-lb

− For each wheel:T = 21,780 in-lb

− Braking normal force:(16-2)

T f = f N r m

N =T f

f r m

N =21,780 in-lb

.4 10 in

N = 5,450 lb

• This is a significant normal force, especially for a disc brake system.



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Rotational Inertia and Brake PowerRotational Inertia and Brake Power

Inertia and frictional horsepower Inertia and frictional horsepower

Energy from rotating torqueEnergy from rotating torque

UUf f = F= Fππ

DD N Ntt == TTf f 22ππ

N Ntt

UUf f = Frictional work = Frictional work

D = effective diameter D = effective diameter

N Ntt = number of turns= number of turns

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Rotational Inertia and Brake PowerRotational Inertia and Brake Power

Power associated with stoppingPower associated with stopping

TTf f inin--lblb

n rpmn rpm

t secondst seconds

t550

U

63,000

nTP f f

f ==



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Energy absorbed can be the potential energy orEnergy absorbed can be the potential energy or

the kinetic energythe kinetic energy Potential energy,Potential energy, ∆∆PE = W (hPE = W (h11 – – hh22))

Potential energy,Potential energy, ∆∆K K E =E = )V(Vg2

W 2

2

2

1 −

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Heat generatedHeat generated

Temperature rise isTemperature rise is

c = specific heatc = specific heat

= 101 ft= 101 ft--lb/lb/lb/lb/°°F for cast ironF for cast iron

= 93 ft= 93 ft--lb/lb/lb/lb/°°F for steelF for steel

= 15 ft= 15 ft--lb/lb/lb/lb/°°F for aluminumF for aluminum

WWmm = weight of brake system that can absorb= weight of brake system that can absorbthe heatthe heat

cW

U∆T

m

f =

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• 3500-pound automobile is traveling 50 mph

and decelerates on flat ground at a rate of

20 ft/sec2.

• Each of the four steel brake drums weighs

10 pounds.

• Assuming all heat is absorbed by the

drums during this period, find energy

absorbed, average frictional power, and

temperature rise of drums.

Example Problem 16-5: Rotational Inertia

and Brake Power

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− Converting 50 mph to ft/sec:

50 mphhr

3600 sec

5280 ft

mile = 73 ft/sec

− Kinetic energy to be absorbed:(16-9)

KE =W V

2

2g

KE =3500 lb (73 ft)

2

2(32.2 ft/sec2) sec2

KE = 289,620 ft-lb

(Energy gain U = KE lost.)


and Brake Power

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(16-12)

U f = W c ∆ T

∆T =U f

W c

∆T =289,620 ft-lb

40 lb 93

ft-lb

lb°F

∆T = 78°

− Finding the stopping time:

∆V = a t

t =∆V

a

t =73 ft/sec

20 ft/sec2

t = 3.7 sec


and Brake Power (cont’d.)

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− Frictional power could then be found:

(16-7)

f hp =U f

550 t =

KE

550 t

f hp =289,620

550 (3.7)

f hp = 142


and Brake Power (cont’d.)



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Automotive brakeAutomotive brake

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Md-16 Clutches and brakes .pdf