16. and Introduction

Objectives „ is a device that connects and disconnects • Recognize the basic geometries of clutch and two collinear shafts. systems. „ Similar to • Calculate the frictional forces and torque capabilities „ and hence heat dissipation in brake systems. „ Purpose of a brake is to stop the rotation of a • Understand the principles of heat generation and shaft. heat removal from brake systems. „ Braking action is produced by friction as a

• Calculate frictional brake horsepower and recognize stationary part bears on a moving part. how to use it. „ Heat dissipation is a problem „ Brake fade during continuous application of braking due to heat generated August 15, 2007 1 August 15, 2007 2

Plate clutch Cone clutch

„Uses Spring loaded flat surfaces „Transmit power in either direction „Uses tapered friction surfaces „Easy to engage

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Caliper

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1 Drum brake Block brake – Wagon brake

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Band brake Fig. 16-7 Disc brake

„Belt wrapped around the

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Friction Materials

„ Asbestos fibers embedded in an epoxy-type material

„ Good thermal properties

„ High friction coefficient (0.35 to 0.50)

„ Environmental concerns

„ Polymer compounds with impregnated material

„ Metal shavings

„ Graphite

„ Sintered iron

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2 Torque and Forces Plate type clutch

„ Sliding friction „ Rotating torque, Tf = ⎛ R + R ⎞ f N⎜ o i ⎟ „ Friction force, Ff = f N „ Ro = outside radius ⎝ 2 ⎠

„ f = coefficient of friction „ Ri = inside radius „ N = normal force

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Plate type clutch Example Problem 16-1: Torques and Forces on Clutches and Brakes

„ Friction power can be calculated as, • A plate-type clutch has the following properties: –Ro = 12 in

– Ri = 9 in T n – engagement force of 120 lb (normal force) „ P = f f 63,000 – turns at 2000 rpm • Friction disc has coefficient of friction of .3. • Determine torque and power that can be transmitted by this system.

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Example Problem 16-1: Torques and Forces on Example Problem 16-2: Torques and Forces on Clutches and Brakes (cont’d.) Clutches and Brakes

• Torque capacity: (16-2) ( r + r ) T = f N o i f 2 • For the short-shoe drum brake shown, determine the braking ⎛ 12 in + 9 in ⎞ Tf = .3 (120 lb) ⎜ ⎟ torque for the following dimensions: ⎝ 2 ⎠ – a = 4 in Tf = 378 in-lb – L = 20 in • Power: (16-3) – D = 12 in T n P = f f 63,000 – f = .4

378 (2000) – W = 100 lb: P = f 63,000

Pf = 12 hp

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3 Example Problem 16-2: Torques and Forces on Cone clutch Clutches and Brakes (cont’d.)

• Find moments to determine normal force:

Σ Mp = WL – a N „ This becomes a bit complicated because of Σ Mp = 100 lb 20 in – 4 in N

N = 500 lb the cone angle

• Torque friction: „ D In this case the frictional force is given by T = f N f 2 „ T = F r = f N r 12 in f f m m T = .4 (500 lb) F f 2 a „ Normal force, N = Normal force, N = sin α + f cosα Tf = 1200 in-lb „ F = axial force • This analysis assumes the lever arms stay approximately horizontal. Fa = axial force „ Α = cone angle August 15, 2007 19 August 15, 2007 20

Fig. 16.8 Cone clutch geometry Cone clutch

„ Combining the above two equations we get

f r F T = m a f sin α + f cosα

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Example Problem 16-3: Torques and Forces on Example Problem 16-3: Torques and Forces on Clutches and Brakes Clutches and Brakes (cont’d.)

(16-5) f r F T = m a f sin α + f cos α • For the cone clutch shown, ⎛ 12 in ⎞ .35 ⎜ ⎟ 75 lb determine the torque- ⎝ 2 ⎠ T = transmitting capacity based on f sin 20° + .35 cos 20° the following parameters: Tf = 235 in-lb – Dmean = 12 inches For α = 10°: – Fa = 75 lb ⎛ 12 in ⎞ .35 ⎜ ⎟ 75 lb ⎝ 2 ⎠ – f = .35 T = f sin 10° + .35 cos 10° – α = 20° Tf = 304 in-lb • Also solve if α = 10° and compare the results. • The smaller angle creates a greater wedging force and, correspondingly, larger torque capacity.

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4 Example Problem 16-4: Torques and Forces on Example Problem 16-4: Torques and Forces on Clutches and Brakes Clutches and Brakes

• A has total weight of 40,000 lb and is •First, find the rate of deceleration. Converting 60 mph to ft/sec: traveling 60 mph. 5280 ft hr 60 mph = 88 ft/sec • The brake design calls for it to be able to stop in mile 3600 sec 400 feet. • Determine stopping force required. • Determine stopping torque required if 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 August 15, 2007 25 August 15, 2007 26 f = .4.

Example Problem 16-4: Torques and Forces on Example Problem 16-4: Torques and Forces on Clutches and Brakes (cont’d.) Clutches and Brakes (cont’d.) • Find the stopping rate: • Find the torque, if the wheels are 36 inches in diameter:

D = Va t T = F r D t = Va 36 in T = 12,100 lb 400 ft 2 t = 88 ft/sec 2 T = 217,800 in-lb

t = 9 sec − For each wheel: T = 21,780 in-lb V = a t

V − Braking normal force: a = t (16-2) Tf = f N rm 88 ft/sec a = 9 sec T N = f f rm a = 9.8 ft/sec2 21,780 in-lb • Find the stopping force: N = W .4 10 in F = a g N = 5,450 lb 40,000 lb F = 9.8 ft/sec2 32.2 ft/sec2 • This is a significant normal force, especially for a disc brake system. August 15, 2007 F = 12,100 lb 27 August 15, 2007 28

Rotational Inertia and Brake Power Rotational Inertia and Brake Power

„ Inertia and frictional horsepower „ Power associated with stopping

„ Energy from rotating torque T n U P = f = f f 63,000 550 t „ Uf = F π D Nt = Tf 2 π Nt

„ Uf = Frictional work „ Tf in-lb „ D = effective diameter „ n rpm

„ Nt = number of turns „ t seconds

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5 Heat generated

U „ Energy absorbed can be the potential energy or „ Temperature rise is ∆T = f the kinetic energy Wm c

„ „ Potential energy, ∆PE = W (h1 – h2) c = specific heat = 101 ft-lb/lb/°F for cast iron W 2 2 = 93 ft-lb/lb/°F for steel „ Potential energy, ∆KE = (V1 − V2 ) 2 g = 15 ft-lb/lb/°F for aluminum

„ Wm = weight of brake system that can absorb the heat

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Example Problem 16-5: Rotational Inertia Example Problem 16-5: Rotational Inertia and Brake Power and Brake Power − Converting 50 mph to ft/sec:

hr 5280 ft • 3500-pound automobile is traveling 50 mph 50 mph = 73 ft/sec and decelerates on flat ground at a rate of 3600 sec mile 2 − Kinetic energy to be absorbed: 20 ft/sec . (16-9) W V 2 KE = • Each of the four steel brake drums weighs 2g 10 pounds. 3500 lb (73 ft)2 KE = • Assuming all heat is absorbed by the 2(32.2 ft/sec2) sec2 drums during this period, find energy KE = 289,620 ft-lb

absorbed, average frictional power, and (Energy gain U = KE lost.) temperature rise of drums.

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Example Problem 16-5: Rotational Inertia Example Problem 16-5: Rotational Inertia and Brake Power (cont’d.) and Brake Power (cont’d.)

(16-12) Uf = W c ∆ T

U ∆T = f − Frictional power could then be found: W c (16-7) 289,620 ft-lb Uf KE ∆T = fhp = = ft-lb 550 t 550 t 40 lb 93 lb°F 289,620 f = ∆T = 78° hp 550 (3.7)

− Finding the stopping time: fhp = 142

∆V = a t

∆V t = a

73 ft/sec t = 20 ft/sec2

t = 3.7 sec August 15, 2007 35 August 15, 2007 36

6 Automotive brake

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7