16. Clutches and Brakes Introduction
Objectives Clutch is a device that connects and disconnects • Recognize the basic geometries of clutch and brake two collinear shafts. systems. Similar to couplings • Calculate the frictional forces and torque capabilities Friction 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 disc brake Drum brake
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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 wheel
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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 truck 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 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 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