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FIGURE 8.17 A leading-shoe trailing-shoe automotive brake.
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2. For either shoe the braking torque can be written in terms of the coefficient of friction, the lining width, and the maximum contact pressure. From Eq. (8.25), T= = pmax fbr2(cos 1 cos 2) (sin )max pmax fb(0.125)2(cos 10 cos 126 ) 1
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= 0.024 57fbpmax The numerical values of T are different for the two shoes, since pmax differs. 3. Now the normal force P must be calculated. Equations (8.28), (8.29), (8.30), and (8.31) are used:
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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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Px = =
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brpmax ( 2 1 + sin 1 cos 1 sin 2 cos 2) 2(sin )max b(0.125)p max 126 10 + sin 10 cos 10 sin 126 cos 126 2(1) 57.296
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= 0.1671bpmax Py = = brp max (cos2 1 cos2 2) 2(sin )max b(0.125)p max (cos2 10 cos2 126 ) = 0.0390bpmax 2(1)
P = (P 2 + P 2 )1/2 = 0.1716bpmax x y p = tan 1 Px 0.1671 = tan 1 = 76.85 Py 0.0390
4. The effective friction radius rf is, by Eq. (8.32), rf = T 0.024 57fbp max = = 0.1432 m = 143.2 mm f P f(0.1716bp max)
5. The moments about pivot point A are found by Eqs. (8.34) to (8.36): Ma = F = F(0.072 + 0.083) = 0.155F Mn = Pa sin p = 0.1716bpmax(0.0867)(sin 76.85 ) = 0.014 49bpmax Mf = fP(rf a cos p) = f(0.1716bpmax)(0.1432 0.0867 cos 76.85 ) = 0.021 19fbpmax 6. For the leading (self-energizing) shoe, the proper form of Eq. (8.37) is Ma Mn + Mf = 0 Therefore
l l 0.155F l 0.014 49bp max + 0.021 19fbp max = 0
The superscript l has been used to designate the leading shoe. 7. For the trailing shoe, Eq. (8.37) has the form Ma Mn Mf = 0 Thus
t t 0.155F t 0.014 49bp max 0.021 19fbp max = 0
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
CLUTCHES AND BRAKES 8.32
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8. Since the same actuating force is used for each shoe, F l = F t. After substituting from the two moment equations, we obtain
l l t t 0.014 49bpmax 0.021 19fbpmax 0.014 49bpmax + 0.021 19fbpmax = 0.155 0.155
After cancellation and substitution of f = 0.32 (the most pessimistic assumption),
l t pmax = 2.759p max
9. The torque capacities of the shoe must sum to 550 N m: T l + T t = 550
l t (0.024 57)(0.32)b(pmax + pmax) = 550
Then
l t b(pmax + p max ) = 69.95 kN m
10. Enough information has been accumulated to begin to specify the design. Since l t pmax = 1000 kPa, p max = 1000/2.759 = 362.4 kPa. Now the distance b can be found. Since
l t b(pmax + p max ) = 69.95
we have b= 69.95 103 = 0.0513 m, or about 51 mm (1000 + 362.4)(103)
11. The actuating force F (F l = F t ) can be found from either shoe s moment equation. For the leading shoe,
l l 0.155F l 0.014 49bpmax + 0.021 19fbpmax = 0 l Substituting values of b, f, and pmax gives F = 2.55 kN. Thus F = F l = F t = 2.55 kN. 12. Now we determine whether the leading shoe is self-locking or too close to selflocking for safety. The moment equation used above in step 11 is used again, but this time with an f value 50 percent higher than the maximum value cited by the lining manufacturer. Use
f = 1.5(0.34 + 0.02) = 0.54 Then by substituting into the moment equation, the corresponding value of F can be found: 0.155F l (0.014 49)(0.0513)(1000) + (0.021 19)(0.54)(0.0513)(1000) = 0 Solving gives F l = F t = 1.001 kN. Since a large positive force is required to activate the leading shoe even for this very high coefficient of friction, the brake is in no danger of self-locking.
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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