2_ch 16 Mechanical Design budynas_SM_ch16

# 2_ch 16 Mechanical Design budynas_SM_ch16 - Ans Eq(16-6 T R...

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Chapter 16 397 RH shoe : F x = 500 sin 30° = 250 lbf, F y = 500 cos 30° = 433 lbf Eqs. (16-8): A = ± 1 2 sin 2 θ ² 120 0 = 0 . 375, B = ± θ 2 1 4 sin 2 θ ² 2 π/ 3 rad 0 = 1 . 264 Eqs. (16-9): R x = 111 . 4(1 . 5)(6) 1 [0 . 375 0 . 28(1 . 264)] 250 =− 229 lbf R y = 111 . 4(1 . 5)(6) 1 [1 . 264 + 0 . 28(0 . 375)] 433 = 940 lbf R = [( 229) 2 + (940) 2 ] 1 / 2 = 967 lbf Ans. LH shoe : F x = 250 lbf, F y = 433 lbf Eqs. (16-10): R x = 57 . 9(1 . 5)(6) 1 [0 . 375 + 0 . 28(1 . 264)] 250 = 130 lbf R y = 57 . 9(1 . 5)(6) 1 [1 . 264 0 . 28(0 . 375)] 433 = 171 lbf R = [(130) 2 + (171) 2 ] 1 / 2 = 215 lbf Ans. 16-2 θ 1 = 15°, θ 2 = 105°, θ a = 90°, sin θ a = 1, a = 5in Eq. (16-2): M f = 0 . 28 p a (1 . 5)(6) 1 ³ 105° 15° sin θ (6 5 cos θ ) d θ = 13 . 06 p a Eq. (16-3): M N = p a (1 . 5)(6)(5) 1 ³ 105° 15° sin 2 θ d θ = 46 . 59 p a c = 2(5 cos 30°) = 8 . 66 in Eq. (16-4): F = 46 . 59 p a 13 . 06 p a 8 . 66 = 3 . 872 p a RH shoe : p a = 500 / 3 . 872 = 129 . 1 psi on RH shoe for cw rotation
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Unformatted text preview: Ans. Eq. (16-6): T R = . 28(129 . 1)(1 . 5)(6 2 )(cos 15° − cos 105°) 1 = 2391 lbf · in LH shoe : 500 = 46 . 59 p a + 13 . 06 p a 8 . 66 ⇒ p a = 72 . 59 psi on LH shoe for ccw rotation Ans. T L = . 28(72 . 59)(1 . 5)(6 2 )(cos 15° − cos 105°) 1 = 1344 lbf · in T total = 2391 + 1344 = 3735 lbf · in Ans. Comparing this result with that of Prob. 16-1, a 2.7% reduction in torque is achieved by using 25% less braking material....
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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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