49_ch 03 Mechanical Design budynas_SM_ch03

# 49_ch 03 Mechanical Design budynas_SM_ch03 - T = f RN = 2...

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62 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 3-71 ν i = 0 . 292, E i = 30 Mpsi, ν o = 0 . 334, E o = 10 . 4 Mpsi δ max = 1 2 ( 2 . 005 2 . 000 ) = 0 . 0025 in δ min = 1 2 ( 2 . 003 2 . 002 ) = 0 . 0005 in 0 . 0025 = ± 1 . 0 10 . 4 ( 10 6 ) ² 2 2 + 1 2 2 2 1 2 + 0 . 334 ³ + 1 . 0 30 ( 10 6 ) ² 1 2 + 0 1 2 0 0 . 292 ³´ p max p max = 11 576 psi Ans. Eq. (3-50) for outer member at r = 1in ( σ t ) o = 1 2 (11 576) 2 2 1 2 ² 1 + 2 2 1 2 ³ = 19 293 psi Ans. Inner member from Prob. 3-52 with r = 1 in ( σ t ) i =− 11 576 psi Ans. For δ min all above answers are 0 . 0005 / 0 . 0025 = 1 / 5 Ans. 3-72 (a) Axial resistance Normal force at ﬁt interface N = pA = p ( 2 π Rl ) = 2 π pRl Fully-developed friction force F ax = fN = 2 π fpRl Ans. (b) Torsional resistance at fully developed friction is
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Unformatted text preview: T = f RN = 2 π f pR 2 l Ans. 3-73 d = 1 in, r i = 1 . 5 in, r o = 2 . 5 in. From Table 3-4, for R = . 5 in, r c = 1 . 5 + . 5 = 2 in r n = . 5 2 2 ( 2 − √ 2 2 − . 5 2 ) = 1 . 968 245 8 in e = r c − r n = 2 . − 1 . 968 245 8 = . 031 754 in c i = r n − r i = 1 . 9682 − 1 . 5 = . 4682 in c o = r o − r n = 2 . 5 − 1 . 9682 = . 5318 in A = π d 2 / 4 = π (1) 2 / 4 = . 7854 in 2 M = Fr c = 1000(2) = 2000 lbf · in...
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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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