21_ch 08 Mechanical Design budynas_SM_ch08

21_ch 08 Mechanical Design budynas_SM_ch08 - a = C P 2 A t...

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224 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Washer: E = 30 Mpsi, t = 0 . 134 in, D = 1 . 125 in, d = 0 . 75 in Eq. (8-20): k 1 = 153 . 3 Mlbf/in Member: E = 16 Mpsi, t = 0 . 75 in, D = 1 . 280 in, d = 0 . 75 in Eq. (8-20): k 2 = 35 . 5 Mlbf/in k m = 1 (2 / 153 . 3) + (2 / 35 . 5) = 14 . 41 Mlbf/in Ans . Bolt: L T = 2(3 / 4) + 1 / 4 = 1 3 / 4 in l = 2(0 . 134) + 2(0 . 75) = 1 . 768 in l d = L L T = 2 . 50 1 . 75 = 0 . 75 in l t = l l d = 1 . 768 0 . 75 = 1 . 018 in A t = 0 . 373 in 2 (Table 8-2) A d = π (0 . 75) 2 / 4 = 0 . 442 in 2 k b = A d A t E A d l t + A t l d = 0 . 442(0 . 373)(30) 0 . 442(1 . 018) + 0 . 373(0 . 75) = 6 . 78 Mlbf/in Ans . C = 6 . 78 6 . 78 + 14 . 41 = 0 . 320 Ans . (c) From Eq. (8-40), Goodman with S e = 18 . 6 kpsi, S ut = 120 kpsi S a = 18 . 6[120 (25 / 0 . 373)] 120 + 18 . 6 = 7 . 11 kpsi The stress components are
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Unformatted text preview: a = C P 2 A t = . 320(6) 2(0 . 373) = 2 . 574 kpsi m = a + F i A t = 2 . 574 + 25 . 373 = 69 . 6 kpsi n f = S a a = 7 . 11 2 . 574 = 2 . 76 Ans . (d) Eq. (8-42) for Gerber S a = 1 2(18 . 6) 120 120 2 + 4(18 . 6) 18 . 6 + 25 . 373 120 2 2 25 . 373 18 . 6 = 10 . 78 kpsi n f = 10 . 78 2 . 574 = 4 . 19 Ans . (e) n proof = 85 2 . 654 + 69 . 8 = 1 . 17 Ans ....
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