13_ch 08 Mechanical Design budynas_SM_ch08

13_ch 08 Mechanical Design budynas_SM_ch08 - = 78 . 57...

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216 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 8-26 Refer to Prob. 8-24 and its solution.Additional information: A = 3 . 5in, D s = 4 . 25 in, static pressure 1500 psi, D b = 6in, C (joint constant) = 0 . 267, ten SAE grade 5 bolts. P = 1 10 π (4 . 25 2 ) 4 (1500) = 2128 lbf From Tables 8-2 and 8-9, A t = 0 . 1419 in 2 S p = 85 000 psi F i = 0 . 75(0 . 1419)(85) = 9 . 046 kip From Eq. (8-28), n = S p A t F i CP = 85(0 . 1419) 9 . 046 0 . 267(2 . 128) = 5 . 31 Ans . 8-27 From Fig. 8-21, t 1 = 0 . 25 in h = 0 . 25 + 0 . 065 = 0 . 315 in l = h + ( d / 2) = 0 . 315 + (3 / 16) = 0 . 5025 in D 1 = 1 . 5(0 . 375) + 0 . 577(0 . 5025) = 0 . 8524 in D 2 = 1 . 5(0 . 375) = 0 . 5625 in l / 2 = 0 . 5025 / 2 = 0 . 251 25 in Frustum 1: Washer E = 30 Mpsi, t = 0 . 065 in, D = 0 . 5625 in k
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Unformatted text preview: = 78 . 57 Mlbf/in (by computer) Frustum 2: Cap portion E = 14 Mpsi, t = . 186 25 in D = . 5625 + 2(0 . 065)(0 . 577) = . 6375 in k = 23 . 46 Mlbf/in (by computer) Frustum 3: Frame and Cap E = 14 Mpsi, t = . 251 25 in, D = . 5625 in k = 14 . 31 Mlbf/in (by computer) k m = 1 (1 / 78 . 57) + (1 / 23 . 46) + (1 / 14 . 31) = 7 . 99 Mlbf/in Ans . 0.8524" 0.5625" 0.25125" 0.8524" 0.6375" 0.18625" 0.5625" 0.6375" 0.065"...
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