25_ch 08 Mechanical Design budynas_SM_ch08

25_ch 08 Mechanical Design budynas_SM_ch08 - s = 2 (0 . 375...

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228 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Eq. (8-40) for Goodman S a = 18 . 6(120 63 . 75) 120 + 18 . 6 = 7 . 55 kpsi n f = S a σ a = 7 . 55 2 . 77 = 2 . 73 Ans . (c) From Eq. (8-42) for Gerber fatigue criterion, S a = 1 2(18 . 6) ± 120 ² 120 2 + 4(18 . 6)(18 . 6 + 63 . 75) 120 2 2(63 . 75)(18 . 6) ³ = 11 . 32 kpsi n f = S a σ a = 11 . 32 2 . 77 = 4 . 09 Ans . (d) Pressure causing joint separation from Eq. (8-29) n = F i (1 C ) P = 1 P = F i 1 C = 4 . 94 1 0 . 102 = 5 . 50 kip p = P A = 5500 π (4 2 ) / 4 6 = 2626 psi Ans . 8-39 This analysis is important should the initial bolt tension fail. Members: S y = 71 kpsi, S sy = 0 . 577(71) = 41 . 0 kpsi. Bolts: SAE grade 8, S y = 130 kpsi, S sy = 0 . 577(130) = 75 . 01 kpsi Shear in bolts A
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Unformatted text preview: s = 2 (0 . 375 2 ) 4 = . 221 in 2 F s = A s S sy n = . 221(75 . 01) 3 = 5 . 53 kip Bearing on bolts A b = 2(0 . 375)(0 . 25) = . 188 in 2 F b = A b S yc n = . 188(130) 2 = 12 . 2 kip Bearing on member F b = . 188(71) 2 . 5 = 5 . 34 kip Tension of members A t = (1 . 25 . 375)(0 . 25) = . 219 in 2 F t = . 219(71) 3 = 5 . 18 kip F = min(5 . 53, 12 . 2, 5 . 34, 5 . 18) = 5 . 18 kip Ans . The tension in the members controls the design....
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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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