27_ch 08 Mechanical Design budynas_SM_ch08

27_ch 08 Mechanical Design budynas_SM_ch08 - 31 Ans Tension...

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230 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Tension in members F t = (3 0 . 875)(3 / 4)(71) 2 . 6 = 43 . 52 kip F = min(35 . 46, 54 . 89, 38 . 83, 43 . 52) = 35 . 46 kip Ans . 8-42 Members: S y = 47 kpsi Bolts: S y = 92 kpsi, S sy = 0 . 577(92) = 53 . 08 kpsi Shear of bolts A d = π (0 . 75) 2 4 = 0 . 442 in 2 τ s = 20 3(0 . 442) = 15 . 08 kpsi n = S sy τ s = 53 . 08 15 . 08 = 3 . 52 Ans . Bearing on bolt σ b =− 20 3(3 / 4)(5 / 8) =− 14 . 22 kpsi n =− S y σ b =− ± 92 14 . 22 ² = 6 . 47 Ans . Bearing on members σ b =− F A b =− 20 3(3 / 4)(5 / 8) =− 14 . 22 kpsi n =− S y σ b =− 47 14 . 22 = 3 .
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Unformatted text preview: 31 Ans . Tension on members σ t = F A = 20 (5 / 8)[7 . 5 − 3(3 / 4)] = 6 . 10 kpsi n = S y σ t = 47 6 . 10 = 7 . 71 Ans . 8-43 Members: S y = 57 kpsi Bolts: S y = 92 kpsi, S sy = . 577(92) = 53 . 08 kpsi Shear of bolts A s = 3 ³ π (3 / 8) 2 4 ´ = . 3313 in 2 τ s = F A = 5 . 4 . 3313 = 16 . 3 kpsi n = S sy τ s = 53 . 08 16 . 3 = 3 . 26 Ans ....
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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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