2_ch 10 Mechanical Design budynas_SM_ch10

# 2_ch 10 Mechanical Design budynas_SM_ch10 -...

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262 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) F | S sy = 45 . 2 lbf Ans . (c) k = 11 . 55 lbf/in Ans . (d) ( L 0 ) cr = 2 . 63 D α = 2 . 63(1 . 120) 0 . 5 = 5 . 89 in Many designers provide ( L 0 ) cr / L 0 5 or more; therefore, plain ground ends are not often used in machinery due to buckling uncertainty. 10-4 Referring to Prob. 10-3 solution, C = 10 . 67, N a = 11, S = 125 . 4 kpsi, ( L 0 ) cr = 5 . 89 in and F = 45 . 2 lbf (at yield) . Eq. (10-18): 4 C 12 C = 10 . 67 O . K . Eq. (10-19): 3 N a 15 N a = 11 O . K . L 0 = 5 . 17 in, L s = 1 . 26 in y 1 = F 1 k = 30 11 . 55 = 2 . 60 in L 1 = L 0 y 1 = 5 . 17 2 . 60 = 2 . 57 in ξ = y s y 1 1 = 5 . 17 1 . 26 2 . 60 1 = 0 . 50 Eq. (10-20): ξ 0 . 15, ξ = 0 . 50 O . K . From Eq. (10-3) for static service τ 1 = K B ± 8 F 1 D π d 3 ² = 1 . 126 ³ 8(30)(1 .
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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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