26_ch 10 Mechanical Design budynas_SM_ch10

# 26_ch 10 Mechanical Design budynas_SM_ch10 - M for the...

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286 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 10-31 For the hook, M = FR sin θ , M /∂ F = R sin θ δ F = 1 EI ± π/ 2 0 FR 2 sin 2 Rd θ = π 2 PR 3 EI The total deﬂection of the body and the two hooks δ = 8 FD 3 N b d 4 G + 2 π 2 FR 3 EI = 8 FD 3 N b d 4 G + π F ( D / 2) 3 E ( π/ 64)( d 4 ) = 8 FD 3 d 4 G ² N b + G E ³ = 8 FD 3 N a d 4 G ± N a = N b + G E QED 10-32 Table 10-4 for A227: A = 140 kpsi · in m , m = 0 . 190 Table 10-5: E = 28 . 5(10 6 ) psi S ut = 140 (0 . 162) 0 . 190 = 197 . 8 kpsi Eq. (10-57): S y = σ all = 0 . 78(197 . 8) = 154 . 3 kpsi D = 1 . 25 0 . 162 = 1 . 088 in C = D / d = 1 . 088 / 0 . 162 = 6 . 72 K i = 4 C 2 C 1 4 C ( C 1) = 4(6 . 72) 2 6 . 72 1 4(6 . 72)(6 . 72 1) = 1 . 125 From σ = K i 32 M π d 3 Solving for
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Unformatted text preview: M for the yield condition, M y = π d 3 S y 32 K i = π (0 . 162) 3 (154 300) 32(1 . 125) = 57 . 2 lbf · in Count the turns when M = N = 2 . 5 − M y d 4 E / (10 . 8 DN ) from which N = 2 . 5 1 + [10 . 8 DM y / ( d 4 E )] = 2 . 5 1 + { [10 . 8(1 . 088)(57 . 2)] / [(0 . 162) 4 (28 . 5)(10 6 )] } = 2 . 417 turns F ± R ± D ± 2...
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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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