14_ch 09 Mechanical Design budynas_SM_ch09

# 14_ch 09 Mechanical Design budynas_SM_ch09 - 1 11 = 18 0...

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252 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Equating τ max to τ all gives h = 0 . 523 in . It follows that I = 60 . 3(0 . 523) = 31 . 5in 4 vol = h 2 l 2 = 0 . 523 2 2 (8 + 8) = 2 . 19 in 3 (eff) V = I vol = 31 . 6 2 . 19 = 14 . 4in (fom ± ) V = I u hl = 85 . 33 0 . 523(8 + 8) = 10 . 2in The ratio of (eff) V / (eff) H is 14 . 4 / 91 . 2 = 0 . 158 . The ratio (fom ± ) V / (fom ± ) H is 10 . 2 / 64 . 5 = 0 . 158 . This is not surprising since eff = I vol = I ( h 2 / 2) l = 0 . 707 hI u ( h 2 / 2) l = 1 . 414 I u hl = 1 . 414 fom ± The ratios (eff) V / (eff) H and (fom ± ) V / (fom ± ) H give the same information. 9-20 Because the loading is pure torsion, there is no primary shear. From Table 9-1, category 6: J u = 2 π r 3 = 2 π (1) 3 = 6 . 28 in 3 J = 0 . 707 hJ u = 0 . 707(0 . 25)(6 . 28) = 1 . 11 in 4 τ = Tr J = 20(1)
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Unformatted text preview: 1 . 11 = 18 . 0 kpsi Ans. 9-21 h = . 375 in, d = 8 in, b = 1 in From Table 9-2, category 2: A = 1 . 414(0 . 375)(8) = 4 . 24 in 2 I u = d 3 6 = 8 3 6 = 85 . 3 in 3 I = . 707 hI u = . 707(0 . 375)(85 . 3) = 22 . 6 in 4 τ ± = F A = 5 4 . 24 = 1 . 18 kpsi M = 5(6) = 30 kip · in c = (1 + 8 + 1 − 2) / 2 = 4 in τ ±± = Mc I = 30(4) 22 . 6 = 5 . 31 kpsi τ max = ± τ ± 2 + τ ±± 2 = ± 1 . 18 2 + 5 . 31 2 = 5 . 44 kpsi Ans....
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