21_ch 09 Mechanical Design budynas_SM_ch09

21_ch 09 Mechanical Design budynas_SM_ch09 -...

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Chapter 9 259 9-27 Use b = d = 4 in. Since h = 5 / 8 in, the primary shear is τ ± = F 1 . 414(5 / 8)(4) = 0 . 283 F The secondary shear calculations, for a moment arm of 14 in give J u = 4[3(4 2 ) + 4 2 ] 6 = 42 . 67 in 3 J = 0 . 707 hJ u = 0 . 707(5 / 8)42 . 67 = 18 . 9in 4 τ ±± x = τ ±± y = Mr y J = 14 F (2) 18 . 9 = 1 . 48 F Thus, the maximum shear and allowable load are: τ max = F ± 1 . 48 2 + (0 . 283 + 1 . 48) 2 = 2 . 30 F F = τ all 2 . 30 = 20 2 . 30 = 8 . 70 kip Ans . From Prob. 9-5 b , τ all = 11 kpsi F all = τ all 2 . 30 = 11 2 . 30 = 4 . 78 kip The allowable load has thus increased by a factor of 1.8 Ans. 9-28 Purchase the hook having the design shown in Fig. P9-28 b . Referring to text Fig. 9-32
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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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