34_ch 08 Mechanical Design budynas_SM_ch08

# 34_ch 08 Mechanical Design budynas_SM_ch08 - 3 = 1650 lbf F...

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Chapter 8 237 8-50 Bearing on members: S y = 54 kpsi, n = 54 9 . 6 = 5 . 63 Ans. Bending of members: Considering the right-hand bolt M = 300(15) = 4500 lbf · in I = 0 . 375(2) 3 12 0 . 375(0 . 5) 3 12 = 0 . 246 in 4 σ = Mc I = 4500(1) 0 . 246 = 18 300 psi n = 54(10) 3 18 300 = 2 . 95 Ans. 8-51 The direct shear load per bolt is F ± = 2500 / 6 = 417 lbf. The moment is taken only by the four outside bolts. This moment is M = 2500(5) = 12 500 lbf · in. Thus F ±± = 12 500 2(5) = 1250 lbf and the resultant bolt load is F = ± (417) 2 + (1250) 2 = 1318 lbf Bolt strength, S y = 57 kpsi; Channel strength, S y = 46 kpsi; Plate strength, S y = 45 . 5 kpsi Shear of bolt: A s = π (0 . 625) 2 / 4 = 0 . 3068 in 2 n = S sy τ = (0 . 577)(57 000) 1318 / 0 . 3068 = 7 . 66 Ans. 2" 3 8 " 1 2 " F ±± A = F ±± B = 4950
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Unformatted text preview: 3 = 1650 lbf F A = 1500 lbf, F B = 1800 lbf Bearing on bolt: A b = 1 2 3 8 = . 1875 in 2 = F A = 1800 . 1875 = 9600 psi n = 92 9 . 6 = 9 . 58 Ans. Shear of bolt: A s = 4 (0 . 5) 2 = . 1963 in 2 = F A = 1800 . 1963 = 9170 psi S sy = . 577(92) = 53 . 08 kpsi n = 53 . 08 9 . 17 = 5 . 79 Ans. F' A 150 lbf A B F' B 150 lbf y x O F&quot; B 1650 lbf F&quot; A 1650 lbf 1 1 2 &quot; 1 1 2 &quot; 300 lbf M 16.5(300) 4950 lbf in V 300 lbf 16 1 2 &quot;...
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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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