42_ch 04 Mechanical Design budynas_SM_ch04

# 42_ch 04 Mechanical Design budynas_SM_ch04 -...

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Chapter 4 111 Use 25 × 5 . 5mm bars Ans. The factor of safety is thus n = 11 010 3786 = 2 . 91 Ans . 4-79 ± F = 0 = 2000 + 10 000 P P = 12 000 lbf Ans. ± M A = 12 000 ² 5 . 68 2 ³ 10 000(5 . 68) + M = 0 M = 22 720 lbf · in e = M P = 22 12 ² 720 000 ³ = 1 . 893 in Ans. From Table A-8, A = 4 . 271 in 2 , I = 7 . 090 in 4 k 2 = I A = 7 . 090 4 . 271 = 1 . 66 in 2 σ c =− 12 000 4 . 271 ´ 1 + 1 . 893(2) 1 . 66 µ 9218 psi Ans. σ t 12 000 4 . 271 ´ 1 1 . 893(2) 1 . 66 µ = 3598 psi 4-80 This is a design problem so the solutions will differ. 4-81 For free fall with y h ± F y m ¨ y = 0 mg m ¨ y = 0, so ¨ y = g Using y = a + bt + ct 2 , we have at t = 0,
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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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