27_ch 03 Mechanical Design budynas_SM_ch03

# 27_ch 03 Mechanical Design budynas_SM_ch03 -...

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40 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 3-25 (a) I = 1 12 (0 . 75)(1 . 5) 3 = 0 . 2109 in 4 A = 0 . 75(1 . 5) = 1 . 125 in M max is at A. At the bottom of the section, σ max = Mc I = 4000(0 . 75) 0 . 2109 = 14 225 psi Ans. Due to V , τ max constant is between A and B at y = 0 τ max = 3 2 V A = 3 2 667 1 . 125 = 889 psi Ans. (b) I = 1 12 (1)(2) 3 = 0 . 6667 in 4 M max is at A at the top of the beam σ max = 8000(1) 0 . 6667 = 12 000 psi Ans. | V max |= 1000 lbf from O to B at y = 0 τ max = 3 2 V A = 3 2 1000 (2)(1) = 750 psi Ans. (c) I = 1 12 . 75)(2) 3 = 0 . 5in 4 M 1 =− 1 2 600(5) 1500 lbf · in = M 3 M 2 1500 + 1 2 (900)(7 . 5) = 1875 lbf · in M max is at span center. At the bottom of the
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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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