28_ch 03 Mechanical Design budynas_SM_ch03

28_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 41 (d) I = 1 12 (1)(2) 3 = 0 . 6667 in 4 M 1 =− 600 2 (6) 1800 lbf · in M 2 1800 + 1 2 750(7 . 5) = 1013 lbf · in At A , top of beam σ max = 1800(1) 0 . 6667 = 2700 psi Ans. At A , y = 0 τ max = 3 2 750 (2)(1) = 563 psi Ans. 3-26 M max = w l 2 8 σ max = w l 2 c 8 I w = 8 σ I cl 2 (a) l = 12(12) = 144 in, I = (1 / 12)(1 . 5)(9 . 5) 3 = 107 . 2in 4 w = 8(1200)(107 . 2) 4 . 75(144 2 ) = 10 . 4 lbf/in Ans. (b) l = 48 in, I = ( π/ 64)(2 4 1 . 25 4 ) = 0 . 6656 in 4 w = 8(12)(10 3 )(0 . 6656) 1(48) 2 = 27 . 7 lbf/in Ans. (c) l = 48 in, I . = / 12)(2)(3 3 ) / 12)(1 . 625)(2 . 625 3 ) = 2 . 051 in 4 w = 8(12)(10 3 )(2 . 051) 1 . 5(48) 2 = 57 . 0 lbf/in Ans. (d) l = 72 in; Table A-6, I = 2(1 . 24) = 2 . 48 in 4 c max = 2 . 158 " w = 8(12)(10 3 . 48) 2 . 158(72) 2 = 21 . 3 lbf/in Ans. (e)
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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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