16_ch 13 Mechanical Design budynas_SM_ch13

16_ch 13 Mechanical Design budynas_SM_ch13 - F 2 = 6426 4(8...

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348 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The compressive loads at A and D are absorbed by the base plate, not the bolts. For W t 32 , the tensions in C and D are ± M AB = 0 1681(4 . 875 + 15 . 25) 2 F (15 . 25) = 0 F = 1109 lbf If W t 32 reverses, 15.25 in changes to 13.25 in, 4.815 in changes to 2.875 in, and the forces change direction. For A and B , 1681(2 . 875) 2 F 1 (13 . 25) = 0 F 1 = 182 . 4 lbf For W r 32 M = 612(4 . 875 + 11 . 25 / 2) = 6426 lbf · in a = p (14 / 2) 2 + (11 . 25 / 2) 2 = 8 . 98 in
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Unformatted text preview: F 2 = 6426 4(8 . 98) = 179 lbf At C and D , the shear forces are: F S 1 = p [153 + 179(5 . 625 / 8 . 98)] 2 + [179(7 / 8 . 98)] 2 = 300 lbf At A and B , the shear forces are: F S 2 = p [153 179(5 . 625 / 8 . 98)] 2 + [179(7 / 8 . 98)] 2 = 145 lbf C a D 153 lbf 153 lbf F 2 F 2 F 2 F 2 612 4 5 153 lbf 4.875 11.25 14 612 lbf 153 lbf B C 1681 lbf 4.875 15.25" F F D F 1 F 1 A...
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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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