23_ch 04 Mechanical Design budynas_SM_ch04

23_ch 04 Mechanical Design budynas_SM_ch04 -...

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92 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 4-45 Place a dummy load Q at the center. Then, M = w x 2 ( l x ) + Qx 2 U = 2 ± l / 2 0 M 2 dx 2 EI , y max = U Q ² ² ² ² Q = 0 y max = 2 ³± l / 2 0 2 M 2 ´ M Q µ Q = 0 y max = 2 ·± l / 2 0 ³ w x 2 ( l x ) + 2 x 2 ¾ Q = 0 Set Q = 0 and integrate y max = w 2 ´ lx 3 3 x 4 4 µ l / 2 0 y max = 5 w l 4 384 Ans . 4-46 I = 2 ( 1 . 85 ) = 3 . 7in 4 Adding weight of channels of 0 . 833 lbf · in, M =− Fx 10 . 833 2 x 2 5 . 417 x 2 M F x δ B = 1 ± 48 0 M M F = 1 ± 48 0 ( + 5 . 417 x 2 )( x ) = (220 / 3)(48 3 ) + (5 . 417 / 4)(48 4 ) 30(10 6 )(3 . 7) = 0 . 1378 in in direction of 220 lbf ± y B 0 . 1378 in
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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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