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 E I , y max = U Q Q = 0 y max = 2 l / 2 0 2 M 2 E I M Q dx Q = 0 y max = 2 E I l / 2 0 w x 2 ( l x ) + Qx 2 x 2 dx Q = 0 Set Q = 0 and integrate y max = w 2 E I lx 3 3 x 4 4 l / 2 0 y max = 5 w l 4 384 E I Ans . 4-46 I = 2 ( 1 . 85 ) = 3 . 7 in 4 Adding weight of channels of 0 . 833 lbf · in, M = − Fx 10 . 833 2 x 2 = − Fx 5 . 417 x 2 M F = − x δ B = 1 E I 48 0 M M F dx = 1 E I 48 0 ( Fx + 5 . 417 x 2 )( x ) dx = (220 / 3)(48 3 ) + (5 . 417 / 4)(48 4 ) 30(10 6 )(3 . 7) = 0 . 1378 in
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