19_ch 04 Mechanical Design budynas_SM_ch04

19_ch 04 Mechanical Design budynas_SM_ch04 - R O = 160 lbf...

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88 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 0 = w 12 l ( l 2 a 2 ) l 3 w 24 l 4 + C 1 l C 1 = w l 24 (2 a 2 l 2 ) y = w 24 EIl [2( l 2 a 2 ) x 3 lx 4 + 2( l + a ) 2 ± x l ² 3 + l 2 (2 a 2 l 2 ) x ] Ans. 4-39 R A = R B = 500 N, and I = 1 12 (9)35 3 = 32 . 156(10 3 )mm 4 For first half of beam, M =− 500 x + 500 ± x 0 . 25 ² 1 where x is in meters EI dy dx =− 250 x 2 + 250 ± x 0 . 25 ² 2 + C 1 At x = 0 . 5 m, dy / dx = 0 0 =− 250(0 . 5 2 ) + 250(0 . 5 0 . 250) 2 + C 1 C 1 = 46 . 875 N · m 2 EIy =− 250 3 x 3 + 250 3 ± x 0 . 25 ² 3 + 46 . 875 x + C 2 y = 0 at x = 0 . 25 m 0 =− 250 3 0 . 25 3 + 46 . 875(0 . 25) + C 2 C 2 =− 10 . 417 N · m 3 EIy =− 250 3 x 3 + 250 3 ± x 0 . 25 ² 3 + 46 . 875 x 10 . 42 Evaluating y at A and the center, y A = 1 207(10 9 )32 . 156(10 9 ) ± 250 3 (0 3 ) + 250 3 (0) 3 + 46 . 875(0) 10 . 417 ² 10 3 =− 1 . 565 mm Ans. y | x = 0 . 5m = 1 207(10 9 )32 . 156(10 9 ) ± 250 3 (0 . 5 3 ) + 250 3 (0 . 5 0 . 25) 3 + 46 . 875(0 . 5) 10 . 417 ² 10 3 =− 2 . 135 mm Ans . 4-40 From Prob. 4-30,
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Unformatted text preview: R O = 160 lbf , F AC = 240 lbf I = . 1667 in 4 M = 160 x + 240 x 6 1 E I dy dx = 80 x 2 + 120 x 6 2 + C 1 E I y = 26 . 67 x 3 + 40 x 6 3 + C 1 x + C 2 y = 0 at x = C 2 = y A = F L AE AC = 240(12) ( / 4)(1 / 2) 2 (10)(10 6 ) = 1 . 4668(10 3 ) in at x = 6 10(10 6 )(0 . 1667)( 1 . 4668)(10 3 ) = 26 . 67(6 3 ) + C 1 (6) C 1 = 552 . 58 lbf in 2...
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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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