5_ch 04 Mechanical Design budynas_SM_ch04

5_ch 04 Mechanical Design budynas_SM_ch04 - So y max = −...

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74 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 4-11 y = Cx (2 lx 2 x 3 l 3 ) where C = w 24 EI dy dx = C (6 lx 2 4 x 3 l 3 ) ± dy dx ² 2 = C 2 (36 l 2 x 4 48 lx 5 12 l 4 x 2 + 16 x 6 + 8 x 3 l 3 + l 6 ) λ = 1 2 l ³ 0 ± dy dx ² 2 dx = 1 2 C 2 l ³ 0 (36 l 2 x 4 48 lx 5 12 l 4 x 2 + 16 x 6 + 8 x 3 l 3 + l 6 ) dx = C 2 ± 17 70 l 7 ² = ´ w 24 EI µ 2 ± 17 70 l 7 ² = 17 40 320 ´ w EI µ 2 l 7 Ans. 4-12 I = 2(5 . 56) = 11 . 12 in 4 y max = y 1 + y 2 =− w l 4 8 EI + Fa 2 6 EI ( a 3 l ) Here w = 50 / 12 = 4 . 167 lbf/in, and a = 7(12) = 84 in, and l = 10(12) = 120 in. y 1 =− 4 . 167(120) 4 8(30)(10 6 )(11 . 12) =− 0 . 324 in y 2 =− 600(84) 2 [3(120) 84] 6(30)(10 6 )(11 . 12) =− 0 . 584 in
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Unformatted text preview: So y max = − . 324 − . 584 = − . 908 in Ans . M = − Fa − ( w l 2 / 2) = − 600(84) − [4 . 167(120) 2 / 2] = − 80 400 lbf · in c = 4 − 1 . 18 = 2 . 82 in σ max = − My I = − ( − 80 400)( − 2 . 82) 11 . 12 (10 − 3 ) = − 20 . 4 kpsi Ans. σ max is at the bottom of the section....
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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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