17_ch 04 Mechanical Design budynas_SM_ch04

# 17_ch 04 Mechanical Design budynas_SM_ch04 - 36 ² 2 −...

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86 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 4-34 Let the load be at x > l / 2. The maximum deﬂection will be in Section AB (Table A-9-10) y AB = Fbx 6 EIl ( x 2 + b 2 l 2 ) dy AB dx = Fb 6 EIl (3 x 2 + b 2 l 2 ) = 0 3 x 2 + b 2 l 2 = 0 x = ± l 2 b 2 3 , x max = ± l 2 3 = 0 . 577 lA n s . For x < l / 2 x min = l 0 . 577 l = 0 . 423 lA n s . 4-35 M O = 50(10)(60) + 600(84) = 80 400 lbf · in R O = 50(10) + 600 = 1100 lbf I = 11 . 12 in 4 from Prob. 4-12 M =− 80 400 + 1100 x 4 . 167 x 2 2 600 ± x 84 ² 1 EI dy dx =− 80 400 x + 550 x 2 0 . 6944 x 3 300 ± x 84 ² 2 + C 1 dy dx = 0 at x = 0 ± C 1 = 0 EIy =− 402 00 x 2 + 183 . 33 x 3 0 . 1736 x 4 100 ± x 84 ² 3 + C 2 y = 0at x = 0 ± C 2 = 0 y B = 1 30(10 6 )(11 . 12) [ 40 200(120 2 ) + 183 . 33(120 3 ) 0 . 1736(120 4 ) 100(120 84) 3 ] =− 0 . 9075 in Ans. 4-36 See Prob. 4-13 for reactions: R O = 860 lbf, R C = 540 lbf M = 860 x 800 ± x 36 ² 1 600 ± x 60 ² 1 EI dy dx = 430 x 2 400 ±
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Unformatted text preview: 36 ² 2 − 300 ± x − 60 ² 2 + C 1 E I y = 143 . 33 x 3 − 133 . 33 ± x − 36 ² 3 − 100 ± x − 60 ² 3 + C 1 x + C 2 y = 0 at x = ⇒ C 2 = y = 0 at x = 120 in ⇒ C 1 = − 1 . 2254(10 6 ) lbf · in 2 Substituting C 1 and C 2 and evaluating at x = 60, E I y = 30(10 6 ) I ² − 1 16 ³ = 143 . 33(60 3 ) − 133 . 33(60 − 36) 3 − 1 . 2254(10 6 )(60) I = 23 . 68 in 4 Agrees with Prob. 4-13. The rest of the solution is the same. 10' 7' R O 600 lbf 50 lbf/ft M O O A B...
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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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