7_ch 04 Mechanical Design budynas_SM_ch04

7_ch 04 Mechanical Design budynas_SM_ch04 - − 216 6(30(10...

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76 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 4-15 I = 1 12 (9)(35 3 ) = 32 . 156(10 3 )mm 4 From Table A-9-10 y C =− Fa 2 3 EI ( l + a ) dy AB dx = Fa 6 EIl ( l 2 3 x 2 ) Thus, θ A = Fal 2 6 EIl = Fal 6 EI y D =− θ A a =− Fa 2 l 6 EI With both loads, y D =− Fa 2 l 6 EI Fa 2 3 EI ( l + a ) =− Fa 2 6 EI (3 l + 2 a ) =− 500(250 2 ) 6(207)(10 9 )(32 . 156)(10 3 ) [3(500) + 2(250)](10 3 ) 2 =− 1 . 565 mm Ans. y E = 2 Fa ( l / 2) 6 EIl ± l 2 ² l 2 ³ 2 ´ = Fal 2 8 EI = 500(250)(500 2 )(10 3 ) 2 8(207)(10 9 )(32 . 156)(10 3 ) = 0 . 587 mm Ans. 4-16 a = 36 in, l = 72 in, I = 13 in 4 , E = 30 Mpsi y = F 1 a 2 6 EI ( a 3 l ) F 2 l 3 3 EI = 400(36) 2 (36
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Unformatted text preview: − 216) 6(30)(10 6 )(13) − 400(72) 3 3(30)(10 6 )(13) = − . 1675 in Ans. 4-17 I = 2(1 . 85) = 3 . 7 in 4 Adding the weight of the channels, 2(5) / 12 = . 833 lbf/in, y A = − w l 4 8 E I − Fl 3 3 E I = − 10 . 833 ( 48 4 ) 8 ( 30 )( 10 6 )( 3 . 7 ) − 220 ( 48 3 ) 3 ( 30 )( 10 6 )( 3 . 7 ) = − . 1378 in Ans. ± A a D C F B a E A...
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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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