13_ch 04 Mechanical Design budynas_SM_ch04

13_ch 04 Mechanical Design budynas_SM_ch04 - Ans. 4-26 x y...

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82 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 4-24 Incorporating a design factor into the solution for d L of Prob. 4-23, d = ± 32 n 3 π El ξ Fb ( l 2 b 2 ) ² 1 / 4 = ³ ³ ³ ³ (mm 10 3 ) kN mm 3 GPa mm 10 3 (10 9 ) 10 9 (10 3 ) ³ ³ ³ ³ 1 / 4 d = 4 ´ 32(1 . 28)(3 . 5)(150) | (250 2 150 2 ) | 3 π (207)(250)(0 . 001) 10 12 = 36 . 4mm Ans. 4-25 The maximum occurs in the right section. Flip beam A-9-6 and use y = Fbx 6 EIl ( x 2 + b 2 l 2 ) where b = 100 mm dy dx = Fb 6 EIl (3 x 2 + b 2 l 2 ) = 0 Solving for x , x = µ l 2 b 2 3 = µ 250 2 100 2 3 = 132 . 29 mm from right y = 3 . 5(10 3 )(0 . 1)(0 . 132 29) 6(207)(10 9 )( π/ 64)(0 . 0364 4 )(0 . 25) [0 . 132 29 2 + 0 . 1 2 0 . 25 2 ](10 3 ) =− 0 . 0606 mm
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Unformatted text preview: Ans. 4-26 x y z F 1 a 2 b 2 b 1 a 1 F 2 3.5 kN 100 250 150 d The slope at x = 0 due to F 1 in the xy plane is xy = F 1 b 1 ( b 2 1 l 2 ) 6 E Il and in the xz plane due to F 2 is xz = F 2 b 2 ( b 2 2 l 2 ) 6 E Il For small angles, the slopes add as vectors. Thus L = 2 xy + 2 xz 1 / 2 = F 1 b 1 ( b 2 1 l 2 ) 6 E Il 2 + F 2 b 2 ( b 2 2 l 2 ) 6 E Il 2 1 / 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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