15_ch 04 Mechanical Design budynas_SM_ch04

15_ch 04 Mechanical Design budynas_SM_ch04 -...

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84 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design At x = 0 , the LH slope is θ L = dy L dx = M B 6 EIl ( 3 a 2 6 al + 2 l 2 ) from which ξ = | θ L | = M B 6 EIl ( l 2 3 b 2 ) Setting I = π d 4 / 64 and solving for d d = ± ± ± ± 32 M B ( l 2 3 b 2 ) 3 π El ξ ± ± ± ± 1 / 4 For a multiplicity of moments, the slopes add vectorially and d L = ± ± ± ± 32 3 π El ξ ² ³ ´ M i ( l 2 3 b 2 i 2 1 / 2 ± ± ± ± 1 / 4 d R = ± ± ± ± 32 3 π El ξ ² ³ ´ M i ( 3 a 2 i l 2 2 1 / 2 ± ± ± ± 1 / 4 The greatest slope is at the LH bearing. So d = ± ± ± ± 32(1200)[9 2 3(4 2 )] 3 π (30)(10 6 )(9)(0 . 002) ± ± ± ± 1 / 4 = 0 . 706 in So use d = 3 / 4in Ans. 4-30 6 F AC = 18(80) F AC = 240 lbf R O = 160 lbf I = 1 12 (0 . 25)(2 3 ) = 0 . 1667 in 4 Initially, ignore the stretch of
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