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14_ch 04 Mechanical Design budynas_SM_ch04

# 14_ch 04 Mechanical Design budynas_SM_ch04 -...

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Chapter 4 83 Designating the slope constraint as ξ , we then have ξ = | θ L | = 1 6 E Il F i b i ( b 2 i l 2 ) 2 1 / 2 Setting I = π d 4 / 64 and solving for d d = 32 3 π El ξ F i b i ( b 2 i l 2 ) 2 1 / 2 1 / 4 For the LH bearing, E = 30 Mpsi, ξ = 0 . 001, b 1 = 12, b 2 = 6, and l = 16 . The result is d L = 1 . 31 in. Using a similar flip beam procedure, we get d R = 1 . 36 in for the RH bearing. So use d = 1 3 / 8 in Ans. 4-27 I = π 64 (1 . 375 4 ) = 0 . 17546 in 4 . For the xy plane, use y BC of Table A-9-6 y = 100(4)(16 8) 6(30)(10 6 )(0 . 17546)(16) [8 2 + 4 2 2(16)8] = − 1 . 115(10 3 ) in For the xz plane use y AB z = 300(6)(8) 6(30)(10 6 )(0 . 17546)(16) [8 2 + 6 2 16 2 ] = − 4 . 445(10 3 ) in δ = ( 1 . 115 j 4 . 445 k )(10 3 ) in | δ | = 4 . 583(10 3 ) in Ans. 4-28 d L = 32 n 3 π El ξ F i b i ( b 2 i l 2 ) 2 1 / 2 1 / 4 = 32(1
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