16_ch 04 Mechanical Design budynas_SM_ch04

# 16_ch 04 Mechanical Design budynas_SM_ch04 -...

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Chapter 4 85 4-31 θ = TL JG = (0 . 1 F )(1 . 5) ( π/ 32)(0 . 012 4 )(79 . 3)(10 9 ) = 9 . 292(10 4 ) F Due to twist δ B 1 = 0 . 1( θ ) = 9 . 292(10 5 ) F Due to bending δ B 2 = FL 3 3 EI = F . 1 3 ) 3(207)(10 9 )( 64)(0 . 012 4 ) = 1 . 582(10 6 ) F δ B = 1 . 582(10 6 ) F + 9 . 292(10 5 ) F = 9 . 450(10 5 ) F k = 1 9 . 450(10 5 ) = 10 . 58(10 3 ) N/m = 10 . 58 kN/m Ans. 4-32 R 1 = Fb l R 2 = Fa l δ 1 = R 1 k 1 δ 2 = R 2 k 2 Spring deﬂection y S =− δ 1 + ± δ 1 δ 2 l ² x k 1 l + ± k 1 l 2 k 2 l 2 ² x y AB = Fbx 6 EIl ( x 2 + b 2 l 2 ) + Fx l 2 ± b k 1 a k 2 ² k 1 l Ans. y BC = ( l x ) 6 ( x 2 + a 2 2 lx ) + l 2 ± b k 1 a k 2 ² k 1 l Ans. 4-33 See Prob. 4-32 for deﬂection due to springs. Replace / l and
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