12_ch 04 Mechanical Design budynas_SM_ch04

# 12_ch 04 Mechanical Design budynas_SM_ch04 - ± 32 Fb b 2...

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Chapter 4 81 Differentiating this equation and solving for the slope at the left bolster gives Thus, dy dx = w 24 EI (6 lx 2 4 x 3 l 3 ) dy dx ± ± ± ± x = 0 =− w l 3 24 EI =− (5000 / 12)(432) 3 24(30)(10 6 )(5450) =− 0 . 008 57 The slope at the right bolster is 0.008 57, so equation at left end is y =− 0 . 008 57 x and at the right end is y = 0 . 008 57( x l ) . Ans. 4-23 From Table A-9-6, y L = Fbx 6 EIl ( x 2 + b 2 l 2 ) y L = Fb 6 EIl ( x 3 + b 2 x l 2 x ) dy L dx = Fb 6 EIl (3 x 2 + b 2 l 2 ) dy L dx ± ± ± ± x = 0 = Fb ( b 2 l 2 ) 6 EIl Let ξ = ± ± ± ± Fb ( b 2 l 2 ) 6 EIl ± ± ± ± And set I = π d 4 L 64 And solve for d L d L = ± ± ±
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Unformatted text preview: ± 32 Fb ( b 2 − l 2 ) 3 π El ξ ± ± ± ± 1 / 4 Ans. For the other end view, observe the ﬁgure of Table A-9-6 from the back of the page, noting that a and b interchange as do x and − x d R = ± ± ± ± 32 Fa ( l 2 − a 2 ) 3 π El ξ ± ± ± ± 1 / 4 Ans. For a uniform diameter shaft the necessary diameter is the larger of d L and d R ....
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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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