17_ch 09 Mechanical Design budynas_SM_ch09

17_ch 09 Mechanical Design budynas_SM_ch09 -...

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Chapter 9 255 9-24 F = 100 lbf all = 3 kpsi F B = 100 ( 16 / 3 ) = 533 . 3 lbf F x B =− 533 . 3 cos 60 266 . 7 lbf F y B 533 . 3 cos 30 462 lbf It follows that R y A = 562 lbf and R x A = 266 . 7 lbf, R A = 622 lbf M = 100(16) = 1600 lbf · in The OD of the tubes is 1 in. From Table 9-1, category 6: A = 1 . 414 hr )( 2 ) = 2 ( 1 . 414 )(π h )( 1 / 2 ) = 4 . 44 h in 2 J u = 2 π r 3 = 2 π( 1 / 2 ) 3 = 0 . 785 in 3 J = 2 ( 0 . 707 ) hJ u = 1 . 414 ( 0 . 785 ) h = 1 . 11 h in 4 τ ± = V A = 622 4 . 44 h = 140 h τ ±± = Tc J = Mc J = 1600 ( 0 . 5 ) 1 . 11 h = 720 . 7 h The shear stresses, τ ± and τ ±±
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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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