27_ch 05 Mechanical Design budynas_SM_ch05

27_ch 05 Mechanical Design budynas_SM_ch05 -...

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Chapter 5 141 Outer radius is plane stress σ x = 11 . 18 kpsi, σ y = 19 . 58 kpsi, τ xy = 7 . 45 kpsi Eq. (5-15) σ ± = [11 . 18 2 (11 . 18)(19 . 58) + 19 . 58 2 + 3(7 . 45 2 )] 1 / 2 = S y n o = 60 n o 21 . 35 = 60 n o n o = 2 . 81 Ans. Inner radius, 3D state of stress From Eq. (5-14) with τ yz = τ zx = 0 σ ± = 1 2 [(9 . 78 22 . 58) 2 + (22 . 58 + 2 . 997) 2 + ( 2 . 997 9 . 78) 2 + 6(6 . 52) 2 ] 1 / 2 = 60 n i 24 . 86 = 60 n i n i = 2 . 41 Ans. 5-39 From Prob. 5-38: p = 2 . 997 kpsi, I = 0 . 5369 in 4 , J = 1 . 0738 in 4 Inner member: Outer radius: ( σ t ) o =− 2 . 997 ± (0 . 875 2 + 0 . 75 2 ) (0 . 875 2 0 . 75 2 ) ² =− 19 . 60 kpsi ( σ r ) o =− 2 . 997 kpsi Inner radius: ( σ t ) i =− 2(2 . 997)(0 . 875 2 ) 0 . 875 2
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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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