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31_ch 05 Mechanical Design budynas_SM_ch05

31_ch 05 Mechanical Design budynas_SM_ch05 -...

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Unformatted text preview: budynas_SM_ch05.qxd 11/29/2006 15:00 FIRST PAGES Page 145 Chapter 5 145 (b) The tangential stress for the outer cylinder at the shrink-ﬁt surface is given by σot = p ¯ ro + R 2 ¯2 ¯ ro − R 2 ¯2 = 18.70(106 ) δ 0.56252 + 0.37552 0.56252 − 0.37552 = 48.76(106 ) δ psi σot = 48.76(106 )(0.001) = 48.76(103 ) psi ¯ σσot = Cδ σot = 0.0707(48.76)(103 ) = 34.45 psi ˆ ¯ σot = N(48 760, 3445) psi Ans. 5-45 From Prob. 5-44, at the ﬁt surface σot = N(48.8, 3.45) kpsi. The radial stress is the ﬁt pressure which was found to be p = 18.70(106 ) δ p = 18.70(106 )(0.001) = 18.7(103 ) psi ¯ σ p = Cδ p = 0.0707(18.70)(103 ) ˆ ¯ = 1322 psi and so p = N(18.7, 1.32) kpsi and σor = −N(18.7, 1.32) kpsi These represent the principal stresses. The von Mises stress is next assessed. σ A = 48.8 kpsi, ¯ σ B = −18.7 kpsi ¯ k = σ B /σ A = −18.7/48.8 = −0.383 ¯¯ σ = σ A (1 − k + k 2 ) 1/2 ¯ ¯ = 48.8[1 − ( −0.383) + ( −0.383) 2 ]1/2 = 60.4 kpsi σσ = C p σ = 0.0707(60.4) = 4.27 kpsi ˆ ¯ Using the interference equation z=− =− ¯¯ S−σ σ S + σσ ˆ2 ˆ2 1/2 95.5 − 60.4 = −4.5 [(6.59) 2 + (4.27) 2 ]1/2 p f = α = 0.000 003 40, or about 3 chances in a million. Ans. ...
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