31_ch 05 Mechanical Design budynas_SM_ch05

31_ch 05 Mechanical Design budynas_SM_ch05 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

Ask a homework question - tutors are online