21_ch 20 Mechanical Design budynas_SM_ch20

21_ch 20 Mechanical Design budynas_SM_ch20 -...

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Unformatted text preview: budynas_SM_ch20.qxd 12/06/2006 19:48 Page 21 FIRST PAGES 21 Chapter 20 x40 − x0 θ − x0 R40 = exp − b 40 − 27.7 = exp − 46.2 − 27.7 p40 = 1 − R40 = 1 − 0.846 = 0.154 = 15.4% = 1 − p40 4.38 = 0.846 Ans. 20-29 x = Sut x0 = 151.9, θ = 193.6, b = 8 µx = 151.9 + (193.6 − 151.9) (1 + 1/8) = 151.9 + 41.7 (1.125) = 151.9 + 41.7(0.941 76) = 191.2 kpsi Ans. σx = (193.6 − 151.9)[ ( 1 + 2/8) − 2 (1 + 1/8)]1/2 ˆ = 41.7[ ( 1.25) − 2 (1.125)]1/2 = 41.7[0.906 40 − 0.941 762 ]1/2 = 5.82 kpsi Ans. 5.82 Cx = = 0.030 191.2 20-30 x = Sut x0 = 47.6, θ = 125.6, b = 11.84 x = 47.6 + (125.6 − 47.6) ( 1 + 1/11.84) ¯ x = 47.6 + 78 ( 1.08) ¯ = 47.6 + 78(0.959 73) = 122.5 kpsi σx = (125.6 − 47.6)[ ( 1 + 2/11.84) − ˆ = 78[ ( 1.08) − 2 2 (1 + 1/11.84)]1/2 (1.17)]1/2 = 78(0.959 73 − 0.936 702 ) 1/2 = 22.4 kpsi From Prob. 20-28 p = 1 − exp − x − x0 θ − θ0 b 100 − 47.6 = 1 − exp − 125.6 − 47.6 = 0.0090 Ans. 11.84 ...
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