23_ch 06 Mechanical Design budynas_SM_ch06

# 23_ch 06 Mechanical Design budynas_SM_ch06 - Given S ut =...

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Chapter 6 169 b 2 = log(1 . 425) log(0 . 064 683) =− 0 . 129 342 a 2 = 48 (15 460) 0 . 129 342 = 167 . 14 kpsi N 2 = ± 38 167 . 14 ² 1 / 0 . 129 342 = 94 110 cycles N R 2 = 94 110 60 000 = 34 110 cycles 0 . 9(76) 38 = a 3 (10 3 ) b 3 a 3 (34 110) b 3 1 . 8 = (0 . 029 317) b 3 b 3 = log 1 . 8 log(0 . 029 317) =− 0 . 166 531, a 3 = 38 (34 110) 0 . 166 531 = 216 . 10 kpsi N 3 = ± 32 216 . 1 ² 1 / 0 . 166 531 = 95 740 cycles Ans . 6-31 Using Miner’s method a = [0 . 9(100)] 2 50 = 162 kpsi b =− 1 3 log 0 . 9(100) 50 =− 0 . 085 091 σ 1 = 70 kpsi, N 1 = ± 70 162 ² 1 / 0 . 085 091 = 19 170 cycles σ 2 = 55 kpsi, N 2 = ± 55 162 ² 1 / 0 . 085 091 = 326 250 cycles σ 3 = 40 kpsi, N 3 →∞ 0 . 2 N 19 170 + 0 . 5 N 326 250 + 0 . 3 N = 1 N = 83 570 cycles Ans . 6-32
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Unformatted text preview: Given S ut = 245 LN (1, 0 . 0508) kpsi From Table 7-13: a = 1 . 34, b = − . 086, C = . 12 k a = 1 . 34 ¯ S − . 086 ut LN (1, 0 . 120) = 1 . 34(245) − . 086 LN (1, 0 . 12) = . 835 LN (1, 0 . 12) k b = 1 . 02 (as in Prob. 6-1) Eq. (6-70) S e = . 835(1 . 02) LN (1, 0 . 12)[107 LN (1, 0 . 139)] ¯ S e = . 835(1 . 02)(107) = 91 . 1 kpsi...
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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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