22_ch 06 Mechanical Design budynas_SM_ch06

22_ch 06 Mechanical Design budynas_SM_ch06 - N 3 = ± 32...

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168 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (a) Miner’s method N 2 = ± 315 1409 . 8 ² 1 / 0 . 141 355 = 40 200 cycles n 1 N 1 + n 2 N 2 = 1 50 000 131 200 + n 2 40 200 = 1 n 2 = 24 880 cycles Ans . (b) Manson’s method Two data points: 0 . 9(590 MPa), 10 3 cycles 266 . 5MPa ,8 1 200 cycles 0 . 9(590) 266 . 5 = a 2 (10 3 ) b 2 a 2 (81 200) b 2 1 . 9925 = (0 . 012 315) b 2 b 2 = log 1 . 9925 log 0 . 012 315 =− 0 . 156 789 a 2 = 266 . 5 (81 200) 0 . 156 789 = 1568 . 4MPa n 2 = ± 315 1568 . 4 ² 1 / 0 . 156 789 = 27 950 cycles Ans . 6-30 (a) Miner’s method a = [0 . 9(76)] 2 30 = 155 . 95 kpsi b =− 1 3 log 0 . 9(76) 30 =− 0 . 119 31 σ 1 = 48 kpsi, N 1 = ± 48 155 . 95 ² 1 / 0 . 119 31 = 19 460 cycles σ 2 = 38 kpsi, N 2 = ± 38 155 . 95 ² 1 / 0 . 119 31 = 137 880 cycles σ 3 = 32 kpsi,
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Unformatted text preview: N 3 = ± 32 155 . 95 ² 1 / − . 119 31 = 582 150 cycles n 1 N 1 + n 2 N 2 + n 3 N 3 = 1 4000 19 460 + 60 000 137 880 + n 3 582 150 = 1 ⇒ n 3 = 209 160 cycles Ans . (b) Manson’s method The life remaining after the ﬁrst cycle is N R 1 = 19 460 − 4000 = 15 460 cycles. The two data points required to deﬁne S ± e ,1 are [0 . 9(76), 10 3 ] and (48, 15 460) . . 9(76) 48 = a 2 (10 3 ) b 2 a 2 (15 460) ⇒ 1 . 425 = (0 . 064 683) b 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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