2_ch 06 Mechanical Design budynas_SM_ch06

# 2_ch 06 Mechanical Design budynas_SM_ch06 - 162 . b = log...

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148 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 6-4 From S f = aN b log S f = log a + b log N Substituting (1, S ut ) log S ut = log a + b log (1) From which a = S ut Substituting (10 3 , fS ut ) and a = S ut log fS ut = log S ut + b log 10 3 From which b = 1 3 log f S f = S ut N (log f ) / 3 1 N 10 3 For 500 cycles as in Prob. 6-3 S f 66 . 2(500) (log 0 . 8949) / 3 = 59 . 9 kpsi Ans. 6-5 Read from graph: (10 3 , 90) and (10 6 , 50). From S = aN b log S 1 = log a + b log N 1 log S 2 = log a + b log N 2 From which log a = log S 1 log N 2 log S 2 log N 1 log N 2 / N 1 = log 90 log 10 6 log 50 log 10 3 log 10 6 / 10 3 = 2 . 2095 a = 10 log a = 10 2 . 2095 =
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Unformatted text preview: 162 . b = log 50 / 90 3 = . 085 09 ( S f ) ax = 162 . 085 09 10 3 N 10 6 in kpsi Ans. Check: 10 3 ( S f ) ax = 162(10 3 ) . 085 09 = 90 kpsi 10 6 ( S f ) ax = 162(10 6 ) . 085 09 = 50 kpsi The end points agree. 6-6 Eq. (6-8): S e = . 5(710) = 355 MPa Table 6-2: a = 4 . 51, b = . 265 Eq. (6-19): k a = 4 . 51(710) . 265 = . 792...
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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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