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1_ch 06 Mechanical Design budynas_SM_ch06

# 1_ch 06 Mechanical Design budynas_SM_ch06 - a k b S ± e =...

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Chapter 6 Note to the instructor: Many of the problems in this chapter are carried over from the previous edition. The solutions have changed slightly due to some minor changes. First, the calculation of the endurance limit of a rotating-beam specimen S e is given by S e = 0 . 5 S ut instead of S e = 0 . 504 S ut . Second, when the fatigue stress calculation is made for deterministic problems, only one approach is given, which uses the notch sensitivity factor, q , together with Eq. (6-32). Neuber’s equation, Eq. (6-33), is simply another form of this. These changes were made to hope- fully make the calculations less confusing, and diminish the idea that stress life calculations are precise. 6-1 H B = 490 Eq. (2-17): S ut = 0 . 495(490) = 242 . 6 kpsi > 212 kpsi Eq. (6-8): S e = 100 kpsi Table 6-2: a = 1 . 34, b = − 0 . 085 Eq. (6-19): k a = 1 . 34(242 . 6) 0 . 085 = 0 . 840 Eq. (6-20): k b = 1 / 4 0 . 3 0 . 107 = 1 . 02 Eq. (6-18):
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Unformatted text preview: a k b S ± e = . 840(1 . 02)(100) = 85 . 7 kpsi Ans . 6-2 (a) S ut = 68 kpsi, S ± e = . 5(68) = 34 kpsi Ans . (b) S ut = 112 kpsi, S ± e = . 5(112) = 56 kpsi Ans . (c) 2024T3 has no endurance limit Ans. (d) Eq. (6-8): S ± e = 100 kpsi Ans . 6-3 Eq. (2-11): σ ± F = σ ε m = 115(0 . 90) . 22 = 112 . 4 kpsi Eq. (6-8): S ± e = . 5(66 . 2) = 33 . 1 kpsi Eq. (6-12): b = − log(112 . 4 / 33 . 1) log(2 · 10 6 ) = − . 084 26 Eq. (6-10): f = 112 . 4 66 . 2 (2 · 10 3 ) − . 084 26 = . 8949 Eq. (6-14): a = [0 . 8949(66 . 2)] 2 33 . 1 = 106 . 0 kpsi Eq. (6-13): S f = aN b = 106 . 0(12 500) − . 084 26 = 47 . 9 kpsi Ans . Eq. (6-16): N = ³ σ a a ´ 1 / b = ± 36 106 . ² − 1 / . 084 26 = 368 250 cycles Ans ....
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