Budynas ch06

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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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 . 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 = 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 ² 0 . 107 = 1 . 02 Eq. (6-18): S e = k a k b S ± e = 0 . 840(1 . 02)(100) = 85 . 7 kpsi Ans . 6-2 (a) S = 68 kpsi, S ± e = 0 . 5(68) = 34 kpsi . (b) S = 112 kpsi, S ± e = 0 . 5(112) = 56 kpsi . (c) 2024T3 has no endurance limit Ans. (d) Eq. (6-8): S ± e = 100 kpsi . 6-3 Eq. (2-11): σ ± F = σ 0 ε m = 115(0 . 90) 0 . 22 = 112 . 4 kpsi Eq. (6-8): S ± e = 0 . 5(66 . 2) = 33 . 1 kpsi Eq. (6-12): b log(112 . 4 / 33 . 1) log(2 · 10 6 ) 0 . 084 26 Eq. (6-10): f = 112 . 4 66 . 2 (2 · 10 3 ) 0 . 084 26 = 0 . 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) 0 . 084 26 = 47 . 9 kpsi Ans . Eq. (6-16): N = ³ σ a a ´ 1 / b = ± 36 106 . 0 ² 1 / 0 . 084 26 = 368 250 cycles Ans .
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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 = log a + b log (1) From which a = S Substituting (10 3 , fS ) and a = S log = log S + b log 10 3 From which b = 1 3 log f S f = S 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 = 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 = 162 . 0 b = log 50 / 90 3 =− 0 . 085 09 ( S f ) ax = 162 0 . 085 09 10 3 N 10 6 in kpsi Ans. Check: 10 3 ( S f ) = 162(10 3 ) 0 . 085 09 = 90 kpsi 10 6 ( S f ) = 162(10 6 ) 0 . 085 09 = 50 kpsi The end points agree. 6-6 Eq. (6-8): S ± e = 0 . 5(710) = 355 MPa Table 6-2: a = 4 . 51, b 0 . 265 Eq. (6-19): k a = 4 . 51(710) 0 . 265 = 0 . 792
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Chapter 6 149 Eq. (6-20): k b = ± d 7 . 62 ² 0 . 107 = ± 32 7 . 62 ² 0 . 107 = 0 . 858 Eq. (6-18): S e = k a k b S ± e = 0 . 792(0 . 858)(355) = 241 MPa Ans . 6-7 For AISI 4340 as forged steel, Eq. (6-8): S e = 100 kpsi Table 6-2: a = 39 . 9, b =− 0 . 995 Eq. (6-19): k a = 39 . 9(260) 0 . 995 = 0 . 158 Eq. (6-20): k b = ± 0 . 75 0 . 30 ² 0 . 107 = 0 . 907 Each of the other Marin factors is unity. S e = 0 . 158(0 . 907)(100) = 14 . 3 kpsi For AISI 1040: S ± e = 0 . 5(113) = 56 . 5 kpsi k a = 39 . 9(113) 0 . 995 = 0 . 362 k b = 0 . 907 (same as 4340) Each of the other Marin factors is unity. S e = 0 . 362(0 . 907)(56 . 5) = 18 . 6 kpsi Not only is AISI 1040 steel a contender, it has a superior endurance strength. Can you see why? 6-8 (a) For an AISI 1018 CD-machined steel, the strengths are Eq. (2-17): S ut = 440 MPa H B = 440 3 . 41 = 129 S y = 370 MPa S su = 0 . 67(440) = 295 MPa Fig. A-15-15: r d = 2 . 5 20 = 0 . 125, D d = 25 20 = 1 . 25, K ts = 1 . 4 Fig. 6-21: q s = 0 . 94 Eq. (6-32): K fs = 1 + 0 . 94(1 . 4 1) = 1 . 376 For a purely reversing torque of 200 N · m τ max = K 16 T π d 3 = 1 . 376(16)(200 × 10 3 N · mm) π (20 mm) 3 τ max = 175 .
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Budynas ch06 - Chapter 6 Note to the instructor Many of the...

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