shi20396_ch07 - Chapter 7 7-1 H B = 490 Eq. (3-17): Eq....

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 7 7-1 H B = 490 Eq. (3-17): S ut = 0 . 495(490) = 242 . 6 kpsi > 212 kpsi Eq. (7-8): S 0 e = 107 kpsi Table 7-4: a = 1 . 34, b =− 0 . 085 Eq. (7-18): k a = 1 . 34(242 . 6) 0 . 085 = 0 . 840 Eq. (7-19): k b = µ 3 / 16 0 . 3 0 . 107 = 1 . 05 Eq. (7-17): S e = k a k b S 0 e = 0 . 840(1 . 05)(107) = 94 . 4 kpsi Ans . 7-2 (a) S = 68 kpsi, S 0 e = 0 . 495(68) = 33 . 7 kpsi . (b) S = 112 kpsi, S 0 e = 0 . 495(112) = 55 . 4 kpsi . (c) 2024T3 has no endurance limit Ans. (d) Eq. (3-17): S 0 e = 107 kpsi . 7-3 σ 0 F = σ 0 ε m = 115(0 . 90) 0 . 22 = 112 . 4 kpsi Eq. (7-8): S 0 e = 0 . 504(66 . 2) = 33 . 4 kpsi Eq. (7-11): b log(112 . 4 / 33 . 4) log(2 · 10 6 ) 0 . 083 64 Eq. (7-9): f = 112 . 4 66 . 2 (2 · 10 3 ) 0 . 083 64 = 0 . 8991 Eq. (7-13): a = [0 . 8991(66 . 2)] 2 33 . 4 = 106 . 1 kpsi Eq. (7-12): S f = aN b = 106 . 1(12 500) 0 . 083 64 = 48 . 2 kpsi Ans . Eq. (7-15): N = ³ σ a a ´ 1 / b = µ 36 106 . 1 1 / 0 . 083 64 = 409 530 cycles Ans . 7-4 From S f = b log S f = log a + b log N Substituting (1, S ) log S = log a + b log (1) From which a = S
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 7 181 Substituting (10 3 , fS ut ) 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. 7-3 500 S f 66 . 2(500) (log 0 . 8991) / 3 = 60 . 2 kpsi Ans. 7-5 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 = 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. 7-6 Eq. (7-8): S 0 e = 0 . 504(710) = 357 . 8MPa Table 7-4: a = 4 . 51, b 0 . 265 Eq. (7-18): k a = 4 . 51(710) 0 . 265 = 0 . 792 Eq. (7-19): k b = µ d 7 . 62 0 . 107 = µ 32 7 . 62 0 . 107 = 0 . 858 Eq. (7-17): S e = k a k b S 0 e = 0 . 792(0 . 858)(357 . 8) = 243 MPa Ans .
Background image of page 2
182 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 7-7 For AISI 4340 as forged steel, Eq. (7-8): S e = 107 kpsi Table 7-4: a = 39 . 9, b =− 0 . 995 Eq. (7-18): k a = 39 . 9(260) 0 . 995 = 0 . 158 Eq. (7-19): k b = µ 0 . 75 0 . 30 0 . 107 = 0 . 907 Each of the other Marin factors is unity. S e = 0 . 158(0 . 907)(107) = 15 . 3 kpsi For AISI 1040: S 0 e = 0 . 504(113) = 57 . 0 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)(57 . 2) = 18 . 7 kpsi Not only is AISI 1040 steel a contender, it has a superior endurance strength. Can you see why? 7-8 (a) For an AISI 1018 CD-machined steel, the strengths are Eq. (3-17): 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. 7-21: q s = 0 . 94 Eq. (7-31): 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 . 2MPa = τ a S 0 e = 0 . 504(440) = 222 MPa The Marin factors are k a = 4 . 51(440) 0 . 265 = 0 . 899 k b = µ 20 7 . 62 0 . 107 = 0 . 902 k c = 0 . 59, k d = 1, k e = 1 Eq. (7-17): S e = 0 . 899(0 . 902)(0 . 59)(222) = 106 . 2.5 mm 20 mm 25 mm
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 7 183 Eq. (7-13): a = [0 . 9(295)] 2 106 . 2 = 664 Eq. (7-14): b =− 1 3 log 0 . 9(295) 106 . 2 0 . 132 65 Eq. (7-15): N = µ 175 . 2 664 1 / 0 . 132 65 N = 23 000 cycles Ans. (b) For an operating temperature of 450°C, the temperature modi±cation factor, from Table 7-6, is k d = 0 . 843 Thus S e = 0 . 899(0 . 902)(0 . 59)(0 . 843)(222) = 89 . 5MPa a = [0 . 9(295)] 2 89 . 5 = 788 b 1 3 log 0 . 9(295) 89 . 5 0 . 157 41 N = µ 175 . 2 788 1 / 0 . 157 41 N = 14 100 cycles Ans. 7-9 f = 0 . 9 n = 1 . 5 N = 10 4 cycles For AISI 1045 HR steel, S ut = 570 MPa and S y = 310 MPa S 0 e = 0 . 504(570 MPa) = 287 .
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 31

shi20396_ch07 - Chapter 7 7-1 H B = 490 Eq. (3-17): Eq....

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online