Tài liệu shi20396 chương 7 pdf

Chapter 7 7-1 H B = 490 Eq. (3-17): S ut = 0.495(490) = 242.6 kpsi > 212 kpsi Eq. (7-8): S  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 =  3/16 0.3  −0.107 = 1.05 Eq. (7-17): S e = k a k b S  e = 0.840(1.05)(107) = 94.4 kpsi Ans. 7-2 (a) S ut = 68 kpsi, S  e = 0.495(68) = 33.7 kpsi Ans. (b) S ut = 112 kpsi, S  e = 0.495(112) = 55.4 kpsi Ans. (c) 2024T3 has no endurance limit Ans. (d) Eq. (3-17): S  e = 107 kpsi Ans. 7-3 σ  F = σ 0 ε m = 115(0.90) 0.22 = 112.4 kpsi Eq. (7-8): S  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 =  σ a a  1/b =  36 106.1  −1/0.083 64 = 409 530 cycles Ans. 7-4 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 shi20396_ch07.qxd 8/18/03 12:35 PM Page 180 Chapter 7 181 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. 7-3 500S 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 ) ax = 162(10 3 ) −0.085 09 = 90 kpsi 10 6 (S f ) ax = 162(10 6 ) −0.085 09 = 50 kpsi The end points agree. 7-6 Eq. (7-8): S  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 =  d 7.62  −0.107 =  32 7.62  −0.107 = 0.858 Eq. (7-17): S e = k a k b S  e = 0.792(0.858)(357.8) = 243 MPa Ans. shi20396_ch07.qxd 8/18/03 12:35 PM Page 181 182 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 =  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  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 fs 16T πd 3 = 1.376(16)(200 ×10 3 N · mm) π(20 mm) 3 τ max = 175.2MPa= τ a S  e = 0.504(440) = 222 MPa The Marin factors are k a = 4.51(440) −0.265 = 0.899 k b =  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) = 106.2MPa 2.5 mm 20 mm 25 mm shi20396_ch07.qxd 8/18/03 12:35 PM Page 182 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 =  175.2 664  1/−0.132 65 N = 23 000 cycles Ans. (b) For an operating temperature of 450°C, the temperature modification 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 =  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  e = 0.504(570 MPa) = 287.3MPa Find an initial guess based on yielding: σ a = σ max = Mc I = M(b/2) b(b 3 )/12 = 6M b 3 M max = (1 kN)(800 mm) = 800 N · m σ max = S y n ⇒ 6(800 × 10 3 N ·mm) b 3 = 310 N/mm 2 1.5 b = 28.5mm Eq. (7-24): d e = 0.808b Eq. (7-19): k b =  0.808b 7.62  −0.107 = 1.2714b −0.107 k b = 0.888 F ϭ Ϯ1 kN b b 800 mm shi20396_ch07.qxd 8/18/03 12:35 PM Page 183 184 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The remaining Marin factors are k a = 57.7(570) −0.718 = 0.606 k c = k d = k e = k f = 1 Eq. (7-17): S e = 0.606(0.888)(287.3MPa)= 154.6MPa Eq. (7-13): a = [0.9(570)] 2 154.6 = 1702 Eq. (7-14): b =− 1 3 log 0.9(570) 154.6 =−0.173 64 Eq. (7-12): S f = aN b = 1702[(10 4 ) −0.173 64 ] = 343.9MPa n = S f σ a or σ a = S f n 6(800 × 10 3 ) b 3 = 343.9 1.5 ⇒ b = 27.6mm Check values for k b , S e , etc. k b = 1.2714(27.6) −0.107 = 0.891 S e = 0.606(0.891)(287.3) = 155.1MPa a = [0.9(570)] 2 155.1 = 1697 b =− 1 3 log 0.9(570) 155.1 =−0.173 17 S f = 1697[(10 4 ) −0.173 17 ] = 344.4MPa 6(800 × 10 3 ) b 3 = 344.4 1.5 b = 27.5mm Ans. 7-10 Table A-20: S ut = 440 MPa, S y = 370 MPa S  e = 0.504(440) = 221.8MPa Table 7-4: k a = 4.51(440) −0.265 = 0.899 k b = 1(axial loading) Eq. (7-25): k c = 0.85 S e = 0.899(1)(0.85)(221.8) = 169.5MPa Table A-15-1: d/w = 12/60 = 0.2, K t = 2.5 12 F a F a 10 60 1018 shi20396_ch07.qxd 8/18/03 12:35 PM Page 184 Chapter 7 185 From Eq. (7-35) and Table 7-8 K f = K t 1 +  2/ √ r  [(K t − 1)/K t ] √ a = 2.5 1 +  2/ √ 6  [(2.5 − 1)/2.5](174/440) = 2.09 σ a = K f F a A ⇒ S e n f = 2.09F a 10(60 − 12) = 169.5 1.8 F a = 21 630 N = 21.6kN Ans. F a A = S y n y ⇒ F a 10(60 − 12) = 370 1.8 F a = 98 667 N = 98.7kN Ans. Largest force amplitude is 21.6 kN. Ans. 7-11 A priori design decisions: The design decision will be: d Material and condition: 1095 HR and from Table A-20 S ut = 120, S y = 66 kpsi . Design factor: n f = 1.6 per problem statement. Life: (1150)(3) = 3450 cycles Function: carry 10 000 lbf load Preliminaries to iterative solution: S  e = 0.504(120) = 60.5 kpsi k a = 2.70(120) −0.265 = 0.759 I c = πd 3 32 = 0.098 17d 3 M(crit.) =  6 24  (10 000)(12) = 30 000 lbf · in The critical location is in the middle of the shaft at the shoulder. From Fig. A-15-9: D/d = 1.5, r/d = 0.10, and K t = 1.68. With no direct information concerning f, use f = 0.9 . For an initial trial, set d = 2.00 in k b =  2.00 0.30  −0.107 = 0.816 S e = 0.759(0.816)(60.5) = 37.5 kpsi a = [0.9(120)] 2 37.5 = 311.0 b =− 1 3 log 0.9(120) 37.5 =−0.1531 S f = 311.0(3450) −0.1531 = 89.3 shi20396_ch07.qxd 8/18/03 12:35 PM Page 185 186 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design σ 0 = M I /c = 30 0.098 17d 3 = 305.6 d 3 = 305.6 2 3 = 38.2 kpsi r = d 10 = 2 10 = 0.2 K f = 1.68 1 +  2/ √ 0.2  [(1.68 − 1)/1.68](4/120) = 1.584 Eq. (7-37): (K f ) 10 3 = 1 − (1.584 − 1)[0.18 −0.43(10 −2 )120 + 0.45(10 −5 )120 2 ] = 1.158 Eq. (7-38): (K f ) N = K 3450 = 1.158 2 1.584 (3450) −(1/3) log(1.158/1.584) = 1.225 σ 0 = 305.6 2 3 = 38.2 kpsi σ a = (K f ) N σ 0 = 1.225(38.2) = 46.8 kpsi n f = (S f ) 3450 σ a = 89.3 46.8 = 1.91 The design is satisfactory. Reducing the diameter will reduce n, but the resulting preferred size will be d = 2.00 in. 7-12 σ  a = 172 MPa, σ  m = √ 3τ m = √ 3(103) = 178.4MPa Yield: 172 + 178.4 = S y n y = 413 n y ⇒ n y = 1.18 Ans. (a) Modified Goodman, Table 7-9 n f = 1 (172/276) + (178.4/551) = 1.06 Ans. (b) Gerber, Table 7-10 n f = 1 2  551 178.4  2  172 276     −1 +  1 +  2(178.4)(276) 551(172)  2    = 1.31 Ans. (c) ASME-Elliptic, Table 7-11 n f =  1 (172/276) 2 + (178.4/413) 2  1/2 = 1.32 Ans. shi20396_ch07.qxd 8/18/03 12:35 PM Page 186 Chapter 7 187 7-13 σ  a = 69 MPa, σ  m = √ 3(138) = 239 MPa Yield: 69 + 239 = 413 n y ⇒ n y = 1.34 Ans. (a) Modified Goodman, Table 7-9 n f = 1 (69/276) + (239/551) = 1.46 Ans. (b) Gerber, Table 7-10 n f = 1 2  551 239  2  69 276     −1 +  1 +  2(239)(276) 551(69)  2    = 1.73 Ans. (c) ASME-Elliptic, Table 7-11 n f =  1 (69/276) 2 + (239/413) 2  1/2 = 1.59 Ans. 7-14 σ  a =  σ 2 a + 3τ 2 a =  83 2 + 3(69 2 ) = 145.5MPa, σ  m = √ 3(103) = 178.4MPa Yield: 145.5 + 178.4 = 413 n y ⇒ n y = 1.28 Ans. (a) Modified Goodman, Table 7-9 n f = 1 (145.5/276) + (178.4/551) = 1.18 Ans. (b) Gerber, Table 7-10 n f = 1 2  551 178.4  2  145.5 276     −1 +  1 +  2(178.4)(276) 551(145.5)  2    = 1.47 Ans. (c) ASME-Elliptic, Table 7-11 n f =  1 (145.5/276) 2 + (178.4/413) 2  1/2 = 1.47 Ans. 7-15 σ  a = √ 3(207) = 358.5MPa, σ  m = 0 Yield: 358.5 = 413 n y ⇒ n y = 1.15 Ans. shi20396_ch07.qxd 8/18/03 12:35 PM Page 187 188 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (a) Modified Goodman, Table 7-9 n f = 1 (358.5/276) = 0.77 Ans. (b) Gerber criterion of Table 7-10 does not work; therefore use Eq. (7-50). n f σ a S e = 1 ⇒ n f = S e σ a = 276 358.5 = 0.77 Ans. (c) ASME-Elliptic, Table 7-11 n f =   1 358.5/276  2 = 0.77 Ans. Let f = 0.9 to assess the cycles to failure by fatigue Eq. (7-13): a = [0.9(551)] 2 276 = 891.0MPa Eq. (7-14): b =− 1 3 log 0.9(551) 276 =−0.084 828 Eq. (7-15): N =  358.5 891.0  −1/0.084 828 = 45 800 cycles Ans. 7-16 σ  a = √ 3(103) = 178.4MPa, σ  m = 103 MPa Yield: 178.4 + 103 = 413 n y ⇒ n y = 1.47 Ans. (a) Modified Goodman, Table 7-9 n f = 1 (178.4/276) + (103/551) = 1.20 Ans. (b) Gerber, Table 7-10 n f = 1 2  551 103  2  178.4 276     −1 +  1 +  2(103)(276) 551(178.4)  2    = 1.44 Ans. (c) ASME-Elliptic, Table 7-11 n f =  1 (178.4/276) 2 + (103/413) 2  1/2 = 1.44 Ans. 7-17 Table A-20: S ut = 64 kpsi, S y = 54 kpsi A = 0.375(1 − 0.25) = 0.2813 in 2 σ max = F max A = 3000 0.2813 (10 −3 ) = 10.67 kpsi shi20396_ch07.qxd 8/18/03 12:35 PM Page 188 Chapter 7 189 n y = 54 10.67 = 5.06 Ans. S  e = 0.504(64) = 32.3 kpsi k a = 2.70(64) −0.265 = 0.897 k b = 1, k c = 0.85 S e = 0.897(1)(0.85)(32.3) = 24.6 kpsi Table A-15-1: w = 1in,d = 1/4in , d/w = 0.25 І K t = 2.45. From Eq. (7-35) and Table 7-8 K f = 2.45 1 +  2/ √ 0.125  [(2.45 − 1)/2.45](5/64) = 1.94 σ a = K f     F max − F min 2A     = 1.94     3.000 − 0.800 2(0.2813)     = 7.59 kpsi σ m = K f F max + F min 2A = 1.94  3.000 + 0.800 2(0.2813)  = 13.1 kpsi r = σ a σ m = 7.59 13.1 = 0.579 (a) DE-Gerber, Table 7-10 S a = 0.579 2 (64 2 ) 2(24.6)   −1 +  1 +  2(24.6) 0.579(64)  2   = 18.5 kpsi S m = S a r = 18.5 0.579 = 32.0 kpsi n f = 1 2  64 13.1  2  7.59 24.6    −1 +  1 +  2(13.1)(24.6) 7.59(64)  2   = 2.44 Ans. (b) DE-Elliptic, Table 7-11 S a =  (0.579 2 )(24.6 2 )(54 2 ) 24.6 2 + (0.579 2 )(54 2 ) = 19.33 kpsi S m = S a r = 19.33 0.579 = 33.40 kpsi shi20396_ch07.qxd 8/18/03 12:35 PM Page 189 [...]... Hole: Fig A-15-1: d/w = 0 .75 /3 .75 = 0.20, K t = 2.5 Kf = 2.5 = 2. 17 √ 1 + 2/ 0 .75 /2 [(2.5 − 1)/2.5](5/64) σmax = 4 = 2. 67 kpsi 0.5(3 .75 − 0 .75 ) σmin = −16 = −10. 67 kpsi 0.5(3 .75 − 0 .75 ) σa = 2. 17 2. 67 − (−10. 67) = 14. 47 kpsi 2 σm = 2. 17 2. 67 + (−10. 67) = −8.68 kpsi 2 Since the midrange stress is negative, ny = Sy 54 = = 5.06 σmin −10. 67 Sa = Se = 24.6 kpsi Sa 24.6 = 1 .70 = σa 14. 47 Thus the design is controlled... 0. 67( 320) = 214.4 MPa Ssy = 0. 577 S y = 0. 577 (180) = 103.9 MPa Se = 0.504(320) = 161.3 MPa ka = 57. 7(320) −0 .71 8 = 0.9 17 de = 0. 370 (20) = 7. 4 mm kb = 7. 4 7. 62 −0.1 07 = 1.003 kc = 0.59 Se = 0.9 17( 1.003)(0.59)(161.3) = 87. 5 MPa shi20396_ ch 07. qxd 198 8/18/03 12:36 PM Page 198 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Modified Goodman, Table 7- 9, nf = 1 1 = = 1.36... = 1 0.30 −0.1 07 = 0. 879 Conservatism is not necessary Se = 0 .76 8[LN(1, 0.058)](0. 879 )(55 .7) [LN(1, 0.138)] ¯ Se = 37. 6 kpsi C Se = (0.0582 + 0.1382 ) 1/2 = 0.150 Se = 37. 6LN(1, 0.150) shi20396_ ch 07. qxd 8/18/03 12:36 PM Page 205 205 Chapter 7 Fig A-15-14: D/d = 1.25, r/d = 0.125 Thus K t = 1 .70 and Eqs (7- 35), (7- 78) and Table 7- 8 give 1 .70 LN(1, 0.15) Kf = √ 1 + 2/ 0.125 [(1 .70 − 1)/(1 .70 )](3/110) =... 0.11) Table 7- 13: = 0 .78 2LN(1, 0.11) Eq (7- 23): de = 0. 37( 1.25) = 0.463 in kb = 0.463 0.30 −0.1 07 = 0.955 Se = 0 .78 2[LN(1, 0.11)](0.955)[29.3LN(1, 0.138)] ¯ Se = 0 .78 2(0.955)(29.3) = 21.9 kpsi C Se = (0.112 + 0.1382 ) 1/2 = 0.150 Table A-16: d/D = 0, a/D = 0.1, A = 0.83 ∴ K t = 2. 27 From Eqs (7- 35) and (7- 78) and Table 7- 8 2.27LN(1, 0.10) = 1 .78 3LN(1, 0.10) Kf = √ 1 + 2/ 0.125 [(2. 27 − 1)/2. 27] (5/58)... 7- 9 σm 1 σa + = Se Sut nf ⇒ 22.14T 1 22.14T + = 321.8 77 0 3 T = 3.42 N · m Ans (b) Gerber, Eq (7- 50) nσa nσm + Se Sut 2 3(22.14)T 3(22.14)T + 321.8 77 0 2 =1 =1 T 2 + 27. 74T − 134.40 = 0 T = 1 − 27. 74 + 2 27. 742 + 4(134.40) = 4.21 N · m Ans (c) To guard against yield, use T of part (b) and the inner stress 420 = 1.91 Ans ny = 52.34(4.21) 7- 25 From Prob 7- 24, Se = 321.8 MPa, S y = 420 MPa, and Sut = 77 0... −6 )(0.1046)(10−3 )( 17. 5)(10−3 ) Aeri 25(10 −Mco T (2.605)(10−3 ) = 44. 27( 106 )T = σo = −6 )(0.1046)(10−3 )(22.5)(10−3 ) Aero 25(10 For fatigue, σo is most severe as it represents a tensile stress 1 σm = σa = (44. 27) (106 )T = 22.14(106 )T 2 Se = 0.504Sut = 0.504 (77 0) = 388.1 MPa ka = 4.51 (77 0) −0.265 = 0 .77 5 de = 0.808[5(5)]1/2 = 4.04 mm kb = 4.04 7. 62 −0.1 07 = 1. 070 Se = 0 .77 5(1. 07) (388.1) = 321.8 MPa... Sut = 77 0 MPa (a) Assuming the beam is straight, 6M 6T = 48(106 )T σmax = 2 = 3 bh 5 [(10−3 ) 3 ] Goodman: 24T 1 24T + = 321.8 77 0 3 ⇒ T = 3.15 N · m Ans shi20396_ ch 07. qxd 8/18/03 12:36 PM Page 1 97 1 97 Chapter 7 3(24)T 3(24)T + 321.8 77 0 (b) Gerber: 2 =1 T 2 + 25.59T − 114. 37 = 1 T = 1 −25.59 + 2 25.592 + 4(114. 37) = 3.88 N · m Ans (c) Using σmax = 52.34(106 )T from Prob 7- 24, ny = 420 = 2. 07 Ans 52.34(3.88)... 0.9(1 671 ) = 1504 MPa Then from Eq (7- 8), Se = 74 0 MPa; Eq (7- 18), ka = 1.58(1 671 ) −0.085 = 0.841; Eq (7- 24) de = 0.808[18(3)]1/2 = 5.938 mm; and Eq (7- 19), kb = (5.938 /7. 62) −0.1 07 = 1.0 27 shi20396_ ch 07. qxd 8/18/03 12:35 PM Page 193 193 Chapter 7 Se = 0.841(1.0 27) (74 0) = 639 MPa (σi ) a = 73 3.2 + 244.4 = 244.4 MPa 2 (σi ) m = At Inner Radius 73 3.2 − 244.4 = −488.8 MPa 2 ␴a 1504 MPa 639 244.4 Ϫ1504... σm 7. 60 (a) Table 7- 10, DE-Gerber nf = 1 2 64 7. 59 2  13.1  −1 + 24.6 1+ 2 (7. 59)(24.6) 64(13.1) 2   = 1 .79 Ans (b) Table 7- 11, DE-Elliptic nf = 7- 19 1 = 1.82 Ans (13.1/24.6) 2 + (7. 59/54) 2 Referring to the solution of Prob 7- 17, for load fluctuations of 800 to −3000 lbf σa = 1.94 0.800 − (−3.000) = 13.1 kpsi 2(0.2813) σm = 1.94 0.800 + (−3.000) = 7. 59 kpsi 2(0.2813) r= σa 13.1 = −1 .72 6 = σm 7. 59... 1.6 = 1 .78 3LN(1, 0.10) σ = Kf Z 0.159 = 17. 95LN(1, 0.10) kpsi shi20396_ ch 07. qxd 8/18/03 12:36 PM Page 2 07 2 07 Chapter 7 σ = 17. 95 kpsi ¯ Cσ = 0.10 z=− Eq (6- 57) : ln (21.9/ 17. 95) (1 + 0.102 )/(1 + 0.152 ) ln[(1 + 0.152 )(1 + 0.102 )] = −1. 07 p f = 0.1423 Table A-10: R = 1 − p f = 1 − 0.1423 = 0.858 Ans For a completely-reversed design load Ma of 1400 lbf · in, the reliability estimate is 0.858 7- 37 For . 0.67S ut = 0. 67( 320) = 214.4MPa S sy = 0. 577 S y = 0. 577 (180) = 103.9MPa S  e = 0.504(320) = 161.3MPa k a = 57. 7(320) −0 .71 8 = 0.9 17 d e = 0. 370 (20) = 7. 4mm k b =  7. 4 7. 62  −0.1 07 =. 57. 7( 570 ) −0 .71 8 = 0.606 k c = k d = k e = k f = 1 Eq. (7- 17) : S e = 0.606(0.888)(2 87. 3MPa)= 154.6MPa Eq. (7- 13): a = [0.9( 570 )] 2 154.6 = 170 2 Eq. (7- 14): b