Baixe cap 7 shigley e outras Notas de estudo em PDF para Engenharia Mecânica, somente na Docsity! Chapter 7 7-1 HB = 490 Eq. (3-17): Sut = 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): ka = 1.34(242.6)−0.085 = 0.840 Eq. (7-19): kb = ( 3/16 0.3 )−0.107 = 1.05 Eq. (7-17): Se = kakbS′e = 0.840(1.05)(107) = 94.4 kpsi Ans. 7-2 (a) Sut = 68 kpsi, S′e = 0.495(68) = 33.7 kpsi Ans. (b) Sut = 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 = σ0εm = 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 · 106) = −0.083 64 Eq. (7-9): f = 112.4 66.2 (2 · 103)−0.083 64 = 0.8991 Eq. (7-13): a = [0.8991(66.2)] 2 33.4 = 106.1 kpsi Eq. (7-12): Sf = aN b = 106.1(12 500)−0.083 64 = 48.2 kpsi Ans. Eq. (7-15): N = (σa a )1/b = ( 36 106.1 )−1/0.083 64 = 409 530 cycles Ans. 7-4 From Sf = aN b log Sf = log a + b log N Substituting (1, Sut ) log Sut = log a + b log (1) From which a = Sut shi20396_ch07.qxd 8/18/03 12:35 PM Page 180 Chapter 7 181 Substituting (103, f Sut ) and a = Sut log f Sut = log Sut + b log 103 From which b = 1 3 log f ∴ Sf = Sut N (log f )/3 1 ≤ N ≤ 103 For 500 cycles as in Prob. 7-3 500Sf ≥ 66.2(500)(log 0.8991)/3 = 60.2 kpsi Ans. 7-5 Read from graph: (103, 90) and (106, 50). From S = aN b log S1 = log a + b log N1 log S2 = log a + b log N2 From which log a = log S1 log N2 − log S2 log N1 log N2/N1 = log 90 log 10 6 − log 50 log 103 log 106/103 = 2.2095 a = 10log a = 102.2095 = 162.0 b = log 50/90 3 = −0.085 09 (Sf )ax = 162−0.085 09 103 ≤ N ≤ 106 in kpsi Ans. Check: 103(Sf )ax = 162(103)−0.085 09 = 90 kpsi 106(Sf )ax = 162(106)−0.085 09 = 50 kpsi The end points agree. 7-6 Eq. (7-8): S′e = 0.504(710) = 357.8 MPa Table 7-4: a = 4.51, b = −0.265 Eq. (7-18): ka = 4.51(710)−0.265 = 0.792 Eq. (7-19): kb = ( d 7.62 )−0.107 = ( 32 7.62 )−0.107 = 0.858 Eq. (7-17): Se = kakbS′e = 0.792(0.858)(357.8) = 243 MPa Ans. shi20396_ch07.qxd 8/18/03 12:35 PM Page 181 184 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The remaining Marin factors are ka = 57.7(570)−0.718 = 0.606 kc = kd = ke = k f = 1 Eq. (7-17): Se = 0.606(0.888)(287.3 MPa) = 154.6 MPa 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): Sf = aN b = 1702[(104)−0.173 64] = 343.9 MPa n = Sf σa or σa = Sf n 6(800 × 103) b3 = 343.9 1.5 ⇒ b = 27.6 mm Check values for kb, Se, etc. kb = 1.2714(27.6)−0.107 = 0.891 Se = 0.606(0.891)(287.3) = 155.1 MPa a = [0.9(570)] 2 155.1 = 1697 b = −1 3 log 0.9(570) 155.1 = −0.173 17 Sf = 1697[(104)−0.173 17] = 344.4 MPa 6(800 × 103) b3 = 344.4 1.5 b = 27.5 mm Ans. 7-10 Table A-20: Sut = 440 MPa, Sy = 370 MPa S′e = 0.504(440) = 221.8 MPa Table 7-4: ka = 4.51(440)−0.265 = 0.899 kb = 1 (axial loading) Eq. (7-25): kc = 0.85 Se = 0.899(1)(0.85)(221.8) = 169.5 MPa Table A-15-1: d/w = 12/60 = 0.2, Kt = 2.5 12Fa Fa 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 = Kt 1 + (2/√r) [(Kt − 1)/Kt ]√a = 2.5 1 + (2/√6) [(2.5 − 1)/2.5](174/440) = 2.09 σa = K f Fa A ⇒ Se n f = 2.09Fa 10(60 − 12) = 169.5 1.8 Fa = 21 630 N = 21.6 kN Ans. Fa A = Sy ny ⇒ Fa 10(60 − 12) = 370 1.8 Fa = 98 667 N = 98.7 kN 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 Sut = 120, Sy = 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 ka = 2.70(120)−0.265 = 0.759 I c = πd 3 32 = 0.098 17d3 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 Kt = 1.68. With no direct information concerning f, use f = 0.9. For an initial trial, set d = 2.00 in kb = ( 2.00 0.30 )−0.107 = 0.816 Se = 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 Sf = 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 σ0 = M I/c = 30 0.098 17d3 = 305.6 d3 = 305.6 23 = 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 )103 = 1 − (1.584 − 1)[0.18 − 0.43(10−2)120 + 0.45(10−5)1202] = 1.158 Eq. (7-38): (K f )N = K3450 = 1.158 2 1.584 (3450)−(1/3) log(1.158/1.584) = 1.225 σ0 = 305.623 = 38.2 kpsi σa = (K f )Nσ0 = 1.225(38.2) = 46.8 kpsi n f = (Sf )3450 σa = 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 = √ 3τm = √ 3(103) = 178.4 MPa Yield: 172 + 178.4 = Sy ny = 413 ny ⇒ ny = 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 = 12 ( 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 7 189 ny = 5410.67 = 5.06 Ans. S′e = 0.504(64) = 32.3 kpsi ka = 2.70(64)−0.265 = 0.897 kb = 1, kc = 0.85 Se = 0.897(1)(0.85)(32.3) = 24.6 kpsi Table A-15-1: w = 1 in, d = 1/4 in, d/w = 0.25 Kt = 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 σa = K f ∣∣∣∣ Fmax − Fmin2A ∣∣∣∣ = 1.94 ∣∣∣∣3.000 − 0.8002(0.2813) ∣∣∣∣ = 7.59 kpsi σm = K f Fmax + Fmin2A = 1.94 [ 3.000 + 0.800 2(0.2813) ] = 13.1 kpsi r = σa σm = 7.59 13.1 = 0.579 (a) DE-Gerber, Table 7-10 Sa = 0.579 2(642) 2(24.6)  −1 + √ 1 + ( 2(24.6) 0.579(64) )2  = 18.5 kpsi Sm = Sa r = 18.5 0.579 = 32.0 kpsi n f = 12 ( 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 Sa = √ (0.5792)(24.62)(542) 24.62 + (0.5792)(542) = 19.33 kpsi Sm = Sa r = 19.33 0.579 = 33.40 kpsi shi20396_ch07.qxd 8/18/03 12:35 PM Page 189 190 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Table 7-16 n f = √ 1 (7.59/24.6)2 + (13.1/54)2 = 2.55 Ans. 7-18 Referring to the solution of Prob. 7-17, for load fluctuations of −800 to 3000 lbf σa = 1.94 ∣∣∣∣3.000 − (−0.800)2(0.2813) ∣∣∣∣ = 13.1 kpsi σm = 1.94 ∣∣∣∣3.000 + (−0.800)2(0.2813) ∣∣∣∣ = 7.59 kpsi r = σa σm = 13.13 7.60 = 1.728 (a) Table 7-10, DE-Gerber n f = 12 ( 64 7.59 )2 (13.1 24.6 )−1 + √ 1 + ( 2(7.59)(24.6) 64(13.1) )2  = 1.79 Ans. (b) Table 7-11, DE-Elliptic n f = √ 1 (13.1/24.6)2 + (7.59/54)2 = 1.82 Ans. 7-19 Referring to the solution of Prob. 7-17, for load fluctuations of 800 to −3000 lbf σa = 1.94 ∣∣∣∣0.800 − (−3.000)2(0.2813) ∣∣∣∣ = 13.1 kpsi σm = 1.94 [ 0.800 + (−3.000) 2(0.2813) ] = −7.59 kpsi r = σa σm = 13.1−7.59 = −1.726 (a) We have a compressive midrange stress for which the failure locus is horizontal at the Se level. n f = Se σa = 24.6 13.1 = 1.88 Ans. (b) Same as (a) n f = Se σa = 24.6 13.1 = 1.88 Ans. shi20396_ch07.qxd 8/18/03 12:35 PM Page 190 Chapter 7 191 7-20 Sut = 0.495(380) = 188.1 kpsi S′e = 0.504(188.1) = 94.8 kpsi ka = 14.4(188.1)−0.718 = 0.335 For a non-rotating round bar in bending, Eq. (7-23) gives: de = 0.370d = 0.370(3/8) = 0.1388 in kb = ( 0.1388 0.3 )−0.107 = 1.086 Se = 0.335(1.086)(94.8) = 34.49 kpsi Fa = 30 − 152 = 7.5 lbf, Fm = 30 + 15 2 = 22.5 lbf σm = 32Mm πd3 = 32(22.5)(16) π(0.3753) (10−3) = 69.54 kpsi σa = 32(7.5)(16) π(0.3753) (10−3) = 23.18 kpsi r = 23.18 69.54 = 0.333 0 (a) Modified Goodman, Table 7-9 n f = 1(23.18/34.49) + (69.54/188.1) = 0.960 Since finite failure is predicted, proceed to calculate N Eq. (7-10): σ ′F = 188.1 + 50 = 238.1 kpsi Eq. (7-11): b = − log(238.1/34.49) log(2 · 106) = −0.133 13 Eq. (7-9): f = 238.1 188.1 (2 · 103)−0.133 13 = 0.4601 Eq. (7-13): a = [0.4601(188.1)] 2 34.49 = 217.16 kpsi σa Sf + σm Sut = 1 ⇒ Sf = σa1 − (σm/Sut ) = 23.18 1 − (69.54/188.1) = 36.78 kpsi Eq. (7-15) with σa = Sf N = ( 36.78 217.16 )1/−0.133 13 = 620 000 cycles Ans. shi20396_ch07.qxd 8/18/03 12:35 PM Page 191 194 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design σm = −4369.7 + √ 4369.72 + 4(3.4577)(106) 2 = 684.2 MPa σa = 684.2 − 152.3 = 531.9 MPa n f = 531.9152.3 = 3.49 Thus, the spring is not likely to fail in fatigue at the outer radius. Ans. 7-22 The solution at the inner radius is the same as in Prob. 7-21. At the outer radius, the yield solution is the same. Fatigue line: σa = ( 1 − σm Sut ) Se = σm − 152.3 639 ( 1 − σm 1671 ) = σm − 152.3 1.382σm = 791.3 ⇒ σm = 572.4 MPa σa = 572.4 − 152.3 = 420 MPa n f = 420152.3 = 2.76 Ans. 7-23 Preliminaries: Table A-20: Sut = 64 kpsi, Sy = 54 kpsi S′e = 0.504(64) = 32.3 kpsi ka = 2.70(64)−0.265 = 0.897 kb = 1 kc = 0.85 Se = 0.897(1)(0.85)(32.3) = 24.6 kpsi Fillet: Fig. A-15-5: D = 3.75 in, d = 2.5 in, D/d = 3.75/2.5 = 1.5, and r/d = 0.25/2.5 = 0.10 ∴ Kt = 2.1 K f = 2.1 1 + (2/√0.25) [(2.1 − 1)/2.1](4/64) = 1.86 σmax = 42.5(0.5) = 3.2 kpsi σmin = −162.5(0.5) = −12.8 kpsi σa = 1.86 ∣∣∣∣3.2 − (−12.8)2 ∣∣∣∣ = 14.88 kpsi shi20396_ch07.qxd 8/18/03 12:35 PM Page 194 Chapter 7 195 σm = 1.86 [ 3.2 + (−12.8) 2 ] = −8.93 kpsi ny = ∣∣∣∣ Syσmin ∣∣∣∣ = ∣∣∣∣ 54−12.8 ∣∣∣∣ = 4.22 Since the midrange stress is negative, Sa = Se = 24.6 kpsi n f = Sa σa = 24.6 14.88 = 1.65 Hole: Fig. A-15-1: d/w = 0.75/3.75 = 0.20, Kt = 2.5 K f = 2.5 1 + (2/√0.75/2)[(2.5 − 1)/2.5](5/64) = 2.17 σmax = 40.5(3.75 − 0.75) = 2.67 kpsi σmin = −160.5(3.75 − 0.75) = −10.67 kpsi σa = 2.17 ∣∣∣∣2.67 − (−10.67)2 ∣∣∣∣ = 14.47 kpsi σm = 2.17 2.67 + (−10.67)2 = −8.68 kpsi Since the midrange stress is negative, ny = ∣∣∣∣ Syσmin ∣∣∣∣ = ∣∣∣∣ 54−10.67 ∣∣∣∣ = 5.06 Sa = Se = 24.6 kpsi n f = Sa σa = 24.6 14.47 = 1.70 Thus the design is controlled by the threat of fatigue at the fillet; the minimum factor of safety is n f = 1.65. Ans. 7-24 (a) M = −T , h = 5 mm, A = 25 mm2 rc = 20 mm, ro = 22.5 mm, ri = 17.5 mm rn = hln ro/ri = 5 ln (22.5/17.5) = 19.8954 mm e = rc − rn = 20 − 19.8954 = 0.1046 mm co = 2.605 mm, ci = 2.395 mm T T shi20396_ch07.qxd 8/18/03 12:35 PM Page 195 196 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design σi = Mci Aeri = −T (0.002 395) 25(10−6)(0.1046)(10−3)(17.5)(10−3) = −52.34(106)T σo = −Mco Aero = T (2.605)(10 −3) 25(10−6)(0.1046)(10−3)(22.5)(10−3) = 44.27(106)T For fatigue, σo is most severe as it represents a tensile stress. σm = σa = 12(44.27)(10 6)T = 22.14(106)T S′e = 0.504Sut = 0.504(770) = 388.1 MPa ka = 4.51(770)−0.265 = 0.775 de = 0.808[5(5)]1/2 = 4.04 mm kb = ( 4.04 7.62 )−0.107 = 1.070 Se = 0.775(1.07)(388.1) = 321.8 MPa Modified Goodman, Table 7-9 σa Se + σm Sut = 1 n f ⇒ 22.14T 321.8 + 22.14T 770 = 1 3 T = 3.42 N · m Ans. (b) Gerber, Eq. (7-50) nσa Se + ( nσm Sut )2 = 1 3(22.14)T 321.8 + [ 3(22.14)T 770 ]2 = 1 T 2 + 27.74T − 134.40 = 0 T = 1 2 [−27.74 + √27.742 + 4(134.40) ] = 4.21 N · m Ans. (c) To guard against yield, use T of part (b) and the inner stress. ny = 42052.34(4.21) = 1.91 Ans. 7-25 From Prob. 7-24, Se = 321.8 MPa, Sy = 420 MPa, and Sut = 770 MPa (a) Assuming the beam is straight, σmax = 6M bh2 = 6T 53[(10−3)3] = 48(106)T Goodman: 24T 321.8 + 24T 770 = 1 3 ⇒ T = 3.15 N · m Ans. shi20396_ch07.qxd 8/18/03 12:36 PM Page 196 Chapter 7 199 Modified Goodman: σ ′a Se + σ ′ m Sut = 1 n 4366P 267.6(106) + 4366P 1000(106) = 1 3 ⇒ P = 16.1(103) N = 16.1 kN Ans. Yield: 1 ny = σ ′ a + σ ′m Sy ny = 800(10 6) 2(4366)(16.1)(103) = 5.69 Ans. (b) If the shaft is not rotating, τm = τa = 0. σm = σa = −2009P kb = 1 (axial) kc = 0.85 (Since there is no tension, kc = 1 might be more appropriate.) Se = 0.723(1)(0.85)(504) = 309.7 MPa n f = 309.7(10 6) 2009P ⇒ P = 309.7(10 6) 3(2009) = 51.4(103) N = 51.4 kN Ans. Yield: ny = 800(10 6) 2(2009)(51.4)(103) = 3.87 Ans. 7-28 From Prob. 7-27, K f = 2.84, K f s = 1.76, Se = 267.6 MPa σmax = −K f 4Pmax πd2 = −2.84 [ (4)(80)(10−3) π(0.030)2 ] = −321.4 MPa σmin = 2080(−321.4) = −80.4 MPa Tmax = f Pmax ( D + d 4 ) = 0.3(80)(103) ( 0.150 + 0.03 4 ) = 1080 N · m Tmin = 2080(1080) = 270 N · m 309.7 m a 800 800 shi20396_ch07.qxd 8/18/03 12:36 PM Page 199 200 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design τmax = K f s 16Tmax πd3 = 1.76 [ 16(1080) π(0.030)3 (10−6) ] = 358.5 MPa τmin = 2080(358.5) = 89.6 MPa σa = 321.4 − 80.42 = 120.5 MPa σm = −321.4 − 80.42 = −200.9 MPa τa = 358.5 − 89.62 = 134.5 MPa τm = 358.5 + 89.62 = 224.1 MPa σ ′a = [ σ 2a + 3τ 2a ]1/2 = [120.52 + 3(134.5)2]1/2 = 262.3 MPa σ ′m = [(−200.9)2 + 3(224.1)2]1/2 = 437.1 MPa Goodman: (σa)e = σ ′ a 1 − σ ′m/Sut = 262.3 1 − 437.1/1000 = 466.0 MPa Let f = 0.9 a = [0.9(1000)] 2 276.6 = 2928 MPa b = −1 3 log [ 0.9(1000) 276.6 ] = −0.1708 N = [ (σa)e a ]1/b = [ 466.0 2928 ]1/−0.1708 = 47 130 cycles Ans. 7-29 Sy = 490 MPa, Sut = 590 MPa, Se = 200 MPa σm = 420 + 1402 = 280 MPa, σa = 420 − 140 2 = 140 MPa Goodman: (σa)e = σa1 − σm/Sut = 140 1 − (280/590) = 266.5 MPa > Se a = [0.9(590)] 2 200 = 1409.8 MPa b = −1 3 log 0.9(590) 200 = −0.141 355 shi20396_ch07.qxd 8/18/03 12:36 PM Page 200 Chapter 7 201 N = ( 266.5 1409.8 )−1/0.143 55 = 131 200 cycles Nremaining = 131 200 − 50 000 = 81 200 cycles Second loading: (σm)2 = 350 + (−200)2 = 75 MPa (σa)2 = 350 − (−200)2 = 275 MPa (σa)e2 = 2751 − (75/590) = 315.0 MPa (a) Miner’s method N2 = ( 315 1409.8 )−1/0.141 355 = 40 200 cycles n1 N1 + n2 N2 = 1 ⇒ 50 000 131 200 + n2 40 200 = 1 n2 = 24 880 cycles Ans. (b) Manson’s method Two data points: 0.9(590 MPa), 103 cycles 266.5 MPa, 81 200 cycles 0.9(590) 266.5 = a2(10 3)b2 a2(81 200)b2 1.9925 = (0.012 315)b2 b2 = log 1.9925log 0.012 315 = −0.156 789 a2 = 266.5(81 200)−0.156 789 = 1568.4 MPa n2 = ( 315 1568.4 )1/−0.156 789 = 27 950 cycles Ans. 7-30 (a) Miner’s method a = [0.9(76)] 2 30 = 155.95 kpsi b = −1 3 log 0.9(76) 30 = −0.119 31 σ1 = 48 kpsi, N1 = ( 48 155.95 )1/−0.119 31 = 19 460 cycles σ2 = 38 kpsi, N2 = ( 38 155.95 )1/−0.119 31 = 137 880 cycles shi20396_ch07.qxd 8/18/03 12:36 PM Page 201 204 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Resulting in a design factor n f of, Eq. (6-59): n f = exp[−(−3.09) √ ln(1 + 0.2972) + ln √ 1 + 0.2972] = 2.56 • Decision: Set n f = 2.56 Now proceed deterministically using the mean values: k̄a = 0.887, kb = 1, k̄c = 0.890, and from Prob. 7-10, K f = 2.09 σ̄a = K̄ f F̄a A = K̄ f F̄a t (60 − 12) = S̄e n̄ f ∴ t = n̄ f K̄ f F̄a (60 − 12) S̄e = 2.56(2.09)(15.10 3) (60 − 12)(175.7) = 9.5 mm Decision: If 10 mm 1018 CD is available, t = 10 mm Ans. 7-34 Rotation is presumed. M and Sut are given as deterministic, but notice that σ is not; there- fore, a reliability estimation can be made. From Eq. (7-70): S′e = 0.506(110)LN(1, 0.138) = 55.7LN(1, 0.138) kpsi Table 7-13: ka = 2.67(110)−0.265LN(1, 0.058) = 0.768LN(1, 0.058) Based on d = 1 in, Eq. (7-19) gives kb = ( 1 0.30 )−0.107 = 0.879 Conservatism is not necessary Se = 0.768[LN(1, 0.058)](0.879)(55.7)[LN(1, 0.138)] S̄e = 37.6 kpsi CSe = (0.0582 + 0.1382)1/2 = 0.150 Se = 37.6LN(1, 0.150) 1.25" M M 1.00" shi20396_ch07.qxd 8/18/03 12:36 PM Page 204 Chapter 7 205 Fig. A-15-14: D/d = 1.25, r/d = 0.125. Thus Kt = 1.70 and Eqs. (7-35), (7-78) and Table 7-8 give K f = 1.70LN(1, 0.15) 1 + (2/√0.125)[(1.70 − 1)/(1.70)](3/110) = 1.598LN(1, 0.15) σ = K f 32M πd3 = 1.598[LN(1 − 0.15)] [ 32(1400) π(1)3 ] = 22.8LN(1, 0.15) kpsi From Eq. (6-57): z = − ln [ (37.6/22.8) √ (1 + 0.152)/(1 + 0.152) ] √ ln[(1 + 0.152)(1 + 0.152)] = −2.37 From Table A-10, pf = 0.008 89 ∴ R = 1 − 0.008 89 = 0.991 Ans. Note: The correlation method uses only the mean of Sut ; its variability is already included in the 0.138. When a deterministic load, in this case M, is used in a reliability estimate, en- gineers state, “For a Design Load of M, the reliability is 0.991.” They are in fact referring to a Deterministic Design Load. 7-35 For completely reversed torsion, ka and kb of Prob. 7-34 apply, but kc must also be con- sidered. Eq. 7-74: kc = 0.328(110)0.125LN(1, 0.125) = 0.590LN(1, 0.125) Note 0.590 is close to 0.577. SSe = ka kb kc S′e = 0.768[LN(1, 0.058)](0.878)[0.590LN(1, 0.125)][55.7LN(1, 0.138)] S̄Se = 0.768(0.878)(0.590)(55.7) = 22.2 kpsi CSe = (0.0582 + 0.1252 + 0.1382)1/2 = 0.195 SSe = 22.2LN(1, 0.195) kpsi Fig. A-15-15: D/d = 1.25, r/d = 0.125, then Kts = 1.40. From Eqs. (7-35), (7-78) and Table 7-8 Kts = 1.40LN(1, 0.15) 1 + (2/√0.125) [(1.4 − 1)/1.4](3/110) = 1.34LN(1, 0.15) τ = Kts 16T πd3 τ = 1.34[LN(1, 0.15)] [ 16(1.4) π(1)3 ] = 9.55LN(1, 0.15) kpsi shi20396_ch07.qxd 8/18/03 12:36 PM Page 205 206 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design From Eq. (6-57): z = − ln [ (22.2/9.55) √ (1 + 0.152)/(1 + 0.1952) ] √ ln [(1 + 0.1952)(1 + 0.152)] = −3.43 From Table A-10, pf = 0.0003 R = 1 − pf = 1 − 0.0003 = 0.9997 Ans. For a design with completely-reversed torsion of 1400 lbf · in, the reliability is 0.9997. The improvement comes from a smaller stress-concentration factor in torsion. See the note at the end of the solution of Prob. 7-34 for the reason for the phraseology. 7-36 Sut = 58 kpsi S′e = 0.506(58)LN(1, 0.138) = 29.3LN(1, 0.138) kpsi Table 7-13: ka = 14.5(58)−0.719LN(1, 0.11) = 0.782LN(1, 0.11) Eq. (7-23): de = 0.37(1.25) = 0.463 in kb = ( 0.463 0.30 )−0.107 = 0.955 Se = 0.782[LN(1, 0.11)](0.955)[29.3LN(1, 0.138)] S̄e = 0.782(0.955)(29.3) = 21.9 kpsi CSe = (0.112 + 0.1382)1/2 = 0.150 Table A-16: d/D = 0, a/D = 0.1, A = 0.83 ∴ Kt = 2.27. From Eqs. (7-35) and (7-78) and Table 7-8 K f = 2.27LN(1, 0.10) 1 + (2/√0.125) [(2.27 − 1)/2.27](5/58) = 1.783LN(1, 0.10) Table A-16: Z = π AD 3 32 = π(0.83)(1.25 3) 32 = 0.159 in3 σ = K f M Z = 1.783LN(1, 0.10) ( 1.6 0.159 ) = 17.95LN(1, 0.10) kpsi M M D 1 4 1 " D Non-rotating 1 8 " shi20396_ch07.qxd 8/18/03 12:36 PM Page 206 Chapter 7 209 Solve for thickness h. To do so we need k̄a = 2.67S̄−0.265ut = 2.67(64)−0.265 = 0.887 kb = 1 k̄c = 1.23S̄−0.078ut = 1.23(64)−0.078 = 0.889 k̄d = k̄e = 1 S̄e = 0.887(1)(0.889)(1)(1)(0.506)(64) = 25.5 kpsi Fig. A-15-5: D = 3.75 in, d = 2.5 in, D/d = 3.75/2.5 = 1.5, r/d = 0.25/2.5 = 0.10 ∴ Kt = 2.1 K̄ f = 2.1 1 + (2/√0.25)[(2.1 − 1)/(2.1)](4/64) = 1.857 h = K̄ f n̄Fa w2 S̄e = 1.857(2.02)(10) 2.5(25.5) = 0.667 Ans. This thickness separates S̄e and σ̄a so as to realize the reliability goal of 0.999 at each shoulder. The design decision is to make t the next available thickness of 1018 CD steel strap from the same heat. This eliminates machining to the desired thickness and the extra cost of thicker work stock will be less than machining the fares. Ask your steel supplier what is available in this heat. 7-39 Fa = 1200 lbf Sut = 80 kpsi (a) Strength ka = 2.67(80)−0.265LN(1, 0.058) = 0.836LN(1, 0.058) kb = 1 kc = 1.23(80)−0.078LN(1, 0.125) = 0.874LN(1, 0.125) S′a = 0.506(80)LN(1, 0.138) = 40.5LN(1, 0.138) kpsi Se = 0.836[LN(1, 0.058)](1)[0.874LN(1, 0.125)][40.5LN(1, 0.138)] S̄e = 0.836(1)(0.874)(40.5) = 29.6 kpsi CSe = (0.0582 + 0.1252 + 0.1382)1/2 = 0.195 1200 lbf 3 4 " 1 4 " 1 2 1 " shi20396_ch07.qxd 8/18/03 12:36 PM Page 209 210 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Stress: Fig. A-15-1; d/w = 0.75/1.5 = 0.5, Kt = 2.17. From Eqs. (7-35), (7-78) and Table 7-8 K f = 2.17LN(1, 0.10) 1 + (2/√0.375)[(2.17 − 1)/2.17](5/80) = 1.95LN(1, 0.10) σa = K f Fa(w − d)t , Cσ = 0.10 σ̄a = K̄ f Fa(w − d)t = 1.95(1.2) (1.5 − 0.75)(0.25) = 12.48 kpsi S̄a = S̄e = 29.6 kpsi z = − ln ( S̄a/σ̄a) √( 1 + C2σ )/( 1 + C2S ) √ ln ( 1 + C2σ ) ( 1 + C2S ) = − ln [ (29.6/12.48) √ (1 + 0.102)/(1 + 0.1952) ] √ ln (1 + 0.102)(1 + 0.1952) = −3.9 From Table A-20 pf = 4.481(10−5) R = 1 − 4.481(10−5) = 0.999 955 Ans. (b) All computer programs will differ in detail. 7-40 Each computer program will differ in detail. When the programs are working, the experi- ence should reinforce that the decision regarding n̄ f is independent of mean values of strength, stress or associated geometry. The reliability goal can be realized by noting the impact of all those a priori decisions. 7-41 Such subprograms allow a simple call when the information is needed. The calling pro- gram is often named an executive routine (executives tend to delegate chores to others and only want the answers). 7-42 This task is similar to Prob. 7-41. 7-43 Again, a similar task. 7-44 The results of Probs. 7-41 to 7-44 will be the basis of a class computer aid for fatigue prob- lems. The codes should be made available to the class through the library of the computer network or main frame available to your students. 7-45 Peterson’s notch sensitivity q has very little statistical basis. This subroutine can be used to show the variation in q , which is not apparent to those who embrace a deterministic q . 7-46 An additional program which is useful. shi20396_ch07.qxd 8/18/03 12:36 PM Page 210