Shi20396 ch07

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Diseno en ingenieria mecanica de Shigley - 8th ---HDes descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html

Transcript of Shi20396 ch07

  • 1. Chapter 77-1 HB = 490Eq. (3-17): Sut = 0.495(490) = 242.6 kpsi > 212 kpsiEq. (7-8): Se= 107 kpsiTable 7-4: a = 1.34, b = 0.085Eq. (7-18): ka = 1.34(242.6)0.085 = 0.840Eq. (7-19): kb =3/160.30.107= 1.05Eq. (7-17): Se = kakbSe= 0.840(1.05)(107) = 94.4 kpsi Ans.7-2(a) Sut = 68 kpsi, Se= 0.495(68) = 33.7 kpsi Ans.(b) Sut = 112 kpsi, Se= 0.495(112) = 55.4 kpsi Ans.(c) 2024T3 has no endurance limit Ans.(d) Eq. (3-17): Se= 107 kpsi Ans.7-3F= 0m = 115(0.90)0.22 = 112.4 kpsiEq. (7-8): Se= 0.504(66.2) = 33.4 kpsiEq. (7-11): b = log(112.4/33.4)log(2 106)= 0.083 64Eq. (7-9): f = 112.466.2(2 103)0.083 64 = 0.8991Eq. (7-13): a = [0.8991(66.2)]233.4= 106.1 kpsiEq. (7-12): Sf = aNb = 106.1(12 500)0.083 64 = 48.2 kpsi Ans.Eq. (7-15): N =aa1/b=36106.11/0.083 64= 409 530 cycles Ans.7-4 From Sf = aNblog Sf = log a + b log NSubstituting (1, Sut )log Sut = log a + b log (1)From which a = Sut

2. Chapter 7 181Substituting (103, f Sut ) and a = Sutlog f Sut = log Sut + b log 103From whichb = 13log f Sf = Sut N(log f )/3 1 N 103For 500 cycles as in Prob. 7-3500Sf 66.2(500)(log 0.8991)/3 = 60.2 kpsi Ans.7-5 Read from graph: (103, 90) and (106, 50). From S = aNblog S1 = log a + b log N1log S2 = log a + b log N2From whichlog a = log S1 log N2 log S2 log N1log N2/N1= log 90 log 106 log 50 log 103log 106/103= 2.2095a = 10log a = 102.2095 = 162.0b = log 50/903= 0.085 09(Sf )ax = 1620.085 09 103 N 106 in kpsi Ans.Check:103(Sf )ax = 162(103)0.085 09 = 90 kpsi106(Sf )ax = 162(106)0.085 09 = 50 kpsiThe end points agree.7-6Eq. (7-8): Se= 0.504(710) = 357.8 MPaTable 7-4: a = 4.51, b = 0.265Eq. (7-18): ka = 4.51(710)0.265 = 0.792Eq. (7-19): kb =d7.620.107=327.620.107= 0.858Eq. (7-17): Se = kakbSe= 0.792(0.858)(357.8) = 243 MPa Ans. 3. 182 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design7-7 For AISI 4340 as forged steel,Eq. (7-8): Se = 107 kpsiTable 7-4: a = 39.9, b = 0.995Eq. (7-18): ka = 39.9(260)0.995 = 0.158Eq. (7-19): kb =0.750.300.107= 0.907Each of the other Marin factors is unity.Se = 0.158(0.907)(107) = 15.3 kpsiFor AISI 1040:Se= 0.504(113) = 57.0 kpsika = 39.9(113)0.995 = 0.362kb = 0.907 (same as 4340)Each of the other Marin factors is unity.Se = 0.362(0.907)(57.2) = 18.7 kpsiNot only is AISI 1040 steel a contender, it has a superior endurance strength. Can you seewhy?7-8(a) For an AISI 1018 CD-machined steel, the strengths areEq. (3-17): Sut = 440 MPa HB = 4403.41= 129Sy = 370 MPaSsu = 0.67(440) = 295 MPa2.5 mm20 mm 25 mmFig. A-15-15:rd= 2.520= 0.125,Dd= 2520= 1.25, Kts = 1.4Fig. 7-21: qs = 0.94Eq. (7-31): Kf s = 1 + 0.94(1.4 1) = 1.376For a purely reversing torque of 200 N mmax = Kf s16Td3= 1.376(16)(200 103 N mm)(20 mm)3max = 175.2 MPa = aSe= 0.504(440) = 222 MPaThe Marin factors areka = 4.51(440)0.265 = 0.899kb =207.620.107= 0.902kc = 0.59, kd = 1, ke = 1Eq. (7-17): Se = 0.899(0.902)(0.59)(222) = 106.2 MPa 4. Chapter 7 183Eq. (7-13): a = [0.9(295)]2106.2= 664Eq. (7-14): b = 13log0.9(295)106.2= 0.132 65Eq. (7-15): N =175.26641/0.132 65N = 23 000 cycles Ans.(b) For an operating temperature of 450C, the temperature modification factor, fromTable 7-6, iskd = 0.843Thus Se = 0.899(0.902)(0.59)(0.843)(222) = 89.5 MPaa = [0.9(295)]289.5= 788b = 13log0.9(295)89.5= 0.157 41N =175.27881/0.157 41N = 14 100 cycles Ans.7-9f = 0.9n = 1.5N = 104 cyclesFor AISI 1045 HR steel, Sut = 570 MPa and Sy = 310 MPaSe= 0.504(570 MPa) = 287.3 MPaFind an initial guess based on yielding:a = max = McI= M(b/2)b(b3)/12= 6Mb3Mmax = (1 kN)(800 mm) = 800 N mmax = Syn 6(800 103 Nmm)b3= 310 N/mm21.5b = 28.5 mmEq. (7-24): de = 0.808bEq. (7-19): kb =0.808b7.620.107= 1.2714b0.107kb = 0.888F1 kNbb800 mm 5. 184 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignThe remaining Marin factors areka = 57.7(570)0.718 = 0.606kc = kd = ke = kf = 1Eq. (7-17): Se = 0.606(0.888)(287.3 MPa) = 154.6 MPaEq. (7-13): a = [0.9(570)]2154.6= 1702Eq. (7-14): b = 13log0.9(570)154.6= 0.173 64Eq. (7-12): Sf = aNb = 1702[(104)0.173 64] = 343.9 MPan = Sfaor a = Sfn6(800 103)b3= 343.91.5 b = 27.6 mmCheck values for kb, Se, etc.kb = 1.2714(27.6)0.107 = 0.891Se = 0.606(0.891)(287.3) = 155.1 MPaa = [0.9(570)]2155.1= 1697b = 13log0.9(570)155.1= 0.173 17Sf = 1697[(104)0.173 17] = 344.4 MPa6(800 103)b3= 344.41.5b = 27.5 mm Ans.7-1010Fa 12 Fa601018Table A-20: Sut = 440 MPa, Sy = 370 MPaSe= 0.504(440) = 221.8 MPa0.265 = 0.899Table 7-4: ka = 4.51(440)kb =1 (axial loading)Eq. (7-25): kc = 0.85Se = 0.899(1)(0.85)(221.8) = 169.5 MPaTable A-15-1: d/w = 12/60 = 0.2, Kt = 2.5 6. Chapter 7 185From Eq. (7-35) and Table 7-8Kf = Kt1 +2/r[(Kt 1)/Kt ]a= 2.51 +62/[(2.5 1)/2.5](174/440)= 2.09a = KfFaA Senf= 2.09Fa10(60 12)= 169.51.8Fa = 21 630 N = 21.6 kN Ans.FaA= Syny Fa10(60 12)= 3701.8Fa = 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: dMaterial and condition: 1095 HR and from Table A-20 Sut = 120, Sy = 66 kpsi.Design factor: nf = 1.6 per problem statement.Life: (1150)(3) = 3450 cyclesFunction: carry 10 000 lbf loadPreliminaries to iterative solution:Se= 0.504(120) = 60.5 kpsika = 2.70(120)0.265 = 0.759Id3= c32= 0.098 17d3M(crit.) =624(10 000)(12) = 30 000 lbf inThe 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 inkb =2.000.300.107= 0.816Se = 0.759(0.816)(60.5) = 37.5 kpsia = [0.9(120)]237.5= 311.0b = 13log0.9(120)37.5= 0.1531Sf = 311.0(3450)0.1531 = 89.3 7. 186 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design0 = MI/c= 300.098 17d3= 305.6d3= 305.623= 38.2 kpsir = d10= 210= 0.2Kf = 1.681 +0.22/[(1.68 1)/1.68](4/120)= 1.584Eq. (7-37):(Kf )103 = 1 (1.584 1)[0.18 0.43(102)120 + 0.45(105)1202]= 1.158Eq. (7-38):(Kf )N = K3450 = 1.15821.584(3450)(1/3) log(1.158/1.584)= 1.2250 = 305.623= 38.2 kpsia = (Kf )N0 = 1.225(38.2) = 46.8 kpsinf = (Sf )3450a= 89.346.8= 1.91The design is satisfactory. Reducing the diameter will reduce n, but the resulting preferredsize will be d = 2.00 in.7-12am= 172 MPa, =3m =3(103) = 178.4 MPaYield: 172 + 178.4 = Syny= 413ny ny = 1.18 Ans.(a) Modified Goodman, Table 7-9nf = 1(172/276) + (178.4/551)= 1.06 Ans.(b) Gerber, Table 7-10nf = 12551178.42 1722761 + 1 +2(178.4)(276)551(172) 8. 2= 1.31 Ans.(c) ASME-Elliptic, Table 7-11nf =1(172/276)2 + (178.4/413)2 9. 1/2= 1.32 Ans. 10. Chapter 7 1877-13am= 69 MPa, =3(138) = 239 MPaYield: 69 + 239 = 413ny ny = 1.34 Ans.(a) Modified Goodman, Table 7-9nf = 1(69/276) + (239/551)= 1.46 Ans.(b) Gerber, Table 7-10nf = 125512392 692761 + 1 +2(239)(276)551(69) 11. 2= 1.73 Ans.(c) ASME-Elliptic, Table 7-11nf =1(69/276)2 + (239/413)2 12. 1/2= 1.59 Ans.7-14a=2a+ 3 2a=832 + 3(692) = 145.5 MPa, m=3(103) = 178.4 MPaYield: 145.5 + 178.4 = 413ny ny = 1.28 Ans.(a) Modified Goodman, Table 7-9nf = 1(145.5/276) + (178.4/551)= 1.18 Ans.(b) Gerber, Table 7-10nf = 12551178.42 145.52761 + 1 +2(178.4)(276)551(145.5) 13. 2= 1.47 Ans.(c) ASME-Elliptic, Table 7-11nf =1(145.5/276)2 + (178.4/413)2 14. 1/2= 1.47 Ans.7-15a=3(207) = 358.5 MPa, m= 0Yield: 358.5 = 413ny ny = 1.15 Ans. 15. 188 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design(a) Modified Goodman, Table 7-9nf = 1(358.5/276)= 0.77 Ans.(b) Gerber criterion of Table 7-10 does not work; therefore use Eq. (7-50).nfaSe= 1 nf = Sea= 276358.5= 0.77 Ans.(c) ASME-Elliptic, Table 7-11nf = 1358.5/2762= 0.77 Ans.Let f = 0.9 to assess the cycles to failure by fatigueEq. (7-13): a = [0.9(551)]2276= 891.0 MPaEq. (7-14): b = 13log0.9(551)276= 0.084 828Eq. (7-15): N =358.5891.01/0.084 828= 45 800 cycles Ans.7-16a=3(103) = 178.4 MPa, m= 103 MPaYield: 178.4 + 103 = 413ny ny = 1.47 Ans.(a) Modified Goodman, Table 7-9nf = 1(178.4/276) + (103/551)= 1.20 Ans.(b) Gerber, Table 7-10nf = 125511032 178.42761 + 1 +2(103)(276)551(178.4) 16. 2= 1.44 Ans.(c) ASME-Elliptic, Table 7-11nf =1(178.4/276)2 + (103/413)2 17. 1/2= 1.44 Ans.7-17 Table A-20: Sut = 64 kpsi, Sy = 54 kpsiA = 0.375(1 0.25) = 0.2813 in2max = FmaxA= 30000.2813(103) = 10.67 kpsi 18. Chapter 7 189ny = 5410.67= 5.06 Ans.Se= 0.504(64) = 32.3 kpsika = 2.70(64)0.265 = 0.897kb = 1, kc = 0.85Se = 0.897(1)(0.85)(32.3) = 24.6 kpsiTable A-15-1: w = 1 in, d = 1/4 in, d/w = 0.25Kt = 2.45. From Eq. (7-35) andTable 7-8Kf = 2.451 +0.1252/[(2.45 1)/2.45](5/64)= 1.94a = KfFmax Fmin2A= 1.943.000 0.8002(0.2813)= 7.59 kpsim = KfFmax + Fmin2A= 1.943.000 + 0.8002(0.2813) 19. = 13.1 kpsir = am= 7.5913.1= 0.579(a) DE-Gerber, Table 7-10Sa = 0.5792(642)2(24.6)1 + 1 +2(24.6)0.579(64)2 = 18.5 kpsiSm = Sar= 18.50.579= 32.0 kpsinf = 126413.12 7.5924.61 + 1 +2(13.1)(24.6)7.59(64)2= 2.44 Ans.(b) DE-Elliptic, Table 7-11Sa = (0.5792)(24.62)(542)24.62 + (0.5792)(542)= 19.33 kpsiSm = Sar= 19.330.579= 33.40 kpsi 20. 190 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignTable 7-16nf = 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 lbfa = 1.943.000 (0.800)2(0.2813)= 13.1 kpsim = 1.943.000 + (0.800)2(0.2813)= 7.59 kpsir = am= 13.137.60= 1.728(a) Table 7-10, DE-Gerbernf = 12647.592 13.124.61 + 1 +2(7.59)(24.6)64(13.1)2 = 1.79 Ans.(b) Table 7-11, DE-Ellipticnf = 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 lbfa = 1.940.800 (3.000)2(0.2813)= 13.1 kpsim = 1.940.800 + (3.000)2(0.2813) 21. = 7.59 kpsir = am= 13.17.59= 1.726(a) We have a compressive midrange stress for which the failure locus is horizontal at theSe level.nf = Sea= 24.613.1= 1.88 Ans.(b) Same as (a)nf = Sea= 24.613.1= 1.88 Ans. 22. Chapter 7 1917-20Sut = 0.495(380) = 188.1 kpsiSe= 0.504(188.1) = 94.8 kpsika = 14.4(188.1)0.718 = 0.335For a non-rotating round bar in bending, Eq. (7-23) gives: de = 0.370d = 0.370(3/8) =0.1388 inkb =0.13