《机械设计-吴昌林》课后习题答案(部分).pdf 26页

'2-4已知某钢制零件受弯曲变应力的作用，其中，最大工作应力max=200MPa，最小工作应力min=－50MPa，危险截面上的应力集中系数k=1.2，尺寸系数=0.85，表明状态系数=1。材料的s=750MPa，0=580MPa，-1=350MPa。试求：（1）绘制材料的简化极限应力图，并在图上标出工作应力点的位置；（2）求材料在该应力状态下的疲劳极限应力r；（3）按疲劳极限应力和安全系数分别校核此零件是否安全（取Smin=1.5）Analloysteelpartsuffersbendingstress：s=750MPa，0=580MPa，-1=350MPaWorkingstress：max=200MPa，min=－50MPak=1.2，=0.85，=1，(1)Makethestressendurancefigure,andfindtheworkingpoint.(2)Findthefatiguestresslimitationr.(3)Checkifthispartissafe,usingthefatiguestresslimitationandthesafetyfactor(ifSmin=1.5).Solution：1)calculatestressamplitude、meanstressandstressratio11(200(50))125MPaamaxmin2211(20050)75MPammaxmin22min50r0.25200max2)judgewheretheworkingpointisk1.2(K)1.4118D0.851((K)1)()D0s1s0r0[((K)1)2]sD01(1.41181)580750350(750580)＝0.46750[(1.41181)5802350]rr0∴workingpointisinAOD a00AB(,)22(0,)1DF1350OG(s,0)m3)calculateendurancelimit21023505800.206958001amr[(K)]Dam1am350(12575)364.6MPar[(K)](1.41181250.206975)Dam4)checkstrengthA.Allowablestressr364.6[]243.1MPa[S]1.5200MPamax[]maxSafe！B.Safetyfactorr364.6S1.82[S]1.5200max3501S1.82[S]1.5[(K)](1.41181250.206975)DamSafe！ 3－3、3－6、3－83－3Atransmissionwithtwopairsofgearsisshownasproblemsfigure3-1.TheaxialforcesofthetwogearsontheaxleII,Fa2andFa3aretobeeliminated.ThehelixangleofthefirstpairofgearsI＝15.Trytodeterminethehelixangleofthesecondpairofgears,andthehelixdirectionsforthegear3andgear4onit.Solution:(1)Thetorqueofgear2andgear3T2=9550000*P2/n2T3=9550000*P3/n3n2=n3,sincethetwogearsareonthesameaxleP2=P3,whenignorethetransferefficiencySothat,T2=T3(2)Theaxialforceofgear2equaltogear3FFtg,FFtga2t22a3t332T2T23F,Ft2t3dd23 mzmzn22n33d,d(P54,Helicalgear,mnzmustbedividedbycos)23coscos33soTT23sinsin23mzmzn22n33mn3z30sinsin4*30*sin15/(3*60)32mzn22sin=0.17254630=9.94=956’24’’33－6Anenclosedtransmissionwithtwopairsofgears.Known:Fig3-22TransmissionwithtwopairofP1=20kW,n1=1430r/min.gearsGearratiou=4.3,onedirectionrotating.16hourperday,andlifeismorethan5years.Withlowerstiffnessshaftandslightimpactload.Material40Cr,withsurfacehardening.Hardness:48～55HRCCompletethedesignforthefirstpairhelicalgear.Solution：1）Determinetherequiredmaterialsandhardnessofthem.Computetheallowablestress.（1）Determinetherequiredmaterialsandhardness.40Cr,withsurfacehardening.Hardness:48～55HRC,anduse55HRC.（2）Computetheallowablestress.①Determinetheendurancelimitstress.σHlimandσFlimCheckthefig3－16(c)，σHlim1=σHlim2=1200MPaCheckthefig3－17(c)，σFlim1=σFlim2=360MPa②LifecyclenumberN.DeterminethelifefactorZN、YN8N60ant6011430(530016)20.05921011 8N120.0592108N4.7888102u4.3Fromfig3－18，ZN1ZN21.Fromfig3－19，YN1YN21.③Computetheallowablestress.Fromthetable3－4,S1.2,S1.5。HminFminZ12001Hlim1N11000MPaHP1S1.2HminHP2HP1Flim1YSTYN136021480MPaFP1S1.5FminFP2FP12）Fatiguetype,anddesigncriteriaHardsurface,enclosedgear.Basicfatiguetypeisteethbreakage.Designusingbendingfatiguestrength,andcheckthecontactfatiguestrength.3）Determinetheinitialdesignparameters.（1）Pinion’snominaltorque6P1620T9.55109.5510133566Nmm1n14301（2）GeartypeHelicalgear.（3）PrecisionclassClass7（4）Initialdesignparametersβ=12°，z1=30，z2=z1u=30×4.3=129，χ1=χ2＝0，Fromtable3－6,ψd＝0.5（5）Designforthebendingstress.Todeterminetheparametermn,theparametersbelowneedtobedecided.YFa1，YFa2，YSa1，YSa2，Yε，Yβ Lowerstiffnessshaftandslightimpactload.Fromtable3—1，K=1.35。z30Equivalenttoothnumberz132.1v133coscos12z129z2137.8v233coscos12Fromfig3－14，YFa1=2.5，YFa2=2.21；Fromfig3－15，YSa1=1.65，YSa2=1.8；Yε=0.7，Yβ=0.9。YFa1YSa12.51.650.00859375480FP1YFa2YSa22.211.80.0082875480FP2Usingthebiggervalueforcalculating:22KT1cosYYYFaYSam3mmn2zd1FP221.35133566cos120.70.930.0085937520.530=1.607Fromtable3－7，thestandardmodulemn=2mmm2na(zz)(30129)162.55mm122cos2cos12Use:a=165mmAdjustthehelixangle：m(zz)2(30129)n12arccosarccos1529"55""2a2165Pitchdiameter：mz230n1d62.264mm1coscos1529"55"mz2129n2d267.746mm2coscos1529"55" o（Notice：βfor××××’××’’；Pitchdiameterfor×××.××（d1＋d2）/2=a=165mm）nd143062.26411v4.6620m/s6000060000Approximatelyequaltotheinitialvalue.Toothwidth：Gear:b2=b=ψdd1=0.5×62.264mm=31.132mmPinion:b1=b2＋（5～10）=（31.132＋6）mm=37.132mm4）CheckforthecontactstressFromfig3－11,Z2.43；Fromtable3－2,Z189.8MPa；Z0.8；HEZcoscos1529"55""0.9816Contactstress:2KTu11ZZZZHHE2bdu121.351335664.312.43189.80.80.9816237.13262.2644.3=695σHP1=σHP2=1200MPaSafe.5）Structuredesign,etc.3－8CheckthepowerP1forapairofbevelgears. Givn：z1＝18，z2＝36，m=2mm,b=13mm,n1=930r/min.Electromotordriven，Withonedirectionrotatingandslightimpactload.Workinglife24000hours.Materialsteel45.Pinionquenchingandtempering.Hardness230～250HBS；Gearnormalizing.Hardness190～210HBSPrecisionclass8.Pinionaxlewithcantileverbeam.Highreliability.Solution：1）Determinetherequiredmaterialsandtreatmentofthem.Computetheallowablestress.（1）Determinetherequiredmaterialsandhardness.Materialsteel45.Pinionquenchingandtempering.Hardness230～250HBS,using240.Gearnormalizing.Hardness190～210HBS,using200.（2）Computetheallowablestress.①Determinetheendurancelimitstress.σHlimandσFlimCheckthefig3－16(b)，σHlim1=600MPa；σHlim2=550MPaCheckthefig3－17(b)，σFlim1=220MPa；σFlim2=210MPa。②LifecyclenumberN.DeterminethelifefactorZN、YN8N60ant601930(24000)13.39210118N113.382108N6.691102u2Fromfig3－18，ZN1ZN21.Fromfig3－19，YN1YN21.③Computetheallowablestress.Fromthetable3－4,S1.1,S1.4。HminFminZ6001Hlim1N1545.45MPaHP1S1.1Hmin550/1.1500HP2 YY22021Flim1STN1314.2857MPaFP1S1.4Fmin210*2*1/1.4300FP22）Fatiguetype,anddesigncriteriaSoftsurface,enclosedgear.Basicfatiguetypeispitting.Designorcheckthecontactfatiguestrengthmainly.(Ifthemoduleissmall,it’snecessarytocheckthebendingfatiguestrength.)3）Determinetheinitialdesignparameters.(known)Pinion’snominaltorquePPT9.5510619.55106110268.8172PNmm11n93014)CheckthecontactfatiguestressTodecideK、ZH、ZE.Electromotordriven，Withonedirectionrotatingandslightimpactload.Pinionaxlewithcantileverbeam.Highreliability.K=1.4.Fromfig3－11,Z2.5；Fromtable3－2,Z189.8MPa；HEBevelgear1:tg=18/36=0.5,sin0.4472,cos0.8944111d1==mz1=2*18=36R=(d1/2)/sin1=18/0.4472=40.2504ψR=b/R=13/40.2504=0.32304KT1ZZMPaHHE23HP0.8510.5duRR141.410268.8172P12.5189.8545.45230.850.323010.50.3230362（T1=4250Nmm） P1=0.4139kW4－1、4－34－1Determinetheworm-geardrivesystem’srotatingdirectionsfortheaxles,wormgearhelixdirection,andtheforceslocationanddirectionforwormgear.Fig.problem4-11、3－worm；2、4－gear4－3DesignanArchimedeanCylindricalwormusinginacrane.Mediumimpactload.Wormaxlewithelectromotordriven.P1=10kW，n1=1470r/min，n2=120r/min.Workingwithintermission.2hoursoneday,andtheworkinglife10years. Solution：1．Determinetherequiredmaterialsandtreatmentforthewormandthegear.Determinetheprecisionclass.Wormmaterial40Cr，surfacehardening.Hardness45～55HRC.GearMaterialZCuSn10P1，Sand-cast.Presetv2≤5m/s，andselecttheprecisionclass8(GB10085－88).2．AllowablecontactstressHPFromtable4－6,"200MPa.HPGearratioi=n1/n2=1470/120=12.2514707LifecyclenumberN60nt602300104.32102h12.25771010LifefactorZ880.833N7N4.3210"AllowablecontactstressZ0.833200166.6MPaHPNHP3．Determinethewormthreadnumberzandthegearteethnumberz.12z4；ziz12.254491214．Designforthecontactfatiguestrength.1）TorqueonthegearT2z4，estimate0.9.1610TTi9.551012.250.9716250Nmm2114702）LoadfactorKMediumimpactloadK1.2。A3）ElasticityfactorZESteelwormVs.Coppergear，Z160MPaE4）Determinem,dandd12 2ZE2md9KT()1A2z2HP1602391.2716250()2971.6mm49166.623Fromtable4－1，md31752971.6mm时，m6.3mm，d80mm，z4，111q12.698.dmz6.349308.7mm.225）Gearvelocityv2dn3.14308.712022v1.9m/s5m/s260100060000Accordwiththepresetvalue.6）Distancebetweenthetwocenters,a.a0.5(dd)0.5(80308.7)194.35mm127）Threadangleγz1m46.30arctan()arctan()1729"4"d8015．Checkthethermalbalance.1）SlidingvelocityvSdn80147011v6.4525m/sS060000cos60000cos1729"4"2）Equivalentfrictionanglev0Fromtable4－9,120"v3）Efficiency0tgtg1729"4"0.950.950.87800tg()tg(1729"4"120")v4）Heattransferareaa1.75194.351.752A0.33()0.33()1.0557m1001005）Workingoilthermal2Environmenttemperaturet0＝20℃.HeatemittingfactorKt＝15W/(m℃1000P1(1)100010(10.878)0tt2097t1opKA151.0557t 6．ResultanalysisWorkingtemperaturetisabovethepermittedvalue.Andsomeextraemittingequipmentmust1beadded.5－45-4DesignaV-beltdriveusinginacrusher.Given:ElectromotorY132S-4,P＝5.5kW.n1＝1440r/min,transmissionratioi＝2.Doubleshift.Thedistancebetweenthetwosheavecentersislimitedto600mm.Solution:1）WorkingpowerPcDoubleshiftmeans16hoursperday.Fromtable5-6,theworkingfactoforcrusher:KA=1.4Pc=KAP=1.4(1.6)×5.5=7.7(8.8)kW2）ThetypeforV-beltPc=7.7kW,n1=1440r/min，Fromfig5-7,selecttypeA.3）Sheavediameterdd1anddd2Fromtable5-7,selectdd1=100mm，ε=0.02did(1)2100(10.02)mm196mmd2d1Fromtable5-7,selectdd2=200mm.4）Beltvelocityvdd1n11001440vm/s7.536m/s601000601000 Between5～25m/s.5）Distancebetweenthetwocentersa,andthebeltlengthLda0,initialvalue210mm≤a0≤600mmSelecta0=400mm(300,500arealsookay).2(dd)L2a(dd)d2d1d00d1d224a023.14(200100)2400(100200)24400=800+471+6.25=1277.25Fromtable5-5，SelectLd=1400(1250)mmTheactualvalueofaaa(LL)2400(14001277.25)2461(385)mm0dd06）Smallsheave’swrapangleα1dd200100180d2d157.318057.31671201a461(165>120)（okay）7）beltnumberd200dd1=100mm，id22.04，v=7.33m/s，d(1)100(10.02)d1Fromtable5-2,P0=1.31kWFromtable5-3,ΔP1=0.1kWα1=167°，Fromtable5-4,Kα=0.968Ld=1400mm，fromtable5-5,KL=0.96(0.93)PP7.7zcc5.80[P](PP)KK(1.310.1)0.9680.96000Lz=6.8）InitialtensionforceF0 F500(2.5K)Pcqv2500(2.50.968)7.70.17.3320Kzv0.96867.33=138.544+5.373=143.917N9）ForceexertingonshaftFQ167F2zFsin126144sin1717NQ02210）Structuredesign.6-4（p152）6-4Atransmissionsystemisshownbelow.Vbelt,horizontal,forceQ=3000NThetransmissioninputaxial,T=510NmoWiththepiniond1=132.992mm,1=1210’38’’,b1=120mmThematerialissteel45.Initiallyselecttherollingbearing7300C.(1)Finishtheshaftstructurefigure.(2)Determinethediameteroftheshaft,usingthestresscombinationofbendingandtorque.(3)Thencheckthesafetyfactorforthefatiguestrength. Notice:1）轴承型号7300C指内径未定；2）按比例画出轴系装配图，不画轴承座及轴承盖；3）注意轴的转向及各分力的方向。Solution：1．SelecttheshaftmaterialSteel#45,quenchingandtempering.FromTable6-1,650MPa,360MPabsFromTable6-4,b650MPa,then[σ-1]b=60MPa.2．Calculatetheforcesandmoments1）Gearforcesod1=132.992mm,T=510Nm,1=1210’38’’ThentheGearforceshouldbe:2T2510000FN7669.6Ntd132.992tgtg20nFF7669.6N2855.8Nrtcoscos1210"38FFtg7669.6tg121038N1655.0Nat2）Forcescalculatingfigure(Forcesdirections) 3）Calculatethebearingreactionforces,andmomentInhorizontalplane,Rha=3096.8N,Rhb=-3241NPointc,Leftside,M"R100/1000=3096.8×100/1000N=309.68NmH(c)HARightside,M"309.68(1655.066.496/1000)309.68110.05419.73NmH(c)(=3000×(250+160)/1000-3241×250/1000=1230-810.25=419.75)Pointb,M3000160/1000480NmH(b)Inverticalplane,Rva=5478.3N,Rvb=2191.3N Pointc,MR100/1000=5478.3×100/1000N=547.83Nmv(c)VA3.Usingthecombinedstrengthtoestimatetheminimumdiameterofshaft1）Calculatethecombinedmoment22BendingmomentMMMHVPointc,leftside2222M"MM309.68547.83Nm629.3Nm(c)H(c)V(c)Pointc,rightside2222M"MM419.73547.83Nmm690.1Nmm(c)H(c)V(c)Torqueisperformedbetweenc-d. 2）Calculatetheequivalentmoment(combiningmomentandtorque)22MM(T)caSet0.6Pointc,rightside2222MM(T)690.1(0.6510)Nmca(c)(c)=754.9NmPointb,2222MM(T)480(0.6510)Nmca(b)(b)=569.24Nm3）Determinetheminimumdiameter,usingthecombinedmomentThepointcandpointdarethepotentialdangerouspoint.3FromTable6-2，W≈0.1d,fortheroundshaft.Thenforpointb(theminimumdiameter)M569.241000d3camm3mm45.6mm0.10.1601bM754.91000(Forpointc,d3camm3mm50.1mm)0.10.1601bThensetthediameterforthesectionb-d(Vbeltsheave)for50mm.4．StructuredesignfortheshaftThemainelementsmountedontheshaftincludethegear,Vbelt,andtherollingbearings.ThediameterfortheVbeltsheave,d=50mm.(lengthislessthan140mm,set136mm.)ThenthelocatingshoulderfortheVbeltsheavemustbe5-10mmmorethan50mm,andset57mm.Thediameterforthebearingisnon-locatingshoulder,2-4mmmorethan57mm,andset60mmforthebearinginnerdiameter.Andselect7312Cbearing.(fromthemanualin机械设计课程设计》，theinnerdiameter60mm,outerdiameter130mm,width31mm.)Thegearinnerdiameter,set65mm.(lengthislessthan120mm,set116mm.)Andtheshoulder75mm. 5015515510116607075656057136503131100250160----------5．CheckthesafetyfactorCrosssectionb-bisillustratedasfollowingBendingstress:M480000a325.9MPaW57/100mTorsionalstress：T5100006.9MPaa32W257/5T6.9MPama650MPa,360MPabsFromtheTable6-1,0.20.1Stressconcentrationstressfromattachedtable6-1k1.825k1.625Surfacequalityfactorandscalefactorfromattachedtable6-5and6-4 0.92(650MPa,R1.6m)0a0.84=0.78Then3001S4.9k1.82525.9am0.840.921551S9.5k1.6256.90.16.9am0.780.92SSS4.3S1.5ca22SSSocrosssectionb-bissafe.7－3（p181）7-3Designaradialhydrodynamicplainsurfacebearinginamotor.RadialloadF=60kN,od=160mm,n=960r/min，wrapangle180,theloadisstable.Thebearingmustbesplithorizontallywheninstallation.Solution:1.BearingstructureHorizontalsplitstructure，wrapangleis180°2.DeterminethewidthFromForm7-4,formotorl/d=0.6~1.5,Specifyl/d＝1.0Sowidthl=(l/d)×d=1.0×160=0.16m3.Calculatep，v,andpvF60000p2.344MPadl0.160.16dn3.14160960v8.04m/s60100060000pv2.3448.0418.84MPam/s4.ChoosematerialofbearingBasedonp,v,andpvfromTable7-2,selectZCuPb30,HBS=25 Table7－25.Chooseparameters0Suppose:tm50CSelectoil:L－AN32SofromFigure14-2Figure14－2γ=20mm2/sη＝γr=20×10－6×900＝30.2530.25(0.6~1)10v(0.6~1)1031.4(0.00142~0.00237) Adopt:ψ＝0.00196.ChecktheminimumoilfilmthicknessSurfacefinishRz1,Rz2Table7-5Table7-5selectRz1＝3.2mm，Rz2＝3.2mmLoadfactorCP22F600000.0019C4.677P2vl20.0188.040.16FromFigure7-16,l/d=1,CP=4.677,soselect＝0.85Figure7-16dh(1)800.001910.850.0228mmmin2Safetyfactor:h0.0228minS3.562RR0.00320.00321Z2ZTherequirementoftheminimumoilfilmthicknessissatisfied.7.Calculatetheoiltemperature FromFigure7-16frictionfactorCf=1.0Figure7-16FromFigure7-17fluxfactorCQ＝0.13Figure7-17CfF1.023440000t10.88C803.14scC17009000.13Qv0.00198.04Averageoiltemperature:t10.8800tt4045.44C50Cm122Sotheoiltemperatureissuitable.Fromtheaboveresultsthedesignispracticable. 8-2(p213)8-2Apairofangularcontactballbearingsareinstalledfacetoface,withthenominalcontactoangle=15.d=35mm，n=1800r/min，mediumimpact.Fr1=3390N,Fr2=1040N.AxialloadFA=870N.Determinetheexpectedworklifeforthebearings.12FAFFr1r2Solutions1．Selectthebearingtypeandcodeinitially.Angularcontactballbearing,withd=35mm.Soselect7307Cinitially.Fromthedesignmanual(<机械设计课程设计>p119,Form12-6)：C0r=26800N，Cr=34200NFromForm8-8：fp=1.5（mediumimpact）；FromForm8-5:ft=1.0（normally）.2．Calculatethederivedaxialforce.FromForm8-9,thederivedaxialforceS=0.5Fr.Thenthederivedaxialforceforthebearing1and2:S1=0.5Fr1=0.5×3390N=1695NS2=0.5Fr2=0.5×1040N=520N3．CalculatetheaxialloadFaforthebearing1and2.Thebearingsareinstalledfacetoface.FaS1S2S2+FA=520+870=1390P2，Usingbearing1forthecalculating.Fortheballbearingε=3.Fromtheequation（8-4）得66310ftCr10134200Lhh2209h60nP16018005514（Ifthelifeisnotlongenough,trytochoose7307AC）'