齿轮作业

Ex.6-1Apairofstandardspurinvolutegearshasamoduleof5mm,pressureangle=20,centredistancea=350mm,transmissionratioi12=9/5.Calculatethenumbersofteeth(Z1andZ2),referencediameters(d1andd2),addendumdiameters(da1andda2),basediameters(db1anddb2),tooththicknesss,spacewidthe,pressureanglesontheaddendumcircles(a1anda2),theradiiofcurvaturesoftoothprofileonthereferencecircles(1and2),andtheradiiofcurvaturesoftoothprofileontheaddendumcircles(a1anda2).Solution：593502'1'212'2'1ddimmddammdmmd450250'2'1mmmdzmmmdzmmddmmdd9054505052504502502211'22'11mmmpesmmddmmddmmmhdhddmmmhdhddobobaaaaaa854.7252286.42220cos*450cos92.23420cos*250cos4605*1*2450222605*1*2250222211*222*111列出公式、待入数据、得出结果三步缺一不可！Ex.6-2Howmanyteethwouldanexternalstandardspurinvolutegearhavewhenitsdedendumcircleanditsbasecirclecoincide?Whichoneisbiggerasthenumberofteethincreases?Ex.6-3ShownareapairofinvoluteprofilesC1andC2withthecommonnormaln-npassingthroughthecontactpointK.Thepinion1isadriver.nK2O1OnM12CC1(1)Drawthetwobasecirclesandtwopitchcircles.nK2O1OnM12CC11NN2rb1rb2r1''2rP(2)LabelthetheoreticallineofactionN1N2andactuallineofactionB1B2.nK2O1OnM12CC11NN22BB1rb1rb2r1''2rP(3)Labeltheworkingpressureangle’andpressureangleKatthepointK.αNK2nV1'rF2rb2O22n1r2KCKB'r1M1α'1CP2b1B1NO(4)LabeltheactualworkingsectionDGoftheprofileC2.D点常出错！nK2O1OnM1αKGD2CC11NN22BB1rb1rb2α'r1''2rP(5)FindoutpointM2ontheprofileC2whichwillengagewithpointM1ontheprofileC1.2ONαnr'12rb22nr1G2KCKB'r1'α1CDPb12BN1OMM1M2Ex.6-4Shownareapairofinvolutepinionandrackwiththeirpitchcircleandpitchline.DeterminegraphicallytheactuallineofactionB1B2,theactualworkingsectionEFonthetoothprofileofthepinion1andtheactualworkingsectionGHonthetoothprofileoftherack.1OPEx.6-4Shownareapairofinvolutepinionandrackwiththeirpitchcircleandpitchline.DeterminegraphicallytheactuallineofactionB1B2,theactualworkingsectionEFonthetoothprofileofthepinion1andtheactualworkingsectionGHonthetoothprofileoftherack.1BBPGFEH2O1G点常出错！1BBPGFEH2O1Ex.6-5Apairofexternalspurgearshavetheparametersas:Z1=10,Z2=27,m=10mm,=20,ha*=1,c*=0.25,workingcenterdistancea’=185mm.Neithergearhascutterinterference.(1)Whattypeofcorrectedgearpairdoesitbelongto?Why?(2)Whataretherangesofmodificationcoefficientsx1andx2,respectively?(3)Ifx1=0.5,calculatera2,rb2,S2,a2,anda2forgear2.列出公式、待入数据、得出结果三步缺一不可！Ex.6-5inexercisebook:Apairofexternalspurgearshavetheparametersas:Z1=10,Z2=27,m=10mm,=20,ha*=1,c*=0.25,workingcenterdistancea’=185mm.Neithergearhascutterinterference.(1)Whattypeofcorrectedgearpairdoesitbelongto?Why?Solution:a=m(Z1+Z2)/2=10*(10+27)/2=185mm=a’Thegeardrivemaybestandardgearpairorheightcorrectiongearpair(等变位传动).SinceZ117,thegeardriveshouldbeheightcorrectiongearpair.(2)Whatistherangeofmodificationcoefficientx1andx2,respectively?17717101717171min1Zx171017271717172min2Zx17717101x1717minZx17101772x(3)Ifx1=0.5,calculatera2,rb2,S2,a2,anda2forgear2.Solution:Sincethegeardriveisheightcorrectiongearpairandx1=0.5,x2=-0.5.r2=mZ2/2=10*27/2=135mmrb2=r2*cos=135*cos20°=126.8585mmForheightcorrectiongearpair,ra2=r2+ha*m+x2m=135+1*10-0.5*10=140mmS2=(/2+2x2tg)m=(/2-2*0.5*tg20°)*10=12.06826mm(3)Ifx1=0.5,calculatera2,rb2,S2,a2,anda2forgear2.219.598585.1261402222222baarra2=cos-1(rb2/ra2)=cos-1(126.8585/140)=25.0238°Ex.6-6Apairofexternalspurgearshavetheparametersas:Z1=20,Z2=40,m=2mm,=20,df1=37.0mm,S2=2.413652185mm.(1)Whatarethemodificationcoefficientsx1andx2?(2)Whattypeofcorrectedgearpairdoesitbelongto?Why?(3)Calculaterb1,ra1,a1,a1,S1,and1forgear1.sinsin1111aaaarrEx.6-7ThereisapairofexternalstandardspurgearsinashapingmachinewithZ1=17,Z2=118,m=5mm,=20,ha*=1,c*=0.25.Thepinioniswornoutandthegearisworntosuchanextentthatthetooththicknessisdecreasedby0.75mm.Thegearistoberepairedbyaddendummodificationandanewpinionistobemanufacturedtomeshwiththerepairedgear.Whatistheminimummodificationcoefficientofthenewpinioniftheoriginalframeisstillused?Ifx1=0.3,calculatera1,rf1,S1,a1,a1,and.X2应该等于－0.3,不应等于－0.2•x1+x2应该等于零。Ex.6-8Inthegeartrainshownbelow,bothgearpairsmusthavethesameworkingcenterdistance.Z1=27,Z2=60,Z2’=63,Z3=25,ha*=1,c*=0.25,m=5mm.Thegearpair2’and3isacorrectedgearpairwithreferencecenterdistance.Themodificationcoefficientx2ofthegear2iszero.(1)Whattypeofcorrectedgearpairshouldthegearpair1and2adopt?Givethebriefreason.(2)Isthegear2astandardgear?Why?Calculatex1,S1,rb1,rf1,a1,anda1.X2＝0，但轮2不是标准齿轮。5.0'21maaxxmxmhrraa1*11Ex.6-9:Givenapairofexternalspurgearswithm,Z1,i12,ha*,c*,,S1,rf2.Writeallformulaetocalculateaccordingtocalculationstep.Z2=Z1i12Solution:2)(21ZZmaS1=m/2+2x1mtg,x1=……rf2=mZ2/2-(ha*+c*)m+x2m,x2=….121122ZZaa(tantan')(tantan')’canbecalculatedfroma’•cos’=a•cos,a’=a•cos/cos’invxxZZinv'()tan21212Note:a’=a+x1m+x2m!!!!!!!ra1=a’-rf2-c*m)cos(cos)(cos1111111aabarrrrr1=mZ1/2r2=mZ2/2r1+ha*m+x1mrf1=r1-(ha*+c*)m+x1mra2=a’-rf1-c*m)cos(cos)(cos2212212aabarrrr121122ZZaa(tantan')(tantan')r2+ha*m+x2m不少人在未求出’的情况下，利用求a’。cos'cos'aaEx