机械原理英文版5齿轮

Geartrainsreviewn1=-n3=120rpmnH=600rpmDeterminenHDifferentialsDirectionofrotationGeartrainsDifferentialaddingmachinez3z1DifferentialsAdifferentialisadevicethatallowsadifferenceinvelocitybetweenintwoelementsGeartrainsdifferentials1zzzz1-nnnni2132434113）（a)DrivingpowerformZ5z1andz3rotateuniformlyb)Ifresistanceismetinz1z3willbefasterthanz1314nn2nGeartrainsDifferentialsGeartrainsDifferentialGeartrainsDifferentialsGeartrainsDifferentialsConstructiondifferentialsGeartrainspowertainCase1:A=1（离合器工作）,B=0nHdirectlyoutput=?n3=0Case2:A=0,B=1n3≠0,2132H3H1zzzznnnnn3Arm54653634zzzznnnninputn4=n1isgiven2132H3H1zzzznnnnn4=n1GeartrainspowertrainGeartrainspowertain3-speedautomobiletransmission1st:z5,z6meshingz3,z4separationA,Bseparation2nd:z3,z4meshingz5,z6separationA,Bseparation3th:A,Bmeshingz3,z4separationz5,z6separation4thBackz6,z8meshingA,Bseparationz3,z4separationz5,z6separationPowerinoutputTransmissioninautomobileBetterchoicethatspurgearsareusedinthecaseGeartrainspowertrainGeartrainspowertrain1’8’Consider:Speedratiointhefirstcaseandthesecondcase?CamDesignGearsinmachinetoolPowertransmissionsysteminmachinetoolCamDesignGearsinmachinetoolSpeedratiodiagramofpowertransmissionsysteminmachinetoolCamDesignGearsinmachinetoolGeartrainsAPPLICATIONSCamDesignEfficiencyThegeneraldefinitionofefficiencyisη=outputpower/inputpowerEfficiencyofaconventionalgeartrainisveryhigh.powerlosspergearset1~2%1)theexternalgearset；η=0.982)Anexternal-internalgearset:η=0.993）twostagesgearsetη=η1*η24)LossestimationEfficiencyofgeartrainsForepicyclictrains,theefficiencyisdependentondesign)(.z21z113f2fisthefrictioncoefficient5)Helicalgearsetη=96~98%,crosshelicalgearsetη=50~90%CamDesignefficiency1;ETTETT01H012;Efficiencyofepicyclegear1)TorquesbalanceT1+T2+TH=02)Powerin=poweroutfromshaft1toshaft2（β=+1）E0T1(ω1-ωH)+T2(ω2-ωH)=0HLHFH2H1H12R1iObtainthetorquesratiothrough(1)(2)1E1TT1;ETT;ETT02H01H012CamDesignefficiency-1i1i1H1HHH1H2H1Letgear2isfixed,ω2=0,thefromgear1toarmHTheefficiencyisthat1H1H11HHiTTTTbecauseand1ETT01Htherefore11EiTT01H1HSeecase5CamDesignefficiency1;ETT01Hb)fromARMtoshaft1（β=-1）Ifω2=0thegear2isfixed，powerfromHto1H1H1HH11iTTTT11ii-i-1H1H1H1HH1H2H1Because000E1)E1E1(Seethecase6CamDesignefficiency•Determinetheefficiencyformgear2toarm1E1TT02HBecauseω1=01-i1i1-2H2HH2HH2H11EiTT02H2Hthereforeρ1β=-1case7CamDesignefficiencyβ=+1Efficiencyfromgear1toarmβ=-1ρCamDesignefficiencyH1H1HH11iTTTT12HZ1=99,Zg=25,Zg’=25,z2=100E0=0.98Determinetheoverallefficiency199100ZgZ1Z2ZgiH12/solutionηH1β=-1000E1)E1E1(11ii-i-1H1H1H1HH1H2H1329099100980199100980..)(.CamDesignGeartrainExampleinp.308ConsiderthesameFerguson’sparadoxtrainasthefollowingfigure,whichhasthefollowingtoothnumbersandinitialconditionsSungear#2N2=100Sungear#3N3=99Sungear#4N4=101PlanetgearN5=20Inputtosungear#20prmInputtoarm100prmcounterclockwiseCamDesignGeartrainCamDesignGeartrain-.ω2=0solutionH2H3H2H3H32i=3CamDesignGeartrain4SOLUTIONTwoteethdifference,butoutputwillbigdifference4CamDesignHOMEWORK1)Givez1=z3=17,z2=z4=39,z5=18,z7=152,n1=1450rpmAisloosing,Bisbraking(switchon)DeterminerotationspeednH2)Given1=3549rpm,Z1=36,z2=60,z3=23,z4=49Z4’=69,z5=31,z6=131,z7=94,z8=36,z9=167DetermineoutputspeednHCamDesignHELICALGEAR3)WritecalculationvelocityformulaintheepicycletraininFigure9-35(l-k)Geartrainshomework3-speedautomobiletransmission1st:z5,z6meshingz3,z4separationA,Bseparation2nd:z3,z4meshingz5,z6separationA,Bseparation3th:A,Bmeshingz3,z4separationz5,z6separation4thBackz6,z8meshingA,Bseparationz3,z4separationz5,z6separationPowerinoutputWriteallspeedratio(1st~4th)inthetransmission4)CamDesignHOMEWORK5)Giveapairofhelicalgearparametersz1=20,z2=40,mn=8,αn=20,β=15oDeterminethedistancebetweengearandpinion,thediameterofaddendumcircle.