汽车悬架的二自由度建模方法及分析

詹长书,吕文超(,150040):采用二自由度1/4车辆模型对汽车悬架进行了建模分析首先建立了微分方程和状态空间模型,其次使用了频域响应法和拉氏变换法分析了模型,并绘制了二者的频率响应曲线图,验证了模型的准确性这为后续的位移速度加速度和力反馈控制奠定了数学模型基础:;;;;Matlab:U463.33:B:1006-0006(2010)06-0009-03ModelingandAnalysisof2DOFVehicleSuspensionZHANChangshu,LUWenchao(TrafficCollege,NortheastForestryUniversity,Harbin150040,China)Abstrac:tWith2DOF(degreeoffreedom)quartercarmode,lautomotivesuspensionismodeledtosimulateinthisresearch.Firstdifferentialequationandstatespacemodelarebuilt,thenthemodelisanalyzedbyfrequencyresponseandLaplacetransform.Finallythefrequencyresponsefigureisdrew.Theresultsshowthemodeliscorrect,whichcanprovidethemathematicalmodelbasisforsubsequentdisplacement,velocity,accelerationandforcefeedbackcontro.lKeywords:Vehiclesuspension;Twodegreefreedommode;lSimulation;Frequencydomainresponse;Matlab,,,,,1/4,[1];Hrovat[2];[3];[4]1/4;GaoHuijun[5]1/41/41/4,11/4=1,,1/4,(1)msxs+ks(xs-xu)+cs(xs-xu)=0muxu-ks(xs-xu)-cs(xs-xu)+kt(xu-xr)=0(1),msmu;kskt;xrxsxu;cs1Fig.1TwoDOFSuspensionModel2:()(),:2010-08-309376Vo.l37No.6201012Tractor&FarmTransporterDec.,2010,(2),f0Zr=-2f0Zr+2G0vw(t)(2),G0;v;w(t);Zr1/4,Matlab/Simulink,312Fig.2PavementInput31/4SmiulinkFig.3SmiulinkModelofTwoDOFQuarterCar1Tab.1MainParametersofSuspensionSystem/kg1010/kg175/(Nm-1)756000/(Nm-1)100000/(Nsm-1)6600,X=[x1,x2,x3,x4]T=[xu,xs,xu,xs]TY=[y1,y2]T=[xu,xs]T(3),ABCDX=AX+Bu(t)Y=CX+Du(t)(3),A=00100001-ks+ktmuksmu-csmucsmuksms-ksmscsms-csms;B=00ktmu0;C=-ks+ktmuksmu-csmucsmuksms-ksmscsms-csms;D=ktmu03Simulink,,Matlabtfestimate,43,muxr,,1062010124Fig.4FrequencyresponseCurvexr(t)=eitxu(t)=H()xu~xreitxu(t)=iH()xu~xreitxu(t)=-2H()xu~xreitxr(t)=eitxs(t)=H()xs~xreitxs(t)=iH()xs~xreitxs(t)=-2H()xs~xreit(1)(-2ms+ics+ks)H()xs~xr=(ics+ks)H()xu~xr(4)(-2mu+ics+kt+ks)H()xu~xr=(ics+ks)H()xs~xr+kt(5)A1=ics+ksA2=-2ms+ics+ksA3=-2mu+ics+kt+ks(4)(5)xu~xrH()xu~xr=A2ktA3A2-A21,,xu~xr|H()|xu~xr=(1-2)2+42212(6),=(1-2)1+-2-12+422(-1+122;=0;0=ksms;=cs2ksms;=ktks;=msmuxs~xrH()xs~xr=A1ktA3A2-A21xs~xr|H()|xs~xr=1+42212(7)5,,,,5Fig.5AmplitudefrequencyCurve41/41/4(1),(mss2+css+ks)xs(s)-(css+ks)xu(s)=0(8)(mus2+css+ks+ku)xu(s)-(css+ks)xs(s)=kuxr(s)(9)()HT=xu(s)xr(s)=kt(mss2+css+ks)s(10)()HS=xs(s)xr(s)=kt(css+ks)s(11),s=mumss4+(mscs+mucs)s3+(msks+mskt+muks)s2+ktcss+kskt(10)(11)Matlabbode(6)6Fig.6LaplaceTransformCurve,(15)11:92Fig.9SectionNephogramofCase2102Fig.10SurfaceNephogramofCase2112Fig.11DeformationMapofCase2,,3ABAQUS,,,,:1)2),,,3),4),,,:[1].[M].:,2001.[2],,,.ABAQUS[M].:,2004.[3].[J].,2009(1):114-116121.()(11)-40dB/-60dB/51/4,,1/4:[1],,.[M].:,2002.[2]HROVATD.SurveyofAdvancedSuspensionDevelopmentsandRelatedOptimalControl[J].ApplicationsAutomatica,1997,33(10):1781-1817.[3]CH.[M].:,2002.[4],,.[J].,2006(11):233-236.[5]GAOHuijun,LAMJames,WANGChanghong.MultiobjectiveControlofVehicleActiveSuspensionSystemsViaLoaddependentControllers[J].JournalofSoundandVibration,2006(290):654-675.():詹长书(1970-),,,,15: