基于Matlab的机构动力学仿真分析

:100320794(2005)0220051203Matlab,,(,324004):MatlabMatlabPSimulink,:;;;Matlab:TH122:B1,,,,,,2,,,,;,(),1770%ANSYSPTC(ParametricTechnologyCorporation)ProPEngineerSolidWorkscosmos,,,,,,,,,,,,,,-;,,,,,4CAE,,,,,,,,,:[1],1Internet[J]1,2002,11(6):214-217.[2],1[J]1(),2000,40(9):93-96.[3],,1Web[J].,2003,2(1):145-148.[4],1ANSYS[M].:,2003.[5],.ANSYS[M].:,2003.:(1973-),,,CADPCAEANSYS1Tel:[phone],E-mail:ningguifeng@1631com1:2004208207ResearchontheApplicationsofCAEonHydraulicPowerSupportNINGGui-feng1,MANCui-hua2(11ChinaCoalResearchInstitute,Beijing100013,China;21ChinaUniversityofMiningandTechnology,Beijing100083,China)Abstract:Thisarticleconcentratesmostlyonthefoundationofhydraulicpowersupportmodelanditsdynamicsimulationanalysis1ThearticlealsointroducesnonlinearsimulationanalysisoneccentricityloadoflegusingANSYSsoftware1Keywords:CAE;hydraulicpowersupport;dynamicsimulationanalysis;finiteelementmethod1520052©1994-2008ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.(FAx,FAy)T1Fig11MechanismsimpledrawingFAx+FBx=m1a1x(1)FAy+FBy=m1a1y+m1gn(2)FAxrC1sin1-FAyrC1cos1-FBx(r1-rC1)sin1+FBy(r1-rC1)cos1+T=J11(3)BC,-FBx+FCx=m2a2x(4)-FBy+FCy=m2a2y+m2gn(5)-FBxrC2sin2+FByrC2cos2-FCx(r2-rC2)sin2+FCy(r2-rC2)cos2=J22(6)-FCx+FPx=m3a3x(7)-FCy+FC1y=m3gn(8)(1)(8),:FAxFAyFBxFByFCxFCyFC1yT,8Matlab,1010000001010000A31A32A33A34000100-101000000-1010000A63A64A65A66000000-100000000110FAxFAyFBxFByFCxFCyFC1yT=m1a1xm1a1y+m1gnJ11m2a2xm2a2y+m2gnJ22m3a3x-Fpxm3gn(9)A31=rC1sin1,A32=-rC1cos1,A33=-(r1-rC1)sin1,A34=(r1-rC1)cos1,A63=-rC2sin2,A64=rC2cos2,A65=-(r2-rC2)sin2,A66=(r2-rC2)cos2212(9)a1xa1ya2xa2ya3x12,a1xa1ya2xa2y=-rC11sin1-rC121cos1rC11cos1-rC121sin1-r11sin1-r121cos1-rC22sin2-rC222cos2r11cos1-r121sin1+rC22cos2-rC222sin2(10)1r2sin20-r2cos2v3x2=r11sin1r11cos1(11)1r2sin20-r2cos2a3x2=-r11sin1-r121cos1-r222cos2r11cos1-r121sin1-r222sin2(12)(10)(11)(12),i,v3x,111,2,(11)v3x2(12)a3x2,(10)a1xa1ya2xa2y,a3x2v3x2,v3x223:r1=0105m,r2=0125m;:rC1=0105m,rC2=0125m;:m1=110kg,m2=012kg,m3=012kg;:J3=01001kgm2;3Fpx=1000N;25Matlab,20052©1994-2008ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.=100radPs;MatlabPSimulink,2sxhf1(9)M,sxhm1sxhm2,(10)(12)M,,sxhFpx51=0,2=0,x3=r1+r2;1=100radPs,(10)1=-20radPs,v3x=00112s,2,simout,simout12Fig12MechanismsimulationmodelMATLAB,34,,3Fig.3Mechanismsinulationresult,MatlabPSimulink,,,,:[1],1[M]1:,20001[2],,1[M]1:,19991[3],()1-MATLABSIMULINK[M]1:,20021[4],1MATLABPSimulink[M]1:,20021[5],1Simulink[J]1,2004,(4):35-371:(1973-),,,,CADPCAM.E-mail:sxh9363@hotmail1com.:2004211221DynamicsSimulationAnalyseofMechanismBasedonMatlabSONGXiao-hua,FANGKun-li,WUJun(WestBranchofZhejiangUniversityofTechnology,Quzhou324004,China)Abstract:ThisarticleintroducestheapplicationofMatlabsoftwareinkinematicanalysisofmechanism1AccordingtoNewtonianmechanicsprincipleandvectormethod,thekinematicsandkineticsmatrixequationofmechanismareestab2lishedandintered1UsingMATLABPSIMAULINK,carriesondynamicssimulationtotheequation,solvestheanti-forceofmechanisminthejointsandthebalanceforcemomentofmechanism1Keywords:dynamics;mechanism;simulation;Matlab3520052Matlab,©1994-2008ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.