平面机构自由度计算Calculationofdegrees

平面机构自由度计算Calculationofdegreesoffreedomforplanarmechanisms主讲人：周会娜123目录CONTENTS平面机构自由度计算机构具有确定运动的条件例题解析CalculationofdegreesoffreedomforplanarmechanismsMechanismshaveconditionsfordeterminingmotionReflecExampleanalysistionquestions教学目标Teachingobjectives1.如何使组合起来的构件产生运动并具有运动确定性?2.机构的自由度3.机构具有确定性运动的条件Howtomakethecombinedcomponentsgeneratemotionandhavemotioncertainty?DegreeofFreedomoftheMechanismConditionsforMechanismtoHaveDeterministicMotion1平面机构自由度计算CalculationofdegreesoffreedomforplanarmechanismsONE平面机构自由度与组成机构的构件数目、运动副的数目及运动副的性质有关。Thedegreeoffreedomofaplanarmechanismisrelatedtothenumberofcomponentsthatmakeupthemechanism,thenumberofmotionpairs,andthepropertiesofthemotionpairs.平面机构自由度计算Calculationofdegreesoffreedomforplanarmechanisms1.低副Lowpair在平面机构中每个平面低副(转动副、移动副等)引入两个约束平面机构自由度计算Calculationofdegreesoffreedomforplanarmechanisms图1转动副Figure1Rotatingpair图2移动副Figure2MobileSub低副提供约束及自由度情况:引入2个约束，保留1个自由度。Lowpairprovidesconstraintsanddegreesoffreedom:introduce2constraintswhileretaining1degreeoffreedom.1.高副SeniorOfficer平面机构自由度计算Calculationofdegreesoffreedomforplanarmechanisms高副提供约束及自由度情况:限制了沿公法线n-n方向的相对移动保留了沿接触处切线t-t方向的相对移动和在平面内的相对转动。引入1个约束，保留2个自由度。Thehighpairprovidesconstraintsanddegreesoffreedom:itrestrictstherelativemovementalongthen-ndirectionofthecommonnormalandretainstherelativemovementalongthetangentt-tdirectionatthecontactpointandtherelativerotationintheplane.Introduceoneconstraintwhileretainingtwodegreesoffreedom.设机构共有N个构件TheestablishmenthasatotalofNcomponents活动构件数:n=N-1Numberofactivecomponents:n=N-1在没有用运动副联接之前，这些可动构件的自由度总数应为3n。当各构件用运动副连接起来之后,由于运动副引入的约束使构件的自由度减少Beforeconnectingwiththemotionpair,thetotalnumberofdegreesoffreedomofthesemovablecomponentsshouldbe3n.Aftereachcomponentisconnectedbyamotionpair,thedegreeoffreedomofthecomponentisreducedduetotheconstraintsintroducedbythemotionpair平面机构自由度计算Calculationofdegreesoffreedomforplanarmechanisms低副数:PLLownumberofpairs:PL高副数:PHHighnumberofpairs:PH机构自由度F:F=3n-(2PL+PH)

MechanismdegreeoffreedomF:F=3n-(2PL+PH)平面机构自由度计算Calculationofdegreesoffreedomforplanarmechanisms例1:计算下图所示的单缸内燃机机构的自由度Example1:Calculatethedegreesoffreedomofthesinglecylinderinternalcombustionenginemechanismshowninthefollowingfigure平面机构自由度计算Calculationofdegreesoffreedomforplanarmechanisms解:活动构件数:n=3Solution:Numberofactivecomponents:n=3低副数:PL=4Lowpairnumber:PL=4高副数:PH=0Highpairnumber:PH=0机构自由度:F=3n-(2PL+PH)=3x3-2x4=1Institutionaldegreeoffreedom:F=3n-(2PL+PH)=3x3-2x4=12机构具有确定运动的条件ThemechanismhascertainconditionsformotionTWO二、机构具有确定运动的条件Themechanismhascertainconditionsformotion机构institution自由度freedom原动件drivinglink结论conclusionF=3×3-2×4=11F=原动件数目，机构有确定运动F=numberoforiginalcomponents,withdefinitemovementofthemechanismF=3×3-2×4=12F<原动件数目，机构有确定运动F<numberoforiginalcomponents,mechanismhasdefinitemotion二、机构具有确定运动的条件Themechanismhascertainconditionsformotion机构institution自由度freedom原动件drivinglink结论conclusionF=3×4-2×5=21F>原动件数目，机构运动不确定F>ThenumberoforiginalcomponentsandthemechanismmotionareuncertainF=3×2-2×3=00F=0，机构为桁架结构，不能运动F=0,themechanismisatrussstructureandcannotmove机构具有确定性运动的条件:F>0，且原动件的数目等于自由度数TheconditionforthemechanismtohavedeterministicmotionisthatF>0andthenumberofactivecomponentsisequaltothenumberofdegreesoffreedom3例题解析ExampleanalysisTHREE例2:计算下图所示的摆动导杆机构的自由度Example2:Calculatethedegreesoffreedomoftheswingingguiderodmechanismshowninthefollowingfigure三、例题解析Exampleanalysis解:活动构件数:n=5Solution:Numberofactivecomponents:n=5低副数:PL=7Lowpairnumber:PL=7高副数:PH=0Numberofhighpairs:PH=0机构自由度:F=3n-(2PL+PH)=3x5-2x7=1Mechanismdegreeoffreedom:F=3n-(2PL+PH)=3x5-2x7=1例3:计算下图所示机构的自由度。如果机构中有1个原动件，请判断机构是否具有确定运动。Example3:Calculatethedegreesoffreedomofthemechanismshowninthefollowingfigure.Ifthereisoneactuatorinthemechanism,pleasedetermineifthemechanismhasadefinitemotion.三、例题解析Exampleanalysis解:活动构件数:n=7Solution:Numberofactivecomponents:n=7低副数：PL=10Lownumberofpairs:PL=10高副数：PH=0Numberofhighpairs:PH=0机构自由度:F=3n-(2PL+PH)=3x7-2x10=1Institutionaldegreeoffreedom:F=3n-(2PL+PH)=3x7-2x10=1F=1，且等于原动件的数目，机构具有确定性运动。F=1andequaltothenumberofactivecomponents,themechanismhasdeterministicmotion.例4:计算下图所示机构的自由度。如果机构中有1个原动件，请判断机构是否具有确定运动。Example4:Calculatethedegreesoffreedomofthemechanismshowninthefollowingfigure.Ifthereisoneactuatorinthemechanism,pleasedetermineifthemechanismhasadefinitemotion.三、例题解析Exampleanalysis解:活动构件数:n=3Solution:Numberofactivecomponents:n=3低副数：PL=4Lowpairnumber:PL=4高副数：PH=1Highpairnumber:PH=1机构自由度:F=3n-(2PL+PH)=3x3-2x4-1=0Institutionaldegreeoffreedom:F=3n-(2PL+PH)=3x3-2x4-1=0F=0，且小于原动件的数目，机构不能动。F=0andlessthanthenumberoforiginalcomponents,themechanismcannotmove.自测题——计算题Selftestquestions-CalculationQuestion1计算下图所示机构的自由度。如果机构中有1个原动件，请判断机构是否具有确定运动。Calculatethedegreesoffreedomofthemechanismshowninthefigurebelow.Ifthere