吉林大学机械原理双语课件3.Kinematics-and-forces-analysis-of--planar-mechanisms

Chapter3Kinematicsandforcesanalysisofplanarmechanisms,(第3章平面机构的运动分析和力分析),Problems：,Positions(轨迹)/Angularpositions(角位移）Velocities(速度)/Angularvelocities(角速度）Accelerations(加速度)/AngularAccelerations(角加速度）.,Purpose：,Itisabaseforforceanalysistounderstandthekinematicspropertiesofexistingmechanisms(了解现有机构的运动性能，为受力分析奠定基础).,Means：,1)Instantcenter(瞬心法)（vand）;2)VectorequationGraphicalMethod(矢量方程图解法)3)Analysismethod(解析法)（上机计算）.,Tasksandmethodsofkinematicanalysisofmechanisms(机构运动分析的目的和方法),3.1Instantcenterofvelocity,(速度瞬心),3.1.1Instantcenterofvelocity(速度瞬心),Thedefinitionofaninstantcenterofvelocityisacoincidentpoint,commontotwolinks/bodiesinplanemotion,whichpointhasthesameabsolutevelocityandnorelativeveloctiy(两个互作平行平面运动的构件上绝对速度相等、相对速度为零的瞬时重合点称为这两个构件的速度瞬心,简称瞬心。瞬心用符号Pij表示).,e.g.P12istheinstantcenterofvelocitybetweenlink1andlink2.,1.Absoluteinstantcenter(绝对瞬心),绝对速度为零的瞬时重合点。,2.Relativeinstantcenter(相对瞬心),绝对速度不为零的瞬时重合点。,3.1.2Numberofinstantcentersofamechanism(机构中瞬心的数目),DenotedasK：,Note:TheframeisincludedinthenumberN.注意:N为包括机架在内的所有构件数。,3.1.3Locationoftheinstantcenteroftwolinksconnectedbyakinematicpair(机构中瞬心位置的确定),1.Observableinstantcenter(直接构成运动副的两构件的瞬心),(1)Revolutepair(转动副连接),Thecenteroftherevolutepairistheinstantcenter(铰链点即为瞬心).,(2)Slidingpair(移动副连接),Infinityineitherdirectionperpendiculartotheguide-way(瞬心在垂直于导路无穷远处),(3)Planarhigherpairs(平面高副)(rollingandslidingpair),Somewhereonthecommonnormaln-nthroughtheconnectingpointA(瞬心在接触点的公法线上；若为纯滚动，瞬心在接触点上),Ifv120,P12liesonthecommonnormaln-n.,Ifv12=0，P12isApoint；,2.TheoremofthreecentersAronhold-KennedyTheorem(三心定理)(unobervableinstantcenters),proof：,SupposeP23isatKpoint，as,Toobtainthesamedirectionoftwovelocity,KpointmustlieonthelinethroughP12andP13points.,So，P12,P23andP13mustlieonthesameline.,vK2vK3,Example1：Forthefour-barmechanismshownasfollow,locateallinstantcentersforthemechanism.,Solution：,P12、P23、P34、P14(observableinstantcenters),P13,P14、P34,P12、P23,P24,P12、P14,P23、P34,P24,P13,Thesolutionofeachlinkinstantcentersrepresentbyapolygongraph.Alinebetweenthedotsrepresentingthelinkpairseachtimewefindaninstantcenter.Thetrianglecomposedbylinesrepresentsthreeinstantcentersandthreecentersarecollinear.,Example：,Example2：Findingallinstantcentersforaslider-crankmechanism。,Solution：,3.1.4Velocityanalysiswithinstantcenters,Example1:w1isgivenintheslider-crankmechanism,findtheratiooftransmissioni13andw3.,Solution：,Solution:,Example2:w1isgiveninthecammechanism,findthevelocityv2ofthefollower2.,Exercise:forthefive-barlinkage,supposethatthedrivinglinks2and3arerotatingwithconstantangularvelocityw2=10rad/s(ccw)andw3=5rad/s(cw),respectively.Determinethecouplersw4andw5.,2,1,4,5,3,3.3.1Goalsandcontentsofforcesanalysisconsideringinfluencesoffriction,Goals：,Analyzetheeffectsoffrictioninkinematicpairsinmechanismsandreducethedisadvantages.,Contents：,1）Introductionoffrictioninseveralkinematicpairs；2）Forceanalysisconsideringinfluencesoffriction；3）Mechanicalefficiencyandself-locking.,3.3Forcesanalysisofmechanisms,(机构的受力分析),1.Frictioninkinematicpairs,1）determinationoffrictionforceinslidingpairs,FR21compositeforce(合外力)FN21normalreactionforce(法向反力)；F21friction(摩擦力)。,F21,direction：directionofF21isoppositetothatofv12.magnitude：F21isproportionaltoFN21withfrictioncoefficientfbeingacertainvalue.F21=fFN21,3.3.2Forceanalysisconsideringinfluencesoffriction,（1）Frictioninslidingpairs,2）Determinationofreactionforceinkinematicpairs,FN21andF21aretheforceslink2actingonlink1andthereactionforceFR21isthecompositionofthem.,direction：anglebetweenFR21andv12is(90+j),magnitude：,supposeincludedanglebetweenFR21andFN21isj,j=arctanf，frictionangle。,FR21,Underdifferentsituations,reactionforceshouldbeasfollows:,当外力F的作用线位于接触表面ab之内时构件1与构件2仅一面受力，FR21如图a所示。,当外力F的作用线位于接触表面ab之外时构件1除了移动之外，还要发生倾转，FR21如图b所示。,当外力F的作用线平行移动轴线并距移动轴线h时，构件1除了移动之外，还要发生倾转，FR21如图c所示。,Example:Link1movesataconstantspeedonaslantoflink2，loadingisFQ,frictioncoefficientisf,drivingforceisF(horizontal)andaisgiven。DetermineFR21=?F=?whenlink1movealongaslantupanddownataconstantspeed.,Solution：1)Forceanalysisonlink1whenitmoveupataconstantspeed.,FQ+F+FR21=0,F=FQtan(a+j),FR21=FQ/cos(a+j),2)Forceanalysisonlink1whenitmovedownataconstantspeed.,Conclusion：thedifferentdirectionofFR21isduetothedifferentdirectionofrelativevelocity.,FQ+F+FR21=0,F=FQtan(a-j),FR21=FQ/cos(a-j),2.Frictioninrevolutepairs（radialshaftfriction径向轴颈摩擦）,1)Fig.a，当构件1与构件2没有相对转动时，其接触点在A点，此时FN21=-FQ，此二力共线，等值反向。,2)Fig.b，在Md的驱动下使轴颈匀速转动，由于摩擦力的存在，构件1在构件2的AB弧段向上爬升，直至B点达到平衡。,平衡条件为,Fy=0FQ=FR21,MO=0Md=Mr,fv(当量摩擦系数)通过理论推导，有,跑合轴颈：fv=1.27f,非跑合轴颈：fv=1.57f,Mr=F21r=FR21r,根据力学观点，若Md与FQ的总合力为FQ，其距离为e，,1)当eMr,构件1作加速转动。,在考虑摩擦受力分析时，常将FQ用FRij表示。,r=fvr为定值。,式中r称为摩擦圆半径。,转动副中总反力的确定：,2)FRAB对轴心取矩的方向与wBA转向相反；,1)FRAB作用线切于摩擦圆；,3)根据整体平衡条件确定FRAB的唯一确切位置。,整体平衡条件包括：若该构件为二力构件，明确其受拉还是受压；若该构件是三力构件，此三力必汇交于一点；若该构件受一个力矩和两个力作用，此时该二力必构成一力偶与力矩达到平衡；等等。,FRAB,步骤如下：,1)明确机构中驱动力（矩）和阻力（矩），搞清运动趋势。,2)画受力图。必须从受力最少构件入手，然后分别画出其他构件受力图。,3)求未知量。从有已知力（矩）的构件入手，根据平衡条件求出未知力（矩），然后以此构件为基准，由近及远分析其它构件，直至求出未知量。,2.考虑摩擦的机构受力分析,Example1:Asshowninfigure,four-barlinkage,positions,sizes,fv,r,drivingforceFandresistanttorqueM3ofthemechanismaregiven,findthepositionanddirectionofreactionforcesineachkinematicpairbytheeffectofF.,Solution：,Link2:,Link1:,Link3:,r=fvr,FR12+FR32=0,F+FR21+FR41=0,FR23+FR43=0M3=FR23h,Solution：,Link2：,Link3：,Link1：,r=fvr，j=arctanf,FR12+FR32=0,FQ+FR23+FR43=0,FR21+FR41=0M1=FR21h,ChoosemN，FR21=M1/hFQ=FQmN,Solution：,Link1：,Link2：,F+FR21+FR31=0,FQ+FR12+FR32=0,FQ,F,1,2,3,4,Example4:Positions,sizes,frictionanglejbetweentheslidingpairandfrictioncirclesaregiveninthefollowingclampingmechanism，PleasedrawallreactionforcelinesofkinematicpairunderthedrivingforceFandt