期末复习课件

Chapter3Kinematicsandforcesanalysisofplanarmechanisms

(第3章平面机构的运动分析和力分析)w14w21w23v34jM1FR21FR12FR32FR23FR43FR41h1234FProblems：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)（vandω）;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)3.1

Instantcenterofvelocity(速度瞬心)3.1.1

Instantcenterofvelocity(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).P1212ABvA1A2vB1B2e.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。3.1.2

Numberofinstantcentersofamechanism(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。3.1.3

Locationoftheinstantcenteroftwolinksconnectedbyakinematicpair(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)12P1212P12Thecenteroftherevolutepairistheinstantcenter(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)v121212v12(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)Ifv12≠0,P12liesonthecommonnormaln-n.Ifv12=0，P12isApoint；nnttA21v122.TheoremofthreecentersAronhold-KennedyTheorem(三心定理)(unobervableinstantcenters)Theorem：

Threeindepentdentlinksingeneralplanemotionhavethreeinstantcentersandthethreeinstantcentersmustlieonastraightline.proof：Suppose

P23

isat

Kpoint，asToobtainthesamedirectionoftwovelocity,KpointmustlieonthelinethroughP12and

P13points.So，P12,P23andP13

mustlieonthesameline.23P12P131vK2vK3w2w3vK2≠vK31234Example1：Forthefour-barmechanismshownasfollow,locateallinstantcentersforthemechanism.Solution：P12、P23、P34、P14(observableinstantcenters)P13{P14、P34P12、P23P24{P12、P14P23、P34P14P34P12P23P24P13Lineargraph/Instantcenterpolygon

框图法(瞬心多边形)

Thesolutionofeachlinkinstantcentersrepresentbyapolygongraph.Alinebetweenthedotsrepresentingthelinkpairseachtimewefindaninstantcenter.Thetrianglecomposedbylinesrepresentsthreeinstantcentersandthreecentersarecollinear.Example：1234P12P24P13P14P34P23Example

2：Findingallinstantcentersforaslider-crankmechanism。Solution：14321234P12P24P13P14P34P233.1.4

VelocityanalysiswithinstantcentersExample1:w1isgivenintheslider-crankmechanism,findtheratiooftransmission

i13and

w3.1234w1P12P34P14P23∞P13Solution：1234P13ω3Solution:123w1nP13P23∞nP12P23∞Example2:

w1

isgiveninthecammechanism,findthevelocityv2ofthefollower2.Exercise:forthefive-barlinkage,supposethatthedrivinglinks2and3arerotatingwithconstantangularvelocityw2

=10rad/s(ccw)and

w3=5rad/s(cw),respectively.Determinethecouplersw4andw5.

214533.3.1GoalsandcontentsofforcesanalysisconsideringinfluencesoffrictionGoals：Analyzetheeffectsoffrictioninkinematicpairsinmechanismsandreducethedisadvantages.Contents：1）Introductionoffrictioninseveralkinematicpairs；2）Forceanalysisconsideringinfluencesoffriction；3）Mechanicalefficiencyandself-locking.3.3

Forcesanalysisofmechanisms(机构的受力分析)1.Frictioninkinematicpairs1）determinationoffrictionforceinslidingpairsFR21—composite

force(合外力)FN21—normal

reactionforce(法向反力)；F21—friction(摩擦力)。F21direction：directionof

F21isoppositetothatof

v12.magnitude

：

F21

isproportionalto

FN21withfrictioncoefficient

fbeingacertainvalue.

F21=fFN213.3.2

Forceanalysisconsideringinfluencesoffriction12FtFN21FR21FFnbF21jv12ab（1）Frictioninslidingpairs2）Determinationofreactionforce

inkinematicpairsFN21

and

F21aretheforceslink2actingonlink1andthereactionforceFR21isthecompositionofthem.direction：anglebetween

FR21and

v12

is

(90°+j)

magnitude

：12FtFN21FR21FFnbF21jv12ab

supposeincludedanglebetween

FR21and

FN21is

j∴j=arctanf，frictionangle。FR21Underdifferentsituations,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。

jFR21v122F1abdc(a)21jcdbajFFR21FR21v12(b)jj21cdbahv12FR21FFR21(c)Example:

Link1movesataconstantspeedonaslantoflink2，loadingisFQ,frictioncoefficientisf,drivingforceisF(horizontal)andaisgiven。Determine

FR21

=?F=?whenlink1movealongaslantupanddownataconstantspeed.Solution：1)Forceanalysisonlink1whenitmoveupataconstantspeed.FQ12aFaj1(a+j)FQFv12FR21FR21FFQFQ+F+FR21=0F=FQtan(a+j)FR21=FQ/cos(a+j)2)Forceanalysisonlink1whenitmovedownataconstantspeed.Conclusion：thedifferentdirectionof

FR21

isdueto

thedifferentdirectionofrelativevelocity.1a(a-j)FQFv12jFR21FQFR21FFQ+F+FR21=0F=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12FQFN21(a)ArFQFN21F21FR21MdrOB12(b)∑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=1.57fMr=F21r

=FR21rrFQFR21rMdO根据力学观点，若Md与FQ的总合力为FQ′，其距离为e，1)当e<r时，FQ′作用在摩擦圆内,此时Md<Mr,机构若原来运动则减速直至静止,若原来静止则自锁。2)当e=

r

时，FQ′切于摩擦圆，此时Md=Mr,构件1达到惯性平衡。3)当e>r时，FQ′作用在摩擦圆外，此时Md>Mr,构件1作加速转动。rFQFR21rFQFR21rrFQFR21MdOBrrFR21OBeFQ′rrFR21OBeFQ′rrFR21OBeFQ′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,drivingforce

F

andresistanttorque

M3

ofthemechanismaregiven,findthepositionanddirectionof

reactionforcesineachkinematicpairbytheeffectofF.Solution：Link

2:Link1:Link3:1234FM3w

21w

23FR12FR21FR23FR32w

14w

34FR41FR43hr=fvrFR12+FR32=0F+FR21+FR41=0FR23+FR43=0M3=FR23hSolution：Link

2：Link

3：Link

1：Example2:Asshowninfigure,guide-barmechanism,positions,sizes,fv

,r,drivingtorqueM1ofthemechanismaregiven.

FQ

isresistance,findtheresistance

FQ

=?M1FQ341232w21v23FR12FR32FR43w34FR23FR21w14FR41h受力面FR43FQFR23FR41FR21FR12FR32r=fvr，j=arctanfFR12+FR32=0FQ

+FR23+FR43=0FR21+FR41=0M1=FR21hChoosemN，FR21=M1/h∴FQ=∣FQ∣mNExample3:Positions,sizes,fv

,r,

drivingforce

FandresistanceFQaregiven.Findthereactionlineoftotalreactionforce.123FFQSolution：Link

1：Link

2：v21w23FR32FR31jv13FR21FR12jF+FR21+FR31=0FQ

+FR12+FR32=0FQF1234Example4:Positions,sizes,frictionanglejbetweentheslidingpair

andfrictioncirclesaregiveninthefollowingclampingmechanism，Pleasedrawallreactionforcelinesofkinematicpairunderthedrivingforce

F

andtheresistanceforce

FQ.Solution

：Link1：Link

3：Link2：jw21w23FR21w14FR41FR23v34jFR43FR12FR32FR12+FR32=0F+FR21+FR41=0FQ

+FR23+FR43=0(1)机械效率机械对能量的有效利用程度，用h

表示。1)简化机械系统，减少运动副。2)减少摩擦，合理选材。3.Mechanicalefficiencyandself-locking(机械效率和自锁)式中：Wd