期末復(fù)習(xí)課件

Chapter3Kinematicsandforcesanalysisofplanarmechanisms

(第3章平面機(jī)構(gòu)的運(yùn)動分析和力分析)w14w21w23v34jM1FR21FR12FR32FR23FR43FR41h1234FProblems：Positions(軌跡)/Angularpositions(角位移）Velocities(速度)/Angularvelocities(角速度）Accelerations(加速度)/AngularAccelerations(角加速度）.Purpose：Itisabaseforforceanalysistounderstandthekinematicspropertiesofexistingmechanisms(了解現(xiàn)有機(jī)構(gòu)的運(yùn)動性能，為受力分析奠定基礎(chǔ)).Means：1)Instantcenter

(瞬心法)（vandω）;2)VectorequationGraphicalMethod(矢量方程圖解法)3)Analysismethod(解析法)（上機(jī)計算）.Tasksandmethodsofkinematicanalysisofmechanisms(機(jī)構(gòu)運(yùn)動分析的目的和方法)3.1

Instantcenterofvelocity(速度瞬心)3.1.1

Instantcenterofvelocity(速度瞬心)Thedefinitionofaninstantcenterofvelocityisacoincidentpoint,commontotwolinks/bodiesinplanemotion,whichpointhasthesameabsolutevelocityandnorelativeveloctiy(兩個互作平行平面運(yùn)動的構(gòu)件上絕對速度相等、相對速度為零的瞬時重合點(diǎn)稱為這兩個構(gòu)件的速度瞬心,簡稱瞬心。瞬心用符號Pij表示).P1212ABvA1A2vB1B2e.g.P12istheinstantcenterofvelocitybetweenlink1andlink2.1.Absoluteinstantcenter(絕對瞬心)

絕對速度為零的瞬時重合點(diǎn)。2.Relativeinstantcenter(相對瞬心)

絕對速度不為零的瞬時重合點(diǎn)。3.1.2

Numberofinstantcentersofamechanism(機(jī)構(gòu)中瞬心的數(shù)目)DenotedasK：Note:TheframeisincludedinthenumberN.注意:N為包括機(jī)架在內(nèi)的所有構(gòu)件數(shù)。3.1.3

Locationoftheinstantcenteroftwolinksconnectedbyakinematicpair(機(jī)構(gòu)中瞬心位置的確定)1.Observableinstantcenter(直接構(gòu)成運(yùn)動副的兩構(gòu)件的瞬心)(1)Revolutepair(轉(zhuǎn)動副連接)12P1212P12Thecenteroftherevolutepairistheinstantcenter(鉸鏈點(diǎn)即為瞬心).(2)Slidingpair(移動副連接)Infinityineitherdirectionperpendiculartotheguide-way(瞬心在垂直于導(dǎo)路無窮遠(yuǎn)處)v121212v12(3)Planarhigherpairs(平面高副)(rollingandslidingpair)Somewhereonthecommonnormaln-nthroughtheconnectingpointA(瞬心在接觸點(diǎn)的公法線上；若為純滾動，瞬心在接觸點(diǎn)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

框圖法(瞬心多邊形)

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(機(jī)構(gòu)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:

當(dāng)外力F

的作用線位于接觸表面ab之內(nèi)時

構(gòu)件1與構(gòu)件2僅一面受力，F(xiàn)R21如圖a所示。

當(dāng)外力F的作用線位于接觸表面ab之外時構(gòu)件1除了移動之外，還要發(fā)生傾轉(zhuǎn)，

FR21如圖b所示。

當(dāng)外力F的作用線平行移動軸線并距移動軸線h時，構(gòu)件1除了移動之外，還要發(fā)生傾轉(zhuǎn)，

FR21如圖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徑向軸頸摩擦）1)Fig.a，當(dāng)構(gòu)件

1與構(gòu)件2沒有相對轉(zhuǎn)動時，其接觸點(diǎn)在A點(diǎn)，此時FN21=-FQ

，此二力共線，等值反向。2)Fig.b，在Md的驅(qū)動下使軸頸勻速轉(zhuǎn)動，由于摩擦力的存在，構(gòu)件1在構(gòu)件2的AB弧段向上爬升，直至B點(diǎn)達(dá)到平衡。平衡條件為12FQFN21(a)ArFQFN21F21FR21MdrOB12(b)∑Fy=0FQ=FR21∑MO=0Md=Mr

fv(當(dāng)量摩擦系數(shù))通過理論推導(dǎo)，有跑合軸頸：fv=1.27f非跑合軸頸：fv=1.57fMr=F21r

=FR21rrFQFR21rMdO根據(jù)力學(xué)觀點(diǎn)，若Md與FQ的總合力為FQ′，其距離為e，1)當(dāng)e<r時，F(xiàn)Q′作用在摩擦圓內(nèi),此時Md<Mr,機(jī)構(gòu)若原來運(yùn)動則減速直至靜止,若原來靜止則自鎖。2)當(dāng)e=

r

時，F(xiàn)Q′切于摩擦圓，此時Md=Mr,構(gòu)件1達(dá)到慣性平衡。3)當(dāng)e>r時，F(xiàn)Q′作用在摩擦圓外，此時Md>Mr,構(gòu)件1作加速轉(zhuǎn)動。rFQFR21rFQFR21rrFQFR21MdOBrrFR21OBeFQ′rrFR21OBeFQ′rrFR21OBeFQ′在考慮摩擦受力分析時，常將FQ′用FRij表示。r=fvr

為定值。式中r稱為摩擦圓半徑。轉(zhuǎn)動副中總反力的確定：2)FRAB對軸心取矩的方向與wBA轉(zhuǎn)向相反；1)FRAB作用線切于摩擦圓；3)根據(jù)整體平衡條件確定FRAB的唯一確切位置。整體平衡條件包括：若該構(gòu)件為二力構(gòu)件，明確其受拉還是受壓；若該構(gòu)件是三力構(gòu)件，此三力必匯交于一點(diǎn)；若該構(gòu)件受一個力矩和兩個力作用，此時該二力必構(gòu)成一力偶與力矩達(dá)到平衡；等等。FRAB步驟如下：1)明確機(jī)構(gòu)中驅(qū)動力（矩）和阻力（矩），搞清運(yùn)動趨勢。2)畫受力圖。必須從受力最少構(gòu)件入手，然后分別畫出其他構(gòu)件受力圖。3)求未知量。從有已知力（矩）的構(gòu)件入手，根據(jù)平衡條件求出未知力（矩），然后以此構(gòu)件為基準(zhǔn)，由近及遠(yuǎn)分析其它構(gòu)件，直至求出未知量。2.考慮摩擦的機(jī)構(gòu)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，F(xiàn)R21=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)機(jī)械效率機(jī)械對能量的有效利用程度，用h

表示。1)簡化機(jī)械系統(tǒng)，減少運(yùn)動副。2)減少摩擦，合理選材。3.Mechanicalefficiencyandself-locking(機(jī)械效率和自鎖)式中：Wd