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1、习题>答案一.概念1.当两构件组成转动副时,其相对速度瞬心在转动副的圆心处;组成移动副时,其瞬心在垂直于移动导路的无穷远处;组成滑动兼滚动的高副时,其瞬心在接触点两轮廓线的公法线上.2.相对瞬心与绝对瞬心相同点是都是两构件上相对速度为零,绝对速度相等的点,而不同点是相对瞬心的绝对速度不为零,而绝对瞬心的绝对速度为零.3.速度影像的相似原理只能用于同一构件上的两点,而不能用于机构不同构件上的各点.4.速度瞬心可以定义为互相作平面相对运动的两构件上,相对速度为零,绝对速度相等的点.5.3个彼此作平面平行运动的构件共有3个速度瞬心,这几个瞬心必位于同一条直线上.含有6个构件的平面机构,其
2、速度瞬心共有15个,其中5个是绝对瞬心,有9个相对瞬心.二.计算题1、2.关键:找到瞬心P366Solution:ThecoordinatesofjointBareyB=ABsinφ=0.20sin45°=0.141mxB=ABsinφ=0.20sin45°=0.141mThevectordiagramoftherightFigisdrawnbyrepresentingtheRTR(BBD)dyad.Thevectorequation,correspondingtothisloop,iswrittenasrB+r-rD=0orr=rD-rBWherer=BDandr=γ.Whent
3、heabovevectorialequationisprojectedonthexandyaxes,twoscalarequationsareobtained:Br*cos(φ3+π)=xD-x=-0.141mr*sin(φ3+π)=yD-yB=-0.541mAngleφ3isobtainedbysolvingthesystemofthetwopreviousscalarequations:0.541tgφ3=0.141φ3=75.36°ThedistancerisxDxBcos(3)r==0.56mThecoordinatesofjointCarexC=CDcosφ3=0.17m
4、yC=CDsinφ3-AD=0.27mForthenextdyadRRT(CEE),therightFig,onecanwriteCecos(π-φ4)=xE-xCCesin(π-φ4)=yE-yCVectordiagramrepresenttheRRT(CEE)dyad.Whenthesystemofequationsissolved,theunknownsφ4andxEareobtained:φ4=165.9°xE=-0.114m7.Solution:TheoriginofthesystemisatA,A≡0;thatis,xA=yA=0.Thecoordinatesofthe
5、RjointsatBarexB=l1cosφyB=l1sinφForthedyadDBB(RTR),thefollowingequationscanbewrittenwithrespecttotheslidinglineCD:mxB-yB+n=0yD=mxD+n1WithxD=d,yD=0fromtheabovesystem,slopemoflinkCDandinterceptncanbecalculated:l1sind1l1sinm=l1cosd1n=d1l1cosCCThecoordinatesxandyofthecenteroftheRjointCresultfromthe
6、systemoftwoequations:l1sind1l1sinxCyC=mxC+n=l1cosd1d1l1cos,222(xC-xD)+(yC-yD)=l3Becauseofthequadraticequation,twosolutionsareabstainedforxCandyC.Forcontinuousmotionofthemechanism,thereareconstraintrelationsfortheChoiceofthecorrectsolution;thatisxC0ForthelastdyadCEE(RRT),apositionfu
7、nctioncanbewrittenforjointE:222(xC-xE)+(yC-h)=l4TheequationproducesvaluesforxE1andxE2,andthesolutionxE>xCisselectedforcontinuousmotionofthemechanism.（注：专业文档是经验性极强的领域，无法思考和涵盖全面，素材和资料部分来自网络，供参考。可复制、编制，期待你的好评与关注）
