机械原理--速度瞬心习题

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:ThecoordinatesofjointBarey=ABsinφ=0.20sin45°=0.141mx=ABsinφ=0.20sin45°=0.141mThevectordiagramoftherightFigisdrawnbyrepresentingtheRTR(BBD)dyad.BBThevectorequation,correspondingtothisloop,iswrittenas+-=0or=-Where=and=γ.Whentheabovevectorialequationisprojectedonthexandyaxes,twoscalarequationsareobtained:r*cos(φ+π)=x-x=-0.141mr*sin(φ+π)=y-y=-0.541mAngleφisobtainedbysolvingthesystemofthetwopreviousscalarequations:tgφ=φ=75.36°Thedistancerisr==0.56mThecoordinatesofjointCarex=CDcosφ=0.17my=CDsinφ-AD=0.27mForthenextdyadRRT(CEE),therightFig,onecanwriteCecos(π-φ)=x-xCesin(π-φ)=y-yrBrrDrrDrBrBDr3DB3DB33141.0541.03)cos(3BDxxC3C34EC4ECVectordiagramrepresenttheRRT(CEE)dyad.Whenthesystemofequationsissolved,theunknownsφandxareobtained:φ=165.9°x=-0.114m7.Solution:TheoriginofthesystemisatA,A≡0;thatis,x=y=0.ThecoordinatesoftheRjointsatBarex=lcosφy=lsinφForthedyadDBB(RTR),thefollowingequationscanbewrittenwithrespecttotheslidinglineCD:mx-y+n=0y=mx+nWithx=d,y=0fromtheabovesystem,slopemoflinkCDandinterceptncanbecalculated:m=n=ThecoordinatesxandyofthecenteroftheRjointCresultfromthesystemoftwoequations:y=mx+n=,(x-x)+(y-y)=l4E4EAAB1B1BBDDD1D111cossindllcossin1111ldldCCCCcossincossin1111111ldldxdllCCD2CD223Becauseofthequadraticequation,twosolutionsareabstainedforxandy.Forcontinuousmotionofthemechanism,thereareconstraintrelationsfortheChoiceofthecorrectsolution;thatisxxxandy0ForthelastdyadCEE(RRT),apositionfunctioncanbewrittenforjointE:(x-x)+(y-h)=lTheequationproducesvaluesforxandx,andthesolutionxxisselectedforcontinuousmotionofthemechanism.CCCBDCCE2C2241E2EEC