機(jī)械原理-速度瞬心習(xí)題

機(jī)械原理--速度瞬心習(xí)題機(jī)械原理--速度瞬心習(xí)題機(jī)械原理--速度瞬心習(xí)題機(jī)械原理--速度瞬心習(xí)題編制僅供參考審核批準(zhǔn)生效日期地址：電話：傳真:郵編:習(xí)題>答案一.概念1.當(dāng)兩構(gòu)件組成轉(zhuǎn)動(dòng)副時(shí),其相對(duì)速度瞬心在轉(zhuǎn)動(dòng)副的圓心處;組成移動(dòng)副時(shí),其瞬心在垂直于移動(dòng)導(dǎo)路的無(wú)窮遠(yuǎn)處;組成滑動(dòng)兼滾動(dòng)的高副時(shí),其瞬心在接觸點(diǎn)兩輪廓線的公法線上.2.相對(duì)瞬心與絕對(duì)瞬心相同點(diǎn)是都是兩構(gòu)件上相對(duì)速度為零,絕對(duì)速度相等的點(diǎn),而不同點(diǎn)是相對(duì)瞬心的絕對(duì)速度不為零,而絕對(duì)瞬心的絕對(duì)速度為零.3.速度影像的相似原理只能用于同一構(gòu)件上的兩點(diǎn),而不能用于機(jī)構(gòu)不同構(gòu)件上的各點(diǎn).4.速度瞬心可以定義為互相作平面相對(duì)運(yùn)動(dòng)的兩構(gòu)件上,相對(duì)速度為零,絕對(duì)速度相等的點(diǎn).個(gè)彼此作平面平行運(yùn)動(dòng)的構(gòu)件共有3個(gè)速度瞬心,這幾個(gè)瞬心必位于同一條直線上.含有6個(gè)構(gòu)件的平面機(jī)構(gòu),其速度瞬心共有15個(gè),其中5個(gè)是絕對(duì)瞬心,有9個(gè)相對(duì)瞬心.二.計(jì)算題1、2.關(guān)鍵:找到瞬心P366Solution:ThecoordinatesofjointBarey=ABsinφ=°=x=ABsinφ=°=ThevectordiagramoftherightFigisdrawnbyrepresentingtheRTR(BBD)dyad.Thevectorequation,correspondingtothisloop,iswrittenas+-=0or=-Where=and=γ.Whentheabovevectorialequationisprojectedonthexandyaxes,twoscalarequationsareobtained:r*cos(φ+π)=x-x=r*sin(φ+π)=y-y=Angleφisobtainedbysolvingthesystemofthetwopreviousscalarequations:tgφ=φ=°Thedistancerisr==ThecoordinatesofjointCarex=CDcosφ=y=CDsinφ-AD=ForthenextdyadRRT(CEE),therightFig,onecanwriteCecos(π-φ)=x-xCesin(π-φ)=y-yVectordiagramrepresenttheRRT(CEE)dyad.Whenthesystemofequationsissolved,theunknownsφandxareobtained:φ=°x=7.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)=lBecauseofthequadraticequation,twosolutionsareabstainedforxandcontinuousmotionofthemechanism,thereareconstraintrelationsfortheChoiceofthecorrectsolution;thatisx<x<xandy>0ForthelastdyadCEE(RRT),apositionfunctioncanbewrittenforjointE:(x-x)+(y-h)=lTheequati