机械原理--速度瞬心习题

1、精品文档习题 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 占八、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

2、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2.关键:找到瞬心P366 Solutio n:The coord inates of joint B are y B =ABsin 0 =0.20sin45 xB=ABsin 0 =0.20sin45 =0.141mThe vector diagram of the right Fig is drawn by representing the RTR (BBD) dyad. The vector equati on, corresp onding to this loop, is writ

3、te n asr b+ r - rD =0 orr = r d- r bWhere= BD and r = 丫 .Whe n the above vectorial equati on is projected on the x and y axes, two scalar equati ons are obta in ed:Br*cos(0+ n )=x D -x =-0.141mr*sin(30+ n )=yD -y B =-0.541mAn gle3 is obta ined by solvi ng the system of the two previous scalar equati

4、 ons:0.541tg 03= 014103 =75.36 The dista nee r isXdXbr= cos( 3) =0.56mThe coord inates of joint C arexC=CDcos 03=O.17m y C =CDsin 03-AD=O.27mFor the next dyad RRT (CEE), the right Fig, one can write-冗04 )= y E - y CCecos( n 04 )=x E - x C Cesin(4 and x E ar0obtained:Vector diagram represe nt the RRT

5、 (CEE) dyad.Whe n the system of equati ons is solved, the unknowns04=165.9 E=-0.114mx A =y A=07. Solution:The origin of the system is at A, A三 0; that is,The coord inates of the R joints at B arex B =l1 cos 0 y = l 1sin 0For the dyad DBB (RTR), the following equations can be written with respect to

6、the sliding line CD: mx B - y B+ n=0 y D=mx D + niWith x D =d , y D =0 from the above system, slope m of link CD and intercept n can be calculated:m=l1 sinl1 cosdid1l1 sind1 l1 cosThe coord inates x two equati ons:C and y C of the center of the R joint C result from the system ofCC -l1 sinl1 cosCd1y

7、 =mx+n=(xC- x D)222+(y C-y D)=l3d1l1 sin d1 l1 cosBecause of the quadratic equation, two solutions are abstained for xC and y C .Forcontinuous motion of the mechanism, there are constraint relations for the Choice ofthe correct solution; that is x C x B 0For the last dyad CEE (RRT), a positi on fun ctio n can be writte n for joi nt E:2 2 2(x C -x E ) +(y C -h) =l 4The equation pro