机械原理--速度瞬心习题

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

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1、 2.关键:找到瞬心P366 Solution: The coordinates of joint B are y=ABsin=0.20sin45=0.141mx=ABsin=0.20sin45=0.141mThe vector diagram of the right Fig is drawn by representing the RTR (BBD) dyad. The vector equation, corresponding to this loop, is written as+

3、 -=0 or =-Where = and =. When the above vectorial equation is projected on the x and y axes, two scalar equations are obtained:r*cos(+)=x-x=-0.141mr*sin(+)=y-y=-0.541mAngle is obtained by solving the system of the two previous scalar equations:tg= =75.36The distance r is r=0.56mThe coordinates of jo

4、int C are x=CDcos=0.17m y=CDsin-AD=0.27mFor the next dyad RRT (CEE), the right Fig, one can write Cecos(- )=x- x Cesin(- )= y- yVector diagram represent the RRT (CEE) dyad.When the system of equations is solved, the unknowns and x are obtained:=165.9 x=-0.114m7. Solution: The origin of the system is

5、 at A, A0; that is, x=y=0. The coordinates of the R joints at B arex=lcos y= lsinFor the dyad DBB (RTR), the following equations can be written with respect to the sliding line CD:mx- y+n=0 y=mx+nWith x=d, y=0 from the above system, slope m of link CD and intercept n can be calculated:m= n=The coord

6、inates x and y of the center of the R joint C result from the system of two equations:y=mx+n=,(x- x)+(y- y)=lBecause of the quadratic equation, two solutions are abstained for x and y.For continuous motion of the mechanism, there are constraint relations for the Choice of the correct solution; that is x x0 For the last dyad CEE (RRT), a position function can be written for joint E:(x-x)+(y-h)=l The equation p