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