機械原理--速度瞬心習題;

1、習題 答案一.概念1.當兩構件組成轉動副時,其相對速度瞬心在 轉動副的圓心 處;組成移動副時,其瞬心在 垂直于移動導路的無窮遠 處;組成滑動兼滾動的高副時,其瞬心在接觸點兩輪廓線的公法線上.2.相對瞬心與絕對瞬心相同點是 都是兩構件上相對速度為零,絕對速度相等的點 ,而不同點是 相對瞬心的絕對速度不為零,而絕對瞬心的絕對速度為零 .3.速度影像的相似原理只能用于 同一構件上的 兩點,而不能用于機構 不同構件上 的各點.4.速度瞬心可以定義為互相作平面相對運動的兩構件上,相對速度為零,絕對速度相等 的點.5.3個彼此作平面平行運動的構件共有 3 個速度瞬心,這幾個瞬心必位于 同一條直線上 .含有

2、6個構件的平面機構,其速度瞬心共有 15 個,其中 5 個是絕對瞬心,有 9 個相對瞬心.二.計算題1、 2.關鍵: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