机械原理--速度瞬心习题讲解学习

1、机械原理-速度瞬心习题精品文档习题 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

2、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 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 .2.关键 :找到瞬心 P36收集于网络，如有侵权请联系管理员删除精品文档xB =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 asr B

3、+ r - r D =0 or r = r D - r BWhere r =BD and r = .收集于网络，如有侵权请联系管理员删除精品文档When the above vectorial equation is projected on the x and y axes, two scalar equations are obtained:Br*cos( 3+)=xD -x =-0.141mr*sin( 3+ )=yD -y B =-0.541mAngle 3 is obtained by solving the system of the two previous scalar equ

4、ations: 0.541tg 3 = 0.1413 =75.36 The distance r isxD xBr= cos( 3 ) =0.56mThe coordinates of joint C arexC =CDcos3 =0.17m yC =CDsin 3 -AD=0.27mFor the next dyad RRT (CEE), the right Fig, one can writeCecos( - 4 )=xE - xC Cesin( - 4 )= yE - yCVector diagram represent the RRT (CEE) dyad.When the syste

5、m of equations is solved, the unknowns 4 and xE are obtained: 4 =165.9 xE =-0.114m7. Solution: The origin of the system is at A, A 0; that is,收集于网络，如有侵权请联系管理员删除精品文档xA=y A=0The coordinates of the R joints at B are x B =l 1 cos By= l1sin For the dyad DBB (RTR), the following equations can be written w

6、ith respect to the sliding line CD: mxB- y B +n=0 yD=mx D+n1With x D=d , yD=0 from the above system, slope m of link CD and intercept n can be calculated:l1 sind1l1 sinm=l1 cosd1 n=d1 l1 cos n=d1l1 sind1 l1 cosThe coordinates xC and yC of the center of the R joint C result from the system of two equ

7、ations:l1 sin xCC Cl1 cosd1yC =mx C +n= 1 12 2 2 (xC- x D)2+(yC- y D)2=l3Because of the quadratic equation, two solutions are abstained forC x and yC .For continuous motion of the mechanism, there are constraint relations for the Choice of the correct solution; that is xC x B 0For the last dyad CEE (RRT), a position function can be written for joint E: 2 2 2(xC-x E) +(y C -h) =l 4The equation produces