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P36
6 Solution:
The coordinates of joint B are yB=ABsinφ=0.20sin45°=0.141m xB=ABsinφ=0.20sin45°=0.141m
The vector diagram of the right Fig is drawn by representing the RTR (BBD) dyad.
The vector equation, corresponding to this loop, is written as
rB+
r-rD=0 or
r=r-rDB
Where
r=BD and r=γ.
When the above vectorial equation is projected on the x and y axes, two scalar equations are obtained: r*cos(φ3+π)=xD-x
B=-0.141m
r*sin(φ3+π)=yD-yB=-0.541m
Angle φ3 is obtained by solving the system of the two previous scalar equations:
0.541tgφ3=0.141 ?φ3=75.36°
The distance r is
xD?xBcos(?3??)=0.56m r=
The coordinates of joint C are
xC=CDcosφ3=0.17m yC=CDsinφ3-AD=0.27m
For the next dyad RRT (CEE), the right Fig, one can write Cecos(π- φ4)=xE- xC Cesin(π- φ4)= yE- yC
Vector diagram represent the RRT (CEE) dyad.
When the system of equations is solved, the unknowns φ4 and xE are obtained: φ4=165.9° xE=-0.114m
7. Solution: The origin of the system is at A, A≡0; that is, xA=yA=0.
The coordinates of the R joints at B are xB=l1cosφ yB= l1sinφ
For the dyad DBB (RTR), the following equations can be written with respect to the sliding line CD:
mxB- yB+n=0 yD=mxD+n
With x, yD=0 from the above system, slope m of link CD and intercept n can be calculated:
D=d
1l1sin?d1l1sin?lcos??d1 n=d1?l1cos? m=1The coordinates xC and yC of the center of the R joint C result from the system of two equations:
l1sin?dlsin?xC?11lcos??d1d1?l1cos?,
yC=mxC+n=1(xC- xD)+(yC- yD)=l
2223Because of the quadratic equation, two solutions are abstained for xC and yC.For continuous motion of the mechanism, there are constraint relations for the Choice of the correct solution; that is xC< xB< xD and yC>0
For the last dyad CEE (RRT), a position function can be written for joint E: (x-x)+(y-h)=l
CE2C224The equation produces values for xE1 and xE2, and the solution xE >xC is selected for continuous motion of the mechanism.
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