机械原理--速度瞬心习题
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.
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
y=ABsinφ=°=
x=ABsinφ=°=
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
+ -=0 or =-
Where = and =γ.
When the above vectorial equation is projected on the x and y axes, two scalar equations are obtained:
r*cos(φ+π)=x-x=
r*sin(φ+π)=y-y=
Angle φ is obtained by solving the system of the two previous scalar equations:
tgφ= φ=°
The distance r is
r==
The coordinates of joint C are
x=CDcosφ= y=CDsinφ-AD=
For the next dyad RRT (CEE), the right Fig, one can write
Cecos(π- φ)=x- x Cesin(π- φ)= y- y
Vector diagram represent the RRT (CEE) dyad.
When the system of equations is solved, the unknowns φ and x are obtained:
φ=° x=
7. Solution:The origin of the system is at A, A≡0; that is, x=y=0.
The coordinates of the R joints at B are
x=lc osφ y= lsinφ
For the dyad DBB (RTR), the following equations can be written with respect to the sliding line CD:
mx- y+n=0 y=mx+n
With x=d, y=0 from the above system, slope m of link CD and intercept n can be calculated:
m= n=
The coordinates 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)=l
Because of the quadratic equation, two solutions are abstained for x and continuous motion of the mechanism, there are constraint relations for the Choice of the correct solution; that is x< x< x and y>0
For the last dyad CEE (RRT), a position function can be written for joint E:
(x-x)+(y-h)=l
The equation produces values for x and x, and the solution x >x is selected
for continuous motion of the mechanism.