机械原理 速度瞬心习题
习题> 答案
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
占
八、、?
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
The coord inates of joint B are y B =ABsin 0 =0.20sin45
° =0.141m
x B =ABsin 0 =0.20sin45
°
=0.141m The vector diagram of the right
Fig is drawn by representing the
RTR (BBD) dyad.
2.关键:找到瞬心
P 36 6 Solutio n:
The vector equati on, corresp onding to this loop, is writte n as
Where r = BD and r = 丫 .
Whe n the above vectorial equati on is projected on the x and y axes, two scalar equati ons are obta in ed:
B
r*cos( $+ n )=x D -x =-0.141m r*sin(
3>+ n )=y D -y B =-0.541m Angle 3 is obtained by solving the system of the two previous scalar equations: 0.541 tg
员=°?1410 3=75.36 °
The dista nee r is
X D X B
A?%
The coord inates of joint C are x C =CDcos 0 3 =0.17m y
C =CDsin 03 -AD=0.27m For the next dyad RRT (CEE), the right Fig, one can write
Cecos( n 0 4)=x E - x C Cesin( -n0 4 )= y E - y
r B
+ r - r D =0 or r = r D - r B
r=
CO s(「3 二)=0.56m
Vector diagram represe nt the RRT (CEE) dyad.
Whe n the system of equati ons is solved, the unknowns 0 4=165.9 ° X=-0.114m
三 0; that
is, x A =y A
=0 The coord inates of the R joints at B are
For the dyad DBB (RTR), the following equations can be written with respect to the sliding line CD: mx B - y B +n=0 y D =mx D +n
i
With x D =d , y D =0 from the above system, slope m of link CD and intercept n can be calculated:
l 1 sin 「
d 1l 1 sin :
m = hcos -d 1 n = d 1 -h cos
111=
n 一 The coord in ates x C and y C of the cen ter of the R joi nt C result from the system of two equati ons:
l 1 si n d 1l 1 si n
x C l 1 cos " -d 1 d^l 1 cos Because of the quadratic equation, two solutions are abstained for x C and y C .For continuous motion of the mechanism, there are constraint relations for the Choice of the correct solution; that is x C < x B < x D and y C >0
4 and x E
are obtained: x B =l 1 cos r y * 3 = l 1sin r
y C =mx C +n= 7. Soluti on: The origi n of the system is at A, A C(x Gl y c )
For the last dyad CEE (RRT), a positi on fun ctio n can be writte n for joi nt E:
2 2 2
(x C -x E ) +(y C -h) =l 4
The equation produces values for x E1 and x E2, and the solution x E >x C is selected for con ti nu ous moti on of the mecha ni sm.
c 2 c 2
(x C - x D ) +(y C - y D) =l