机械原理--速度瞬心习题讲解学习

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.
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
y B=ABsinφ=0.20sin45°=0.141m
x B=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
r B+ r-r D=0 or r=D-B
r= and r=γ.
Where
When the above vectorial equation is projected on the x and y axes, two scalar equations are obtained:
r*cos(φ3+π)=x D -x B =-0.141m
r*sin(φ3+π)=y D -y B =-0.541m
Angle φ3 is obtained by solving the system of the two previous scalar equations:
tgφ3=141.0541
.0 ⇒φ3=75.36°
The distance r is
r=)cos(3πϕ+-B
D x x =0.56m
The coordinates of joint C are x C =CDcosφ3=0.17m y C =CDsinφ3-AD=0.27m
For the next dyad RRT (CEE), the right Fig, one can write
Cecos(π- φ4)=x E - x C Cesin(π- φ4)= y E - y C
Vector diagram represent the RRT (CEE) dyad.
When the system of equations is solved, the unknowns φ4 and x E are obtained: φ4=165.9° x E =-0.114m
7. Solution: The origin of the system is at A, A≡0; that is,
x A =y A =0.
The coordinates of the R joints at B are x B =l 1cosφ y B = l 1sinφ
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
With x D =d 1, y D =0 from the above system, slope m of link CD and intercept n can be calculated:
m=111cos sin d l l -ϕϕ n=ϕϕ
cos sin 1111l d l d - The coordinates x C and y C of the center of the R joint C result from the system of two equations:
y C =mx C +n=ϕϕϕϕcos sin cos sin 111
1111l d l d x d l l C -+-,
(x C - x D )2+(y C - y D )2=l 2
3
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
For the last dyad CEE (RRT), a position function can be written for joint E:
(x C -x E )2+(y C -h)2=l 2
4
The equation produces values for x 1E and x 2E , and the solution x E >x C is selected for continuous motion of the mechanism.。