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P 36

6 Solution:

The coordinates of joint B are

y =ABsinφ=0.20sin45°=0.141m

x =ABsinφ=0.20sin45°=0.141m

The vector diagram of the right Fig is drawn by representing the RTR (BBD) dyad. B B

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 =-0.141m

r*sin(φ+π)=y -y =-0.541m

Angle φ is obtained by solving the system of the two previous scalar equations:

tgφ= φ=75.36°

The distance r is

r=

=0.56m

The coordinates of joint C are

x =CDcosφ=0.17m y =CDsinφ-AD=0.27m

For the next dyad RRT (CEE), the right Fig, one can write

Cecos(π- φ)=x

- x Cesin(π- φ)= y - y

r B r r D r D B r 3D B 3D B 33141.0541

.0⇒3)cos(3πϕ+-B

D x x C 3C 34

E C 4E C

Vector diagram represent the RRT (CEE) dyad.

When the system of equations is solved, the unknowns φ and x are obtained: φ=165.9° x =-0.114m

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 =l cosφ y = l sinφ

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 4E 4E A A B 1B 1B B D D D 1D 11

1cos sin d l l -ϕϕϕϕ

cos sin 1111l d l d -C C C C ϕϕϕϕcos sin cos sin 111

111

1l d l d x d l l C -+-C D 2C D 22

3

Because of the quadratic equation, two solutions are abstained for x and y .For 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.

C C C B

D C C

E 2C 22

4

1E 2E E C