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 =r D -r B

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)=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(3π?+-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 -?? n=??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=????cos 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 x1E and x2E, and the solution x E >x C is selected for continuous motion of the mechanism.

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