机械原理--速度瞬心习题

机械原理--速度瞬心习题(共5页)

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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

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2.关键:找到瞬心P36

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6 Solution:

The coordinates of joint B are yB=ABsinφ=°=

xB=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

rB+

r-rD=0 or

r=r-rDB

Where

r=BD and r=γ.

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When the above vectorial equation is projected on the x and y axes, two scalar equations are obtained: r*cos(φ3+π)=xD-x

B=

r*sin(φ3+π)=yD-yB=

Angle φ3 is obtained by solving the system of the two previous scalar equations:

0.1tgφ3=0.141 φ3=° The distance r is

xDxBr=cos(3)=

The coordinates of joint C are

xC=CDcosφ3= yC=CDsinφ3-AD=

For the next dyad RRT (CEE), the right Fig, one can write Cecos(π- φ4)=xE- xC Cesin(π- φ4)= yE- yC Vector diagram represent the RRT (CEE) dyad.

When the system of equations is solved, the unknowns φ4 and xE are obtained:

φ4=° xE=

7. Solution: The origin of the system is at A, A≡0; that is,

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xA=yA=0.

The coordinates of the R joints at B are

xB=l1cosφ yB= l1sinφ

For the dyad DBB (RTR), the following equations can be written with respect to the sliding line CD: mxB- yB+n=0 yD=mxD+n

1With xD=d, yD=0 from the above system, slope m of link CD and intercept n can be calculated:

l1sind1l1sinlcosd1 n=d1l1cos m=1The coordinates xC and yC of the center of the R joint C result from the system of two equations:

l1sindlsinxC11lcosd1d1l1cos, yC=mxC+n=1(xC- xD)

2+(yC- yD)

223=l

Because of the quadratic equation, two solutions are abstained for xC and

C continuous motion of the mechanism, there are constraint relations for

the Choice of the correct solution; that is xC< xB< xD and yC>0

For the last dyad CEE (RRT), a position function can be written for joint E: (xC-xE)

2+(yC-h)

22=l4

The equation produces values for xE1 and xE2, and the solution xE >xC is selected for continuous motion of the mechanism.

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