哈尔滨工业大学机械原理大作业-连杆机构运动分析

1、第嚼it矣求衣孝Harbin I nstituteof Technology大作业设计说明书课程名称：机械原理设计题目：连杆机构运动分析院系：机电学院班级：设计者：一学号：指导教师：设计时间：2012/6/20哈尔滨工业大学运动分析题目如图所示的甘草压缩机中，已知lAB = 150mm , lBC = 600mm , lCE = 120mm ,Icd = 500mm , Ief = 600mm , xd = 400mm , y = 500mm , yF = 600mm ,曲柄 1 作等 速转动，其转速山=50r/min 。求在一个运动循环中活塞5的位移、速度和加速度变化曲线。4机构的机构分析、

2、组成机构的基本杆组划分图1:1级机构图2: RRRI级杆组F图3: RRPd级杆组三、 各基本杆组的运动分析数学模型1、由“同一构件上点的运动分析”B点的运动参数，以A点为原点建立坐标系，可求得图4A点的坐标方程A点的速度A点的加速度曲柄转角 = 0360 (1) 位置分析原动构件1的长度则可求出B点的坐标方程(2) 速度和加速度分析B点的速度B点的加速度aBxaByXa = 0yA = 0VAx = 0VAy = 0aAx = 0aAy = 0li = 150mmXB = XA + 11 COS yB = yA + 11 sin vbx = xb = Xa - l1 sin VBy = yB

3、 = yA + l1 COS 2xB = xA - l1 cos - l1sin yB = yA - 2l1 sin + l 1 cos D点的坐标方程D点的速度D点的加速度（1）位置方程构件2、构件3的长度xD = 400yD = 500VDx = 0VDy = 0aDx = 0aDy = 012 = 600mm13 = 500mmAo = 21 2(Xd - Xb)Bq = 212 (y d - yB)其中2 2Co = 12 + 1 BD -131bd =(Xb- xd)2 + (yB- y。)2)则可求得2Bq +2=2 arctan _Aq2 + Bq2 - Cq2Aq + Cq则可

4、求出c点的坐标方程xc = xB + 12 cos 2 yc = yB + 12 sin 2求得33 = 2arctan(xc- Xd(2)速度方程构件2、3的角速度C3 Xd - Xb + S3(y D - yB) co 2 =$2 =-Gi_ C2 xD - xB + S2(y D - yB)O 3$33屮3G1式中：C2 = I2 ? COS $ 252 = I2 ? sin $2C3 = I3 ? COS $353 = I3 ? sin $3G1 = C2S3 - C3S2C点的速度XC = Xb -$2l2Sin $2yc = yB + $212COS $2(3)加速度方程构件2、3

5、的角加速度G2C3 + G3S3 2 = $2 =G1G2C2 + G3S2 3 =$3 =G1式中：22G2 = Xd -Xb+$2C2-$3C322G3 = yD -yB+$2S2-$3S33、由“同一构件上点的运动分析” 点的运动参数，已知B点的运动参数和构件 2的运动参数，可求得 E(1)位置分析Ibe = 480mmB点和E点间的距离则可求出E点的坐标方程xe = xb + Ibe cos $2 yE = yB + lBE sin $2(2)速度和加速度分析E点的速度Vex = Xe = Xb - $2lBESin $2VEy = yE = yB + O2IBECOS $2E点的加速

6、度a ex = xe =xb-$2 I becos$2 -$2Ibes in$2aEy = yE =yB-$22Ibes in$2 +$21 be cos$24、由“ RRPH级杆组”，已知E点的运动参数，在滑块导路上任取一点K作为参考点，可求得构件5上F点的运动参数图6K点的坐标方程Xk = 0yK = 600K点的速度VKx = 0VKy = 0K点的加速度aKx = 0aKy = 0（1）位置方程构件4长度l4= 600mm则可求得$4, .A0$4 = - arcs in - l4式中Ao = yK- yE则可求出F点的坐标方程XF = xe + l4 cos $4yF = yE+ I

7、beSin $4F相点对于参考点 K点的位移s = xk - xf(2)速度方程构件4的角速度co 4-Q2 4 =4 =Q3F点的速度Q3(Q1 l4cos4 + Q2I4sin 4) vF = xf =-式中:Qi = Xk- XbQ2 = yK - yBQ3 = - 14COS4(3)加速度方程F点加速度Q3-Q4I4COS 4 - Q5l4sin 4 aF = xf =式中:Q4 = xk - XB + 42l4 cos 4Q5 = yK - yB + 4214 sin 4四、建立坐标系建系，以A为原点，如图7所示图7五、 计算编程 xA=0;yA=0;vAx=0;vAy=0;aAx=

8、0;aAy=0; xD=400;yD=500;vDx=0;vDy=0;aDx=0;aDy=0; xK=0;yK=600;vKx=0;vKy=0;aKx=0;aKy=0; l1=150;l2=600;l3=500;lBE=480;l4=600;fi5=pi;n1=50;w1=2*pi*n1/60; fi1=linspace(0,2*pi,1000);for i=1:1000% 求出B点的位置、速度和加速度 xB(i)=xA+l1*cos(fi1(i);yB(i)=yA+l1*sin(fi1(i);vBx(i)=vAx-w1*l1*sin(fi1(i); vBy(i)=vAy+w1*l1*cos(

9、fi1(i);aBx(i)=aAx-w1A2*l1*cos(fi1(i);aBy(i)=aAy-wU2*l1*si n(fi1(i);%求出构件2、 3的运动参数A0(i)=2*l2*(xD-xB(i);B0(i)=2*l2*(yD-yB(i);C0(i)=l2A2+(xD-xB(i)A2+(yD-yB(i)A2-l3A2;fi2(i)=2*ata n( (B0(i)+sqrt(A0(i)A2+B0(i)A2-C0(i)A2)/(A0(i)+C0(i); xC(i)=xB(i)+l2*cos(fi2(i);yC(i)=yB(i)+l2*sin(fi2(i); fi3(i)=atan(yC(i)

10、-yD)/(xC(i)-xD);C2(i)=l2*cos(fi2(i);S2(i)=l2*sin(fi2(i);C3(i)=l3*cos(fi3(i);S3(i)=l3*sin(fi3(i);G1(i)=C2(i)*S3(i)-C3(i)*S2(i); w2(i)=(C3(i)*(vDx-vBx(i)+S3(i)*(vDy-vBy(i)/G1(i);w3(i)=(C2(i)*(vDx-vBx(i)+S2(i)*(vDy-vBy(i)/G1(i);G2(i)=aDx-aBx(i)+w2(i)A2*C2(i)-w3(i)A2*C3(i);G3(i)=aDy-aBy(i)+w2(i)A2*S2(i)

11、-w3(i)A2*S3(i); a2(i)=(G2(i)*C3(i)+G3(i)*S3(i)/G1(i); a3(i)=(G2(i)*C2(i)+G3(i)*S2(i)/G1(i);%求E点的运动参数xE(i)=xB(i)+lBE*cos(fi2(i); yE(i)=yB(i)+lBE*sin(fi2(i);vEx(i)=vBx(i)-w2(i)*lBE*sin(fi2(i);vEy(i)=vBy(i)+w2(i)*lBE*cos(fi2(i);aEx(i)=aBx(i)-w2(iF2*IBE*cos(fi2(i)-a2(i)*IBE*si n(fi2(i); aEy(i)=aBy(i)-w2

12、(iF2*IBE*si n(fi2(i)+a2(i)*IBE*cos(fi2(i);%求F点的运动参数A1(i)=(yE(i)-yK)*cos(fi5)-(xE(i)-xK)*sin(fi5); fi4(i)=fi5-asin(A1(i)/I4);xF(i)=xE(i)+I4*cos(fi4(i);yF(i)=yE(i)+I4*cos(fi4(i);sF(i)=(xF(i)-xK)*cos(fi5);Q1(i)=vKx-vEx(i);Q2(i)=vKy-vEy(i);Q3(i)=I4*sin(fi4(i)*sin(fi5)+I4*cos(fi4(i)*cos(fi5); w4(i)=(-Q1(i)*sin(fi5)+Q2(i)*cos(fi5)/Q3(i); vF(i)=-(Q1(i)*I4*cos(fi4(i)+Q2(i)*I4*sin(fi4(i)/Q3(i);Q4(i)=aKx-aEx(i)+w4(i)A2*I4*cos(fi4(i);Q5(i)=aKy-aEy(i)+w4(i)A2*I4*si n( fi4(i); aF(i)=(-Q4(i)*I4*cos(fi