机械原理大作业1的第12题
1、运动分析题目
如图所示的六杆机构中，各构件的尺寸分别为：lAB=100mm ，lBC=500mm ，lCD=800mm ,xF=400mm ，xD=350mm ，yD=350mm ，ω=100rad s ，求构件5上点F的位移，速度和加速度。
2、建立坐标系
建立以点A为原点的固定平面直角系A x,y
机械原理大作业1的第12题
3、对机构进行结构分析
该机构由I级杆组RR（原动件1）、II级杆组RRR（杆2、杆3）和II级杆组PRP（滑块4及滑块5）组成。I级杆组RR，如图2所示；II级杆组RRR，如图2所示；II级杆组PRP，如图3所示。
机械原理大作业1的第12题
4、确定已知参数和求解流程
图1所示，规定当φ=10° 时，F点纵坐标为0 （1）如图2所示，已知原动件杆1的转角，
φ=0~360° x=lAB×cosφ B yB=lAB×sinφ
（2）如图3所示，已知B，D两点坐标 lAB lCD
（x xB）+(y yB)2=lAB2利用方程组 可以求解出C点坐标
222
(x (xD+xF))+(y yF)=lCD
（3）如图4所示，已知C点坐标、xF 、xD 、yD
利用几何关系可以求解出E点坐标
同时，当φ=10° 时，可以求出杆EF长，记为 lEF
进而，可以求出F点坐标，即F点位移
（4）利用导数的定义与其物理意义
2
sn+h sn h
vn=
2h
sn+h 2sn+sn h
an=利用上述公式，选取适当的步长h，利用F点位移就可以得出速度与加速度
机械原理大作业1的第12题
5、用VC编程
#include <stdio.h> #include <math.h>
#define pi 3.14159265358979323846 //定义全局变量
double Lab,Lbc,Lcd,Xf,Xd,Yd;//定义已知位置量 double Wab;//定义角速度量
//定义被调用函数
void Ccorner (double *a,double *b,double c);//声明C点坐标函数求解函数
void Pcorner (double *a,double *b,double c,double d,double e);//声明P点坐标函数求解函数 double Kcd (double a,double b);//声明CD直线倾斜角求解函数
//主函数 main () {
Lab=0.200;Lbc=0.500;Lcd=0.800;Xf=0.400;Xd=0.350;Yd=0.350;//赋位置量值 Wab=100;//赋角速度值
//未知几何与位置参量 double Xb,Yb;//定义B点坐标 double Xc,Yc;//定义C点坐标 double Ye;//定义E点纵坐标 double Xp,Yp;//定义瞬心p点坐标 double Lef;//定义bp,cp,ef,bd杆长 double Yf[720];//定义F点纵坐标
//未知速度参量
double Vf[720];//定义EF杆速度
//未知加速度参量
double Af[720];//定义加速度
//其余参量
double o=10*pi/180,k,k1;//主动杆角度变量与CD杆倾斜角 double t=1*pi/180/100;//时间参量，用定义法求速度与加速度 int i;//循环控制变量
机械原理大作业1的第12题
//主函数主体 //求位移量
for (i=0;i<=361;i++) {
//准备几何量
Xb=Lab*cos(o), Yb=Lab*sin(o); Ccorner (&Xc,&Yc,o);//求C点坐标
Pcorner (&Xp,&Yp,Xc,Yc,o);//求瞬心P点坐标 k=Kcd (Xp,Yp);//求CD杆倾斜角
//求解位移量（规定主动杆10度为Yf零点） if (i==0) {
k1=k; }
Lef=tan(k1)*Xd+Yd;//ef杆长 Ye=tan(k)*Xd+Yd; Yf[i]=Ye-Lef; o=o+1*pi/180; }
//用定义求速度 for (i=1;i<=361;i++) {
Vf[i]=(Yf[i+1]-Yf[i-1])/(2*t); }
//用定义求加速度 for (i=1;i<=361;i++) {
Af[i]=(Yf[i+1]-2*Yf[i]+Yf[i-1])/pow(t,2); }
//输出语句
for (i=1;i<=180;i++) {
printf ("%d,%lf,%lf,%lf,\t",i+10,Yf[i]*1000,Vf[i],Af[i]);
printf ("%d,%lf,%lf,%lf\n",i+190,Yf[i+180]*1000,Vf[i+180],Af[i+180]);
机械原理大作业1的第12题
} }
//C点坐标值函数
void Ccorner(double *a,double *b,double c)//&Xc,&Yc,o {
double i,j,x,y,z;//中间参数
i=(pow(Lcd,2)-pow(Lbc,2)+pow(Lab*sin(c),2)-pow(Yd,2)+pow(Lab*cos(c),2)-pow((Xd+Xf),2))/(2*(Lab*cos(c)-Xd-Xf));
j=(Lab*sin(c)-Yd)/(Lab*cos(c)-Xf-Xd); x=pow(j,2)+1;
y=2*(Xf+Xd)*j-2*i*j-2*Yd;
z=pow(i,2)-2*(Xf+Xd)*i+pow((Xf+Xd),2)+pow(Yd,2)-pow(Lcd,2); *b=(-y+sqrt(pow(y,2)-4*x*z))/(2*x); *a=i-j*(*b); }
//克拉默法求瞬心P点坐标函数
void Pcorner(double *a,double *b,double c,double d,double e)//&Xp,&Yp,Xc,Xp,o {
double a1,b1,c1,a2,b2,c2,i,j,k;//中间参量 a1=tan(e),a2=(d-Yd)/(c-Xd-Xf); b1=b2=-1;
c1=0,c2=(c*(d-Yd)/(c-Xd-Xf))-d; i=a1*b2-b1*a2; j=c1*b2-b1*c2; k=a1*c2-c1*a2; *a=j/i; *b=k/i; }
//CD杆倾斜角
double Kcd(double a,double b)//Xp,Yp {
double k1;//中间参量
k1=-atan ((b-Yd)/(a-Xd-Xf)); return k1; }
机械原理大作业1的第12题
6、计算结果（程序计算结果附在图像之后）
6.1位移、速度、加速度的图像
s（mm）） 点F的位移线图如图5所示。（φ ° v（m/s）） 点F的速度线图如图6所示。（φ ° a（m/s2）） 点F的加速度线图如图7所示。（φ °
机械原理大作业1的第12题
6.2程序计算结果
机械原理大作业1的第12题
哈尔滨工业大学机械原理课程设计
29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67
38.003246 40.075766 42.147127 44.215946 46.280818 48.340312 50.392976 52.437340 54.471916 56.495203 58.505688 60.501850 62.482163 64.445096 66.389119 68.312705 70.214332 72.092489 73.945675 75.772404 77.571207 79.340638 81.079273 82.785715 84.458595 86.096578 87.698362 89.262682 90.788315 92.274076 93.718826 95.121473 96.480972 97.796327 99.066593 100.290881 101.468352 102.598226 103.679777
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机械原理大作业1的第12题
哈尔滨工
业大学机械原理课程设计
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机械原理大作业1的第12题
哈尔滨工业大学机械原理课程设计
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机械原理大作业1的第12题
哈尔滨工业大学机械原理课程设计
146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184
53.811264 52.200455 50.585214 48.966112 47.343709 45.718555 44.091189 42.462138 40.83191
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机械原理大作业1的第12题
哈尔滨工业大学机械原理课程设计
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