机械原理大作业1连杆机构(18题)
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1、运动分析题目
如图1-18所示机构,已知机构各构件的尺寸为ß = 90°,AB = 108mm,AD = 266mm,DG = 278mm,EF = FG = 114mm,BC = CE = CD = 200mm,构件1的角速度为w1 = 10rad/s,试求构件2上的E的轨迹及构件5的角位移、角速度和角加速度,并对计算结果进行分析。
图1-18
2、建立坐标系
以A点为坐标原点,杆AD所在的直线为X轴,垂直于AD的直线为Y轴,建立直角坐标系,如下图2所示:
图2
3、对机构进行结构分析
该机构由1个I级杆组RR(原动件1)、II级杆组RRR(杆2、杆3)和II级杆组RRR(杆4、杆5)组成。
4、各基本杆组的运动分析数学模型
(1)原动件杆1(Ⅰ级杆组RR)
如图3所示
图3 Ⅰ级杆组RR
已知原动件杆1 的转角φ = 0~360°,角速度ω1 = 10rad/s,角加速度α1 = 0 转动副A
的位置坐标X A=0,Y A=0;速度X A’= 0,Y A’= 0;加速度X A”= 0,Y A”= 0 原动件杆1 的
长度L AB = 100mm,
可求转动副B 的位置坐标(X B,Y B),速度(X B’, Y B’)和角加速度(X B”, Y B”);
(2)杆2 和杆3(Ⅱ级杆组RRR)
如图4所示:
图4 Ⅱ级杆组RRR
由于BC = CE = CD = 200mm,所以,可以根据几何关系求出杆2与X轴之间的夹角Ψ,B 的位置坐标(X B,Y B);速度(X B’,Y B’);角加速度(X B’’,Y B’’)杆2 的长度L BE = 400mm,杆3的长度为L CD = 200mm。
可求点E的位置坐标(X E,Y E),速度(X E’,Y E’)和角加速度(X E’’,Y E’’);(3)杆4 和杆5(Ⅱ级杆组RRR)
如图5所示
图5 Ⅱ级杆组RRR
已知点E的位置坐标(X E,Y E),速度(X E’,Y E’)和角加速度(X E’’,Y E’’),EF = FG = 114mm,G点的坐标为(-266,278);速度X G’=0,Y G’=0;加速度X G’’=0,Y G’’=0,可求杆5 的角位移θ,角速度ω 及角加速度α。
5、计算编程
LAB = 108;
LAD = 266;
LDG = 278;
LEF = 114;
LFG = 114;
LBC = 200;
LCE = 200;
LCD = 200;
w1 = 10;
syms t;
F = w1 * t;
XB = LAB * cos(w1 * t);
YB = LAB * sin(w1 * t);
X1B = diff(XB);
Y1B = diff(YB);
X2B = diff(X1B);
Y2B = diff(Y1B);
XD = -266;
YD = 0;
X1D = 0;
Y1D = 0;
X2D = 0;
Y2D = 0;
A0 = 2 * LBC * (XD - XB);
B0 = 2 * LBC * (YD - YB);
LBD = sqrt((XB-XD)^2+(YB-YD)^2);
C0 = LBC ^ 2 + LBD ^ 2 - LCD ^2;
Fi = 2 * atan((B0 - sqrt(A0 ^ 2 + B0 ^ 2 - C0 ^ 2))/(A0 + C0)); XC = XB + LBC * cos(Fi);
YC = YB + LBC * sin(Fi);
X1C = diff(XC);
Y1C = diff(YC);
X2C = diff(X1C);
Y2C = diff(Y1C);
XE = 2*XC - XB;
YE = 2*YC - YB;
X1E = diff(XE);
Y1E = diff(YE);
X2E = diff(X1E);
Y2E = diff(Y1E);
XG = -266;
YG = 278;
X1G = 0;
Y1G = 0;
X2G = 0;
Y2G = 0;
A00 = 2 * LEF * (XG - XE);
B00 = 2* LEF * (YG - YE);
LEG = sqrt((XE - XG)^2+(YE - YG)^2);
C00 = LEF ^ 2 + LEG ^ 2 - LFG ^2;
Fii = 2 * atan((B00 + sqrt(A00 ^ 2 + B00 ^ 2 - C00 ^ 2))/(A00 + C00)); XF = XE + LEF * cos(Fii);
YF = YE + LEF * sin(Fii);
X1F = diff(XF);
Y1F = diff(YF);
X2F = diff(X1F);
Y2F = diff(Y1F);
O1 = atan((YF - YG)/(XF - XG));
w = diff(O1);
a = diff(w);
fprintf('结果集\n');
fprintf('t\t\tφ\t\tXE\t\tYE\t\tθ\t\tω\t\tα\t\t\n');
for alpha=0:pi/500:pi/5;%角度
%fprintf('%6.5f\t%6.5f\t%6.5f\t%6.5f\t%6.5f\t%6.5f\t\n',alpha,s subs(F,alpha),subs(XE,alpha),subs(YE,alpha),subs(O1,alpha),subs(w ,alpha));
%fprintf('%6.5f\t%6.5f\t%6.5f\t%6.5f\t%6.5f\t%6.5f\t%6.5f\t\n' ',alpha,subs(F,alpha),subs(XE,alpha),subs(YE,alpha),subs(O1,alpha ),subs(w,alpha),subs(a,alpha));
end
alpha2=0:pi/500:pi/5;%角度
x = subs(XE,alpha2);
y = subs(YE,alpha2);
plot(x,y,'r');
6、计算结果
t φXE YE θω
0.00000 0.00000 -266.00000 141.85909 0.63989 -3.00899