机械原理大作业二、题目（平面机构的力分析）在图示的正弦机构中,已知l AB =100 mm,h1=120 mm,h2 =80 mm,W1 =10 rad/s(常数)，滑块2和构件3的重量分别为G2 =40 N和G3 =100 N,质心S2 和S3 的位置如图所示，加于构件3上的生产阻力Fr=400 N，构件1的重力和惯性力略去不计。

试用解析法求机构在Φ1=60°、150°、220°位置时各运动副反力和需加于构件1上的平衡力偶M。

b Array二、受力分析图三、算法(1)运动分析AB l l =1滑块222112112/,/s m w l a s m w l v c c ==滑块3 21113113/cos ,sin s m l w v m l s ϕϕ==212113/sin s m w l a ϕ-=(2)确定惯性力N w l g G a m F c 21122212)/(==N w l g G a m F 121133313sin )/(ϕ-==(3)受力分析i F F i F F x R D R x R C R 43434343,=-= j F j F F R R R 232323-==j F i F j F i F F R x R y R x R R 2121121212--=+= j F F F y R x R R 414141+= 取移动副为首解副① 取构件3为分离体，并对C 点取矩由0=∑yF 得 1323F F F r R -= 由0=∑xF得 C R D R F F 4343=由∑=0CM得 2112343/cos h l F F R D R ϕ=②取构件2为分离体由0=∑x F 得 11212cos ϕR x R F F = 由0=∑y F 得 1123212sin ϕF F F R y R -=③取构件1为分离体，并对A 点取矩由0=∑x F 得 x R x R F F 1241= 由0=∑y F 得 y R y R F F 1241=由0=A M 得 1132cos ϕl F M R b =四、根据算法编写Matlab 程序如下：%--------------已知条件---------------------------------- G2=40; G3=100; g=9.8; fai=0; l1=0.1; w1=10; Fr=400; h2=0.8;%--------分布计算，也可将所有变量放在一个矩阵中求解------------------- for i=1:37a2=l1*(w1^2);a3=-l1*(w1^2)*sin(fai); F12=(G2/g)*a2;F13=-(G3/g)*l1*(w1^2)*sin(fai);FR23=Fr-F13;FR43D=FR23*l1*cos(fai)/h2;FR43C=-FR43D;FR12x=F12*cos(fai);FR12y=(-FR23)-F12*sin(fai);FR41x=FR12x;FR41y=FR12y;Mb=(-FR23)*l1*cos(fai);fai=fai+pi/18;fai1=fai*180/pi;f1(i,:)=[fai1,FR23,FR43D,FR43C,FR12x,FR12y,FR41x,FR41y,Mb];end%------------------输出结果---------------------------------------disp ' fai1 FR23 FR43D FR43C FR12x FR12y FR41x FR41y Mb 'con=[f1(:,1),f1(:,2),f1(:,3),f1(:,4),f1(:,5),f1(:,6),f1(:,7),f1(:,8),f1(:,9)];disp(con);%-------------------绘制线图--------------------------------------figure %输出构件3所受反力与角位移fai1的函数图线plot(f1(:,1),f1(:,2),'c-',f1(:,1),f1(:,3),'r-',f1(:,1),f1(:,4),'m-')grid ontitle('构件3所受各运动副反力')xlabel('fai1角的角位移rad')ylabel('各运动副约束反力N')text(100,520,'反力FR23')text(20,70,'反力FR43D')text(150,70,'反力FR43C')figure %输出构件2所受反力与角位移fai1的函数图线plot(f1(:,1),f1(:,5),'c-',f1(:,1),f1(:,6),'r-')grid ontitle('构件2所受各运动副反力')xlabel('fai1角的角位移rad')ylabel('各运动副约束反力N')text(200,0,'反力FR12x')text(100,-400,'反力FR12y')figure %输出构件2所受反力与角位移fai1的函数图线plot(f1(:,1),f1(:,7),'c-',f1(:,1),f1(:,8),'r-',f1(:,1),f1(:,9),'m-')grid ontitle('构件1所受各运动副反力和应加平衡力矩Mb')xlabel('fai1角的角位移rad')ylabel('各运动副约束反力N')text(150,0,'反力FR41x')text(100,-400,'反力FR41y')text(150,70,'平衡力矩Mb')五、输出结果主动件1旋转一周，即一个周期内各反力和平衡力Mb的大小变化fai1 FR23 FR43D FR43C FR12x FR12y FR41x FR41y Mb10.0000 400.0000 50.0000 -50.0000 40.8163 -400.0000 40.8163 -400.0000 -40.000020.0000 417.7192 51.4216 -51.4216 40.1962 -424.8069 40.1962 -424.8069 -41.137330.0000 434.9000 51.0840 -51.0840 38.3548 -448.8600 38.3548 -448.8600 -40.867240.0000 451.0204 48.8244 -48.8244 35.3480 -471.4286 35.3480 -471.4286 -39.059550.0000 465.5906 44.5829 -44.5829 31.2671 -491.8268 31.2671 -491.8268 -35.666360.0000 478.1678 38.4200 -38.4200 26.2362 -509.4349 26.2362 -509.4349 -30.736070.0000 488.3699 30.5231 -30.5231 20.4082 -523.7179 20.4082 -523.7179 -24.418580.0000 495.8870 21.2004 -21.2004 13.9600 -534.2418 13.9600 -534.2418 -16.960390.0000 500.4906 10.8637 -10.8637 7.0877 -540.6868 7.0877 -540.6868 -8.6909100.0000 502.0408 0.0000 -0.0000 0.0000 -542.8571 0.0000 -542.8571 -0.0000110.0000 500.4906 -10.8637 10.8637 -7.0877 -540.6868 -7.0877 -540.6868 8.6909120.0000 495.8870 -21.2004 21.2004 -13.9600 -534.2418 -13.9600 -534.2418 16.9603130.0000 488.3699 -30.5231 30.5231 -20.4082 -523.7179 -20.4082 -523.7179 24.4185140.0000 478.1678 -38.4200 38.4200 -26.2362 -509.4349 -26.2362 -509.4349 30.7360150.0000 465.5906 -44.5829 44.5829 -31.2671 -491.8268 -31.2671 -491.8268 35.6663160.0000 451.0204 -48.8244 48.8244 -35.3480 -471.4286 -35.3480 -471.4286 39.0595170.0000 434.9000 -51.0840 51.0840 -38.3548 -448.8600 -38.3548 -448.860040.8672180.0000 417.7192 -51.4216 51.4216 -40.1962 -424.8069 -40.1962 -424.8069 41.1373190.0000 400.0000 -50.0000 50.0000 -40.8163 -400.0000 -40.8163 -400.0000 40.0000200.0000 382.2808 -47.0591 47.0591 -40.1962 -375.1931 -40.1962 -375.1931 37.6473210.0000 365.1000 -42.8852 42.8852 -38.3548 -351.1400 -38.3548 -351.1400 34.3082220.0000 348.9796 -37.7781 37.7781 -35.3480 -328.5714 -35.3480 -328.5714 30.2225230.0000 334.4094 -32.0216 32.0216 -31.2671 -308.1732 -31.2671 -308.1732 25.6172240.0000 321.8322 -25.8587 25.8587 -26.2362 -290.5651 -26.2362 -290.5651 20.6870250.0000 311.6301 -19.4769 19.4769 -20.4082 -276.2821 -20.4082 -276.2821 15.5815260.0000 304.1130 -13.0016 13.0016 -13.9600 -265.7582 -13.9600 -265.7582 10.4013270.0000 299.5094 -6.5012 6.5012 -7.0877 -259.3132 -7.0877 -259.3132 5.2009280.0000 297.9592 -0.0000 0.0000 -0.0000 -257.1429 -0.0000 -257.1429 0.0000290.0000 299.5094 6.5012 -6.5012 7.0877 -259.3132 7.0877 -259.3132 -5.2009300.0000 304.1130 13.0016 -13.0016 13.9600 -265.7582 13.9600 -265.7582 -10.4013310.0000 311.6301 19.4769 -19.4769 20.4082 -276.2821 20.4082 -276.2821 -15.5815320.0000 321.8322 25.8587 -25.8587 26.2362 -290.5651 26.2362 -290.5651 -20.6870330.0000 334.4094 32.0216 -32.0216 31.2671 -308.1732 31.2671 -308.1732 -25.6172340.0000 348.9796 37.7781 -37.7781 35.3480 -328.5714 35.3480 -328.5714 -30.2225350.0000 365.1000 42.8852 -42.8852 38.3548 -351.1400 38.3548 -351.1400 -34.3082360.0000 382.2808 47.0591 -47.0591 40.1962 -375.1931 40.1962 -375.1931 -37.6473370.0000 400.0000 50.0000 -50.0000 40.8163 -400.0000 40.8163 -400.0000 -40.0000六、绘出的函数图线①构件3所受各运动副反力与的函数关系曲线1②构件2所受各运动副反力与ϕ的函数关系曲线1③构件1所受各运动副反力和平衡力矩Mb与ϕ的函数关系曲线1七、结题结果挑选出220150601、、=ϕ时各反力和平衡力矩Mb 制表如下：附：参考资料1、《机械原理》（第七版）孙桓陈作模葛文杰主编高等教育出版社2、《机械原理同步辅导及习题全解》（第七版）唐亚楠主编中国矿业大学出版社3、《数学实验初步》（MATLAB）肖海军编科学出版社。