机械原理大作业正弦机构力分析.docx

1、机械原理大作业正弦机构力分析机械原理大作业正弦机构力分析姓名： 郑豪 学号： 887 班级： 072125 专业： 机械设计制造及其自动化 联系方式： 指导老师： 王玉丹 完成时间： 在图示的正弦机构中，已知，（为常数），滑块2和构件3的重量分别为和，质心和的位置如图所示，加于构件3上的生产阻力，构件1的重力和惯性力略去不计。试用解析法求机构在原动件的一个回转周期内各运动副反力和需加于构件1上的平衡力偶矩。（图像表示）解：（1）分别对三个构件进行受力分析如下图：（2）运动分析： 滑块2： 构件3： （3）确定惯性力： （4）各构件的平衡方程： 构件3： 构件2： 构件1： （5）根据式可以进行

2、编程，以10为步幅可以算得各个约束处的反力。结果如表一所示。表一 各约束处反力表1Fr23/Fr32Fr43_1Fr43_2Fr12_xFr12_yFr41_xFr41_yMbradNNNNNNNN*m00.17950.35900.53860.71810.89761.07711.25661.43621.61571.79521.97472.15422.33382.51332.69282.87233.05183.23143.41093.59043.76993.94944.12904.30854.48804.66754.84705.02655.20615.38565.56515.74465.9241

3、6.10376.2832400.0000382.1443364.8625348.7101334.2061321.8169311.9404304.8943300.9050300.1007302.5072308.0472316.5427327.7205341.2215356.6116373.3963391.0361408.9639426.6037443.3884458.7785472.2795483.4573491.9528497.4928499.8993499.0950495.1057488.0596478.1831465.7939451.2899435.1375417.8557400.0000

4、500.0000470.0039426.9963374.1872314.6014250.8119184.7735117.771950.4893-16.8300-84.1427-151.3378-217.9780-283.0943-345.0675-401.6200-449.9252-486.8273-509.1470-514.0376-499.3489-463.9495-407.9684-332.9190-241.6872-138.3782-28.034983.7439191.2451289.0952372.6779438.4701484.2616509.2386513.9257500.000

5、0500.0000470.0039426.9963374.1872314.6014250.8119184.7735117.771950.4893-16.8300-84.1427-151.3378-217.9780-283.0943-345.0675-401.6200-449.9252-486.8273-509.1470-514.0376-499.3489-463.9495-407.9684-332.9190-241.6872-138.3782-28.034983.7439191.2451289.0952372.6779438.4701484.2616509.2386513.9257500.00

6、0040.000039.357237.449434.338030.122924.939618.954712.36075.3693-1.7946-8.9008-15.7210-22.0359-27.6425-32.3607-36.0388-38.5585-39.8390-39.8390-38.5585-36.0388-32.3607-27.6425-22.0359-15.7210-8.9008-1.79465.369312.360718.954724.939630.122934.338037.449439.357240.0000400.0000375.0020350.8075328.194130

7、7.8886290.5436276.7166266.8521261.2670260.1410263.5101271.2661283.1597298.8087317.7101339.2563362.7548387.4505412.5495437.2452460.7437482.2899501.1913516.8403528.7339536.4899539.8590538.7330533.1479523.2834509.4564492.1114471.8059449.1925424.9980400.000040.000039.357237.449434.338030.122924.939618.9

8、54712.36075.3693-1.7946-8.9008-15.7210-22.0359-27.6425-32.3607-36.0388-38.5585-39.8390-39.8390-38.5585-36.0388-32.3607-27.6425-22.0359-15.7210-8.9008-1.79465.369312.360718.954724.939630.122934.338037.449439.357240.0000400.0000375.0020350.8075328.1941307.8886290.5436276.7166266.8521261.2670260.141026

9、3.5101271.2661283.1597298.8087317.7101339.2563362.7548387.4505412.5495437.2452460.7437482.2899501.1913516.8403528.7339536.4899539.8590538.7330533.1479523.2834509.4564492.1114471.8059449.1925424.9980400.000040.000037.600334.159729.935025.168120.065014.78199.42184.0391-1.3464-6.7314-12.1070-17.4382-22

10、.6475-27.6054-32.1296-35.9940-38.9462-40.7318-41.1230-39.9479-37.1160-32.6375-26.6335-19.3350-11.0703-2.24286.699515.299623.127629.814235.077638.740940.739141.114140.0000（6）源程序代码如下：Lab=0.1;H1=0.12;H2=0.08;w=10; %曲柄角速度g=9.8; %重力加速度G2=40;m2=G2/g;G3=100;m3=G3/g;Fr=400; %构件3上的生产阻力fai=0:pi/36:2*pi;a2=Lab

11、*w*w; %滑块2加速度a3=-Lab*w*w*sin(fai); %推杆3加速度F2=m2*a2; %滑块2惯性力F3=m3*a3; %滑块3惯性力Fr23=Fr-F3; %滑块2与推杆3之间的反力%Fr43_1=Fr43_2; %支座4对推杆3的反力Fr43_1=Lab*Fr23.*cos(fai)/H2;%支座4对推杆3的反力Fr12_x=F2*cos(fai); %铰支座2的水平分力Fr12_y=Fr23-F2*sin(fai);%铰支座2的竖直分力Fr41_x=Fr12_x; %铰支座1的水平分力Fr41_y=Fr12_y; %铰支座1的竖直分力Mb=Lab*Fr23.*cos(f

12、ai); %附加于曲柄1的平衡力偶figure(1);plot(fai,Fr23,w);title(构件2、3之间反力图);xlabel(角度fai/rad);ylabel(反力Fr23(Fr32)/N);figure(2);plot(fai,Fr43_1,w);title(构件3、4之间反力图);xlabel(角度fai/rad);ylabel(反力Fr43(Fr34)/N);figure(3);plot(fai,Fr12_x,-w);hold on;plot(fai,Fr12_y,w);legend(Fr12-x,Fr12-y);title(构件1、2之间反力图);xlabel(角度fai/rad);ylabel(反力Fr12(Fr21)/N);figure(4);plot(fai,Fr41_x,-w);hold on;plot(fai,Fr41_y,w);legend(Fr14-x,Fr14-y);title(构件1、4之间反力图);xlabel(角度fai/rad);ylabel(反力Fr14(Fr41)/N);figure(5);plot(fai,Mb,w);title(构件1上的平衡力偶);xlabel(角度fai/rad);ylabel(Mb/Nm);（7）运行结果分别如图15所示。图1 Fr23图图2 Fr34图图3 Fr12图图4 Fr14图图5 Mb图