机械原理大作业——正弦机构力分析

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1、提供全套，各专业毕业设计机械原理大作业——正弦机构力分析姓名：学号：班级：专业：机械设计制造及其自动化联系方式：指导老师：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~~可以进行编程，以10°为步

2、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Φ1Fr23/Fr32Fr43_1Fr43_2Fr12_xFr12_yFr41_xFr41_yMbradNNNNNNNN*m00.17950.35900.53860.71810.89761.07711.25661.43621.61571.79521.97472.15422.33382.51332.69282.87233.05183.23143.41093.59043.76993.94944.12904.30854.48804.66754.84705.02655.20615.3856

3、5.56515.74465.92416.10376.2832400.0000382.1443364.8625348.7101334.2061321.8169311.9404304.8943300.9050300.1007302.5072308.0472316.5427327.7205341.2215356.6116373.3963391.0361408.9639426.6037443.3884458.7785472.2795483.4573491.9528497.4928499.8993499.0950495.1057488.0596478.

4、1831465.7939451.2899435.1375417.8557400.0000500.0000470.0039426.9963374.1872314.6014250.8119184.7735117.771950.4893-16.8300-84.1427-151.3378-217.9780-283.0943-345.0675-401.6200-449.9252-486.8273-509.1470-514.0376-499.3489-463.9495-407.9684-332.9190-241.6872-138.3782-28.0349

5、83.7439191.2451289.0952372.6779438.4701484.2616509.2386513.9257500.0000500.0000470.0039426.9963374.1872314.6014250.8119184.7735117.771950.4893-16.8300-84.1427-151.3378-217.9780-283.0943-345.0675-401.6200-449.9252-486.8273-509.1470-514.0376-499.3489-463.9495-407.9684-332.919

6、0-241.6872-138.3782-28.034983.7439191.2451289.0952372.6779438.4701484.2616509.2386513.9257500.000040.000039.357237.449434.338030.122924.939618.954712.36075.3693-1.7946-8.9008-15.7210-22.0359-27.6425-32.3607-36.0388-38.5585-39.8390-39.8390-38.5585-36.0388-32.3607-27.6425-22.

7、0359-15.7210-8.9008-1.79465.369312.360718.954724.939630.122934.338037.449439.357240.0000400.0000375.0020350.8075328.1941307.8886290.5436276.7166266.8521261.2670260.1410263.5101271.2661283.1597298.8087317.7101339.2563362.7548387.4505412.5495437.2452460.7437482.2899501.191351

8、6.8403528.7339536.4899539.8590538.7330533.1479523.2834509.4564492.1114471.8059449.