完整版利用传递矩阵法和Riccati传递矩阵法分析转子临界转速VIP免费

利用传递矩阵法和Riccati传递矩阵法分析转子临界转速一、所需求解转子参数将转子简化为如下所示：ABC1m2m3m1J2J3J三个盘的参数为：1232221232221233.5730.0160.050.0160.0120.0250.012PPPdddmkgmkgmkgIkg mIkg mIkg mIkg mIkg mIkg m另，阶梯轴的三段轴的截面惯性矩分别为：4142431.73.20.9JcmJcmJcm三段轴的单位长度轴段的质量分别为：1232.45/3.063/1.587/mkgmmkgmmkgm二、试算转轴的传递矩阵取试算转速1200/prads ; 则，各轴段的传递矩阵分别为：第 1 段840.061.7 102.45/lmJmmkgm1 1.0006e+000 6.0007e-002 5.2943e-007 1.0588e-008 3.7356e-002 1.0006e+000 1.7649e-005 5.2943e-007 6.3506e+003 1.2701e+002 1.0006e+000 6.0007e-002 2.1170e+005 6.3506e+003 3.7356e-002 H1.0006e+000第 2 段840.153.2 103.063/lmJmmkgm2 1.0145e+000 1.5044e-001 1.7595e-006 8.7927e-008 3.8782e-001 1.0145e+000 2.3506e-005 1.7595e-006 4.9669e+004 2.4821e+003 1.0145e+000 1.5044e-001 6.6353e+005 4.9669e+004 3.8782e-001 H1.0145e+000第 3 段840.053.2 103.063/lmJmmkgm3 1.0002e+000 5.0002e-002 1.9531e-007 3.2552e-009 1.4358e-002 1.0002e+000 7.8128e-006 1.9531e-007 5.5135e+003 9.1890e+001 1.0002e+000 5.0002e-002 2.2054e+005 5.5135e+003 1.4358e-002 H1.0002e+000第 4 段840.033.2103.063/lmJmmkgm4 1.0000e+000 3.0000e-002 7.0313e-008 7.0313e-010 3.1013e-003 1.0000e+000 4.6875e-006 7.0313e-008 1.9848e+003 1.9848e+001 1.0000e+000 3.0000e-002 1.3232e+005 1.9848e+003 3.1013e-003 H1.0000e+000第 5 段840.10.9 101.587/lmJmmkgm5 1.0053e+000 1.0011e-001 2.7788e-006 9.2607e-008 2.1163e-001 1.0053e+000 5.5614e-005 2.7788e-006 1.1430e+004 3.8094e+002 1.0053e+000 1.0011e-001 2.2877e+005 1.1430e+004 2.1163e-001 H1.0053e+000第 6 段840.060.9 101.587/lmJmmkgm6 1.0007e+000 6.0008e-002 1.0000e-006 2.0000e-008 4.5706e-002 1.0007e+000 3.3338e-005 1.0000e-006 4.1137e+003 8.2272e+001 1.0007e+000 6.0008e-002 1.3714e+005 4.1137e+003 4.5706e-002 H1.0007e+000此 6 段传递矩阵均采用MATLAB编程求解， MATLAB的源文件为 H.m三、采用传递矩阵法 进行各段轴的状态参...