# Vibration Responses of a Coaxial Dual-Rotor System with Supporting Misalignment
## 2 Modeling of a Dual-Rotor System with Supporting Misalignment
### Bearing Load of the Inner Rotor  
The force analysis of the inner rotor is shown in Figure 5. The inner rotor is a redundantly supported beam. In order to obtain the bearing load, the fixed constraint of Bearing 2 is replaced by bearing load \(Y_{2}\) and a pair of bending moments \(M_{2}\), and then the redundantly supported beam is decomposed into two simply supported beams. Finally, bearing loads of the inner rotor can be calculated by the deformation compatibility principle and the moment balance equation.  
According to the deformation compatibility principle, the deflection angles caused by both various loads and bearing misalignment are equal on the left and right end faces of Bearing 2. It is stipulated that the counterclockwise deflection angles are positive and the clockwise deflection angles are negative. Thus, the deflection angle equations can be obtained as follows:  
\[\theta_{2}^{R}=\frac{-L_{23}M_{2}}{3\text{E}I_{23}}+\frac{Y_{5}a_{21}(L_{23}- a_{21})(2L_{23}-a_{21})}{6L_{23}\text{E}I_{23}}-\frac{m_{3}ga_{22}(L_{23}-a_{22})(2L_{23}- a_{22})}{6L_{23}\text{E}I_{23}}-\frac{q_{23}(L_{23})^{3}}{24\text{E}I_{23}}- \frac{\Delta_{3}}{L_{23}}+\frac{\Delta_{2}}{L_{23}} \tag{3}\]  
\[\theta_{2}^{L}=\frac{L_{12}M_{2}}{3\text{E}I_{12}}+\frac{m_{1}ga_{11}(L_{12}- a_{11})(L_{12}+a_{11})}{6L_{12}\text{E}I_{12}}+\frac{m_{2}ga_{12}(L_{12}-a_{12})(L_{12}+ a_{12})}{6L_{12}\text{E}I_{12}}+\frac{q_{12}(L_{12})^{3}}{24\text{E}I_{12}}- \frac{\Delta_{2}}{L_{12}}+\frac{\Delta_{1}}{L_{12}} \tag{4}\]  
where \(\theta_{2}^{R}\) and \(\theta_{2}^{L}\) are the deflection angles of the right and left ends of Bearing 2, respectively; \(L_{12}\) and \(L_{23}\) are the bearing spacing; \(M_{2}\) is the bending moment caused by the over-constraint of Bearing 2; \(E\) is the elastic modulus of the shaft; \(I_{12}\) and \(I_{23}\) are the moment of inertia; \(a_{11}\), \(a_{12}\), \(a_{21}\), and \(a_{22}\) represent the distances from the different loads to Bearings 1 and 2, respectively; \(\Delta_{1}\), \(\Delta_{2}\), and \(\Delta_{3}\) represent the misalignment of the three bearings, respectively. It is specified that the misalignment is negative if the bearing is padded up.  
Set \(\theta_{2}^{R}\) equal to \(\theta_{2}^{L}\), and obtain the bending moment M2. Then, take each simply supported beam of the inner rotor as the research object, and then the moment balance equations are obtained as follows:  
\[\left\{\begin{array}{l}Y_{2}^{L}L_{12}+M_{2}-m_{1}ga_{11}-m_{2}ga_{12}- \frac{1}{2}(L_{12})^{2}q_{12}=0\\ -Y_{1}L_{12}+M_{2}+m_{1}g(L_{12}-a_{11})+m_{2}g(L_{12}-a_{12})+\frac{1}{2}(L_{ 12})^{2}q_{12}=0\end{array}\right. \tag{5}\]  
\[\left\{\begin{array}{l}-Y_{2}^{R}L_{23}-M_{2}-Y_{5}(L_{23}-a_{21})+m_{3}g(L_ {23}-a_{22})+\frac{1}{2}(L_{23})^{2}q_{23}=0\\ Y_{3}L_{23}-M_{2}+Y_{5}a_{21}-m_{3}ga_{22}-\frac{1}{2}(L_{23})^{2}q_{23}=0 \end{array}\right. \tag{6}\]  
where \(m_{1}\), \(m_{2}\), and \(m_{3}\) represent the masses of Disks 1, 2, and 3, respectively; \(Y_{2}^{L}\) and \(Y_{2}^{R}\) are the shear forces on the left and right faces of Bearing 2, respectively; \(q_{12}\) is the uniform distributed mass of the inner rotor.