## 2 Misalignment and Rub-Impact Coupling Dynamic Modeling  
### Misalignment Model  
The misalignment of the two ends of the rotor supports leads to an angle between the driving torque and the shaft, which brings an additional bending moment to the rotor system. The schematic diagram of the misalignment model is shown in Figure 1. The airflow excitation force moment \(T\) is horizontally left, the misalignment angle is \(\alpha\); and the angle between the \(T_{s}\) moment and the \(y\) axis is \(\beta\). \(T\) can be decomposed into \(T_{a}\) that coincides with the shaft and \(T_{s}\) that is perpendicular to the shaft, and \(T_{s}\) can be decomposed into \(T_{x}\) and \(T_{y}\). According to the geometric relationship,  
\[T_{a} =T\cos\alpha \tag{1}\] \[T_{s} =T\sin\alpha \tag{2}\](3)  
\[T_{x}=T_{s}\cos\beta=T\sin\alpha\sin\beta \tag{4}\]  
Supposing that the displacement difference of the bearing misalignment at both ends of the rotor is \(\Delta x\) and \(\Delta y\), respectively, then \(\alpha\) and \(\beta\) are  
\[\alpha=arct\frac{\sqrt{\Delta x^{2}+\Delta y^{2}}}{l} \tag{5}\]  
\[\beta=arct\frac{\Delta y}{\Delta x} \tag{6}\]  
Since the radial displacement of the rotor at the bearing is much smaller than the shaft length, \(\alpha\) and \(\beta\) can be regarded as fixed values. According to [PERSON]'s Equation for the rotation around a fixed point, \(T_{x}\), \(T_{y}\), \(T_{z}\) are expressed as  
\[\begin{cases}T_{x}=I_{x}\dot{\omega}_{x}+\omega_{y}\omega_{z}(T_{y}-T_{z})\\ T_{y}=I_{y}\omega_{y}+\omega_{x}\omega_{z}(T_{z}-T_{x})\\ T_{z}=I_{z}\omega_{z}+\omega_{x}\omega_{y}(T_{x}-T_{y})\end{cases} \tag{7}\]  
where \(\omega_{i}\) is the angular velocity of the rotor; \(I_{i}\) is the principal moments of the inertia of the rotor shaft, \(i=x\), \(y\), \(z\); due to the rotation of the rotor around the \(z\) axis, \(\omega_{x}\), \(\omega_{y}\), \(\dot{\omega}_{x}\), \(\dot{\omega}_{y}\) are all zero. Therefore, [PERSON]'s Equation of rotating around a fixed point can be simplified as  
\[T_{z}=I_{z}\varepsilon_{z} \tag{8}\]  
where \(\varepsilon_{z}\) is the rotor angular acceleration. After deriving, the angular velocity and the misalignment angle satisfy the following relationship  
\[\omega/\omega_{d}=M/(1+N\cos 2\theta_{d}) \tag{9}\]  
where \(\omega\) is the rotational angular velocity of the rotor; \(\omega_{d}\) is the angular velocity of the air-excited turbine disk; and \(\theta_{d}\) is the angular displacement of the turbine disk. Under the small angle misalignment, the turbine disk and the rotor speed are equal, i.e., \(\omega_{d}\approx\omega\). \(M\) and \(N\) are the coefficients related to the misalignment angle. \(M=4\text{cos}\alpha/(3+\text{cos}2\alpha)\), \(N=(1-\text{cos}2\alpha)/(3+\text{cos}2\alpha)\). After differentiating Equation (9), one can get  
\[\varepsilon_{z}=\frac{2M\cdot N\sin(2\omega t)}{[1+N\cos(2\omega t)]^{2}} \omega^{2} \tag{10}\]  
Figure 1: Schematic diagram of the misalignment model.  
Substituting Equation (10) into Euler's Equation, one can get  
\[T=\frac{2I_{z}\cdot\omega^{2}\cdot M\cdot N\sin(2\omega t)}{\cos\alpha\left[1+N \cos(2\omega t)\right]^{2}} \tag{11}\]  
By decomposing the misalignment torque, it can be obtained  
\[T_{y}=\frac{2I_{z}\cdot\omega^{2}\cdot M\cdot N\sin(2\omega t)}{\cos\alpha \cdot\left[1+N\cos(2\omega t)\right]^{2}}\sin\alpha\cos\beta \tag{12}\]  
\[T_{x}=\frac{2I_{z}\cdot\omega^{2}\cdot M\cdot N\sin(2\omega t)}{\cos\alpha \cdot\left[1+N\cos(2\omega t)\right]^{2}}\sin\alpha\sin\beta \tag{13}\]