## 2 Mathematical Model
### Mechanical Characteristics and Sudden Unbalance Excitation Caused by Blade Off
#### 2.1.2 Inertial Asymmetry of the Blade-Disk System  
As mentioned above, the blade off event will make the disk become asymmetric in structure. The motion equation of the damaged blade-disk system can be deduced by  
\[\frac{d}{dt}\bigg{(}\frac{\partial T_{\text{d}}}{\partial\dot{\mathbf{q}}_{\text{ d}}}\bigg{)}-\bigg{(}\frac{\partial T_{\text{d}}}{\partial\dot{\mathbf{q}}_{\text{ d}}}\bigg{)}=0 \tag{2}\]  
where \(\mathbf{q}=[x_{\text{d}},y_{\text{d}},\theta_{\text{xd}},\theta_{\text{yd}}]^{ \text{T}}\). The subscript d is used to identify the disk parameters. \(x_{\text{d}}\) and \(y_{\text{d}}\) are the translational displacements in the \(x\) and \(y\) directions, respectively. \(\theta_{\text{xd}}\) and \(\theta_{\text{yd}}\) are the rotational displacements around the \(x\) and \(y\) directions, respectively. When the blade loss occurs, the polar moment of inertia of the blisk system is reduced to  
\[J_{\text{loss}}=J_{\text{p}}-m_{\text{loss}}R^{2} \tag{3}\]  
where \(J_{\text{p}}\) is the polar moment of the inertia of the intact blisk, and \(m_{\text{loss}}\) denotes the lost mass of the blade.  
The diameter moments of inertia around the \(X\)-axis and \(Y\)-axis change to \(J_{x}=J_{\text{p}}/2-m_{\text{loss}}R^{2}\) and \(J_{y}=J_{\text{p}}/2\), respectively. Then, the kinetic energy of the damaged blade-disk system can be expressed in the form  
\[\begin{array}{rl}T_{\text{d}}=&\frac{1}{2}m_{\text{b}}\Big{(}\dot{x}_{ \text{d}}^{2}+\dot{y}_{\text{d}}^{2}\Big{)}+\frac{1}{2}J_{\text{x}}\Big{(} \dot{\theta}_{\text{xd}}\cos\omega t+\dot{\theta}_{\text{yd}}\sin\omega t \Big{)}^{2}+\\ &\frac{1}{2}J_{\text{y}}\Big{(}\dot{\theta}_{\text{xd}}\sin\omega t-\dot{ \theta}_{\text{yd}}\cos\omega t\Big{)}^{2}+\frac{1}{2}J_{\text{loss}}\Big{(} \omega^{2}+\omega\Big{(}\dot{\theta}_{\text{xd}}\theta_{\text{yd}}-\dot{ \theta}_{\text{yd}}\theta_{\text{xd}}\Big{)}\Big{)}\end{array} \tag{4}\]  
where \(m_{\text{b}}\) is the mass of the disk after the blade is lost. The average diameter moment of inertia (\(J_{\text{m}}\)) and the asymmetrical increment of inertia (\(\Delta J_{\text{yx}}\)) of the blisk system can be expressed as  
\[J_{\text{m}}=\frac{1}{2}\big{(}J_{\text{x}}+J_{\text{y}}\big{)}=\frac{1}{2}J_ {p}-\Delta J \tag{5}\]  
\[\Delta J=\frac{1}{2}\big{(}J_{\text{y}}-J_{\text{x}}\big{)}=\frac{1}{2}m_{ \text{loss}}R^{2} \tag{6}\]  
Equation (6) indicates the asymmetry between \(J_{y}\) and \(J_{x}\). By substituting Equation (4) into Equation (2), the motion equation of the damaged blade-disk system can be derived as  
\[(M_{\text{ds}}+M_{\text{dv}}(t))\ddot{\mathbf{q}}+(G_{\text{ds}}+G_{\text{dv}}(t) )\dot{\mathbf{q}}=0 \tag{7}\]  
where \(M_{\text{ds}}\) and \(G_{\text{ds}}\) represent the constant components of mass and gyroscopic matrices, respectively, and \(M_{\text{dv}}(t)\) and \(G_{\text{dv}}(t)\) denote the corresponding time-varying componentsof mass and gyroscopic matrices, respectively, of which the time-varying period \(T=\pi/\omega\). Those matrices are expressed as follows.  
\[M_{\text{ds}}=\left[\begin{array}{cccc}m_{\text{b}}&0&0&0\\ 0&m_{\text{b}}&0&0\\ 0&0&J_{\text{m}}&0\\ 0&0&0&J_{\text{m}}\end{array}\right] \tag{8}\]  
\[M_{\text{dv}}(t)=-\Delta J\left[\begin{array}{cccc}0&0&0&0\\ 0&0&0&0\\ 0&0&\cos 2\omega t&\sin 2\omega t\\ 0&0&\sin 2\omega t&-\cos 2\omega t\end{array}\right] \tag{9}\]  
\[G_{\text{ds}}=\left[\begin{array}{cccc}0&0&0&0\\ 0&0&0&0\\ 0&0&0&J_{\text{loss}}\omega\\ 0&0&-J_{\text{loss}}\omega&0\end{array}\right] \tag{10}\]  
\[G_{\text{dv}}(t)=2\omega\Delta J\left[\begin{array}{cccc}0&0&0&0\\ 0&0&0&0\\ 0&0&\sin 2\omega t&-\cos 2\omega t\\ 0&0&-\cos 2\omega t&-\sin 2\omega t\end{array}\right] \tag{11}\]