# Study on Coupled Vertical Vehicle-Bridge Dynamic Performance of Medium and Low-Speed Maglev Train
## 2 Coupled Vehicle-Bridge System Considering Elastic Beam
The first mode function \(\sigma_{1}(y)\) can be expressed as:  
Figure 2.0  
where \(m_{r}\) is a constant, and its value is the mass of beam, which can be obtained by \(m_{r}=\rho_{r}L\). Equations (5) and (6) are introduced into Equation (4), and the two sides of Equation (4) are integrated from 0 to \(L\). The following formula is finally obtained:  
Figure 2.1  
Since the beam span is much longer than the length of the electromagnet, the electromagnet is regarded as a point on the beam, and the center of the electromagnet is located in the center of the track, that is \(y_{0}=L/2\). Where \(\tau_{1}\) is the damping ratio of the first mode, \(\omega_{1}\) is the natural frequency of the first mode, it can be calculated as:  
Figure 2.2Both sides of Equation (8) are multiplied by \(\sigma_{1}(y)\), let \(K_{1}=2\sigma_{1}^{2}(y_{0})\), and Equation (8) can then be transformed into:  
Figure 2.3  
Finally, the coupled vehicle-bridge system model considering elastic beam in the first mode is obtained by combining the elastic beam model with the single-point suspension model:  
\[\left\{ \begin{array}{l}x=x_{c}-x_{R},\\ F_{m}=K_{2}\frac{\dot{\sigma}^{2}}{x^{2}},\\ m\ddot{x}_{c}=(m+M)g-F_{m},\\ u=Ri+\frac{2K_{2}}{x}i-\frac{2K_{2}i}{x^{2}}\dot{x},\\ \ddot{x}_{R}(t)+2\tau_{1}\omega_{1}\dot{x}_{R}(t)+\omega_{1}^{2}x_{R}(t)=K_{1}F _{m},\end{array} \right. \tag{11}\]  
where:  
\[K_{1}=\left(\sqrt{\frac{2}{m_{r}}}\sin\Bigl{(}\frac{\pi}{L}y_{0}\Bigr{)} \right)^{2},\,K_{2}=\frac{\mu_{0}N^{2}S}{4}.\]  
The state vector is \(Z=(z_{1},z_{2},z_{3},z_{4},z_{5})^{\mathrm{T}}=\left(x_{c},\dot{x}_{c},x_{R}, \dot{x}_{R},i\right)^{\mathrm{T}}\), where \(z_{1}\) is the vertical displacement of the electromagnet, \(z_{2}\) is the vertical velocity of the electromagnet, \(z_{3}\) is the vertical displacement of the beam, \(z_{4}\) is the vertical velocity of the beam, and \(z_{5}\) is the current of electromagnet coil. The state-space equation of open-loop coupled vehicle-bridge levitation system under the first mode is as follows:  
\[\left\{ \begin{array}{l}\dot{z}_{1}=z_{2},\\ \dot{z}_{2}=\frac{m+M}{m}g-\frac{K_{2}}{m}\Bigl{(}\frac{z_{5}}{z_{1}-z_{3}} \Bigr{)}^{2},\\ \dot{z}_{3}=z_{4},\\ \dot{z}_{4}=K_{1}K_{2}\Bigl{(}\frac{z_{5}}{z_{1}-z_{3}}\Bigr{)}^{2}-2\tau_{1} \omega_{1}z_{4}-\omega_{1}^{2}z_{3},\\ \dot{z}_{5}=\Bigl{(}\frac{z_{5}}{z_{1}-z_{3}}\Bigr{)}(z_{2}-z_{4})-\frac{R}{2K _{2}}(z_{1}-z_{3})z_{5}+\frac{z_{1}-z_{3}}{2K_{2}}u.\end{array} \right. \tag{12}\]  
The premise of stability analysis of Equation (12), based on the Hurwitz criterion, is to solve the singularities of the system. Let \(\dot{z}=\left( \begin{array}{cccc}\dot{z}_{1}&\dot{z}_{2}&\dot{z}_{3}&\dot{ z}_{4}&\dot{z}_{5}\end{array} \right)^{\mathrm{T}}=0\):  
\[\left\{ \begin{array}{l}z_{2}=0,\\ \frac{m+M}{m}g-\frac{K_{2}}{m}\Bigl{(}\frac{z_{5}}{z_{1}-z_{3}}\Bigr{)}^{2}=0, \\ z_{4}=0,\\ K_{1}K_{2}\Bigl{(}\frac{z_{5}}{z_{1}-z_{3}}\Bigr{)}^{2}-2\tau_{1}\omega_{1}z_{4 }-\omega_{1}^{2}z_{3}=0,\\ \Bigl{(}\frac{z_{5}}{z_{1}-z_{3}}\Bigr{)}(z_{2}-z_{4})-\frac{R}{2K_{2}}(z_{1}- z_{3})z_{5}+\frac{z_{1}-z_{3}}{2K_{2}}u=0.\end{array} \right. \tag{13}\]  
The singularity of Equation (13) can be shown as two solutions, \(z^{0}=\left(z_{1}^{0},z_{2}^{0},z_{3}^{0},z_{4}^{0},z_{5}^{0}\right)^{\mathrm{T }}=:\)  
\[\left(1\right)\left(\frac{K_{1}(m+M)g}{\omega_{1}^{2}}+\frac{u}{R}\sqrt{\frac{ K_{2}}{(m+M)g}},0,\frac{K_{1}(m+M)g}{\omega_{1}^{2}},0,\frac{u}{R}\right)^{\mathrm{T}}\]  
\[\left(2\right)\left(\frac{K_{1}(m+M)g}{\omega_{1}^{2}}-\frac{u}{R}\sqrt{\frac{ K_{2}}{(m+M)g}},0,\frac{K_{1}(m+M)g}{\omega_{1}^{2}},0,\frac{u}{R}\right)^{\mathrm{T}}\]  
Since the two solutions have the same characteristics corresponding to the equilibrium point, either of them can be chosen for stability analysis, and solution (1) was selected here. At the equilibrium point \(z^{0}\), the system is linearized as \(\dot{z}=A\big{(}z^{0}\big{)}\big{(}z-z^{0}\big{)}\), where \(A\big{(}z^{0}\big{)}\) is the Jacobian matrix of the system:  
\[A(z^{0})=\left[