# Numerical Study on Track-Bridge Interaction of Integral Railway Rigid-Frame Bridge
## 2 Track-Bridge Interaction of Integral Rigid-Frame Bridge
### Differential Equation  
Taking a small length of rail segment \(dx\), under the temperature variation condition, the longitudinal force of the rail microbody is shown in Figure 2.  
Figure 1: Basic mechanism of longitudinal additional force.  
Figure 2: Longitudinal mechanical properties of rail microbody (under temperature variation).  
In light of the obvious nonlinear characteristics of longitudinal resistance between CWRs and bridge (shown in Figure 3), the basic equations of rail stress and displacement can be divided into the following two conditions.  
**I.**\(|\Delta u|\leq u_{s}\) (the longitudinal resistance is in the linear zone).  
The equilibrium equation of rail segment \(dx\) can be expressed as follows:  
\[-N+N+N^{\prime}dx-k\Delta udx=0 \tag{1}\]  
(a) For rails located on the subgrade  
\[\Delta u=u \tag{2}\]  
Equation (1) can be derived as:  
\[-EAu^{\prime\prime}+ku=0 \tag{3}\]  
The general solution of the displacement and the additional force can be obtained as follows:  
\[u(x)=B_{i}e^{\lambda x}+C_{i}e^{-\lambda x} \tag{4}\]  
\[N(x)=EA(\lambda B_{i}e^{\lambda x}-\lambda C_{i}e^{-\lambda x}-\alpha\Delta t) \tag{5}\]  
where \(u\) is the longitudinal displacement of rail; \(\Delta\) is the longitudinal displacement of the bridge; the relative displacement \(\Delta u=u-\Delta\); \(u_{S}\) is the critical point of the displacement in the resistance model; \(N\) is the additional longitudinal force of the rail; \(E\) is the elastic modulus of the rail; \(A\) is the cross-sectional area of the rail; \(k\) is the stiffness coefficient of the springs; and \(\lambda=(k/EA)^{1/2}\).  
(b) For rails located on the bridge  
Equation (3) can be derived as:  
\[-EAu^{\prime\prime}+k(u-\Delta)=0 \tag{6}\]  
The general solution of the displacement and the additional force can be obtained as follows:  
\[u(x)=B_{i}e^{\lambda x}+C_{i}e^{-\lambda x}+\alpha_{0}\Delta Tx \tag{7}\]  
Figure 3: Bilinear model of longitudinal resistance.  
\[N(x)=EA(\lambda B_{i}\epsilon^{\lambda x}-\lambda C_{i}\mathrm{e}^{-\lambda x}+ \alpha_{0}\Delta T-\alpha\Delta t) \tag{8}\]  
**II.**\(|\Delta u|>u_{s}\) (the longitudinal resistance is in the plastic zone and the value is constant). The equilibrium equation of rail segment dx can be expressed as follows:  
\[-N+N+N^{\prime}dx-rdx=0 \tag{9}\]  
(a) For rails located on the subgrade  
Equation (1) can be derived as:  
\[-EAu^{\prime\prime}\pm r=0 \tag{10}\]  
The general solution of the displacement and the additional force can be obtained as follows:  
\[u(x)=\pm px^{2}+B_{i}x+C \tag{11}\]  
\[N(x)=EA(\pm 2 px+B_{i}-\alpha\Delta t) \tag{12}\]  
where \(r\) is the constant of resistance and \(p\) = \(r\)/\(2 EA\).  
(b) For rails located on the bridge  
Equation (3) can be derived as:  
\[-EAu^{\prime\prime}\pm r=0 \tag{13}\]  
The general solution of the displacement and the additional force can be obtained as follows:  
\[u(x)=\pm px^{2}+B_{i}x+C_{i}+\alpha_{0}\Delta Tx \tag{14}\]  
\[N(x)=EA(\pm 2 px+B_{i}+\alpha_{0}\Delta T-\alpha\Delta t) \tag{15}\]  
For the above analysis, the rail tensile stress is considered as positive and compressive stress as negative, whilst the rise of temperature is defined as positive. The fixed end of the bridge is set as the coordinate origin and the rightward \(x\)-axis is defined as positive. The beam displacement \(\Delta\), the rail displacement \(u\) and the relative displacement \((u-\Delta)\) are all regarded as positive when rightward.