## 2 Research Method
### Numerical Model
#### 2.1.2 Dynamic Train Load  
Two approaches are usually used to determine the dynamic train load in numerical simulations: the excitation force function and field measurements. Research indicates that it is feasible to simulate the high-speed train load with the excitation force function [28; 29]. The following is the expression of the dynamic train load when doing this research [26]:  
\[P(t)=k_{1}k_{2}(P_{0}+P_{1}\sin\omega_{1}t+P_{2}\sin\omega_{2}t+P_{3}\sin\omega_ {3}t) \tag{1}\]  
\begin{table}
\begin{tabular}{c c c c} \hline \hline
**Condition** & \multicolumn{2}{c}{**Computing Parameter**} & **Notes** \\ \hline \hline
\begin{tabular}{c} Rock level is V, rock thickness between two tunnels \\ is 1 m, and cross angle is 90\({}^{\circ}\) \\ \end{tabular} & Static analysis \\ \hline \hline
2 & \begin{tabular}{c} III \\ 3 \\ 4 \\ \end{tabular} & \
\begin{tabular}{c} Train passes through the upper tunnel, \\ IV \\ V \\ \end{tabular} &
\begin{tabular}{c} Train passes through the upper tunnel, \\ tunnel cross angle is 90\({}^{\circ}\), and rock thickness \\ between two tunnels is 1 m. \\ \end{tabular} \\ \hline \hline
5 & \
\begin{tabular}{c} Upper tunnel \\ Lower tunnel \\ Simultaneously \\ \end{tabular} &
\begin{tabular}{c} Train speed is 350 km/h, rock level is V, \\ tunnel cross angle is 90\({}^{\circ}\), and rock thickness \\ between two tunnels is 1 m. \\ \end{tabular} \\ \hline
8 & \begin{tabular}{c} 250 km/h \\ 9 \\ 10 \\ \end{tabular} & \
\begin{tabular}{c} Train passes through the upper tunnel, rock \\ 350 km/h \\ \end{tabular} &
\begin{tabular}{c} Train passes through the upper tunnel, rock \\ level is V, tunnel cross angle is 90\({}^{\circ}\), and rock \\ thickness between two tunnels is 1 m. \\ \end{tabular} \\ \hline
11 & \
\begin{tabular}{c} 0\({}^{\circ}\) \\ 30\({}^{\circ}\) \\ 60\({}^{\circ}\) \\ 90\({}^{\circ}\) \\ \end{tabular} &
\begin{tabular}{c} Train passes through the upper tunnel at \\ a speed of 350 km/h, rock level is V, and rock \\ thickness between two tunnels is 3 m. \\ \end{tabular} \\ \hline
15 & \begin{tabular}{c} 1 m \\ 3 \\ 16 \\ 17 \\ \end{tabular} & \
\begin{tabular}{c} Train passes through the upper tunnel at \\ 3 \\ the two tunnels \\ \end{tabular} &
\begin{tabular}{c} Train passes through the upper tunnel at \\ 3 \\ 10 m \\ \end{tabular} \\ \hline \hline \end{tabular}
\end{table}
Table 1: Conditions for the numerical model.  
\begin{table}
\begin{tabular}{c c c c c c} \hline \hline
**Rock Level** & \
\begin{tabular}{c} **Volumetric Weight** \\ **(k/m3)** \\ \end{tabular} & **Elastic Modus (GPa)** &
\begin{tabular}{c} **Passion Ratio** \\ **\(\mu\)** \\ \end{tabular} & **Cohesion (kPa)** & **Internal Friction Angle (\(\varphi\)\({}^{\circ}\))** \\ \hline III & 25 & 6 & 0.30 & 700 & 39 \\ IV & 22 & 0.6 & 0.35 & 100 & 30 \\ V & 19.5 & 0.2 & 0.35 & 50 & 25 \\ \hline \hline \end{tabular}
\end{table}
Table 2: Mechanical parameters of strata.  
Figure 2: Schematic diagram of cross tunnels.  
where \(k_{1}\) and \(k_{2}\) are the adjacent wheel-rail force superposition coefficient and dispersion coefficient, which are in the range of 1.2-1.7 and 0.6-0.9, respectively; \(P_{0}\) is the static vehicle load; \(P_{1}\), \(P_{2}\), and \(P_{3}\) are the vibration load. The mass of the train is set as \(M_{0}\). The following equation is used to calculate the amplitude of the vibration load:  
\[P_{i}=M_{0}a_{i}\omega_{i}^{2} \tag{2}\]  
where \(\omega_{\rm i}\) is the track irregularity rise; \(\omega_{\rm i}=2\pi\upsilon/L_{i}\), \(\upsilon\) is the speed of the high-speed train; and \(L_{i}\) is the typical wavelength of the geometric irregularity curve; \(M_{0}\) is the mass under the lower spring. The value is 750 kg in this paper.  
Referring to the Code [29], wave length and vector height under the three conditions of ride flatness, the additional load acting on railway and wave loss were \(L_{1}=10\) m, \(L_{1}=2\) m, \(L_{3}=0.5\) m, \(a_{1}=3.5\) mm, \(a_{2}=0.4\) mm, and \(a_{3}=0.08\) mm, respectively.