# Study on the Disturbance of Existing Subway Tunnels by Foundation Sloping Excavation
## 3 The Foundations of Theory Calculation
### Exavate on Both Sides Soil and Reserve the Middle Core Soil
In our study, the discount factor method is used to calculate the tunnel flexural stiffness \(EI\) based on the calculation method of tunnel flexural stiffness \(EI\); \(k\) is the stiffness of [PERSON]'s elastic foundation beam [26], and \(Gc\) is the stiffness of the shear layer, which is calculated as follows [27;28]:  
\[k=\frac{1.3E_{t}}{d(1-\mu^{2})}\sqrt[12]{\frac{E_{t}d^{4}}{EI}} \tag{22}\]  
\[G_{c}=\frac{E_{t}h_{t}}{6(1+\mu)} \tag{23}\]  
where \(E_{t}\) and \(\mu\) are the elastic modulus and Poisson's ratio of the soil, respectively, \(E_{t}\) = 1.3\(\sim\)5.0 \(E_{s}\); \(E_{s}\) is the compression modulus of the soil; \(h_{t}\) is the thickness of the shear layer related to the soil characteristics, and \(h_{t}\) is typically considered to be 2.5 times equal to the pipe diameter. Equation (21) is a high-order differential equation, and is difficult to analyze using conventional calculation methods. The finite difference method is one of the most important tools for solving high-order differential equations. The continuous tunnel is discretized into \(2N+5\) elements, and each element length is equals to t. This includes \(2N+1\) elements within the calculation range and two virtual elements on each side of the calculation range. The displacement magnitude of each element is determined by the boundary conditions and the coordination of the displacements between the elements. Figure 8 shows the discretized existing tunnel. Figure 7: Pasternak foundation beam model. According to the finite difference equation, the high-order differential equation can be transformed into:  
\[\left\{ \begin{array}{l}\frac{\partial^{4}w_{y}(z)}{\partial y^{4}}=\frac{6w_{N }(z)-4(w_{N+1}(z)+w_{N-1}(z))+(w_{N+2}(z)+w_{N-2}(z))}{t^{4}}\\ \frac{\partial^{2}w_{y}(z)}{\partial y^{2}}=\frac{w_{N+1}(z)-2w_{N}(z)+w_{N-1 }(z)}{t^{2}}\end{array} \right. \tag{24}\]  
The locations (point 0 and point \(N\)) on the boundary of the calculation range \(B\) are considered to be unaffected by the excavation of the shield tunnel. The bending moment \(M\) and the shear force \(Q\) at the corresponding points are equal to 0. Accordingly, we fit the following:  
\[\left\{ \begin{array}{l}M_{0}=M_{N}=-EI\frac{d^{2}w_{0}(z)}{dy^{2}}=-EI\frac{ d^{2}w_{0}(z)}{dy^{2}}=0\\ Q_{0}=Q_{N}=-EI\frac{d^{2}w_{0}(z)}{dy^{3}}=-EI\frac{d^{2}w_{0}(z)}{dy^{2}}=0 \end{array} \right. \tag{25}\]  
Combining the boundary conditions and the finite difference equation, the displacement expressions for the four virtual nodes outside the calculation range B can be obtained as:  
\[\left\{ \begin{array}{l}w_{-1}=4w_{0}-4w_{1}+w_{2}\\ w_{-2}=2w_{-1}-w_{0}\\ w_{2N+1}=2w_{2N}-w_{2N-1}\\ w_{2N+2}=4w_{2N}-4w_{2N-1}+w_{2N-2}\end{array} \right. \tag{26}\]  
Based on Equations (24) and (26), iterative operations can determine the displacement relationship of all elements after discretization. Substituting it into Equation (21), then \(2N+1\) sets of equations can be obtained, and the corresponding sets of equations are transformed into matrix vectors as follows:  
\[\left[q(z)\right]_{(2N+1)(2N+1)}=\left\{\left[k\right]_{(2N+1)(2N+1)}-\left[G \right]_{(2N+1)(2N+1)}+\left[K\right]_{(2N+1)(2N+1)}\right\}\left[w_{G}(z) \right]_{(2N+1)(2N+1)} \tag{27}\]  
where \([q(z)]_{2N+1}\) is the additional load matrix acting on the discrete element; \([k]_{2N+1}\) is the soil bed coefficient matrix; \([G]_{2N+1}\) is the shear layer stiffness matrix; \([K]_{2N+1}\) is the tunnel stiffness matrix; and \([w_{G}(z)]_{2N+1}\) is the deformation matrix of the discrete element. \[\left[k\right]=k \begin{bmatrix}1&&0\\ &\ddots&&\\ &&\ddots&\\ 0&&&&1\end{bmatrix} _{(2N+1)\times(2N+1)} \tag{28}\]  
Figure 8: Discrete model of the tunnel. \[[G_{c}]=\frac{G_{c}}{t^{2}} \begin{bmatrix}-2&2\\ 1&-2&1\\ &\ddots&\ddots&\ddots\\ &&1&-2&1\\ &&&2&-2\end{bmatrix} _{(2N+1)\times(2N+1)} \tag{29}\]  
\[[K]=\frac{EI}{dt^{4}} \begin{bmatrix}6&-8&2\\ -4&7&-4&1\\ 1&-4&6&-4&1\\ &\ddots&\ddots&\ddots&\ddots\\ &&\ddots&\ddots&\ddots&\ddots\\ &&&&1&-4&6&-4&1\\ &&&1&-4&7&-4\\ &&&&2&-8&6\end{bmatrix}