## 2 Theory of Tunnel Deformation under Surcharge Loading
### Responses of Tunnels to External Loads
Then, the bending moment \(M\) and the shearing force are 0 at both ends, and they are determined, respectively, as follows:  
\[M_{0}=M_{n}=-(EI)_{eq}\left.\frac{d^{2}w}{dx^{2}}\right|_{x=0,n}=0 \tag{6}\]  
\[Q_{0}=Q_{n}=-(EI)_{eq}\left.\frac{d^{3}w}{dx^{3}}\right|_{x=0,n}=0 \tag{7}\]  
According to the finite difference principle,  
\[M_{0}=-(EI)_{eq}\left.\frac{d^{2}w}{dx^{2}}\right|_{x=0}=-(EI)_{eq}\frac{w_{1} -2w_{0}+w_{-1}}{l^{2}}=0 \tag{8a}\]  
\[M_{n}=-(EI)_{eq}\left.\frac{d^{3}w}{dx^{3}}\right|_{x=n}=-(EI)_{eq}\frac{w_{n+ 1}-2w_{n}+w_{n-1}}{l^{2}}=0 \tag{8b}\]  
\[Q_{0}=-(EI)_{eq}\left.\frac{d^{3}w}{dx^{3}}\right|_{x=0}=-(EI)_{eq}\frac{w_{2 }-2w_{1}+2w_{-1}-w_{-2}}{2l^{3}}=0 \tag{8c}\]  
\[(EI)_{eq}\frac{\mathrm{d}^{4}w}{\mathrm{d}x^{4}}-G_{c}D_{t}\frac{\mathrm{d}^ {2}w}{\mathrm{d}x^{2}}+kD_{t}w=q(x)D_{t} \tag{4}\]0  
The horizontal displacement expressions of the four newly added virtual elements can be obtained with (8a), (8b), (8c), and (8d):  
\[(EI)_{eq}\frac{\mathrm{d}^{4}w}{\mathrm{d}x^{4}}-G_{c}D_{t}\frac{\mathrm{d}^ {2}w}{\mathrm{d}x^{2}}+kD_{t}w=q(x)D_{t} \tag{4}\]1  
\[(EI)_{eq}\frac{\mathrm{d}^{4}w}{\mathrm{d}x^{4}}-G_{c}D_{t}\frac{\mathrm{d}^ {2}w}{\mathrm{d}x^{2}}+kD_{t}w=q(x)D_{t} \tag{4}\]2  
\[(EI)_{eq}\frac{\mathrm{d}^{4}w}{\mathrm{d}x^{4}}-G_{c}D_{t}\frac{\mathrm{d}^ {2}w}{\mathrm{d}x^{2}}+kD_{t}w=q(x)D_{t} \tag{4}\]3  
\[(EI)_{eq}\frac{\mathrm{d}^{4}w}{\mathrm{d}x^{4}}-G_{c}D_{t}\frac{\mathrm{d}^ {2}w}{\mathrm{d}x^{2}}+kD_{t}w=q(x)D_{t} \tag{4}\]4  
Equation (5) is expressed in matrix form as  
\[(EI)_{eq}\frac{\mathrm{d}^{4}w}{\mathrm{d}x^{4}}-G_{c}D_{t}\frac{\mathrm{d}^ {2}w}{\mathrm{d}x^{2}}+kD_{t}w=q(x)D_{t} \tag{4}\]5  
After sorting and merging similar items, the following results are obtained:  
\[(EI)_{eq}\frac{\mathrm{d}^{4}w}{\mathrm{d}x^{4}}-G_{c}D_{t}\frac{\mathrm{d}^ {2}w}{\mathrm{d}x^{2}}+kD_{t}w=q(x)D_{t} \tag{4}\]6  
where \(\{w\}\) is the column vector of the longitudinal displacement of the tunnel, \(\{w\}=\{w(x_{0}),w(x_{1}),\cdots w(x_{i}),\cdots w(x_{n-1}),w(x_{n}),\}^{T}\); \(\{p\}\) is the column vector of the external load in the longitudinal direction caused by basement excavation; and  
\[(EI)_{eq}\frac{\mathrm{d}^{4}w}{\mathrm{d}x^{4}}-G_{c}D_{t}\frac{\mathrm{d}^ {2}w}{\mathrm{d}x^{2}}+kD_{t}w=q(x)D_{t} \tag{4}\]7 Finite difference form of the existing tunnel. \(\{Q\}=\{q(x_{0}),q(x_{1}),\cdots q(x_{i}),\cdots q(x_{n-1}),q(x_{n})\}^{\text{T}}\). The expressions of \([K_{1}]\), \([K_{2}]\), and \([K_{3}]\) are obtained from the following virtual cell expression:  
\[[K_{1}]=\frac{\left(EI\right)_{eq}}{l^{4}}\left[ \begin{array}{ccccccccc}2&-4&2 &&&&&&&\\ -2&5&-4&1&&&&\\ 1&-4&6&-4&1&&&\\ &\ddots&\ddots&\ddots&\ddots&\ddots&&\\ &&&1&-4&6&-4&1&\\ &&&&1&-4&6&-4&1\\ &&&&1&-4&5&-2\\ &&&&2&-4&2\end{array} \right]_{(n+1)\times(n+1)} \tag{12}\]  
\[[K_{2}] =\text{D}_{k}\mathbf{k}\left[ \begin{array}{ccccccccc}1&&&&&&&\\ &1&&&&&&\\ &&1&&&&\\ &&&\ddots&&\\ &&&&1&&\\ &&&&1&\\ &&&&&&&1\end{array} \right]_{(n+1)\times(n+1)} \tag{13}\]  
\[[K_{3}]=G_{e}\text{D}_{t}\left[ \begin{array}{ccccccccc}0&0&0&&&&\\ 1&-2&1&&&\\ &\ddots&\ddots&\ddots&&\\ &&1&-2&1&\\ &&&1&-2&1\\ &&&1&-2&1\\ &&&0&0&0\end{array} \right]_{(n+1)\times(n+1)} \tag{14}\]  
Let \([K]=[K_{1}]+[K_{2}]-[K_{3}]\); then, Equation (11) can be further rewritten:  
\[\{w\}=[K]^{-1}\{Q\} \tag{15}\]  
where \([K]^{-1}\) is the inverse of \([K]\). By solving (1)-(3), the external load \(q(x)\) at the existing tunnel position caused by the surcharge can be obtained. The longitudinal displacement \(\{w\}\) of the existing tunnel caused by ground surcharge can be obtained by combining Equations (12)-(15).