# Analytical Prediction of Tunnel Deformation Beneath an Inclined Plane: Complex Potential Analysis
## 3 Mathematical Analysis
### Complex Potential Analysis
#### 3.2.2 Approximate Analysis of the Tunnel Beneath the Slope
Using \(\omega(\xi)/\overline{\omega^{\prime}(\xi)}\), the displacement is obtained by Equation (8). At the boundary conditions, the new displacement on the slope can be expressed as \(\mu\big{(}u_{x}+iu_{y}\big{)}=2(1-v)/(3-4v)\varphi(\xi)\). When \(p_{1}\) is applied to the limit \(|\xi|=1\), the displacement can be established as follows:  
\[u_{x}(|\xi|=1)=0.61 xu_{\delta}\frac{R}{3-4v}.\frac{y\big{(}(3-4v)h^{2}+R^{2} \big{)}}{\left(xR\right)^{2}+0.71\big{(}y(h+R)\big{)}^{2}} \tag{41}\]  
\[u_{y}(|\xi|=1)=-2.86u_{\delta}\frac{1}{3-4v}.\frac{y^{2}(3-4v)\big{(}h+R\big{)} ^{2}+0.9\left(xR\right)^{2}}{\left(xR\right)^{2}+0.71\big{(}y(h+R)\big{)}^{2}} \tag{42}\]  
The initial vertical translation can be developed as follows:  
\[\Delta u_{y|i}(|\xi|=1)=-3.67\frac{u_{\delta}}{3-4v}\Bigg{\{}\frac{\left(1-3. 53v\right)\Big{(}\frac{R}{h}\Big{)}^{2}+\left(1.15-1.18v\right)\Big{(}\frac{R }{h}\Big{)}^{4}}{\Big{(}\frac{R}{h}\Big{)}^{2}+4.05}\Bigg{\}} \tag{43}\]  
Resulting from Equations (41) and (42), the approximate displacement at the boundary \(|\xi|=1\) can be obtained as:  
\[u_{x}(|\xi|=1)=-\frac{4.07u_{\delta}x(0.45(y-h)(h+R)-h)R^{4}}{\left(3-4v \right)\Big{(}0.26\left(xR\right)^{2}+\left(0.45(y-h)(h+R)-h\right)^{2}\Big{)} h^{2}}\left( \begin{array}{c}0.28(1-v)\big{(}1.07R^{2}+0.75\big{)} \left(xR\right)^{2}+\ldots\\ +\left(1+R^{2}\right)\left(0.45(y-h)(h+R)-h\right)^{2}\\ \hline 0.30\left(xR\right)^{2}+\left(0.45(y-h)(h+R)-h\right)^{2}\end{array} \right) \tag{44}\]\[u_{y}(|\xi|=1)=-\frac{1.07u_{\delta}x(0.45(y-h)(h+R)-h)R^{4}}{(3-4v)\Big{(}0.26( xR)^{2}+(0.45(y-h)(h+R)-h)^{2}\Big{)}h^{2}}\left( \begin{array}{c}(1-v)\big{(}1+2.30R^{ 2})(xR)^{2}+\ldots\\ +(3.11+7.69R^{2})(0.45(y-h)(h+R)-h)^{2}\end{array} \right) \tag{45}\]  
Thus, for \(y=0\) and \(x=\pm h\), the maximum displacement can be obtained by:  
\[u_{y}(|\xi|=1)max=-\frac{0.49u_{\delta}}{(3-4v)}\frac{\Big{(}3.22+\frac{R}{h} \Big{)}\Big{(}\frac{R}{h}\Big{)}}{0.18\Big{(}\frac{R}{h}\Big{)}^{2}-\Big{(}1+ 0.31\frac{R}{h}\Big{)}^{2}}\left\{ \begin{array}{c}0.07(1-v)\bigg{(}0.44+ \Big{(}\frac{R}{h}\Big{)}^{2}\bigg{)}-\ldots\\ -\bigg{(}0.10+\Big{(}\frac{R}{h}\Big{)}^{2}\bigg{)}\big{(}0.78+0.24\frac{R}{h }\Big{)}^{2}\\ 0.20\Big{(}\frac{R}{h}\Big{)}^{2}-\Big{(}1+0.30\frac{R}{h}\Big{)}^{2}\end{array}  \right\} \tag{46}\]  
Therefore, the vertical translation (\(\Delta u_{y}\)) can be rewritten as:  
\[\Delta u_{y}(|\xi|=1)=\frac{4u_{\delta}}{(3-4v)}\left\{ \begin{array}{c}(1-v )+(1+v)\Big{(}\frac{R}{h}\Big{)}^{2}-(1-4v)\Big{(}\frac{R}{h}\Big{)}^{3}+2 \Big{(}\frac{R}{h}\Big{)}^{4}+(2-v)\Big{(}\frac{R}{h}\Big{)}^{6}+2\Big{(} \frac{R}{h}\Big{)}^{7}+\ldots\\ +(4-v)\Big{(}\frac{R}{h}\Big{)}^{8}+5\Big{(}\frac{R}{h}\Big{)}^{9}+2\Big{(} \frac{R}{h}\Big{)}^{10}\\ \Big{(}\frac{R}{h}\Big{)}^{2}-2\end{array} \right\} \tag{47}\]  
With \(R=h\),\(\Delta u_{y}max(|\xi|=1)=\pm(0.7+0.31v)u_{\delta}/(3-4v)\) (maximum vertical translation). Consequently, the approximate solution resulting from the final shape of the displacement can be obtained as follows (\(|\xi|=1\)):  
\[U_{f}(|\xi|=1)=\frac{u_{\delta}}{(3-4v)}\left\{\frac{(1196-3220v)\Big{(}\frac{ R}{h}\Big{)}^{2}+(162-160v)\Big{(}\frac{R}{h}\Big{)}^{4}-\Big{(}\frac{R}{h} \Big{)}^{5}-(5-4v)\Big{(}\frac{R}{h}\Big{)}^{6}-10,580+12,696v}{\Big{(}\Big{(} \frac{R}{h}\Big{)}^{2}-23\Big{)}^{2}}\right\} \tag{48}\]  
Considering \(R=h\), the maximum final shape of displacement can be obtained by \(U_{f}(|\xi|=1)max=\pm 19u_{\delta}(1-v)/(3-4v)\). Comparing the maximum displacement resulting from Equation (48) and the condition \(2\mu(u_{x}+iu_{y})=2(1-v)/(3-4v)\varphi(\varsigma)\), the approximate results are roughly equal to the exact solution.