# 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
To the limit, \(|\varsigma|=\alpha\), Equation (8) can be re-expressed as:  
\[2\mu(u_{x}+iu_{y})(|\xi|=\alpha)=\left\{\frac{0.04x(h+R)^{2}y\;h\;R^{2})}{0.19 y^{2}(h+R)^{2}+0.07x^{2}R^{4}}+i\Bigg{\{}(0.45y(h+R)-h)+\frac{6.8(1-v)h^{2}}{y(h+R) ^{2}}\Bigg{\}}\right\} \tag{49}\]  
Separating the real from the imaginary, the displacement can be rewritten as:  
\[u_{x}(|\varsigma|=\alpha)=1.79x\frac{u_{\delta}}{3-4v}R\left\{\frac{0.04(y+h)( 3-4v)(h+R)^{2}-R^{2}}{(1-v)\Big{(}0.71(y+h)^{2}(h+R)^{2}\Big{)}+(xR)^{2}}\right\} \tag{50}\]  
\[u_{y}(|\varsigma|=\alpha)=\frac{u_{\delta}}{3-4v}\Bigg{\{}\frac{(3-4v)(0.45(y+h )(h+R)-h)}{4h(1-v)}+\frac{0.75R^{2}h}{(y+h)^{\prime}h+R)}\Bigg{\}} \tag{51}\]  
\[u_{x}(|\varsigma|=\alpha)=1.79x\frac{u_{\delta}}{3-4v}R\left\{\frac{0.04(y-h)(3 -4v)(h+R)^{2}-R^{2}}{(1-v)\Big{(}0.71(y-h)^{2}(h+R)^{2}\Big{)}+(xR)^{2}}\right\} \tag{52}\]\[u_{y}(|\varsigma|=\alpha)=\frac{u_{\delta}}{3-4v}\bigg{\{}\frac{(3-4v)(0.45(y-h)(h+ R)-h)}{4h(1-v)}+\frac{0.75R^{2}h}{(y-h)(h+R)}\bigg{\}} \tag{53}\]  
Hence, for \(|\varsigma|=\alpha\), the displacement acting on the inner tunnel can be given by:  
\[u_{x}(|\varsigma|=\alpha)=\frac{u_{\delta}}{3-4v}\frac{R}{h}\left\{ \begin{array} []{c}\frac{(3-4v)\big{\{}0.02((y-h)(h+R)-2.5h)(xR)^{2}-2.20(1-v)(h^{2}+0.25(xR )^{2}\big{\}}R^{2}\big{\}}{0.42(h+R)(y-h)^{2}+0.28(xR)^{2}}+\ldots\\ +\left(\begin{array}{c}0.15(y-h)^{2}(h+R)^{2}-\ldots\\ -(y-h)(h+R)+(2.48-v)h^{2}\end{array}\right)^{R^{2}}\\ (0.45(y-h)(h+R)-h)h\end{array} \right)\right\} \tag{54}\]  
\[u_{y}(|\varsigma|=\alpha)=0.55x\frac{u_{\delta}}{3-4v}\frac{R^{2}}{h}\left\{ \begin{array}{c}\frac{0.23(3-4v)(0.21((y-h)(h+R))^{2}-h^{2})}{1.61(y-h)^{2} (h+R)+(xR)^{2}}+\ldots\\ +\left(\begin{array}{c}(y-h)(h+R)\big{\{}0.01(y-h)(h+R)-0.45h(0.17+R^{4}(1-v ))\big{\}}+\ldots\\ +0.11h^{2}+h(0.04+(1-v)h)R^{4}\end{array}\right)\end{array} \right\} \tag{55}\]  
Using Equations (55) and (56), for \(x=0\) and \(y=\pm h\), \(u_{x}\text{max}=0\), \(y=0\) and \(x=\pm h\), \(u_{y}(|\varsigma|=\alpha)max=\pm u_{\delta}(0.64+0.33v)/(3-4v)\). The vertical translation can be written as follows:  
\[\Delta u_{y}(|\varsigma|=\alpha)=-\frac{0.25u_{\delta}}{(3-4v)}=\left\{\frac{ (3v-2)\Big{(}\frac{R}{h}\Big{)}+(2+9v)\Big{(}\frac{R}{h}\Big{)}^{2}+4(1-v) \Big{(}\frac{R}{h}\Big{)}^{3}+2(3-2v)\Big{(}\frac{R}{h}\Big{)}^{4}+2}{4+\Big{(} \frac{R}{h}\Big{)}^{2}}\right\} \tag{56}\]  
Therefore, the final shape of the displacement can be given by:  
\[U_{f}(|\varsigma|=\alpha)=5.33\frac{u_{\delta}}{(3-4v)}\left\{\frac{(23.5-46v )\Big{(}\frac{R}{h}\Big{)}^{2}-2(1.9+5.81)\Big{(}\frac{R}{h}\Big{)}^{3}-(0.1+ 12.12v)\Big{(}\frac{R}{h}\Big{)}^{4}}{\Big{(}1+\frac{R}{h}\Big{)}^{4}}\right\} \tag{57}\]  
For \(R=h\), Equations (57) and (58) become \(U_{f}\Big{|}_{|\varsigma|=1}\text{max}=\pm u_{\delta}/(3-4v)\) and \(U_{f}(|\varsigma|=\alpha)max=\pm u_{\delta}/(3-4v)\). Due to the polar nature of the deformation being evaluated, the radial and tangential stresses and radial and tangential displacement are calculated to estimate the tunnel rotation when downstream loads are low. Particular attention is paid to the pressures and displacements resulting from \"settlement\" and \"buoyancy\" on the ground surface. According to [PERSON] et al. [41], although complex geometry poses many geotechnical design problems, using numerical modelling to obtain more realistic results than conventional analytical solutions is necessary to get an idea of the general nature of the solution.