## Appendix A  
CFG I: In this configuration, both the reinforced zone and the natural zone remain elastic. Hence, the boundary conditions read:  
\[ \begin{array}{l}\sigma_{tr=r_{0}}=p_{i}\\ \sigma_{tr=\infty}=p_{0}+p_{e}\\ u_{tr=r_{a}-\mathrm{d}r}=u_{tr=r_{a}+\mathrm{d}r}\end{array} , \tag{10}\]  
The solution of the tunnel radial displacement at \(r=r_{0}\) would be:  
\[u_{r_{0}}=-\frac{h^{\prime}r_{0}}{(r_{a}^{2}-r_{0}^{2})}\Big{\{}p_{r_{a}}r_{a} ^{2}(2-2\mu^{\prime})-p_{i}\Big{[}(1-2\mu^{\prime})r_{0}^{2}+r_{a}^{2}\Big{]} \Big{\}}, \tag{11}\]  
where \(h^{\prime}=(1+\mu^{\prime})/E^{\prime}\), \(E^{\prime}\) and \(\mu^{\prime}\) are the elastic modulus and Poisson's ratio of the reinforced ground, respectively, \(p_{r_{a}}\) is the radial stress at \(r=r_{a}\) which can be formulated by:  
\[p_{r_{a}}=-\frac{2E(1-\mu^{\prime 2})r_{0}{}^{2}p_{i}-E^{\prime}(1+\mu)(r_{a}{}^ {2}-r_{0}{}^{2})(p_{0}+p_{e})+\eta}{E^{\prime}(1+\mu)(r_{a}{}^{2}-r_{0}{}^{2} )+E(1+\mu^{\prime})[(1-2\mu^{\prime})r_{a}{}^{2}+r_{0}{}^{2}]}, \tag{12}\]  
where \(\eta=EE^{\prime}(r_{a}{}^{2}-r_{0}{}^{2})\Big{(}u_{r_{a}}^{S}-u_{r_{a}}^{n} \Big{)}/r_{a}\), \(E\) and \(\mu\) in order are the elastic modulus and Poisson's ratio of the natural ground, \(u_{r_{a}}^{S}\) and \(u_{r_{a}}^{n}\) are the radial displacement caused by the seepage effect at \(r=r_{a}-\mathrm{d}r\) (reinforced zone) and \(r=r_{a}+\mathrm{d}r\) (natural zone), respectively. These are given by:  
\[ \begin{array}{l}u_{r_{a}}^{n}=np_{a}r_{a}\Big{[}\Big{(}\frac{R_{a}^{2}}{r_{ a}^{2}}-2\mathrm{ln}\frac{R_{0}}{r_{a}}-1\Big{)}+2\Big{(}\frac{R_{a}^{2}}{r_{a} ^{2}}-1\Big{)}\Big{]}\\ u_{r_{a}}^{S}=n^{\prime}l^{\prime}p_{a}r_{a}\Big{(}\frac{r_{a}^{2}}{r_{a}^{2}}-2 \mathrm{ln}\frac{r_{0}}{r_{a}}-1\Big{)}\end{array} , \tag{13}\]  
where  
\[ \begin{array}{l}n=h(1-2\mu)/\big{[}4(1-\mu)\big{]},h=(1+\mu)/E,n^{\prime}=h^ {\prime}(1-2\mu^{\prime})\big{/}\big{[}4(1-\mu^{\prime})\big{]}\\ h^{\prime}=(1+\mu^{\prime})/E^{\prime},l^{\prime}=\alpha k/\Big{(}k^{\prime} \mathrm{ln}\frac{R_{0}}{r_{a}}+k\mathrm{ln}\frac{r_{a}}{r_{0}}\Big{)},l=\alpha k ^{\prime}/\Big{(}k^{\prime}\mathrm{ln}\frac{R_{0}}{r_{a}}+k\mathrm{ln}\frac{r_ {a}}{r_{0}}\Big{)}\end{array} , \tag{14}\]  
CFG II: In this configuration, the plastic zone only appears in the reinforced ground while the natural zone remains elastic. In addition to Equation (10), the following conditions should be satisfied:  
\[ \begin{array}{l}(\sigma_{r}+\sigma_{\theta})_{|r=R_{1}-dr}=(\sigma_{r}+ \sigma_{\theta})_{|r=R_{1}+dr}\\ \varepsilon_{\theta|r=R_{1}}^{p}=0\end{array} , \tag{15}\]  
where \(R_{1}\) is the plastic zone radius in the reinforced zone. The solution of the tunnel radial displacement at \(r=r_{0}\) yields:  
\[u_{r_{0}}=h^{\prime}r\bigg{[}nr^{\prime}(p_{i}+\beta^{\prime})+\frac{A^{ \prime}{}_{1}-A^{\prime}{}_{2}}{N^{\prime}{}_{\psi}+2}r_{0}+C^{\prime}r_{0}^{-N {}_{\psi}-1}+A^{\prime}{}_{2}\bigg{]}, \tag{16}\]where  
\[ \begin{array}{l}{A^{\prime}}_{1}=\frac{1}{{\mathbb{C}}_{1}-1}\bigg{[}\frac{2{ \mu}^{\prime}-1}{1-\mu}{\mathbb{C}}^{\prime}2-\frac{{\mu}^{\prime}({\mathbb{C}} _{1}^{\prime}+1)-1}{1-\mu}{\mathbb{I}}^{\prime}p_{a}\bigg{]},\,\beta^{\prime}= \frac{C_{2}+{\mathbb{I}}p_{a}}{{\mathbb{C}}_{1}-1},C^{\prime}{}_{1}=\frac{1+ \sin\varphi^{\prime}}{1-\sin\varphi^{\prime}},C^{\prime}{}_{2}=\frac{2{\cal c }^{\prime}\cos\varphi^{\prime}}{1-\sin\varphi^{\prime}},\\ {A^{\prime}}_{2}=\frac{1}{{\mathbb{C}}_{1}-1}\bigg{[}\frac{2{\mu}^{\prime}-1}{ 1-\mu}{\mathbb{C}}^{\prime}2+\Big{(}\frac{{\mu}^{\prime}}{1-\mu}{\mathbb{I}}^ {\prime}-{\cal C}_{1}\Big{)}{\mathbb{I}}^{\prime}p_{a}\bigg{]},m^{\prime}= \frac{(1-\mu)\big{(}1-{\mathbb{C}}_{1}^{2}\big{)}}{N^{\varphi}+{\mathbb{C}} _{1}}+{\cal C}_{1}-\mu^{\prime}({\mathbb{C}}_{1}+1),N^{\varphi}_{\psi}=\frac {1+\sin\psi^{\prime}}{1-\sin\psi^{\prime}}\end{array}  \tag{10}\]  
The plastic zone radius \(R_{1}\) can be calculated from the following set of equations:  
\[