## 3 Solutions for Stress and Displacement of Rock Mass
### Solution in the Residue Region  
In the residue region, Equations (7)-(9) are simplified as Equations (13)-(15).  
\[e_{\rm\theta}^{\rm P}=\frac{(1-\sin\psi_{\rm r})\gamma^{\rm P}}{2} \tag{13}\]  
\[e_{\rm r}^{\rm P}+\frac{1+\sin\psi_{\rm r}}{1-\sin\psi_{\rm r}}e_{\rm\theta}^{ \rm P}=0 \tag{14}\]  
\[\sigma_{\rm\theta}=\frac{1-\sin\varphi_{\rm r}}{1+\sin\varphi_{\rm r}}  
Using Equations (5), (6), and (9), the differential formula \(\mathrm{d}r/r\) can be rewritten as:  
\[\frac{\mathrm{d}r}{r}=\frac{\mathrm{d}R}{R}\exp\left[\frac{1}{2G}\left(\frac{-2 \sin\varphi_{\mathrm{r}}}{1+\sin\varphi_{\mathrm{r}}}\sigma_{\mathrm{r}}+\frac {2c_{\mathrm{r}}\cos\varphi_{\mathrm{r}}}{1-\sin\varphi_{\mathrm{r}}}\right)+ \gamma^{\mathrm{p}}\right] \tag{16}\]  
By substituting radial stress and \(\mathrm{d}r/r\) in Equation (16) into Equation (2), the first differential equation of the stress equilibrium equation can be obtained, expressed as Equation (17).  
\[\frac{\mathrm{d}\sigma_{\mathrm{r}}}{\mathrm{d}R}=\frac{1}{R}\left(\frac{-2 \sin\varphi_{\mathrm{r}}}{1+\sin\varphi_{\mathrm{r}}}\sigma_{\mathrm{r}}+\frac {2c_{\mathrm{r}}\cos\varphi_{\mathrm{r}}}{1-\sin\varphi_{\mathrm{r}}}\right) \exp\left[\frac{1}{2G}\left(\frac{-2\sin\varphi_{\mathrm{r}}}{1+\sin\varphi_{ \mathrm{r}}}\sigma_{\mathrm{r}}+\frac{2c_{\mathrm{r}}\cos\varphi_{\mathrm{r} }}{1-\sin\varphi_{\mathrm{r}}}\right)+\gamma^{\mathrm{p}}\right] \tag{17}\]  
The differential form of the tangential strain can be derived as Equation (18) by differentiating the tangential strain regarding the initial radius in Equation (3) and substituting it into Equation (16).  
\[\frac{\mathrm{d}\varepsilon_{\mathrm{\varphi}}}{\mathrm{d}R}=\frac{1}{R}- \frac{\mathrm{d}r}{r\mathrm{d}R}=\frac{1}{R}\left\{1-\exp\left[\frac{1}{2G} \left(\frac{-2\sin\varphi_{\mathrm{r}}}{1+\sin\varphi_{\mathrm{r}}}\sigma_{ \mathrm{r}}+\frac{2c_{\mathrm{r}}\cos\varphi_{\mathrm{r}}}{1-\sin\varphi_{ \mathrm{r}}}\right)+\gamma^{\mathrm{p}}\right]\right\} \tag{18}\]  
Seeking the derivative of each term in Equation (6) regarding the initial radius \(r\), and another partial differential of the tangential strain can be expressed as Equation (19).  
\[\frac{\mathrm{d}\varepsilon_{\mathrm{\varphi}}}{\mathrm{d}R}=\frac{1}{2G} \left[(1-v)\frac{1-\sin\ \varphi_{\mathrm{r}}}{1+\sin\ \varphi_{\mathrm{r}}}-v\right]\frac{\mathrm{d}\sigma_{ \mathrm{r}}}{\mathrm{d}R}+\frac{1-\sin\psi_{\mathrm{r}}}{2}\frac{\mathrm{d} \gamma^{\mathrm{p}}}{\mathrm{d}R} \tag{19}\]  
From Equations (18) and (19), the second differential equation for the plastic shear strain regarding the initial radius is as follows:  
\[\frac{\mathrm{d}\gamma^{\mathrm{p}}}{\mathrm{d}R}=\frac{k_{1}}{G}\left(v-(1-v )\frac{1}{k}\right)\frac{\mathrm{d}\sigma_{\mathrm{r}}}{\mathrm{d}R}+\frac{2k _{1}}{R}\left\{1-\exp\left[\frac{1}{2G}\left(\frac{-2\sin\varphi_{\mathrm{r}} }{1+\sin\varphi_{\mathrm{r}}}\sigma_{\mathrm{r}}+\sigma_{\mathrm{c}}\right)+ \gamma^{\mathrm{p}}\right]\right\} \tag{20}\]  
where \(k_{1}=1/(1-\sin\psi_{\mathrm{r}})\).  
The tangential stress can be easily obtained by substituting the radial stress into Equation (7) and the displacement of rock mass material point can be calculated using Equations (1), (3), (6), and (13).  
\[u=R\left\{1-\exp\left[\frac{-(1-v)(\sigma_{\mathrm{r}}-P_{0})+v(\sigma_{ \mathrm{\varphi}}-P_{0})}{2G}-\frac{(1-\sin\psi_{\mathrm{r}})\gamma^{\mathrm{ p}}}{2}\right]\right\} \tag{21}\]