## 4 Model Solving  
### Solution of Stress Field in the Elastic State  
Under the elastic state, according to the theoretical solution of a thick-walled cylinder in elasticity [34], the analytical solution of the frozen wall stress is as follows:  
\[\left\{\begin{array}{l}\sigma_{r}=\frac{\sigma_{r_{1}}[a_{1}(-\frac{1}{2})+ \frac{b_{1}}{2}(1-\frac{1}{2})]}{\left(1-\frac{1}{r_{1}}\right)a_{1}+\left(1- \frac{1}{2}\right)\frac{b_{1}}{2}}\\ \sigma_{\theta}=\frac{\sigma_{r_{1}}[a_{1}+\frac{b_{1}}{2}(1+\frac{1}{2})]}{ \left(1-\frac{1}{r_{1}}\right)a_{1}+\left(1-\frac{1}{2}\right)\frac{b_{1}}{2} }\end{array}\right.\quad r\in[1,r_{1}] \tag{5}\]  
\[\left\{\begin{array}{l}\sigma_{r}=\frac{1}{r_{2}^{2}-r_{1}^{2}}\left[\frac{ r_{1}^{2}r_{2}^{2}(\sigma_{r_{2}}-\sigma_{r_{1}})}{r^{2}}+\left(r_{2}^{2} \sigma_{r_{2}}-r_{1}^{2}\sigma_{r_{1}}\right)\right]\\ \sigma_{\theta}=\frac{1}{r_{2}^{2}-r_{1}^{2}}\left[\frac{r_{1}^{2}r_{2}^{2}( \sigma_{r_{2}}-\sigma_{r_{1}})}{r^{2}}+\left(r_{2}^{2}\sigma_{r_{2}}-r_{1}^{2} \sigma_{r_{1}}\right)\right]\end{array}\right.\quad r\in[r_{1},r_{2}] \tag{6}\]  
\[\left\{\begin{array}{l}\sigma_{r}=\frac{(P-\sigma_{r_{2}})\left(\frac{a_{1}} {7}-\frac{b_{2}}{2}\right)-a_{1}\left(\frac{P}{2}-\frac{\sigma_{r_{2}}}{7r_{e }}\right)+\frac{b_{2}}{2}\left(\frac{P}{r_{2}^{2}}-\frac{\sigma_{r_{2}}}{r_{e }^{2}}\right)}{\left(\frac{1}{r_{1}}-\frac{1}{2}\right)a_{1}+\frac{b_{2}}{2} \left(\frac{1}{r_{2}^{2}}-\frac{1}{r_{1}^{2}}\right)}\\ \sigma_{\theta}=\frac{\frac{b_{2}}{2r_{2}}(P-\sigma_{r_{2}})-a_{1}\left(\frac {P}{2}-\frac{\sigma_{r_{2}}}{7r_{e}}\right)+\frac{b_{2}}{2}\left(\frac{P}{r_{ 2}^{2}}-\frac{\sigma_{r_{2}}}{r_{e}^{2}}\right)}{\left(\frac{1}{r_{1}}-\frac{1 }{2}\right)a_{1}+\frac{b_{2}}{2}\left(\frac{1}{r_{2}^{2}}-\frac{1}{r_{2}^{2}} \right)}\end{array}\right.\quad r\in[r_{2},r_{e}]\right. \tag{7}\]  
where \(\sigma_{r}\) is the radial stress, MPa; \(\sigma_{\theta}\) is the tangential stress, MPa; \(P\) is the external load; \(\sigma_{r_{1}}\) is the radial stress of the inner frozen pipe, MPa; \(\sigma_{r_{2}}\) is the radial stress of the outer frozen pipe, MPa; and \(P\) is the external load, MPa.  
According to the displacement continuity condition, we solve \(\sigma_{r_{1}}\) and \(\sigma_{r_{2}}\), where \(u_{1}^{-}\) is the displacement of the inner boundary when the relative radius \(r=r_{1}\), m; \(u_{1}^{+}\) is the displacement of the outer boundary when the relative radius \(r=r_{1}\), m; \(u_{2}^{-}\) is the displacement of the inner boundary when the relative radius \(r=r_{2}\), m; and \(u_{2}^{+}\) is the displacement of the outer boundary when the relative radius \(r=r_{2}\), m.  
According to \(u_{1}^{-}=u_{1}^{+}\), we get  
\[-\mu\sigma_{r_{1}}+(1-\mu)\frac{\sigma_{r_{1}}[a_{1}+\frac{b_{1}}{2}\left(1+ \frac{1}{r^{2}}\right)]}{\left(1-\frac{1}{r_{1}}\right)a_{1}+\left(1-\frac{1}{ r^{2}}\right)\frac{b_{1}}{2}}=\frac{(2-2\mu)r_{2}^{2}}{r_{2}^{2}-r_{1}^{2}} \sigma_{r_{2}}-\frac{(1-2\mu)r_{1}^{2}+r_{2}^{2}}{r_{2}^{2}-r_{1}^{2}}\sigma_{r _{1}} \tag{8}\]  
where \(\mu\) is [PERSON]'s ratio.  
According to \(u_{2}^{-}=u_{2}^{+}\), we get:  
\[\begin{array}{l}\frac{r_{1}^{2}+(1-2\mu)r_{2}^{2}}{r_{2}^{2}-r_{1}^{2}} \sigma_{r_{2}}-\frac{(2-2\mu)r_{2}^{2}}{r_{2}^{2}-r_{1}^{2}}\sigma_{r_{1}}= \frac{-\mu\left[(P-\sigma_{r_{2}})\left(\frac{a_{1}}{7}-\frac{b_{2}}{2r_{2}^{2} }\right)-a_{1}\left(\frac{P}{2}-\frac{\sigma_{r_{2}}}{7r_{e}}\right)+\frac{b_{2 }}{2}\left(\frac{P}{r_{2}^{2}}-\frac{\sigma_{r_{2}}}{r_{e}^{2}}\right)\right]} {\left(\frac{1}{r_{e}}-\frac{1}{r_{2}}\right)a_{1}+\frac{b_{2}}{2}\left(\frac{1 }{r_{2}^{2}}-\frac{1}{r_{e}^{2}}\right)}\end{array} \tag{9}\]  
By simultaneous Equations (8) and (9), we get \(\sigma_{r_{1}}\) and \(\sigma_{r_{2}}\).