## 6 Discussion  
The main content of this paper is based on the fact that a double-row-pipe frozen wall is equivalent to a trapezoidal heterogeneous frozen wall, and the existing research has already compared the double-row-pipe frozen wall to a homogeneous frozen wall [17]. Now we will introduce them briefly, and compare and analyze the two equivalent methods. According to unified strength theory (Equation (18)), the general solution of the elastic zone of the homogeneous frozen wall is  
\[\left\{ \begin{array}{l}\sigma_{r}^{e}=\frac{A}{r^{2}}+2D\\ \sigma_{\theta}^{e}=-\frac{A}{r^{2}}+2D\end{array} \right. \tag{29}\]  
where \(A\) and \(D\) are coefficients. According to the boundary conditions, when \(r=r_{\rho}\), \(\sigma_{r_{\rho}}^{e}=\sigma_{r_{\rho}}\); when \(r=r_{\rho}\), \(\sigma_{r_{\rho}}^{e}=P\). Substituting the boundary conditions into Equation (29), we get:  
\[\left\{ \begin{array}{l}A=\frac{r_{\rho}^{2}r_{\rho}^{2}(P-\sigma_{r_{\rho}}) }{r_{\rho}^{3}-r_{\rho}^{2}}\\ 2D=\frac{r_{\rho}^{2}r_{\rho}-r_{\rho}^{2}P}{r_{\rho}^{2}-r_{\rho}^{2}}\end{array} \right. \tag{30}\]  
The equation of the stress field in the elastic zone is:  
\[\left\{ \begin{array}{l}\sigma_{r}^{e}=\frac{r_{\rho}^{2}}{r^{2}}\frac{r_{ \rho}^{2}(P-\sigma_{r_{\rho}})}{r_{\rho}^{2}-r_{\rho}^{2}}+\frac{r_{\rho}^{2} r_{\rho}-r_{\rho}^{2}P}{r_{\rho}^{2}-r_{\rho}^{2}}\\ \sigma_{\theta}^{e}=-\frac{r_{\rho}^{2}}{r^{2}}\frac{r_{\rho}(P-\sigma_{r_{ \theta}})}{r_{\rho}^{2}-r_{\rho}^{2}}+\frac{r_{\rho}^{2}r_{\rho}-r_{\rho}^{2}P} {r_{\rho}^{2}-r_{\rho}^{2}}\end{array} \right. \tag{31}\]  
The equation of the stress field in the plastic zone is:  
\[\left\{ \begin{array}{l}\sigma_{r}^{p}=\frac{C\cos\varphi}{\sin\varphi}\bigg{[} r_{\rho}^{\frac{4(1+b)\sin\varphi}{(2+b)(1-\sin\varphi)}}-1\bigg{]}\\ \sigma_{\theta}^{p}=\frac{4(1+b)\big{(}C\cos\varphi+c_{\theta}^{p}\sin\varphi \big{)}}{(2+b)(1-\sin\varphi)}+\sigma_{r}^{p}\end{array} \right. \tag{32}\]  
At the interface of elastic-plastic zone, i.e., \(r=r_{\rho}\); according to Equation (31), the radial stress and tangential stress at the interface are:  
\[\left\{ \begin{array}{l}\sigma_{r_{\rho}}^{e}=\frac{r_{\rho}^{2}}{r^{2}} \frac{r_{\rho}^{2}(P-\sigma_{r_{\rho}}^{e})}{r_{\rho}^{2}-r_{\rho}^{2}}+\frac{ r_{\rho}^{2}\sigma_{\rho}-r_{\rho}^{2}P}{r_{\rho}^{2}-r_{\rho}^{2}}=\sigma_{r_{ \rho}}\\ \sigma_{\theta_{r_{\rho}}}^{e}=-\frac{r_{\rho}^{2}}{r_{\rho}^{2}}\frac{r_{\rho} ^{2}(P-\sigma_{\rho})}{r_{\rho}^{2}-r_{\rho}^{2}}+\frac{r_{\rho}^{2}\sigma_{ \rho}-r_{\rho}^{2}P}{r_{\rho}^{2}-r_{\rho}^{2}}=\frac{2r_{\rho}^{2}r_{\rho}(r_ {\rho}^{2}+r_{\rho}^{2})\sigma_{\rho}}{r_{\rho}^{2}-r_{\rho}^{2}}\end{array} \right. \tag{33}\]  
Equation (33) should satisfy strength theory Equation (18), namely:  
\[\sigma_{\theta_{r_{\rho}}}^{e}-\sigma_{r_{\rho}}^{e}=\frac{2r_{\epsilon}^{2}(P -\sigma_{r_{\rho}}^{e})}{r_{\rho}^{2}-r_{\rho}^{2}}=\frac{4(1+b)\Big{[}C\cos \varphi+\sigma_{r_{\rho}}^{e}\sin\varphi\Big{]}}{(1-\sin\varphi)(2+b)} \tag{34}\]  
At the same time, according to the stress continuity conditions at the interface between the elastic zone and the plastic zone, it can be obtained that  
\[\sigma_{r_{\rho}}^{e}=\sigma_{r_{\rho}}^{p}=\frac{C\cos\varphi}{\sin\varphi} \bigg{[}r_{\rho}^{\frac{4(1+b)\sin\varphi}{(2+b)(1-\sin\varphi)}}-1\bigg{]} \tag{35}\]  
Substituting Equation (35) into Equation (34) obtains the relative radius of the plastic zone. In the zone \([1,r_{\rho})\), the radial stress and tangential stress are plotted by Equation (32), respectively. In the zone \([r_{\rho},r_{e}]\), the radial stress and tangential stress are plotted by Equation (31), respectively. These stress fields are compared with the stress field of a heterogeneous frozen wall.