## 3 Analytical Solutions
### Analytical Solution Calculation  
Each ring steel corrugated plate-mold bag concrete structure can be regarded as a closed cylinder, and its force is shown in Figure 5a. According to the symmetry of load, structure, and boundary conditions, the internal force of the structure can be calculated by the uniform pressure calculation model of the thin-walled cylinder in classical elastic mechanics theory. Figure 5b shows the force model diagram of the mold bag concrete subjected to internal pressure \(q_{2}\) (inter-structure internal force) and external pressure \(q_{1}\) (active earth pressure). Figure 5c shows the force model diagram of the steel corrugated plate with the outer radius \(R_{2}\) and the inner radius \(R_{3}\) subjected to the external pressure \(q_{2}\). In this paper, the actual excavation radius of the field process test is 3.12 m (\(R_{1}\)), and the design radius of the circular shaft is 3.0 m (\(R_{3}\)). The study was carried out assuming a large and negligible difference in the amplitude of the wave surface compared to the radius of the corrugated steel shell-like structure and the lack of major interfacial contact traction between the two main bodies mentioned above. It can be known from elastic mechanics that in the axisymmetric problem, the stress component in the polar coordinate plane is only a function of \(\rho\), which does not change with \(\varphi\), and the shear stress \(\tau_{\rho\varphi}\) is zero. Due to the different material properties of the inner and outer cylinders, the  
Figure 4: Introduction of the cross-section and load. Figure 5: Cylindrical structure plane force model. (**a**) Closed cylinder (**b**) the force model of inter-structure force (**c**) the force model of external pressure. assumption of uniformity is not met, so the solution cannot be expressed by the same function. The force analysis of the inner and outer cylinders in Figure 5 is carried out, respectively, and their respective stress boundary conditions are as follows:  
\[\text{Mold bag concrete}:\ \left\{ \begin{array}{l}\left(\tau_{\varphi 1} \right)_{\rho=R_{1}}=0\\ \left(\tau_{\varphi 1}\right)_{\rho=R_{2}}=0\end{array} \right.\text{ and }\left\{ \begin{array}{l}\left( \sigma_{\rho 1}\right)_{\rho=R_{1}}=-q_{1}\\ \left(\sigma_{\rho 1}\right)_{\rho=R_{2}}=-q_{2}\end{array} \right. \tag{1}\]  
\[\text{Steel corrugated plate}:\ \left\{ \begin{array}{l}\left(\tau_{\varphi \theta 2}\right)_{\rho=R_{2}}=0\\ \left(\tau_{\varphi 2}\right)_{\rho=R_{3}}=0\end{array} \right.\text{ and }\left\{ \begin{array}{l}\left( \sigma_{\rho 2}\right)_{\rho=R_{2}}=-q\prime_{2}\\ \left(\sigma_{\rho 2}\right)_{\rho=R_{3}}=0\end{array} \right. \tag{2}\]  
Substituting Equations (1) and (2) into the general solution of axisymmetric stress, the calculated solutions of the inner and outer cylinders are as follows:  
\[\text{Mold bag concrete}:\ \left\{ \begin{array}{l}\sigma_{\rho 1}=\frac{R_{1}^{2}R_{2 }^{2}(q_{1}-q_{2})}{\rho^{2}\left(R_{1}^{2}-R_{2}^{2}\right)}+\frac{q_{2}R_{2} ^{2}-q_{1}R_{1}^{2}}{R_{1}^{2}-R_{2}^{2}}\\ \sigma_{\varphi 1}=-\frac{R_{1}^{2}R_{2}^{2}(q_{1}-q_{2})}{\rho^{2}\left(R_{1}^{2}- R_{2}^{2}\right)}+\frac{q_{2}R_{2}^{2}-q_{1}R_{1}^{2}}{R_{1}^{2}-R_{2}^{2}}\\ \tau_{\varphi\varphi 1}=\tau_{\varphi\varphi 1}=0\end{array} \right. \tag{3}\]  
\[\text{Steel corrugated plate}:\ \left\{ \begin{array}{l}\sigma_{\rho 2}=\frac{q_{2}R_{2}^{2}R_{ 3}^{2}}{\rho^{2}\left(R_{2}^{2}-R_{3}^{2}\right)}-\frac{q_{2}R_{2}^{2}}{R_{2} ^{2}-R_{3}^{2}}\\ \sigma_{\varphi 2}=-\frac{q_{2}R_{2}^{2}R_{3}^{2}}{\rho^{2}\left(R_{2}^{2}-R_{3}^ {2}\right)}-\frac{q_{2}R_{2}^{2}}{R_{2}^{2}-R_{3}^{2}}\\ \tau_{\varphi\varphi 2}=\tau_{\varphi\rho 2}=0\end{array} \right. \tag{4}\]  
It is assumed that the two elastomers remain in 'full contact' on the contact surface, is neither disengage nor slide against each other. In this way, on the contact surface, the contact condition of the stress direction is that the normal stress and shear stress of the two elastic bodies are equal on the contact surface.