## 3 Results
### Failure Criteria
According to the Mohr-Coulomb criteria, the radius of the surrounding rock plastic zone can be calculated as follows [55; 56; 57]:  
\[R_{p}=a\left[\frac{(\sigma_{0}+c\cot\varphi)(1-\sin\varphi)}{c\cot\varphi} \right]^{\frac{1-\sin\varphi}{2\sin\varphi}} \tag{4}\]  
where \(\sigma_{0}\) is the original stress field, and \(\sigma_{0}=0\) in this study. The surrounding rock pressure in a uniform stress field can be calculated using the Fenner formula [55; 56; 57], as follows:  
\[\sigma_{r}=[(1-\sin\varphi)(\sigma_{0}+c\cot\varphi)]\left(\frac{a}{R_{p}} \right)^{\frac{2\sin\varphi}{1+\sin\varphi}}-c\cot\varphi \tag{5}\]  
Figure 13: Plastic zone of circular tunnel. By combining Problems 1 and 2, the anisotropy strength criteria of a different interface dip with the tunnel can be expressed as follows:  
\[\sigma_{1m}=\left\{ \begin{array}{ll}\sigma_{3}\frac{1+\sin\varphi}{(1-\sin \varphi)}\cos\alpha+c\cdot\frac{2\cos\varphi}{(1-\sin\varphi)}\cos\alpha+c_{i} \cdot\frac{1}{\tan\varphi_{i}}-\sigma_{r}&(\alpha<\alpha_{1})\\ \sigma_{3}\cdot\frac{\sin(2a-\varphi_{i})+\sin\varphi_{i}}{\sin(2a-\varphi_{ i})-\sin\varphi_{i}}+c_{i}\cdot\frac{2\cos\varphi_{i}}{\sin(2a-\varphi_{i})- \sin\varphi_{i}}-\sigma_{r}&(\alpha_{1}<\alpha<\alpha_{2})\\ \sigma_{3}\cdot\frac{1-\mu}{\mu}+\sigma_{c}-\sigma_{r}&(\alpha<\alpha_{2}) \end{array} \right. \tag{6}\]  
\[\sigma_{1m}=\frac{2a}{2a+b}\sigma_{1}+\frac{b}{2a+b}\sigma_{1i}-\sigma_{r} \tag{7}\]  
When the cell pressure is zero, the first principal stress is expressed as follows:  
\[\sigma_{1}=\sigma_{c}+\frac{c_{i}}{\sin\beta\cos(\varphi_{i}+\beta)} \tag{8}\]  
The strength can be calculated using Equations (5)-(7):  
In Figure 14, the fitting degree is lower than 45\({}^{\circ}\) and 90\({}^{\circ}\) (Figure 14a,b). In this study, when the inclination angle was 45\({}^{\circ}\), under the uniaxial condition, the tip of the sample was first subjected to force, which resulted in a small contact area; therefore, the failure mode was fragmentation (Figure 7e). Shearing slip does not occur until the contact area of the rock mass is sufficient to bear the vertical load. There are two failure forms, which were explained in detail in the previous section. The calculated result for the strength of the interface reveals that the strength is lower than that obtained by testing. \begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline \(\alpha\) & **0** & **15** & **30** & **45** & **60** & **75** & **90** \\ \hline a/(cm) & \(\backslash\) & \(\backslash\) & \(\backslash\) & \(\backslash\) & 2.12 & 3.66 & 5.00 \\ b/(cm) & \(\backslash\) & \(\backslash\) & \(\backslash\) & \(\backslash\) & 5.77 & 2.68 & 5.00 \\ \(\sigma_{1m}/\)(MPa) & 17.47 & 17.24 & 15.60 & 5.70 & 3.10 & 16.26 & 19.236 \\ \hline \hline \end{tabular} a and b are the size of the rock mass after being cut by the structural plane, as shown in Figure 12b.  
\end{table} 
Table 2: Calculation results. Figure 14: Theory versus test. The calculation results were obtained by data fitting in Table 2.  
In this study, because the interface with an inclination of 90\({}^{\circ}\) divided the sample into two parts, the test results reflect the strength of the individual parts; therefore, the UCS of the sample was larger than the calculated result.