## 4 Discussions  
### Mechanical Analysis of Fracture Mode  
In this paper, the method proposed by [PERSON] et al. (Formulas (3) and (4)) to calculate the stress magnitude at the center of the transversely isotropic Brazilian disc is used to analyze the mechanical mechanism of fracture modes of samples with different layer orientations [12]. The tensile stress at the center point of the disc is:  
\[\sigma_{\rm t}=\frac{2P}{\pi D_{\rm t}}\bigg{[}(\sqrt[4]{E/E^{\prime}})^{\cos(2 \theta)}-\frac{\cos(4\theta)}{4}(b-1)\bigg{]} \tag{3}\]  
The compressive stress at the center point of the disc is:  
\[\sigma_{\rm P}=\frac{6P}{\pi D_{\rm t}}\bigg{[}(\sqrt[4]{E^{\prime}/E})^{\cos( 2\theta)}+\frac{\cos(4\theta)}{4}(b-1)\bigg{]} \tag{4}\]  
where \(b=\frac{\sqrt{EE}}{2}\Big{(}\frac{1}{G^{\prime}}-\frac{2\
u^{\prime}}{E^{ \prime}}\Big{)}\). Where \(E^{\prime}\) is the elastic modulus of the vertical isotropic plane; \(E\) is the elastic modulus of the parallel isotropic surface; \(\
u^{\prime}\) is the Poisson's ratio perpendicular to the isotropic plane; and \(G^{\prime}\) is the shear modulus perpendicular to the isotropic plane. Through the uniaxial compression test, the values of \(E^{\prime}\), \(E\), and \(\
u^{\prime}\) can be obtained. \(G^{\prime}\) and the characteristic parameters of shear strength of the lamination and matrix are obtained according to the literature [13], and the results are shown in Table 6.  
The normal stress and shear stress on the lamination can be calculated by Formulas (5) and (6), respectively. \[\sigma_{\rm n}=\frac{1}{2}(\sigma_{\rm P}+\sigma_{\rm t})+\frac{1}{2}(\sigma_ {\rm P}-\sigma_{\rm t})\cos 2\theta \tag{5}\]  
\[\tau=\frac{1}{2}(\sigma_{\rm P}-\sigma_{\rm t})\sin 2\theta \tag{6}\]  
Assuming that the strength of lamination and matrix conforms to the Mohr- Coulomb criterion, and the compressive and tensile stresses are specified as positive and negative, respectively, the fracture conditions of the lamination and matrix are as follows:  
\[\sigma_{\rm n}^{b}\geq R_{\rm t}^{b} \tag{7}\]  
\[\sigma_{\rm n}^{m}\geq R_{\rm t}^{m} \tag{8}\]  
where \(\sigma_{n}^{b}\), \(\tau^{b}\), \(\sigma_{\rm n}^{m}\), and \(\tau^{m}\) are the normal stress and shear stress on the matrix and lamination, respectively; \(R_{\rm t}^{b}\) is the tensile strength value of the lamination calculated by Formula (1), \(R_{\rm t}^{b}\) = 1.31 MPa; and \(R_{\rm t}^{m}\) is the matrix tensile strength value calculated by Formula (1), \(R_{\rm t}^{m}\) = 5.154 Mpa. \(c^{b}\), \(\varphi^{b}\), \(c^{m}\) and \(\varphi^{m}\) are the cohesive force and internal friction angle of the lamination and the matrix, respectively. Figure 10 shows the relationship between stress and strength of samples with different layer orientations, when \(0^{\circ}\leq\theta\leq 90^{\circ}\). \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline \(E\) (GPa) & \(E^{\prime}\) (Gpa) & \(\
u^{\prime}\) & \(G^{\prime}\) (Gpa) & \multicolumn{2}{c}{**Matrix**} & \multicolumn{2}{c}{**Lamina**} \\ \cline{4-7}  & & & & \(\varphi\) (\({}^{\circ}\)) & \(c\) (Mpa) & \(\varphi\) (\({}^{\circ}\)) & \(c\) (Mpa) \\ \hline
31.3 & 15.8 & 0.2 & 8.7 & 35.94 & 16.88 & 27.02 & 5.01 \\ \hline \hline \end{tabular}
\end{table} 
Table 6: Transversely isotropic mechanical parameter values of continental shale. It can be seen from Figure 10 that when \(\theta=90^{\circ}\), \(\sigma_{\rm n}=\sigma_{\rm t}\) and \(\tau=0\) can be calculated from Formulas (5) and (6), respectively, indicating that there is no shear stress on the lamination, and the sample will not undergo shear failure along the lamination. The tensile stress at the center of the sample was calculated to be 4.54 MPa, which was larger than the tensile strength of the lamination (Formula (7)). Therefore, tensile failure occurred along the lamination. When \(\theta=0^{\circ}\), there is also only normal stress but no shear stress on the lamination.