## 4 Analysis of Specimen Fragmentation
### Crack Propagation Mechanism  
According to the sliding crack model theory and rock fracture mechanics [44; 45], effective shear stress on the crack surface is described as follows:  
\[\tau_{eff}=\tau-\mu\sigma_{n} \tag{9}\]  
where \(\tau_{eff}\) represents the shear stress of sliding between fracture surfaces, which is the main factor causing initiation and propagation at the crack tip; and \(\tau\), \(\mu\), and \(\sigma_{n}\) represent the shear stress, friction coefficient, and normal stress on the fracture surface, respectively. According to the superposition principle of stress field, under the action of one-way compression, the horizontal compressive \(\sigma_{x}\), normal compressive \(\sigma_{y}\) and shearing stresses \(\tau_{xy}\) are expressed as follows:  
\[\left. \begin{array}{l}\sigma_{x}=\sigma_{1}\sin^{2}\alpha\\ \sigma_{y}=\sigma_{1}\cos^{2}\alpha\\ \tau_{xy}=\sigma_{1}\sin\alpha\cos\alpha\end{array} \right\} \tag{10}\]  
where \(\sigma_{1}\) represents the maximum principal stress, and \(\alpha\) represents the angle between the prefabricated crack and the vertical direction. According to the above equation, when the inclination angle of the prefabricated crack is 45\({}^{\circ}\), the stress intensity on the crack surface is the largest, and crack initiation at the tip is sufficient. By combining Formulas (9) and (10), the following equation is obtained:  
\[\tau_{eff}(\alpha)=\frac{\sigma_{1}}{2}\sin(2\alpha)-\mu\sigma_{1}\cos^{2}\alpha \tag{11}\]  
Under the condition that the length of the prefabricated crack is constant, the bearing capacity of the crack body is the lowest when the shear stress on its surface is the largest. By deriving the above formula with respect to the inclination angle \(\alpha\), which is equal to 0, we obtain the following Equation (12):  
\[\alpha_{0}=\frac{1}{2}\text{arc }\cot(-\mu)=\frac{1}{2}(\pi-\text{arc }\cot\mu) \tag{12}\]  
when \(\mu=0\), crack propagation is in an ideal situation, the fracture surface closes, there is no friction, and the angle is \(\alpha=45^{\circ}\). This result is the same as the result of Equation (11). According to the theory of elliptical cracking, the stress \(\sigma_{\theta}\) at the boundary end of elliptical crack under uniaxial compression is given as follows:  
\[\left. \begin{array}{l}\sigma_{\theta}=\sigma_{y}\frac{3+m}{3-m}-\sigma_{x} \;,\;\text{at the end of the long axis}\\ \sigma_{\theta}=\sigma_{x}\frac{1}{1+m}-\sigma_{y}\;,\;\text{at the end of the minor axis}\end{array} \right\} \tag{13}\]  
where \(\text{m}=(\text{c}-\text{b})/(\text{c}+\text{b})\). The long axis of the crack is given by 2c and the short axis is given by 2b. By substituting Equation (10) into the above equation, we obtain the Equation (14):  
\[\left. \begin{array}{l}\sigma_{\theta}=\sigma_{1}(\cos^{2}\alpha\times\frac{3 +m}{3-m}-\sin^{2}\alpha)\;,\;\text{at the end of the long axis}\\ \sigma_{\theta}=\sigma_{1}(\sin^{2}\alpha\times\frac{3-m}{1+m}-\cos^{2}\alpha )\;,\;\text{at the end of the minor axis}\end{array} \right\} \tag{14}\]  
According to the analysis of Equation (14), with the increase in the thickness of the prefabricated crack, the transverse pressure of the crack surface increases gradually, and a crack gradually initiates in the middle position. However, Figure 7 shows that there was no tensile crack in the middle of the prefabricated crack of the sandstone specimen; the reason for this may be that with the increase in prefabricated fracture thickness, the peak strength of fractured rock mass decreases gradually although the tensile stress in the middle of the crack theoretically equates to the generation of the middle crack. However, because of the low strength of the crack, the crack surface closes quickly under the action of compressive stress, so deformation is not fully reflected, resulting in tensile stress in the middle of the prefabricated crack that does not reach the threshold value.