# Study on Explosive Fracturing Mechanism of Qianjiang Shale Oil
## 3 Results and Discussion
### Material Model  
#### 3.2.1 Rock [32]  
The primary mineral composition of the shale [31] in the reservoir examined in this chapter consists of albite (69.37%), dolomite (9.78%), illite (8.27%), and calcite (6.89%). The fundamental physical and mechanical properties of the reservoir are as follows: density \(\rho_{0}\) = 2370 kg/m\({}^{3}\), uniaxial compressive strength \(f_{c}\) = 192.5 MPa, elastic modulus \(E\) = 112.02 GPa, Poisson's ratio \(v\) = 0.29, and porosity \(q\) = 15.9% [33]. The maximum tensile strength \(T\), shear modulus \(G\), and bulk modulus \(K\) can be calculated using Equations (3)-(5), respectively. \[T=0.62\sqrt{f_{c}} \tag{3}\]  
\[G=\frac{E}{2(1+V)} \tag{4}\]  
\[K=\frac{E}{3(1-2V)} \tag{5}\]  
Obtain \(T\) = 8.602 MPa, \(G\) = 4.659 GPa, \(K\) = 8.465 GPa  
Figure 1: Coupled modeling of dynamic and static loads (Unit: cm). The strength model is described by the dimensionless equivalent stress \(\sigma^{*}\), which is formulated as:  
\[\sigma^{*}=\frac{\sigma}{f_{c}} \tag{6}\]  
where \(\sigma\) represents the actual equivalent stress and \(f_{c}\) denotes the uniaxial compressive strength. Additionally, \(\sigma^{*}\) can be defined in terms of damage, which may be expressed as:  
\[\sigma^{*}=\left[A(1-D)+BP^{*N}\right]\left[1+Cln\left(\dot{\epsilon}^{*} \right)\right] \tag{7}\]  
where \(D\) represents the damage parameter, with values ranging from 0 to 1. Specifically, \(D=1\) signifies complete failure of the rock, while \(D=0\) indicates an undamaged state. The parameters \(A\), \(B\), and \(N\) are associated with the yield surface of the HJC model. In the absence of damage (\(D=0\)) and without considering strain rate effects, Equation (7) can be simplified as follows:  
\[\sigma^{*}=A+BP^{*N} \tag{8}\]  
Based on the principles of plasticity theory and without accounting for damage evolution or strain rate effects, both the HJC constitutive model and the Mohr-Coulomb constitutive model intersect at the corresponding points of pure shear and uniaxial compression on the compressive meridian plane. Consequently, the relationship between the dimensionless viscous strength coefficient \(A\), cohesion \(c\), and uniaxial compressive strength \(f_{c}\) can be derived. \[A=\frac{c}{f_{c}} \tag{9}\]  
Based on the Mohr-Coulomb strength criterion, the cohesion \(c\) can be determined using the following equation:  
\[\sigma_{1}=\sigma_{3}\frac{1+sin\varphi}{1-sin\varphi}+\frac{2 ccos\varphi}{1- sin\varphi} \tag{10}\]  
In the formula, \(\sigma_{1}\) and \(\sigma_{3}\) represent the maximum and minimum principal stressors at the point of static compressive failure, respectively, while \(\varphi\) denotes the angle of internal friction. The Hoek-Brown criterion is widely recognized for its utility in assessing the strength index of rock under triaxial static compression. In scenarios where static triaxial compression test data are unavailable, [PERSON] et al. (2018) and [PERSON] et al. (2018) employed this criterion to calibrate parameters for the RHT model and JH-2 model, respectively, achieving satisfactory results in numerical simulations. Consequently, this study adopted this method to obtain static compressive strength data of rocks under varying confining pressure conditions using the following Hoek-Brown empirical formula:  
\[\sigma_{1}=\sigma_{3}+\sigma_{ci}\left(m_{b}\frac{\sigma_{3}}{\sigma_{ci}}+s \right)^{a} \tag{11}\]  
where \(\sigma_{ci}\) represents the static uniaxial compressive strength, with \(\sigma_{ci}=f_{c}\); \(m_{b}\), \(s\), and \(a\) denote constants, which are typically 24, 1, and 0.5, respectively, for intact rock materials. The data presented in Table 2 were derived by applying various confining pressures \(\sigma_{3}\) in conjunction with Equation (11). Through linear regression analysis based on Equation (10), as illustrated in Figure 2, the following expression was obtained:  
\[G=\frac{E}{2(1+V)} \tag{4}\]0In conjunction with Equation (10), the values \(\varphi=45.585^{\circ}\) and \(c=48.266\) MPa can be derived from Equation (13). \[\left\{