## 3 Parameter Determination Method for the JH-2 Model
### Determination of Strength
#### 3.3.1 Determination of Intact Strength  
In this section, the constants \(A\), \(N\), and \(C\) will be calculated. Referring to previous experimental data [29], we set the following parameter values: uniaxial compressive strength (UCS) \(\sigma_{ci}=167.1\) MPa; triaxial compressive strength (TCS) \(\sigma_{3}=5\) MPa, \(\sigma_{1}=220.2\) MPa; \(\sigma_{3}=10\) MPa, \(\sigma_{1}=274.9\) MPa; and equivalent strain rate \(\dot{\epsilon}=2.59\times 10^{-5}\). Reliable estimates for the triaxial compressive strengths of rocks at high confining pressures can be achieved using empirical equations, such as the Drucker-Prager yield criterion [45]. For intact rock pieces that make up the rock mass, the Drucker-Prager yield criterion has the form  
\[f(I_{1},I_{2})=\alpha I_{1}+\sqrt{I_{2}}-k=\alpha(\sigma_{1}+\sigma_{2}+\sigma _{3})+\sqrt{\frac{1}{6}\Big{[}(\sigma_{1}-\sigma_{2})^{2}+(\sigma_{2}-\sigma _{3})^{2}+(\sigma_{1}-\sigma_{3})^{2}\Big{]}}-k=0 \tag{24}\]  
where \(I_{1}\) is the first stress invariant (compression is negative and tension is positive) and \(I_{2}\) is the second invariant of the deviatoric stress. The relationship between the principal stresses at failure for a given rock is defined by two constants, which are \(\alpha=2\sin\phi/\sqrt{3}(3-\sin\phi)\) and \(k=6c\cdot\cos\phi/\sqrt{3}(3-\sin\phi)\). Due to the shortage of the values of cohesion, \(c\), and the internal friction angle, \(\phi\), the values of \(\alpha\) and \(k\)  
Figure 3: Hydrostatic pressure against volumetric strain response of Barre granite. cannot be directly calculated, but can be obtained by substituting the data of the principal stresses from the previous UCS and TCS experiments into Equation (24). Then Equation (24) can be transformed as  
\[f(I_{1},J_{2})=0.44(\sigma_{1}+\sigma_{2}+\sigma_{3})+\sqrt{\frac{1}{6}\Big{[} \big{(}\sigma_{1}-\sigma_{2}\big{)}^{2}+\big{(}\sigma_{2}-\sigma_{3}\big{)}^{2 }+\big{(}\sigma_{1}-\sigma_{3}\big{)}^{2}\Big{]}}-22.9=0(\text{MPa}) \tag{25}\]  
In order to depict the relationship between the hydrostatic pressure and equivalent stress of intact granite, a series of data for pressure, \(P\), and equivalent stress, \(\sigma_{i}\), of intact rock should be obtained. According to the fitting formula in Equation (25), it can be assumed that \(\sigma_{2}=\sigma_{3}\), the value of which ranges from 0 to 3200 MPa, and can then be substituted into Equation (25) to achieve the value of \(\sigma_{1}\). The value of \(P\) can be derived from \(P=\frac{\sigma_{1}+\sigma_{2}+\sigma_{3}}{3}\)[46]. The value of \(\sigma_{i}\) can be calculated using Equation (3). The corresponding data are shown in Figure 4. In these uniaxial and triaxial compression tests, the equivalent strain rate in Equation (6) is \(\dot{\epsilon}=2.59\times 10^{-5}=\dot{\epsilon}^{*}\). According to previous studies [4; 26; 47], the maximum tensile hydrostatic pressure \(T\) is calculated from the dynamic tensile strength considered as the spall strength by planar impact tests. The dynamic tensile strength of granite can be determined by flyer plate impact tests and is equal to the tensile stress while the incipient tensile cracks begin to happen accompanied by the decrease onset of longitudinal wave. [PERSON] and [PERSON] [26] determined the dynamic tensile strength of granite to be 0.13 GPa. Additionally, in plate impact experiments to investigate shock-induced inelasticity in Westerly granite, the mechanical property is weaker than the general granite, and the spall strength is 0.05 GPa [48]. Because it is short of the specific value of the spall strength (dynamic tensile strength), \(T_{spall}\), for Barre granite, in this study the intermediate approximate value of (0.05 GPa, 0.13 GPa) was chosen to ensure the magnitude of the spall strength, and a value of \(T_{spall}=0.10\) GPa was assumed.