## 2 New Extended Semi-Weight Function Method for Calculating SIFs under Hydro-Mechanical Coupling
### Determination of Weight Function and Semi-Weight Function for Hydro-Mechanical Coupling Crack  
According to the derivation of the new hydro-mechanical coupling stress function \(F(r,\,\theta,\,P_{h})\) in Section 2.1, the corresponding stress and displacement fields are shown as follows:  
\[\begin{array}{l}\sigma_{r}=-\lambda r^{\lambda-1}[(\lambda+1)(A\sin(\lambda+ 1)\theta+B\cos(\lambda+1)\theta)+(\lambda-3)(Cin(\lambda-1)\theta+D\cos( \lambda-1)\theta)]+P_{h}\\ \sigma_{\theta}=\lambda(\lambda+1)r^{\lambda-1}[A\sin(\lambda+1)\theta+B\cos( \lambda+1)\theta+C\sin(\lambda-1)\theta+D\cos(\lambda-1)\theta]+P_{h}\\ \tau_{r\theta}=-\lambda r^{\lambda-1}[(\lambda+1)The stress in Equation (9) also satisfies the boundary conditions. By substituting Equation (9) into Equation (3), the following can be obtained:  
\[ \begin{cases}A\sin(\lambda+1)\pi+B\cos(\lambda+1)\pi+C\sin(\lambda-1)\pi+D\cos( \lambda-1)\pi=0\\ (\lambda+1)(A\cos(\lambda+1)\pi-B\sin(\lambda+1)\pi)+(\lambda-1)(C\cos( \lambda-1)\pi-D\sin(\lambda-1)\pi)=0\\ -A\sin(\lambda+1)\pi+B\cos(\lambda+1)\pi-C\sin(\lambda-1)\pi+D\cos(\lambda-1 )\pi=0\\ (\lambda+1)(A\cos(\lambda+1)\pi+B\sin(\lambda+1)\pi)+(\lambda-1)(C\cos( \lambda-1)\pi+D\sin(\lambda-1)\pi)=0\end{cases}  \tag{11}\]  
In order to solve the undetermined weight function (\(A\), \(B\), \(C\), and \(D\)), the sub-matrix method is used [25]. First, by rewriting Equation (9) into a matrix form, the following can be obtained:  
\[ \begin{split}-\mathbf{N}_{1}\mathbf{S}\left\{\begin{array}{c}C \\ D\end{array}\right\}=\mathbf{N}_{2}\mathbf{S}\left\{\begin{array}{c}A\\ B\end{array}\right\}\\ \mathbf{N}_{1}\mathbf{T}\mathbf{T}\left\{\begin{array}{c}C\\ D\end{array}\right\}=-\mathbf{N}_{2}\mathbf{T}\mathbf{T}\left\{\begin{array}[] {c}A\\ B\end{array}\right\}\end{split}  \tag{12}\]  
where \(\mathbf{N}_{i}\) (\(i=1\), \(2\)), \(\mathbf{S}\), and \(\mathbf{T}\) are second-order coefficient matrices, and their definitions are as follows:  
\[ \begin{split}\mathbf{N}_{1}=\left[\begin{array}{cc}1&0\\ 0&\lambda-1\end{array}\right]\ \ \mathbf{N}_{2}=\left[\begin{array}{cc}1&0\\ 0&\lambda+1\end{array}\right]\\ \mathbf{S}=\left[\begin{array}{cc}\sin\lambda\pi&\cos\lambda\pi\\ \cos\lambda\pi&-\sin\lambda\pi\end{array}\right]\ \ \ \mathbf{T}=\left[\begin{array}{cc}1&0\\ 0&-1\end{array}\right]\end{split}  \tag{13}\]  
Then, by selecting \(\{A,B\}^{\mathrm{T}}\) as the basic undetermined weight function and combining the two equations in Equation (12), the following can be calculated:  
\[ \begin{split}\left(-\mathbf{N}_{1}\mathbf{T}\mathbf{S}\mathbf{T }^{-1}\mathbf{N}_{1}^{-1}\mathbf{N}_{2}\mathbf{S}+\mathbf{N}_{2}\mathbf{T} \mathbf{T}\right)\left\{\begin{array}{c}A\\ B\end{array}\right\}=0\end{split}  \tag{14}\]  
To calculate the non-zero solutions of the basic undetermined weight function (\(A\) and \(B\)), the determinant for the coefficient matrix of Equation (14) must satisfy the condition of being equal to \(0\), as follows:  
\[4\sin(2\lambda\pi)^{2}=0 \tag{15}\]  
Obviously, Equation (15) is the characteristic equation fused to solve the singularity index (\(\lambda\)) of the crack tip. Finally, according to the definition of the stress-intensity factor, there are  
\[K_{1}=\lim_{r\to 0}(r^{1-\lambda}\sigma_{\theta}\Big{|}_{\theta=0}),\ K_{2}= \lim_{r\to 0}(r^{1-\lambda}\tau_{r\theta}\Big{|}_{\theta=0}) \tag{16}\]  
Thus, the basic weight functions \(A\) and \(B\) can be obtained by substituting Equations (9) and (12) into Equation (16):  
\[A=-K_{2},\ \ B=\frac{K_{1}}{3} \tag{17}\]  
and \(\{C,D\}^{\mathrm{T}}\) can be solved using Equation (12). Similar to the weight function (\(A\), \(B\), \(C\), and \(D\)), the key to calculating the semi-weighted function (\(A^{\prime}\), \(B^{\prime}\), \(C^{\prime}\), and \(D^{\prime}\)) is to replace \(-\lambda\) with \(\lambda\) in Equation (8) to obtain the corresponding stress field and displacement field, as follows:\[