## 2 Deformation Characteristics of Rock Joint
### Elastic Solution of Displacement and Stress Distributions of Rock Joints  
After the exact solutions of \(\sigma_{r}\), \(\sigma_{\theta}\), and \(\tau_{r\theta}\) of the rock joint are obtained from Section 2.1, the strain can be deduced from the physical Equation (3) of the rock joint. Substituting the exact stress solution (Equation (5)) into physical Equation (3), we obtain  
\[\left\{ \begin{array}{l}\varepsilon_{r}=-\frac{R^{2}F\cos\theta}{2\pi Gr^{3} }\\ \varepsilon_{\theta}=\frac{R^{2}F\cos\theta}{2\pi Gr^{3}}\\ \gamma_{r\theta}=-\frac{R^{2}F\sin\theta}{\pi Gr^{3}}\end{array} \right. \tag{8}\]  
According to the geometric Equations (2) and (8), Equation (9) can be obtained as follows:  
\[\left\{ \begin{array}{l}\varepsilon_{r}=\frac{\partial u_{r}}{\partial r}=- \frac{R^{2}F\cos\theta}{2\pi Gr^{3}}\\ \varepsilon_{\theta}=\frac{u_{r}}{r}+\frac{1}{r}\frac{\partial u_{ \theta}}{\partial\theta}=\frac{R^{2}F\cos\theta}{2\pi Gr^{3}}\\ \gamma_{\theta r}=\frac{1}{r}\frac{\partial u_{r}}{\partial\theta}+\frac{ \partial u_{\theta}}{\partial r}-\frac{u_{\theta}}{r}=-\frac{R^{2}F\sin\theta }{\pi Gr^{3}}\end{array} \right. \tag{9}\]  
From the integral of the expression (9) \(\varepsilon_{r}\), the radial displacement \(u_{r}\) of the compressive-shear side rock joint can be obtained as  
\[u_{r}=\frac{R^{2}F\cos\theta}{4\pi Gr^{2}}+f_{1}(\theta) \tag{10}\]  
Figure 4: Microelement. where \(f_{1}(\theta)\) is a function of \(\theta\). Substituting Equation (10) into the second expression of Equation (9) (\(\varepsilon_{\theta}\)) yields  
\[\frac{\partial u_{\theta}}{\partial\theta}=\frac{R^{2}F\cos\theta}{4\pi Gr^{2}}- f_{1}(\theta) \tag{11}\]  
According to the aforementioned integral, the displacement \(u_{\theta}\) of the rock joint at the compression-shear side along the polar angle can be expressed as  
\[u_{\theta}=\frac{R^{2}F\sin\theta}{4\pi Gr^{2}}-\int f_{1}(\theta)d\theta+f_{2 }(r) \tag{12}\]  
where \(f_{2}(r)\) is a function of \(r\) and then substituting Equations (10) and (12) into Equation (9) and multiplying both sides by \(r\) yields  
\[ \begin{split}-\frac{R^{2}F\sin\theta}{4\pi Gr^{2}}+\frac{df_{1}( \theta)}{d\theta}-\frac{R^{2}F\sin\theta}{2\pi Gr^{2}}+r\frac{df_{2}(r)}{dr} \\ -\frac{R^{2}F\sin\theta}{4\pi Gr^{2}}+\int f_{1}(\theta)d\theta- f_{2}(r)=-\frac{R^{2}F\sin\theta}{\pi Gr^{2}}\end{split}  \tag{13}\]  
After sorting out Equation (13), we obtain  
\[\frac{df_{1}(\theta)}{d\theta}+\int f_{1}(\theta)d\theta=0 \tag{14}\]  
\[r\frac{df_{2}(r)}{dr}-f_{2}(r)=0 \tag{15}\]  
From Equation (14), we obtain  
\[f_{1}(\theta)=m\cos\theta+n\sin\theta \tag{16}\]  
where \(m\) and \(n\) are undetermined coefficients. From Equation (15), we can obtain  
\[f_{2}(r)=\lambda r \tag{17}\]  
where \(\lambda\) is the undetermined coefficient. Substituting Equations (16) and (17) into Equations (10) and (12), we can obtain Equation (18), which is the displacement component of the joint on the compression-shear side expressed in polar coordinates in the bolted joint system. The three undetermined coefficients \(m\), \(n\), and \(\lambda\) have to be determined according to the boundary conditions of the specific problem:  
\[\left\{ \begin{array}{l}u_{r}=\frac{R^{2}F\cos\theta}{4\pi Gr^{2}}+m\cos \theta+n\sin\theta\\ u_{\theta}=\frac{R^{2}F\sin\theta}{4\pi Gr^{2}}-m\sin\theta+n\cos\theta+ \lambda r\end{array} \right. \tag{18}\]  
For the shear bolt in Figure 2, if the bolt on the rock joint, i.e., the rigid disk, is subjected to shear load, its displacement only occurs in the \(x\) direction and is \(0\) in the \(y\) direction such that  
\[u=x_{0},v=0 \tag{19}\]  
In the formula, \(u\) and \(v\) are displacements in the \(x\) and \(y\) directions, respectively, and \(x_{0}\) is the displacement of the bolt in the \(x\) direction and is the known value. The displacement transformation relationship between the polar coordinate and Cartesian coordinate systems can be expressed as  
\[\left\{ \begin{array}{l}u_{r}=u\cos\theta+v\sin\theta\\ u_{\theta}=-u\sin\theta+v\cos\theta\end{array} \right.