## 3 Stress Analysis of the Wellbore Surrounding Rocks in Broken Coal Seams  
### Stress Analysis of the Wellbore Surrounding Rocks as a Continuous Body  
In order to obtain the stress distribution of broken coal seams, firstly we assume that the coal rock is a continuous body. Then, the stress field around the wall of the well can be obtained under non-uniform, in-situ stress and wellbore pressure \(P_{0}\):  
\[\left\{\begin{array}{l}\sigma^{\prime}r=-\sigma_{\text{m}}\Big{[}1-\left( \frac{r}{R_{0}}\right)^{-2}\Big{]}-\sigma_{\text{n}}\Big{[}1-3\left(\frac{r}{R _{0}}\right)^{-2}+3\left(\frac{r}{R_{0}}\right)^{-4}\Big{]}\cos 2\theta-\left( \frac{r}{R_{0}}\right)^{-2}\!\!P_{0}-\alpha_{\text{Bilot}}P_{0}\\ \sigma^{\prime}{}_{\theta}=-\sigma_{\text{m}}\Big{[}1+\left(\frac{r}{R_{0}} \right)^{-2}\Big{]}+\sigma_{\text{n}}\Big{[}1+3\left(\frac{r}{R_{0}}\right)^{ -4}\Big{]}\cos 2\theta+\left(\frac{r}{R_{0}}\right)^{-2}\!\!P_{0}-\alpha_{\text{ Bilot}}P_{0}\\ \tau^{\prime}{}_{r\theta}=\sigma_{\text{n}}\Big{[}1+2\left(\frac{r}{R_{0}} \right)^{-2}-3\left(\frac{r}{R_{0}}\right)^{-4}\Big{]}\cdot\sin 2\theta\end{array} \right., \tag{1}\]  
where \(\sigma_{r}\), \(\sigma_{\theta}\), \(\tau_{r\theta}\), are radial stress, cyclic stress, and shear stress under polar coordinates, respectively. \(2\sigma_{\text{m}}=\sigma_{\text{H}}+\sigma_{\text{h}}\), \(2\sigma_{\text{n}}=\sigma_{\text{H}}-\sigma_{\text{h}}\); \(R_{0}\) is the wellbore radius; \(r\) is the length of the calculation point to the wellbore central axis; \(\theta\) is the wellbore angle; \(\alpha_{\text{Bilot}}\) is Biot coefficient.  
Then wellbore rock displacement [23] can be expressed as follows, and the specific derivation process of the formula can be referred to paper 23:  
\[\left\{\begin{array}{ll}u^{\prime}r=&\frac{1+\mu}{r\cdot E}\frac{R_{2}^{2}R_ {0}^{2}(P_{0}-(1-2\mu)\sigma_{\text{m}})+(1-2\mu)P_{0}r^{2}-R_{2}^{2}\sigma_{ \text{m}}r^{2}}{R_{2}^{2}-R_{0}^{2}}+\\ &\frac{2\cos 2\theta}{5E}\Big{[}(12-6\mu)K_{1}r^{3}-(1+\mu)K_{2}r+(6+4\mu)K_{3}/ r+9(1+\mu)K_{4}/r^{3}\Big{]}\\ u^{\prime}{}_{\theta}=&\frac{6\sin 2\theta}{5E}\Big{[}(3+\mu)K_{1}r^{3}+(1+\mu)K_{ 2}r-(1-\mu)K_{3}/r+(1+\mu)K_{4}/r^{3}\Big{]}\end{array}\right., \tag{2}\]  
where \(E\), \(\mu\) are the elastic modulus and Poisson's ratio of coal seam under plane strain conditions, respectively; \(u^{\prime}r\), \(u^{\prime}{}_{\theta}\) are the radial displacement and the cyclic displacement, respectively; \(R_{2}\) is the radius of the circular outer boundary set for the solution, \(R_{2}\gg R_{0}\); \(K\) is a constant coefficient determined by the geometric size of the model. Its specific calculation formula can be written as Equation (3):  
\[\left[\begin{array}{ll}0&1&\frac{2}{R_{0}^{2}}&\frac{3}{R_{0}^{4}}\\ 3R_{0}^{2}&1&\frac{1}{R_{0}^{2}}&\frac{-3}{R_{0}^{4}}\\ 0&2&\frac{4}{R_{0}^{2}}&\frac{6}{R_{0}^{4}}\\ 6R_{2}^{2}&2&\frac{22}{R_{0}^{2}}&\frac{-6}{R_{0}^{2}}\end{array}\right]\left( \begin{array}{l}K_{1}\\ K_{2}\\ K_{3}\\ K_{4}\end{array}\right)=\left\{\begin{array}{l}0\\ 0\\ \sigma_{\text{n}}\\ \sigma_{\text{n}}\end{array}\right\}, \tag{3}\]  
The displacement field under the polar coordinates obtained by Equation (2) can be changed to the displacement field under the orthogonal Cartesian coordinate system by the coordinate transformation method. The transformed displacement component can be expressed as \(u^{\prime}{}_{x}\) and \(u^{\prime}{}_{y}\), and the transformation process is shown in Equation (4).  
\[\left\{\begin{array}{l}u^{\prime}{}_{x}\\ u^{\prime}{}_{y}\end{array}\right\}=\left[\begin{array}{ll}\cos\theta&-\sin \theta\\ \sin\theta&\cos\theta\end{array}\right]\left\{\begin{array}{l}u^{\prime}{}_{r} \\ u^{\prime}{}_{\theta}\end{array}\right\}, \tag{4}\]