## 3 Analysis of the Coal Pillar Width under the Influence of the Water-Logged Goal
### Calculation of the Width of the Elastic Zone of the Waterproof Coal Pillar  
The coal seam in the elastic zone is influenced by lateral water pressure and mining stress at the mining face in waterlogged mining areas. Therefore, a mechanical model was developed to convert the problem into a plane stress scenario to determine the width of the coal pillar. Meanwhile, the mechanical model makes the following assumptions. The mechanical calculation model of coal pillar width in the elastic core area is shown in Figure 6.  
1. The coal pillar is a uniform, continuous, and isotropic elastomer. 2. The stress \(\sigma_{x0}\) on both sides of the coal body in the elastic area is symmetric and uniformly distributed. 3. The stress \(K_{2}\gamma H\) above the coal body in the elastic area is uniformly distributed and symmetric about the \(x\)-axis. 4. The lateral water pressure \(p\) suffered by the coal body in the elastic area is uniformly distributed, ignoring the blocking effect of the plastic zone on the water pressure, and the size of the water pressure is calculated in accordance with the maximum value measured in the field. The elastic area coal pillar model boundary conditions are:  
\[\left\{ \begin{array}{l}\left(\sigma_{y}\right)_{y=-\frac{1}{2}}=-K_{2}\gamma H \\ \left(\sigma_{x}\right)_{x=0}=-p-\sigma_{x0}\\ \left(\sigma_{x}\right)_{x=l_{2}}=-\sigma_{x0}\end{array} \right. \tag{21}\]  
The semi-inverse solution method was used to solve the model stress components, assuming that each stress component satisfies the stress function \(\Phi=\Phi(x,y)\); therefore, the stress function was solved. It was assumed that \(\sigma_{y}\) does not vary with \(x\), and \(\sigma_{y}\) is a function of y. Therefore, the expression of the assumed function is as follows:  
\[\sigma_{y}=f(y) \tag{22}\]  
In the direction of the coal pillar width, the mine pressure \(K_{2}\gamma H\) above the coal pillar is uniformly distributed, so the following formula can be obtained:  
\[f(y)=-K_{2}\gamma H \tag{23}\]  
Based on the relationship between the Airy stress function and the stress components:  
\[\left\{ \begin{array}{l}\sigma_{x}=\frac{\partial^{2}\Phi}{\partial y^{2}}\\ \sigma_{y}=\frac{\partial^{2}\Phi}{\partial x^{2}}\\ \tau_{yx}=-\frac{\partial^{2}\Phi}{\partial x\partial y}\end{array} \right. \tag{24}\]  
The stress function representation of the compatibility equation is:  
\[\frac{\partial^{4}\[PERSON]}{\partial x^{4}}+2\frac{\partial^{4}\[PERSON]}{\partial x^{ 2}\partial y^{2}}+\frac{\partial^{4}\[PERSON]}{\partial y^{4}}=0 \tag{25}\]  
Figure 6: Mechanical calculation model of the coal pillar width in the elastic area. Substituting \(\sigma_{y}\) into Equation (24) yields:  
\[\frac{\partial^{2}\[PERSON]}{\partial x^{2}}=f(y)=-K_{2}\gamma H \tag{26}\]  
Integrating over \(x\) yields:  
\[\left\{ \begin{array}{l}\frac{\partial\Phi}{\partial x}=xf(y)+f_{1}(y)\\ \Phi=\frac{x^{2}}{2}f(y)+xf_{1}(y)+f_{2}(y)\end{array} \right. \tag{27}\]  
where \(f_{1}(y)\) and \(f_{2}(y)\) are both y functions to be determined. Substituting Equation (27) into the compatibility Equation (25) yields:  
\[\frac{1}{2}\frac{d^{4}f(y)}{dy^{4}}x^{2}+\frac{d^{4}f_{1}(y)}{dy^{4}}x+\frac{d^ {4}f_{2}(y)}{dy^{4}}+2\frac{d^{2}f(y)}{dy^{2}}=0 \tag{28}\]  
Since the stress function must satisfy the compatibility equation in the region, the coefficients and the free term of Equation (28) must be equal to zero, and the solution of the equation is polynomial:  
\[f_{1}(y)=Ay^{3}+By^{2}+Cy+D \tag{29}\]  
\[f_{2}(y)=Ey^{3}+Fy^{2}+Gy+I \tag{30}\]  
Substituting Equations (23), (29), and (30) into Equation (27) yields:  
\[\Phi(x,y)=-\frac{K_{2}\gamma H}{2}x^{2}+(Ay^{3}+By^{2}+Cy+D)x+(Ey^{3}+Fy^{2}+Gy +I) \tag{31}\]  
Substituting Equation (31) into Equation (24) yields the stress component as follows:  
\[\left\{ \begin{array}{l}\sigma_{x}=\frac{y^{2}\Phi}{\partial y^{2}}=x(6 Ay+2 B)+(6 Ey+2F)\\ \sigma_{y}=\frac{y^{2}\Phi}{\partial x^{2}}=f(y)=-K_{2}\gamma H\\ \tau_{xy}=-\frac{y^{2}\Phi}{\partial x\partial y}=-(3 Ay^{2}+2 By+C)\end{array} \right.