Comment
Author: Admin | 2025-04-28
(Table 1). The loading conditions considered in different sections of the roof by Xu (2009) are high in the cantilever section and oversimplistic in the abutment stress section. Correspondingly, the roof deflection values are significantly different from those previously developed due to the change in the boundary conditions as Xu (2009) allowed excessive roof deflection at the excavation boundary. The bending moment developed by Xu (2009) does not consider the influence of \(L\), which has been incorporated into the new model.Table 1 Different scenarios in longwall mining and corresponding stress, boundary conditions, deflection, and bending momentFull size table2.2 Strain EnergyThe external work done on the rock by overburden stress is stored in the elastically stressed rock as strain energy (Young et al. 2012; Agrawal et al. 2019). The strain energy stored in the roof and the coal seam can be calculated assuming that the stress–strain relationship from Hooke’s law follows (Wu 1995; Young et al. 2012; Fedotova et al. 2017; Dong et al. 2018; Xue et al. 2021),$${W}_{i}=\frac{1}{2}\left({\sigma }_{x}{\varepsilon }_{x}+{\sigma }_{y}{\varepsilon }_{y}+{\sigma }_{z}{\varepsilon }_{z}+{\tau }_{xy}{\gamma }_{xy}+{\tau }_{yz}{\gamma }_{yz}+{\tau }_{xz}{\gamma }_{xz}\right)$$ (5) where \({W}_{i}\) is the strain energy density, \({\sigma }_{x}, {\sigma }_{y}{, \sigma }_{z}, {\tau }_{xy}, {\tau }_{yz}, {\tau }_{xz}\) are stress components and \({\varepsilon }_{x}, {\varepsilon }_{y}, {\varepsilon }_{z}, {\gamma }_{xy}, {\gamma }_{yz}, {\gamma }_{xz}\) are strain components.For a bending beam, the non-zero stress components are flexural stress and shear stress. The flexural stress and strain can be calculated as (Young et al. 2012),$$\begin{array}{c}{\sigma }_{x}= My/I \\ {\varepsilon }_{x}={\sigma }_{x}/{E}_{{\text{roof}}} \\ I={\int }_{A}{y}^{2} {\text{d}}A \end{array}$$ (6) where \(A\) is the area of cross-section (m2).The strain energy occurring in the cantilevering and supporting roof due to bending can be calculated as,$${W}_{i}=\frac{1}{2}{\int }_{V}{\sigma }_{x}{\varepsilon }_{x}{\text{d}}V=\frac{1}{2}{\int }_{V}{M}^{2}{y}^{2}/{E}_{{\text{roof}}}{I}^{2}{\text{d}}V\to \frac{1}{2}{\int }_{{x}_{1}}^{{x}_{2}}{\int }_{A}{M}^{2}{y}^{2}/{E}_{{\text{roof}}}{I}^{2}{\text{d}}A{\text{d}}x=\frac{1}{2}{\int }_{{x}_{1}}^{{x}_{2}}{M}^{2}/{E}_{{\text{roof}}}I {\text{d}}x$$ (7) where \(V\) is the volume of the roof rock (m3).For the coal seam, following Hooke’s law, neglecting shear stresses and strains, the stress–strain components are related as,$${\varepsilon }_{x}=\left[{\sigma }_{x}-{\upsilon }_{{\text{coal}}} \left({\sigma }_{y}+{\sigma }_{z}\right)\right]/{E}_{{\text{coal}}}$$ (8) $${\varepsilon }_{y}=\left[{\sigma }_{y}-{\upsilon }_{coal} \left({\sigma }_{x}+{\sigma }_{z}\right)\right]/{E}_{{\text{coal}}}$$ (9) $${\varepsilon }_{z}=\left[{\sigma }_{z}-{\upsilon }_{{\text{coal}}} \left({\sigma }_{x}+{\sigma }_{y}\right)\right]/{E}_{{\text{coal}}}$$ (10) Putting these values in Equ. (5), and solving, one gets,$${W}_{{\text{coal}}}=\left[\left({\sigma }_{x}^{2}+{\sigma }_{y}^{2}+{\sigma }_{z}^{2}\right)-2{\upsilon }_{{\text{coal}}}\left({\sigma }_{x}{\sigma }_{y}+{\sigma }_{y}{\sigma }_{z}+{\sigma }_{x}{\sigma }_{z}\right)\right]/2{E}_{{\text{coal}}}$$ (11) The horizontal beam deflects to produce a continuously distributed reaction force \(q=Cy(x)\) in the coal seam, which acts vertically and opposes the deflection of the beam. As it is assumed that the beam is loaded normal to its faces, the loaded beam may deflect beside the vertical reaction, and there may be some horizontal forces originating along the surface between the beam and the elastic supports, but the horizontal forces would be low in magnitude and thus neglected in the current analysis (Stephansson 1971). The stresses due to the Poisson’s effect make the stress in the coal seam biaxial,$$\begin{array}{c} {\sigma }_{z}={(P}_{o}{e}^{-\alpha x}-Cy) \\ { \sigma }_{x}={\sigma }_{y}={\upsilon }_{{\text{coal}}}{\sigma }_{z}/\left(1-{\upsilon }_{{\text{coal}}}\right) \rightarrow{{\upsilon }_{{\text{coal}}}(P}_{o}{e}^{-\alpha x}-Cy)/\left(1-{\upsilon }_{{\text{coal}}}\right) \end{array} $$ (12) Substituting the values of stress components in the coal seam (Equ. (12)) into Equ. (11), one gets,$${W}_{{\text{coal}}}=\left[\left(1+{\upsilon }_{{\text{coal}}}\right)(1-2{\upsilon }_{{\text{coal}}})/2{E}_{{\text{coal}}}(1-{\upsilon }_{{\text{coal}}})\right]{\int }_{{x}_{1}}^{{x}_{2}}{{(P}_{o}{e}^{-\alpha
Add Comment