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Author: Admin | 2025-04-28
On both sides of the mining opening is finally obtained by solving the equation:$$Y = \frac{{XH_{{\text{p}}} \left( {1 - \sin \varphi } \right)}}{{2\left( {1 + \sin \varphi } \right)\tan \varphi_{0} }}$$ (25) 3.2 Calculation of the rib pillar compression under the dynamic load of mining haul trucks3.2.1 Theoretical model of the beam on Winkler elastic foundationTo establish the differential equation of the deflection curve satisfied by the beam (in Fig. 5), a section of the beam with a length of l, a width of b1 and a height of h is taken, and the foundation coefficient of the elastic foundation is presented as k and the width as b2. Under the load q(x), the compression of the foundation is y(x), the reaction of the foundation to the beam is qf(x), a micro section dx is intercepted in the beam, the bending moment is M and the shear force is Q.Fig. 5Elastic foundation beam model composed of the rib pillar and the roofFull size imageAccording to Winkler's assumption (Soon et al. 2021; Li et al. 2021), the base reaction force is:$$q_{{\text{f}}} (x) = kb_{2} y(x)$$ (26) If the influence of shear force on beam deflection is not considered, according to the balance condition of the force, α = (kb2/4E1I)1/4 and αx can be used to replace x, then:$$\frac{{{\text{d}}^{4} y(x)}}{{{\text{d}}\left( {\alpha x} \right)^{4} }} + 4y(x) = \frac{4}{{kb_{2} }}q(\alpha x)$$ (27) The additional term caused by uniformly distributed load q(x) = q0 is introduced, Eq. (28) can be obtained:$$y(x) = \left\{ \begin{gathered} y_{0} \varphi_{1} (x) + \theta_{0} \frac{1}{2\alpha }\varphi_{2} (x) - M_{0} \frac{{2\alpha^{2} }}{{kb_{2} }}\varphi_{3} (x) - Q_{0} \frac{\alpha }{{kb_{2} }}\varphi_{4} (x) + \frac{{q_{0} }}{{kb_{2} }}\left[ {\varphi_{1} (x - x_{i + 1} ) - \varphi_{1} (x - x_{i} )} \right] \, x \in \left[ {x_{i + 1} ,l} \right] \hfill \\ y_{0}
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