## 6 Results  
For each acceleration pulse, dynamic numerical simulations using Abaqus/Explicit solvers were carried out for the impact of the sliding plate on the guide. Twelve simulations were carried out in total for the pulses corresponding to three winding velocities (Figure 12), two masses of the working load, and two position points of the sliding plate impacting the guide. The first position corresponded to an impact in the mid-span of the guide (Figure 13a), while the second was in the vicinity of the button (Figure 13b). Figure 13 shows the horizontal displacement in the slider plate's action direction and von Mises stress maps for two force application points. The interaction of the sliding plate at the middle of the guide span caused it to bend, while in the case of action near the button, shearing occurred. In order to take into account the effect of the mass of the excavated material being pulled on the value of the force, an analysis was carried out for the loading of the conveyance with a nominal mass--20 tons--and a halved mass.  
Figure 12: Extreme pulses of the conveyance acceleration.  
Figure 13: Position of the conveyance relative to the guide. Proximity to the button (**a**) and guide mid-span (**b**). Horizontal displacement (m) for the entire model and the details of the von Mises stress (Pa).  
The impact on the conveyance guide induced transverse bending in the guide, resulting in normal and shear stresses. The distribution of shear stresses in the cross-section of the guide in the direction of the sliding plate's action is shown in Figure 14.  
The shear force in the guide under the action of the guiding force is equal to  
\[T=\iint\limits_{A}\tau_{12}dA\]  
where \(T\) is the shear force in section \(A\) of the guide in the direction of the sliding plate's action; and \(\tau_{12}\) represents the shear stresses in the section of the guide in the direction of the sliding plate's action.  
By calculating the lateral force above and below the guide's action, it was possible to determine the leading force as the difference in the lateral forces above and below the guide's action.  
\[F_{c}=T_{Above}-T_{Below}\]  
where \(F_{c}\) is the face guiding force; and \(T_{Above}\) and \(T_{Below}\) are the shear forces above and below the sliding plate's action, respectively (Figure 14).  
The results of the numerical analysis for three winding velocities, two conveyance positions, and two excavated masses are shown in Table 2.  
\begin{table}
\begin{tabular}{c c c c c} \hline \hline \multicolumn{2}{c}{**Excavated Material Mass**} & \multicolumn{2}{c}{**10 t**} & \multicolumn{2}{c}{**20 t**} \\ \hline \multirow{2}{*}{**Sliding Plate Position**} & \multirow{2}{*}{\begin{tabular}{c} **Bunton** \\ **Connection** \\ \end{tabular} } & \multirow{2}{*}{\begin{tabular}{c} **Guide** \\ **Mid-Span** \\ \end{tabular} } & \multirow{2}{*}{\
\begin{tabular}{c} **Bunton** \\ **Connection** \\ \end{tabular} } & \multirow{2}{*}{
\begin{tabular}{c} **Guide** \\ **Mid-Span** \\ \end{tabular} } \\ \hline \multirow{3}{*}{**Guiding speed**} & **4 m/s** & \(10.61\) & 2.50 & 10.84 & 2.54 \\ \cline{2-6}  & **8 m/s** & \(11.04\) & 2.66 & 11.12 & 2.69 \\ \cline{2-6}  & **16 m/s** & \(29.84\) & 6.22 & 29.90 & 6.24 \\ \hline \hline \end{tabular}
\end{table}
Table 2: Face force for different simulation variants. The force values are given in kN.  
Figure 14: Shear stress distribution in the guide (Pa).