## 4 Shear Energy Dissipation of Fiber-Reinforced Iron Tailings Powder
### F-\(s\) Curve Fitting Based on a BP Neural Network  
In this paper, a BP neural network algorithm was used to fit the \(F\)-\(s\) curve. The network structure contains one input layer node, two hidden layer nodes, and one output layer node. The transfer functions of the hidden layer and output layer are shown in Equations (3) and (4), and the shear displacement \(s\) obtained by the direct shear test represents the input variable, and the shear force \(F\) represents the output variable [33; 39].  
Figure 7: Cohesion and internal friction angle of fiber-reinforced iron tailings powder with different polypropylene fiber dosages.  
\[f(v)=\frac{1}{e^{-v}+1} \tag{3}\]  
\[f(v)=v \tag{4}\]  
In order to ensure the fitting accuracy, the input variables and output variables need to be normalized [38; 39]. The fitting function of \(F\)-\(s\) is obtained as  
\[F(s)=\frac{a_{1}}{e^{-a_{2}s-a_{3}}+1}+\frac{a_{4}}{e^{-a_{5}s-a_{6}}+1}+a_{7}, \tag{5}\]  
where \(a_{1}=\frac{F_{\max}\omega_{21}}{2},a_{2}=\frac{2\omega_{11}}{s_{\max}},a_{3}= b_{11}-\omega_{11},a_{4}=\frac{F_{\max}\omega_{22}}{2},a_{5}=\frac{2\omega_{12}}{s_{ \max}},a_{6}=b_{12}-\omega_{12},a_{7}=\frac{F_{\max}}{2}(b_{2}+1)\), among them, \(\omega_{11}\), \(\omega_{12}\), \(\mathrm{b_{11}}\), \(\mathrm{b_{12}}\), \(\omega_{21}\), \(\omega_{22}\), and \(\mathrm{b_{2}}\) are BP neural network calculation parameters, the specific meaning of which can be found in reference [39]. Table 2 shows the calculation results of BP neural network parameters and the average fitting error. Figure 8a,b shows the comparison of the F-s fitting curve and the test curve of polypropylene-reinforced iron tailings powder under normal stresses of 100, 200, 300, and 400 kPa with 0.5% and 1% fiber dosages.  
It can be seen from Table 2 and Figure 8 that the BP neural network can better fit the \(F\)-\(s\) function relationship with a maximum fitting error of 0.4 N. Therefore, the BP neural network algorithm can be used to fit the \(F\)-\(s\) function relationship in the direct shear test.  
\begin{table}
\begin{tabular}{c c c c c c c c c c} \hline \hline \begin{tabular}{c} **Fiber** \\ **Dosage** \\ **(\%)** \\ \end{tabular} & \begin{tabular}{c} **Normal** \\ **Stress** \\ **(kPa)** \\ \end{tabular} & \begin{tabular}{c} **\(\mathbf{a_{1}}\)** \\ **(\%)** \\ \end{tabular} & \begin{tabular}{c} **\(\mathbf{a_{2}}\)** \\ **(\%)** \\ \end{tabular} & \begin{tabular}{c} **\(\mathbf{a_{3}}\)** \\ **(\%)** \\ \end{tabular} & \begin{tabular}{c} **\(\mathbf{a_{4}}\)** \\ **(\%)** \\ \end{tabular} & \begin{tabular}{c} **\(\mathbf{a_{6}}\)** \\ **(\%)** \\ \end{tabular} & \
\begin{tabular}{c} **\(\mathbf{a_{7}}\)** \\ **(\%)** \\ \end{tabular} &
\begin{tabular}{c} **Average** \\ **Error (N)** \\ \end{tabular} \\ \hline \multirow{4}{*}{0} & 100 & \(-\)49.93 & 0.0023 & \(-\)9.71 & 727,324 & 0.00077 & 7.87 & \(-\)727,051 & 0.03 \\  & 200 & \(-\)19.77 & 0.0055 & \(-\)29.97 & \(-\)4,557,279 & \(-\)0.00091 & \(-\)9.36 & 399 & 0.06 \\  & 300 & \(-\)14.06 & 0.0033 & \(-\)17.77 & 317,668 & 0.00037 & 6.33 & \(-\)317,062 & 0.05 \\  & 400 & 24.61 & \(-\)0.0062 & 34.9 & \(-\)38,062 & \(-\)0.00067 & \(-\)3.79 & 823 & 0.02 \\ \hline \multirow{4}{*}{0.25} & 100 & 111.6 & 0.0022 & \(-\)4.46 & 189,740 & 0.0017 & 6.99 & \(-\)189,573 & 0.05 \\  & 200 & \(-\)8044 & \(-\)0.00056 & \(-\)3.26 & 668 & 0.0053 & 1.41 & \(-\)248 & 0.02 \\  & 300 & 1230 & 0.001 & \(-\)0.03 & \(-\)94 & \(-\)0.02 & 6.82 & \(-\)514 & 0.07 \\  & 400 & \(-\)1634 & \(-\)