# Failure of Rock Slope with Heterogeneous Locked Patches: Insights from Numerical Modelling
## 3 Numerical Results
### ANOVA Analysis  
To determine the weighting contribution of each input (e.g., strength, length, spacing, depth, height, and homogeneity in this study) to output (e.g., shear strength and peak shear displacement), analysis of variance (ANOVA) was employed to quantify the effects. The total sum of squares, \(SST\), which shows the total variability of response, is calculated as follows:  
\[SST=\sum_{i=1}^{i=r}\sum_{j=1}^{i=m}\left(y_{ij}-\overline{y}\right)^{2} \tag{7}\]  
Figure 9: Relationships between (**a**) the depth of right locked patch, (**b**) the height of right locked patch, (**c**) the homogeneity of right locked patch vs. shear strength and peak displacement of two locked patches. LP denotes locked patch, and SS and PSD represent shear strength and peak shear displacement, respectively.  
where \(y_{ij}\) denotes the response of \(ij\)th experiment; \(\overline{y}\) denotes the mean response of the total experiments [34]. The total sum of squares is composed of two sources of the sum of squares due to each factor and the error sum of squares \(SSE\), which can be given by:  
\[SSE=\sum_{i=1}^{i=r}\sum_{j=1}^{j=m}(y_{ij}-\overline{y_{i}})^{2} \tag{8}\]  
\[SSF=SST-SSE=m\sum_{i=1}^{i=r}(\overline{y_{i.}}-\overline{y})^{2} \tag{9}\]  
where \(\overline{y_{i.}}\) denotes the mean response of the levels of the factor \(i\).  
The total mean square, error mean square, and each factor mean square can be calculated by dividing the corresponding degree of freedom (DOF). DOF can be given by:  
\[dft=r\times m-1 \tag{10}\]  
\[dfe=r\times(m-1) \tag{11}\]  
\[diff=r-1 \tag{12}\]  
\[F=\frac{MSSF}{MSSE}=\frac{SSF/diff}{SSE/dfe} \tag{13}\]  
The \(F\) value of each factor can be obtained by the ratio of mean square of factor to the mean square of error:  
The \(F\)-test is undertaken to evaluate the significance of each parameter. The calculated \(F\) values are compared with \(F\)-distribution values with specific confidence levels. If the \(F\) value is greater than the \(F\)-distribution value, it means that the factor has a significant effect on the response with that specified level of confidence. Namely, a larger value of \(F\) value means the effect is more pronounced. Therefore, the weighting contributions of different parameters are based on the \(F\) value.  
The results of ANOVA are calculated and presented in Tables 2 and 3. The effect of different parameters on the shear strength and peak shear displacement can be divided into three groups. The length of locked patch has the greatest effect on the shear strength and peak shear displacement where the weightings are more than 46%, and 86%. In the case of the strength and spacing of the locked patch, they have the similar effect where the average weightings of the shear strength and peak shear displacement are 17.6% and 4.1%, respectively. In the case of the locked patch depth, height, and homogeneity, the results indicate that they have limited effect on the rock slope. The sum of the three weights in peak shear displacement is only 5.8%. Compared with the other parameters, the influence of these three parameters on the peak shear displacement can be ignored. The average weight of these three parameters in shear strength is only 6.2%. Therefore, the attention should be focused on the strength, length, and spacing for the locked rock slope with two locked patches.  
\begin{table}
\begin{tabular}{c c c c c c} \hline \hline \multirow{2}{*}{**Parameters**} & **Sum of** & \multirow{2}{*}{**DOF**} & \multirow{2}{*}{**Mean Square**} & \multirow{2}{*}{**F Value**} & **Weighting** \\  & **Squares** & & & & **Contribution** \\ \hline Strength (MPa) & 68.64 & 4 & 17.16 & 1.63 & 17.17 \\ Length (m) & 184.91 & 4 & 46.23 & 4.38 & 46.26 \\ Spacing (m) & 72.03 & 4 & 18.01 & 1.71 & 18.02 \\ Depth (m) & 27.52 & 5 & 5.50 & 0.52 & 5.51 \\ Height (m) & 32.92 & 5 & 6.58 & 0.62 & 6.59 \\ Homogeneity, \(m\) & 32.27 & 4 & 8.07 & 0.61 & 6.46 \\ \hline \hline \end{tabular}
\end{table}
Table 2: Summary of ANOVA for shear strength.