# Failure of Rock Slope with Heterogeneous Locked Patches: Insights from Numerical Modelling
## 5 Mechanical Model of Locked Patch  
The above analysis shows that the failure patterns of the locked patches differ, which makes it difficult to represent the mechanical behaviour by a simple model. However, it should be noted that most of the numerical and experimental results yield similar failure patterns where the crack expands along the interface between the locked patch and rock slope, whereas little damage occurs within the bedrock, the slide body, and the locked patch; thus, they can be simplified as rigid bodies. Figure 12 illustrates the simplified model to illuminate the failure process of the locked patch. Since the crack propagation on the upper and lower interfaces of the locked patch is almost synchronous and symmetric, only the interface between the bedrock and the locked patch is analysed. For a locked patch confined by a normal stress subjected to shear, its shear resistance (\(R_{s}\)) is:  
\[R_{s}=R_{n}tan\varnothing+cl_{c} \tag{14}\]  
Figure 11: Comparisons between numerical results and experimental results. (**a**) Total length vs. shear strength, (**b**) total length vs. peak shear displacement, (**c**) spacing vs. shear strength, (**d**) spacing vs. peak shear displacement, (**e**) total height vs. shear strength, (**f**) total height vs. peak shear displacement. where \(R_{n}\) denotes the normal stress, \(\varnothing\) is the friction angle, \(c\) is the cohesion per unit length, and \(l_{c}\) is the contact length between the locked patch and the bedrock. The locked patch is represented by a slider and a spring in series (Figure 12). There are three possible types of contacts between the bedrock and the slider. When the shear loading is less than the shear resistance of the slider, the slider remains stationary (Figure 12a). When the shear loading exceeds \(R_{s}\), the slider begins to slide along the interface (Figure 12b). The effective contact area between the slider and bedrock decreases gradually. When \(l_{c}\) decreases to 0, the slider is separated from the bedrock and the slider completely loses its resistance (Figure 12c). For the model with two locked patches, the shear displacement of the slide body and the locked patches are identical. The two locked patches are parallel connected. The stiffness of the first and second locked patch are \(k_{1}\) and \(k_{2}\), respectively. The force of the first and second spring \(F_{s1}\) and \(F_{s2}\) can be calculated by:  
\[u=u_{1}=u_{2} \tag{15}\]  
Figure 12: Schematic of the contact states: (**a**) stationary state, (**b**) sliding state, (**c**) separated state, (**d**) simplified model with two locked patches, and (**e**) fitting curves of the model with two identical locked patches. \[F_{s1} =ku_{11} \tag{16}\] \[F_{s2} =ku_{21} \tag{17}\]  
where \(u\) is the shear displacement of the slide body; \(u_{1}\) and \(u_{2}\) are shear displacements of first and second locked patch, respectively. \(u_{11}\) and \(u_{21}\) are displacements of the first and second spring, respectively. When the shear displacement is small, the force of the spring is also small. Therefore, the slider remains stationary. In this stage, the locked patch can be simplified into a spring. Later, the increased displacement increases the force of the spring as well. When the force of the spring is larger than the shear resistance of the slider, the slider begins to slide. The sliding of the slider decreases the contact length. Correspondingly, spring elongation decreases and the force of spring decreases. The force of spring \(F_{s}\) and slider in this stage is calculated by:  
\[F_{s}=k\big{(}u_{spmax}-\Delta u_{sp}\big{)} \tag{18}\]  
\[R_{s}=R_{n}tan\varnothing+c(l_{c}-\Delta u_{sl}) \tag{19}\]  
where \(u_{spmax}\) is the maximum elongation displacement of the spring; \(\Delta u_{sp}\) and \(\Delta u_{sl}\) are the variations of the shear displacement of spring and slider, respectively.