# On Two Approaches to Slope Stability Reliability Assessments Using the Random Finite Element Method
## 5 Slope Reliability Assessment--The Two Approaches
### Approach 1  
The approach will be explained in the case of an undrained clay slope characterised by the spatially variable undrained shear strength \(S_{u}\).  
The mean (i.e., \(\mu\)) of \(S_{u}\) for \(F_{d}=1.0\), based on deterministic finite element analysis (FEA), is found first. Next, using this value of \(\mu\) (e.g., \(\mu=16.3\) kPa for the slope analysed in Figure 6) and the corresponding values of \(\sigma\) and \(\theta\), the \(S_{u}\) distribution in the slope is generated (note that, due to the relative positioning of weak and strong zones in the slope, the realised \(F_{R}\) could be larger or smaller than 1.0). The \(S_{u}\) values (for the realisation generated in the last step) are then scaled up by progressively larger values of \(F_{d}\), and, for each value of \(F_{d}\), the slope is analysed to see if it is stable (i.e., the FEA is conducted only for one trial SRF = 1.0, if the plastic iteration count reaches 500, then the slope fails; that is \(F_{R}\leq 1\). The algorithm does not try to find the exact \(F_{R}\); it could be any value smaller than 1.0 if the slope failed). By repeating this process for a large number of realisations (\(N\)) while counting the number of failed slopes (\(N_{f}\)) for each \(F_{d}\), the reliability of the slope may be computed for a range of values of \(F_{d}\), thereby resulting in a reliability versus \(F_{d}\) curve.  
Note that if the FEA converges within \(C=500\) iterations (i.e., less than 500) with a trial SRF = 1.0, then it means that the realised \(F_{R}\) for that realisation must be larger than 1.0 (i.e., an SRF larger than 1.0 is needed to allow the plastic algorithm to hit the 500 iteration ceiling); that is, the slope is stable for that \(F_{d}\) that is used to scale up that realisation.  
The pseudocode for approach I is as follows (\(Iter_{p}\) is the plastic iteration count):  
```
10 Initialisation: \(N_{f}(j)=0\) and input \(\mu_{F_{d}=1.0}\)
20 For \(N\) realisations (loop \(i\)):
30 Generate a slope and shear strength profile (\(S_{u}\)) with \(\mu_{F_{d}=1.0},\sigma,\theta\);
40 For a series of values of \(F_{d}\) (e.g., \(F_{d}=0.9,1.0,1.1,1.2,1.3,1.4\) with \(\Delta F_{d}=0.1\)) (loop \(j\)):
50 Scale up \(S_{u}\) profile by \(F_{d}\)
60 Carry out FEA, let \(SRF=1.0\), if \(Iter_{p}\geq 500\), then \(N_{f}(j)=N_{f}(j)+1\)
70 Output: \(N_{f}\) for each \(F_{d}\)
********************************block 1************************************************

or alternatively,
``` 10 Initialisation: \(N_{f}(j)=0\) and input \(\mu_{F_{d}=1.0}\)
20 For a series of values of \(F_{d}\) (e.g., \(F_{d}=0.9,1.0,1.1,1.2,1.3,1.4\) with \(\Delta F_{d}=0.1\)) (loop \(j\)):
30 For \(N\) realisations (loop \(i\)):
40 Generate a slope and shear strength profile (\(S_{u}\)) with \(\mu_{F_{d}=1.0},\sigma,\theta\);
50 Scale up \(S_{u}\) profile by \(F_{d}\)
60 Carry out FEA, let \(SRF=1.0\), if \(Iter_{p}\geq 500\), then \(N_{f}(j)=N_{f}(j)+1\)
70 Output: \(N_{f}\) for each \(F_{d}\)
********************************block 2************************************************  
The algorithm presented in block 2 is not efficient compared to that shown in block 1 as line 30 needs to be regenerated for each \(F_{d}\) considered.  
The probability of failure, \(p_{f}\), for a given \(F_{d}\), is  
\[p_{f}=\frac{N_{f}^{1}}{N} \tag{5}\]  
in which \(N\) is the total number of realisations and \(N_{f}^{1}\) (in which superscript 1 denotes the approach I) is the number of realisations in which slope failure has occurred (i.e., \(F_{R}\leq 1.0\)).