# On Two Approaches to Slope Stability Reliability Assessments Using the Random Finite Element Method
## 5 Slope Reliability Assessment--The Two Approaches
### Approach II  
The \(S_{u}\) (i.e., undrained shear strength) distribution in the slope is generated, based on the statistics (i.e., mean \(\mu\), variance \(\sigma^{2}\) and scales of fluctuation \(\theta\)) of \(S_{u}\). The factor of safety for each realisation of the slope is then found by scaling down the values of \(S_{u}\) by progressively larger values of strength reduction factor (SRF), until the limiting value of SRF (i.e., the realised \(F_{R}\)) is obtained. This process is repeated for a large number of realisations, to obtain a distribution of \(F_{R}\). The pseudocode for approach II is as follows:  
``` 10 Input statistics corresponding to \(F_{d}^{i}\)
20 For \(N\) realisations (loop \(k\)):
30 Generate a slope and shear strength profile (\(S_{u}\)) with \(\mu_{F_{d}^{i}},\sigma,\theta\);
40 Carry out FEA using progressively larger SRF, if \(Iter_{p}=500\), then \(F_{R}^{k}=SRF^{k}\)
50 Output: a distribution of \(F_{R}^{i}\)
********************************block 3************************************************

Note the similarity of block 3 to block 2; block 3 has one input \(F_{d}\) and tries to find the whole distribution of \(F_{R}\) for all realisations, whereas block 2 works on a series values of \(F_{d}\) and does not try to find the exact \(F_{R}\) for all realisations (only to see if it is stable or not). Now, define a unified factor \(F_{i}\) as

\[F_{i}=\frac{F_{d}^{i}}{F_{R}^{i}}=\frac{\alpha_{1}F_{d}^{1}}{\alpha_{1}F_{R}^{ i}}=\frac{\alpha_{2}F_{d}^{2}}{\alpha_{2}F_{R}^{2}} \tag{6}\]

Note that, for a particular realisation, if \(F_{d}^{i}\) is proportional to \(F_{d}^{1}\) (or any other value \(F_{d}^{2}\)) by a factor of \(\alpha_{1}\) (or \(\alpha_{2}\)), the realised \(F_{R}^{i}\) is also propotional to \(F_{R}^{1}\) (or \(F_{R}^{2}\)) by the same factor. This is because, whether one starts the strength reduction process (to get the realised factor of safety for that realisation) from a high mean shear strength (high \(F_{d}\)) or a low mean shear strength (low \(F_{d}\)), in both cases the slope falls down when the mean strength along the failure surface reaches the same critical value (for the same spatial variation of soil properties). For example, if \(F_{d}^{1}=1.4\) and \(F_{R}^{1}=1.3\) for a particular realisation (the critical value of strength is then \(\mu_{critical}=f(1.4\times(s_{u}^{1},s_{u}^{2},\ldots,s_{u}^{n})_{\mu_{d-10}}/1.3)\), where \(s_{u}^{i}(i=1,\ldots,n)\) is the random cell values along the failure path, generated based on \(\mu_{F_{d}=1.0}\), and \(f\) is a function that determines the mean strength along the failure path); then, for the case of \(F_{d}^{i}=1.8\), to reach the same critical value, the realised \(F_{R}^{i}\) must be \(1.3\times(1.8/1.4)\). In this case, \(\alpha_{1}=1.8/1.4\). For multiple realisations, the same holds. The probability of failure, \(p_{f}\), at a given \(F_{d}^{i}\), is

\[p_{f}=\frac{N_{f}^{2}}{N} \tag{7}\]

in which \(N\) is the total number of realisations and \(N_{f}^{2}\) (in which superscript 2 denotes the approach II) is the number of realisations in which slope failure has occurred (i.e., \(F_{i}\geq F_{d}^{i}\)). The failure is defined as a slope failed at \(F_{R}^{i}\leq 1.0\) for any \(F_{d}^{i}\), e.g., \(F_{d}^{i}=1.3\). Upon transformation to \(F_{i}\) (Equation (6)), the definition of slope failure is preserved by defining \(F_{i}\geq F_{d}^{i}\) as failure (i.e., when \(F_{R}^{i}\leq 1.0\) in Equation (6), \(F_{i}\geq F_{d}^{i}\)). As shown in Figure 7, an analysis based on \(\mu=40\) kPa and \(\sigma=12\) kPa (\(F_{d}=2.46\)) gives virtually the same results as the analysis based on \(\mu=20\) kPa and \(\sigma=6\) kPa (\(F_{d}=1.23\)), where \(R=1-p_{f}\). The slight difference is due to the different settings (initial FOS, initial FOS step and the three cut values of plastic iteration counts) for the AutoFOS implementation. Note that it does not matter whether the input statistics are \(\mu=40\) kPa and \(\sigma=12\)kPa (\(F_{d}=2.46\)) or \(\mu=20\)kPa and \(\sigma=6\)kPa (\(F_{d}=1.23\)).