## 2 Materials and Methods
### Reliability Analysis  
Before estimating the reliability of the whole structure, it is important to compute the reliability of single elements. First, the safety margin \(M\), which is a difference between the bearing capacity (\(N\)) and the effect of action (\(E\)), should be estimated:  
\[M_{i}=N_{i}-E_{i} \tag{5}\]  
where index \(i\) is the number of the element. The subsequent steps include the computation of the following values:  
* Standard deviation of safety margin: \[\sigma_{Mi}=\sqrt{\sigma_{Ni}^{2}+\sigma_{Ei}^{2}}\] (6) where \(\sigma_{Ni}\) and \(\sigma_{Ei}\) are standard deviations of bearing capacity and effect of action, respectively.
* Reliability index \(t_{i}\): \[t_{i}=\frac{\mu_{Mi}}{\sigma_{Mi}}\] (7) where \(\mu_{Mi}=\mu_{Ni}-\mu_{Ei}\) is the expected value of the safety margin, defined as the difference between expected values of the bearing capacity (\(\mu_{Ni}\)) and effect of actions (\(\mu_{Ei}\)).  
It is worth noting that reliability index \(t\) corresponds to an element, while the reliability index \(\beta\) refers to a whole structure.  
* Probability of failure of an element: \[P_{f_{i}}=\Phi(-t_{i})\] (8)
* Reliability of the element: \[R_{i}=1-P_{f_{i}}\] (9)  
\begin{table}
\begin{tabular}{c c c} \hline \hline \multicolumn{2}{c}{**Reduction Factors at Temperature 0a Relative to the Value of f\({}_{y}\) or Ea at 20 \({}^{\circ}\)C**} \\ \hline \multirow{2}{*}{**Steel Temperature**} & \multicolumn{2}{c}{**Reduction Factor (Relative to E\({}_{a}\)) for**} \\  & \multicolumn{2}{c}{**\(\mathbf{f_{y}}\)) for Effective Yield Strength**} & \multicolumn{2}{c}{**Reduction Factor (Relative to E\({}_{a}\)) for**} \\  & \multicolumn{2}{c}{**\(\mathbf{f_{y}}\)) for Effective Yield Strength**} & \multicolumn{2}{c}{**the Slop of the Linear Elastic Range.**} \\ \cline{2-3}  & \multicolumn{2}{c}{\(k_{y,\theta}\)=\(\frac{f_{eff}}{\gamma}\)} & \multicolumn{2}{c}{\(k_{E,\theta}\)=\(\frac{E_{eff}}{k_{a}}\)} \\ \hline
20 & 1.0 & 1.0 \\ \hline
100 & 1.0 & 1.0 \\ \hline
200 & 1.0 & 0.9 \\ \hline
300 & 1.0 & 0.8 \\ \hline
400 & 1.0 & 0.7 \\ \hline
500 & 0.780 & 0.6 \\ \hline
600 & 0.470 & 0.31 \\ \hline
700 & 0.230 & 0.13 \\ \hline
800 & 0.110 & 0.09 \\ \hline
900 & 0.06 & 0.0675 \\ \hline
1000 & 0.04 & 0.045 \\ \hline
1100 & 0.02 & 0.0225 \\ \hline
1200 & 0.00 & 0.00 \\ \hline \hline \end{tabular}
\end{table}
Table 1: Reduction factors for carbon steel at elevated temperatures [24].  
It is essential to know reliabilities (\(R\)) of all elements in order to find reduced cut-sets, i.e., to analyze in the algorithm only these elements, that reliabilities are different from 1. This is easy to explain, when considering Equations (10) and (11). There are two basic structural systems: series (for statically determinate structures) and parallel (for some simply statically indeterminate structures). Most of the statically indeterminate structures correspond to the mixed system, but there are always some combinations of two basic systems: series and parallel ones. Thus, even in these cases, Equations (10) and (11) are useful. Equation (10) corresponds to the series system, the reliability of the whole structure is just a product of reliabilities of all elements. Therefore, it is clear that any \(i\)-th element, whose reliability is equal to 1 (Ri = 1), has no influence on the reliability of the structure.  
Equation (11) corresponds to the parallel system. If the reliability of any \(i\)-th element in this system is equal to 1 (\(R_{i}\) = 1), then the expression in brackets equals to 0, so it has no influence on the whole sum in Equation (11)  
\[R=\prod_{i=1}^{n}R_{i} \tag{10}\]  
\[R=1-\sum_{i=1}^{n}(1-R_{i}) \tag{11}\]