## 3 Results
### Computational Study
#### 3.1.1 Example 1: Statically Indeterminate Truss with Two Degrees of Freedom  
The aim of presenting such a simple truss (Figure 2) is to introduce the idea of the method. \begin{table}
\begin{tabular}{c c c} \hline \hline
**Variable** & **Symbol** & **Coefficient of Variation v [\%]** \\ \hline Effect of action & \(E\) & 6 \\ Cross-sectional area & \(A\) & 6 \\ Yield strength & \(f_{y}\) & 8 \\ Load bearing capacity & \(N\) & 10 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Coefficients of variation for random variables. Figure 2: Statically indeterminate truss with two degrees of freedom. With the results of the FEM computation, it is possible to perform a reliability analysis. It is comfortable to use the table (Table 2) to compute the reliability of single elements. The values in the subsequent columns were computed according to Equations (13)-(18). For the following minutes of fire duration, the tables similar to this above were prepared in an Excel program. With the knowledge of the reliability of single elements, it is possible to conduct system analysis. This task was realized by using the author-prepared console application, written in the C++ language. In the Figure 3, the full cut-sets for the analyzed structure are presented. It is divided into two- and three-element cut-sets, in the both cases single cut-sets are connected with each other in a parallel-serial way (parallel-series system). \begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline \multirow{2}{*}{**No. **} & \multirow{2}{*}{**Profiles**} & – & \(\sigma_{E}\) & – & \(\sigma_{N}\) & – & \(\sigma_{M}\) & \multirow{2}{*}{**\(t\) [-]**} \\  & & \(E\) [kN] & [kN] & \(N\) [kN] & [kN] & \(M\) [kN] & [kN] & \(t\) [-] \\ \hline \multirow{2}{*}{1} & IPE & \multirow{2}{*}{8.93} & \multirow{2}{*}{0.54} & \multirow{2}{*}{221.45} & \multirow{2}{*}{22.15} & \multirow{2}{*}{212.52} & \multirow{2}{*}{22.15} & \multirow{2}{*}{9.59} \\  & IPE & & & & & & & \\  & 100 & & & & & & \\  & IPE & & & & & & & \\  & 160 & & & & & & & \\  & 4 & IPE160 & & & & & & \\  & CHS & & & & & & & \\  & 51 \(\times\) 4 & & & & &  
In Figure 3, some elements are in a red color; these are elements whose reliability is different then 1.0 (see Table 2). Two cut-sets are in a green frame; in fact, they decide about the reliability of the whole structure. Please note that according to Equation (10), the single cut-set, which is represented by the parallel system, has a reliability different than one only if all its causative elements' reliabilities are different than one. This is the way in which reduced cut-sets were searched. In Figure 4, the reduced cut-sets until 15 min of fire duration are presented. They changed, because under the temperature load, the effect of action in single elements changed and the load-bearing capacity of the elements decreased. Based on the cut-sets above, the reliability for whole structures in particular minutes can be computed as for a parallel-series system. Parallel cut-sets are connected in a series way, so firstly it should be computed parallelly and then in a series way. The examples corresponding to the subsequent minutes of fire duration are presented below. _The reliability in 0 min (without temperature) of fire duration:_  
\[R_{0 min}=[1-(1-R_{5})(1-R_{8})]\cdot[1-(1-R_{7})(1-R_{11})] \tag{20}\]  
Figure 4: Reduced cut-sets for the structure of the initial static scheme of the truss (Figure 1). Figure 3: Full cut-sets for the truss with two degrees of freedom (Figure 1). The reliability in the 5 th min of fire duration can be computed in a few steps, as follows:  
\[ \begin{array}{ll}R_{5\min,2 el}=[1-(1-R_{5})(1-R_{8})]\cdot[1-(1-R_{7})(1-R_{11})] \\ R_{5\min,3 el}=[1-(1-R_{5})(1-R_{6})(1-R_{7})]\cdot[1-(1-R_{6})(1-R_{7})(1-R_{8}) ]\\ \cdot[1-(1-R_{6})(1-R_{8})(1-R_{11})]\\ R_{5\min}=R_{5\min,2 el}\cdot R_{5\min,3 el}\end{array}