# Calculation of Safety Factors of the Eurocodes
## 3 Results
### Two Loads  
The permanent load \(G(x),\ g(x)\) and the variable load \(Q(x),\ q(x,)\) are combined independently using the convolution equation to obtain the cumulative distribution of the combination load \(GQ(x)\)[7, 8]. \[GQ(x)=\int_{-\infty}^{\infty}G(x-r)q(r)dr, \tag{13}\]  
\[GQ(x)=\int_{-\infty}^{\infty}G(r)q(x-r)dr, \tag{14}\]  
\[GQ(x)=\int_{-\infty}^{\infty}g(x-r)Q(r)dr, \tag{15}\]  
and  
\[GQ(x)=\int_{-\infty}^{\infty}g(r)Q(x-r)dr. \tag{16}\]  
One equation applicable in calculating the combination load in load proportion \(\alpha\) is  
\[GQ(x,\ \alpha)=\int_{-\infty}^{\infty}G(x-r,\ \mu_{G}(1\ -\alpha),\sigma_{G}(1\ - \alpha))q(r,\ \mu_{Q}\alpha,\sigma_{Q}\alpha)dr \tag{17}\]  
The failure probability or the safety factors are calculated independently in the combination load using Equations (7)-(17). In Figure 3, the reliability index \(\beta_{50}\) for \(V_{M}=0.15,\ \gamma_{M}=1.15\) is given. The curve is presented in Figure 5 of [3] and the curves are the same within graphical and calculation accuracy, which means that the equations given above yield the same result as the current calculation algorithms. In Figure 4, a complementary calculation is given, i.e., the material safety factor is given as a function of the target reliability \(\beta_{50}=3.83\). The equations above give the same result as given in [12]. In Figure 5, the material factors of the current Eurocodes are given using the combination rule (8.12) of the Eurocodes [1]. The curve for concrete matches well with the recommended value of safety factor \(\gamma_{M}=1.4\). The recommended value for timber \(\gamma_{M}=1.3\) looks high and the steel \(\gamma_{M}=1.0\), low for the loads where the variable load is dominating. Figure 3: The reliability index calculated for \(V_{M}=0.15,\ \gamma_{M}=1.15\). The horizontal dashed line denotes the target reliability index \(\beta_{50}=3.83\). In some recent calculations [6], the reliability and the safety factors of the Eurocodes were calculated for one-year loads \(\beta_{1}=4.7\). Such calculation necessitates other sensitivity factors than disclosed in the Eurocodes (\(\alpha_{\text{E}}=0.7\)). Here \(\alpha_{\text{E}}=0.89\), i.e., \(\beta_{1}=4.2\), is selected as it results in about the same outcome as presented above and the safety factors are given in dotted lines Figure 5.  
When the material safety factor is used as a reference, both calculation methods lead to about the same outcome for variable loads, and the one-year calculation results in higher safety factors for the permanent loads. Figure 6 discloses the complementary calculation i.e., the reliability indexes are given as a function of the load ratio. Figure 4: The material factor calculated for \(\text{V}_{\text{M}}=0.15\), \(\beta_{\text{S0}}=3.83\). The horizontal dashed line denotes \(\gamma_{\text{M}}=1.15\). Figure 5: Material factors \(\gamma_{\text{M}}\) of the current Eurocodes using the combination rule (8.12). Solid line, steel; dashed line, timber; and dash-dotted line, concrete. The dotted lines denote the safety factors of the one-year calculation. The current safety factor for steel is \(\gamma_{\text{M}}=1\). Figure 5 shows that the calculated value is 27% higher. However, Figure 6 shows that the reliability index for the current reliability calculation is 3.01 and for the one-year calculation 3.50, i.e., the relative deviations are 26% and 20%, respectively. Using the safety factor vs. the reliability index as the reference leads to a different outcome. In this case, the reliability index option results in a lower safety factor for the variable load. It is obvious that using the safety factor as the reference is more correct which confirms the earlier observation by the authors [7] that the reliability index is a biased abstraction, when it is used as the reference in the safety factor setting.