## 4 Numerical Examples
### Numerical Example 3  
In Figure 8, a seven-bar truss problem [32] is modified and used as example 3. Each bar has the same length \(L\), and their cross-section areas \(A_{i}(i=1,2,\cdots,7)\) are 1820 mm\({}^{2}\), 1641 mm\({}^{2}\), 1722 mm\({}^{2}\), 1695 mm\({}^{2}\), 2095 mm\({}^{2}\), 3275 mm\({}^{2}\), and 2678 mm\({}^{2}\), respectively. Three concentrated loads \(F_{1}\), \(F_{2}\), and \(F_{3}\) are uncertain parameters whose marginal intervals are \(F_{1}^{1}=\left[2.7\times 10^{5}\text{N},3.3\times 10^{5}\text{N}\right]\), \(F_{2}^{1}=\left[0.855\times 10^{5}\text{N},1.045\times 10^{5}\text{N}\right]\), and \(F_{3}^{1}=\left[0.9\times 10^{5}\text{N},1.1\times 10^{5}\text{N}\right]\), respectively. Neglecting the buckling failure mode, this statically determinate structure can be regarded as a series system consisting of seven performance functions (details on the performance functions' determination can be found in Appendix A):  
\[\begin{array}{l}g_{1}=S-\left(\frac{F_{1}}{2\sqrt{3}}+\frac{3F_{2}}{4}+\frac {F_{3}}{4\sqrt{3}}\right)/A_{1}\\ g_{2}=S-\left(\frac{F_{1}}{2\sqrt{3}}+\frac{F_{2}}{4}+\frac{\sqrt{3}F_{3}}{4} \right)/A_{2}\\ g_{3}=S-\left(\frac{F_{1}}{\sqrt{3}}-\frac{F_{2}}{2}+\frac{F_{3}}{2\sqrt{3}} \right)/A_{3}\\ g_{4}=S-\left(\frac{F_{1}}{\sqrt{3}}-\frac{F_{2}}{2}+\frac{F_{3}}{2\sqrt{3}} \right)/A_{4}\\ g_{5}=S-\left(\frac{F_{1}}{\sqrt{3}}+\frac{F_{2}}{2\sqrt{3}} \right)/A_{5}\\ g_{6}=S-\left(\frac{F_{1}}{\sqrt{3}}+\frac{F_{2}}{2}+\frac{\sqrt{3}F_{3}}{2} \right)/A_{6}\\ g_{7}=S-\left(\frac{F_{1}}{\sqrt{3}}+\frac{F_{2}}{2}+\frac{F_{3}}{2\sqrt{3}} \right)/A_{7}\end{array} \tag{26}\]  
where the yield strength \(S\) is 100 Mpa. The uncertainty domain can be expressed as follows:  
\[\begin{bmatrix}F_{1}-F_{1}^{c}\\ F_{2}-F_{2}^{c}\\ F_{3}-F_{3}^{c}\end{bmatrix}\begin{bmatrix}1.1317\text{e}-9&-3.2489\text{e}-10 &-3.0864\text{e}-10\\ -3.2489\text{e}-10&1.0186\text{e}-8&-9.7466\text{e}-10\\ -3.0864\text{e}-10&-9.7466\text{e}-10&1.0185\text{e}-8\end{bmatrix}\begin{bmatrix} F_{1}-F_{1}^{c}\\ F_{2}-F_{2}^{c}\\ F_{3}-F_{3}^{c}\end{bmatrix}\leq 1 \tag{27}\]  
\begin{table}
\begin{tabular}{c c c} \hline \hline
**Methods** & **Non-Probabilistic Failure Degree** & **Alternatives** \\ \hline NRBM [30] & 0.0123710609 \(\leq f_{8}\leq\) 0.0123710611 & 4! = 24 \\ The proposed method & 0.0123710609 \(\leq f_{8}\leq\) 0.0123710611 & 3! = 6 \\ MCS [26] & 0.01237106107628 & - \\ \hline \hline \end{tabular}
\end{table}
Table 4: Reliability analysis results for numerical example 2.  
Figure 8: A seven-bar truss.  
The redundant failure modes and the identification criteria are listed in Table 5. Reliability analysis results for the proposed method, NRBM, and MCS are given in Table 6.  
In Table 5, \(G_{2}(g_{2})\), \(G_{3}(g_{3})\), and \(G_{7}(g_{7})\) are redundant failure modes and, hence, can be removed. In Table 6, both the proposed method and the NRBM yield reliability bounds containing the reference solution provided by the MCS, and the former has narrower bounds than the latter. For the number of possible ordering alternatives, as listed in Table 6, the proposed method and the NRBM have 24 and 5040, respectively. Thus, for this problem, the proposed method is superior to the NRBM in terms of accuracy and efficiency.