# Consistency of Imperfections in Steel Eurocodes
## 2 Equivalent Member Method
### Geometrical Interpretations of the above Quantities  
The quantities \(\eta_{\text{tot,c}}/j\), \(U_{\text{N}}\), \(U_{\text{M}}\) and \(U\) are functions of only three parameters: \(\alpha\), \(\overline{\lambda}\), and \(\overline{\lambda}_{0}\). The quantities \(\alpha_{\text{cr}}\), \(k\), \(U_{\text{M},\alpha_{\text{b,d}}}\), \(U_{\text{0,d}}^{\text{II}}\), \(\varepsilon_{0,\text{d}}/j\), \(\eta_{\text{tot,c},\alpha_{\text{b,d}}}\), and \(\eta_{\text{add,c}}/j\) (not directly in all diagrams) depended on four parameters: \(\alpha\), \(\overline{\lambda}\), \(\overline{\lambda}_{0}\), and \(\gamma_{\text{M}1}\). This fact enabled the distributions of the above quantities to be drawn in the following diagrams, which were created for the values \(\overline{\lambda}_{0}=0.2\), \(\gamma_{\text{M}1}=1.0\); \(1.1\) and \(\alpha=0.13\); \(0.21\); \(0.34\); \(0.49\); and \(0.76\). The values in diagrams were valid for \(N_{\text{Ed}}=N_{\text{b,Rd}}=\chi Af_{\text{y}}/\ \gamma_{\text{M}1}\) and, therefore, for both utility factors \(U_{\text{EMM}}=1.0\) and \(U^{\text{II}}=1.0\) if \(\omega_{0,\text{d}}\) was used in the calculation. Note that the diagrams have two vertical axes with different scales. The right one is valid for \(\alpha_{\text{cr}}\) and \(k\). The diagrams enable the determination of: (a) the amplitudes of all kinds of deformations, and (b) the influence of the second-order theory. The first-order analysis may be used for the structure if the increase in the relevant internal forces or moments, or any other change in structural behavior caused by deformations, can be neglected. This condition may be assumed fulfilled if the following criterion is met: \(\alpha_{\text{cr}}=N_{\text{cr}}/N_{\text{Ed}}\geq 10\). This condition is fulfilled in the case of \(\gamma_{\text{M}1}=1.0\) [ \(\gamma_{\text{M}1}=1.1\) ] and for \(\alpha=0.13\) when \(\overline{\lambda}\leq 0.319\) [\(0.335\)]; for \(\alpha=0.21\) when \(\overline{\lambda}\leq 0.321\) [\(0.337\)]; for \(\alpha=0.34\) when \(\overline{\lambda}\leq 0.323\) [\(0.341\)]; for \(\alpha=0.49\) when \(\overline{\lambda}\leq 0.327\) [\(0.345\)]; and for \(\alpha=0.76\) when \(\overline{\lambda}\leq 0.334\) [\(0.353\)]. The contribution of the bending moment \(U_{\text{M}}\) was greater than that of the normal force \(U_{\text{N}}\) for \(\alpha=0.13\) when \(\overline{\lambda}\geq 1.309\); for \(\alpha=0.21\) when \(\overline{\lambda}\geq 1.248\); for \(\alpha=0.34\) when \(\overline{\lambda}\geq 1.16\); for \(\alpha=0.49\) when \(\overline{\lambda}\geq 1.07\); and for \(\alpha=0.76\) when \(\overline{\lambda}\geq 0.938\). The design values of the relative amplitudes of additional deformations \(\eta_{\text{add}}(x)\) can be obtained from the diagrams as differences \(\eta_{\text{add,c}}/j=\eta_{\text{tot,c}}/j-e_{0,\text{d}}/j\). The program FE-STAB [1] provided the following results for the input values from Numerical Example 1: (a) for the buckling perpendicular to the weak axis z-z, with the safety factor \(\gamma_{\text{M}1}=1.1\), the value of axial force \(N_{\text{Ed}}=N_{\text{b,Rd}}=859.584\) kN, and \(e_{0,\text{d}}=10.51\) mm : \(\alpha_{\text{cr}}=1.43499\), \(\eta_{\text{add,d,c}}=24.19\) mm, \(M_{\text{c}}^{\text{II}}=29.83\) kNm. Compare these values to those in Table 4: (b) for \(\varepsilon_{0,\text{d}}=8.09\) mm, FE-STAB gives: \(\alpha_{\text{cr}}=1.43439\), \(\eta_{\text{add,d,c}}=18.62\) mm, and \(M_{\text{eff}}^{\text{II}}=22.96\) kNm. Compare these values to those in Equations (42) and (43). Consequently, \(U_{\text{e}_{0,\text{k}}}^{\text{II}}=0.85\) (Equation (45)). The FE-STAB results confirm the correctness of the values in Table 4 and in Equations (41)-(45); (c) for safety factor \(\gamma_{\text{M}1}=1.0\), a value of the axial force \(N_{\text{Ed}}=N_{\text{b,Rd}}=945.543\) kN, and \(e_{0,\text{k}}=8.09\) mm, FE-STAB gives: \(\alpha_{\text{cr}}=1.30402\), \(\eta_{\text{add,d,c}}=26.608\) mm, and \(M_{\text{eff}}^{\text{II}}=32.81\) kNm. Consequently, \(U^{\text{II}}=U_{\text{N}}+U_{\text{M}}=0.348+0.652=1.0\).