Check for Axial Force

Need to satisfy the following equation:

Thus, the ratio is $\frac{1,500}{122.7\times {10}^{-1}\times 239\times 1}=0.51<1$

Check the slenderness per Cl. 10.4.1. Limit the slenderness to 120.

Thus, the ratio is 89.92 / 120 = 0.75

Check Cl. 7.1.3

Therefore, $\overline{\lambda }=3.063$

Need to satisfy the following equation:

where

ϕ	=	$0.5\frac{\delta -\sqrt{{\delta }^2-39.48\times {\overline{\lambda }}^2}}{{\overline{\lambda }}^2}$, per Eq. 8
δ	=	$9.87(1-\alpha +\beta \overline{\lambda })+{\overline{\lambda }}^2$, per Eq. 9

From Table 7, note 2, α = 0.04 and β = 0.14.

Thus, the ratio is $\frac{1,500}{0.551\times 122.7\times {10}^{-1}\times 239\times 1}=0.93<1$

Check Cl. 7.3.2

Design height of the cross-section wall, h_ef = 238 mm (Cl. 7.3.1)

Since $\overline{\lambda }=3.063>2$, ${\overline{\lambda }}_{uw}=1.20+0.35\overline{\lambda }=2.27<2.3$ (per F.(24), Table 9)

${\overline{\lambda }}_w<{\overline{\lambda }}_{uw}$, OK

Check Cl. 7.3.8

Design width of the cross-section wall, b_ef = b/2 - 2×t_w = 130 mm

Since $\overline{\lambda }=3.063>2$, ${\overline{\lambda }}_{uf}=0.36+0.10\overline{\lambda }=0.67$ (per F.(37), Table 10)

${\overline{\lambda }}_f<{\overline{\lambda }}_{uf}$, OK