Question 21.P.11: A mild steel column is 6 m long, is fixed at both ends and h...

The column will buckle about an axis along the line of symmetry through the web (see Section 21.4). The second moment of area of the column cross section about this axis is

I=2 \times \frac{10 \times 150^{3}}{12}+\frac{230 \times 8^{3}}{12}=5.6 \times 10^{6} \mathrm{~mm}^{4}

The area of the cross section is

A = 2 × 150× 10 + 230 × 8 = 4840 mm²

The radius of gyration of the cross section is then

r=\sqrt{5.6 \times 10^{6} / 4840}=34.0 \mathrm{~mm}

The equivalent length, L_{e} , of the column is 0.5 × 6 × 10³ = 3000 mm. Then, from Eq. (21.25) \sigma_{\mathrm{CR}}=\frac{\pi^{2} E}{\left(L_{e} / r\right)^{2}}

\sigma_{\mathrm{CR}}=\frac{\pi^{2} \times 200000}{(3000 / 34.0)^{2}}=253.7 \mathrm{~N} / \mathrm{mm}^{2}

The yield stress of the material of the column is given as 300 N/mm². Then, substituting these values in Eq. (21.46) \sigma=\frac{1}{2}\left[\sigma_{\mathrm{Y}}+\left(1+0.003 \frac{L}{r}\right) \sigma_{\mathrm{CR}}\right]-\sqrt{\frac{1}{4}\left[\sigma_{\mathrm{Y}}+\left(1+0.003 \frac{L}{r}\right) \sigma_{\mathrm{CR}}\right]^{2}-\sigma_{\mathrm{Y}} \sigma_{\mathrm{CR}}} gives

σ = 168.1 N/mm²

Applying a factor of safety of 2 the safe working load is then

P_{\text {safe }}=\frac{168.1 \times 4840 \times 10^{-3}}{2}=406.8 \mathrm{\ kN}