Question 21.P.14: A mild steel pin-ended column is 2.5 m long and has the cros...

The column will buckle about an axis parallel to its web. The second moment of area of the column is then given by

I=\frac{2 \times 8 \times 130^{3}}{12}+\frac{184 \times 6^{3}}{12}=2.93 \times 10^{6} \mathrm{~mm}^{4}

The area of cross section is

A = 2 × 130 × 8 + 184 × 6 = 3184 mm²

Then

r^{2}=2.93 \times \frac{10^{6}}{3184}=920.2

so that

r = 30.3 mm

Then

\frac{L}{r}=\frac{2.5 \times 10^{3}}{30.3}=82.5

Also

\sigma_{\mathrm{CR}}=\frac{\pi^{2} \times 200000}{82.5^{2}} (see Eq. (21.25) \sigma_{\mathrm{CR}}=\frac{\pi^{2} E}{\left(L_{e} / r\right)^{2}})

i.e.

\sigma_{\mathrm{CR}} = 290.0 N/mm²

Substituting these values etc. 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 = 180.9 N/mm²

Then the maximum load the column can withstand is given by

P=180.9 \times 3184 \times 10^{-3}=576 \mathrm{\ kN}

The Euler buckling load is

P_{\mathrm{CR}}=290 \times 3184 \times 10^{-3}=923.4 \mathrm{\ kN}

Then

\frac{P}{P_{\mathrm{CR}}}=\frac{576}{923.4}=0.62.