Question 21.P.13: A column is fabricated from two 250 mm × 75 mm channel secti...
A column is fabricated from two 250 mm × 75 mm channel sections attached to two flange plates each 300 mm × 12 mm as shown in Fig. P.21.13; the column height is 10 m and the ends of the column are fixed. If the yield stress of mild steel in compression is 250 N/mm² use the Perry-Robertson formula to calculate the crippling load for the column. Take E = 205000 N/mm². The properties of a single 250 mm × 75 mm channel are:
Area = 36.6 cm², I_{z}=3440 \mathrm{~cm}^{4}, I_{y}=166 \mathrm{~cm}^{4} ..
By inspection and by comparison with S.21.11 the column will buckle about its axis of symmetry along the web. The second moment of area of the section about this axis is then
I=2\left(166+36.6 \times 6.9^{2}+\frac{1.2 \times 30^{3}}{12}\right)=9217 \mathrm{~cm}^{4}
The area of the cross section is
A = 2(36.6 + 30 × 1.2) = 145.2 cm²
The radius of gyration of the section is
r=\sqrt{9217 / 145.2}=7.97 \mathrm{~cm}
The effective length of the column is = 0.5 × 10 × 10² = 500 cm so that its slenderness ratio is equal to 500/7.97 = 62.8. 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}{62.8^{2}}=513 \mathrm{~N} / \mathrm{mm}^{2}
The yield stress of the material of the column is 250 N/mm². Therefore, 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= 192.1 N/mm²
The crippling load for the column is then
P=\frac{192.1 \times 145.2 \times 10^{2}}{10^{3}}=2789 \mathrm{\ kN}