Question 21.P.12: A column is fabricated from two 305 mm × 305 mm × 158 kg Uni...
A column is fabricated from two 305 mm × 305 mm × 158 kg Universal Column sections placed side by side as shown in Fig. P.21.12. The column has fixed ends and an overall height of 12 m. Given that the yield stress in compression of mild steel is 250 N/mm² calculate the maximum allowable load for the column using the Perry-Robertson formula. Take E = 200000 N/mm² and assume a factor of safety of 2. The properties of a single 305 mm × 305 mm × 158 kg UC are:
Area = 201.2 \mathrm{~cm}^{2}, I_{z}=38740 \mathrm{~cm}^{4}, I_{y}=12524 \mathrm{~cm}^{4}
The second moment of area about the axis of symmetry parallel to the flanges of the composite section is
I_{x}=2 \times 38740=77480 \mathrm{~cm}^{4}
and about the axis of symmetry parallel to the webs of the section
I_{y}=2\left(12524+201.2 \times 19^{2}\right)=170314 \mathrm{~cm}^{4}
The column will therefore buckle about the x axis.
The corresponding radius of gyration is then given by
r=\sqrt{77480 / 2 \times 201.2}=13.88 \mathrm{~cm}
The slenderness ratio = \frac{0.5 \times 12 \times 10^{3}}{13.88 \times 10}=43.2
and
\sigma_{\mathrm{CR}}=\frac{\pi^{2} \times 200000}{(43.2)^{2}}=1058 \mathrm{~N} / \mathrm{mm}^{2} (Eq. (21.25) \sigma_{\mathrm{CR}}=\frac{\pi^{2} E}{\left(L_{e} / r\right)^{2}})
The yield stress in compression of the material of the column is 250 N/mm². 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 = 215.1 N/mm²
The maximum allowable load is then, taking the factor of safety as 2
P_{\max }=\frac{215.1 \times 2 \times 201.2 \times 10^{2} \times 10^{-3}}{2}=4328 \mathrm{\ kN}