Question 7 of 8: Buckling of a Slit-Tube Column — Flexural and Torsional Modes

Question 7: Buckling of a Slit-Tube Column — Flexural and Torsional Modes (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A pin-ended, warping-free slit circular tube (open thin-walled section).

Find. The axial buckling load in (a) the flexural (bending) mode and (b) the torsional mode.

Approach. For flexure, apply the Euler formula with the section’s bending inertia. For torsion, use the torsional-buckling load with the open-section torsion constant J, the warping constant C_w, and the polar radius of gyration about the shear centre (which for a slit tube lies a distance 2R from the centroid).

1. (a) Flexural (Euler) buckling. For pin-ended ends, effective length = L: $$P_{cr}=\frac{\pi^2 EI}{L^2}=\frac{\pi^2(50\,000)(1.532\times10^{5})}{2000^{2}}.$$ $$\boxed{P_{flex}\approx 18.9\ \text{kN}}$$
2. Open-section constants. For a slit tube of mean radius R and thickness t, $$J=\frac{2\pi R t^{3}}{3}=486\ \text{mm}^4,\qquad C_w=\frac{2}{3}\pi(\pi^2-6)\,tR^{5}=3.32\times10^{8}\ \text{mm}^6.$$
3. Polar radius about the shear centre. The shear centre is a distance $e=2R$ from the centroid, so $$r_0^{2}=\frac{I_x+I_y}{A}+e^{2}=R^{2}+4R^{2}=5R^{2}=4205\ \text{mm}^2.$$
4. (b) Torsional buckling. With warping-free ends, $$P_{\theta}=\frac{1}{r_0^{2}}\Big(GJ+\frac{\pi^2 E C_w}{L^{2}}\Big)=\frac{7.29\times10^{6}+4.10\times10^{7}}{4205}.$$ $$\boxed{P_{tor}\approx 11.5\ \text{kN}}$$
5. Governing mode. Because the open (slit) section is torsionally very weak, the torsional load (11.5 kN) is below the flexural load (18.9 kN): the column buckles first in pure torsion.