Question 6 of 7: Buckling of a Slit (Open) Thin Tube — Flexural vs Torsional

Question 6: Buckling of a Slit (Open) Thin Tube — Flexural vs Torsional (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 thin circular tube, slit longitudinally so it is an open section; ends free to warp.

Find. (a) the Euler flexural buckling load; (b) the pure-torsional buckling load.

Approach. An open thin section has very low torsional stiffness, so torsional buckling can precede flexural (Euler) buckling. Compute the Euler load from I, and the torsional-buckling load from the St.-Venant term GJ plus the warping term π²EC_w/L², divided by the polar radius r₀² about the shear centre.

1. Flexural (Euler) load — part a. With I = πR³t = π(24)³(2) = 8.686×10⁴ mm⁴, $$\boxed{P_{\text{flex}}=\frac{\pi^{2}EI}{L^{2}}=\frac{\pi^{2}(70\,000)(8.686\times10^{4})}{1750^{2}}=19.6\ \text{kN}.}$$
2. Open-section torsion & warping constants. $$J=\frac{2\pi Rt^{3}}{3}=402\ \text{mm}^{4},\qquad C_w=\tfrac{2}{3}\pi(\pi^{2}-6)tR^{5}=1.291\times10^{8}\ \text{mm}^{6}.$$ The polar radius about the shear centre is r₀² = 5R² = 2880 mm² (shear centre offset e = 2R adds 4R² to R²).
3. Torsional-buckling load — part b. With ends free to warp, $$P_\theta=\frac{1}{r_0^{2}}\!\left(GJ+\frac{\pi^{2}EC_w}{L^{2}}\right)=\frac{1}{2880}\!\left(6.03\times10^{6}+2.91\times10^{7}\right),$$ $$\boxed{P_\theta=\frac{3.515\times10^{7}}{2880}=12.2\ \text{kN}.}$$
4. Governing mode. Because P_θ = 12.2 kN < P_flex = 19.6 kN, the slit column buckles torsionally first. The longitudinal slit removes the closed-tube shear path, collapsing the torsional rigidity so that twisting, not bending, sets the critical load — a hallmark of thin open sections.