Buckling of a slit thin-walled column: 22-Mec-A7 Advanced Strength of Materials, May 2016, Q5

Question 5: Buckling of a slit thin-walled column (equal value)

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. Length L = 2200 mm; outer diameter 55 mm, wall t = 2.0 mm, so mean radius R = (55−2)/2 = 26.5 mm; slit (open) section; pin-ended with warping free; E = 50 GPa, G = 15 GPa.

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

Open (slit) thin tube: the longitudinal cut destroys the closed-torsion path, so the section carries torque only through St. Venant plus warping stiffness.

Approach. Compute the flexural Euler load from the ring’s bending inertia. For torsion of the open section, use the thin-strip torsion constant J and the slit-tube warping constant C_w, with the polar radius r₀ taken about the shear centre; combine St. Venant and warping stiffness in the torsional-buckling formula.

Flexural (bending) buckling — part (a). The thin ring’s second moment is I = πR³t: $$I=\pi R^3 t=\pi(26.5)^3(2)=1.169\times10^{5}\ \text{mm}^4$$ $$P_E=\frac{\pi^2 E I}{L^2}=\frac{\pi^2(50\,000)(1.169\times10^{5})}{2200^2}=\boxed{11.9\ \text{kN}}$$
Open-section torsion properties. For a slit thin tube the St. Venant constant is that of an unrolled strip of length 2πR, and the warping constant follows the standard slit-tube result: $$J=\frac{2\pi R t^3}{3}=444\ \text{mm}^4,\qquad C_w=\frac{2}{3}\pi(\pi^2-6)\,t R^5=2.118\times10^{8}\ \text{mm}^6$$
Polar radius about the shear centre. The slit-tube shear centre lies e = 2R from the centroid, so $$r_0^2=\frac{I_x+I_y}{A}+e^2=R^2+(2R)^2=5R^2=3511\ \text{mm}^2$$
Pure-torsional buckling — part (b). With warping free over the length L, $$P_\theta=\frac{1}{r_0^2}\!\left(GJ+\frac{\pi^2 E C_w}{L^2}\right)=\frac{6.66\times10^{6}+2.160\times10^{7}}{3511}$$ $$\boxed{P_\theta=8.05\ \text{kN}}$$
Governing mode. Because the slit removes the closed-tube torsional stiffness, P_θ = 8.05 kN < P_E = 11.9 kN: the column buckles first in pure torsion. (A stress check confirms the Euler stress, 34 MPa, is well below yield, so elastic buckling is valid.)

Question 5 — buckling loads
Mode	Critical load
Pure bending (Euler)	11.9 kN
Pure torsion	8.05 kN (governs)

Topic. Flexural vs. torsional buckling of a thin-walled open (slit) column.

Key relations. Euler P_E = π²EI/L²; open-strip J = 2πRt³/3; slit warping C_w = (2/3)π(π²−6)tR⁵; torsional P_θ = (GJ + π²EC_w/L²)/r₀², r₀² = 5R².

Why this works. A closed circular tube has an enormous torsion constant (J = 2πR³t), so torsional buckling never governs. Cutting a longitudinal slit collapses J by a factor of order (t/R)² to the thin-strip value, leaving warping as the main torsional stiffness. The open section’s shear centre also shifts to e = 2R from the centroid, enlarging r₀ and further lowering P_θ. The result is that a slit tube of these proportions buckles torsionally below its Euler load — the classic warning against treating an open section as if it were closed.

Common pitfall. Reusing the closed-tube J = 2πR³t (overstates torsional stiffness by ~(R/t)²), taking r₀² = R² instead of 5R² (ignoring the shear-centre offset), or omitting the warping term for a warping-free end condition.

Source. Boresi & Schmidt, Advanced Mechanics of Materials Ch. 12 (torsion of thin-walled open sections & torsional buckling); Timoshenko & Gere, Theory of Elastic Stability Ch. 5.

Question 5 of 8: Buckling of a slit thin-walled column

Question 5: Buckling of a slit thin-walled column (equal value)