Question 5 of 8: Open Thin-Walled Circular Tube Column — Flexural and Torsional Buckling
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, May 2016 — 98-Mar-A5 Advanced Strength of Materials, 3 hours, open book, non-communicating calculator permitted (any five of the eight problems constitute a complete paper, all equal value; all eight answered below for full study coverage).
Reference texts: Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed.; Hibbeler, Mechanics of Materials, 10th ed.; Timoshenko & Goodier, Theory of Elasticity, 3rd ed.
It is solved as the strength-of-materials exam it actually is.
Question 5: Open Thin-Walled Circular Tube Column — Flexural and Torsional Buckling (equal value)
Given. Thin circular tube, split by a narrow longitudinal slit (open section), ends pinned and free to warp.
Given data
Quantity
Value
Column height $L$
2.2 m
External diameter
55 mm
Wall thickness $t$
2.0 mm
$E$
50 GPa
$G$
15 GPa
Find. $P_{cr}$ in (a) pure flexural (bending) buckling and (b) pure torsional buckling.
Pin-ended column (left) of an open, slit thin circular tube (right); ends free to warp.
Approach. Compute the section's bending ($I$), St Venant torsion ($J$) and warping ($C_w$) constants for a thin circular tube opened by a slit, then apply the Euler formula for pure flexural buckling and the standard warping-torsion formula for pure torsional buckling of a pin-ended column free to warp.
Section constants (thin-wall, mean radius $R$). Mean radius $R=(55-2)/2=26.5\text{ mm}$. For a thin tube slit along one generator:
$$I=\pi R^3t=1.169\times10^{-7}\text{ m}^4\qquad J=\tfrac23\pi Rt^3=4.440\times10^{-10}\text{ m}^4$$
$$C_w=\frac{2\pi R^5t(\pi^2-6)}{3}=2.118\times10^{-10}\text{ m}^6$$
(the warping constant follows from integrating the principal sectorial coordinate $\omega_n$ of the slit tube about its shear centre, located $2R$ from the tube's own centre — a standard open-section result).
Part (a) — pure flexural (Euler) buckling. Pin-ended, $K=1$:
$$\boxed{P_{cr,\text{bend}}=\frac{\pi^2EI}{L^2}=\frac{\pi^2(50\times10^9)(1.169\times10^{-7})}{2.2^2}=11.92\text{ kN}}$$
Part (b) — pure torsional buckling. For a thin ring, the polar radius of gyration about the section's own centre is exactly $r_0=R$ (since $I_x=I_y=I$ and $A=2\pi Rt$, so $r_0^2=(I_x+I_y)/A=2\pi R^3t/(2\pi Rt)=R^2$). With warping free at both ends:
$$P_{cr,\text{tors}}=\frac{1}{r_0^2}\left(GJ+\frac{\pi^2EC_w}{L^2}\right)=\frac{1}{(0.0265)^2}\left[15\times10^9(4.440\times10^{-10})+\frac{\pi^2(50\times10^9)(2.118\times10^{-10})}{2.2^2}\right]$$
$$\boxed{P_{cr,\text{tors}}=\frac{6.660+21.60}{7.0225\times10^{-4}}=40.24\text{ kN}}$$
Question 5 — final results
Buckling mode
Critical axial load
Pure bending (Euler)
$P_{cr}=11.92\text{ kN}$
Pure torsion (warping)
$P_{cr}=40.24\text{ kN}$
Check: this open, singly-symmetric section has its shear centre offset $2R$ from the centroid, so a centrally-loaded column strictly buckles in a coupled flexural-torsional mode about the symmetry axis, not "pure torsion." Following the question's explicit request for a separate "pure torsion mode" value, $P_{cr,\text{tors}}$ above uses the standard idealized torsional-buckling formula referenced to the section's own centroidal polar radius of gyration ($r_0=R$), the usual simplified treatment taught for this classic problem; because $P_{cr,\text{bend}}$ is smaller than $P_{cr,\text{tors}}$ here, flexural buckling governs the actual (lower) failure load regardless.
