Question 1 of 8: Buckling of the tubular compression strut BC

Question 1: Buckling of the tubular compression strut BC

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 bracket frame with beam AB pinned to the wall at A and diagonal two-force strut BC pinned at C (wall) and B; the tube is D_o = 40 mm, t = 4 mm.

Find. (a) the factor of safety of BC against elastic (Euler) buckling; (b) the wall thickness giving a buckling factor of safety of 3.

Approach. Resolve the axial force in the two-force strut BC by taking moments about A on beam AB, then compare it with the Euler critical load of a pin-ended tube.

1. Axial compression in BC. The strut is a two-force member, so its reaction on beam AB acts along BC. Taking moments about A (span AB = 2.0 m, load at mid-span 1.0 m), the vertical component of the strut reaction is $V_B\cdot 2.0 = 30000\cdot 1.0$, so $V_B = 15\,000\ \text{N}$. With the strut inclined at $\tan\theta = 1.0/2.0$ ($\sin\theta = 1/\sqrt5$), the axial force is $$P_{BC}=\frac{V_B}{\sin\theta}=15000\sqrt5 = \boxed{33.54\ \text{kN (compression)}}$$
2. Section properties of the tube. With $D_o=40$ mm and $D_i=40-2(4)=32$ mm, $$I=\frac{\pi}{64}\left(D_o^4-D_i^4\right)=\frac{\pi}{64}\left(40^4-32^4\right)=7.42\times10^{4}\ \text{mm}^4,\quad A=\frac{\pi}{4}\left(40^2-32^2\right)=452\ \text{mm}^2$$
3. Euler critical load. The strut length is $L_{BC}=\sqrt{2.0^2+1.0^2}=2.236$ m and both ends are pinned ($K=1$): $$P_{cr}=\frac{\pi^2 E I}{L_{BC}^2}=\frac{\pi^2 (205\,000)(7.42\times10^4)}{(2236)^2}=\boxed{30.0\ \text{kN}}$$ The buckling stress $\sigma_{cr}=P_{cr}/A=66.4\ \text{MPa}\lt 380\ \text{MPa}$, so the column is long and Euler (elastic) buckling governs.
4. Factor of safety (part a). $$\text{FoS}=\frac{P_{cr}}{P_{BC}}=\frac{30.0}{33.54}=\boxed{0.90}$$ Because FoS < 1, the strut as detailed cannot carry the load — it buckles elastically.
5. Thickness for FoS = 3 (part b). Requiring $P_{cr}=3P_{BC}=100.6$ kN gives a required second moment $$I_{req}=\frac{P_{cr,req}L_{BC}^2}{\pi^2 E}=2.49\times10^{5}\ \text{mm}^4.$$ Keeping $D_o=40$ mm, the largest achievable value is that of a solid bar, $I_{solid}=\tfrac{\pi}{64}(40)^4=1.26\times10^{5}\ \text{mm}^4$, which is only half of $I_{req}$. Hence no wall thickness at $D_o=40$ mm reaches FoS = 3 (a solid 40 mm bar gives FoS ≈ 1.52). The section must be enlarged: a solid diameter of $D=\left(64I_{req}/\pi\right)^{1/4}=\boxed{47.4\ \text{mm}}$ (or a larger-diameter tube) is required.