Buckling of Plates and Columns Questions and Answers

Answer: A

The Southwell relation shows the deflection of an imperfect column growing hyperbolically towards the critical load, so the member yields before reaching it. Design curves are therefore fitted to test data rather than taken from the ideal theory.

Answer: B

Plotting deflection against deflection over load gives a straight line whose slope is the critical load, so the value can be found without ever damaging the specimen. It is one of the more elegant pieces of experimental structural mechanics.

Answer: C

The critical load is proportional to the second moment of area, so the column takes the cheapest route and buckles about its weakest axis. This is why an angle or a channel, with very different second moments about its two axes, is inefficient as a strut.

Answer: D

The compression flange behaves rather like a column and moves sideways, dragging the section into twist because it is restrained by the tension flange. Deep, narrow beams are most susceptible, and lateral bracing of the compression flange is the usual remedy.

Answer: A

Correct answer: Reduces its buckling stress as the square of the width

\[ \sigma_{cr} = \frac{k\pi^2 E}{12(1-\nu^2)}\left(\frac{t}{b}\right)^2. \] The stress goes as the inverse square of the width, so widening the panel between stringers by half reduces its buckling stress to about 44 per cent. This is exactly why closing up the stringer pitch is such an effective way of strengthening a compression skin.

Answer: B

The Euler load for pinned ends follows directly: \[ P_{cr} = \frac{\pi^{2} EI}{L^{2}} = \frac{9.8696 \times 7 \times 10^{10}}{(1000)^{2}} = \frac{6.909 \times 10^{11}}{10^{6}} = 6.91 \times 10^{5} \ \mathrm{N} = 691 \ \mathrm{kN} \] Note that the material's strength never enters the expression: buckling is governed by stiffness \( EI \) and length alone, which is why a high-strength alloy buys nothing here.

Answer: C

Fully fixed ends give an effective length \( L_{e} = 0.5L \), and the critical load varies inversely with \( L_{e}^{2} \): \[ P_{cr} = \frac{\pi^{2} EI}{(0.5L)^{2}} = \frac{\pi^{2} EI}{0.25 L^{2}} = 4 \times \frac{\pi^{2} EI}{L^{2}} \] \[ = 4 \times 691 = 2764 \ \mathrm{kN} \] Real end fittings never achieve full fixity, so the whole fourfold gain is not realised in practice; design codes use an effective-length factor above 0.5 for that reason.

Answer: D

A cantilever, fixed at one end and free at the other, has effective length \( L_{e} = 2L \): \[ P_{cr} = \frac{\pi^{2} EI}{(2L)^{2}} = \frac{\pi^{2} EI}{4L^{2}} = \frac{1}{4} \times \frac{\pi^{2} EI}{L^{2}} \] \[ = \frac{691}{4} = 173 \ \mathrm{kN} \] This is the weakest of the standard end conditions, a quarter of the pinned value and one sixteenth of the fixed-fixed value.

Answer: A

Correct answer: 44.7 mm

\[ r = \sqrt{\frac{I}{A}} = \sqrt{\frac{10^6}{500}} = \sqrt{2000} = 44.7\ \mathrm{mm}. \] The radius of gyration measures how far the material sits from the neutral axis, which is why a tube is so much better than a solid bar of the same area - the same material at a greater radius gives a larger \( r \).

Answer: B

Correct answer: 22.4

\[ \frac{L_e}{r} = \frac{1000}{44.7} = 22.4. \] This is a low slenderness, so the column is short and the Euler formula would overpredict its capacity - failure would be by yielding or inelastic buckling rather than by elastic instability.