Buckling of Plates and Columns Questions and Answers

Answer: C

Correct answer: 69 MPa

\[ \sigma_{cr} = \frac{\pi^2 E}{(L_e/r)^2} = \frac{9.87 \times 70{,}000}{100^2} = 69\ \mathrm{MPa}. \] Well below the yield stress of any structural aluminium alloy, which confirms that at a slenderness of 100 the column is genuinely in the elastic buckling regime and Euler’s formula applies.

Answer: D

The critical stress varies inversely with the square of slenderness, so doubling it divides the stress by four. It is why a strut's length is such an expensive dimension and why bracing that halves the unsupported length is so effective.

Answer: A

Correct answer: 103 MPa

\[ \sigma_{cr} = \frac{k\pi^2 E}{12(1-\nu^2)}\left(\frac{t}{b}\right)^2 = \frac{4 \times 9.87 \times 70{,}000}{12(1 - 0.33^2)} \times \left(\frac{2}{100}\right)^2 \] \[ = \frac{2{,}763{,}000}{10.69} \times 0.0004 = 103\ \mathrm{MPa}. \] The factor \( (t/b)^2 \) dominates: halving the panel width would quadruple this figure.

Answer: B

The stress varies with the square of thickness over width, so halving the width multiplies it by four, giving about 414 MPa. In practice the panel would yield or cripple first, which is why very close stringer pitches bring diminishing returns.

Answer: C

Plate buckling stress varies as the square of the thickness-to-width ratio: \[ \sigma_{cr} = k \frac{\pi^{2} E}{12(1-\nu^{2})} \left(\frac{t}{b}\right)^{2} \] With \( b \) held constant, doubling \( t \) gives \[ \frac{\sigma_{2}}{\sigma_{1}} = \left(\frac{2t}{t}\right)^{2} = 2^{2} = 4 \] so the buckling stress quadruples. Compare the column case, where doubling a cross-sectional dimension raises the second moment of area far more sharply still.

Answer: D

Local buckling arrives well before overall, so the stiffeners are stronger than they need to be relative to the skin. Redistributing material towards the skin, or reducing the stringer pitch, would give a lighter panel for the same capacity.

Answer: A

The Euler formula has returned a stress the material cannot sustain elastically, which signals that the column is too short for the elastic theory. A Johnson or tangent modulus treatment applies instead, and the true capacity lies below yield.

Answer: B

Treated as a short strut between fasteners, the inter-rivet buckling stress varies inversely with the square of the pitch \( p \): \[ \sigma_{cr} \propto \frac{1}{p^{2}} \] Halving the pitch therefore gives \[ \frac{\sigma_{2}}{\sigma_{1}} = \left(\frac{p}{p/2}\right)^{2} = 2^{2} = 4 \] a fourfold rise. Rivet pitch is thus the direct lever on inter-rivet buckling, which is why closer pitch is the standard fix when a skin buckles between fasteners below limit load.

Answer: C

A brace at mid-span halves the unsupported length, and the critical load varies inversely with the square of that length: \[ \frac{P_{2}}{P_{1}} = \left(\frac{L}{L/2}\right)^{2} = 2^{2} = 4 \] so the capacity is multiplied by four. A single well-placed brace is often the cheapest structural improvement available, since it adds almost no weight while quadrupling the buckling capacity.

Answer: D

The stress is directly proportional to the coefficient, so the factor is \( 7/4 = 1.75 \). Real riveted edges provide partial restraint and fall between the simply supported and clamped ideals, so an intermediate value is normally assumed.