Bending of Thin Walled Beams Questions and Answers

Answer: C

Correct answer: 22.5 degrees

\[ \tan 2\theta = \frac{2I_{xy}}{I_{xx}-I_{yy}} = \frac{4\times10^6}{4\times10^6} = 1, \] so \( 2\theta = 45^\circ \) and \( \theta = 22.5^\circ \). The factor of two is the trap: forgetting to halve the angle gives 45 degrees, which is offered here.

Answer: D

Using \( I_{1,2} = (I_{xx}+I_{yy})/2 \pm \sqrt{((I_{xx}-I_{yy})/2)^2 + I_{xy}^2} \), the mean is \( 6 \times 10^6 \) and the radius is \( \sqrt{2^2 + 2^2} = 2.83 \times 10^6 \), giving \( 8.83 \times 10^6 \) mm\( ^4 \). The smaller principal value is \( 3.17 \times 10^6 \).

Answer: A

Setting the direct stress to zero gives \( \tan\alpha = I_{xy}/I_{yy} = 2/4 = 0.5 \), so \( \alpha = 26.6^\circ \). The neutral axis is therefore not horizontal even though the moment is applied about the horizontal axis, which is the signature of unsymmetric bending.

Answer: B

Either use \( bh^3/3 = 50 \times 3.375 \times 10^6/3 = 5.63 \times 10^7 \), or add the transfer term to the centroidal value: \( 1.406 \times 10^7 + 7500 \times 75^2 = 1.406 \times 10^7 + 4.22 \times 10^7 \). Both routes agree, which is a useful check on the parallel axis theorem.

Answer: C

Correct answer: 3.33 mm

\[ \delta = \frac{PL^3}{3EI} = \frac{5000 \times 2^3}{3 \times 4\times10^6} = \frac{40{,}000}{1.2\times10^7} = 3.33\times10^{-3}\ \mathrm{m}. \] The cube on the length is what makes cantilever deflection so sensitive to span - doubling the length multiplies the tip deflection by eight.

Answer: D

Tip deflection goes as the cube of the length, so doubling it multiplies the deflection by \( 2^3 = 8 \). Answering four confuses this with a uniformly distributed load's dependence, which is a fourth power rather than a cube.

Answer: A

Each cap lies \( 150 \) mm from the neutral axis, so \( I = 2 \times 1200 \times 150^2 = 2 \times 1200 \times 22\,500 = 5.4 \times 10^7 \) mm\( ^4 \). Adding the webs would raise this only slightly, since they sit close to the neutral axis where they contribute little.

Answer: B

Correct answer: 1.78

With the cap areas unchanged the second moment goes as the square of the separation: \[ \left(\frac{400}{300}\right)^2 = 1.78. \] A third more depth buys nearly four fifths more stiffness for no extra material, which is why depth is the first thing a designer reaches for.

Answer: C

The ratio is \( 200/70 = 2.86 \), so the steel's width is multiplied by that factor to give an equivalent area of aluminium with the same stiffness. Stresses computed in the transformed material are scaled back by the same ratio at the end.

Answer: D

Correct answer: 200 mm^2

A panel carrying uniform stress contributes equally to the booms at its two ends, so each takes half of the 400 square millimetres. The general formula weights the two ends according to the stress ratio between them; uniform stress is the special case in which the split is even.