Bending of Thin Walled Beams Questions and Answers

Answer: A

The familiar result follows from integrating the square of the distance across the depth. About the base rather than the centroid the divisor becomes three, which is exactly what the parallel axis theorem gives when the transfer term is added.

Answer: B

As the axes rotate the second moments trace out a Mohr's circle, and the principal values sit at its extremes where the product of inertia vanishes. Every other orientation gives values between them, so bending is stiffest about one principal axis and softest about the other.

Answer: C

The Z has point symmetry but no axis of symmetry, so its product of inertia about the obvious reference axes does not vanish. It therefore deflects sideways under a purely vertical load, which makes it the standard worked example of unsymmetric bending.

Answer: D

The angle is unsymmetric, so its neutral axis inclines under a vertical moment, and its shear centre lies at the intersection of the two legs rather than at the centroid, so a load applied elsewhere twists it as well. Both effects have to be handled separately.

Answer: A

Most of the direct-stress-carrying area is concentrated in the stringers and caps, and the skin contributes a share that can be lumped into the adjacent booms. The idealisation is an accounting convenience that reflects where the material actually is.

Answer: B

Using \( bh^3/12 \), the numerator is \( 50 \times 150^3 = 50 \times 3.375 \times 10^6 = 1.6875 \times 10^8 \), and dividing by 12 gives \( 1.406 \times 10^7 \) mm\( ^4 \). Taking the axis at the base instead would divide by 3 and give four times as much.

Answer: C

The extreme fibre is \( 75 \) mm from the neutral axis, so \( \sigma = My/I = (10 \times 10^6 \times 75)/(1.406 \times 10^7) = 53.3 \) MPa, working in newtons and millimetres throughout. Mixing metres and millimetres is the usual source of error by a factor of a thousand.

Answer: D

Correct answer: 6.28e6 mm^4

For a thin tube, \[ I = \pi r^3 t = \pi \times 100^3 \times 2 = 6.28\times10^6\ \mathrm{mm^4}. \] The cube on the radius is what makes a tube so efficient: doubling the radius at constant wall thickness multiplies the stiffness eightfold for twice the material.

Answer: A

Correct answer: 5.0e6 mm^4

The transfer term is \[ Ay^2 = 500 \times 100^2 = 5.0\times10^6\ \mathrm{mm^4}. \] The square on the distance is why the booms furthest from the neutral axis dominate the section, and why the boom’s own centroidal second moment is negligible beside it.

Answer: B

Correct answer: 5.4e7 mm^4

All four booms are at the same distance, so all four contribute equally: \[ 4 \times 600 \times 150^2 = 2400 \times 22{,}500 = 5.4\times10^7\ \mathrm{mm^4}. \] Whether a boom is above or below the axis makes no difference, since the distance is squared.