Torsion of Thin Walled Sections Questions and Answers

Answer: A

Correct answer: 0.57 degrees

\[ \theta = \frac{TL}{GJ} = \frac{5000 \times 3}{1.5 \times 10^6} = 0.01\ \text{radians}, \] which converted gives \[ 0.01 \times \frac{180}{\pi} = 0.57\ \text{degrees}. \] The option offering 0.01 degrees is the trap: it is the answer in radians presented as though it were degrees.

Answer: B

Both are true and the second explains the first. Material near the axis is lightly stressed and contributes little, so removing it and placing the same mass at a larger radius raises the polar second moment substantially. Drive shafts are tubular for exactly this reason.

Answer: C

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the cause. In a circular shaft every point of the section is equivalent under rotation about the axis, so there is no reason for one part to displace axially more than another - the axial displacement is uniform, which is to say the section stays plane. Every non-circular section lacks that symmetry and warps.

Answer: D

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason explains why. The polar second moment of area is the correct torsion constant only when the section stays plane. A non-circular section warps, so part of the deformation goes into axial displacement rather than into resisting twist, and the effective constant is smaller - sometimes very much smaller, as for a thin open section.

Answer: A

Both statements are true and the second is the reason for the first. The slope of the film gives the shear stress and twice the volume beneath it the torque, so awkward sections can be understood qualitatively without solving anything.

Answer: B

Both are true and the second explains the first. It is a counter-intuitive result that the analogy makes immediately obvious. A re-entrant corner is the opposite case, with the film folding sharply and the stress concentrating.

Answer: C

Both statements are true and the second is the reason for the first. The stresses are largest at the restraint and decay along the member, and they are entirely absent from the St Venant solution, which assumes free warping throughout.

Answer: D

The assertion is true but the reason is false and irrelevant. Area has nothing to do with it: the effect is large because an open section wants to warp a great deal, so restraining it forces large axial strains. A closed section warps very little in the first place.

Answer: A

Both are true and the second explains the first. Without the circulating flow only the St Venant mechanism remains, and its cube on a very small thickness makes it feeble. Bending stiffness, by contrast, is barely affected.

Answer: B

Both statements are true and the second is the reason for the first. The pairing GJ plays the same part in torsion that EI plays in bending, and confusing the two moduli is a common source of answers wrong by a factor of about 2.6 for metals.