Torsion of Thin Walled Sections Questions and Answers

Answer: A

Plane sections remain plane and radii remain straight in a circular shaft, so the shear strain and hence the stress grow in direct proportion to radius. Material near the axis is therefore barely worked, which is exactly why a hollow shaft is so much more efficient.

Answer: B

The polar value is twice the diametral second moment of area, so where the diametral figure carries sixty-four in the denominator the polar one carries thirty-two. The value over sixteen is the polar section modulus rather than the second moment.

Answer: C

Axial symmetry means every point on the section moves identically along the axis, so the section stays plane. Every non-circular section warps under torsion, which is why the simple circular theory cannot be carried across to other shapes.

Answer: D

Correct answer: Torque times length, over rigidity

The relation is \[ \theta = \frac{TL}{GJ}, \] which is the torsional counterpart of \( \delta = PL/AE \) in axial loading. Twist grows with the torque applied and with the length over which it acts, and falls as the torsional rigidity rises - so rigidity appears in the denominator, not multiplied.

Answer: A

Correct answer: The shear modulus and the torsion constant

Torsional rigidity is \( GJ \), where \( G \) is the shear modulus and \( J \) the torsion constant of the section. It plays exactly the role that \( EI \) plays in bending: the material property multiplied by a purely geometric one. Young’s modulus does not appear, because torsion loads the material in shear rather than in direct stress.

Answer: B

Only for a circular section do the two coincide. For any other shape the torsion constant is smaller, sometimes very much smaller, because warping relieves part of the section's ability to resist twist. Substituting the polar value for a non-circular section badly overstates the stiffness.

Answer: C

The governing equations for the stress function and for a lightly stretched membrane under pressure are identical. The slope of the film gives the shear stress and twice the volume beneath it gives the torque, which makes the analogy a powerful way of reasoning about awkward sections without solving anything.

Answer: D

Steep slopes mean high stress, which is why the analogy shows immediately that a re-entrant corner is a stress raiser and that the film is steepest at the edge of a narrow strip. Twice the volume under the film gives the torque the section carries.

Answer: A

A long thin hole constrains the film to a low ridge, so the volume beneath, and hence the torque carried at a given twist, is small. Joining the ends of the strip to enclose an area transforms the picture completely, which is the closed-section case.

Answer: B

This expression follows from combining the constant shear flow with the rate of twist integral. For a thin circular tube it reduces to twice pi r cubed t, and it shows the enclosed area entering squared, which is why closing a section pays so handsomely.