Torsion of Thin Walled Sections Questions and Answers

Answer: C

An open section such as an I has a torsion constant orders of magnitude below its polar second moment of area, so the calculation would predict far less twist than actually occurs. The error is unconservative, which makes it a dangerous one.

Answer: D

Correct answer: It produces direct stresses only when it is restrained

Warping itself is a displacement, and a section free to warp does so without any direct stress arising. Direct stresses appear only where the warping is prevented - at a built-in end, a heavy rib or a joint - because the axial displacements the section wants to make are resisted. This is why restraint locations govern where such stresses are checked.

Answer: A

Correct answer: It is proportional to the torsional rigidity

This is the false statement: \( \theta = TL/GJ \) puts the rigidity in the denominator, so twist is inversely proportional to it. The other three are all correct - twist rises with torque and with length, and falls as the shear modulus rises, the last being the same fact as the first stated for the material alone.

Answer: B

Correct answer: Increasing the shear modulus of the material

Shear stress under a given torque depends on the geometry alone, through \( \tau = Tr/J \); the modulus governs how much the shaft twists, not how highly it is stressed. The other three all reduce the stress: a smaller torque directly, a larger diameter through the fourth power in \( J \), and a thicker-walled hollow section by adding material where the stress is highest.

Answer: C

An open section resists torque only by the feeble St Venant mechanism, so it twists enormously and reaches high stresses at modest torque. Closing the section, even with a light panel, transforms its behaviour and is almost always the right answer.

Answer: D

Correct answer: A tube warps less than a solid shaft of the same weight

Neither warps: both are circular, and a circular section does not warp at all, so this cannot be a reason for choosing one over the other. The genuine reasons are the other three, which are three statements of one fact - that stress and stiffness both grow with radius, so material is best placed at the outside.

Answer: A

Correct answer: Both cells must twist at the same rate

The two cells share a web and belong to one structure, so compatibility requires them to have the same rate of twist. That condition, together with equilibrium of the total torque, is what determines how the shear flow divides between them. The flows are generally unequal, and the shared web carries their difference, not their sum.

Answer: B

Correct answer: 64

An open section’s torsion constant carries a cube on wall thickness, so \[ 4^3 = 64. \] Thickening is therefore remarkably effective for an open section - and remarkably expensive in weight, since the mass rises only fourfold but so does the material bill. Closing the section would usually achieve far more for far less.

Answer: C

Direct stress from bending and shear stress from torsion act together, and a criterion such as maximum shear stress or von Mises resolves them into an equivalent stress. Simply adding them is dimensionally reasonable but physically wrong.

Answer: D

Tresca's criterion works on the difference between the extreme principal stresses, which for combined bending and torsion produces the familiar equivalent stress expression. Von Mises gives a slightly less conservative result and is more commonly used for ductile metals.