Shear Flow in Open and Closed Sections Questions and Answers

Answer: C

The redundant flow restored after the cut compensates exactly for it, so the final distribution is independent of where the cut was made. A discrepancy therefore points to an error, most often in the compatibility integral.

Answer: D

Correct answer: Their shear flow can be found from equilibrium alone

An open section has a free edge where the shear flow must be zero, and integrating from there determines the flow everywhere by equilibrium. A closed section has no free edge, is one degree redundant, and requires a compatibility condition - which is the extra step the cut-and-restore method supplies.

Answer: A

Correct answer: It occurs only in closed sections

Warping occurs in both, and it is generally more pronounced in open sections; only a few special shapes, such as a circular tube, do not warp at all. The other three statements are correct: warping is an axial displacement, it leaves the section non-planar, and it produces direct stresses only where it is restrained.

Answer: B

Correct answer: Reducing the enclosed area

Since \( \tau = T/(2At) \), the stress falls when the torque falls, when the enclosed area grows, or when the wall is thickened - and it rises when the enclosed area is reduced. Enclosed area is the cheapest of these levers in weight terms, which is why designers reach for it first.

Answer: C

Cutting is only half the method; the redundant constant must be restored using a compatibility condition. Omitting it leaves a distribution that satisfies equilibrium but not compatibility, and it misplaces the flows around the section, unconservatively in some walls.

Answer: D

Correct answer: Shear flow is easier to measure experimentally

The other three are genuine reasons: constancy along a tapering wall, simpler equilibrium equations, and easy recovery of the stress at the end. Measurement is not among them - what a gauge senses is strain and hence stress at a point, and the shear flow has to be inferred from it by multiplying by the thickness.

Answer: A

Correct answer: Doubles; the flow is unchanged

The shear flow depends on the torque and the enclosed area, neither of which the designer has changed, so \( q \) stays as it was and \( \tau = q/t \) doubles. Thinning one face of a torsion box therefore concentrates the penalty entirely on that face, and this is the standard way a weight saving turns into a strength problem.

Answer: B

The torsional stiffness gained is two or three orders of magnitude for a few per cent of added weight, and wing torsional stiffness governs both aeroelastic divergence and control reversal. Closing the section is among the cheapest structural decisions available.

Answer: C

A wing that twists readily under aerodynamic load can reach a speed at which the twist reinforces itself and diverges, and a lower one at which aileron deflection twists the wing enough to cancel its own effect. Both speeds rise with torsional stiffness, which is why the torsion box is defended so carefully.

Answer: D

The downward deflection produces a nose-down pitching moment on the section, and if the wing is torsionally soft the resulting twist reduces the incidence more than the aileron adds. Above the reversal speed the roll response is in the wrong direction entirely.