Shear Flow in Open and Closed Sections Questions and Answers

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.

Answer: A

A spoiler dumps lift locally by separating the flow and produces very little pitching moment on the section, so it does not suffer reversal. Many transports lock the outboard ailerons at high speed and roll on spoilers and inboard ailerons for exactly this reason.

Answer: B

Above the divergence speed a small twist raises the incidence, which raises the lift, which twists the wing further, with no equilibrium available. It is a static aeroelastic instability and it is avoided by keeping the torsion box stiff and the aerodynamic centre near the flexural axis.

Answer: C

Loads applied along the flexural axis bend the wing without twisting it, since each section is loaded through its own shear centre. The relative positions of the flexural axis and the aerodynamic centre determine whether an increase in lift twists the wing nose-up or nose-down.

Answer: D

The flow is \( 300\,000/(2 \times 1.5) = 100 \) kN/m, and dividing by the thickness gives \( 100\,000/0.002 = 50 \) MPa. Checking the thinnest wall is the essential step, since it is always the critical one under torsion.

Answer: A

The flow is unchanged at 100 kN/m, so the stress becomes \( 100\,000/0.003 = 33 \) MPa. Thickening the wall reduces the stress in proportion but does nothing to the flow, which is fixed by the torque and the enclosed area.

Answer: B

Both are true and the second explains the first. The outboard aileron has the longest moment arm but acts on the most flexible part of the wing, so it reverses first. Rolling on spoilers and inboard ailerons at speed avoids the problem entirely.