Shear Flow in Open and Closed Sections Questions and Answers

Answer: A

Since the flow is constant round the cell and stress is flow over thickness, the thinnest wall carries the highest stress. That wall is therefore the critical one, which is a useful check when a designer is tempted to thin one face of a box to save weight.

Answer: B

The added panel allows a circulating shear flow to form, replacing the feeble St Venant mechanism with the Bredt-Batho one. Gains of two or three orders of magnitude for a few per cent of weight are typical, which is why aerofoil sections are closed wherever possible.

Answer: C

Points on the section move axially by differing amounts, so the section no longer remains plane. If that movement is restrained, for instance at a built-in end, additional direct stresses appear which simple torsion theory does not predict.

Answer: D

Warping only generates direct stress when it is prevented. Where the member is free to warp, the axial displacements occur without restraint and pure torsion theory applies unmodified. Restraint at a root fitting is the usual place the extra stresses arise.

Answer: A

The flow accumulates the first moment of area as one integrates down from the flange, which produces a parabolic distribution peaking at the neutral axis. In practice the web is often taken to carry a uniform average flow, an approximation good to a few per cent for a deep section.

Answer: B

Correct answer: 100 kN/m

By the Bredt-Batho relation, \[ q = \frac{T}{2A} = \frac{120}{2 \times 0.6} = 100\ \mathrm{kN/m}. \] Note that \( A \) is the area enclosed by the median line of the wall, not the cross-sectional area of the material - confusing the two is the standard error in this calculation.

Answer: C

Shear stress is flow over thickness, \( 100\,000/0.0025 = 40 \times 10^6 \) Pa. Because the flow is constant round a cell under torsion, a wall half this thickness would carry twice the stress and would be the critical one.

Answer: D

Using \( e = b^2 h^2 t/(4I_{xx}) \), the numerator is \( 2500 \times 10\,000 \times 2 = 5 \times 10^7 \) and the denominator is \( 4 \times 6.67 \times 10^5 = 2.67 \times 10^6 \), giving \( 18.8 \) mm. The point lies outside the section, on the far side of the web from the flanges.

Answer: A

Rearranging Bredt-Batho, \( T = 2Aq = 2 \times 0.8 \times 150 = 240 \) kN m. Note how directly the enclosed area buys torque capacity, which is why a deep, wide box is so much better in torsion than a slender one of the same material.

Answer: B

Correct answer: 80 N/mm

The step at a boom is \[ \Delta q = \frac{V B y}{I} = \frac{40{,}000 \times 800 \times 200}{8 \times 10^{7}} = 80\ \mathrm{N/mm}. \] The step is proportional to the boom’s first moment of area about the neutral axis, so a boom lying on that axis produces no step at all.