Two cells of a box carry shear flows of \( 140 \) and \( 90 \) kN/m in the same circulating sense. The net flow in the web they share is
Answer: C
The two circulating flows oppose one another in the common wall, so the net value is \( 140 - 90 = 50 \) kN/m. Interior webs are often the most lightly loaded part of a multi-cell box under pure torsion for exactly this reason.
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An open section is built from three walls each \( 100 \) mm long and \( 2 \) mm thick. Its torsion constant is nearest to
Answer: D
Applying \( J = \tfrac{1}{3}\sum s t^3 \), each wall gives \( 100 \times 8/3 = 267 \) mm\( ^4 \), and three of them total about 800 mm\( ^4 \). Compare that with a closed cell of the same walls, whose effective value would be larger by a factor of many thousands.
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Halving the wall thickness of an open section changes its torsion constant by a factor of
Answer: A
The St Venant torsion constant carries the cube on the thickness, so halving it divides the stiffness by \( 2^3 = 8 \). Contrast the closed-section case, where the stiffness depends on thickness only to the first power.
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A closed box \( 0.5 \) m by \( 0.3 \) m in its wall centrelines carries \( 90 \) kN m of torque. Its shear flow is
Answer: B
The enclosed area is \( 0.5 \times 0.3 = 0.15 \) m\( ^2 \), so \( q = 90\,000/(2 \times 0.15) = 300 \) kN/m. Using the outer dimensions rather than the centrelines is a small but systematic error that always overstates the area.
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A transverse load of \( 20 \) kN acts \( 60 \) mm from the shear centre of a section. The equivalent torque is
Answer: C
Correct answer: 1.2 kN m
\[ T = P e = 20 \times 0.06 = 1.2\ \mathrm{kN\,m}. \] Any transverse load not passing through the shear centre is statically equivalent to the same load at the shear centre plus this torque, and the section is then analysed for bending and torsion separately and the results superposed.
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A doubly symmetric box has walls of \( 3 \) mm and \( 1.5 \) mm thickness carrying the same shear flow. The ratio of shear stresses in them is
Answer: D
Correct answer: 1 to 2
Since the shear flow is the same in both walls, \[ \frac{\tau_3}{\tau_{1.5}} = \frac{q/3}{q/1.5} = \frac{1.5}{3} = \frac{1}{2}. \] The thin wall carries twice the stress of the thick one. This is the whole reason the thinnest wall of a torsion box governs its strength.
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A web \( 250 \) mm deep and \( 2 \) mm thick carries \( 50 \) kN of shear. Taking the flow as uniform, the shear stress is
Answer: A
Correct answer: 100 MPa
\[ \tau = \frac{V}{dt} = \frac{50{,}000}{250 \times 2} = 100\ \mathrm{MPa}. \] Taking the web to carry the shear uniformly over its full depth is the standard idealisation, justified when the flanges or booms carry the bending and the web carries the shear.
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A single cell of enclosed area \( 0.45 \) m\( ^2 \) is replaced by one of \( 0.90 \) m\( ^2 \) at the same torque. The shear flow becomes
Answer: B
Shear flow varies inversely with the enclosed area, so doubling the area halves the flow. Enlarging the cell is therefore a far more effective way of reducing shear stress than thickening the walls, though it is usually constrained by the aerofoil shape.
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A boomed section has booms of \( 500 \) mm\( ^2 \) at \( \pm 150 \) mm and \( 500 \) mm\( ^2 \) on the neutral axis. The neutral-axis boom's contribution to the shear flow step is
Answer: C
The step is proportional to the boom's first moment about the neutral axis, and a boom lying on that axis has a zero first moment. It contributes nothing to the shear flow, though it still adds area and so affects the centroid position.
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A closed section requires \( 240 \) kN m of torque capacity at a shear flow limit of \( 200 \) kN/m. The minimum enclosed area is
Answer: D
Correct answer: 0.6 m^2
Rearranging the Bredt-Batho relation, \[ A = \frac{T}{2q} = \frac{240}{2 \times 200} = 0.6\ \mathrm{m}^2. \] Enclosed area is bought cheaply - it costs only the perimeter of skin needed to enclose it - which is why torsion boxes are made as deep and wide as the aerodynamic shape allows.
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