Question 8 of 8: Torsion of a three-cell thin-walled wing box

Question 8: Torsion of a three-cell thin-walled wing box (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A rectangular three-cell box in pure torsion. Each cell is a closed loop bounded by an upper panel, a lower panel and two vertical webs; the two interior webs are shared between adjacent cells.

Find. (a) the three cell shear flows, and (b) the largest shear stress in the section together with the wall in which it occurs.

Approach. Write the rate of twist of each cell in terms of the three unknown cell flows, impose that all three cells twist at the same rate because the section is rigid in its own plane, and close the system with torque equilibrium; four equations, four unknowns.

1. Cell areas and the torque equation. The enclosed areas are $A_1 = A_3 = 160\times200 = 3.2\times10^{4}$ and $A_2 = 300\times200 = 6.0\times10^{4}$ mm². For a multi-cell section the applied torque is shared according to $$T = 2\,(A_1q_1 + A_2q_2 + A_3q_3)$$with $T = 10\,000$ N·m $= 1.0\times10^{7}$ N·mm. This is one equation in three unknowns; the other two come from compatibility.
2. Rate of twist of a single cell. For cell $i$, $$\frac{d\theta}{dx} = \frac{1}{2A_iG}\oint \frac{q}{t}\,ds$$where the loop integral runs round the cell boundary and each wall carries the flow appropriate to it: an exterior wall carries $q_i$, while an interior web shared with cell $j$ carries the difference $q_i - q_j$.
3. Part (a) — assemble the compatibility equations. Writing out the loop integrals with the given gauges, cell 1 has $160/2 + 160/1.5 + 200/1 + 200/1 = 586.7$ mm of $ds/t$ on its own account and shares $200/1$ with cell 2. Doing the same for cells 2 and 3 and setting all three rates equal to a common $d\theta/dx$ gives three equations which, with the torque equation, form a four-by-four linear system in $q_1$, $q_2$, $q_3$ and $d\theta/dx$.
4. Solve the system. The box is symmetric about its vertical centre-line, so $q_1 = q_3$ before any arithmetic is done, which halves the work. Solving, $$\boxed{q_1 = q_3 = 34.72\ \text{N/mm},\qquad q_2 = 46.30\ \text{N/mm}}$$with a common rate of twist $d\theta/dx = 1.157\times10^{-5}$ rad/mm, equivalent to $0.663$ degrees per metre and an effective torsion constant $J = T/(G\,d\theta/dx) = 5.76\times10^{7}$ mm⁴.
5. Check the solution before using it. Substituting back, $2(A_1q_1+A_2q_2+A_3q_3) = 2(1.111\times10^{6}+2.778\times10^{6}+1.111\times10^{6}) = 1.000\times10^{7}$ N·mm, the applied torque exactly; and recomputing each cell’s twist rate independently returns the same $1.157\times10^{-5}$ rad/mm for all three. Both checks are worth doing because a sign error on a shared web is the classic failure mode of this calculation.
6. Part (b) — wall-by-wall shear stresses. Dividing each wall’s flow by its own thickness: the four outer webs and covers of cells 1 and 3 give $34.72/1.0 = 34.72$, $34.72/1.5 = 23.15$ and $34.72/2.0 = 17.36$ MPa; the cell-2 covers give $46.30/1.5 = 30.86$ and $46.30/2.0 = 23.15$ MPa; and the two interior webs carry only the difference $46.30-34.72 = 11.57$ N/mm, hence 11.57 MPa.
7. The governing wall. The largest value is therefore $$\boxed{\tau_{\max} = 34.72\ \text{MPa, in the two outer vertical webs (t = 1.0 mm)}}$$The result is worth a sentence of interpretation: the most heavily loaded flow is in the wide middle cell, but the most heavily stressed wall is an outer web, because it combines the full cell flow with the thinnest gauge. The interior webs, made of the same 1 mm sheet, are only a third as highly stressed because they see the flow difference rather than the flow itself.