Question 4 of 8: Three-cell wing box in pure torsion

Question 4: Three-cell wing box in pure torsion (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 three-cell thin-walled wing box in pure torsion.

Find. (a) the three cell shear flows $q_1$, $q_2$, $q_3$; (b) the magnitude and location of the maximum shear stress.

Approach. Three unknown cell flows plus the unknown rate of twist give four unknowns; supply one torque equation and two equal-twist (compatibility) equations, exploit the left–right symmetry, then tabulate every distinct wall to find the peak stress.

1. Enclosed areas of the three cells. Each cell is a rectangle of the full box depth: $$A_1 = 150 \times 220 = 33,000,\quad A_2 = 250 \times 220 = 55,000,\quad A_3 = A_1 = 33,000 \text{ mm}^2.$$
2. Torque equilibrium (Bredt–Batho, summed over the cells). Each circulating flow contributes twice its enclosed area: $$T = 2(A_1 q_1 + A_2 q_2 + A_3 q_3) = 2(33,000\,q_1 + 55,000\,q_2 + 33,000\,q_3) = 58,000,000 \text{ N}\cdot\text{mm}.$$
3. Compatibility: every cell twists at the same rate. For cell $i$, $$\frac{d\theta}{dz} = \frac{1}{2A_i G}\oint \frac{q\,ds}{t},$$ where the interior webs carry the difference of the adjacent cell flows. Evaluating the line integrals from the wall lengths and gauges: for cell 1, $\oint ds/t = 150/3.5 + 150/2.5 + 2(220/2.0) = 322.857$ mm/mm, and for cell 2, $150$ becomes $250$ so $\oint ds/t = 391.429$ mm/mm.
4. Write the two independent compatibility equations. Taking the interior-web term across explicitly, $$\frac{1}{2A_1}\left[322.857\,q_1 - 110.0\,q_2\right] = G\frac{d\theta}{dz},$$ $$\frac{1}{2A_2}\left[391.429\,q_2 - 110.0\,q_1 - 110.0\,q_3\right] = G\frac{d\theta}{dz}.$$ The third cell repeats the first. Because the box is symmetric about its centre web line and the gauges are the same on both sides, $q_1 = q_3$ before any arithmetic is done, which reduces the system to two equations in $q_1$ and $q_2$.
5. Part (a) — solve the system. Solving the torque equation together with the two compatibility equations gives $$\boxed{\;q_1 = q_3 = 209.32 \text{ N/mm},\qquad q_2 = 276.09 \text{ N/mm}\;}$$ Check by back-substitution into the torque equation: $2(33,000 \times 209.32 + 55,000 \times 276.09 + 33,000 \times 209.32) = 5.8e+07$ N·mm, the applied torque. The interior webs each carry only $q_2 - q_1 = 66.77$ N/mm.
6. Part (b) — convert flow to stress, wall by wall. Shear stress is $\tau = q/t$, and because the gauges differ the largest flow is not necessarily in the most highly stressed wall. Tabulating all six distinct wall types:

7. Identify the maximum and say where it is. The winner is the lower panel, cell 2: $$\boxed{\;\tau_{\max} = \frac{276.09}{2.5} = 110.44 \text{ MPa, in the lower skin of the centre cell}\;}$$ It beats the outer webs (104.66 MPa) because the centre cell carries the largest flow and the lower skin is the second-thinnest gauge; the interior webs, carrying only the difference of two similar flows, are by far the quietest walls at 33.38 MPa.
8. Rate of twist (a free by-product). From either compatibility equation, $G\,d\theta/dz = 0.5638$ N/mm² per mm, so with $G = 40$ GPa $$\frac{d\theta}{dz} = 1.41e-05 \text{ rad/mm} = 0.808 \text{ degrees per metre of span,}$$ a useful sanity check on the whole solve — a wing box twisting more than a degree or two per metre under limit torque would be aeroelastically unacceptable.