Question 7 of 8: Torsion of a three-cell thin-walled box

Question 7: Torsion of a three-cell thin-walled 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 box 200 mm deep, divided by two interior webs into three cells 160, 300 and 160 mm wide, carrying pure torque.

Find. (a) the constant shear flow circulating in each of the three cells; (b) the largest shear stress anywhere in the box and the wall in which it occurs.

Approach. Write the rate of twist of each cell in terms of its own circulating flow and those of its neighbours, set the three rates equal (the section is rigid in its own plane, so every cell twists together), add the torque equilibrium equation and solve the resulting four-by-four system.

1. Part (a) — set up the cell areas and line integrals. The enclosed areas are $A_1 = A_3 = 160(200) = 32\,000$ mm$^{2}$ and $A_2 = 300(200) = 60\,000$ mm$^{2}$. The line integral $\oint \mathrm{d}s/t$ around each cell's own boundary is $$\delta_1 = \delta_3 = \frac{160}{2.0}+\frac{160}{1.5}+\frac{2(200)}{1.0} = 586.7, \qquad \delta_2 = \frac{300}{2.0}+\frac{300}{1.5}+\frac{2(200)}{1.0} = 750,$$ while each interior web contributes $200/1.0 = 200$ to the coupling between adjacent cells.
2. Write the rate of twist of each cell. For cell $i$, $$\frac{\mathrm{d}\theta}{\mathrm{d}x} = \frac{1}{2A_iG} \left[\delta_i q_i - \sum_{j\ \text{adjacent}} \frac{h}{t_w} q_j\right],$$ in which the negative terms account for the neighbouring cell's flow running the opposite way through the shared web.
3. Impose compatibility and equilibrium. All three cells must twist at the same rate, giving two independent equations, and the torque is carried by the Bredt sum $$T = 2\sum_i A_i q_i = 2\left(32\,000\,q_1 + 60\,000\,q_2 + 32\,000\,q_3\right).$$ Symmetry of both geometry and thickness about the centre line requires $q_1 = q_3$ before any arithmetic is done, which is a useful check on the solution.
4. Solve the system. Equating the twist of cell 1 to that of cell 2 reduces to $24\,000\,q_1 = 18\,000\,q_2$, i.e. $q_2 = \tfrac{4}{3}q_1$. Substituting into the torque equation, $128\,000\,q_1 + 120\,000\left(\tfrac43 q_1\right) = 1.5\times10^{7}$, so $$\boxed{q_1 = q_3 = 52.08\ \text{N/mm}, \qquad q_2 = 69.44\ \text{N/mm}}$$ Back-substituting gives $2\sum A_iq_i = 1.5\times10^{7}\ \text{N}\cdot\text{mm}$ exactly, which confirms the solution.
5. Recover the rate of twist. Substituting into any one cell equation, $\mathrm{d}\theta/\mathrm{d}x = 1.302\times10^{-5}$ rad/mm, that is $0.746^{\circ}$ per metre of span — a useful figure because torsional stiffness, not strength, often governs a wing box.
6. Part (b) — tabulate the stress in every wall type. Shear stress is $\tau = q/t$ evaluated wall by wall, and the six distinct wall types give $$\begin{aligned} \text{outer webs (1.0 mm)} &: 52.08\ \text{MPa}, & \text{centre lower skin (1.5 mm)} &: 46.30\ \text{MPa}, \\ \text{centre upper skin (2.0 mm)} &: 34.72\ \text{MPa}, & \text{end lower skins (1.5 mm)} &: 34.72\ \text{MPa}, \\ \text{end upper skins (2.0 mm)} &: 26.04\ \text{MPa}, & \text{interior webs (1.0 mm)} &: 17.36\ \text{MPa}. \end{aligned}$$
7. Identify the maximum. The largest value is $$\boxed{\tau_{\max} = 52.08\ \text{MPa, in the two outer vertical webs}}$$ because those walls combine the full flow of an end cell with the thinnest gauge in the section. The centre cell carries the higher flow, but its thinnest wall is the 1.5 mm lower skin, which reaches only 46.30 MPa.
8. Note where the section is least stressed. The interior webs carry only $q_2-q_1 = 17.36$ N/mm, a third of the outer webs, because the two adjacent circulations largely cancel there. This is the standard argument for why interior webs in a multi-cell box are sized by buckling and by their role in carrying bending shear, not by torsion.