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

Question 7: 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, with three different wall gauges and a single shear modulus.

Find. (a) the three cell shear flows, and (b) the magnitude and location of the greatest shear stress in the box.

Approach. Write the rate of twist for each cell in terms of its own circuit integral, set the three rates equal because the cells share the same cross-section, add the torque equation, and solve the four-by-four system.

1. Exploit the symmetry first. Cells 1 and 3 have the same width, the same depth and the same three wall gauges, and each shares one interior web with the centre cell, so the structure is mirror-symmetric about the centre of cell 2. It follows immediately that $$q_{1} = q_{3}$$ which reduces the problem from four unknowns to three and halves the arithmetic. The two interior webs must then carry the same difference flow.
2. Compute the cell areas. Each cell is a rectangle of depth 200 mm: $$\begin{aligned} A_{1} &=A_{3}=160\times200 = 32\,000\ \text{mm}^{2}, \\ A_{2} &=300\times200 = 60\,000\ \text{mm}^{2} \end{aligned}$$ The total enclosed area is 124 000 mm², but that number never appears on its own in a multi-cell calculation — each cell contributes through its own area.
3. Assemble the wall flexibility terms $\int ds/t$. These are the only place the thicknesses enter: $$\begin{aligned} \frac{b_{1}}{t_{u}} &= \frac{160}{2.5} = 64.0, \\ \frac{b_{1}}{t_{l}} &= \frac{160}{1.75} = 91.429, \\ \frac{h}{t_{w}} &= \frac{200}{1.5} = 133.333 \end{aligned}$$ $$\begin{aligned} \frac{b_{2}}{t_{u}} &= 120.0, \\ \frac{b_{2}}{t_{l}} &= 171.429 \end{aligned}$$ so a full circuit of an outer cell is $64.0+91.429+2\left(133.333\right) = 422.095$, and of the centre cell $120.0+171.429+2\left(133.333\right) = 558.095$.
4. Write the twist-rate equation for each cell. For cell $i$ the Bredt–Batho rate of twist is $$\frac{d\theta}{dx} = \frac{1}{2A_{i}G}\oint_{i}q\,\frac{ds}{t}$$ where the interior webs carry the difference between the flows of the two cells they separate. For the left cell, $$2A_{1}G\frac{d\theta}{dx} = 422.095\,q_{1} - 133.333\,q_{2}$$ and for the centre cell, with an interior web on each side, $$2A_{2}G\frac{d\theta}{dx} = 558.095\,q_{2} - 133.333\left(q_{1}+q_{3}\right) = 558.095\,q_{2} - 266.667\,q_{1}$$
5. Impose compatibility between the cells. All three cells belong to one cross-section and must twist at the same rate, so equating the two expressions above after dividing by their respective $2A_{i}G$, $$\frac{422.095\,q_{1}-133.333\,q_{2}}{64\,000} = \frac{558.095\,q_{2}-266.667\,q_{1}}{120\,000}$$ Cross-multiplying and collecting terms gives $67.718\times10^{6}\,q_{1} = 51.718\times10^{6}\,q_{2}$, that is $$\frac{q_{2}}{q_{1}} = 1.3094$$ The centre cell carries the larger flow because it encloses nearly twice the area for only a third more perimeter.
6. Apply the torque equation. The applied torque is the sum of the Bredt–Batho contributions of the three cells: $$T = 2\sum A_{i}q_{i} = 2\left(32\,000\,q_{1}+60\,000\,q_{2}+32\,000\,q_{3}\right) = 12\,000\times10^{3}\ \text{N}\cdot\text{mm}$$ Substituting $q_{3}=q_{1}$ and $q_{2}=1.3094\,q_{1}$, $$\left(128\,000 + 157\,125\right)q_{1} = 12.0\times10^{6}$$
7. Solve for the three cell flows. $$\begin{aligned} q_{1} &= q_{3} = \boxed{42.09\ \text{N/mm}}, \\ q_{2} &= \boxed{55.11\ \text{N/mm}} \end{aligned}$$ Substituting back, $2\left(32\,000\times42.087+60\,000\times55.107+32\,000\times42.087\right) = 12.00\times10^{6}$ N·mm, so the torque equation is satisfied exactly. The two interior webs each carry the difference $q_{2}-q_{1} = 13.02$ N/mm.
8. Report the rate of twist as a by-product. Substituting into the cell-1 equation, $$\frac{d\theta}{dx} = \frac{422.095\left(42.087\right)-133.333\left(55.107\right)}{2\left(32\,000\right)\left(18\,000\right)} = 9.043\times10^{-6}\ \text{rad/mm} = 0.518\ \text{deg/m}$$ which corresponds to a torsional stiffness $GJ = T/\left(d\theta/dx\right) = 1.327\times10^{12}$ N·mm², or $J = 7.373\times10^{7}$ mm⁴.

Part (b) asks not for the largest flow but for the largest stress, and because the gauges differ those are in different walls.

1. Divide each wall flow by that wall’s own thickness. Working through the six distinct wall types, $$\begin{aligned} \tau_{u,\text{outer}} &= \frac{42.09}{2.5} = 16.83, \\ \tau_{l,\text{outer}} &= \frac{42.09}{1.75} = 24.05, \\ \tau_{\text{end webs}} &= \frac{42.09}{1.5} = 28.06 \end{aligned}$$ $$\begin{aligned} \tau_{u,\text{centre}} &= \frac{55.11}{2.5} = 22.04, \\ \tau_{l,\text{centre}} &= \frac{55.11}{1.75} = 31.49, \\ \tau_{\text{interior webs}} &= \frac{13.02}{1.5} = 8.68 \end{aligned}$$ all in MPa.
2. Identify the critical wall. The largest value is $$\tau_{\max} = \boxed{31.49\ \text{MPa}}$$ in the lower skin of the centre cell, where the largest cell flow meets the second-thinnest gauge. The end webs are a close second at 28.06 MPa, since they combine the smaller flow with the thinnest gauge.
3. Note what the interior webs are doing. They carry only 13.02 N/mm, less than a third of the flow in either adjacent cell, and at 8.68 MPa they are the least stressed walls in the box. That is the standard signature of a multi-cell section in torsion: adjacent cells circulate in the same sense, so their flows partly cancel in the wall they share, and the interior webs are structurally quiet in pure torsion even though they are essential in shear and bending.