Torsion of a three-cell thin-walled wing box: 22-Mec-B9 Advanced Engineering Structures, December 2018, Q5

Question 5: 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.

Item	Value
Cell widths (left, centre, right)	120 / 250 / 120 mm
Box depth	180 mm
Upper skin thickness	2.75 mm
Lower skin thickness	2.25 mm
Vertical web thickness (all four)	1.75 mm
Shear modulus $G$	10 GPa
Applied torque $T$	15 000 N·m clockwise

Find. (a) the shear flow circulating in each of the three cells, and (b) the magnitude and location of the largest shear stress in the section.

Question 5 — the three-cell box. Each cell carries its own circulating shear flow $q_i$ (shown clockwise, matching the applied torque); the interior webs carry the difference between the flows of the cells they separate.

Approach. Three unknown cell flows plus one unknown rate of twist need four equations: one torque equilibrium statement and three compatibility statements requiring all cells to twist at the same rate.

Set up the cell areas. Each cell is rectangular, so$$A_1 = A_3 = 120(180) = 21\,600\ \text{mm}^2,\qquad A_2 = 250(180) = 45\,000\ \text{mm}^2$$The outer cells are mirror images, so symmetry demands $q_1 = q_3$ and the problem is effectively two unknown flows.
Write torque equilibrium. Each cell contributes $2A_iq_i$ to the resisting torque:$$T = 2\sum A_iq_i = 2\big(21\,600q_1 + 45\,000q_2 + 21\,600q_3\big) = 1.50\times10^{7}\ \text{N}\cdot\text{mm}$$With $q_1 = q_3$ this reduces to $86\,400q_1 + 90\,000q_2 = 1.50\times10^{7}$.
Write the rate-of-twist equation for each cell. For cell $i$, with $\delta = 2G\,\mathrm{d}\theta/\mathrm{d}z$,$$\oint_i \frac{q}{t}\,\mathrm{d}s = A_i\,\delta$$where the interior webs carry the flow difference. Evaluating the wall compliances $L/t$ — $120/2.75 = 43.64$, $120/2.25 = 53.33$, $180/1.75 = 102.86$, $250/2.75 = 90.91$, $250/2.25 = 111.11$ — gives for cells 1 and 2$$302.68\,q_1 - 102.86\,q_2 = 21\,600\,\delta,\qquad407.73\,q_2 - 205.71\,q_1 = 45\,000\,\delta$$with the cell-3 equation identical to the first by symmetry.
Eliminate the rate of twist. Equating the two expressions for $\delta$ gives a single relation between the flows,$$1.8064\times10^{7}\,q_1 = 1.3436\times10^{7}\,q_2\quad\Longrightarrow\quad q_1 = 0.74377\,q_2$$so the outer cells run about three quarters as hard as the centre one. That ratio is set by geometry alone, not by the size of the torque.
Solve for the flows (part a). Substituting into the torque equation, $86\,400(0.74377q_2)+90\,000q_2 = 1.50\times10^{7}$, and therefore$$\boxed{q_1 = q_3 = 72.32\ \text{N/mm},\qquad q_2 = 97.24\ \text{N/mm}}$$Back-substituting recovers $T = 1.50\times10^{7}$ N·mm exactly, and both rate-of-twist equations return the same $\delta = 0.5504$, confirming compatibility.
Convert every wall to a shear stress (part b). Since $\tau = q/t$ and both $q$ and $t$ vary from wall to wall, the peak can sit anywhere; the only safe method is to tabulate all six wall types:

Wall	$q$ (N/mm)	$t$ (mm)	$\tau$ (MPa)
Outer-cell upper skins	72.32	2.75	26.30
Outer-cell lower skins	72.32	2.25	32.14
The two end webs	72.32	1.75	41.33
Centre-cell upper skin	97.24	2.75	35.36
Centre-cell lower skin	97.24	2.25	43.22
The two interior webs	97.24 − 72.32 = 24.92	1.75	14.24

The largest value is$$\boxed{\tau_{\max} = 43.22\ \text{MPa, in the lower skin of the \textbf{centre} cell}}$$and the runner-up is 41.33 MPa in the two end webs, only 4.6 per cent lower. Which of those two wins is decided by the gauges rather than by anything structural, and on a box with slightly thinner webs the answer would flip — which is exactly why the full table is worth writing out rather than guessing at the thinnest wall.

The interior webs are the quietest walls in the section, at 14.24 MPa, because they see only the 24.92 N/mm difference between two flows that are circulating the same way. This is the characteristic signature of multi-cell torsion and it explains why adding interior webs is such an efficient way to stiffen a box: they contribute their full share to the enclosed area and hence to the torsional stiffness, while carrying very little stress themselves. The resulting rate of twist is $\mathrm{d}\theta/\mathrm{d}z = \delta/(2G) = 2.752\times10^{-5}$ rad/mm, or 1.577 degrees per metre of span.

Quantity	Value
(a) $q_1 = q_3$ (outer cells)	72.32 N/mm
(a) $q_2$ (centre cell)	97.24 N/mm
Flow in each interior web	24.92 N/mm
(b) Maximum shear stress	43.22 MPa
(b) Its location	lower skin of the centre cell
Rate of twist	$2.752\times10^{-5}$ rad/mm = 1.577 °/m

Topic. Pure torsion of a multi-cell closed thin-walled section.

Key relations. Torque from circulating flows, $T = 2\sum A_iq_i$; equal rate of twist for every cell, $\dfrac{\mathrm{d}\theta}{\mathrm{d}z} = \dfrac{1}{2A_iG}\oint_i \dfrac{q}{t}\,\mathrm{d}s$; interior webs carry $q_i - q_j$; wall stress $\tau = q/t$.

Why this works. A single closed cell in torsion is statically determinate — Bredt–Batho gives $q = T/(2A)$ directly. With $n$ cells there are $n$ unknown flows but still only one equilibrium equation, so the section is indeterminate to degree $n-1$ and compatibility must supply the rest. The compatibility statement is geometric and almost self-evident once seen: the cells are rigidly joined along their common webs, so a cross-section rotates as one body and every cell must twist at the same rate. That gives $n$ rate-of-twist equations in $n+1$ unknowns (the flows plus the rate of twist itself), which with torque equilibrium closes the system exactly.

Common pitfall. Signing the interior-web terms wrongly. In cell 1's circuit the shared web is traversed in the sense of $q_1$, so it contributes $(q_1-q_2)L/t$; in cell 2's circuit the same web is traversed the other way and contributes $(q_2-q_1)L/t$. Getting one of those backwards produces flows that satisfy torque equilibrium but not compatibility, and the error is invisible unless you check that all three cells return the same $\delta$. The second pitfall is quoting the maximum shear stress as $q_{\max}/t_{\min}$ — here that would pair 97.24 N/mm with the 1.75 mm web and give a spurious 55.6 MPa, in a wall that never carries that flow.

Source. Megson, Aircraft Structures for Engineering Students, 6th ed., §23.2 (torsion of multi-cell wing sections); Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed., ch. 6 (torsion of thin-walled hollow members).

Question 5 of 8: Torsion of a three-cell thin-walled wing box

Question 5: Torsion of a three-cell thin-walled wing box (20 marks)