Three-cell wing box in pure torsion: 22-Mec-B9 Advanced Engineering Structures, December 2016, Q7

Question 7: 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 rectangular three-cell box 620 mm wide overall and 200 mm deep, divided into bays of 160, 300 and 160 mm, carrying pure torque:

Given data — three-cell box
Quantity	Value
Bay widths (cells 1, 2, 3)	160, 300, 160 mm
Depth of every cell	200 mm
Upper skin thickness	3.0 mm
Lower skin thickness	2.5 mm
All vertical webs (outer and interior)	1.5 mm
Applied torque (clockwise)	28 000 N·m = $28\times10^{6}$ N·mm
Shear modulus	$G$ = 70 GPa

Find. The three cell shear flows, and then the largest shear stress anywhere in the section together with the wall in which it acts.

Figure 7.1 — the three-cell box. Cells 1 and 3 are identical, so $q_1 = q_3$ by symmetry, and the two interior webs carry only the difference $q_2 - q_1$.

Approach. Write one equal-twist-rate equation per cell, close the set with torque equilibrium, solve the resulting four equations for $q_1$, $q_2$, $q_3$ and $G\theta'$, then tabulate $\tau = q/t$ over every distinct wall type.

Exploit the symmetry before doing any arithmetic. Cells 1 and 3 are geometrically identical and identically gauged, and the loading is pure torque, so $q_1 = q_3$ must hold. That halves the work and gives an immediate check on the numerical solution. The enclosed areas are$$A_1 = A_3 = 160(200) = 32\,000\ \text{mm}^2,\qquad A_2 = 300(200) = 60\,000\ \text{mm}^2 .$$
Assemble the wall compliances. For each cell the outer path is the sum of $\mathrm{d}s/t$ over its own skins and outer web, and every interior web contributes $200/1.5 = 133.33$:$$\delta_1 = \delta_3 = \frac{160}{3.0} + \frac{160}{2.5} + \frac{200}{1.5} = 53.33 + 64.00 + 133.33 = 250.67,$$$$\delta_2 = \frac{300}{3.0} + \frac{300}{2.5} = 100.0 + 120.0 = 220.0 .$$
Impose equal twist rate on all three cells. With $u = G\theta'$ and each interior web carrying the difference between the flows it separates,$$\text{cell 1:}\quad q_1(\delta_1 + 133.33) - 133.33\,q_2 = 2A_1 u,$$$$\text{cell 2:}\quad q_2(\delta_2 + 266.67) - 133.33(q_1 + q_3) = 2A_2 u,$$$$\text{cell 3:}\quad q_3(\delta_3 + 133.33) - 133.33\,q_2 = 2A_3 u .$$These three, taken alone, are homogeneous in the four unknowns; one more equation is needed.
Close with torque equilibrium and solve (part a). The fourth equation is$$T = 2\left(A_1q_1 + A_2q_2 + A_3q_3\right) = 28\times10^{6}\ \text{N}\cdot\text{mm}.$$Solving the four simultaneously,$$\boxed{q_1 = q_3 = 96.98\ \text{N/mm},\qquad q_2 = 129.89\ \text{N/mm}} ,$$and the interior webs carry only the difference, $q_2 - q_1 = 32.91$ N/mm. Substituting back, $2[32\,000(96.98)\times2 + 60\,000(129.89)] = 28.00\times10^{6}$ N·mm, so the torque is recovered exactly. The rate of twist follows as $\theta' = u/G = 4.447\times10^{-6}$ rad/mm, or 0.255° per metre of span.
Tabulate the stress in every distinct wall (part b). There are six wall types, and each must be divided by its own gauge:$$\tau = \frac{96.98}{3.0} = 32.33,\ \frac{96.98}{2.5} = 38.79,\ \frac{96.98}{1.5} = 64.65,\ \frac{129.89}{3.0} = 43.30,\ \frac{129.89}{2.5} = 51.96,\ \frac{32.91}{1.5} = 21.94\ \text{MPa},$$for the cell-1/3 upper skin, cell-1/3 lower skin, outer webs, cell-2 upper skin, cell-2 lower skin and interior webs respectively.
Identify the peak. The largest of the six is$$\boxed{\tau_{\max} = 64.7\ \text{MPa, in the two outer (end) webs}} .$$The centre cell carries the larger flow, but its skins are the thick ones; the outer webs pair the smaller flow with the thinnest gauge in the section and win. The interior webs, carrying only 21.9 MPa, are by far the quietest walls — that is a general feature of a multi-cell box, because an interior web sees only the difference between two adjacent circulations.

Final results — Question 7
Wall	Thickness (mm)	Shear flow (N/mm)	Shear stress (MPa)
Outer webs (ends of cells 1 and 3)	1.5	96.98	64.65
Lower skin, cell 2	2.5	129.89	51.96
Upper skin, cell 2	3.0	129.89	43.30
Lower skin, cells 1 and 3	2.5	96.98	38.79
Upper skin, cells 1 and 3	3.0	96.98	32.33
Interior webs	1.5	32.91	21.94
Rate of twist			0.255° per metre

Topic. Pure torsion of a three-cell thin-walled box.

Key relations. One equal-twist-rate equation per cell, $2A_iG\theta' = \oint_i q\,\mathrm{d}s/t$ with interior webs carrying $q_i - q_j$, plus $T = 2\sum A_i q_i$; wall stress $\tau = q/t$.

Why this works. An $n$-cell box is $n$ times statically indeterminate in torsion: torque equilibrium supplies one equation for $n$ unknown circulating flows. The missing $n-1$ conditions come from compatibility — the cells are joined along their common webs and so must all twist at the same rate. Writing the system with $G\theta'$ as an extra unknown turns the whole problem into one $(n+1)\times(n+1)$ linear solve, which is far cleaner than eliminating variables by hand and gives the twist rate as a by-product.

Common pitfall. Assuming that the cell with the largest flow contains the largest stress, and reporting $q_{\max}/t_{\text{web}}$. Flow and gauge vary independently, so every distinct wall must be divided separately — on this paper the peak is in an outer web fed by the smaller flow, whereas on a similar box with different gauges the centre cell’s lower skin can govern instead. The other classic slip is giving an interior web the full flow of one cell rather than the difference between the two it separates.

Source. Megson, Aircraft Structures for Engineering Students, 6th ed., Ch. 18 (torsion of beams, multi-cell sections); Boresi and Schmidt, Advanced Mechanics of Materials, 6th ed., Ch. 6.

Question 7 of 8: Three-cell wing box in pure torsion

Question 7: Three-cell wing box in pure torsion (20 marks)