Two-segment circular torsion box: 22-Mec-B9 Advanced Engineering Structures, December 2019, Q6

Question 6: Two-segment circular torsion 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 circular thin-walled tube of median radius 200 mm divided into two equal cells by a diametral wall, in pure torsion, with a limiting shear stress imposed on the material.

Torsion box data
Quantity	Value
Median radius R	200 mm
Outer skin thickness t₁	10 mm
Dividing wall thickness t₂	5 mm
Shear modulus G	80 GPa
Allowable shear stress τ	20 MPa

Find. The largest torque the section can carry without exceeding 20 MPa anywhere, and the corresponding rate of twist.

Figure 6.1 — The two cells are mirror images, so they carry identical flows and the wall between them carries their difference, which is zero. The section behaves exactly as the undivided tube.

Approach. Exploit the mirror symmetry to collapse the two-cell problem to one unknown flow, set that flow from the stress limit in the governing wall, then obtain the torque from the Bredt–Batho sum and the twist rate from either cell.

Set out the cell geometry. The diametral wall splits the tube into two semicircular cells, each enclosing $$A_1=A_2=\frac{\pi R^{2}}{2}=\frac{\pi(200)^2}{2}=62\,832\ \text{mm}^2$$ Each cell is bounded by a semicircular arc of length $\pi R = 628.3$ mm at thickness $t_1$ and by the straight dividing wall of length $2R = 400$ mm at thickness $t_2$.
Use symmetry to eliminate one unknown. For a multi-cell section in pure torsion every cell must twist at the same rate. The two cells here are geometric mirror images with identical wall thicknesses, so the equal-twist condition is satisfied by $$q_1=q_2=q$$ and the shear flow carried by the wall they share is the difference of the two cell flows: $$\boxed{q_{\text{wall}} = q_1 - q_2 = 0}$$ The dividing wall is unloaded, and $t_2$ cannot appear in any answer. Recognising this at the outset replaces a two-equation system with a single line.
Set the flow from the stress limit. With the wall carrying nothing, the governing stress is in the outer skin, so $$\tau_1=\frac{q}{t_1}\le 20\ \text{MPa}\quad\Longrightarrow\quad q=\tau_1t_1=20(10)=200\ \text{N/mm}$$
Sum the cell torques. Each cell contributes $2A_iq_i$ by Bredt–Batho, so $$T=2\left(A_1q_1+A_2q_2\right)=2\left(2\times62\,832\times200\right)$$ $$\boxed{T_{\max} = 5.027{\times}10^{7}\ \text{N}\!\cdot\!\text{mm} = 50.27\ \text{kN}\!\cdot\!\text{m}}$$ Because the flows are equal, this is identical to $2\pi R^2q$ for the undivided tube, which is a one-line independent check.
Compute the rate of twist. Applying the twist-rate integral to cell 1 and remembering that the dividing wall carries zero flow and so contributes nothing, $$\frac{d\theta}{dx}=\frac{1}{2A_1G}\oint\frac{q\,ds}{t}=\frac{1}{2(62\,832)(80\,000)}\left[200\left(\frac{628.3}{10}\right)+0\left(\frac{400}{5}\right)\right]$$ $$\boxed{\frac{d\theta}{dx} = 1.25{\times}10^{-6}\ \text{rad/mm} = 0.0716^{\circ}\text{ per metre}}$$ Cell 2 returns the same number, as symmetry requires.
Cross-check with the torsion constant. Writing $T=GJ\,d\theta/dx$ gives $J=5.027{\times}10^{8}\ \text{mm}^4$, which is exactly the thin-ring result $2\pi R^{3}t_1$. This confirms that the dividing wall adds nothing to the torsional stiffness of a circular tube.
Say what the dividing wall is actually for. Its irrelevance here is a consequence of the perfect symmetry and of the loading being pure torsion. Make the cells unequal, change either skin gauge, or introduce a transverse shear load, and the wall immediately picks up the difference between the two cell flows. Its real structural purpose is to stabilise the skin against buckling and to carry shear when the section is bent, neither of which is examined by this question.

Question 6 — results
Quantity	Value
Enclosed area of each cell	62 832 mm²
Shear flow in each cell	200 N/mm
Shear flow in the dividing wall	0
Shear stress in the outer skin / dividing wall	20 MPa / 0
Maximum torque	50.27 kN·m
Rate of twist	1.25 × 10⁻⁶ rad/mm = 0.0716°/m
Equivalent torsion constant J	5.027 × 10⁸ mm⁴ (= 2πR³t₁)

Topic. Torsion of multi-cell thin-walled closed sections, and the symmetry argument that collapses them.

Key relations. Bredt–Batho $T=\sum 2A_iq_i$; equal twist rate $\dfrac{d\theta}{dx}=\dfrac{1}{2A_iG}\oint\dfrac{q\,ds}{t}$ for every cell; interior wall flow $q_1-q_2$; shear stress $\tau=q/t$; thin ring $J=2\pi R^{3}t$.

Why this works. A section of $n$ cells is $n$ times redundant in torsion, and the compatibility statement that closes it is that all cells twist together. When the section has a plane of symmetry that maps one cell onto another and the loading respects that symmetry, the two flows must be equal, which is a solution of the compatibility equations obtainable by inspection. The interior wall then carries the difference of two equal quantities. This is the torsional analogue of the familiar result that no current flows through the bridge in a balanced Wheatstone bridge.

Common pitfall. Assuming the dividing wall is the most highly stressed member because it is the thinnest, and sizing the section on $q/t_2$. It is in fact the least stressed. The opposite pitfall appears the moment the symmetry is broken: with unequal cells the interior wall carries a genuine difference flow, and it must then be checked like any other wall. A third error is adding the two cell areas and treating the tube as a single cell of area $\pi R^2$ with the wall included in the perimeter integral; that double-counts the wall and gives too stiff a section.

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 6 of 7: Two-segment circular torsion box

Question 6: Two-segment circular torsion box (20 marks)