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

Question 5: Torsion of a three-cell thin-wall 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 three-cell rectangular thin-wall box, 180 mm deep, with cell widths 120, 250 and 120 mm, carrying a constant clockwise torque.

Find. (a) the three cell shear flows q₁, q₂, q₃, and (b) the magnitude and location of the largest shear stress in the box.

Approach. Pure torsion of a multi-cell box is one equilibrium equation short, so write the torque equation plus the compatibility statement that all three cells twist at the same rate, and solve the three equations for the three circulating flows.

1. Enclosed area of each cell. With a uniform depth of 180 mm, $$A_1 = A_3 = 120(180) = 21\,600\ \text{mm}^2, \qquad A_2 = 250(180) = 45\,000\ \text{mm}^2$$
2. Torque equilibrium. Each circulating flow contributes twice its enclosed area: $$T = 2(A_1q_1 + A_2q_2 + A_3q_3) = 10\times10^6\ \text{N}\cdot\text{mm}$$
3. Rate of twist of one cell. For cell i, $$\frac{\mathrm{d}\theta}{\mathrm{d}z} = \frac{1}{2A_iG}\oint_i \frac{q}{t}\,\mathrm{d}s$$ where the interior webs carry the difference of the two adjacent flows — for the web between cells 1 and 2 the term is \((q_1-q_2)(180/1.5)\).
4. Compatibility. The three cells belong to one rigid cross-section, so $$\left(\frac{\mathrm{d}\theta}{\mathrm{d}z}\right)_1 = \left(\frac{\mathrm{d}\theta}{\mathrm{d}z}\right)_2 = \left(\frac{\mathrm{d}\theta}{\mathrm{d}z}\right)_3$$ Two independent equations come out of that, and with the torque equation they close the problem.
5. Solve the 3 × 3 system (part a). Substituting the panel lengths and gauges and eliminating, $$q_1 = q_3 = \boxed{48.11\ \text{N/mm}}, \qquad q_2 = \boxed{64.93\ \text{N/mm}}$$ The symmetry \(q_1 = q_3\) is guaranteed by the symmetry of the box, and is a free check on the algebra. Substituting back, \(2(21\,600 \times 48.11 + 45\,000 \times 64.93 + 21\,600 \times 48.11) = 10\times10^6\) N·mm, so the torque closes exactly.
6. Shear stress in every wall (part b). The stress in a wall is its flow divided by its own thickness, and the interior webs carry only \(q_2 - q_1 = 16.82\) N/mm: $$\tau = \frac{q}{t}$$

   Wall	q (N/mm)	t (mm)	τ (MPa)
   Centre lower panel	64.93	2.0	32.46
   End webs	48.11	1.5	32.07
   Centre upper panel	64.93	2.5	25.97
   Outer lower panels	48.11	2.0	24.05
   Outer upper panels	48.11	2.5	19.24
   Interior webs	16.82	1.5	11.21

   $$\tau_{\max} = \boxed{32.5\ \text{MPa}}\quad \text{in the lower panel of the centre cell}$$
7. Rate of twist, as a check on the whole solution. Using cell 1 and \(G = 15\,000\) MPa, $$\frac{\mathrm{d}\theta}{\mathrm{d}z} = 1.381\times10^{-5}\ \text{rad/mm} = 0.79\,{}^{\circ}\text{ per metre}$$ and cells 2 and 3 return the same figure to ten significant digits, which is the compatibility condition being satisfied rather than assumed.

The result is a useful piece of design intuition. The thinnest walls in the box are the 1.5 mm webs, yet the interior webs are the quietest walls in the section at 11.2 MPa, because they see only the small difference between two similar circulating flows. The end webs, by contrast, carry the whole of \(q_1\) and are within 1.2 per cent of the peak. In a multi-cell box it is the outer skin and the end webs that size the section, not the internal ones.