22-Mec-B9 Advanced Engineering Structures · May 2017
Question 6 of 8: Three-cell thin-walled box in pure torsion
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2017 — 16-Mec-B9 Advanced Engineering Structures. Three hours, open book, any non-communicating calculator permitted. Eight questions of equal total value (20 marks each); the rubric states that any five constitute a complete paper. All eight are solved here, because the complete set is the study resource.
Reference texts.
T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. — unsymmetrical bending, shear flow in open and closed thin-walled sections, shear centre, structural idealisation, single- and multi-cell torsion (Ch. 15, 16, 17, 20, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — three-dimensional stress states, the Tresca and von Mises yield criteria, torsion of non-circular prismatic bars (Ch. 1, 4, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of solid rectangular sections and of multiply connected cells (Ch. 4, 6).
R. M. Jones, Mechanics of Composite Materials, 2nd ed. — orthotropic lamina stiffness, the reduced stiffness matrix and its transformation (Ch. 2).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — strain-life fatigue, the Coffin–Manson relation, the Palmgren–Miner rule, and Paris-law crack growth (Ch. 9, 11, 14).
Question 6: Three-cell thin-walled box in pure torsion (20 marks)
Given. A rectangular three-cell box of constant depth carrying pure torque, with three different wall gauges.
Given data
Quantity
Value
Cell widths (left, centre, right)
150, 250, 150 mm
Box depth
220 mm
Upper panels
$t = 3.5$ mm
Lower panels
$t = 2.5$ mm
All vertical panels
$t = 2.0$ mm
Applied torque
$T = 48\,000$ N·m $= 48 \times 10^6$ N·mm
Shear modulus
$G = 90$ GPa
Find. (a) the three cell shear flows; (b) the magnitude and location of the largest shear stress in the box.
The three-cell box. Each cell carries a constant circulating shear flow; the two interior webs carry only the difference between the flows of the cells they separate.
Approach. Assign one circulating shear flow to each cell, impose the condition that all three cells twist at the same rate (they share the same rigid ribs), add the torque equilibrium equation, and solve the resulting $3 \times 3$ system; then divide each wall flow by its own gauge to find the stresses.
Set out the cell areas. Each cell is a rectangle of the full 220 mm depth, so $$A_1 = A_3 = 150 \times 220 = 33\,000\ \text{mm}^2, \qquad A_2 = 250 \times 220 = 55\,000\ \text{mm}^2.$$ Because cells 1 and 3 are geometrically identical and carry identical gauges, symmetry requires $q_1 = q_3$ before any arithmetic is done.
Write the rate of twist of a general cell. For cell $i$, $$\frac{\mathrm{d}\theta}{\mathrm{d}x} = \frac{1}{2A_i G}\oint \frac{q}{t}\,\mathrm{d}s,$$ where the circuit runs around the walls bounding that cell and each interior web carries the difference of the two adjacent cell flows. The wall compliances $s/t$ are $150/3.5 = 42.86$ and $150/2.5 = 60.0$ for the outer cells' skins, $250/3.5 = 71.43$ and $250/2.5 = 100.0$ for the centre cell's skins, and $220/2.0 = 110.0$ for every vertical panel.
Assemble the compatibility equations. Writing the loop integrals out, $$\frac{1}{A_1}\left[212.86\,q_1 + 110(q_1-q_2)\right] = \frac{1}{A_2}\left[171.43\,q_2 + 220(q_2-q_1)\right] = \frac{1}{A_3}\left[212.86\,q_3 + 110(q_3-q_2)\right].$$ With $q_1 = q_3$ these three expressions collapse to the single relation $2274.3\,q_1 = 1724.3\,q_2$, that is $q_1 = 0.7582\,q_2$.
Close the system with torque equilibrium. The applied torque is carried by the three circulating flows as $$T = 2\sum A_i q_i = 2\left(A_1 q_1 + A_2 q_2 + A_3 q_3\right) = 48 \times 10^6.$$ Substituting $q_1 = q_3 = 0.7582\,q_2$ gives $105\,039\,q_2 = 24 \times 10^6$, hence $$\boxed{q_1 = q_3 = 173.2\ \text{N/mm}, \qquad q_2 = 228.5\ \text{N/mm}}.$$
Check the solution before using it. Substituting back, $2[33\,000(173.23) + 55\,000(228.49) + 33\,000(173.23)] = 48.0 \times 10^6$ N·mm, which closes on the applied torque, and all three loop integrals return the same rate of twist, $\mathrm{d}\theta/\mathrm{d}x = 5.184 \times 10^{-6}$ rad/mm (0.297° per metre). Both checks are worth the thirty seconds they cost.
Part (b) — convert every distinct wall to a stress. The shear stress in a wall is $\tau = q/t$ evaluated wall by wall, because the flow and the gauge both change around the box: $$\tau = \frac{173.2}{3.5} = 49.5, \quad \frac{173.2}{2.5} = 69.3, \quad \frac{173.2}{2.0} = 86.6\ \text{MPa (outer cells)},$$ $$\tau = \frac{228.5}{3.5} = 65.3, \quad \frac{228.5}{2.5} = 91.4\ \text{MPa (centre cell)}, \qquad \tau = \frac{228.5-173.2}{2.0} = 27.6\ \text{MPa (interior webs)}.$$
Identify the governing wall. The largest value is $$\boxed{\tau_{\max} = 91.4\ \text{MPa, in the LOWER panel of the CENTRE cell}},$$ ahead of the two end webs at 86.6 MPa. It is the centre cell's larger flow meeting the second-thinnest gauge that wins, not the thinnest gauge on its own — which is why all six distinct wall types have to be tabulated rather than guessed at.
The one invariant of a multi-cell box in torsion is that the interior webs are the quietest walls: they carry only the difference $q_2 - q_1 = 55.3$ N/mm and reach barely a third of the peak stress. That is the structural argument for the multi-cell arrangement in the first place — the internal webs cost little in shear duty. They add only about 5 % to the torsional stiffness of a single cell of the same outline ($J = 1.029 \times 10^8$ against $0.981 \times 10^8$ mm4); their real value is in carrying the shear from bending and in stabilising the skins.
