Question 8 of 8: Two-cell wing torsion box: shear flows and maximum shear stress

Question 8: Two-cell wing torsion box: shear flows and maximum shear stress (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 wing torsion box symmetric about the x-axis, in pure torsion. Wall 5 divides it into a semicircular nose cell and a trapezoidal rear cell.

Find. (a) the shear flow in each of walls 1 to 5; (b) the maximum shear stress and the wall in which it occurs.

Approach. Two cells means two unknown circulating flows plus an unknown twist rate; supply the Bredt–Batho torque equation and the equal-twist condition between the two cells, then convert flow to stress wall by wall.

1. Part (a) — geometry: enclosed areas and wall lengths. The nose is a semicircle of radius 100 mm and the rear cell is a trapezoid 300 mm long between parallel sides of 200 and 100 mm: $$A_I = \tfrac{1}{2}\pi R^2 = 15,708 \text{ mm}^2,\qquad A_{II} = \tfrac{1}{2}(200+100)(300) = 45,000 \text{ mm}^2.$$ The wall lengths follow directly: $s_1 = \pi R = 314.16$ (arc, not chord), $s_5 = 200$, $s_3 = 100$, and the sloping skins $s_2 = s_4 = \sqrt{300^2 + 50^2} = 304.14$ mm.
2. Torque equilibrium. Each cell contributes twice its area times its own flow: $$T = 2A_I q_I + 2A_{II} q_{II} = 31,416\,q_I + 90,000\,q_{II} = 37,000,000 \text{ N}\cdot\text{mm.}$$
3. Equal twist rate of the two cells. The cells share wall 5 and must twist together, so $$\frac{1}{2A_I}\oint_I \frac{q\,ds}{t} = \frac{1}{2A_{II}}\oint_{II}\frac{q\,ds}{t} = G\frac{d\theta}{dz}.$$ Wall 5 carries $q_I - q_{II}$ when walked with cell I and $q_{II} - q_I$ when walked with cell II. The line integrals are $\oint_I ds/t = 314.16/2.0 + 200/3.0 = 223.746$ and $\oint_{II} ds/t = 304.14/2.5 + 100/2.0 + 304.14/1.5 + 200/3.0 = 441.081$ mm/mm.
4. Solve the three equations. Written out, $$\frac{1}{2A_I}\left[223.746\,q_I - 66.667\,q_{II}\right] = \frac{1}{2A_{II}}\left[441.081\,q_{II} - 66.667\,q_I\right],$$ together with the torque equation. Solving simultaneously, $$\boxed{\;q_I = 279.92 \text{ N/mm (nose cell)},\qquad q_{II} = 313.40 \text{ N/mm (rear cell)}\;}$$ Back-substitution recovers the torque: $2(15,708 \times 279.92 + 45,000 \times 313.40) = 3.7e+07$ N·mm.
5. Shear flow in each of the five walls. Walls 1 belongs to cell I alone and walls 2, 3, 4 to cell II alone, so each carries its own cell flow; the dividing wall 5 carries only the difference: $$q_5 = q_{II} - q_I = 313.40 - 279.92 = 33.47 \text{ N/mm,}$$ only 11–12 % of either cell flow. This is the characteristic result for an interior web in torsion and the reason such webs are sized by shear, buckling and bending rather than by torque.
6. Part (b) — convert to stresses. With $\tau = q/t$ and five different gauges, every wall must be tabulated:

7. Name the maximum. $$\boxed{\;\tau_{\max} = \frac{313.40}{1.5} = 208.9 \text{ MPa, in wall 2 — the lower skin}\;}$$ Wall 2 wins not because it carries the largest flow (walls 2, 3 and 4 all carry $q_{II}$) but because it is the thinnest gauge in the higher-flow cell. The nose skin, despite belonging to the smaller cell, still reaches 140.0 MPa because its 2.0 mm gauge is thin, while the 3.0 mm front spar is the quietest wall in the section at only 11.2 MPa.
8. Note what the torque does not depend on. The two cell flows above required no shear modulus, because $G$ appears on both sides of the equal-twist equation and cancels. It would be needed only to convert the common $G\,d\theta/dz = 1.329$ N/mm² per mm into an actual rate of twist, and no value of $G$ is quoted for this section.