Question 6 of 8: Shear centre and shear flow of an idealised wing box with a semicircular nose

Question 6: Shear centre and shear flow of an idealised wing box with a semicircular nose (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 horizontally symmetric idealised wing box with a semicircular nose, four booms carrying all the direct stress, and walls effective in shear only.

Find. (a) the location of the shear centre; (b) the shear flow around the box when the upward shear force acts 100 mm to the left of the shear centre.

Approach. Because the walls are shear-ineffective in bending, the open shear flow steps by a constant at each boom and is constant between booms. Cut the rear spar, build the open flows, close the cell by the zero-twist condition to locate the shear centre, then superpose a pure-torque flow for the offset load.

1. Part (a) — second moment of area of the boom idealisation. Only the booms carry direct stress, and all four sit at $y = \pm 100$ mm: $$I_{xx} = \sum B_r y_r^2 = 2(700)(100)^2 + 2(550)(100)^2 = 25.00 \times 10^{6} \text{ mm}^4.$$
2. Open shear flows from a cut in the rear spar. With $S_x = 0$ the open-section flow steps at each boom by $$\Delta q = -\frac{S_y}{I_{xx}}B_r y_r,$$ and is constant between booms. Walking $1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1$ (top skin right to left, nose downward, bottom skin left to right, rear spar upward) and cutting in wall 4–1: $$q_{b,12} = -39.20,\quad q_{b,23} = -70.00,\quad q_{b,34} = -39.20,\quad q_{b,41} = 0.00 \text{ N/mm.}$$ The walk closes on zero, as it must.
3. Close the cell by requiring zero twist. By definition the shear centre is the point at which a transverse load produces no twist, so $$\oint \frac{(q_b + q_{s,0})}{t}\,ds = 0 \;\Longrightarrow\; q_{s,0} = -\frac{\displaystyle\oint q_b\,ds/t}{\displaystyle\oint ds/t}.$$ The four wall lengths are 500, $\pi R = 314.16$ (the nose arc), 500 and 200 mm at a common thickness, so the thickness cancels and $$q_{s,0} = +40.413 \text{ N/mm.}$$
4. Total flows with the load at the shear centre. Adding $q_{s,0}$ to each open flow, $$q_{12} = +1.213,\quad q_{23} = -29.587,\quad q_{34} = +1.213,\quad q_{41} = +40.413 \text{ N/mm.}$$ Equilibrium check: resolving these along the wall chord vectors returns $\sum F_x = 0$ and $\sum F_y = 14000$ N, the applied shear.
5. Locate the shear centre by moment equivalence. Taking moments about the centre of the nose semicircle, the moment of a constant flow along a wall is $q$ times twice the area swept by the radius vector, $\oint p\,ds = 2A_{\text{swept}}$. For the curved nose this is exact and needs no chording: $2A = \pi R^2 = 31,416$ mm². Summing over the four walls and equating to $S_y \xi_S$: $$\xi_S = \frac{\sum q_i (2A_i)}{S_y} = 230.93 \text{ mm aft of the front spar,}$$ i.e. $$\boxed{\;\text{the shear centre lies } 269.1 \text{ mm forward of the rear spar, on the axis of symmetry}\;}$$ A free check: the four swept areas sum to 231,416 mm², which is exactly twice the enclosed cell area $A = 500 \times 200 + \pi R^2/2 = 115,708$ mm².
6. Part (b) — resolve the offset load into shear plus torque. Moving the 14 000 N force 100 mm to the left of the shear centre is statically the same as leaving it at the shear centre and adding a torque $$T = S_y \times 100 = 1.4 \times 10^{6} \text{ N}\cdot\text{mm},$$ applied in the sense that carries the load forward (i.e. opposite to the positive walking sense used above).
7. Superpose the pure-torque flow. A torque on a single closed cell produces a constant circulating flow $q_T = T/2A$ with $A = 115,708$ mm²: $$q_T = \frac{-1.4\times 10^{6}}{2 \times 115,708} = -6.050 \text{ N/mm.}$$ Adding it to the shear-centre flows gives the final answer $$\boxed{\;q_{12} = -4.84,\; q_{23} = -35.64,\; q_{34} = -4.84,\; q_{41} = +34.36 \text{ N/mm}\;}$$ where a positive value acts in the walking sense $1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1$.
8. Check vertical equilibrium the right way. The resultant of a constant flow along a wall is $q$ times the wall’s chord vector, so the curved nose skin carries vertical load even though it is not a web: its chord is 200 mm, giving $35.64 \times 200 = 7127$ N, while the rear spar carries $34.36 \times 200 = 6873$ N. These sum to 14000 N, the applied shear; the top and bottom skins cancel horizontally. Checking with the spar alone would return only 6873 N and look like a gross error — it is the check that is wrong, not the flows.