Question 6 of 8: Shear centre and panel shear flows of an idealised six-boom wing box

Question 6: Shear centre and panel shear flows of an idealised six-boom wing 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 single-cell rectangular box idealised into six direct-stress booms joined by shear-only panels of uniform thickness.

Find. (a) the horizontal position of the shear centre measured from boom 6; (b) the shear flow in every panel for the load as actually applied.

Approach. Cut the closed cell to make it statically determinate, walk the open-section shear flow from boom to boom, then restore the cut with a constant redundant flow — fixed by zero twist to locate the shear centre in part (a), and by moment equilibrium about boom 6 for the load as actually applied in part (b).

1. Part (a) — second moment of area of the boom idealisation. All six booms lie at $y = \pm100$ mm from the horizontal centroidal axis, so $$I_{xx} = \sum B_r y_r^{2} = 2\bigl(500+300+800\bigr)(100)^{2} = \boxed{32\times10^{6}\ \text{mm}^{4}}$$ The panels themselves are assumed to carry no direct stress, which is the whole point of the idealisation.
2. Open the cell and walk the basic shear flow. Cutting panel 1-2 and walking in the sense $1\!\to\!2\!\to\!3\!\to\!4\!\to\!5\!\to\!6\!\to\!1$, the flow steps at each boom by $$\Delta q_r = -\frac{S_y}{I_{xx}}B_r y_r = -\frac{5000}{32\times10^{6}}B_r y_r = -1.5625\times10^{-4}B_r y_r$$ so, starting from $q_{b,12} = 0$ at the cut, $$\begin{aligned}q_{b,23} &= -4.6875, \\ q_{b,34} &= -17.1875, \\ q_{b,45} &= -4.6875, \\ q_{b,56} &= 0, \\ q_{b,61} &= 7.8125\ \ \text{N/mm}\end{aligned}$$ Returning to boom 1 brings the flow back to zero, which confirms the walk.
3. Close the cell with the zero-twist condition. A load through the shear centre causes no twist, so $$\oint\frac{q_b + q_{s,0}}{t}\,\mathrm{d}s = 0 \;\Rightarrow\; q_{s,0} = -\frac{\sum q_{b,i}s_i}{\sum s_i}$$ because the thickness is the same in every panel and cancels. With panel lengths 400, 400, 200, 400, 400 and 200 mm, $\sum q_{b,i}s_i = -5625$ N and $\sum s_i = 2000$ mm, so $q_{s,0} = 2.8125$ N/mm.
4. Shear-centre flows, and the check that they carry the load. Adding the redundant flow gives $$\begin{aligned}q_{12} &= 2.8125, \\ q_{23} &= -1.8750, \\ q_{34} &= -14.3750, \\ q_{45} &= -1.8750, \\ q_{56} &= 2.8125, \\ q_{61} &= 10.6250\ \ \text{N/mm}\end{aligned}$$ Only the two vertical webs carry vertical force: $14.3750(200) + 10.6250(200) = 2875 + 2125 = 5000$ N $= S_y\ \checkmark$, and the horizontal flange forces sum to zero as they must.
5. Moments about boom 6 locate the shear centre. Taking the moment of every panel flow about boom 6 (the two panels meeting at boom 6 contribute nothing, and the rear web dominates), $$S_y\,\xi_S = \sum q_i\,(2A_i) = 2\,225\,000\ \text{N}\!\cdot\!\text{mm} \;\Rightarrow\; \xi_S = \frac{2\,225\,000}{5000} = \boxed{445.0\ \text{mm from boom 6}}$$ The shear centre sits aft of mid-span because the rear booms are the largest, so the rear web attracts the greater share of the vertical shear.
6. Part (b) — the applied load is offset from the shear centre. The 5 kN acts on the line of booms 3 and 4, that is 800 mm from boom 6, so it is equivalent to the same force at the shear centre plus a torque $$T = S_y\,(800 - \xi_S) = 5000\,(800 - 445.0) = 1\,775\,000\ \text{N}\!\cdot\!\text{mm}$$ about the shear centre, in the sense that lifts the rear of the box.
7. Superpose the torsional flow. A pure torque on a single cell of enclosed area $A = 800(200) = 160\,000$ mm$^{2}$ produces the constant flow $$\Delta q = \frac{T}{2A} = \frac{1\,775\,000}{320\,000} = 5.5469\ \text{N/mm}$$ circulating in the opposite sense to the assumed walking direction, so it is subtracted from the shear-centre flows. Equivalently, the same answer follows by replacing the zero-twist condition with moment equilibrium about boom 6, which gives $q_{s,0} = -2.7344$ N/mm directly.
8. Final panel shear flows and their statical checks. $$\begin{aligned}q_{12} &= q_{56} = -2.734, \\ q_{23} &= q_{45} = -7.422, \\ q_{34} &= -19.922, \\ q_{61} &= 5.078\ \ \text{N/mm}\end{aligned}$$ Vertical equilibrium: $19.922(200)+5.078(200) = 3984 + 1016 = 5000$ N $\checkmark$. Moment about boom 6: 4,000,000 N·mm $= 5000(800)\ \checkmark$. The largest flow is in the rear web, giving $\tau = 9.96$ MPa.