Question 7 of 8: Panel shear flows and stringer loads in a cantilever wing box

Question 7: Panel shear flows and stringer loads in a cantilever 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 prismatic six-stringer box of wedge cross-section, 3 m long, built in at the root and loaded only at the free end.

Find. (a) the shear flow in each of the six panels, and (b) the axial load carried by each stringer, both at the station 2 m from the free end.

Approach. The section is unsymmetrical, so first fix the centroid and all three second moments. Part (b) then follows directly from the linear stress field; part (a) follows from differentiating that field along the span to get the rate of change of each stringer load, which the panel shear flows must balance, closing the cell by moments.

1. Locate the centroid. With total stringer area $\sum B = 2300$ mm$^{2}$,$$\bar{Z} = \frac{\sum B_{r}Z_{r}}{\sum B_{r}} = 347.8\ \text{mm},\qquad \bar{Y} = 123.9\ \text{mm}$$The centroid sits low and aft, pulled there by the two 550 mm$^{2}$ stringers at the deep end and by the fact that four of the six stringers lie on the flat lower surface.
2. Evaluate the second moments about the centroid. Because the walls carry no direct stress, every term is simply $\sum B_{r}(\;)^{2}$:$$I_{YY} = 3.8935 \times 10^{7}, \qquad I_{ZZ} = 3.2174 \times 10^{8}, \qquad I_{YZ} = -2.4130 \times 10^{7}\ \text{mm}^{4}$$The product of inertia is far from negligible, so the two load components cannot be treated independently.
3. Find the moments carried at $x$ = 2 m. Both loads act at the free end, 2000 mm away:$$M_{Z} = \int\sigma Y\,\mathrm{d}A = -(20\,000)(2000) = -4.00 \times 10^{7}, \qquad M_{Y} = \int\sigma Z\,\mathrm{d}A = -(17\,000)(2000) = -3.40 \times 10^{7}\ \text{N}\cdot\text{mm}$$Both are negative: the 20 kN acts upward so the upper stringers are compressed, and the 17 kN acts in $+Z$ so the stringers on the $+Z$ side are compressed too.
4. Solve for the stress field. Substituting into $aI_{ZZ}+bI_{YZ} = M_{Y}$ and $aI_{YZ}+bI_{YY} = M_{Z}$,$$a = -0.19164, \qquad b = -1.14613\ \text{N/mm}^{3}$$
5. Evaluate the stringer loads (part b). Each stringer carries $P_{r} = \sigma_{r}B_{r} = \left(aZ_{r} + bY_{r}\right)B_{r}$ with coordinates measured from the centroid. Working through the six stringers gives the loads tabulated below; the largest is $\boxed{P_{5} = 114.8\ \text{kN (tension)}}$ in the deep lower corner. The six loads sum to 0.000 N, which is the required zero net axial force and the arithmetic check on the whole calculation.
6. Differentiate along the span to get the load gradients (part a). Because the box is prismatic and both loads act at the tip, $a$ and $b$ grow linearly with distance from the free end, so $a' = a/2000$ and $b' = b/2000$ per millimetre and$$\frac{\mathrm{d}P_{r}}{\mathrm{d}x} = \left(a'Z_{r} + b'Y_{r}\right)B_{r}$$giving, in N/mm, $\mathrm{d}P_{5}/\mathrm{d}x = 57.39$, $\mathrm{d}P_{6}/\mathrm{d}x = 22.57$, $\mathrm{d}P_{1}/\mathrm{d}x = 1.70$, $\mathrm{d}P_{2}/\mathrm{d}x = -15.49$, $\mathrm{d}P_{3}/\mathrm{d}x = -29.00$, $\mathrm{d}P_{4}/\mathrm{d}x = -37.17$.
7. Build the open-section shear flow. Longitudinal equilibrium of each stringer requires the shear flows on either side of it to differ by its load gradient. Cutting panel 5–6 and walking 5→6→1→2→3→4→5, the accumulated basic flows are, in N/mm, $q_{b,56} = 0.00$, $q_{b,61} = 22.57$, $q_{b,12} = 24.28$, $q_{b,23} = 8.79$, $q_{b,34} = -20.22$, $q_{b,45} = -57.39$. The circuit closes back to zero after stringer 5, confirming the gradients.
8. Close the cell by moments. Taking moments about stringer 5, the applied loads give$$M_{\text{ext}} = (500)(20\,000) - (150)(17\,000) = 7.4500 \times 10^{6}\ \text{N}\cdot\text{mm}$$while the basic flows give $\sum q_{b}(2A) = 1.9271 \times 10^{6}$. The enclosed area of the cell is $A_{\text{cell}} = 225\,000$ mm$^{2}$ (a trapezium of base 1000 mm and depths 300 mm and 150 mm), so$$q_{s,0} = \frac{M_{\text{ext}} - \sum q_{b}(2A)}{2A_{\text{cell}}} = 12.273\ \text{N/mm}$$
9. Assemble and check the panel flows. Adding $q_{s,0}$ to each basic value gives the six answers tabulated below; the most heavily loaded panel is the deep front spar, $\boxed{q_{45} = -45.11\ \text{N/mm}}$, the negative sign meaning the flow runs from 5 up to 4 rather than in the walk direction. Resolving all six flows returns 20000 N vertically and 17000 N horizontally, reproducing the applied 20 kN and 17 kN exactly.