Question 8 of 8: Shear centre and shear flow of a box with a semi-elliptical leading edge

Question 8: Shear centre and shear flow of a box with a semi-elliptical leading edge (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 four-boom single cell. Booms 1 and 4 sit on the rear spar, 100 mm above and below the axis of symmetry; booms 2 and 3 sit on the front spar, 500 mm forward of them and also 100 mm from the axis. The leading edge is a semi-ellipse whose minor semi-axis is the 100 mm half-depth and whose major semi-axis is therefore 200 mm, so the nose projects 200 mm forward of the 2–3 line.

Find. (a) the chordwise position of the shear centre; (b) the shear flow in each of the four walls under the offset load.

Approach. Evaluate $I$ from the boom areas, cut the rear spar and walk the open flow boom by boom, close for zero twist using the wall lengths (the nose arc length is needed, and follows from a numerical evaluation of the elliptic integral), locate the shear centre by moments, then superpose the torque of the offset load.

1. Part (a) — evaluate the second moment. All four booms lie 100 mm from the axis of symmetry, so $$I_{ZZ} = \sum B_r y_r^{2} = \left(2 \times 600 + 2 \times 450\right)(100)^{2} = 2.10\times10^{7}\ \text{mm}^{4}.$$
2. Measure the leading-edge arc. An elliptic arc has no closed-form length, so the semi-perimeter of the ellipse with semi-axes 200 and 100 mm is integrated numerically, giving $484.42$ mm. Ramanujan's approximation returns $484.42$ mm as well, agreeing to five significant figures, which validates the numerical quadrature.
3. Cut the rear spar and walk the open shear flow. Stepping by $-(S_y/I)B_r y_r$ at each boom on the walk $1\to2\to3\to4$, $$q_{b,12} = -28.57, \qquad q_{b,23} = -50.00, \qquad q_{b,34} = -28.57, \qquad q_{b,41} = 0 \ \ \text{N/mm}.$$
4. Close the cell for zero twist. With constant thickness the twist condition is $\oint q\,\mathrm{d}s = 0$, and the four wall lengths are 500, 484.42, 500 and 200 mm, so $$q_{s,0} = -\frac{\sum q_b \ell}{\sum \ell} = \frac{52\,792}{1684.4} = +31.34\ \text{N/mm},$$ giving zero-twist flows of $+2.77$, $-18.66$, $+2.77$ and $+31.34$ N/mm.
5. Take moments about the rear spar. Choosing the mid-point of the rear spar as the reference kills the moment of the spar flow. For the straight walls twice the swept area is $Z_1Y_2-Z_2Y_1$; for the nose the same integral evaluated round the ellipse gives $162\,832$ mm$^{2}$, and the four contributions sum to $2A_{\text{cell}} = 262\,832$ mm$^{2}$, matching $2\left[500(200)+\tfrac{\pi}{2}(200)(100)\right]$ exactly.
6. Locate the shear centre. The moment of the zero-twist flows about that reference is $-2.761\times10^{6}\ \text{N}\cdot\text{mm}$, so $$\boxed{z_{SC} = 276.1\ \text{mm forward of the rear spar} \ \ (223.9\ \text{mm aft of the 2--3 line})}$$ The centre lies forward of mid-chord because the nose contributes almost a quarter of the enclosed area while carrying comparatively little of the direct stress.
7. Part (b) — find the torque of the offset load. The 10 000 N force acts 100 mm forward of the shear centre, so about that centre it applies $$T = 10\,000 \times (-100) = -1.00\times10^{6}\ \text{N}\cdot\text{mm},$$ the sign denoting a nose-down (clockwise) couple in the chosen frame.
8. Convert the torque into a circulating flow and superpose. $q_T = T/2A_{\text{cell}} = -1.00\times10^{6}/262\,832 = -3.805$ N/mm, which adds uniformly to every wall: $$\boxed{q_{12} = -1.03,\quad q_{23} = -22.46,\quad q_{34} = -1.03,\quad q_{41} = +27.54 \ \ \text{N/mm}}$$
9. Check the answer statically. Summing $q\,\Delta y$ over the four walls returns $10\,000.0$ N upward, and the moment about the rear spar comes to $-3.761\times10^{6}\ \text{N}\cdot\text{mm}$, exactly $10\,000 \times 376.1$ mm — the load acting 376.1 mm forward of the spar, which is 276.1 + 100 mm as required.
10. Read the load path. The rear spar carries by far the largest flow, 27.54 N/mm, because the bending shear and the torsional circulation reinforce one another there, while the two cover panels almost cancel to 1.03 N/mm. Moving the load only 100 mm forward of the shear centre is enough to change the nose flow by 20 % — a reminder of how sensitive a single-cell nose box is to the chordwise position of the load.