Question 8 of 8: Shear centre and shear flow of a four-boom box with a semi-elliptical nose

Question 8: Shear centre and shear flow of a four-boom box with a semi-elliptical 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 single-cell idealised box whose leading edge is a semi-ellipse, symmetric about its horizontal centreline.

Find. (a) the position of the shear centre, and (b) the shear flow in each of the four walls when the load acts 100 mm forward of it.

Approach. Set the origin at the mid-point of the 2–3 spar, take the section constants from the four booms, walk anticlockwise from a cut, close for zero twist to find the shear centre, then close again by moment equivalence with the load on its stated line.

1. Fix the nose geometry from the stated ratio. The semi-ellipse spans from boom 2 at $Y=+100$ to boom 3 at $Y=-100$, so its vertical semi-axis is the minor one at 100 mm and the major radius, being twice that, is $$a_{e} = 200\ \text{mm}$$ projecting forward of the 2–3 spar. This is the one place the ratio matters, and it fixes both the arc length and the swept area used later.
2. Compute the section constants. Booms 1 and 4 are equal and so are booms 2 and 3, so the centroidal axis is the horizontal centreline and the product term vanishes: $$\begin{aligned} I_{YY} &= \sum B_{r}Y_{r}^{2} = \left(600+450+450+600\right)\left(100\right)^{2} = \boxed{2.10\times10^{7}\ \text{mm}^{4}}, \\ I_{YZ} &= 0 \end{aligned}$$ The nose walls contribute nothing to $I_{YY}$ because, by the idealisation, the walls carry no direct stress at all.
3. Measure the nose arc length and the enclosed area. An elliptic arc has no elementary length, so it is integrated numerically; the half-perimeter comes to $$s_{23} = 484.42\ \text{mm}$$ which Ramanujan’s approximation $\pi\left(a_{e}+b_{e}\right)\left[1+3h/\left(10+\sqrt{4-3h}\right)\right]/2$ with $h=\left(a_{e}-b_{e}\right)^{2}/\left(a_{e}+b_{e}\right)^{2}$ reproduces to five figures. The cell area is the rectangle plus the half ellipse, $$A_{\text{cell}} = 500\times200 + \tfrac{1}{2}\pi\left(200\right)\left(100\right) = 100\,000+31\,416 = 131\,416\ \text{mm}^{2}$$ so the nose adds nearly a quarter of the whole enclosed area — chording it off would be a serious error.
4. Cut the cell and walk the booms. Walking $1\to2\to3\to4\to1$ takes the circuit anticlockwise as drawn. With the cut in panel 1–2 and $S_{Y}/I_{YY} = 10\,000/2.10\times10^{7} = 4.7619\times10^{-4}$, the steps $\Delta q_{r}=-\left(S_{Y}/I_{YY}\right)B_{r}Y_{r}$ at booms 2, 3 and 4 are $-21.429$, $+21.429$ and $+28.571$ N/mm, giving $$q_{b} = \left\lbrace 0,\ -21.429,\ 0,\ +28.571\right\rbrace\ \text{N/mm}$$ for walls 1–2, 2–3, 3–4 and 4–1. The step at boom 1 closes the walk back to zero.
5. Close for zero twist to find the shear centre. The gauge is uniform at 1.8 mm, so $\oint q\,ds/t = 0$ becomes $\sum q_{i}L_{i}=0$ with wall lengths 500, 484.42, 500 and 200 mm: $$q_{s,0} = -\frac{\left(-21.429\right)\left(484.42\right)+\left(28.571\right)\left(200\right)}{500+484.42+500+200} = -\frac{-4666.2}{1684.42} = \boxed{+2.770\ \text{N/mm}}$$ so the no-twist flows are $\left\lbrace 2.770,\ -18.658,\ 2.770,\ 31.342\right\rbrace$ N/mm.
6. Take moments about the mid-point of the 2–3 spar. Each wall contributes its flow times twice the area it sweeps about that point. The straight walls give $50\,000$, $50\,000$ and $100\,000$ mm², and the nose sweeps twice the half-ellipse area, $2\times31\,416 = 62\,832$ mm² — the exact swept-area rule handles the curve with no chording. Hence $$M = 2.770\left(50\,000\right)+\left(-18.658\right)\left(62\,832\right)+2.770\left(50\,000\right)+31.342\left(100\,000\right) = 2.2388\times10^{6}$$ $$z_{SC} = \frac{M}{S_{Y}} = \frac{2.2388\times10^{6}}{10\,000} = \boxed{223.9\ \text{mm aft of the 2--3 spar}}$$ equivalently 276.1 mm forward of the rear spar, and 423.9 mm aft of the nose tip.
7. Place the actual load and find the torque. The 10 kN acts 100 mm to the left of the shear centre, that is at $z = 223.9-100 = 123.9$ mm aft of the 2–3 spar. An upward force forward of the shear centre twists the section nose-down, so $$T = -100\times10\,000 = -1.00\times10^{6}\ \text{N}\cdot\text{mm}$$ $$q_{T} = \frac{T}{2A_{\text{cell}}} = \frac{-1.00\times10^{6}}{262\,832} = -3.805\ \text{N/mm}$$ on the anticlockwise convention of the walk.
8. Superpose and convert to stresses. Adding the torsional flow to the no-twist flows, $$q = \left\lbrace \boxed{-1.03},\ \boxed{-22.46},\ \boxed{-1.03},\ \boxed{+27.54}\right\rbrace\ \text{N/mm}$$ for walls 1–2, 2–3, 3–4 and 4–1. Dividing by the uniform 1.8 mm gauge gives shear stresses of 0.57, 12.48, 0.57 and 15.30 MPa, so the rear spar is the critical wall at $$\tau_{\max} = \boxed{15.30\ \text{MPa}}$$ The two skins are almost unloaded because the bending contribution and the torsional contribution very nearly cancel in them.
9. Check the result against statics. The two vertical walls must carry the whole shear: the rear spar gives $27.54\times200 = 5507$ N upward and the nose, whose end-to-end vertical projection is also 200 mm, gives $22.46\times200 = 4493$ N upward, totalling 10 000 N. Taking moments about the 2–3 spar mid-point with the same flows returns $1.2388\times10^{6}$ N·mm, which divided by 10 000 N places the resultant at 123.9 mm — exactly the stated line of action. Those two checks close the question.