Question 8 of 8: Shear centre and wall shear flows of a four-boom box with an elliptical leading edge

Question 8: Shear centre and wall shear flows of a four-boom box with an 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 single-cell box of four booms and four shear-only walls, one of which is a semi-ellipse.

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

Approach. The section is symmetric about the horizontal axis, so a vertical shear alone drives the flow; cut a wall, step the flow at each boom, close the cell by zero twist to find the shear centre, and then add the constant torsional flow that the 100 mm offset demands.

1. Part (a) — geometry of the cell. The nose is half an ellipse of semi-axes $a = 200$ mm (streamwise) and $b = 100$ mm (vertical), so $$\begin{aligned}A_{\text{nose}} &= \tfrac12\pi a b = \tfrac12\pi(200)(100) = 31\,416\ \text{mm}^{2}, \\ A_{\text{rect}} &= 500(200) = 100\,000\ \text{mm}^{2}\end{aligned}$$$$\begin{aligned}A &= 131\,416\ \text{mm}^{2}, \\ s_{23} &= \int_0^{\pi}\sqrt{a^{2}\cos^{2}\phi + b^{2}\sin^{2}\phi}\;\mathrm{d}\phi = 484.4\ \text{mm}\end{aligned}$$ The elliptic arc length has no elementary closed form; Ramanujan’s approximation gives 484.4 mm, which agrees with the numerical value to four figures.
2. Second moment of area of the boom idealisation. All four booms are 100 mm from the axis of symmetry: $$I_{xx} = \sum B_r y_r^{2} = 2(500)(100)^{2} + 2(400)(100)^{2} = \boxed{18\times10^{6}\ \text{mm}^{4}}$$ The walls contribute nothing, by the terms of the idealisation.
3. Cut the cell and walk the basic flow. Cutting wall 1-2 and walking $1\!\to\!2\!\to\!3\!\to\!4\!\to\!1$ (forward along the top, round the nose, aft along the bottom, up the rear spar), the flow steps at each boom by $-\dfrac{S_y}{I_{xx}}B_r y_r$ with $S_y = 10\,000$ N: $$\begin{aligned}q_{b,12} &= 0, \\ q_{b,23} &= -\frac{10\,000(400)(100)}{18\times10^{6}} = -22.222, \\ q_{b,34} &= 0, \\ q_{b,41} &= 27.778\ \ \text{N/mm}\end{aligned}$$ The walk closes on itself at boom 1, confirming the arithmetic.
4. Close the cell by zero twist. The shear centre is by definition the point at which the load produces no rotation, so $$q_{s,0} = -\frac{\sum q_{b,i}s_i/t_i}{\sum s_i/t_i} = -\frac{-22.222(484.4) + 27.778(200)}{500 + 484.4 + 500 + 200} = 3.0927\ \text{N/mm}$$ The thickness is uniform, so it cancels; with different wall gauges each length would be divided by its own thickness.
5. Shear-centre wall flows and the vertical check. $$\begin{aligned}q_{12} &= q_{34} = 3.093, \\ q_{23} &= -19.130, \\ q_{41} &= 30.870\ \ \text{N/mm}\end{aligned}$$ The rear spar carries $30.870(200) = 6174$ N and the nose wall the remaining 3 826 N (its vertical resultant is $q_{23}$ times the 200 mm change in height), giving 10 000 N in total $\checkmark$.
6. Moments about the forward spar locate the shear centre. Take moments about the mid-point of the line joining booms 2 and 3. The moment of a constant flow along any wall is $q$ times twice the area swept by the radius from that point, so the two straight skins contribute $q_{12}$ and $q_{34}$ times $2(500)(100)$, the rear spar contributes $q_{41}$ times $2(500)(100)$, and the curved nose contributes $q_{23}\!\left(2A_{\text{nose}}\right)$ — exactly, with no chordwise approximation: $$S_y\,\xi_S = \sum q_i(2A_i) = 2\,194\,370\ \text{N}\!\cdot\!\text{mm} \;\Rightarrow\; \xi_S = \boxed{219.4\ \text{mm aft of the 2-3 spar}}$$ equivalently 280.6 mm forward of the 1-4 spar.
7. Part (b) — the offset load adds a torque. The 10 kN acts 100 mm to the left of (that is, forward of) the shear centre, so about the shear centre it produces $$T = S_y(-100) = -1\,000\,000\ \text{N}\!\cdot\!\text{mm}$$ a nose-down torque on the cell. Its line of action is at $219.4 - 100 = 119.4$ mm aft of the 2-3 spar.
8. Superpose the constant torsional flow. For a single cell, $$\Delta q = \frac{T}{2A} = \frac{-1\,000\,000}{2(131\,416)} = -3.8047\ \text{N/mm}$$ that is 3.805 N/mm circulating opposite to the assumed walking direction, and it is added to every wall alike.
9. Final wall shear flows and their checks. $$\begin{aligned}q_{12} &= q_{34} = -0.712, \\ q_{23} &= -22.934, \\ q_{41} &= 27.066\ \ \text{N/mm}\end{aligned}$$ Vertical equilibrium still returns 10 000 N, because the added flow is self-equilibrating; the moment of the flows about the 2-3 spar is now 1,194,370 N·mm, which equals $10\,000(119.4)\ \checkmark$. With $t = 1$ mm the shear stresses are numerically equal to the flows, the largest being $27.07$ MPa in the rear spar.