Question 4 of 8: Shear centre and shear flow in an idealised nose box

Question 4: Shear centre and shear flow in an idealised nose 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.

Measure $x$ positive forward from the rear spar and $y$ positive upward from the axis of symmetry. The nose semicircle is centred at $(500,0)$ and so reaches forward to $x = 600$ mm.

Find. (a) the position of the shear centre, and (b) the shear flow in each of the four walls when the 15 kN load acts 120 mm forward of that point.

Approach. Cut the closed cell to get a statically determinate open flow, restore continuity with a constant closing flow found from the zero-twist condition, then locate the resultant by taking moments; part (b) superposes the pure torque produced by moving the load.

1. Compute the section second moment from the booms. The walls are effective in shear only, so all direct stress is carried by the four booms:$$I_{xx} = \sum B_r y_r^2 = 2(500)(100)^2 + 2(450)(100)^2 = 1.90\times10^{7}\ \text{mm}^4$$Because every boom sits at the same distance from the neutral axis, this is simply the total boom area times $100^2$.
2. Walk the open shear flow from a cut in the rear spar. Cutting wall 4–1 makes $q_b = 0$ there, and for a section with $I_{xy} = 0$ and $S_x = 0$ each boom crossed changes the flow by a fixed step,$$q_b^{(n+1)} = q_b^{(n)} - \frac{S_y}{I_{xx}}\,B_n y_n$$Because the walls carry no direct stress the flow is piecewise constant — there is no running integral to evaluate. Starting past boom 1 and walking 1→2→3→4:$$q_b(1\text{-}2) = -39.47,\quad q_b(2\text{-}3) = -75.00,\quad q_b(3\text{-}4) = -39.47,\quad q_b(4\text{-}1) = 0\ \ \text{N/mm}$$The walk closes exactly on zero at the cut, which is the arithmetic check on this step.
3. Restore continuity with the zero-twist condition (part a). For the load to pass through the shear centre the cell must not twist, so $\oint q/(Gt)\,\mathrm{d}s = 0$. With $G$ and $t$ both constant this collapses to a length-weighted mean and needs neither value:$$q_{s0} = -\frac{\oint q_b\,\mathrm{d}s}{\oint \mathrm{d}s} = -\frac{-63\,036}{1514.16} = +41.63\ \text{N/mm}$$The wall lengths are 500, $\pi(100) = 314.16$, 500 and 200 mm, summing to the perimeter 1514.16 mm.
4. Assemble the shear-centre flow set and check equilibrium. Adding $q_{s0}$ to each open flow gives$$q(1\text{-}2) = +2.16,\quad q(2\text{-}3) = -33.37,\quad q(3\text{-}4) = +2.16,\quad q(4\text{-}1) = +41.63\ \ \text{N/mm}$$A constant flow along a wall has a resultant equal to $q$ times the wall's chord, so the rear spar carries $41.63(200) = 8326$ N upward and the nose arc carries $33.37(200) = 6674$ N upward through its own chord; together exactly 15 000 N, while the top and bottom skins cancel horizontally. Overlooking the arc's chord is the easiest way to lose this check.
5. Locate the shear centre by moments. Taking moments about the mid-point of the rear spar, each wall contributes $q\times 2A_{\text{swept}}$, where the swept areas are $-50\,000$, $-131\,416$, $-50\,000$ and $0$ mm$^2$ (their sum is twice the enclosed area, $2\times115\,708$ mm$^2$, which confirms the geometry). Hence$$M_O = \sum q_i\,(2A_i) = 4.1695\times10^{6}\ \text{N}\cdot\text{mm}\quad\Longrightarrow\quad\boxed{e = \frac{M_O}{S_y} = 278.0\ \text{mm forward of the rear spar}}$$By the horizontal symmetry of the box the shear centre lies on the centreline, so no second coordinate is needed.
6. Superpose the torque for part (b). Shifting the same 15 kN load 120 mm further forward adds a pure torque about the shear centre,$$T = S_y e_{\text{off}} = 15\,000(120) = 1.80\times10^{6}\ \text{N}\cdot\text{mm}$$which a single cell carries as a uniform circulating flow$$q_T = \frac{T}{2A} = \frac{1.80\times10^{6}}{231\,416} = 7.778\ \text{N/mm}$$acting anticlockwise, i.e. $-7.778$ N/mm in the walk sense used above.
7. Add the two flow sets. Superposing gives the answer to part (b):$$\boxed{q(1\text{-}2) = -5.62,\quad q(2\text{-}3) = -41.15,\quad q(3\text{-}4) = -5.62,\quad q(4\text{-}1) = +33.85\ \ \text{N/mm}}$$The largest magnitude is 41.15 N/mm in the nose skin, a shear stress of $41.15/2 = 20.57$ MPa. Moving the load forward has relieved the rear spar by 19 per cent and increased the nose flow by 23 per cent.

It is worth pausing on why the shear centre sits so far forward — 278 mm ahead of the spar, more than half way to the nose. The rear spar is short (200 mm) and therefore an inefficient place to carry vertical load, whereas the nose arc presents the same 200 mm chord while also enclosing a large area. The vertical load therefore splits roughly evenly between the two, and the resultant of that pair acts well forward of the spar. This is the structural reason real wing torsion boxes are closed at a nose spar: the closed nose cell both carries shear and pulls the shear centre forward towards the aerodynamic centre, reducing the torsion the structure has to resist in flight.