Shear centre and shear flow in an idealised wing box with a semicircular nose: 22-Mec-B9 Advanced Engineering Structures, May 2017, Q8

Question 8: Shear centre and shear flow in an idealised wing box with a semicircular 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, symmetric about the horizontal axis, with four booms carrying all the direct stress and walls effective in shear only.

Given data
Quantity	Value
Boom areas 1 and 4	750 mm² each
Boom areas 2 and 3	600 mm² each
Half-depth (all booms)	100 mm above or below the axis
Spar spacing 2–3 to 1–4	500 mm
Nose wall 2–3	semicircular, $R = 100$ mm
Wall thickness	1.75 mm throughout
Vertical shear force	$S_y = 15\,000$ N upward

Find. (a) the position of the shear centre; (b) the shear flow in every wall when the 15 kN acts 100 mm to the left of that point.

The idealised box. Booms 1 and 4 (750 mm²) sit on the rear spar, booms 2 and 3 (600 mm²) at the ends of the semicircular nose. The computed shear centre is marked.

Approach. Because the walls carry no direct stress, the second moment of area comes from the booms alone. Cut the cell to get the open-section flows, close it with the zero-twist condition to locate the shear centre, then add the pure-torque flow produced by moving the load off that point.

Part (a) — compute the second moment of area from the booms. All four booms lie 100 mm from the horizontal axis of symmetry, so $$I_{xx} = \sum B_r y_r^2 = 2(750)(100)^2 + 2(600)(100)^2 = 2.70 \times 10^7\ \text{mm}^4.$$ The walls contribute nothing because they are assumed effective in shear only.
Cut the cell and walk the open-section flows. Making a cut in the rear spar 4–1 and walking the circuit 1→2→3→4→1 (which is counter-clockwise as drawn), the flow steps by $\Delta q = -(S_y/I_{xx})B_r y_r$ at each boom: $$q_{b,12} = -41.67, \quad q_{b,23} = -75.00, \quad q_{b,34} = -41.67, \quad q_{b,41} = 0\ \text{N/mm}.$$ The walk returning to zero at the cut is the arithmetic check on this step. Note that $q_b$ is constant between booms — there is no running integral, because the walls carry no direct stress to shed.
Close the cell with the zero-twist condition. The shear centre is the point at which the applied shear produces no twist, so $\oint (q_b + q_{s,0})/t\ \mathrm{d}s = 0$. With the thickness constant this reduces to a length-weighted average and needs neither $t$ nor $G$: $$q_{s,0} = -\frac{\sum q_b s}{\sum s}, \qquad \sum s = 500 + \pi(100) + 500 + 200 = 1514.2\ \text{mm},$$ giving $q_{s,0} = +43.08$ N/mm.
Add the two parts and check vertical equilibrium. The total flows when the load acts at the shear centre are $$q_{12} = q_{34} = +1.41, \quad q_{23} = -31.92, \quad q_{41} = +43.08\ \text{N/mm}.$$ Only the rear spar and the nose have a vertical extent, so $43.08(200) + (-31.92)(-200) = 8616 + 6384 = 15\,000$ N, which recovers the applied shear exactly.
Take moments to locate the shear centre. Working about the mid-point of the rear spar, the moment of a wall flow is $q$ times twice the area it sweeps. For the semicircular nose the swept area must be integrated, not chorded: $$2A_{23} = \oint (x\,\mathrm{d}y - y\,\mathrm{d}x) = \pi R^2 + 2 L R = 31\,416 + 100\,000 = 131\,416\ \text{mm}^2,$$ while the top and bottom skins each sweep $2A = LR = 50\,000\ \text{mm}^2$ and the spar sweeps nothing. Summing, $\sum q\,(2A) = -4.054 \times 10^6$ N·mm, so $$\boxed{x_{SC} = \frac{-4.054 \times 10^6}{15\,000} = 270.2\ \text{mm forward of the rear spar 1–4}},$$ that is 229.8 mm aft of the line joining booms 2 and 3, on the axis of symmetry.
Part (b) — convert the offset into a torque. Moving the 15 kN load 100 mm to the left (forward) of the shear centre applies, in addition to the shear, a torque about the shear centre of $$T = -100 \times 15\,000 = -1.50 \times 10^6\ \text{N}\cdot\text{mm},$$ the minus sign meaning it acts clockwise in the sense of the circuit chosen above.
Add the constant torsional flow. The enclosed area of the cell is $A_{\text{cell}} = 500(200) + \tfrac{1}{2}\pi(100)^2 = 115\,708\ \text{mm}^2$, so the Bredt–Batho flow is $$q_T = \frac{T}{2A_{\text{cell}}} = \frac{-1.50 \times 10^6}{231\,416} = -6.48\ \text{N/mm},$$ and superposing it on the shear-centre flows gives the answer: $$\boxed{q_{12} = q_{34} = -5.07, \quad q_{23} = -38.40, \quad q_{41} = +36.60\ \text{N/mm}}.$$
Verify the final field. Vertical equilibrium still holds, because a pure torque adds no net force: $36.60(200) + (-38.40)(-200) = 15\,000$ N. Taking moments again about the mid-spar point returns $-5.554 \times 10^6$ N·mm, which equals $15\,000 \times (-370.2)$ mm — the load line 370.2 mm forward of the spar, as specified. With $t = 1.75$ mm throughout, the largest shear stress is $38.40/1.75 = 21.9$ MPa in the nose.

The two statical checks used above — resolving the wall flows vertically and taking their moment about a fixed point — are worth applying to every shear-flow problem of this kind. They cost very little and they catch the sign errors that formula matching alone will not, particularly on the curved wall, where chording the semicircle instead of integrating it would understate its swept area by 24 % and move the shear centre by more than 60 mm.

Final results
Quantity	Symbol	Value
Second moment of area	$I_{xx}$	$2.70 \times 10^7$ mm⁴
Open-section flows (cut at 4–1)	$q_b$	$-41.67$, $-75.00$, $-41.67$, 0 N/mm
Closing flow (zero twist)	$q_{s,0}$	$+43.08$ N/mm
(a) Shear centre	$x_{SC}$	270.2 mm forward of spar 1–4
Cell area	$A_{\text{cell}}$	115 708 mm²
Torque about the shear centre	$T$	$-1.50 \times 10^6$ N·mm
Torsional flow	$q_T$	$-6.48$ N/mm
(b) Wall 1–2 and 3–4	$q$	$-5.07$ N/mm
(b) Nose wall 2–3	$q$	$-38.40$ N/mm
(b) Rear spar 4–1	$q$	$+36.60$ N/mm
(b) Maximum shear stress	$\tau_{\max}$	21.9 MPa in the nose

Topic. Shear flow and shear centre of an idealised single-cell box.

Key relations. $I_{xx} = \sum B_r y_r^2$; $\Delta q_b = -(S_y/I_{xx})B_r y_r$ at each boom; zero twist gives $q_{s,0} = -\oint (q_b/t)\mathrm{d}s / \oint (\mathrm{d}s/t)$; moment equivalence $S_y x_{SC} = \sum q(2A)$; pure torque adds $q_T = T/2A_{\text{cell}}$.

Why this works. Structural idealisation lumps the direct-stress capacity of the skins into discrete booms and leaves the walls carrying shear only. That makes $q_b$ piecewise constant, so the open-section shear flow can be walked boom by boom instead of integrated. A closed cell is once redundant in shear, and the redundancy is resolved by compatibility: at the shear centre, by definition, the section does not twist. Once the shear centre is known, any other line of action is handled by superposing a Bredt–Batho torsional flow, because shear and torsion are uncoupled about that point.

Common pitfall. Chording a curved wall when computing its swept area. The exact result for a circular arc, $\oint p\,\mathrm{d}s = \pi R^2 + 2 z_O R$ about a centre a distance $z_O$ away, has no approximation in it and is no harder to use. A second trap is using the zero-twist flows for part (b) instead of adding the torsional flow — here that would flip the sign of the two skin flows.

Source. T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed., Ch. 17.3 (shear of closed section beams), Ch. 17.4 (shear centre) and Ch. 20 (structural idealisation).

Question 8 of 8: Shear centre and shear flow in an idealised wing box with a semicircular nose

Question 8: Shear centre and shear flow in an idealised wing box with a semicircular nose (20 marks)