Shear centre and shear flow in an idealised wing box: 22-Mec-B9 Advanced Engineering Structures, May 2018, Q3

Question 3: Shear centre and shear flow in an idealised wing 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. A horizontally symmetric idealised wing box of constant wall thickness 1 mm. Booms 1 and 4 sit on the rear spar at ±120 mm from the horizontal axis of symmetry; booms 2 and 3 sit 600 mm forward of them at the same heights, and wall 2–3 is a semicircular nose of radius 120 mm. The walls carry shear only, so the booms carry all of the direct stress.

Quantity	Value
Boom areas B₁ = B₄	500 mm²
Boom areas B₂ = B₃	400 mm²
Box depth (2 × 120 mm)	240 mm
Spar-to-spar distance	600 mm
Nose radius R	120 mm
Wall thickness t	1 mm
Applied vertical force S_y	10,000 N upward

Find. (a) the chordwise position of the shear centre, and (b) the shear flow in every wall when the 10 kN acts 100 mm to the left of (that is, forward of) the shear centre.

Question 3 — the idealised box. The circuit 1–2–3–4 is walked anticlockwise and the section is cut in the rear spar 4–1, which is the wall whose flow is wanted last.

Approach. Idealise, cut, close. Compute $I_{xx}$ from the booms alone, walk the open-section flows round the cut box, restore continuity with a constant flow $q_{s,0}$ chosen for zero twist (that locates the shear centre), then superpose the pure torque produced by moving the load off the shear centre.

Second moment of area from the booms. All four booms lie at $y = \pm 120$ mm from the axis of symmetry, so $$I_{xx} = \sum B_i y_i^2 = 2(500)(120)^2 + 2(400)(120)^2 = 2.592\times10^7\ \text{mm}^4$$ The walls contribute nothing because they are declared effective in shear only.
Open-section shear flows. Cut the box in the rear spar 4–1 and accumulate the boom terms anticlockwise from boom 1: $$q_b = -\frac{S_y}{I_{xx}}\sum B_i y_i$$ Boom 1 gives $q_{b,12} = -(10\,000/2.592\times10^7)(500)(120) = -23.15$ N/mm; adding boom 2 gives $q_{b,23} = -41.67$ N/mm; adding boom 3 returns $q_{b,34} = -23.15$ N/mm, and boom 4 closes the walk at zero as it must at a cut.
Close the section for zero twist. With one constant thickness the rate-of-twist condition $\oint (q/t)\,\mathrm{d}s = 0$ reduces to a length-weighted average: $$q_{s,0} = -\frac{\sum q_b L}{\sum L}, \qquad L_{23} = \pi R = 377\ \text{mm}, \quad \sum L = 1817\ \text{mm}$$ $$q_{s,0} = +23.93\ \text{N/mm}$$
Shear flows when the load passes through the shear centre. Adding $q_{s,0}$ to each open flow, $$q_{12} = q_{34} = 0.78,\quad q_{23} = -17.73,\quad q_{41} = 23.93\ \text{N/mm}$$ Vertical equilibrium checks the set: the rear spar and the nose between them carry $(23.93 + 17.73)(240) = 10\,000$ N.
Locate the shear centre (part a). Take moments of those flows about the mid-point of the rear spar, using the swept area $2A$ of each wall — for the nose, $2A_{23} = \pi R^2 + 2 L R = 189\,239$ mm²: $$x_{sc} = \frac{\sum q\,(2A)}{S_y} = \boxed{324\ \text{mm forward of the rear spar}}$$ It lies 24 mm forward of the mid-point between the boom stations (300 mm), pulled forward by the closed semicircular nose, but still 36 mm aft of mid-chord (360 mm from the rear spar to the nose tip).
Add the torque for part (b). Moving the 10 kN 100 mm forward of the shear centre applies a pure torque $T = 10\,000 \times 100 = 10^6\ \text{N}\cdot\text{mm}$. Bredt’s formula spreads it as a constant flow round the single cell, whose enclosed area is $A = \tfrac{1}{2}(333\,239) = 166\,619$ mm²: $$q_T = \frac{T}{2A} = \frac{-10^6}{333\,239} = -3.00\ \text{N/mm}$$ the sign following the sense of the offset.
Superpose (part b). $$q_{12} = q_{34} = -2.22,\qquad q_{23} = -20.73,\qquad q_{41} = \boxed{20.93\ \text{N/mm}}$$ and with $t = 1$ mm the largest wall shear stress is 20.9 MPa in the rear spar. The vertical resultant is unchanged at 10 kN, because a pure torque adds no net force.

Quantity	Value
I_xx (booms only)	2.592 × 10⁷ mm⁴
Closing flow q_s,0 for zero twist	23.93 N/mm
(a) Shear centre	324 mm forward of the rear spar 1–4, on the axis of symmetry
Torsional flow from the 100 mm offset	−3.00 N/mm
(b) q₁₂ = q₃₄	−2.22 N/mm
(b) q₂₃ (nose)	−20.73 N/mm
(b) q₄₁ (rear spar)	20.93 N/mm
Maximum wall shear stress (t = 1 mm)	20.9 MPa

Topic. Shear flow, shear centre and torsion in a single-cell idealised box.

Key relations. $q_b = -(S_y/I_{xx})\sum B_i y_i$ for a cut section; zero-twist closure $\oint (q/t)\,\mathrm{d}s = 0$; moment equivalence $S_y x_{sc} = \sum q\,(2A)$; Bredt torsion $q_T = T/2A$; nose swept area $2A = \pi R^2 + 2LR$.

Why this works. A closed cell is once redundant in shear, so the flow is written as a determinate part found by cutting the cell plus an unknown constant that restores continuity. Choosing that constant to give zero twist produces the flow set that a load through the shear centre would cause, and taking moments of that set locates the shear centre itself. Any load applied elsewhere is then the same force through the shear centre plus a pure torque, and Bredt’s formula handles the torque in one line. The whole method is superposition twice over.

Common pitfall. Forgetting that the swept area of a curved wall is not the area of the chord — for the semicircular nose $2A = \pi R^2 + 2LR$, not $2R \times L$. A second trap is closing the cell for zero shear flow at the cut rather than for zero twist; only the twist condition locates the shear centre.

Source. Megson, Aircraft Structures for Engineering Students, 6th ed., §17.2–17.3 (shear of closed idealised sections) and §18.1 (torsion of closed sections); Boresi & Schmidt, 6th ed., ch. 7.

Question 3 of 8: Shear centre and shear flow in an idealised wing box

Question 3: Shear centre and shear flow in an idealised wing box (20 marks)