Shear centre and panel flows of an idealised six-boom wing box: 22-Mec-B9 Advanced Engineering Structures, December 2016, Q4

Question 4: Shear centre and panel flows of an idealised six-boom 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 rectangular single-cell box 800 mm wide and 200 mm deep, idealised into six booms at the corners and mid-span of each cover, with the 5 kN upward load applied at boom 3 (the top right-hand corner).

Given data — boom areas and stations
Boom	Area (mm$^2$)	$z$ from boom 6 (mm)	$y$ from mid-depth (mm)
1	600	0	+100
2	400	400	+100
3	850	800	+100
4	850	800	−100
5	400	400	−100
6	600	0	−100
Applied shear $S_y$ = 5 000 N upward, acting at boom 3

Find. The spanwise station of the shear centre measured from boom 6, and then the actual shear flow in each of the six panels for the load as applied.

Figure 4.1 — the idealised six-boom box. The 5 kN load is applied at boom 3; the shear centre (SC) lies 432.4 mm aft of boom 6, so the load is eccentric by 367.6 mm and a torque is superposed.

Approach. Cut one panel to make the section statically determinate, walk the open-section flow from boom to boom, then close the cut twice: once with the zero-twist condition, which locates the shear centre, and once with moment equivalence about boom 6, which gives the true flows for the load at its actual station.

Compute the section constant. All six booms lie at $y = \pm100$ mm, so$$I = \sum B_r y_r^{2} = (600+400+850+850+400+600)(100)^{2} = 37.0\times10^{6}\ \text{mm}^4 .$$The section is symmetric about the horizontal axis, so $I_{yz} = 0$ and the simple form of the shear-flow equation applies. The panels carry no direct stress by definition of the idealisation, so they contribute nothing to $I$.
Walk the open-section flow from a cut. Cutting panel 6–1 (the front web) and stepping clockwise $1\!\to\!2\!\to\!3\!\to\!4\!\to\!5\!\to\!6$, each boom sheds$$\Delta q_r = -\frac{S_y}{I}B_r y_r = -\frac{5000}{37\times10^{6}}B_r y_r .$$Accumulating, $q_{12} = -8.108$, $q_{23} = -13.514$, $q_{34} = -25.000$, $q_{45} = -13.514$, $q_{56} = -8.108$ and $q_{61} = 0$ N/mm. The walk closes exactly back on zero at boom 6, which is the first check that the boom bookkeeping is right.
Close the cut with zero twist to find the shear centre (part a). With no thickness data supplied, the panels are taken to be of uniform gauge, so the no-twist condition $\oint q\,\mathrm{d}s/t = 0$ reduces to $\sum q_i L_i = 0$. With panel lengths 400, 400, 200, 400, 400 and 200 mm,$$q_{s,0} = -\frac{\sum q_{b,i}L_i}{\sum L_i} = -\frac{-22\,297}{2000} = +11.149\ \text{N/mm},$$giving the shear-centre flow set $+3.041$, $-2.365$, $-13.851$, $-2.365$, $+3.041$ and $+11.149$ N/mm.
Take moments about boom 6. The moment of a panel flow about a point is $q(z_a y_b - z_b y_a)$ with coordinates measured from that point. Summing over the six panels gives $2.162\times10^{6}$ N·mm, and equating that to the moment of the resultant 5 kN,$$z_{SC} = \frac{2.162\times10^{6}}{5000} = \boxed{432.4\ \text{mm from boom 6}} .$$The shear centre sits aft of mid-span (400 mm) because booms 3 and 4 are the largest, so the rear web is called on to carry more of the vertical shear than the front web.
Close the cut again by moment equivalence for the real load (part b). The 5 kN acts at boom 3, that is at $z = 800$ mm, so the closing constant follows from$$q_{s,0}\sum(z_a y_b - z_b y_a) = S_y z_{\text{load}} - \sum q_{b}(z_a y_b - z_b y_a).$$The walk is clockwise, so its signed swept area is negative, $\sum(z_a y_b - z_b y_a) = -320\,000$ mm$^2$ — this sign matters, and using $+2A$ here would reverse $q_{s,0}$ to $-5.405$ N/mm, flip the front web to $-5.41$ N/mm and inflate every other panel, so the moment check would miss by $2q_{s,0}(2A) = 3.46\times10^{6}$ N·mm. Substituting, $q_{s,0}(-320\,000) = 4.00\times10^{6} - 5.73\times10^{6}$, so $q_{s,0} = +5.405$ N/mm.
Superpose and check. Adding the closing constant to the open-section walk gives the final panel flows$$\boxed{q_{12}=-2.70,\ q_{23}=-8.11,\ q_{34}=-19.59,\ q_{45}=-8.11,\ q_{56}=-2.70,\ q_{61}=+5.41\ \text{N/mm}} .$$Two independent checks confirm the result. Vertically, only the two webs carry load: $19.59(200) + 5.41(200) = 5\,000$ N — both webs push upward — which is the applied shear exactly. And superposing the shear-centre solution with a pure torque $T = S_y(800-432.4) = 1.838\times10^{6}$ N·mm, for which $q_T = T/2A = -5.743$ N/mm, reproduces the same six numbers — the offset load is simply bending through the shear centre plus torsion.

Final results — Question 4
Quantity	Value
Second moment of area $I$	$37.0\times10^{6}$ mm$^4$
Shear centre from boom 6	432.4 mm
Closing constant $q_{s,0}$ (load at boom 3)	+5.405 N/mm
Panel 1–2 (top, front half)	−2.70 N/mm
Panel 2–3 (top, rear half)	−8.11 N/mm
Panel 3–4 (rear web)	−19.59 N/mm
Panel 4–5 (bottom, rear half)	−8.11 N/mm
Panel 5–6 (bottom, front half)	−2.70 N/mm
Panel 6–1 (front web)	+5.41 N/mm

Check: the paper gives boom areas but no panel gauges, so the zero-twist closure in step 3 assumes all six panels are of the same thickness. That assumption affects part (a) only; part (b) is closed by moment equivalence and is therefore independent of thickness. If gauges were supplied, replace $\sum q_i L_i = 0$ by $\sum q_i L_i / t_i = 0$ and the shear centre would shift toward the thinner webs.

Topic. Shear of an idealised single-cell box: open-section flow, cell closure, and the shear centre.

Key relations. $\Delta q_r = -(S_y/I_{zz})B_r y_r$ for a section symmetric about the horizontal axis; zero twist $\oint q\,\mathrm{d}s/t = 0$ locates the shear centre; moment equivalence $\sum q(z_ay_b - z_by_a) = S_y z_{\text{load}}$ closes the cell for an arbitrary load position; and $q_T = T/2A$ superposes a pure torque.

Why this works. Idealisation concentrates all direct-stress capacity into the booms, so the shear flow is piecewise constant and changes only where a boom sheds load. Cutting the cell removes the single redundancy and makes that walk possible; the closing constant restores it. Which physical condition supplies the closing constant depends on what is being asked — zero twist defines the shear centre, moment equivalence gives the true flows — and it is worth recognising that these are two different questions rather than two routes to one answer.

Common pitfall. Dividing the extra torque by $+2A$ regardless of the direction of the walk. If the boom sequence traverses the cell clockwise in a $z$-right, $y$-up frame, the signed swept area is negative, and using $+2A$ reverses the closing constant — here it flips the front web and inflates the other five panels — while leaving the vertical check apparently satisfied, because a constant circulating flow has no net force. Always recompute $\sum(z_ay_b-z_by_a)$ from the actual walk, and always finish with the two statical checks — vertical resultant and moment — before reporting.

Source. Megson, Aircraft Structures for Engineering Students, 6th ed., Ch. 20 (structural idealisation) and Ch. 17 (shear of closed section beams, shear centre).

Question 4 of 8: Shear centre and panel flows of an idealised six-boom wing box

Question 4: Shear centre and panel flows of an idealised six-boom wing box (20 marks)