22-Mec-B9 Advanced Engineering Structures · May 2016
Question 6 of 8: Shear centre and panel flows of a six-boom idealised wing box
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2016 — 07-Mec-B9 Advanced Engineering Structures. Three hours, open book, any non-communicating calculator permitted. Eight questions of equal total value (20 marks each); the rubric states that any five constitute a complete paper. All eight are solved here, because the complete set is the study resource.
Reference texts.
T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. — open and closed thin-walled sections, shear flow, shear centre, structural idealisation, multi-cell torsion (Ch. 16, 17, 18, 20, 21, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — yield criteria, torsion of thin-walled members, unsymmetrical bending (Ch. 2, 4, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of multiply connected cells, columns and stability (Ch. 6, 12).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — linear elastic fracture mechanics, the Paris law, strain-life fatigue and Miner’s rule (Ch. 8, 11, 14).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — Euler buckling of pin-ended struts and combined loading (Ch. 8, 13).
Axes and sign convention used throughout. A right-handed set is used for every thin-walled question: the span axis runs along the member, $Y$ is vertical (upward) in the cross-section and $Z$ is horizontal (to the right). Section constants are written $I_{ZZ}=\int Z^{2}\,dA$, $I_{YY}=\int Y^{2}\,dA$ and $I_{YZ}=\int ZY\,dA$ about the centroid, and the bending stress is carried as the linear field $\sigma = aZ + bY$, whose coefficients follow from the pair
with $M_{Y}\equiv\int\sigma Z\,dA$ and $M_{Z}\equiv\int\sigma Y\,dA$. This one pair replaces the Megson fraction and handles symmetric and unsymmetrical sections alike. Shear flows are counted positive in the direction of the walk stated with each figure; a negative answer simply means the flow runs the other way round the circuit. Moments and torques are anticlockwise-positive in the $Z$–$Y$ plane.
Question 6: Shear centre and panel flows of a six-boom idealised wing box (20 marks)
Given. A single-cell rectangular box idealised into six booms and six shear panels, loaded vertically at one corner boom.
Given data — Question 6
Boom
Position $\left(Z,Y\right)$ from boom 6, mm
Area, mm2
1
(0, 200)
600
2
(400, 200)
400
3
(800, 200)
850
4
(800, 0)
850
5
(400, 0)
400
6
(0, 0)
600
Applied load: 5 000 N acting vertically upward at boom 3
Find. (a) the shear centre measured from boom 6, and (b) the shear flow in each of the six panels under the load as applied.
Figure 6.1 — the idealised box with the computed panel flows and the shear centre. Arrows show the sense of each flow; the load at boom 3 lies well outboard of the shear centre, so the section twists.
Approach. Because the boom pattern is mirror-symmetric about the horizontal centreline the section constants are trivial; cut one panel, walk round accumulating the boom contributions, close the cell twice — once with the no-twist condition to locate the shear centre, once with moment equivalence to place the real load.
Locate the centroidal axis and compute the second moment. Each top boom has an identical partner directly below it, so the horizontal centroidal axis lies at mid-depth, $Y = 100$ mm, and the product term $\sum BZY$ vanishes. Only the vertical bending stiffness is needed:
$$I_{YY} = \sum B_{r}Y_{r}^{2} = 2\left(600+400+850\right)\left(100\right)^{2} = \boxed{3.70\times10^{7}\ \text{mm}^{4}}$$
With $I_{YZ}=0$ a vertical shear produces no horizontal bending, which is what makes the walk below a single running sum.
Cut the cell and walk the booms. In an idealised section the shear flow is constant in each panel and steps at each boom by
$$\Delta q_{r} = -\frac{S_{Y}}{I_{YY}}B_{r}Y_{r}$$
with $Y_{r}$ measured from the centroidal axis. Putting the cut in panel 1–2 and walking $1\to2\to3\to4\to5\to6\to1$, the coefficient is $S_{Y}/I_{YY} = 5000/3.70\times10^{7} = 1.35135\times10^{-4}$, so the steps at booms 2, 3, 4, 5, 6 are $-5.405$, $-11.486$, $+11.486$, $+5.405$ and $+8.108$ N/mm.
Accumulate the open flows. Running those steps from the cut gives
$$q_{b} = \left\lbrace 0,\ -5.405,\ -16.892,\ -5.405,\ 0,\ +8.108\right\rbrace\ \text{N/mm}$$
for panels 1–2, 2–3, 3–4, 4–5, 5–6 and 6–1 in turn. The step at boom 1 closes the circuit back to zero, which is the arithmetic check. Notice that the two half-panels of the bottom skin carry equal and opposite steps to their top-skin partners, as the symmetry demands.
Close the cell for zero twist to locate the shear centre. The paper gives no panel thicknesses, so a uniform wall gauge and a single shear modulus are assumed throughout; on that basis the no-twist condition $\oint q\,ds/Gt = 0$ reduces to $\sum q_{i}L_{i} = 0$. With panel lengths 400, 400, 200, 400, 400 and 200 mm,
$$q_{s,0} = -\frac{\sum q_{b,i}L_{i}}{\sum L_{i}} = -\frac{-6081.1}{2000} = \boxed{+3.041\ \text{N/mm}}$$
and the shear-centre flows become $\left\lbrace 3.041,\ -2.365,\ -13.851,\ -2.365,\ 3.041,\ 11.149\right\rbrace$ N/mm.
Take moments about boom 6 to place the shear centre. Each panel contributes $q_{i}$ times twice the area it sweeps about the chosen point, and the two panels that pass through boom 6 sweep nothing. The remaining swept areas are $-80\,000$ mm2 for each half of the top skin and $-160\,000$ mm2 for the right-hand web, so
$$M_{6} = 3.041\left(-80\,000\right)+\left(-2.365\right)\left(-80\,000\right)+\left(-13.851\right)\left(-160\,000\right) = 2.162\times10^{6}\ \text{N}\cdot\text{mm}$$
$$z_{SC} = \frac{M_{6}}{S_{Y}} = \frac{2.162\times10^{6}}{5000} = \boxed{432.4\ \text{mm from boom 6}}$$
The shear centre sits outboard of mid-width (400 mm) because booms 3 and 4, at 850 mm2 each, are the stiffest pair and pull the resultant toward them.
Convert the eccentric load into a shear at the shear centre plus a torque. The 5 kN acts at boom 3, that is at $z = 800$ mm, so the eccentricity is $800 - 432.4 = 367.6$ mm and
$$T = 5000\times367.6 = 1.838\times10^{6}\ \text{N}\cdot\text{mm}\ \text{(anticlockwise)}$$
The cell encloses $800\times200 = 160\,000$ mm2, so this torque needs a uniform flow of
$$q_{T} = \frac{T}{2A_{\text{cell}}} = \frac{1.838\times10^{6}}{320\,000} = 5.743\ \text{N/mm anticlockwise}$$
The walk $1\to2\to\cdots\to6\to1$ runs clockwise as the box is drawn, so this contributes $-5.743$ N/mm to every panel on our sign convention.
Superpose to get the answer to part (b). Adding the torsional flow to the shear-centre flows,
$$q = \left\lbrace \boxed{-2.70},\ \boxed{-8.11},\ \boxed{-19.59},\ \boxed{-8.11},\ \boxed{-2.70},\ \boxed{+5.41}\right\rbrace\ \text{N/mm}$$
for panels 1–2, 2–3, 3–4, 4–5, 5–6 and 6–1. The right-hand web, directly under the load, carries 19.59 N/mm, three and a half times the left-hand web at 5.41 N/mm; a symmetric box loaded at its shear centre would have shared the shear far more evenly.
Check the answer against statics. The vertical components must return the applied load: the right web gives $19.59\times200 = 3919$ N and the left web $5.41\times200 = 1081$ N, and both act upward, totalling 5 000 N. Taking moments about boom 6 with the same flows returns $4.00\times10^{6}$ N·mm, exactly $800\times5000$, so the resultant passes through boom 3 as required. Those two checks together confirm the closing constant and the sign convention.
