22-Mec-B9 Advanced Engineering Structures · December 2014
Question 7 of 8: Shear centre and panel shear flows of an idealised wing box
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 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 (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, statically indeterminate axial members (Ch. 2, 4, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of open and closed sections, multiply connected cells (Ch. 6).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — linear elastic fracture mechanics, the Paris law, strain-life and Miner’s rule (Ch. 8, 11, 14).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — thermal stress in restrained members and combined loading (Ch. 4, 8).
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. 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 $\sigma = aZ + bY$. 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.
Question 7: Shear centre and panel shear flows of an idealised wing box (20 marks)
Given. A single-cell idealised box: the booms carry all the direct stress, the 1 mm panels carry only shear.
Quantity
Symbol
Value
Boom stations (measured from boom 6)
—
booms 1 and 6 at 0 mm, 2 and 5 at 500 mm, 3 and 4 at 1000 mm
Box depth
$h$
300 mm
Boom areas
$B_1 = B_6$
600 mm2
$B_2 = B_5$
400 mm2
$B_3 = B_4$
900 mm2
Panel thickness
$t$
1 mm (uniform)
Applied load
$S_Y$
4 000 N upward, at boom 3
Find. (a) the distance from boom 6 to the shear centre, and (b) the shear flow carried by each of the six panels under the load as actually applied.
Figure Q7 — the idealised box with boom areas and the final panel shear-flow magnitudes. The shear centre lies 542.5 mm from boom 6, so the load applied at boom 3 is 457.5 mm eccentric and adds a torsional flow.
Approach. Cut one panel to make the cell open and walk the boom-by-boom shear flow round it; restore continuity with the constant flow that gives zero twist, which is the shear-centre loading; take moments of that flow system about boom 6 to locate the shear centre; then add the uniform torsional flow demanded by the actual load position.
Section constants. The boom areas are symmetric top to bottom, so the centroid sits at mid-depth, $\bar{Y} = 150$ mm, and the product of inertia vanishes. Summing $B_r(Y_r-\bar{Y})^{2}$ over all six booms, each at 150 mm from the axis, $$I_{YY} = 3\,800 \times 150^{2} = 85.5\times10^{6}\ \text{mm}^{4}$$With $I_{YZ}=0$ the shear-flow step across a boom is simply $\Delta q_r = -(S_Y/I_{YY})B_r(Y_r-\bar{Y})$.
Part (a) — open the cell and walk the basic flow. Cut panel 6–1 and walk 1→2→3→4→5→6, taking that direction as positive. Each boom subtracts $(4000/85.5\times10^{6})B_r(\pm150)$, giving in turn $$q_b = -4.211,\ -7.018,\ -13.333,\ -7.018,\ -4.211,\ 0.000\ \text{N/mm}$$for panels 1–2, 2–3, 3–4, 4–5, 5–6 and 6–1. The return to zero at the cut is the arithmetic check on the walk.
Restore the cut with the zero-twist flow. By definition the shear centre is the load position that produces no twist, so the constant $q_{s,0}$ added round the cell must satisfy $\oint (q_b + q_{s,0})\,ds/t = 0$. With uniform thickness the perimeter integral is just the perimeter, 2 600 mm, and $\oint q_b\,ds/t = -15\,228$, hence $$q_{s,0} = \frac{15\,228}{2\,600} = +5.857\ \text{N/mm}$$Adding it gives the shear-centre panel flows $+1.646$, $-1.161$, $-7.476$, $-1.161$, $+1.646$ and $+5.857$ N/mm.
Locate the shear centre by taking moments. Summing $q\,\times$ twice the swept area of each panel about boom 6 gives a moment of $2.170\times10^{6}$ N·mm, which must equal $S_Y z_{SC}$: $$\boxed{z_{SC} = \frac{2.170\times10^{6}}{4\,000} = 542.5\ \text{mm from boom 6}}$$The result sits outboard of mid-span because booms 3 and 4 are the largest, which stiffens the right-hand web and draws more of the vertical shear into it; the flange-panel couple pulls the answer back from 560 mm to 542.5 mm.
Part (b) — the torque the actual load applies. The 4 kN acts at boom 3, at station 1 000 mm, so about the shear centre it applies $$T = 4\,000\,(1\,000 - 542.5) = 1.830\times10^{6}\ \text{N}\cdot\text{mm}$$which a closed cell carries as the uniform Bredt–Batho flow $q_T = T/2A$. With a cell area $A = 1000\times300 = 3.0\times10^{5}$ mm2, $|q_T| = 3.050$ N/mm, acting in the sense opposite to the assumed walk.
Total panel shear flows. Superposing the two systems gives, in the walk direction, $$q = -1.404,\ -4.211,\ -10.526,\ -4.211,\ -1.404,\ +2.807\ \text{N/mm}$$for panels 1–2, 2–3, 3–4, 4–5, 5–6, 6–1. In plain terms the circuit runs up both webs and closes across the covers, with the outboard web carrying by far the most: $$\boxed{|q|_{\max} = 10.53\ \text{N/mm in web 3--4},\qquad \tau_{\max} = 10.53\ \text{MPa}}$$since the walls are 1 mm thick.
Two statical checks. Resolving the final flows vertically returns exactly 4 000 N and horizontally zero, and their moment about boom 6 is $4\,000 \times 1\,000 = 4.00\times10^{6}$ N·mm — the moment of the load at its true station. Solving the constant directly from that moment equation gives the same $+2.807$ N/mm offset, an independent route to the same answer.
