22-Mec-B9 Advanced Engineering Structures · May 2013
Question 4 of 8: Shear centre and shear flow of an idealised wing box
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2013 — 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, structural idealisation, shear centre (Ch. 16, 17, 20, 21, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — unsymmetrical bending, yield criteria, torsion of non-circular bars (Ch. 4, 5, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of rectangular sections, columns and elastic stability (Ch. 6, 12).
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. — Euler buckling and combined loading (Ch. 8, 13).
Axes and sign convention used throughout. A right-handed set $(X, Y, Z)$ is used for every thin-walled question: $X$ runs along the span from the free end toward the root, $Y$ is vertical (upward) in the cross-section, and $Z$ is horizontal, positive to the right when the section is viewed looking along $+X$. Direct stress is written $\sigma = aZ + bY$ with $Z$ and $Y$ measured from the section centroid, so that the two constants follow from the stress resultants the section must carry, $M_{Y} = \int \sigma Z\,\mathrm{d}A$ and $M_{Z} = \int \sigma Y\,\mathrm{d}A$:
$$a I_{ZZ} + b I_{YZ} = M_{Y}, \qquad a I_{YZ} + b I_{YY} = M_{Z}$$
This form is used in preference to the memorised unsymmetrical-bending fraction because it is self-checking: the physical sign of every answer can be confirmed from the simple rule that, for a cantilever carrying a tip load, the fibres on the side the load points toward go into compression.
Question 4: Shear centre and shear flow of an idealised wing box (20 marks)
Given. A single-cell closed box, symmetric about its horizontal axis, with a semicircular nose, four booms carrying all the direct stress, and walls carrying shear only.
Quantity
Symbol
Value
Boom areas 1 and 4 (rear spar)
$B_{1}, B_{4}$
500 mm$^{2}$ each
Boom areas 2 and 3 (nose)
$B_{2}, B_{3}$
400 mm$^{2}$ each
Box depth (booms 1–4 spar)
$2R$
240 mm
Nose radius (wall 2–3 semicircular)
$R$
120 mm
Straight-wall length
$L$
600 mm
Wall thickness
$t$
1 mm throughout
Applied vertical shear (upward)
$S_{Y}$
10 000 N
Find. (a) the horizontal position of the shear centre, and (b) the shear flow in each of the four walls when the same shear force acts 100 mm to the left of that point.
The idealised box. Booms are numbered 1 (top right), 2 (top left), 3 (bottom left), 4 (bottom right); the circuit is walked in the order 1→2→3→4→1, which is counter-clockwise, and shear flows are reckoned positive in that direction.
Approach. Cut the cell to make it statically determinate, build the open-section shear flow from the boom areas, then close it twice over: once with the no-twist condition, which locates the shear centre, and once with a constant torsional flow that accounts for the offset load.
Compute the second moment of area. Only the booms carry direct stress, and all four sit 120 mm from the horizontal axis of symmetry:$$I_{YY} = \sum B_{r}Y_{r}^{2} = 2(500)(120)^{2} + 2(400)(120)^{2} = 2.5920 \times 10^{7}\ \text{mm}^{4}$$Symmetry about the horizontal axis makes $I_{YZ} = 0$, so the vertical shear produces no horizontal bending and the analysis stays uncoupled.
Cut the cell and build the open-section (basic) shear flow. Cutting wall 1–2 and walking counter-clockwise, the flow steps by $-\left(S_{Y}/I_{YY}\right)B_{r}Y_{r}$ as each boom is passed:$$q_{b,12} = 0, \quad q_{b,23} = -18.519, \quad q_{b,34} = 0, \quad q_{b,41} = 23.148\ \text{N/mm}$$The flow returns to zero after boom 1, which is the arithmetic check that the boom areas and coordinates have been entered correctly.
Impose zero twist to close the cell (part a). The shear centre is the point through which the load produces no twist, so$$\oint \frac{q_{b}+q_{s,0}}{t}\,\mathrm{d}s = 0 \;\Longrightarrow\; q_{s,0} = -\frac{\oint q_{b}\,\mathrm{d}s}{\oint \mathrm{d}s}$$because the thickness is constant. With wall lengths 600, $\pi R = 377.0$, 600 and 240 mm (perimeter 1817.0 mm),$$q_{s,0} = 0.7847\ \text{N/mm}$$
Assemble the no-twist flows. Adding $q_{s,0}$ to each basic value gives $q_{12} = 0.785$, $q_{23} = -17.734$, $q_{34} = 0.785$ and $q_{41} = 23.933$ N/mm. These resolve to $10000$ N vertically and zero horizontally, recovering the applied shear exactly — the statical check that the closure is right.
Take moments to locate the shear centre. Taking moments about boom 4, each wall contributes $q\times 2A$, where $2A$ is twice the area swept by the radius from boom 4. For the straight walls $2A_{12} = 144\,000$ mm$^{2}$ while walls 3–4 and 4–1 pass through the moment centre and contribute nothing; for the semicircular nose, integrating $\int(Z\,\mathrm{d}Y - Y\,\mathrm{d}Z)$ round the arc gives the closed form $2A_{23} = \pi R^{2} + 2Z_{O}R = 189\,239$ mm$^{2}$. Hence$$\sum q\,(2A) = -3.2429 \times 10^{6}\ \text{N}\cdot\text{mm} = S_{Y}\left(Z_{S} - 600\right)$$$$\boxed{Z_{S} = 275.7\ \text{mm from the boom 2--3 line, i.e. 324.3\ \text{mm forward of the spar 1--4}}}$$
Convert the offset load into a torque (part b). Moving the same 10 000 N upward force 100 mm to the left of the shear centre adds a pure torque about the shear centre of$$T = -100 \times 10\,000 = -1.00 \times 10^{6}\ \text{N}\cdot\text{mm}\ (\text{clockwise})$$
Add the constant torsional flow. The Bredt–Batho relation gives a shear flow uniform round the cell. The enclosed area is the rectangle plus the semicircle,$$A_{\text{cell}} = 600(240) + \tfrac{1}{2}\pi(120)^{2} = 166\,619\ \text{mm}^{2}, \qquad q_{T} = \frac{T}{2A_{\text{cell}}} = -3.001\ \text{N/mm}$$
Superpose. Adding $q_{T}$ to the no-twist flows gives the answer to part (b), tabulated below. The rear spar 4–1 still carries the bulk of the vertical shear at $\boxed{q_{41} = 20.93\ \text{N/mm}}$, but the torque has reversed the sense of the flow in both straight walls, and the check $240\left(q_{41} - q_{23}\right) = 10000$ N confirms that the resultant is still the applied 10 000 N.
Quantity
Result
$I_{YY}$
$2.592 \times 10^{7}\ \text{mm}^{4}$
(a) Shear centre
324.3 mm forward of the spar 1–4 (275.7 mm aft of the boom 2–3 line)
Torsional flow for the 100 mm offset
-3.001 N/mm (clockwise)
(b) $q$ in wall 1–2 (upper straight)
-2.22 N/mm
(b) $q$ in wall 2–3 (semicircular nose)
-20.73 N/mm
(b) $q$ in wall 3–4 (lower straight)
-2.22 N/mm
(b) $q$ in wall 4–1 (rear spar)
20.93 N/mm
Maximum shear stress ($t$ = 1 mm)
20.9 MPa in the nose wall
Check
The sign convention is that shear flow is positive in the walk direction 1→2→3→4→1 (counter-clockwise as drawn), so a negative value means the flow runs the other way. The final sentence of part (b) is truncated in the source paper (“… to the left of the shear center and the.”); it is answered as a pure offset-load problem, which is the only reading consistent with the 10 marks allotted.
