22-Mec-B9 Advanced Engineering Structures · December 2017
Question 5 of 8: Shear centre and panel shear flows of a six-boom idealised box
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2017 — 16-Mec-B9
Advanced Engineering Structures. Three hours, open book, any non-communicating
calculator permitted. The paper prints eight questions of equal value (20 marks each) and
states that any five constitute a complete exam paper. All eight are solved below,
because the set is a study resource rather than a sitting.
Reference texts.
T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. —
unsymmetrical bending, shear flow in open and closed thin-walled sections, shear centre,
structural idealisation, single- and multi-cell torsion (Ch. 15, 16, 17, 20, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed.
— three-dimensional states of stress, principal stresses, the Tresca and von Mises yield
criteria (Ch. 1, 4).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. —
shear centre, torsion of multiply connected cells, column stability (Ch. 6, 12).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — strain-life
fatigue, the Coffin–Manson relation, the Palmgren–Miner rule and Paris-law crack
growth (Ch. 9, 11, 14).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — Euler buckling of
pin-ended columns, slenderness ratio and the transition slenderness (Ch. 13).
Sign convention used throughout. A single right-handed frame is used
for every thin-walled question: x runs along the span from the root to the free end,
Y is vertically upward and Z horizontally to the right, both measured from the
section centroid. Direct stress is written $\sigma_x = aZ + bY$, and the two coefficients follow
from $aI_{ZZ} + bI_{YZ} = M_Y$ and $aI_{YZ} + bI_{YY} = M_Z$, in which
$I_{ZZ}=\int Z^{2}\,dA$, $I_{YY}=\int Y^{2}\,dA$, $I_{YZ}=\int ZY\,dA$,
$M_Y \equiv \int \sigma_x Z\,dA$ and $M_Z \equiv \int \sigma_x Y\,dA$. This one pair of
equations replaces the memorised unsymmetrical-bending fraction and is self-checking: for a
cantilever carrying a tip load, the fibres on the side the load points toward must go into
compression.
Question 5: Shear centre and panel shear flows of a six-boom idealised box (20 marks)
Given. A rectangular single-cell box idealised into six
direct-stress-carrying booms joined by six shear-only panels. Booms 1, 2 and 3 lie along the top
cover and booms 6, 5 and 4 directly beneath them. The overall width from boom 1 to boom 3 is 800
mm, boom 2 sits 400 mm from boom 3 (hence at mid-span), and the box is 200 mm deep.
Given data — boom coordinates and areas
Boom
$z$ from boom 6 (mm)
$y$ from the bottom cover (mm)
Area (mm$^2$)
1
0
200
500
2
400
200
300
3
800
200
750
4
800
0
750
5
400
0
300
6
0
0
500
Applied load: 7000 N vertical, acting upward in the plane of boom 3.
Find. (a) the spanwise position of the shear centre measured from boom 6;
(b) the shear flow in each of the six panels under the applied load.
The idealised section. Because the booms carry all the direct stress, the shear
flow is constant in each panel and steps by $-(S_y/I)B_r y_r$ at every boom, which is what makes
the walk around the cell purely arithmetic.
Check: the printed figure and the printed text disagree on the
load. The question text states a vertical force of 7000 N while the arrow on the figure
is annotated 5 kN. The prose is the question's own statement of the datum and the figure only
fixes where the load acts, so 7000 N is used throughout. Both parts of a linear-elastic
shear-flow calculation scale exactly with the load, so the last column of the results table gives
the panel flows at the figure value (multiply by $5000/7000 = 0.7143$). The shear centre itself
is a property of the geometry alone and is unaffected.
Approach. Take the second moment of the boom areas about the horizontal
centroidal axis, cut one panel and walk the open shear flow boom by boom, close the cell for
zero twist to locate the shear centre, then superpose the torque the offset load applies about
that centre.
Locate the centroidal axis and evaluate $I$. The booms are arranged
symmetrically about mid-depth and the paired areas are equal top and bottom, so the centroid
sits at $\bar y = 100$ mm without arithmetic. Hence
$$I_{ZZ} = \sum B_r (y_r-\bar y)^{2} = 2(500+300+750)(100)^{2}
= 3.10\times10^{7}\ \text{mm}^{4}.$$
Cut the cell and walk the open shear flow. With the booms carrying all the
direct stress, the flow is constant in each panel and steps at each boom by
$\Delta q = -(S_y/I)B_r y_r'$. Cutting panel 6–1 and walking
$1\to2\to3\to4\to5\to6$ gives
$$q_{b} = -11.29,\ -18.06,\ -35.00,\ -18.06,\ -11.29,\ 0 \ \ \text{N/mm}$$
for panels 1–2, 2–3, 3–4, 4–5, 5–6 and 6–1. The walk
returns to zero at the cut, which verifies the arithmetic.
Part (a) — close the cell for zero twist. The shear centre is the
point through which the load must pass to produce no twist. For constant wall thickness and
shear modulus the rate-of-twist condition reduces to $\oint q\,\mathrm{d}s = 0$, so with the
panel lengths 400, 400, 200, 400, 400 and 200 mm,
$$q_{s,0} = -\frac{\oint q_b\,\mathrm{d}s}{\oint \mathrm{d}s}
= \frac{30\,484}{2000} = +15.242\ \text{N/mm}.$$
Neither $t$ nor $G$ appears, because both cancel when they are uniform.
Take moments to locate the shear centre. Adding $q_{s,0}$ gives the
zero-twist flows $+3.95$, $-2.82$, $-19.76$, $-2.82$, $+3.95$ and $+15.24$ N/mm. Taking moments
of these about boom 6, using twice the swept area of each panel,
$$z_{SC} = \frac{\sum q\,(2A_r)}{S_y} = \frac{3.0710\times10^{6}}{7000}
\quad \Longrightarrow \quad \boxed{z_{SC} = 438.7\ \text{mm from boom 6}}$$
The centre lies aft of the mid-width of 400 mm because the largest booms, $B_3 = B_4 = 750$
mm$^2$, sit at the right-hand end and pull the stiffness distribution that way.
Part (b) — find the torque the applied load exerts. The 7000 N force
acts in the plane of boom 3, at $z = 800$ mm, which is $800-438.7 = 361.3$ mm outboard of the
shear centre. It therefore applies a torque
$$T = S_y\,(800 - z_{SC}) = 7000 \times 361.3 = 2.529\times10^{6}\ \text{N}\cdot\text{mm}.$$
Convert the torque into a uniform circulating flow. Walking
$1\to2\to3\to4\to5\to6\to1$ is clockwise in the chosen frame, so the signed swept area of the
walk is $-3.20\times10^{5}$ mm$^{2}$ and
$$q_T = \frac{T}{\sum 2A_r} = \frac{2.529\times10^{6}}{-3.20\times10^{5}}
= -7.903\ \text{N/mm}.$$
Dividing by $+2A$ instead of by the signed area of the actual walk is the classic sign trap here
and would flip the sense of most of the panels.
Superpose to obtain the final panel flows. Adding $q_T$ to the zero-twist
flows,
$$\boxed{q_{12} = -3.95,\ q_{23} = -10.73,\ q_{34} = -27.66,\ q_{45} = -10.73,\
q_{56} = -3.95,\ q_{61} = +7.34 \ \ \text{N/mm}}$$
Check the flows statically. Summing $q\,\Delta y$ over the panels returns
$7000.0$ N vertically and $\sum q\,\Delta z = 0.0$ N horizontally, and the moment about boom 6
comes to $5.600\times10^{6}\ \text{N}\cdot\text{mm}$, which is exactly $7000 \times 800$. The
resultant of the shear flows is therefore the applied load acting on its true line of action, in
both magnitude and position.
Read the physical picture. The rear web 3–4 carries by far the largest
flow because it both sits deepest in the bending-stress gradient and is where the load is
introduced; the front web 6–1 carries a much smaller flow of opposite sign, since the
torsional circulation there opposes the bending shear.
