22-Mec-B9 Advanced Engineering Structures · May 2013
Question 3 of 8: Bending stress in an unsymmetrical thin-walled channel
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 3: Bending stress in an unsymmetrical thin-walled channel (20 marks)
Given. A channel with unequal flanges, both on the same side of the web, built in at one end and loaded at the free end by a 1000 N vertical force and a 500 N horizontal force, both applied at the shear centre.
Quantity
Symbol
Value
Web depth and thickness
$h \times t_{w}$
150 mm × 2.0 mm
Upper flange width and thickness
$b_{t} \times t_{t}$
60 mm × 2.0 mm
Lower flange width and thickness
$b_{b} \times t_{b}$
120 mm × 1.0 mm
Vertical tip load (upward)
$P_{Y}$
1000 N
Horizontal tip load (away from the flange tips)
$P_{Z}$
500 N
Distance from the loads to the section
$L$
2000 mm
Find. The direct stress at point A, the free tip of the lower flange, on a section 2000 mm inboard of the loads.
Median-line idealisation of the channel, looking along $+X$ (from the free end toward the built-in end). The centroid C sits 20 mm from the web and at mid-depth; the neutral axis is inclined at 61.7° to the $Z$ axis, nowhere near either load direction.
Approach. Because the flanges are unequal the section has no axis of symmetry, so $I_{YZ}$ does not vanish and the elementary $My/I$ formula is invalid. Locate the centroid, evaluate all three second moments about it, express the direct stress as a general linear field and fix its two constants from the moments the section carries.
Idealise the section by its median line. The three walls are treated as thin strips of area $t \times$ length located on the mid-thickness lines, with the origin at the foot of the web: web $2.0 \times 150 = 300$ mm$^{2}$ at $(0, 75)$, upper flange $2.0 \times 60 = 120$ mm$^{2}$ at $(30, 150)$, and lower flange $1.0 \times 120 = 120$ mm$^{2}$ at $(60, 0)$, giving $A = 540$ mm$^{2}$.
Locate the centroid. Taking first moments about the origin,$$\bar{Z} = \frac{120(30) + 120(60)}{540} = 20.0\ \text{mm},\qquad \bar{Y} = \frac{300(75) + 120(150)}{540} = 75.0\ \text{mm}$$The vertical centroid lands exactly at mid-depth because the two flange areas happen to be equal — the thinner lower flange is exactly twice as wide.
Evaluate the three second moments about the centroid. Each strip contributes its own second moment about its own mid-line plus the transfer term; the product term needs the signed offsets of each strip:$$I_{YY} = 1.9125 \times 10^{6}\ \text{mm}^{4},\qquad I_{ZZ} = 5.0410 \times 10^{5}\ \text{mm}^{4},\qquad I_{YZ} = -2.700 \times 10^{5}\ \text{mm}^{4}$$The product of inertia is negative and large — roughly $-0.28\sqrt{I_{YY}I_{ZZ}}$ — because the upper flange sits above and to the right of the centroid while the lower flange sits below and further to the right.
Find the moments carried by the section. The loads act at the shear centre, so the section carries pure bending with no twist. Working from the free-end free body, 2000 mm long,$$M_{Z} = \int\sigma Y\,\mathrm{d}A = -(1000)(2000) = -2.00 \times 10^{6}\ \text{N}\cdot\text{mm}, \qquad M_{Y} = \int\sigma Z\,\mathrm{d}A = +(500)(2000) = 1.00 \times 10^{6}\ \text{N}\cdot\text{mm}$$The signs follow from the rule stated at the head of this paper: the 1000 N load points up, so the upper fibres are compressed and $M_{Z}$ is negative; the 500 N load points away from the flange tips, so the flange-tip side is in tension and $M_{Y}$ is positive.
Solve for the stress field. Substituting into the pair of equations $aI_{ZZ} + bI_{YZ} = M_{Y}$ and $aI_{YZ} + bI_{YY} = M_{Z}$ and inverting the $2\times2$ system gives$$a = 1.5401\ \text{N/mm}^{3}, \qquad b = -0.8283\ \text{N/mm}^{3}$$so that $\sigma = 1.5401Z -0.8283Y$ MPa with $Z$ and $Y$ in millimetres from the centroid.
Evaluate the stress at point A. Point A is the free tip of the lower flange, at $(120, 0)$ in section coordinates, hence at $(Z, Y) = (100.0, -75.0)$ mm from the centroid:$$\sigma_{A} = 1.5401(100.0) + (-0.8283)(-75.0) = 154.01 + 62.12 = \boxed{216.1\ \text{MPa (tension)}}$$Both loads pull point A into tension and the smaller horizontal load supplies the larger share, 71 per cent of the total, because the section is far weaker about its vertical axis.
Check the result against the neutral axis. Setting $\sigma = 0$ gives $Y/Z = -a/b = 1.859$, a neutral axis through the centroid inclined at 61.7 degrees to the $Z$ axis. Point A lies on the far side of that line from the compressed upper flange corner, confirming the tensile sign, and it is the point of the section furthest from the neutral axis — so it is also the most highly stressed point.
Quantity
Result
Centroid position $(\bar{Z}, \bar{Y})$
(20.0, 75.0) mm from the foot of the web
$I_{YY}$ (about the horizontal centroidal axis)
$1.913 \times 10^{6}\ \text{mm}^{4}$
$I_{ZZ}$ (about the vertical centroidal axis)
$5.041 \times 10^{5}\ \text{mm}^{4}$
$I_{YZ}$ (product of inertia)
$-2.700 \times 10^{5}\ \text{mm}^{4}$
Contribution of the 500 N load at A
154.0 MPa
Contribution of the 1000 N load at A
62.1 MPa
Bending stress at point A
216.1 MPa (tension)
Neutral-axis inclination to $Z$
61.7°
Check
The section is idealised on its median lines, so the web is taken as 150 mm deep and the flanges as 60 mm and 120 mm wide measured from the web centreline; using outside dimensions instead changes the answer by under two per cent. The horizontal 500 N load is read from the isometric view as acting away from the flange tips. Reversing that direction leaves the magnitude of the horizontal contribution unchanged but makes point A the most compressed point instead of the most tensioned one, so state the assumed direction on the answer paper.
