Unsymmetrical bending of a thin-walled channel: 22-Mec-B9 Advanced Engineering Structures, May 2017, Q5

Question 5: Unsymmetrical bending of a thin-walled channel (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A thin-walled channel cantilever carrying a vertical and a horizontal tip load at its shear centre, both applied 1200 mm from the section of interest.

Given data — section and loading
Item	Dimension	Thickness
Web (overall depth)	200 mm	4.0 mm
Upper flange (from the web centre-line)	80 mm	4.0 mm
Lower flange (from the web centre-line)	160 mm	2.0 mm
Vertical tip load	1000 N upward	—
Horizontal tip load	500 N, directed away from the flange tips	—
Distance from the loads to the section	1200 mm	—

Find. The direct (bending) stress at point A, the outer extremity of the lower flange.

The cantilever as drawn on the paper. The span axis projects to the right and up; the flanges project to the right and down, toward the viewer. The 500 N arrow is drawn up and to the left, so it acts away from the flange tips — the sign that decides whether point A ends up in tension or compression.

The section, with the centroid C, the axes $Z$ (toward the flange tips) and $Y$ (upward), and the computed neutral axis. Point A is the outer extremity of the lower flange. Because the two flange areas are equal ($80 \times 4 = 160 \times 2$), the centroid sits at exact mid-depth.

Approach. Because the section has no axis of symmetry, the direct stress must be written as a general linear field $\sigma = aZ + bY$ and the two coefficients found from the two moment equations that involve $I_{ZZ}$, $I_{YY}$ and the product moment $I_{YZ}$. Sectional properties come first, then the coefficients, then the stress at A.

Fix a right-handed axis system and the sign of each load. Take $X$ along the span from the loaded end toward the root, $Y$ vertically upward and $Z$ horizontally toward the flange tips; this triad is right-handed for the isometric as drawn. The vertical load is then $S_Y = +1000$ N and, because the 500 N arrow points away from the flange tips, $S_Z = -500$ N. Getting this sign right is the whole question: it decides whether A is the most tensioned or the most compressed point on the section.
Idealise the section and locate its centroid. Treating each of the three walls as a thin rectangle about its own centre-line, the areas are $200 \times 4 = 800$, $80 \times 4 = 320$ and $160 \times 2 = 320\ \text{mm}^2$, giving $A = 1440\ \text{mm}^2$. Measuring $Z$ from the web centre-line and $Y$ from the lower flange, $$\bar{Z} = \frac{320(40) + 320(80)}{1440} = 26.67\ \text{mm}, \qquad \bar{Y} = 100.0\ \text{mm}.$$ The two flange areas happen to be equal, so $\bar{Y}$ falls at exact mid-web without any arithmetic — a useful check that the idealisation was set up correctly.
Compute the three second moments about the centroid. Using the parallel-axis theorem wall by wall, with $I_{ZZ} = \int Z^2 \mathrm{d}A$ (about the vertical axis) and $I_{YY} = \int Y^2 \mathrm{d}A$ (about the horizontal axis), $$I_{ZZ} = 2.390 \times 10^6\ \text{mm}^4, \quad I_{YY} = 9.067 \times 10^6\ \text{mm}^4, \quad I_{YZ} = -1.280 \times 10^6\ \text{mm}^4.$$ The product moment is non-zero and negative because the wide lower flange sits below the centroid on the positive-$Z$ side, so the section is 3.8 times weaker about the vertical axis than about the horizontal one.
Write the bending moments at the section. With the loads a distance $L = 1200$ mm from the section, equilibrium of the free-end segment gives $$M_Y \equiv \int \sigma Z\, \mathrm{d}A = -S_Z L = -(-500)(1200) = +6.00 \times 10^5 \ \text{N}\cdot\text{mm},$$ $$M_Z \equiv \int \sigma Y\, \mathrm{d}A = -S_Y L = -(1000)(1200) = -1.20 \times 10^6 \ \text{N}\cdot\text{mm}.$$
Solve the coupled pair for the stress-field coefficients. Substituting $\sigma = aZ + bY$ into the two definitions above gives the linear system $$a I_{ZZ} + b I_{YZ} = M_Y, \qquad a I_{YZ} + b I_{YY} = M_Z,$$ and solving it, $$a = 0.19487\ \text{MPa/mm}, \qquad b = -0.10484\ \text{MPa/mm}.$$ Writing the field this way avoids having to memorise the Megson fraction, and it is self-checking: with the vertical load acting alone the top of the section comes out in compression, which is what an upward tip load must do to a cantilever.
Evaluate the stress at point A. Point A lies at the outer tip of the lower flange, so relative to the centroid $Z_A = 160 - 26.67 = 133.3$ mm and $Y_A = 0 - 100 = -100$ mm. Then $$\sigma_A = aZ_A + bY_A = (0.19487)(133.3) + (-0.10484)(-100) = 25.98 + 10.48,$$ $$\boxed{\sigma_A = +36.5\ \text{MPa} \ \text{(tensile)}}.$$
Check where the answer comes from and where the neutral axis lies. Solving the same system with each load acting alone gives 32.4 MPa from the 500 N horizontal load and only 4.10 MPa from the 1000 N vertical load: the smaller load supplies 89 % of the stress, because the section is far weaker about the vertical axis. The neutral axis passes through the centroid with slope $Y = -(a/b)Z = 1.859\,Z$, and it happens to pass almost exactly through the tip of the upper flange, where the stress is only $-0.09$ MPa.

Point A is therefore the most highly stressed fibre in the section, at 36.5 MPa tension against $-15.7$ MPa at the web-to-upper-flange junction. The result is a useful illustration of why unsymmetrical sections are treated with the full product-moment formulation rather than by resolving the loads onto the drawn axes: doing the latter here would have missed the $I_{YZ}$ coupling and returned 46.7 MPa at A, overstating the answer by 28 % — and the shortcut is not reliably on the safe side either: at the upper-flange tip it even reverses the sign of the stress.

Final results
Quantity	Symbol	Value
Section area	$A$	1440 mm²
Centroid from the web centre-line	$\bar{Z}$	26.67 mm
Centroid from the lower flange	$\bar{Y}$	100.0 mm
Second moments	$I_{ZZ}$, $I_{YY}$	$2.390 \times 10^6$, $9.067 \times 10^6$ mm⁴
Product moment	$I_{YZ}$	$-1.280 \times 10^6$ mm⁴
Stress-field coefficients	$a$, $b$	0.19487, $-0.10484$ MPa/mm
Neutral-axis slope	$Y/Z$	1.859
Bending stress at A	$\sigma_A$	$+36.5$ MPa (tension)

Topic. Unsymmetrical (biaxial) bending of a thin-walled open section.

Key relations. $\sigma = aZ + bY$ with $aI_{ZZ} + bI_{YZ} = M_Y$ and $aI_{YZ} + bI_{YY} = M_Z$, where $M_Y = \int\sigma Z\,\mathrm{d}A$ and $M_Z = \int\sigma Y\,\mathrm{d}A$; for a cantilever tip load at distance $L$, $M_Y = -S_Z L$ and $M_Z = -S_Y L$. Neutral axis: $Y = -(a/b)Z$.

Why this works. The Euler–Bernoulli assumption — plane sections remain plane — makes the direct stress a linear function of position, $\sigma = aZ + bY$, whatever the shape of the section. What symmetry buys is only that $I_{YZ}$ vanishes and the two coefficients uncouple. When it does not vanish, a purely vertical load still produces horizontal curvature, and the neutral axis is no longer perpendicular to the plane of loading. Solving the $2 \times 2$ system directly is both shorter and less error-prone than substituting into the fully expanded closed-form expression.

Common pitfall. Assuming the neutral axis is horizontal for a vertical load, and reading the sign of the horizontal load off the isometric by guesswork. On this section, flipping the 500 N load turns the answer from $+36.5$ MPa tension into $-28.3$ MPa compression: trace the two horizontal axes of the isometric before assigning signs.

Source. T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed., Ch. 15.2 (unsymmetrical bending; direct stress distribution and the neutral axis).

Question 5 of 8: Unsymmetrical bending of a thin-walled channel

Question 5: Unsymmetrical bending of a thin-walled channel (20 marks)