Question 3 of 7: Direct stress at a point of an unsymmetrical thin-walled channel

Question 3: Direct stress at a point of an unsymmetrical 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 cantilever 2000 mm long whose cross-section is a channel with both flanges on the same side of the web and with three different wall gauges, loaded at the free end by two mutually perpendicular forces applied at the shear centre.

Find. The direct (bending) stress at point A, the free tip of the lower flange, on the cross-section at mid-span.

Approach. Idealise the walls to their median lines, locate the centroid, compute the three second moments including the product term, convert the tip loads into the two internal moments at mid-span, and solve the unsymmetrical-bending equations for a linear stress field.

1. Set up a consistent axis system. Use $X$ along the span from the free end towards the root, $Y$ vertically up and $Z$ horizontally towards the flange tips, and write the direct stress as a plane $$\sigma = aZ + bY$$ measured from the centroid. Two constants describe the whole field, and two moment equations will fix them. Because there is no axial load, no constant term is needed.
2. Locate the centroid. Treating each wall as a line of area $t\times\text{length}$, the areas are 360, 90 and 90 mm² for web, upper flange and lower flange, so $A = 540\ \text{mm}^2$. First moments about the web and about the lower flange give $$\bar Z=\frac{\sum A_iZ_i}{A}=10.0\ \text{mm},\qquad \bar Y=\frac{\sum A_iY_i}{A}=60.0\ \text{mm}$$ The centroid sits exactly at mid-depth because the two flanges happen to have equal areas, but it is pulled 10 mm out from the web by them.
3. Compute the second moments, product term included. Summing $A_i\Delta Z_i^2$, $A_i\Delta Y_i^2$ and $A_i\Delta Z_i\Delta Y_i$ and adding each wall's own local term, $$I_{ZZ}=216\,270\ \text{mm}^4,\qquad I_{YY}=1\,080\,075\ \text{mm}^4,\qquad I_{YZ}=-162\,000\ \text{mm}^4$$ where $I_{ZZ}=\int Z^2\,dA$ resists horizontal bending and $I_{YY}=\int Y^2\,dA$ resists vertical bending. The section is five times stiffer vertically than horizontally, and the non-zero product term confirms that neither axis is principal.
4. Convert the tip loads into moments at mid-span. The examined section is 1000 mm from the free end, so taking moments of the tip forces about that section, $$M_Y=\int Z\sigma\,dA = (1000)(400)=4.00{\times}10^{5}\ \text{N}\!\cdot\!\text{mm},\qquad M_Z=-\int Y\sigma\,dA = -(1000)(800)=-8.00{\times}10^{5}\ \text{N}\!\cdot\!\text{mm}$$ Both grow linearly from zero at the tip, so mid-span carries exactly half the root values.
5. Solve the two unsymmetrical-bending equations. Substituting $\sigma=aZ+bY$ into the definitions of $M_Y$ and $M_Z$ produces the pair $$aI_{ZZ}+bI_{YZ}=-M_Y,\qquad aI_{YZ}+bI_{YY}=M_Z$$ whose solution is $$a=-2.7087\ \text{MPa/mm},\qquad b=-1.1470\ \text{MPa/mm}$$ Solving the pair rather than quoting a memorised fraction keeps the sign convention visible, which is where this question is usually lost.
6. Evaluate the stress at point A. Point A is the outer tip of the lower flange, at $Z_A=90-10=80$ mm and $Y_A=0-60=-60$ mm from the centroid, so $$\sigma_A=a Z_A+b Y_A=(-2.7087)(80)+(-1.1470)(-60)$$ $$\boxed{\sigma_A = -147.9\ \text{MPa}\ \ (\text{compressive})}$$
7. See where that stress comes from. The two terms are $-216.7$ MPa from the 400 N horizontal load and $+68.8$ MPa from the 800 N vertical load. The smaller force therefore contributes three times as much stress, because the section is five times weaker about the vertical axis and point A lies far out along that weak direction. Recognising this before computing anything is the practical lesson of the question.
8. Locate the neutral axis and the extreme fibres. Setting $\sigma=0$ gives $Y=-(a/b)Z$, a line through the centroid inclined at $112.95^{\circ}$ to the $+Z$ axis. Evaluating the stress plane at each extremity, the upper flange tip carries $-123.0$ MPa, the web top $-41.7$ MPa and the web foot $+95.9$ MPa. Point A is thus the most compressed fibre in the whole section and the web foot the most tensioned, so the pair $-147.9$ and $+95.9$ MPa are the values a strength check would use.
9. Confirm that no torsion is present. The question specifies loading at the shear centre, which is what allows the section to be treated in pure bending. Walking the open section and taking moments of the resulting shear flow places the shear centre 11.40 mm outside the web on the side away from the flanges and 16.03 mm above the lower flange. Loads applied there produce no twist, so no warping stresses are superposed on the answer above.