22-Mec-B9 Advanced Engineering Structures · December 2016
Question 6 of 8: Unsymmetrical bending of a thin-walled channel
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2016 — 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. — thin-walled open and closed sections, shear flow, shear centre, structural idealisation, single- and multi-cell torsion (Ch. 16, 17, 18, 20, 21, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — yield criteria, torsion of non-circular and thin-walled members, unsymmetrical bending (Ch. 2, 4, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of multiply connected cells and of solid non-circular shafts (Ch. 6).
R. M. Jones, Mechanics of Composite Materials, 2nd ed. — orthotropic lamina stiffness, the reduced stiffness matrix and its transformation (Ch. 2).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — strain-life fatigue, the Coffin–Manson relation and the Palmgren–Miner rule (Ch. 9, 14).
Question 6: Unsymmetrical bending of a thin-walled channel (20 marks)
Given. A 2 000 mm cantilever of unequal-flange channel section, loaded at the free end by 800 N vertically upward and 400 N horizontally, both acting through the shear centre. The stress is wanted 1 200 mm inboard of the load point. Taking $Z$ positive toward the flange tips and $Y$ positive upward, with the web mid-plane at $Z = 0$:
Given data — section and loading
Element
Length (mm)
Thickness (mm)
Position
Web
100
2.0
$Z=0$, $Y = 0$ to 100
Upper flange
40
2.0
$Y = 100$, $Z = 0$ to 40
Lower flange
80
1.0
$Y = 0$, $Z = 0$ to 80
Vertical tip load
$F_Y = +800$ N (upward)
at the shear centre
Horizontal tip load
$F_Z = -400$ N (away from the flange tips)
at the shear centre
Distance from load to section
$x = 1200$ mm
—
Point A
tip of the lower flange
$Z = 80$, $Y = 0$
Find. The direct bending stress at point A, the outer extremity of the 80 mm lower flange, 1 200 mm from the loaded end.
Figure 6.1 — the unequal-flange channel, with the centroid C, point A at the tip of the lower flange, and the inclined neutral axis that the non-zero product of inertia produces.
Approach. Locate the centroid, evaluate all three second moments including the product term, resolve the two tip loads into bending moments at the section, and use the general unsymmetrical-bending stress field $\sigma = aZ + bY$.
Locate the centroid. The three element areas are $A_{\text{web}} = 100(2.0) = 200$, $A_{\text{top}} = 40(2.0) = 80$ and $A_{\text{bot}} = 80(1.0) = 80$ mm$^2$, totalling 360 mm$^2$. Because the two flange areas happen to be equal, the vertical centroid sits at mid-web:$$\bar{Y} = \frac{200(50) + 80(100) + 80(0)}{360} = 50.0\ \text{mm},\qquad \bar{Z} = \frac{200(0) + 80(20) + 80(40)}{360} = 13.33\ \text{mm}.$$The centroid lies 13.33 mm out from the web, pulled that way by the wide lower flange.
Evaluate the three second moments about the centroid. Using the parallel-axis theorem element by element,$$I_{YY} = \int Y^{2}\mathrm{d}A = 166\,667 + 200\,027 + 200\,007 = 566\,700\ \text{mm}^4,$$$$I_{ZZ} = \int Z^{2}\mathrm{d}A = 35\,622 + 14\,222 + 99\,556 = 149\,400\ \text{mm}^4,$$$$I_{YZ} = \int YZ\,\mathrm{d}A = 0 + 80(6.667)(50) + 80(26.667)(-50) = -80\,000\ \text{mm}^4 .$$The product of inertia is not zero, so the principal axes are inclined and the two loads cannot be treated independently. Note also how weak the section is about the vertical axis: $I_{ZZ}$ is less than a third of $I_{YY}$.
Convert the tip loads into moments at the section. Working from the free-body of the tip segment, the internal stress resultants at a cut $x$ from the loads are$$M_Y \equiv \int \sigma Z\,\mathrm{d}A = -xF_Z = -1200(-400) = +480\,000\ \text{N}\cdot\text{mm},$$$$M_Z \equiv \int \sigma Y\,\mathrm{d}A = -xF_Y = -1200(+800) = -960\,000\ \text{N}\cdot\text{mm}.$$The signs carry the physics: the upward load must compress the top of the section, and the load pointing away from the flange tips must put those tips into tension.
Solve for the stress field. Writing the stress as the linear field $\sigma = aZ + bY$ and substituting into the two definitions above gives the pair$$aI_{ZZ} + bI_{YZ} = M_Y,\qquad aI_{YZ} + bI_{YY} = M_Z,$$that is $149\,400a - 80\,000b = 480\,000$ and $-80\,000a + 566\,700b = -960\,000$. Solving,$$a = 2.4943\ \text{MPa/mm},\qquad b = -1.3419\ \text{MPa/mm}.$$A quick sanity check: $b$ is negative, so the top of the web is in compression, exactly as an upward tip load demands.
Evaluate the stress at point A. Point A lies at $Z = 80 - 13.33 = 66.67$ mm and $Y = 0 - 50 = -50$ mm relative to the centroid, so$$\sigma_A = a Z + b Y = 2.4943(66.67) + (-1.3419)(-50) = 166.29 + 67.10,$$$$\boxed{\sigma_A = +233.4\ \text{MPa (tension)}} .$$The striking feature is the split: the 400 N horizontal load, half the size of the vertical one, supplies 166 MPa or 71 % of the total, because the section is nearly four times weaker about the vertical axis than about the horizontal one. Designers of channel sections meet this repeatedly — a modest side load dominates the stress at the flange tips.
Locate the neutral axis as a check. Setting $\sigma = 0$ gives $Y = -(a/b)Z$, a line through the centroid of slope $\mathrm{d}Y/\mathrm{d}Z = 1.859$. Point A lies well below that line on the tension side, and the top of the web lies above it on the compression side, which is consistent with the signs obtained. The inclination of the neutral axis — some 62° from the horizontal — is itself the visible consequence of the non-zero product of inertia.
