Module 9 · Lesson 9.6

Unsymmetrical bending and principal axes

What happens when the section has no axis of symmetry, and the beam refuses to bend the way you pushed it.

Why this matters

Every bending calculation so far has quietly assumed the section is symmetric about the plane of loading. Angles, channels, zeds and cold-formed purlins are not. Load one vertically and it deflects sideways as well — and if you use σ = My/I you will underestimate the stress, sometimes badly. The correction is not difficult, but you have to know it is needed.

By the end of this lesson you should be able to

Calculate the product of inertia I_xy of a built-up section
Find the principal axes and principal second moments
Use the general bending formula for a section with I_xy ≠ 0
Locate the neutral axis and explain why it is not perpendicular to the loading
Describe anticlastic curvature

What you should already know

Second moment of area and the parallel-axis theorem (this module)
Bending stress σ = My/I (this module)
Mohr's circle for stress (Module 14) — the algebra is identical

Alongside I_xx and I_yy there is a third quantity, the product of inertia:

I_xy = ∫ xy dA

Unlike the second moments it can be negative, because x and y carry signs. And it vanishes whenever the section has an axis of symmetry — for every element at (x, y) there is a mirror element at (−x, y) or (x, −y), and the contributions cancel in pairs.

That is why you have never needed it. Rectangles, circles and I-sections all have at least one axis of symmetry, so I_xy = 0 and σ = My/I is exact. Angles, zeds and plain channels loaded off-axis do not, and it is not.

For a built-up section of rectangles with sides parallel to the axes, each rectangle's own product of inertia is zero, so the whole of I_xy comes from the parallel-axis transfer:

I_xy = Σ A_i·x̄_i·ȳ_i

with x̄ and ȳ measured from the centroid of the whole section.

General bending formula

σ = \frac{( M _{x} I _{y y} - M _{y} I _{x y} ) y + ( M _{y} I _{xx} - M _{x} I _{x y} ) x}{I _{xx} I _{y y} - I _{x y}^{2}}

What it calculates: direct stress at any point of any section, under moments about both axes

M_x, M_y: bending moments about the centroidal x and y axes (N·mm)
x, y: coordinates of the point, from the centroid (mm)
I_xx, I_yy: second moments about those axes (mm⁴)
I_xy: product of inertia (mm⁴)

This assumes

Linearly elastic material and plane sections remaining plane
The axes pass through the centroid
No axial force and no twisting — the load acts through the shear centre

In plain terms: Set I_xy = 0 and M_y = 0 and everything collapses to σ = M_x y/I_xx. The extra terms are exactly the correction that symmetry usually hands you for free.

There is always an orientation in which the product of inertia disappears. Rotate the axes and I_xx, I_yy and I_xy transform in exactly the same way as σ_x, σ_y and τ_xy — the algebra is identical, with −I_xy playing the part of the shear stress. So the results are Mohr's circle results:

I₁, I₂ = (I_xx + I_yy)/2 ± √( ((I_xx − I_yy)/2)² + I_xy² )

These are the principal second moments, and the axes on which they act are the principal axes. About those axes I_xy = 0, so ordinary bending theory applies again.

That gives you two equivalent routes: use the general formula in your original axes, or resolve the moment onto the principal axes and use σ = My/I twice. They give the same answer. The second is more intuitive; the first is less work.

For an equal angle the principal axes are at 45° — which is exactly why an equal angle, loaded vertically, tries to bend and deflect diagonally.

Predict first

An unequal angle is loaded by a purely vertical force through its centroid. Which way does the neutral axis lie?

Setting the general formula to zero gives the neutral axis. Under M_x alone:

tan α = I_xy / I_yy

where α is measured from the x axis. With I_xy = 0 this gives α = 0 and the neutral axis is horizontal, as expected. Otherwise it is tilted — and the extreme fibre is then whichever corner is furthest from that tilted line, which is often not the corner you would have guessed.

This is the practical trap. Using σ = M y/I_xx on an angle means measuring y from the wrong axis to the wrong corner, and the error is unconservative.

One further effect completes the picture. Bending a beam about one axis produces a curvature across its width too, of opposite sign, because the compression face expands laterally through the Poisson effect while the tension face contracts. The saddle shape this produces is anticlastic curvature:

transverse curvature = −ν × longitudinal curvature

It is negligible for a narrow beam and matters for wide plates and slabs, where it is one reason a plate is stiffer than beam theory alone suggests.

Worked example

An equal angle in vertical bending

Given

Equal angle 100 × 100 × 10 mm, made of two rectangles: a vertical leg 10 wide × 100 deep, and a horizontal leg 90 wide × 10 deep
Bending moment M_x = 5.00 kNm applied about the horizontal centroidal axis, M_y = 0

Find

The section properties, the principal axes, the neutral axis and the stress at the top of the vertical leg.

Practice

An equal angle 100 × 100 × 10 mm has I_xx = I_yy = 1.80 × 10⁶ mm⁴ and I_xy = −1.066 × 10⁶ mm⁴. What is the major principal second moment I₁, in mm⁴?

Practice

For the same angle under a moment about the horizontal axis only, at what angle to the horizontal does the neutral axis lie? Give the magnitude in degrees.

Practice

A beam is bent to a longitudinal curvature of 2.5 × 10⁻⁶ per mm. With ν = 0.30, what is the magnitude of the anticlastic curvature across its width, in per mm?

Summary

I_xy = ∫xy dA can be negative, and vanishes whenever the section has an axis of symmetry
For built-up rectangles, I_xy = ΣA·x̄·ȳ about the section centroid
Second moments transform exactly like stresses, so principal axes follow Mohr's-circle algebra
About the principal axes I_xy = 0 and simple bending theory applies again
Under M_x alone the neutral axis is tilted at tan α = I_xy/I_yy
Equal angles have principal axes at 45°, so they bend diagonally
Anticlastic curvature = −ν × primary curvature: negligible for narrow beams, real for wide plates

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint