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Structural Analysis Fundamentals / Bending of beams

Unsymmetrical bending and principal axes

Resolving bending about centroidal principal axes and locating the neutral axis.

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Why this matters

Simple principal-axis bending assumes the loading and section axes are compatible. For sections or load directions that do not satisfy that assumption, calculate the principal axes and resolve the bending components. Check whether loading away from the shear centre also introduces torsion.

By the end of this lesson you should be able to

  • Calculate the product of inertia Ixy of a built-up section
  • Find the principal axes and principal second moments
  • Use the general bending formula for a section with Ixy ≠ 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

The product of inertia is Ixy = ∫xy dA. Its sign depends on the chosen axes. It vanishes about centroidal principal axes. A centroidal axis aligned with an axis of section symmetry is one sufficient condition, because the paired area contributions cancel.

Rotating the axes of a non-square rectangle can give a non-zero Ixy. Plain channels and equal angles also have a symmetry axis; their behaviour depends on the selected axes and loading direction. Resolve bending about principal axes where appropriate. Loading away from the shear centre can introduce torsion, which is a separate effect.

For rectangles whose sides are parallel to the chosen axes, each rectangle’s own centroidal product of inertia is zero, so the built-up section has Ixy = ΣAi x̄i ȳ_i, with offsets measured from the section centroid.

General bending formula
σ=(MxIyy−MyIxy) y+(MyIxx−MxIxy) xIxxIyy−Ixy2\sigma = \frac{(M_x I_{yy} - M_y I_{xy})\,y + (M_y I_{xx} - M_x I_{xy})\,x}{I_{xx}I_{yy} - I_{xy}^2}

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

Mx, My
bending moments about the centroidal x and y axes (N·mm)
x, y
coordinates of the point, from the centroid (mm)
Ixx, Iyy
second moments about those axes (mm⁴)
Ixy
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 Ixy = 0 and My = 0 and everything collapses to σ = Mx y/Ixx. 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 Ixx, Iyy and Ixy transform in exactly the same way as σx, σy and τxy — the algebra is identical, with −Ixy playing the part of the shear stress. So the results are Mohr's circle results:

I₁, I₂ = (Ixx + Iyy)/2 ± √( ((Ixx − Iyy)/2)² + Ixy² )

These are the principal second moments, and the axes on which they act are the principal axes. About those axes Ixy = 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 carries Mx alone about its horizontal centroidal axis, with no torsion or axial force. Where does the elastic neutral axis lie?

Under Mx alone, setting the stress expression to zero gives tan α = Ixy/Iyy for the stated coordinates. A non-zero product of inertia tilts the neutral axis. Evaluate all section vertices to find the largest stress; using Mx y/Ixx alone may overestimate or underestimate a particular value.

A freely deforming isotropic strip in uniaxial bending can also develop transverse anticlastic curvature, −ν times its longitudinal curvature. Restraint of transverse deformation changes this response and can increase plate bending stiffness. Apply the free-curvature relation only with compatible boundary conditions.

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 Mx = 5.00 kNm applied about the horizontal centroidal axis, My = 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 Ixx = Iyy = 1.80 × 10⁶ mm⁴ and Ixy = −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

    • Ixy = ∫xy dA can be negative, and vanishes when the chosen centroidal axes are principal axes, including an axis aligned with section symmetry
    • For built-up rectangles, Ixy = 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 Ixy = 0 and simple bending theory applies again
    • Under Mx alone the neutral axis is tilted at tan α = Ixy/Iyy
    • Equal angles have principal axes at 45°, so they bend diagonally
    • Free anticlastic curvature is −ν times longitudinal curvature in the stated isotropic strip model; transverse restraint changes it