Unsymmetrical Bending & Shear Centre Notes

Last reviewed 16 Sept 2026 · 6 min read

Symmetrical and unsymmetrical bending

The simple bending formula $σ = M y / I$ assumes the load acts in a plane of symmetry, so the beam bends in that same plane. Bending is unsymmetrical when:

the section has no axis of symmetry (unequal angle, Z-section), or
the section is symmetric but the plane of loading is inclined to the axes of symmetry — a purlin on a sloping roof, a rectangular beam loaded diagonally.

Then the neutral axis is not perpendicular to the plane of loading, the beam deflects out of the load plane, and stresses must be found about the principal axes.

Product of inertia and principal axes

For axes $x$ and $y$ through the centroid:

I_{xx} = \int y^{2} d A, I_{y y} = \int x^{2} d A, I_{x y} = \int x y d A

$I_{x y}$ may be positive, negative or zero.
If either axis is an axis of symmetry, $I_{x y} = 0$ .
Principal axes are the pair of perpendicular centroidal axes about which $I_{x y} = 0$ ; the moments of inertia about them are the maximum and minimum.

FormulaPrincipal axes and principal moments of inertia

tan 2 α = \frac{2 I _{x y}}{I _{y y} - I _{xx}}

I_{u, v} = \frac{I _{xx} + I _{y y}}{2} \pm (\frac{I _{xx} - I _{y y}}{2})^{2} + I_{x y}^{2}

$I_{xx} + I_{y y} = I_{u} + I_{v}$ (invariant, equal to the polar moment $J$ ).

These equations have the same form as the stress transformation equations, so Mohr's circle can also be drawn for moments of inertia.

Bending stress for a moment inclined to the principal axes

Let $U$ and $V$ be the principal axes, with $I_{u}$ and $I_{v}$ . A moment $M$ acts in a plane making angle $θ$ with the $V$ -axis. Resolve it:

M_{u} = M cos θ (about U), M_{v} = M sin θ (about V)

FormulaStress at a point (u, v)

σ = \frac{M _{u} v}{I _{u}} + \frac{M _{v} u}{I _{v}} = M (\frac{v cos θ}{I _{u}} + \frac{u sin θ}{I _{v}})

with signs chosen by inspection (tension or compression) for each component. The maximum stress occurs at the point farthest from the neutral axis — for a rectangle, at a corner.

Neutral axis

Setting $σ = 0$ gives a straight line through the centroid inclined at angle $ϕ$ to the $U$ -axis:

tan ϕ = \frac{I _{u}}{I _{v}} tan θ

Since $I_{u} \neq = I_{v}$ in general, $ϕ \neq = θ$ : the neutral axis is not perpendicular to the plane of loading. It swings towards the axis of minimum moment of inertia. The beam deflects perpendicular to the neutral axis.

General formula (non-principal axes)

If stresses are wanted directly in $x$ – $y$ axes with $I_{x y} \neq = 0$ :

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

(signs depend on the convention for $M_{x}$ and $M_{y}$ ; with $I_{x y} = 0$ it reduces to the principal-axis form).

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Unsymmetrical Bending & Shear Centre