Statics & strength · Polygon contour

Unsymmetrical (biaxial) bending of an arbitrary cross-section

Draw any polygon contour and calculate the biaxial bending stress (Mx, My) directly from its real cross-section geometry, including the product moment of inertia Ixy.

Mx, My2D

The calculation runs locally in your browser. Coordinates are neither transmitted nor stored.

Cross-section contour

Place or drag points along the outer boundary. The y-axis points upwards; contours close automatically.

P1P2P3P4

Coordinates

Pointx (mm)y (mm)Action
P1
P2
P3
P4

Only single-axis bending about one axis? Go to the bending stress calculator for arbitrary cross-sections →

What is unsymmetrical / biaxial bending?

Unsymmetrical (skew) bending occurs when the bending moment does not act along a principal axis of inertia. For an unsymmetric cross-section, Ixy can be nonzero, coupling both bending directions. The neutral axis is therefore not simply one of the x/y axes in the general case, but a tilted line through the centroid.

Simplified formula: only when Ixy = 0

If the x/y axes are already principal axes of inertia (Ixy = 0), the stress can be found by simply superimposing two single-axis bending cases:

σ = -Mx·y’/Ix + My·x’/Iy

This formula applies strictly only when Ixy = 0. For a general unsymmetric cross-section it would be incorrect.

General formula for unsymmetrical bending

In the general case (also valid for Ixy != 0), the inertia-tensor approach gives:

D = IxIy - Ixy²σ = -(IyMx+IxyMy)/D · y’ + (IxyMx+IxMy)/D · x’

with x' = x - x̄ and y' = y - ȳ. For Ixy = 0 this equation reduces exactly to the simplified formula above.

Why evaluate at the vertices?

Since sigma is a linear (affine) function of x and y, it has no interior extrema over the cross-section area. The maximum tensile and compressive stress therefore always occur at a vertex of the polygon contour -- even for concave cross-sections.

Example: 100 × 50 mm rectangle with Mx and My

For the axis-aligned rectangle, Ixy = 0, so the simplified formula applies. At Mx = 1,000 N·m and My = 500 N·m, the vertex (100, 0) carries a tensile stress of sigma = -1,000,000*(-25)/1,041,666.667 + 500,000*50/4,166,666.667 = 24 + 6 = 30 N/mm^2, the plain superposition of the two individual bending stresses.

Scope and limitations

The calculator supports one closed, non-self-intersecting outer contour and one polygonal inner contour under pure bending without axial force. Disconnected regions, arcs, DXF, shear stresses and plastic cross-section utilization are currently excluded.

FAQ

Common questions

What is unsymmetrical bending?

Unsymmetrical bending occurs when the resultant bending moment does not act along a principal axis of inertia of the cross-section. For an unsymmetric section, Ixy can then be nonzero, coupling bending about x with bending about y.

What is the difference between unsymmetrical and biaxial bending?

Biaxial bending just means Mx and My act simultaneously. Unsymmetrical bending is the special case where this combination produces a tilted neutral axis -- when Ixy = 0, a biaxial load can be handled by simply superimposing two independent single-axis bending cases.

Why is the product moment of inertia Ixy needed?

Ixy couples the two bending directions. If Ixy != 0, a moment purely about the x-axis already produces a stress distribution that also depends on the x-coordinate -- the simplified formula sigma = -Mx*y/Ix would give the wrong stress.

How do you calculate bending stress for Mx and My?

In the general case with sigma = -(Iy*Mx + Ixy*My)/D*y' + (Ixy*Mx + Ix*My)/D*x', where D = Ix*Iy - Ixy^2 and x', y' are coordinates relative to the centroid. When Ixy = 0 this simplifies to sigma = -Mx*y'/Ix + My*x'/Iy.

Where does the neutral axis lie for unsymmetrical bending?

Wherever sigma = 0 -- in the general case a line through the centroid tilted relative to the x/y axes, not simply one of the two coordinate axes.

When may the simplified formula with Mx/Ix and My/Iy be used?

Only when Ixy = 0, i.e. when x and y are already the section's principal axes of inertia. Otherwise the full formula with D = Ix*Iy - Ixy^2 must be used.

Where does the maximum bending stress occur?

Since the stress distribution is linear, the extreme values always occur at a vertex of the polygon contour -- the one farthest from the neutral axis on the respective side.

References

Engineering and mathematical sources

Calculation basis: engineering mechanics and area integrals for polygonal sections; this is not a normative design rule.