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Structural Analysis

Unsymmetric Bending Simulator

Apply a bending moment to a rectangular beam section at an angle offset from its principal axes. Change the load angle and watch the moment split into strong- and weak-axis components, and see the corner bending stress, neutral-axis tilt and deflection direction update in real time. Experience how a beam deflects away from the direction of the load.

Parameters
Section width b
mm
Section dimension in the z (horizontal) direction
Section depth h
mm
Section dimension in the y (vertical) direction
Bending moment M
kN·m
Load angle φ
°
Tilt of the moment plane measured from the strong (z) axis
Beam length L
m
Young's modulus E
GPa
Steel ≈ 200, aluminium ≈ 70 GPa
Results
—
Strong-axis moment M_z (kN·m)
—
Weak-axis moment M_y (kN·m)
—
Corner bending stress σ (MPa)
—
Neutral-axis tilt α (deg)
—
Deflection direction β (deg)
—
Deviation from load (deg)
—
Cross-section — load, neutral axis & stress field

Apply the load (orange arrow) at angle φ to the rectangular section and the neutral axis (cyan line) tilts by angle α. Colour shows the bending stress (blue = compression / red = tension). The circle marks the most-stressed corner.

Corner stress σ vs load angle φ
Neutral-axis angle α & deflection angle β vs load angle φ
Theory & Key Formulas

$$\sigma=\frac{M_z\,y}{I_z}+\frac{M_y\,z}{I_y}$$

Stress in unsymmetric bending. Mz and My are the strong- and weak-axis components of the moment, Iz and Iy are the second moments of area, and y, z are the coordinates of the point. For a rectangle the product of inertia is zero, so this is a simple biaxial superposition.

$$\tan\alpha=\frac{M_y\,I_z}{M_z\,I_y}, \qquad I_z=\frac{b\,h^{3}}{12}, \quad I_y=\frac{h\,b^{3}}{12}$$

Neutral-axis tilt α and the second moments of area of a rectangle. When Iz ≠ Iy the neutral axis tilts away from the load plane.

$$\tan\beta=\frac{M_y/I_y}{M_z/I_z}$$

Deflection-direction angle β. Because curvature is proportional to M/I, β does not equal the load angle φ when Iz ≠ Iy.

What is Unsymmetric Bending?

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I've never heard the term "unsymmetric bending" before. How is it different from ordinary beam bending?
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In plain terms, it is the bending you get when the direction of bending does not line up with the section's "principal axes". The beam bending in your first textbook usually applies the load straight down, exactly along the section's strong axis. But in real structures the load can come in at an angle, or the section itself is tilted. Then the beam bends about two axes at once, not just one. That is unsymmetric bending, also called "oblique bending".
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Bending about two axes at once... When I move the load angle φ from 0° with the slider, the strong-axis moment M_z drops and the weak-axis moment M_y grows. Is that the "resolution" you mean?
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Exactly that. A moment is a vector, so it splits into a strong-axis component Mz = M·cosφ and a weak-axis component My = M·sinφ. At φ = 0° it is all strong-axis bending — the same as ordinary beam bending. The higher φ goes, the larger the weak-axis component, and the section starts bending sideways too. For a rectangle the y and z axes are already principal, so the two bending stresses simply add: σ = Mz·y/Iz + My·z/Iy.
🙋
In the cross-section diagram the orange load arrow and the cyan neutral axis are not at right angles. Why is that?
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Good catch. The neutral axis is the line where the stress is zero. In symmetric bending it is perpendicular to the load plane. But in unsymmetric bending its tilt is set by tan α = (My·Iz)/(Mz·Iy). Here is the key point: a rectangle has different Iz and Iy. For a tall section Iz is far larger than Iy. So the neutral axis is pulled toward the stiff strong axis and tilts well away from the load plane. With the default values the neutral axis is tilted by about 66.6°.
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There's also a "deviation from load" card. Doesn't a beam deflect in the direction you load it?
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That is the most interesting — and most dangerous — part of unsymmetric bending. Deflection is proportional to curvature, and curvature to M/I. The weak-axis direction has a small I, so it bends more for the same moment component. So the deflection direction β shifts toward the weak axis relative to the load direction φ. With the defaults the load is at 30° but the beam deflects at 66.6° — a deviation of about 36.6°. A roof purlin sagging diagonally downward is exactly this effect.
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So what should I watch out for most when designing?
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Always check the corner stress. In unsymmetric bending the strong-axis and weak-axis stresses pile up at the corner of the section. Estimating with a simple M/Z misses it badly on the unsafe side. A practical fix is to choose a section where Iz and Iy are close. With a near-square section, or a smartly chosen orientation, even an oblique load keeps the neutral-axis tilt and the deflection deviation small. Using a tall, thin section at an angle is the worst-case pattern.

Frequently Asked Questions

Unsymmetric bending occurs when the plane of the bending moment is not aligned with a principal axis (strong or weak axis) of the cross-section, so the beam bends about two axes at once. When the load is tilted by an angle phi from the strong axis, the moment resolves into a strong-axis component Mz and a weak-axis component My, each bending the section. The neutral axis is then no longer perpendicular to the load plane, and the beam deflects in a direction different from the load.
For a rectangle the y and z axes are principal, so the product of inertia is zero and the stress is a simple superposition: sigma = Mz*y/Iz + My*z/Iy, with Iz = b*h^3/12 and Iy = h*b^3/12. The maximum stress is at the extreme corner (y = h/2, z = b/2), where the strong-axis and weak-axis bending stresses add together, giving a value much higher than a simple M/Z estimate.
The neutral axis is the line where the stress is zero. For biaxial bending, setting sigma = Mz*y/Iz + My*z/Iy = 0 gives the neutral-axis inclination tan(alpha) = (My*Iz)/(Mz*Iy). If Iz = Iy the neutral axis is perpendicular to the load plane, but for a rectangle with Iz different from Iy it is pulled toward the stiffer (strong) axis and tilts away from the load plane. The taller and narrower the section, the larger this tilt.
Deflection is proportional to curvature, and curvature is proportional to M/I. Because I differs between the strong and weak axes, the weak-axis direction (lower stiffness) deflects more for the same moment component. As a result the deflection direction shifts toward the weak axis relative to the load angle phi, and the deflection-direction angle beta = atan((My/Iy)/(Mz/Iz)) does not equal phi. This is why a slender purlin sags sideways and an L-angle twists.

Real-World Applications

Roof purlins and rafters on a sloped roof: A purlin laid horizontally on a pitched roof carries a vertical load (roof weight or snow) that does not line up with the section's principal axes, making it a textbook case of unsymmetric bending. The purlin deflects not only vertically but also along the roof plane, and that sideways deflection cannot be ignored. Designers add purlin braces and sag rods to restrain the weak-axis deflection and limit the effect of unsymmetric bending.

L-shaped equal/unequal angle sections: The principal axes of an angle section are tilted by roughly 45° from the leg directions. Loading an angle along a leg always produces oblique bending offset from the principal axes, so the neutral axis tilts strongly and the angle deflects as if twisting. This is why an angle is hard to use as a stand-alone bending member: excellent as a truss tension or compression member, but a bending member needs careful checking.

Crane rails and guide rails: A crane runway rail carries a side thrust (lateral load) in addition to the vertical wheel load. The combined vertical-plus-horizontal load is offset from the section's strong axis, so the rail head is in a state of unsymmetric bending. Stress concentrates at the corner of the rail head, which over time becomes the starting point of head wear and cracking, so evaluation with the combined stress is essential.

Machine parts and cantilever structures: Machine-tool arms and robot links carry their own weight and reaction forces from arbitrary directions. When an oblique load acts on a part with a rectangular or I-section whose principal-axis direction is fixed, an unexpected lateral deflection appears that affects machining or positioning accuracy. Using a hand calculation like this tool to estimate the unsymmetric-bending contribution before FEM saves you from chasing mysterious deflections.

Common Misconceptions and Pitfalls

The biggest misconception is "bending stress can just be calculated with M/Z". The formula σ = M/Z using the section modulus Z holds only when the load is cleanly aligned with a principal axis — that is, only for symmetric bending. If the load is tilted even slightly from the principal axis, the strong-axis and weak-axis stresses add at the corner, and you must evaluate it with the biaxial superposition σ = Mz·y/Iz + My·z/Iy. Even with this tool's default values, the corner stress is the strong-axis part (64.95 MPa) plus the weak-axis part (74.99 MPa), totalling 139.9 MPa. An M/Z estimate alone is nearly half on the unsafe side.

Next, assuming the neutral axis is always perpendicular to the load plane. This carries over the intuition of symmetric bending. In unsymmetric bending the neutral-axis tilt α is set by tan α = (My·Iz)/(Mz·Iy), and it is perpendicular to the load plane only in the special case Iz = Iy (square or circular sections). For a rectangle, the more Iz and Iy differ, the more the neutral axis tilts toward the strong axis, and which corner of the section is most critical departs from intuition. Get the location of the maximum stress wrong and you put the reinforcement — and the inspection — in the wrong place.

Finally, the naive assumption that "a beam deflects in the direction it is loaded". In unsymmetric bending the deflection direction β does not match the load direction φ. The lower-stiffness weak-axis side bends more, so the deflection shifts toward the weak axis relative to φ. This "lateral deflection shift" creates unexpected clearance loss or interference in guide mechanisms and precision positioning. Moreover, for a section like an L-angle whose principal axes are far from the leg directions, oblique bending can come with twisting, so lateral-torsional buckling must also be checked. With unsymmetric bending, it is essential to look not only at "stress" but also at "the direction of deflection".

How to Use

  1. Enter rectangular beam width (b) in mm and height (h) in mm to define the cross-section geometry.
  2. Input total applied moment (M) in kN·m and its angle (φ) in degrees measured counterclockwise from the z-axis.
  3. The simulator decomposes the moment into principal components M_z and M_y, calculates maximum corner bending stress σ in MPa, determines neutral-axis tilt α and deflection direction β, then displays deviation from the applied load angle.

Worked Example

A 150 mm × 250 mm rectangular steel beam section (E = 200 GPa) receives an applied moment of 8.5 kN·m at φ = 35° from the strong z-axis. The simulator computes: M_z = 6.96 kN·m, M_y = 4.87 kN·m. Using I_z = 195.3 × 10^6 mm^4 and I_y = 42.2 × 10^6 mm^4, the maximum corner stress reaches 89.4 MPa. The neutral axis tilts α = 22.8° from vertical, deflection direction β = 35° tracks the load angle, and deviation is 0.0°. Principal stresses and deflection paths remain orthogonal to the inclined neutral axis.

Practical Notes

  1. When φ = 0° or 90°, unsymmetric bending reduces to standard symmetric bending; verify M_y = 0 or M_z = 0 respectively.
  2. Corner points opposite the load direction experience maximum tensile stress; locate them at ±b/2, ±h/2 relative to the centroid.
  3. Neutral-axis tilt α ≠ φ because the axis aligns with principal axes weighted by stiffness ratios I_y/I_z, not load angle.
  4. For I-sections or channels, use this tool only after computing accurate principal moments of inertia; default rectangular formulas do not apply.