---
title: "Bending stress in unsymmetrical bending"
subject: "Mechanics of Solids"
chapter: 7
key: "semester_5/mechanics_of_solids:unsymmetrical-bending-flexure-formula"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/unsymmetrical-bending-flexure-formula
site: BCT Engineering
---

# Bending stress in unsymmetrical bending

Mechanics of Solids · Chapter 7: Non Symmetrical Bending of Straight Beams

- Asked in: year not given, 6 marks
- Video: [Unsymmetric Bending with Example](https://www.youtube.com/watch?v=UwYpeATR6Xg) — Structure Pro
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/unsymmetrical-bending-flexure-formula
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-7

## Question

What is unsymmetrical (non-symmetrical) bending? Derive the flexure formula for a beam of arbitrary section with product of inertia $I_{xy}$ and obtain the equation of the neutral axis. Explain why the neutral axis is not perpendicular to the plane of the load.

## Short answer

Unsymmetrical bending arises when loading is not about a principal axis or the section is asymmetric. The general flexure formula is sigma_z = ((M_x I_y + M_y I_xy) y - (M_y I_x + M_x I_xy) x) / (I_x I_y - I_xy^2). The neutral axis (tan alpha = (M_y I_x + M_x I_xy)/(M_x I_y + M_y I_xy)) is not perpendicular to the load plane because the ratio I_x/I_y tilts it towards the weaker axis.

## Answer

#### Meaning

Bending is **unsymmetrical** when (a) the load plane does not coincide with a principal axis of the cross-section (e.g. an inclined load on a rectangular beam), or (b) the section has no axis of symmetry (angle, Z-section). The beam then deflects in a plane different from the plane of loading.

#### Derivation

Let $x, y$ be centroidal axes, $z$ the beam axis, $M_x$ the moment about $x$ (positive when it causes tension at $+y$) and $M_y$ the moment about $y$ (positive moment vector along $+y$, causing compression at $+x$). Plane sections remain plane, so the longitudinal strain varies linearly and

$$
\sigma_z = a + bx + cy
$$

Conditions of equilibrium:

$$
\begin{aligned}
\int \sigma_z\,dA &= 0 \ \Rightarrow\ a = 0 \quad (x, y \text{ centroidal}) \\
\int \sigma_z\,y\,dA &= M_x \ \Rightarrow\ b\,I_{xy} + c\,I_x = M_x \\
-\int \sigma_z\,x\,dA &= M_y \ \Rightarrow\ b\,I_y + c\,I_{xy} = -M_y
\end{aligned}
$$

Solving for $b$ and $c$ with $D = I_xI_y - I_{xy}^2$:

$$
\boxed{\sigma_z = \frac{(M_xI_y + M_yI_{xy})\,y - (M_yI_x + M_xI_{xy})\,x}{I_xI_y - I_{xy}^2}}
$$

Special cases:

- Principal axes ($I_{xy} = 0$): $\sigma_z = \dfrac{M_x y}{I_x} - \dfrac{M_y x}{I_y}$ (the two bending stresses simply superpose).
- Only one moment, symmetric section: the ordinary $\sigma = My/I$.

#### Neutral axis

Put $\sigma_z = 0$:

$$
\frac{y}{x} = \tan\alpha = \frac{M_yI_x + M_xI_{xy}}{M_xI_y + M_yI_{xy}}
$$

This is a straight line through the centroid. In the principal-axes form ($I_{xy}=0$): $\tan\alpha = \dfrac{M_y I_x}{M_x I_y}$. If the axis of the applied moment makes angle $\phi$ with the x-axis (so $M_y/M_x = \tan\phi$, and the load plane makes angle $\phi$ with the y-axis), then

$$
\tan\alpha = \tan\phi\,\frac{I_x}{I_y}
$$

#### Why the neutral axis is not perpendicular to the load plane

The neutral axis would be perpendicular to the load plane only if it coincided with the axis of the applied moment, that is $\alpha = \phi$. This needs $I_x = I_y$ (a circle or square), or a load along a principal axis. Otherwise the ratio $I_x/I_y$ turns the neutral axis towards the axis of **least** moment of inertia. The deflection is perpendicular to the neutral axis, so the beam deflects in a plane different from the loading plane.

```text
     y
     |   load plane
     |  /
     | / phi
  ---+/------- x
    /|
   / | neutral axis (alpha > phi if Ix > Iy)
```

The maximum stress occurs at the point of the section farthest from the neutral axis.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Bending is unsymmetrical when
   A. The load is applied at the free end only
   B. The load plane does not coincide with a principal axis, or the section has no axis of symmetry
   C. The beam is curved
   D. The material is not isotropic
   Answer: B. Either an inclined load plane relative to the principal axes or an unsymmetrical section (angle, Z) makes the beam deflect out of the load plane. (Key point: Meaning of unsymmetrical bending)

2. The derivation starts from sigma_z = a + bx + cy. Which assumption justifies this linear form?
   A. Stress is uniform over the section
   B. Plane sections remain plane so strain varies linearly
   C. The product of inertia is zero
   D. The neutral axis passes through the load
   Answer: B. Plane sections staying plane gives a linear strain, hence a linear stress in x and y; a = 0 from zero net force with centroidal axes. (Key point: Starting assumption)

3. For principal axes (Ixy = 0), the neutral axis inclination satisfies
   A. tan(alpha) = tan(phi)
   B. tan(alpha) = tan(phi) Ix/Iy
   C. tan(alpha) = tan(phi) Iy/Ix
   D. tan(alpha) = Ix Iy
   Answer: B. Setting sigma = 0 gives y/x = My Ix/(Mx Iy) = tan(phi) Ix/Iy. (Key point: Neutral axis angle)

4. Why is the neutral axis generally not perpendicular to the plane of the load?
   A. Because Ix differs from Iy, so alpha differs from phi
   B. Because shear force rotates it
   C. Because the neutral axis does not pass through the centroid
   D. Because the load acts at an angle to the beam axis
   Answer: A. Only when Ix = Iy (or load along a principal axis) does alpha equal phi; otherwise the axis rotates towards the axis of least moment of inertia. (Key point: Why NA is not perpendicular to load)

## Related in Mechanics of Solids

- [Stress in a section with product of inertia](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aunsymmetrical-section-product-inertia) — Mechanics of Solids, chapter 7
- [Winkler-Bach formula for curved beams](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Awinkler-bach-curved-beam-derivation) — Mechanics of Solids, chapter 6
- [Stress in a crane hook section](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Acrane-hook-curved-beam-stress) — Mechanics of Solids, chapter 6
- [Plane stress: principal stresses and oblique plane](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aplane-stress-principal-numerical) — Mechanics of Solids, chapter 2

## Related in other subjects

- [Derivation of the flexure formula](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Aflexure-formula-derivation-assumptions) — Strength of Materials, chapter 3
- [Pure bending, neutral layer, neutral axis](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Apure-bending-vs-simple-bending) — Strength of Materials, chapter 3
- [Neutral axis ratio in WSM beam](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aneutral-axis-ratio-in-wsm-beam) — Design of RCC Structure, chapter 2
- [Parallel axis theorem for product of inertia](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aparallel-axis-theorem-product-of-inertia) — Strength of Materials, chapter 2