---
title: "Derivation of the flexure formula"
subject: "Strength of Materials"
chapter: 3
key: "semester_4/strength_of_materials_mechanical:flexure-formula-derivation-assumptions"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/flexure-formula-derivation-assumptions
site: BCT Engineering
---

# Derivation of the flexure formula

Strength of Materials · Chapter 3: Pure Bending

- Asked in: year not given, 8 marks
- Video: [Derivation of Flexure Formula (Bending Stress in Beams) | Strength of Materials](https://www.youtube.com/watch?v=5LXEHkeL8Ck) — MATHalino
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/flexure-formula-derivation-assumptions
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-3

## Question

State the assumptions made in the theory of simple bending and derive the flexure formula M/I = sigma/y = E/R. Show that the neutral axis passes through the centroid of the cross-section.

## Short answer

Assumptions: straight beam, homogeneous, isotropic, Hooke's law, plane sections remain plane, symmetry, large curvature radius. Derivation: strain ε = y/R, stress σ = Ey/R. Zero net axial force gives ∫y dA = 0, so neutral axis through centroid. Moment M = EI/R, thus M/I = σ/y = E/R.

## Answer

#### Assumptions
1. The beam is initially straight, with constant cross-section, and the material is homogeneous and isotropic.
2. The material obeys Hooke's law and $E$ is the same in tension and compression.
3. Plane sections remain plane after bending (Bernoulli's assumption).
4. The beam bends in a plane of symmetry and the loads act in that plane.
5. The radius of curvature is large compared with the depth of the beam.
6. Each layer is free to expand or contract, with no stress between layers.

#### Derivation
```text
       ___ M                 M ___
   ---/    \_____N_____A_____/    \---
        top: compression      O = centre of curvature
        NA : unchanged length
        bottom: tension
```

Consider a length $dx$ of beam bent to a radius $R$ (measured to the neutral layer) that subtends an angle $d\theta$ at the centre of curvature. A layer at distance $y$ from the neutral layer, measured away from the centre of curvature, has original length $dx = R\,d\theta$ and new length $(R+y)d\theta$.

$$
\varepsilon = \frac{(R+y)d\theta - R\,d\theta}{R\,d\theta} = \frac{y}{R}
$$

By Hooke's law:

$$
\sigma = E\varepsilon = \frac{E\,y}{R} \qquad (1)
$$

So the stress varies linearly with $y$: tensile on the side away from the centre of curvature and compressive on the side towards it.

**Pure bending means no net axial force**, so over the section

$$
\int \sigma\,dA = \frac{E}{R}\int y\,dA = 0 \;\Rightarrow\; \int y\,dA = 0
$$

This says the first moment of area about the neutral axis is zero, so the **neutral axis passes through the centroid**.

The **moment of resistance** equals the applied moment $M$:

$$
M = \int \sigma\,y\,dA = \frac{E}{R}\int y^2\,dA = \frac{E\,I}{R} \qquad (2)
$$

where $I = \int y^2\,dA$ is the second moment of area about the neutral axis. From (1) and (2):

$$
\boxed{\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}}
$$

#### Use
The maximum stress is at the extreme fibre $y_{max}$:

$$
\sigma_{max} = \frac{M\,y_{max}}{I} = \frac{M}{Z}, \qquad Z = \frac{I}{y_{max}}
$$

$Z$ is the **section modulus**. For a rectangle $Z = bh^2/6$; for a circle $Z = \pi d^3/32$.

## Check yourself

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

1. In the derivation, what is the strain of a layer at distance y from the neutral layer?
   A. epsilon = R/y
   B. epsilon = y/R
   C. epsilon = E y / R
   D. epsilon = y R
   Answer: B. Original length R dtheta, new length (R+y)dtheta, so epsilon = y/R; then sigma = E y/R. (Key point: Strain in a layer)

2. Why does the neutral axis pass through the centroid?
   A. Pure bending gives zero net axial force, so integral of y dA = 0
   B. The moment of inertia is least there
   C. Shear force is zero there
   D. The section is symmetric
   Answer: A. Integral of sigma dA = (E/R) integral y dA = 0, so the first moment of area about the NA is zero, i.e. the NA is at the centroid. (Key point: NA through centroid)

3. How is M/I = E/R obtained?
   A. M = integral of sigma dA
   B. M = integral of sigma y dA = (E/R) integral y^2 dA = E I / R
   C. M = integral of sigma dA / y
   D. M = E R / I
   Answer: B. Equating the applied moment to the moment of the stresses gives M = EI/R, hence M/I = sigma/y = E/R. (Key point: Moment of resistance)

4. Which is one of the assumptions of simple bending theory?
   A. Plane sections remain plane after bending
   B. Plane sections warp after bending
   C. E is different in tension and compression
   D. Layers exert stress on each other
   Answer: A. Bernoulli's assumption: plane sections remain plane. Also E is the same in tension and compression and layers act freely. (Key point: Assumptions of bending)

## Related in Strength of Materials

- [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
- [Derivation of shear stress formula in beams](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Abeam-shear-stress-formula-derivation) — Strength of Materials, chapter 5
- [Bending stresses in a T-section beam](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Acast-iron-t-beam-bending-stresses) — Strength of Materials, chapter 3

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