---
title: "Derive the bending equation"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:derive-bending-equation"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/derive-bending-equation
site: BCT Engineering
---

# Derive the bending equation

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2074 Asoj, 6 marks; 2073 Shrawan, 8 marks (asked 2 times)
- Video: [BENDING EQUATION OF BEAM](https://www.youtube.com/watch?v=QaAIjr-64bU) — TECHNICAL CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/derive-bending-equation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

Derive the bending equation $\left[\frac{\sigma}{y} = \frac{M}{I} = \frac{E}{R}\right]$

## Short answer

From pure bending, strain ε=y/R and σ=Ey/R, so σ/y=E/R. Neutral axis at centroid. Moment equilibrium M=∫σy dA gives M/I=E/R. Hence σ/y=M/I=E/R.

## Answer

**Bending equation (flexure formula):** $\dfrac{\sigma}{y} = \dfrac{M}{I} = \dfrac{E}{R}$, where $\sigma$ is the bending stress at distance $y$ from the neutral axis, $M$ the bending moment, $I$ the moment of inertia of the section about the neutral axis, $E$ Young's modulus and $R$ the radius of curvature of the beam.

#### Assumptions
1. The beam is initially straight, of uniform section, symmetrical about the plane of loading, and of homogeneous, isotropic material.
2. Plane sections remain plane after bending (Bernoulli).
3. The material obeys Hooke's law; stress is within the elastic limit.
4. $E$ is the same in tension and compression.
5. The beam bends in a circular arc (pure bending) and the radius of curvature is large compared with the depth.

#### Derivation
```text
    M ( __________________ ) M
        N______________A        before bending
                  
         bent into an arc:
            R
        O   .  (centre of curvature)
          \ | /
    ----- -N----A-- neutral layer NA (unchanged length)
          layer EF at distance y
```

Consider a short length $dx$ of the beam bent by moment $M$ into an arc of radius $R$ subtending angle $d\theta$ at the centre O. The neutral layer NA does not change in length:

$$
NA = R\,d\theta = dx
$$

A layer EF at distance $y$ from the neutral layer has the length $(R+y)\,d\theta$ after bending (original length $dx = R\,d\theta$). Its strain is

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

By Hooke's law, $\sigma = E\varepsilon$:

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

So the stress varies linearly with $y$.

#### Neutral axis
For equilibrium of axial forces there is no net force on the section: $\int\sigma\,dA = \frac{E}{R}\int y\,dA = 0$. Hence $\int y\,dA = 0$, so the neutral axis passes through the centroid.

#### Moment of resistance
The force on a strip of area $dA$ is $\sigma\,dA$ and its moment about the neutral axis is $\sigma\,dA\cdot y$. The total moment of resistance equals the applied bending moment:

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

$$
\frac{M}{I} = \frac{E}{R} \qquad (2)
$$

Combining (1) and (2):

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

The maximum stress occurs at the extreme fibre: $\sigma_{max} = \dfrac{M}{I}y_{max} = \dfrac{M}{Z}$, where $Z = \dfrac{I}{y_{max}}$ is the section modulus.

## Check yourself

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

1. What is true of the neutral layer when a beam bends?
   A. It has maximum compression
   B. It has maximum tension
   C. It becomes longer
   D. Its length does not change (NA = R dtheta = dx)
   Answer: D. The neutral layer is neither stretched nor compressed. (Key point: Neutral layer)

2. What is the strain of a layer at distance y from the neutral layer?
   A. y/R
   B. R/y
   C. y R
   D. y/(R+y)
   Answer: A. Strain = ((R+y)dtheta - R dtheta)/(R dtheta) = y/R. (Key point: Strain at distance y)

3. Why does the neutral axis pass through the centroid?
   A. Because M = 0 there
   B. Net axial force is zero, so integral of y dA = 0
   C. Because I is minimum there
   D. Because sigma is the same everywhere
   Answer: B. Zero net axial force with sigma = E y/R requires integral y dA = 0, the centroid. (Key point: Neutral axis location)

4. How does M = E I/R arise?
   A. M = integral of sigma dA
   B. M = sigma/y
   C. M = integral of sigma y dA = (E/R) integral of y^2 dA
   D. M = E y/R
   Answer: C. Moment of strip forces about the NA gives M = (E/R) I, the second moment of area. (Key point: Moment of resistance)

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