Derive the bending equation

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

    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.