Derivation of the flexure formula

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

       ___ 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 $Z = \pi d^3/32$.