Bending stress and the flexure formula, explained

When a beam bends, its top fibres shorten and its bottom fibres lengthen (for a sagging moment). Somewhere in between is a layer that does neither: the neutral axis.

Assumptions of simple bending

1. Plane sections remain plane after bending.
2. The material is homogeneous, isotropic and obeys Hooke's law.
3. E is the same in tension and compression.
4. The beam bends in a circular arc with radius R much larger than its depth.

The flexure formula

MI = σy = ER

where M is the bending moment, I the second moment of area about the neutral axis, σ the bending stress at distance y from the neutral axis, and R the radius of curvature.

Maximum stress and section modulus

Stress is largest at the extreme fibre, y_max. Writing Z = I / y_max (the section modulus):

σ_max = MZ

For a rectangle b × d:

I = b d³12 and Z = b d²6

For a solid circle of diameter D:

Z = π D³32

Worked example

A timber beam 150 mm wide and 300 mm deep carries M = 18 kN·m.

Z = 150 × 300²6 = 2.25 × 10⁶ mm³

σ_max = 18 × 10⁶2.25 × 10⁶ = 8 N/mm²

Why I-sections work

Since stress grows with distance from the neutral axis, material near the neutral axis does little work. An I-section moves most of its area into the flanges, giving a large I for little weight. Doubling depth multiplies Z by four, while doubling width only doubles it.