Theory of Simple Bending
Why I depends on the cube of depth, why that makes an I-beam an I, and what the flexure formula quietly assumes away.
Skip to the animationIf plane sections remain plane, strain varies linearly with distance from the neutral axis, and integrating the resulting stresses gives M/I = σ/y = E/R — from which the fourth-power dependence of I on depth explains the I-beam, the joist on edge, and most of structural design.
The neutral axis
Sagging shortens the top fibres and lengthens the bottom ones. Between them, continuity requires a layer whose length is unchanged — the neutral axis. It is forced by the geometry, not assumed for convenience.
With no axial load, the compressive resultant must equal the tensile one, which places the neutral axis at the section's centroid. For a symmetric section that is mid-depth. For a T-section it is not, so the top and bottom fibre stresses differ and the section has a strong way up.
From assumption to formula
- 1Assume a plane cross-section remains plane as the beam bends.
- 2Geometry then gives ε = y/R — strain proportional to distance from the neutral axis.
- 3Hooke's law converts that to a linear stress distribution, σ = Ey/R.
- 4Multiply each fibre's stress by its area and its distance, and sum over the section.
- 5The result is M/I = σ/y = E/R.
The third term, E/R, relates moment to curvature and is what the deflection calculation integrates. It is easy to overlook because the first two terms are what size the beam.
Why shape beats quantity
I = bd³/12 for a rectangle — the cube of depth. Turning a 50 × 150 timber joist on edge multiplies its I by nine with no extra material at all.
Because stress grows with distance from the neutral axis, material near the axis is barely stressed and barely useful. Move it outward and it earns its weight. That is the entire reason an I-beam is an I: flanges far out where y is large, and a thin web only to hold them apart and carry shear.
| Section | Relative I for the same area | Comment |
|---|---|---|
| Solid square | 1.0 | Baseline |
| Rectangle, 1:3, on edge | ≈ 3 | Free, just by orientation |
| I-section | ≈ 5 – 10 | Material moved to the flanges |
| Hollow tube | ≈ 4 – 8 | Equal in every direction, which matters for columns |
Section modulus
Defining Z = I/y_max reduces the design check to σ_max = M/Z. Since Z is tabulated for every rolled section, sizing a beam becomes three steps: find M_max, divide by the allowable stress, and pick a section with a larger Z.
An unsymmetric section has two values of Z, one for each extreme fibre, and the smaller one governs. Steel tables list Z for both axes and both fibres for exactly this reason.
The assumptions, and what each rules out
| Assumption | What breaks it | Consequence |
|---|---|---|
| Plane sections remain plane | Short, deep beams | Warping; beam theory does not apply |
| Linear elastic material | Loading past yield | Plastic bending gains capacity the elastic theory cannot see |
| Load in a plane of symmetry | A channel loaded off its shear centre | It twists as well as bends — unpredicted by the formula |
| Bending only | Any real beam | Shear stress is a separate calculation entirely |
The twisting case is the nastiest, because the flexure formula gives no warning at all — it assumed the load was symmetric, so a violation simply produces a confident wrong answer.
The numbers you will be asked for
- Flexure formula
M / I = σ / y = E / R
- Bending stress
σ = M·y / I
- Section modulus
Z = I / y_max · σ_max = M / Z
- Rectangle
I = bd³ / 12 · Z = bd² / 6
- Circle
I = πd⁴ / 64
- Parallel axis theorem
I = I_G + A·h²
for building up composite sections
Watch it work
Check yourself
question 1 / 4
One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.