Why is an I-beam shaped like an I?

01

The idea

Same steel. Very different beams.

Each shape below uses exactly the same amount of steel.

× 1Square bar

× 18Tall plate

× 40I-beam

× 1Square bar

× 16Tall plate

× 35I-beam

It's not how much steel you have. It's where you put it.

02

Play with it

Rearrange the steel

Start from the square bar. Make it deeper, then push steel into the flanges. Watch the sag.

03

The rule

Distance counts squared

Rectangle

I = \dfrac{b\,\len{h}^3}{12}

Any shape: add up the parts

I = \sum \left( I_0 + A\,\len{d}^{2} \right)

Stiffer means less sag

\delta = \dfrac{\force{P}\,\len{L}^3}{48\,E\,I}

Double the depth and a rectangle gets 8× stiffer. Steel at distance $\len{d}$ from the middle counts $\len{d}^2$ times. So the flanges do most of the work and the web just holds them apart.

04

Your turn

Prove it

◆ Pick1/4

A plank across a gap: which way up?

To do

Same 50 × 200 mm2 × 8 in plank, same gap. Which sags less?

▲ Build2/4

Same steel, at least × 30 stiffer

To do

Depth is capped at 300 mm (the floor above has to fit). Shape the section, then check.

= Solve3/4

Work out I for a rectangle

To do

A solid rectangle, $b = 100\,\mathrm{mm}$ wide and $\len{h} = 200\,\mathrm{mm}$ deep. What is $I$ ?

A solid rectangle, $b = 4\,\mathrm{in}$ wide and $\len{h} = 8\,\mathrm{in}$ deep. What is $I$ ?

◆ Pick4/4

When the beam bends, where is the steel working hardest?

To do

05 · Takeaway

Put the steel far from the middle.

I = \sum \left( I_0 + A\,\len{d}^{2} \right)

Go deeper in the path: Strength of materials covers bending stress and deflection properly.