The material far from the middle does nearly all the work

Take a steel ruler and try to bend it flat-on. It bends easily. Turn it on edge and it will not bend at all by hand.

Nothing about the steel changed. The same amount of the same material is present, and the difference in stiffness is a factor of several hundred. It comes entirely from where the material sits relative to the axis it is being bent about.

Why the square, and not the distance #

Bending a beam makes one face longer and the other shorter, and somewhere between them is a surface whose length does not change — the neutral axis.

A strip of material at distance $y$ from that axis is stretched in proportion to $y$. Being elastic, its stress is proportional to its strain, so the stress is also proportional to $y$. That is one factor.

The second factor is the lever arm. The strip’s force acts at distance $y$ from the axis, so its contribution to the resisting moment is force times $y$.

Stress proportional to $y$, times area, times a lever arm of $y$, gives a contribution proportional to $y^2$. That is the whole derivation, and the quantity summed over the section is the second moment of area:

$$I = \int y^2 \, dA.$$

The two factors are worth keeping separate, because each on its own would give a linear relationship. It is the coincidence of the outer material being both more highly stressed and further from the axis that produces the square, and the square is what makes the whole of structural shaping worthwhile.

What it buys #

A rectangle of breadth $b$ and depth $d$ has

$$I = \frac{bd^3}{12},$$

which grows as the cube of the depth. Doubling the depth multiplies the stiffness by eight and the strength by four, at twice the material.

That is the ruler. On edge, the depth is the ruler’s width and the breadth is its thickness; flat-on, the two swap. For a ruler 25 mm by 1 mm the ratio of the two second moments is $(25/1)^2 = 625$.

The I-section is the conclusion drawn from the strip diagram, and it is the same argument as truss depth at the scale of a cross-section. If the outer material does nearly all the work, the middle can be reduced to whatever holds the outer material apart and carries the shear. What remains is two flanges and a thin web — a shape that would be structurally absurd for any purpose except this one.

The theorem that makes it computable #

Real sections are assembled from pieces, and the second moment of a piece about the section’s axis is not the same as about its own. The parallel-axis theorem converts between them:

$$I = I_{\text{own}} + A d^2,$$

where $d$ is the distance from the piece’s own centroid to the section’s neutral axis.

For a flange of an I-section, the two terms are wildly unequal. A flange 110 mm wide and 12 mm thick has $I_{\text{own}} = 15{,}800\,\text{mm}^4$ about its own axis, and sitting 94 mm from the section’s neutral axis it contributes $A d^2 = 1320 \times 94^2 = 11.7\,\text{million}$. The transfer term is seven hundred times the flange’s own contribution.

That ratio is the argument in a number. A flange’s own bending stiffness is irrelevant; everything it does comes from being somewhere else. Which is why a flange can be thin and wide, and why moving it further out is worth far more than making it thicker.

Every figure on this site that quotes a second moment computed it this way — summing $I_{\text{own}} + Ad^2$ over the rectangles that make up the profile — rather than looking it up.

From stiffness to strength #

The second moment governs stiffness. Strength needs one more step, because failure happens at the most stressed fibre rather than on average.

The stress at distance $y$ is $\sigma = My/I$, so the largest stress is at the extreme fibre, $y = c$. Setting that to the material’s limit gives the moment the section can carry:

$$M = \frac{\sigma I}{c} = \sigma Z, \qquad Z = \frac{I}{c}.$$

$Z$ is the section modulus, and it is the strength property while $I$ is the stiffness property. For a rectangle, $Z = bd^2/6$ — the square rather than the cube, because dividing by $c$ removes one power of depth.

So depth is worth more for stiffness than for strength: doubling it gives eight times the stiffness and four times the strength. That difference is the reason deflection and strength do not fail at the same span, and the reason long-span beams are almost always governed by how far they move.

Where the neutral axis is #

For a symmetric section the neutral axis is at mid-depth. For an asymmetric one it is at the centroid of the area, and it is not obvious by eye.

That matters because $c$ — the distance to the extreme fibre — is then different on the two faces, so the section has two section moduli. A tee-section has a large $Z$ for the flange side and a small one for the stem side, and it is efficient when the moment always acts in the direction that puts the flange in compression and hopeless when it reverses.

This is why a tee is used for a beam that always sags and an I-section for one that might do either, and why the reversal of load direction during construction has caused more than one failure in a member sized for the other sign.

The same quantity, doing a different job #

The second moment of area appears in two completely different failure calculations, and it is worth noticing that it is the same number.

In bending, $I$ sets how far a beam moves under load. In buckling, it sets the load at which a column stops being able to stay straight. A section chosen for bending stiffness about one axis may have very little about the other, and as a column it is judged by the worse of the two.

Span dominates everything and $I$ is the only term available to fight it, which is why the entire catalogue of structural sections is an argument about where to put the material.

Where the model stops #

Plane sections stay plane. The whole derivation assumes a flat cross-section remains flat as the beam bends, which makes the strain proportional to distance. That is very accurate for slender beams and progressively wrong for deep ones. A beam whose span is less than about twice its depth does not obey it at all, and needs a different theory.

Linear elastic material. Stress is taken as proportional to strain everywhere. Past yield it is not, and the section then has a plastic capacity larger than $\sigma Z$ — for a rectangle, half as much again, because the stress block becomes rectangular rather than triangular.

Bending about one axis. A section has a second moment about every axis through its centroid, and the two principal values can differ enormously. Loading a section about an axis that is not principal produces bending in two directions at once, which for an angle section is the normal case and a well-known trap.

Shear ignored. The whole calculation concerns direct stress from bending, and shear is the other thing a cut reveals. Shear stress has its own distribution, peaks at the neutral axis where the bending stress is zero, and governs short deep members.

The figures have a distortion worth stating: the strips in the first figure are drawn at equal thickness and their contribution bars at true relative length, but there are only eleven of them. The real integral is continuous, and the outer contributions are consequently even more dominant than eleven strips make them look — the top strip of a genuine integration carries a larger share than the diagram can show.

The ladder from here #

Later rungs on this anchor: the parallel-axis theorem derived. Section modulus, elastic and plastic. Principal axes and unsymmetric bending. The shear centre, which is where load must act to avoid twisting. Shear stress distribution and shear flow. Composite sections and the transformed-area method. Cracked sections in concrete. Torsional constants, which follow a completely different rule. And the radius of gyration, which is the same quantity again in the language of buckling.

The $bd^3/12$ result is in Parent’s work of 1713 and in Navier’s of 1826, and it took the intervening century for the profession to believe that the neutral axis sits at the centroid rather than at the bottom face.