Deliberately the wrong shape

Plane sections fixes the strain across a bending section: linear, zero at the neutral axis, largest at the face. The material then decides what stress goes with each strain, and for concrete that relationship is a curve — rising to a peak at about 0.2% strain, flat or falling from there, crushing at 0.35%.

Read that curve off the linear strain profile and the compression zone of a beam carries a stress distribution shaped like this:

Nobody integrates it. Every code in the world replaces the curve with a rectangle — reduced intensity, reduced depth — and gets the same answer. The interesting question is not whether that works, since it has worked for a century, but why it works, because the reason is not that the shapes are similar. They are not.

The two questions, which is the whole argument #

A bending calculation interrogates a stress distribution exactly twice:

$$C = \int \sigma\, dA \qquad\qquad \bar{y} = \frac{\int \sigma y \, dA}{\int \sigma \, dA}$$

How much compression is there, and where does its resultant act. Nothing else about the distribution enters the moment. The lever arm is $d - \bar{y}$, the moment is $C(d - \bar{y})$, and two distributions agreeing on those two numbers give identical moments whatever else they differ in.

So the equivalent rectangle is not chosen to look like the curve. It is chosen to have the same two integrals, and once that is done the resemblance is beside the point.

The construction takes two lines. Write the curve’s resultant as a fraction of the box it sits in and its centroid as a fraction of the depth:

$$\alpha = \frac{C}{f_{cd}\, b\, x}, \qquad \beta = \frac{\bar{y}}{x}$$

For the standard parabolic-rectangular curve taken to 0.35% strain these are

$$\alpha = 0.8095, \qquad \beta = 0.4160$$

A rectangle of intensity $\eta f_{cd}$ over a depth $\lambda x$ has resultant $\eta \lambda f_{cd} b x$ and centroid $\lambda x / 2$. Matching both gives

$$\lambda = 2\beta = 0.832, \qquad \eta = \alpha/\lambda = 0.973$$

and that is the derivation in its entirety. $\lambda = 2\beta$ is not a coincidence and not a fit; it is what “same centroid” means for a rectangle.

Which free body produced the number #

The compression zone above the neutral axis, taken as a free body with the tension steel’s force below it. Horizontal equilibrium fixes the neutral axis depth, and the moment about the steel gives the capacity.

For the section drawn above — 300 by 550 effective, C30, 1,800 mm² of grade 500 steel — the steel pulls 783 kN, the neutral axis settles at 189.6 mm, and the moment comes out 368.7 kNm. Solved with the curve, integrated over two thousand strips: 189.56 mm and 368.73 kNm. Solved with the rectangle, in three lines of arithmetic: 189.56 mm and 368.73 kNm.

Not close. Identical, to the precision the numbers are printed at, because the two distributions were built to have the same two integrals and the two integrals are all that equilibrium and moment use.

Two shapes that do not work, and by how much #

The claim that this is about the integrals rather than about smoothness is testable, so it is worth testing. Take the same compression zone and put two other plausible distributions on it.

A triangle — the elastic distribution, which is what the section carried before it yielded. Resultant $0.5f_{cd}bx$, centroid at $x/3$ from the face. It gives 383 kN at a lever arm of 500 mm: 36.7% low.

A full rectangle — the crudest plastic idealisation, everything at $f_{cd}$. Resultant $f_{cd}bx$, centroid at $x/2$. It gives 765 kN at 475 mm: 20.3% high.

Both are smooth, both are simple, both sit over exactly the same compression zone, and both are useless. The equivalent rectangle is not in a different class because it is a better-shaped approximation. It is in a different class because it is the only one of the three that was constructed from the curve’s own integrals.

How the neutral axis is found, which is where the rectangle earns its keep #

The exactness above is a statement about a given neutral axis. In a real calculation the neutral axis is not given: it is the unknown, and it is found from horizontal equilibrium.

With the curve, that means guessing $x$, integrating the compression over the zone, comparing with the steel’s tension, and iterating. It is four lines of code and it was never four lines of anything else. With the rectangle the compression is $\eta f_{cd} b \lambda x$ — linear in $x$ — so equilibrium is a single division:

$$x = \frac{A_s f_{yd}}{\eta \lambda f_{cd} b}$$

and the moment follows directly. That is the actual saving, and it is larger than the saving on the moment: the moment was one integral either way, while the neutral axis was a root-finding problem and is now not.

It is also why the block survived the arrival of computers. A closed form for the neutral axis is what lets a designer see how the answer moves when the steel changes, and a section calculation that has to be re-run to answer “what if” is a different tool from one that can be read.

Where the numbers came from, which was not a derivation #

The construction above is how the factors are justified now. It is not how they arrived.

Charles Whitney proposed the equivalent rectangle in 1937 as a fit: he had a body of test results on beams failing in compression, he wanted an arithmetic that reproduced them without an integration, and he found a pair of factors that did. The two-integral argument came afterwards, as the explanation of why the fit was so good.

That order of events is worth knowing for a general reason. A rule that is a fit and a rule that is a theorem behave identically inside their range and completely differently outside it — and the whole of the difference shows up at the edges. Whitney’s factors held for sixty years and then began to move, above C50, exactly where the tests he fitted them to had no specimens. The theorem-shaped account says why: the shape of the normalised curve had changed, and a factor that describes a shape cannot survive the shape changing.

What the factors are properties of, which is not the concrete #

Compute $\alpha$ and $\beta$ at C20, C25, C30, C35, C40, C45 and C50 and they come out identical: 0.8095 and 0.4160 at every grade.

That is not a rounding. The design strength $f_{cd}$ multiplies the whole curve, and both integrals are normalised by it — one divided by $f_{cd}bx$ and one a ratio of two integrals — so a uniform scaling of the stress axis cancels out completely. The factors describe the shape of the normalised curve and nothing else.

What does move them is the shape itself:

Both of those are real variations. High-strength concretes above about C50 are less ductile, crush at a lower strain and have a less pronounced plateau — which is exactly why every code that goes above C50 starts varying $\lambda$ and $\eta$ with the grade at that point, and not before. The variation is not the strength arriving in the formula; it is the strength changing the shape of the curve.

What the rounding costs, which is the only real error #

The codes do not print 0.832 and 0.973. They print 0.8 and 1.0, and for a good reason — nobody wants a lever-arm calculation with four significant figures in it when the concrete strength is known to one.

That rounding is a second approximation on top of a construction that was exact, and it is the only place in this whole procedure where an error actually lives. On the section above it is 0.69%, and it is on the safe side. It stays under one per cent across the useful range of neutral axis depths, and it changes sign nowhere — because $\lambda$ was rounded down and $\eta$ up, and the two errors partly cancel in the resultant while adding in the lever arm.

Which is worth saying plainly, because the received account has it backwards. The stress block is usually described as “an approximation that works well”. It is an identity that has been rounded, and the rounding is the approximation. If the printed factors were 0.832 and 0.973 there would be no error at all.

Where the model stops #

Four places, and the first two are ordinary while the third is the one that catches people.

Above C50. The curve’s shape changes with the grade and the factors have to vary with it, as described above.

Under axial load. Everything here assumed the strain profile runs from zero at the neutral axis to $\varepsilon_{cu}$ at the face. A section carrying substantial axial compression may be entirely in compression, with the strain profile a trapezium rather than a triangle, and the factors for that are different ones — which is where the second branch of every column interaction diagram comes from.

Where the compression zone is not rectangular. The construction gives $\alpha$ and $\beta$ for a stress distribution, and turns them into a rectangle for a section of constant width. A tee beam whose neutral axis falls in the web, a circular column, a section with a chamfer — for all of these the equivalent rectangle has to be applied to the actual width at each depth, and quoting a single $\lambda x$ without checking the width over that depth is a common and quiet error.

In the serviceability calculation, where it does not apply at all. The block is an ultimate-limit-state device. At working load the concrete is nowhere near crushing, the distribution really is close to linear, and the section to use is the cracked transformed one. Using a stress block to compute a deflection is using the wrong end of the material’s curve.

The third of those deserves an illustration rather than a sentence, because it is the one that produces wrong numbers in practice rather than merely imprecise ones. A tee beam with a 150 mm flange and a neutral axis at 190 mm has a stress block that is 150 mm of full width and 40 mm of web — and a designer who applies $\lambda x = 158$ mm at the flange width has assumed 158 mm of a section that is only 150 mm wide there. The error is in the direction of over-estimating the capacity, and it grows as the neutral axis approaches the flange boundary from below.

What the picture cannot show #

The figure draws a stress distribution as though the compression zone were a continuum with a stress at every point. It is not. Concrete at 0.35% strain in a real beam is a network of cracks and crushed patches, and the “stress” plotted is a smeared quantity over a region a good deal larger than the aggregate. The curve is a description of a specimen’s average behaviour, transferred to a region of a member on the assumption that the region behaves like a small specimen — which is precisely the assumption the size effect says is not safe in general.

It happens to be safe here, because the compression zone of a bending member is confined by the material around it and fails in a much more ductile way than a cylinder in a testing machine does. But the reason is a physical one about confinement rather than anything in the arithmetic, and it is why the same block cannot be used for a plain concrete member with nothing holding it together sideways. Confining the compression zone changes both the peak and the crushing strain, and so changes the factors — which is the one material intervention that moves them at all.

The thing the block quietly decides, which is not the moment #

There is one output of a section calculation that the equivalent rectangle does not reproduce, and it matters more than any of the small errors above.

The rectangle gets the resultant and the centroid right. It does not get the strain distribution right, because it does not have one — it is a stress shape with no material behind it. So anything that depends on how far the concrete has actually been strained is outside what the block can say: the curvature at failure, the rotation capacity of a hinge, whether the section is ductile enough to redistribute, and how much warning it gives.

Those are exactly the properties a plastic analysis depends on, which is why the two calculations are usually run with different machinery. The moment comes from the block in three lines; the rotation capacity comes from a full integration of the real curve, and the two live in different parts of the same design.

It is also why a section that “just passes” on a block calculation can be a poor section. The block will report the same moment for a heavily reinforced section with its neutral axis at $0.6d$ as any other method would, and will say nothing whatever about the fact that such a section crushes before its steel yields.

The generalisation, which is worth more than the stress block #

The habit is this: before approximating a distribution, find out what the calculation actually asks of it.

A moment asks two integrals. A shear check asks a different pair — a first moment over a second moment — which is why the shear stress distribution cannot be replaced by a rectangle even though the direct stress can. A deflection asks for the whole curvature field along the member, which is why it needs the real stiffness rather than an equivalent one. A fatigue check asks for the peak, which no smeared distribution contains at all. And a stability check asks whether the compression zone can reach any of these stresses without buckling, which is a question about the geometry of the region rather than about the stress in it.

Once the question is named, the approximation is usually obvious and often exact. The stress block is the best-known instance and it is a good one to carry, because it is the case where the approximation looks least like the truth and reproduces it best — which is a useful corrective to the instinct that a picture which looks right is a picture that computes right.