What is left after the first fibre yields

Elastic bending theory produces one number for a section and stops: the section modulus $Z$, and with it the moment $M_y = f_y Z$ at which the outermost fibre reaches yield. Everything on this site that computes a bending stress computes it against that.

The section, however, has not stopped. One fibre out of a few thousand has reached its limit, and the rest are carrying somewhere between nothing and $f_y$. Put more moment on and the yielded zone eats inwards from the top and the bottom, the elastic core between them shrinks, and the section keeps taking load until every fibre is at $\pm f_y$ and there is nothing left to recruit. The moment then is the plastic moment $M_p = f_y S$, and the ratio

$$\nu = \frac{S}{Z}$$

is the shape factor. The remarkable thing about it is not that it exceeds one. It is what it contains.

Two neutral axes, not one #

The first surprise is not the ratio at all, it is the axis.

The elastic neutral axis is the centroid. That follows from the stress being proportional to distance: for the axial force on the section to vanish, $\int y \, dA$ must vanish about the axis, which is the definition of the centroid.

The plastic neutral axis is not. When every fibre carries $\pm f_y$ and nothing in between, the axial force is $f_y (A_{tension} - A_{compression})$, and for that to vanish the two areas must be equal. The plastic axis is therefore the equal-area axis, and the first moment of area has nothing to do with it.

For a section symmetric about the axis of bending the two coincide, which is why the distinction almost never comes up: rectangles, circles, I-sections about their major axis, boxes. For anything else they do not. A tee of 200 mm flange and 400 mm depth has its centroid well down in the web and its equal-area axis up inside the flange — sixty-nine millimetres apart, a sixth of the depth.

So a yielding asymmetric section migrates its own neutral axis, continuously, from the centroid at first yield to the equal-area axis at full plasticity. No elastic calculation contains that motion, and neither does the plastic one — each of them describes an end of it. What happens in between is a family of stress distributions with a shrinking elastic wedge that is not centred on either axis.

The elastic end of that motion has a second consequence that the plastic end does not, and it is worth drawing before leaving the axis behind. If the neutral axis is not at mid-depth, the two extreme fibres are at different distances from it, so the section has two elastic section moduli and they are not the same number.

That is the cleanest statement of what the equal-area axis does. An asymmetric section is two members elastically and one member plastically, and the shape factor is therefore two numbers as well — a large one for the direction that was weak and a small one for the direction that was strong.

Four closed forms with nothing in them #

For simple shapes the ratio comes out exactly, and it is worth writing the numbers down because they are so bare.

A rectangle of breadth $b$ and depth $d$: $Z = bd^2/6$ and $S = bd^2/4$, since the plastic stress block is two rectangles of area $bd/2$ with their centroids $d/4$ either side. So $\nu = 3/2$, exactly, whatever $b$ and $d$ are.

A circle of diameter $D$: $Z = \pi D^3/32$ and $S = D^3/6$, giving $\nu = 16/3\pi = 1.6977$.

A diamond — a rhombus on its point: $Z = bd^2/24$ and $S = bd^2/12$, so $\nu = 2$ exactly.

A thin circular tube: $\nu \to 4/\pi = 1.2732$ as the wall gets thin.

Not one of those contains a length, a stress, a modulus or a material. That is the sense in which the shape factor is a property of shape: it is a pure number attached to a form, in the way that a rectangle’s ratio of area to bounding box is, and it survives every scaling of the section that keeps the form.

The agreement is the check. Two independent routes to the same four numbers — a pair of centroids on one side and a numerical sum over the area on the other — and nothing in either of them knows what the other did.

The ordering, which is backwards #

Read the six values on the figure above and something uncomfortable falls out. The sections that are best at elastic bending have the least left afterwards.

A rolled I-section — the shape this whole subject exists to justify, the one that gets the material as far from the middle as it can — has a shape factor of about 1.13. A diamond, which is the worst conceivable arrangement of a given area for bending, has 2.00. A rectangle, halfway between, has 1.50.

The reason is exactly the reason the I-section is good. Its material is already at the extreme fibre, so when the extreme fibre yields there is very little further in that was under-stressed and can be recruited. The diamond’s material is nearly all near the middle, doing almost nothing elastically, and the plastic stress block puts every bit of it to work at full stress.

Efficiency and reserve are the same quantity read twice. A section that wastes nothing elastically has nothing to recover; a section that wastes most of itself has a great deal. That symmetry is worth carrying, because it means the phrase “plastic design is worth thirteen per cent” is a statement about I-sections specifically and not about plastic design.

The reserve is approached and not reached #

The second uncomfortable fact is that $M_p$ is an asymptote.

For a rectangle the moment–curvature relation past first yield is exact and short. With the elastic core reaching to $\pm y_e$ and the rest yielded,

$$\frac{M}{M_y} = \frac{3}{2} - \frac{1}{2}\left(\frac{\kappa_y}{\kappa}\right)^2$$

which reaches $3/2$ only as $\kappa \to \infty$. At three times the yield curvature the section is at 96.3 per cent of $M_p$; at ten times, 99.5.

Two consequences follow, and the second is the one that decides real design.

The first is that the plastic moment is never actually developed. Every plastic hinge in every collapse mechanism on this site is carrying a few per cent less than $M_p$, and the calculation is conservative by that amount — which is a small comfort next to how large the modelling assumptions are.

The second is that the reserve has to be paid for in curvature, and the section has to survive supplying it. A stocky section can rotate a long way past first yield with its compression flange still flat; a slender one buckles locally somewhere on the way and stops. That is the entire content of section classification: a class 1 section can reach $M_p$ and hold it long enough for a mechanism to form, a class 2 can reach it and not hold it, a class 3 can reach $M_y$ only, and a class 4 cannot even reach that.

The classification limits are a statement about how much curvature a section can supply before its compression parts stop being flat, expressed as width-to-thickness ratios on the flange and the web and scaled by the square root of the yield stress — so the same profile is a lower class in higher-strength steel, because a higher yield stress asks for more strain before it is reached. The shape factor says how much reserve there is; the class says whether it can be collected.

So the shape factor and the classification are two halves of one question, and they pull in opposite directions on the same variable. Making a section more efficient — thinner flanges further out — lowers the shape factor and pushes the section towards a higher class. The reserve gets smaller and harder to reach at the same time.

Which free body produced the number #

The plastic modulus deserves the same treatment every other quantity here gets: a free body, and an equation that could have come out wrong.

Cut the section on its axis of bending and take the piece above. Every fibre in it carries $f_y$ in compression, so the total force on it is $f_y A_c$, and it acts through the centroid of that area. The piece below carries $f_y A_t$ in tension through its own centroid. Axial equilibrium demands $A_c = A_t$, which fixes the axis; moment equilibrium about any point then gives

$$M_p = f_y (A_c \bar{y}_c + A_t \bar{y}_t) = f_y \cdot \frac{A}{2}(\bar{y}_c + \bar{y}_t)$$

so $S$ is half the area times the distance between the two half-centroids, which is a lever arm in the ordinary sense. For a rectangle that is $(bd/2)(d/2) = bd^2/4$ in one line, and it is the same construction a stress block in a concrete section uses.

The check that matters is that the same number can be reached from the other end. Sweeping the curvature of a fibre model to infinity and reading off the moment gives $S$ by a completely different route — an integration rather than a pair of centroids — and the two agree to the seventh figure. Neither derivation contains the other.

What the picture cannot show #

Everything above is a section under pure bending, and three things are missing from it.

Axial force. The equal-area axis is where it is because the net force is zero. Put an axial load on the section and the axis moves off it, the two areas stop being equal, and the plastic moment falls — which is the whole of the N–M interaction diagram. For a section carrying half its squash load, the plastic moment can be down by a quarter.

Read that as a statement about the number rather than about the member: the shape factor quoted anywhere is the value at zero axial load, which is one corner of a curve rather than a property of the section. A column carrying bending is being asked for a plastic modulus the catalogue does not print.

Shear. The web is carrying shear as well, and where the shear is a large fraction of the plastic shear capacity the material in the web is using part of its yield criterion up on it. The standard treatment is to reduce the web’s yield stress, which reduces $S$ — and the reduction only bites above about half the shear capacity.

Which axis. Every number here is about one bending axis. A section bent about a diagonal has a different equal-area axis, a different plastic modulus and a different shape factor, and for an angle or a channel that axis is nowhere near either principal direction. The shape factor is not one number attached to a section — it is one number attached to a section and a direction.

The forty years it took to be allowed #

The arithmetic in this essay is not hard and it was not new when it became legal. Every term in it was available to a competent engineer in 1900, and plastic design entered British practice in the 1950s.

Gábor Kazinczy tested fixed-ended beams in Budapest in 1914 and reported what the moment–curvature curve above predicts: they carried roughly twice the load at which the first section reached yield, and they did so by forming hinges in a sequence. Hermann Maier-Leibnitz did the same on continuous beams in the 1920s and made the further point that the collapse load did not depend on the support settlements — which is the statement that a plastic calculation does not need the compatibility a continuous elastic analysis spends most of its effort on.

What took the remaining decades was not evidence but the theorems. Until the upper- and lower-bound theorems were stated and proved, a plastic calculation was a plausible estimate with no way of telling which side of the answer it fell on, and no code can be written on that. Once the two bounds existed, a mechanism became a demonstrable overestimate and an equilibrium field a demonstrable underestimate, and the collapse load became a quantity that could be bracketed rather than guessed.

John Baker’s work at Cambridge supplied the third thing needed, which was a reason to care: wartime air-raid shelters designed elastically were heavy, and designed plastically they were not. The Morrison shelter is a plastic design, and the argument that got the method accepted was made out of steel that had to be saved rather than out of theorems.

It is worth naming the assumption all of this rests on, because it is the one that survives yielding unchanged. Sections stay plane. What stops being true when the material yields is the proportionality between strain and stress; the linearity of the strain is untouched, which is why the yielded section can still be described by a single curvature and why the moment-curvature relation above is a curve rather than a surface.

Where it changes what gets built #

The place plastic reserve is worth most is a continuous member, and it is not because of the shape factor.

A continuous beam gets its economy from redistributing moment: the support section yields first, forms a hinge, and passes further load to midspan until a second hinge turns the span into a mechanism. That is worth far more than 13 per cent — a uniformly loaded fixed-ended beam has a collapse load twice its first-yield load — and it needs the shape factor only to be non-zero.

The shape factor is worth a few per cent of one hinge’s capacity; the redistribution between hinges is worth a great deal more, and the two are usually credited to the same idea because they arrive in the same calculation.

There is a third place it matters and it is a detail rather than a member. A bolt group, a base plate and a beam-to-column endplate are all sections in the sense used here — a set of areas resisting a moment — and each of them has an elastic distribution and a plastic one with a ratio between them. The endplate’s is often the largest shape factor in the whole structure, because the plate’s material is spread evenly rather than concentrated at the extremes, and it is the reason a connection designed elastically and tested to failure so consistently outperforms its calculation.

A simply supported beam gets nothing but the shape factor, because there is only one hinge and it is the mechanism. For a rolled I-section that is 13 per cent, on a member whose deflection is usually the thing that governs anyway. That is why plastic design changed continuous frames and left simply supported floor beams almost exactly as they were.

There is a further reason the 13 per cent so rarely decides anything, and it has nothing to do with the mechanics. The section is chosen from a catalogue.

A reserve of 13 per cent is smaller than a single step in that staircase for most of its length. So on a simply supported beam the shape factor very often buys nothing at all: it moves the required modulus down by less than the distance to the next section below, and the same rolled section is ordered either way. What it buys is utilisation on a section already chosen, which is a different and much smaller kind of saving than the arithmetic suggests.

Turn the section over and the reserve triples #

The shape factor is a property of a form and a direction, and the difference between the two directions of the commonest structural shape is larger than anything else on this page.

Bend an I-section about its major axis and the material is nearly all in the flanges, at nearly the same distance from the axis — so the elastic and plastic distributions are almost the same and the ratio is about 1.13. That is the number everybody quotes.

Bend the same section about its minor axis and it is a completely different object. Each flange is now a rectangle bent about its own centre-line, and a rectangle’s shape factor is exactly 1.5. The web is a thin rectangle bent about its own centre-line too, and its shape factor is also 1.5. Both components contribute 1.5, so

$$\frac{S_z}{Z_z} \approx 1.50$$

The same piece of steel has a plastic reserve of 13 per cent one way and 50 per cent the other.

The reason is the one this essay keeps returning to. A shape factor measures how much material was under-stressed at first yield, and an I-section bent the strong way has almost none — every fibre that matters is already near the extreme. Bent the weak way, half of every flange is close to the neutral axis and doing very little, and that half is what the plastic distribution recruits.

Two consequences follow. Plastic design is worth much more about the minor axis, which is the case where it is least often used, because a member bent about its weak axis is usually a member somebody would rather not have. And a member bent about both axes has two different reserves at once, which is why the biaxial interaction is a curve rather than a line — the two axes are not being asked the same question and do not answer it the same way.

The generalisation #

Two sentences are worth separating before the general one, because they are the practical residue of everything above.

The shape factor is a property of a form and a direction, and it is a pure number. It does not scale, it does not depend on the steel, and it can be written down for any section anybody can draw by finding one axis and taking two centroids. Where a section is symmetric the axis is the centroid; where it is not, the axis moves as the section yields and neither end of that motion is where an elastic calculation put it.

And the reserve is a trade rather than a gift. It is collected in curvature, the curvature has to be survivable, and the survivability is a separate calculation with its own limits — which is why the two most efficient shapes in the collection have the smallest reserves and the hardest time reaching them.

The habit this essay is really about is the difference between a limit set by a point and a limit set by a body.

$Z$ is a point limit: it is decided entirely by the single most stressed fibre, and every other fibre in the section is irrelevant to it. $S$ is a body limit: every fibre appears, weighted by area and lever arm, and no single one of them decides anything. The same pair recurs everywhere on this site — a bolt group’s elastic and plastic distributions, a weld group’s, a frame’s first-yield load and its collapse load, and a soil bearing pressure’s peak against its resultant.

In every one of those pairs, the ratio between the two answers is a pure number attached to the geometry, and reaching the second requires the material to redistribute. The shape factor is the simplest member of the family and the only one where the number can be written down exactly for a shape a reader can draw.