The force that arrives along a length

Every transfer of force in this collection so far has had an address. A bolt takes its share at a hole and hands it over through bearing on a plate. A weld takes its share along a line and hands it over through shear on a throat. A bearing takes its share under a plate. In each case there is a place where the force crosses from one piece of material to another, and the design question is how big that place has to be.

A reinforcing bar has no such place. It is a rod cast into a hole of its own shape, and the only thing preventing it from sliding straight out is the ribbing bearing on the concrete between the ribs — a shear stress smeared over the bar’s surface, with no beginning and no end. The force in the bar is therefore not a number at all. It is a function of position, and the thing every code in the world calls a development length is where that function reaches zero.

The arithmetic everybody quotes, which is one line #

Assume the bond stress is uniform at $f_{bd}$ over the whole embedded length. Then equilibrium of the embedded bar is a single statement — the force the bar carries has to equal the force the surface can pass:

$$\sigma_s \cdot \frac{\pi \phi^2}{4} = f_{bd} \cdot \pi \phi \cdot l \qquad\Longrightarrow\qquad \frac{l}{\phi} = \frac{\sigma_s}{4 f_{bd}}$$

and π, the area and the perimeter have all cancelled. What is left is a multiple of the bar’s own diameter and of nothing else: not its length, not the beam it sits in, not the cover over it. A 500-grade bar at its design stress in ordinary concrete wants about forty diameters, and that is why every code states development length in diameters and why the same table serves a footing and a bridge pier.

The free body is worth naming explicitly, because it is the smallest one on this site. Cut the bar at the point where it is fully stressed and again at its end; the body between is a cylinder of steel with $\sigma_s A_s$ pulling on one face, nothing on the other, and a shear traction over its curved surface. Nothing else touches it. Two forces, one equation.

The arithmetic nobody quotes, which has a length in it #

A uniform bond stress is what the interface looks like after it has yielded along its whole length. Before that it has not, and the difference is the whole of this essay.

Take the simplest possible bond law — a stress proportional to slip, $\tau = k s$, which is the secant to the real bond–slip curve at its peak. The same free body, now taken as a differential element, gives

$$\frac{d^2 s}{dx^2} = \frac{\pi \phi k}{E_s A_s} s = \alpha^2 s, \qquad \alpha = \sqrt{\frac{4k}{E_s \phi}}$$

whose solution with a free far end is $s \propto \cosh \alpha(L - x)$. The bond stress is therefore largest at the loaded end and falls away over a distance $1/\alpha$, which for a 20 mm bar with a slip of 0.6 mm at peak bond is about 470 millimetres. The far end of a bar embedded forty diameters is doing very little.

Set the peak bond stress to its limit and the anchorable force comes out as

$$F(L) = \tau_{max}\,\pi \phi L \cdot \frac{\tanh \alpha L}{\alpha L}$$

so the efficiency of an embedded length is $\tanh(\alpha L)/(\alpha L)$. At the code’s own development length that is 55 per cent — a little over half the surface is being used, and the rest is waiting.

The same function, in a problem that looks nothing like it #

That expression is not new to this collection, and the place it has already appeared is worth the detour.

A line of bolts in a lap splice obeys it too. The two plates strain at different rates — at the leading end one carries everything and the other nothing — so the slip between them is largest at the ends and almost nothing in the middle, and the bolt forces are a hyperbolic cosine with its minimum at the centre. The efficiency of a bolted lap of length $L$ is $\tanh(\beta L/2) \big/ (\beta L/2)$, with $\beta^2 = k(1/EA_1 + 1/EA_2)$.

Two mechanisms with no material, no geometry and no physical quantity in common share one equation, and the reason is that both are the same abstract problem: something passed across an interface between two members whose stiffnesses are not equal. A cooling fin’s temperature does it as well. Saint-Venant’s principle is the same argument again in a third setting, where what dies away is a self-equilibrating stress rather than a shear flow.

And the consequence that makes the code’s answer true #

If the efficiency falls as $1/\alpha L$ for long bars, the anchorable force approaches a ceiling. For the 20 mm bar drawn above that ceiling is 80 kN, and the bar itself carries 137 kN at its design stress. An elastic, brittle bond therefore cannot anchor the bar at all, however long it is made.

Real anchorages work, so something must be wrong with that model — and what is wrong with it is the word brittle. A real bond–slip curve rises to a peak and then holds a substantial residual, and it is that plateau which lets the loaded end stop taking more, slip, and hand its share along. Once the bond has yielded from the loaded end all the way to the far end, the stress along it is uniform, and the uniform-stress answer becomes exactly right.

So the code’s development length is not an approximation to the elastic solution. It is a statement about ductility. The arithmetic at the top of this essay is the fully plastic limit, and it is reachable only because the interface can deform. That is the same relationship the long bolted joint has with its own code rule: the reduction applied to long joints exists because the end bolt runs out of deformation capacity, not because the elastic distribution is uneven. Where the fastener is brittle — a resin anchor, a grouted socket in a hard grout — the elastic answer is the one that governs, and lengthening the anchorage stops helping.

Which free body produced the number, when the beam is cracked #

Everything above concerns a bar being pulled out of a block. A bar in a beam is being pulled by something else: the bending moment, through the internal lever arm. And the moment at a section is not what decides the tension in the bar at that section.

The reason is the diagonal crack. In a truss model of a cracked beam, the compression in a diagonal strut running at $\theta$ to the axis is balanced at the bottom node by tension in the chord — and that node is not below the section the moment was computed at, it is further along. Working the equilibrium of the free body cut by the crack rather than by a vertical plane gives the chord tension as that of a section

$$a_l = \frac{z(\cot\theta - \cot\alpha)}{2}$$

further along the span, where $\alpha$ is the inclination of the links. For vertical links and struts at 45 degrees, that is half the lever arm; for the flatter struts a variable-angle truss allows, it is more.

This is the tension shift, and it is the reason a bar cannot simply be stopped where the moment says it is no longer needed. The development length has to start from a station the moment diagram does not mark, and it is a station that moves when the strut angle is chosen. A designer who flattens the struts to reduce the shear reinforcement has, without noticing, lengthened every bar in the beam.

The place that matters most is the one where a bar is deliberately stopped. Curtailment is the only economy available in a reinforced beam — the moment falls away from mid-span and the steel provided is allowed to fall with it — and the shift and the development length are the two taxes on it, applied in that order.

The tail that produces is 1,420 mm at each end, which is 16 per cent of the span carrying steel the moment diagram said was finished with. That is the whole cost of the two rules, and it is why curtailment saves so much less than the moment diagram promises.

How much it promises is a separate question with an answer in closed form, because a parabola is flat where it is largest.

So the first steel curtailed buys almost nothing and only the last of it buys much. On the beam drawn the ideal saving is 20 per cent and what survives the shift and the anchorage is 2.2 — a number small enough that the decision to curtail at all is usually made on bar-fixing labour rather than on steel weight.

Where the bar is not in a beam at all #

The far side of this subject is the region where beam theory has never applied. A disturbed region — an end block, a corbel, a wall with an opening, the anchorage zone behind a prestressing plate — has no lever arm to compute a tension from, and the bar’s force comes from a strut-and-tie model instead.

The tie force there is a result of the node geometry rather than of a section calculation, and there is no moment diagram to shift because there is no moment diagram. What replaces the shift is the node itself: the bar has to be developed from the far side of the node it enters, and the node’s position is a decision the designer made when the truss was drawn.

The development rule changes with it, and in a way that is easy to get wrong. A tie in a strut-and-tie model must be developed from the point where it enters the node, not from where the tie is drawn as beginning — because inside the node the bar is confined by the strut bearing on it, the bond is better, and the anchorage is partly mechanical. A hook or a bend at that point is worth a substantial part of the straight length, which is why the end of a beam looks the way it does.

The other end of the same argument is the anchorage zone behind a prestressing plate, where the force arrives all at once through a bearing and the bar’s job is to resist the bursting that follows. There the bond length is irrelevant and the geometry of the flow is everything.

What a lap splice actually needs, and why it is longer #

A lap is two bars overlapping, and it looks as though each of them needs its development length and no more. It needs about half as much again, and the reason is arithmetic rather than caution.

Over a lap, one bar is shedding its force while the other is picking it up. If every bar in a section is spliced at the same place — which is what happens when a whole layer is lapped over a support — then the bond demand per unit length of member is doubled, because both the incoming and the outgoing bars are transferring their full force through the same volume of concrete. The concrete between them has to carry a splitting force that no single anchored bar produces.

A lap is the same interface picture the first half of this essay drew, with the surface running through the concrete between two bars rather than between a bar and a block, and with the demand on it counted twice over.

Staggering the laps so that no more than half the bars are spliced at one section brings the factor back towards one, and that is why splices are staggered rather than because it is tidy. The rule is a statement about how much bond demand a given volume of concrete can take, and it is one of very few places in a code where the answer depends on what the other bars are doing.

The century it took to stop being a mystery #

Bond was for a long time the least theorised quantity in reinforced concrete, and the reason is that it is the only one that cannot be measured without deciding first what is being measured.

Duff Abrams pulled bars out of blocks at Illinois from 1913 and reported an average bond stress — which is the uniform-stress answer, arrived at not as a plastic limit but because a pull-out test measures a total force and a length and can report nothing else. The number that came out was useful and the mechanism behind it was invisible: an average of a distribution nobody had drawn. Plain round bars were still normal then, and their bond really was mostly adhesion and friction, so the average was not badly wrong.

Deformed bars changed the mechanism completely and the arithmetic not at all. Once the ribs bear on the concrete, bond is a bearing problem with a radial component, and the failure is a splitting of the cover rather than a shearing along the bar. That is why the modern rule multiplies $f_{bd}$ by factors for cover, spacing and confinement — three things that have no place at all in an adhesion model and decide everything in a bursting one. The equation at the top of this essay survived the change of mechanism because it never contained the mechanism.

The distribution itself was measured properly only when it became possible to instrument a bar along its length, and what the strain gauges showed was the hyperbolic cosine: crowded at the loaded end, flat in the middle of a long embedment, migrating along the bar as the near end yielded. The plastic redistribution that makes the code’s answer true is visible in those records as it happens.

What a bend costs the concrete inside it #

A hook or a bend shortens the straight length required, and it does so by putting the force somewhere else. Where it puts it is worth a number.

A bar carrying $F$ round a bend of radius $r$ presses on the concrete inside the bend with a transverse force per unit length of $F/r$ — the deviation force any curved tension produces — and that force is delivered over the bar’s own diameter, so the bearing stress on the concrete is

$$\sigma_b = \frac{F}{r\,\phi}$$

For the 20 mm bar above at its design stress, $F = 137$ kN, and a standard mandrel of four diameters gives $r = 80$ mm and $\sigma_b = 86$ N/mm². On a concrete whose design compressive strength is 20, that is more than four times the cylinder value.

It works because the concrete inside a bend is squeezed from every side by the material around it, and confined concrete carries multiples of its unconfined strength. Which is exactly why the permitted mandrel diameter is not a constant: where the confinement is missing — a bar near a free edge, a row of bars at close spacing so that each one’s bearing zone overlaps its neighbour’s — the required radius grows, and a bend that would have been fine in the body of a member is not fine at the corner of one.

Where the model stops #

Three limits are worth naming, because each of them is invisible in the arithmetic above.

The bond strength is not a material property. It is a property of the bar, the cover, the transverse reinforcement and the state of the concrete around it, and it is fundamentally a splitting problem: the ribs bear on the concrete at an angle, the radial component of that bearing tries to burst the cover open like a pipe under internal pressure, and what limits the bond is usually the cover’s ring tension rather than any shearing of the concrete. Increase the cover, or wrap the bar in links, and the bond strength rises — which is why $f_{bd}$ in a code is a table rather than a number.

The same cover decides a second failure of the same interface, on a timescale nothing in a strength calculation reaches.

Cover appears in both terms and the concrete’s own tensile strength appears in one of them, linearly, which is the asymmetry that puts durability clauses in terms of cover and crack width and hardly at all in terms of strength. It is also the same asymmetry the bond strength has: the quantity that decides both is the ring of concrete around the bar, and it is being asked to carry tension in a material chosen for compression.

The elastic model has a linear bond law and real bond is not linear. The decay length quoted here comes from a secant taken at the peak, and a stiffer initial branch would make the crowding worse and the ceiling lower. Nothing in this essay compares the linear answer with a full non-linear bond–slip integration, and the honest statement is that the linear model gets the shape right and the numbers approximately.

And the picture cannot show what the bar does under load reversal. Everything drawn here is a bar pulled once. Under cycling, the ribs grind the concrete in front of them, the bond–slip loop pinches, and the plateau that made the plastic answer true starts to disappear. That is why a seismic detail asks for longer anchorages than a gravity one for the same bar at the same stress — the length is not being asked to carry more force, it is being asked to survive having its ductility spent.

The generalisation #

What this essay is really about is a class of problem, not a material. Whenever a force has to cross between two members that strain at different rates, the transfer is not uniform: it crowds at the ends, obeys a hyperbolic cosine, has a decay length of its own, and reaches a ceiling if the interface cannot yield. A shear connector in a composite beam does it, a bolted lap does it, a bonded plate does it, and a bar in concrete does it.

The engineering consequence is always the same and always slightly counter-intuitive: the interface works because it is ductile, not because it is long. Design it to be strong and brittle and the length stops buying anything. That is the reverse of nearly every other rule on this site, where more material in the right place is the answer, and it is the reason the least glamorous property in the whole subject — the ability of an interface to slip a millimetre without letting go — is the one holding up the arithmetic.