The beam that becomes a truss

An uncracked web carries shear as a stress field: a continuous distribution, parabolic through the depth, computed from $VQ/Ib$ and requiring nothing of anybody. Once the web has cracked, that field does not exist. There is no continuum left to carry a stress through — there is a set of concrete fingers between the cracks, a compression chord along the top, a tension chord along the bottom, and whatever steel happens to cross a crack.

That is a truss. Not an analogy in the loose sense of a helpful picture: a genuine pin-jointed assembly whose members are identifiable pieces of material and whose forces follow from cutting it and summing.

And it has a free parameter that no other structure on this site has. The angle of the cracks is a choice.

Which free body produced the number #

Cut the beam along a plane parallel to the cracks and take the piece to the right.

Vertically, the only thing crossing the cut that can carry a vertical force is the stirrups — the concrete struts run along the cut and the chords run across it horizontally. So the shear is carried entirely by the stirrups the cut severs, and how many it severs is a matter of geometry: over a lever arm $z$, at an angle $\theta$, the cut has a horizontal reach of $z\cot\theta$, and at a spacing $s$ that is

$$n = \frac{z\cot\theta}{s}$$

stirrups. Each carries $A_{sw} f_{ywd}$, so

$$V = \frac{A_{sw}}{s}\, z\, f_{ywd} \cot\theta$$

For the beam drawn — 300 mm wide, $z = 495$ mm, two 10 mm legs at 150 mm centres, so 1,047 mm² of stirrup per metre — that is 225 kN at 45° and 563 kN at $\cot\theta = 2.5$. The same steel, the same beam, two and a half times the shear.

Nothing was gained from nowhere. Flattening the crack makes the cut longer, and a longer cut crosses more stirrups.

What the web was doing before any of this is a continuous shear flow through an uncracked section, with no discrete anything anywhere in it. The truss replaces that picture rather than correcting it, and the two share no arithmetic — which is why a cracked beam’s shear capacity cannot be got by adjusting the uncracked one.

The other end of the trade #

Something has to push back. Horizontal equilibrium on the same cut needs a compression along the struts, and the strut force divided by the area it acts over is

$$\sigma_c = \frac{V(\cot\theta + \tan\theta)}{b_w z}$$

which has a minimum at 45° and rises in both directions. At $\cot\theta = 2.5$ the factor $(\cot + \tan)$ is 2.9 rather than 2, so the same shear produces 45% more compression in the concrete.

That compression is being carried by concrete that is cracked across the strut and reinforced through it, which is weaker than the same concrete in a cylinder test — the usual allowance is a strength reduction to about $0.6(1 - f_{ck}/250)$ of the design value, which here is $0.528 \times 20 = 10.6$ N/mm². Set the strut stress equal to that and there is a ceiling on the shear:

$$V_{max} = \frac{b_w z\, \nu f_{cd}}{\cot\theta + \tan\theta}$$

784 kN at 45° and 541 kN at $\cot\theta = 2.5$ — falling exactly where the stirrup capacity is rising.

The angle is chosen, and choosing it is the design #

Fix the shear at 450 kN and ask what the stirrups have to be. At 45° the answer is 2,091 mm² per metre; at $\cot\theta = 2.5$ it is 836. Sixty per cent less steel for the same beam, and the only thing that changed is which angle the calculation assumed the cracks would form at.

The price is the 45% rise in strut stress, which is free until it is not: at 450 kN the strut is at 8.8 N/mm² against 10.6 available, so it is affordable; at 550 kN it is not, and the choice disappears.

This is unlike anything else in the subject. A bending calculation has no free parameter — the neutral axis is where equilibrium puts it. A shear calculation in a cracked web has one, and the designer sets it.

The general form of the same freedom is a strut-and-tie model: a load path chosen by the designer rather than found by an analysis, justified by the lower-bound theorem that any equilibrium system the material can sustain is safe. The variable-angle truss is a strut-and-tie model for the part of a beam where sections still make sense, which is why it looks like a formula rather than like a drawing.

The thing it has actually become #

It is worth drawing the truss the beam has turned into beside a truss somebody built on purpose, because the comparison is exact rather than suggestive.

A Warren truss has a top chord in compression, a bottom chord in tension, and diagonals that alternate. A cracked web has a compression chord, a tension chord, concrete diagonals in compression and steel diagonals — the stirrups, vertical rather than inclined — in tension. The difference is that the concrete truss has an infinite number of diagonals of infinitesimal width instead of a countable few, and that its geometry is decided after the load arrives rather than before.

Every quantity transfers. The chord force is the moment over the depth; the diagonal force is the shear over the sine of its angle; the panel length is $z\cot\theta$. A designer who can read a truss can read a cracked beam.

Why a lower bound is enough #

The justification for choosing an angle at all is the lower-bound theorem, and it is worth stating plainly because it is what makes the whole business legitimate rather than optimistic.

If a set of internal forces can be found that is in equilibrium with the applied load and nowhere exceeds the material’s strength, the structure will not collapse at that load. It does not matter whether the real beam distributes its forces that way. It only matters that it could.

So the designer picks an angle, provides stirrups sufficient for the equilibrium that angle implies, checks that the strut it implies is not overstressed, and is safe — provided the beam has enough ductility to reach the chosen distribution, which is why the flattest permitted angle is capped rather than left open.

The other half of what happens across a crack is left out on purpose. Shear across a crack that is already there is carried by aggregate interlock and by dowel action in the longitudinal bars, and the variable-angle truss assumes none of it. That omission is the deliberate conservatism that pays for the freedom to choose $\theta$: a lower bound that discards two real mechanisms has room to be wrong about a third.

The bar is longer than the moment diagram says #

The same cut that severs the stirrups also crosses the tension chord, and it carries something there too.

Take moments about the point where the cut meets the compression chord. The applied shear has a lever arm; the stirrups crossed have their own; and what balances the difference is a force in the bottom chord of

$$\Delta F = \frac{V \cot\theta}{2}$$

At the support of the beam drawn, with 450 kN of shear and $\cot\theta = 2.5$, that is 563 kN of tension in the bottom bars at a section where the bending moment is zero.

The general statement is tidier than the special case. The chord force everywhere is $M/z + V\cot\theta/2$, which for a beam under a uniform load is exactly the ordinary $M/z$ diagram translated $z\cot\theta/2 = 619$ mm toward the support. A bar curtailed where the moment diagram says it is no longer needed is 619 mm short of where the truss says it is.

The cut that reveals the internal forces is taken perpendicular to the axis on every other page of this collection. The truss’s cut is the same operation at a different angle, and the extra chord force is precisely what the perpendicular cut cannot see — because a perpendicular cut crosses no crack, and therefore crosses no stirrups whose vertical force has to be balanced by a horizontal one.

The horizontal cut, which the same truss also answers #

The cut taken so far runs parallel to the cracks. Take a horizontal one instead, along the interface between the web and a flange cast at a different time, and the truss answers that question too — with the same struts and the same angle.

The compression chord force changes along the span at a rate $dF/dx = V/z$, and that change has to be delivered to the flange by shear across the interface. Which is the same $V/z$ that any longitudinal shear calculation gives, arrived at from a truss rather than from a stress field — and it is the same number, which is the check.

What the truss adds is the strut angle at the interface. The compression is delivered by inclined struts in the flange, not perpendicular to it, so the interface reinforcement is designed for $V\cot\theta_f/z$ with its own flange angle, and a flange that is flatter than the web is cheaper for the same reason the web is.

The shear across a joint between two pours is where this arithmetic meets a real interface. The truss and the stress field give the same longitudinal shear, and they disagree about nothing except how the load gets from one side of the joint to the other.

Why the same steel per metre is not the same beam #

The capacity formula contains $A_{sw}/s$ and nothing else about the stirrups, which says that two 10 mm legs at 150 mm centres and two 20 mm legs at 600 are the same beam. They are not, and the reason is in the free body rather than in the formula.

The cut has a horizontal reach of $z\cot\theta$ — 1,238 mm on this beam at $\cot\theta = 2.5$, and 495 mm at 45°. At 150 mm centres that cut crosses 8.25 stirrups, and the fractional part does not matter: whether it happens to catch eight or nine changes the answer by a few per cent either way, and the average is what the formula computes.

At 600 mm centres the same cut crosses 2.06. Now the fractional part is most of the answer. A crack that starts just past a stirrup crosses two; one that starts just before it crosses three; and the difference between them is 50% of the capacity. The formula returns the average of a quantity whose variation is as large as itself.

Worse, at 45° the reach is 495 mm and the cut crosses 0.83 stirrups — a crack can form that crosses none at all, in a beam the calculation says is adequately reinforced.

So the maximum spacing rules are not a detailing nicety layered on top of the model; they are the condition under which the model’s own free body means anything. The stirrups have to be close enough that “the number the cut crosses” is a smooth function of where the cut is, which in practice means the spacing must be a modest fraction of $z$. Below that, the smeared quantity $A_{sw}/s$ describes the beam. Above it, the beam is a set of discrete ties and the crack chooses where to go.

Inclined stirrups, and why they stopped being worth it #

Stirrups do not have to be vertical. A bar inclined at $\alpha$ to the axis crosses the crack more nearly at right angles, and the general form of the capacity is

$$V = \frac{A_{sw}}{s}\,z\,f_{ywd}\,(\cot\theta + \cot\alpha)\sin\alpha$$

which reduces to the vertical case at $\alpha = 90°$. Put $\alpha = 45°$ into it and compare, at the same steel per metre.

At a 45° crack — the fixed angle the truss analogy was originally written with — vertical stirrups give 225 kN and inclined ones give 319 kN, a gain of 42%. That is a large return for bending some bars up, and it is why every reinforced concrete beam detailed before about 1970 has bent-up bars in it: the diagonal tension was assumed to run at 45°, and inclining the steel to meet it was the obvious economy.

At $\cot\theta = 2.5$ the same comparison gives 564 kN vertical and 558 kN inclined — no gain at all, and marginally a loss. Flattening the crack has already done what inclining the steel was doing, and the two do not stack.

What inclined stirrups still buy at a flat angle is the ceiling. The web crushing limit becomes $\nu f_{cd} b_w z (\cot\theta + \cot\alpha)/(1 + \cot^2\theta)$, which at $\cot\theta = 2.5$ is 760 kN with 45° stirrups against 543 with vertical ones — a 40% rise in the maximum the section can ever carry, on a beam where the ceiling was the binding constraint.

So the historical shift is explained by the arithmetic. Under a fixed 45° truss, bent-up bars were the way to get more capacity from the same steel. Under a variable-angle one, they are the way to get past a web that is already at its limit — a much narrower use, on much heavier members, which is where they are still found.

Ductility is what the choice is bought with #

The flattest angle is capped, and the cap is not a strength limit. It is a rotation limit.

For the whole of a long cut’s worth of stirrups to be at yield together, the ones nearest the crack mouth have to have strained well past yield while the ones near its tip are only just reaching it. A flatter crack means a longer cut, more stirrups, and a wider spread of strain across them — so the flatter the assumed angle, the more plastic deformation the steel has to deliver before the assumed distribution exists.

That is the same argument that governs every plastic analysis on this site: a collapse mechanism computed from equilibrium alone is only available to a structure ductile enough to reach it. The cap on $\cot\theta$ at about 2.5 is where the required rotation stops being something a reinforced web will reliably provide.

It is the same demand a plastic mechanism makes on a bent member before its collapse load exists: a hinge has to rotate far enough for the last section to reach its plastic moment while the first is well past it. Every stirrup in a flat-angle truss is making that demand on a smaller scale, and the cap on the angle is the same limit read in a different member.

Where the model stops #

It says the concrete carries no shear at all across the crack, and it does. Aggregate interlock across a narrow crack, dowel action in the longitudinal bars and the shear carried by the uncracked compression zone are all real and all omitted. The omission is what pays for the freedom to flatten the angle, and it is why a beam with no stirrups at all — which this model says has no shear capacity — plainly does have some.

The stirrups are all at yield. They are not: the ones near the tip of the crack are barely strained and the ones near its mouth are past yield, and the model’s uniform distribution is a plastic idealisation that needs the steel to be ductile enough to reach it. That is the real reason the angle is capped, and it is a ductility limit rather than a strength one.

And the strut is a strut. In a real web the compression fans near the support and near a load, so there is no single angle even in the region where the model says there is one. The fan is a strut-and-tie region and the parallel-strut field is the interior of a B-region.

What the pictures cannot show #

The cracks in the drawing are parallel and evenly spaced. Real ones are not: they start at different loads, curve as they climb toward the compression zone, and change angle where a longitudinal bar happens to be. What is drawn is the model’s crack pattern, and it is the model’s rather than a photograph’s.

Nor can they show the sequence. The truss does not exist until the beam has cracked, so at low load the picture is wrong and the uncracked stress field is right. Somewhere between the two there is a beam that is partly one and partly the other, and no figure on this page is about that beam.

The assumption the figure rests on #

The lever arm $z$ is taken as $0.9d$ and held constant along the span. That single number sets the height of the truss, and therefore the number of stirrups any cut crosses, the strut area, and the chord shift — every quantity on this page is proportional to it. In a real member it varies with the moment, and near a support where the moment is small the compression chord is not where $0.9d$ says it is.

It comes from the neutral axis of a cracked section, which puts the compression resultant somewhere near a tenth of the depth below the top face. That is a fair average and it is not a constant, and the truss is built on it.

The number that decides whether any of this is needed #

There is a threshold below which none of it applies, and it is worth being clear that the threshold is about cracking rather than about strength.

A web that has not cracked in shear is a continuum, its shear stress is $VQ/Ib$, and its principal tension is a diagonal stress of the same order. The web cracks when that reaches the concrete’s tensile strength, and for the section drawn — 300 wide, 495 lever arm, an average shear stress of $V/b_w z$ — that happens somewhere near 200 kN.

Below it the beam does not need the truss and does not have one. Above it the beam has cracked, the stress field is gone, and the only description available is the truss. There is no intermediate model on this page and there is not much of one anywhere: the transition is a change of mechanism rather than a change of degree, which is why the shear capacity of a beam is discontinuous in a way its bending capacity is not.

The practical consequence is that minimum shear reinforcement is provided even where the calculation asks for none. It is not a strength requirement — the calculation genuinely says zero. It is there so that when the web does crack, the truss it needs exists.

The ladder from here #

Later rungs on this anchor: the fan regions near supports and point loads, where the angle is not constant and the model becomes a genuine strut-and-tie one. Beams with no shear reinforcement, where the whole of the resistance is the mechanisms this model discards. Shear and torsion together, where the same links serve two demands and the two add rather than compete. The compression field theory that replaces the chosen angle with a computed one, by requiring compatibility as well as equilibrium. Prestressed webs, where an axial compression rotates the cracks flatter and the angle is smaller before anybody chooses it. And the shear at a construction joint, where the crack’s position is decided by the pour rather than by the stress field.