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.

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 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 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 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.

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.

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.

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.