Internal forces

The angle is a choice, not a property

The truss inside a cracked concrete web has a strut angle, and nothing measures it. The designer picks it, the stirrup requirement falls as it flattens, the web stress rises, and every choice in between is a different structure that carries the same load.

Assumes The strength with no mechanism in it, When there is no section to design and Two ways of being wrong.

Once a concrete web has cracked diagonally it stops being a continuum and becomes a truss: a compression chord, a tension chord, concrete struts between the cracks, and stirrups crossing them. That truss has a strut angle, and the remarkable thing about the angle is that nothing in the calculation measures it.

Flattening the truss saves stirrups and crushes the web. Two capacities against the angle of the cracks, for a web 350 mm wide with a lever arm of 630 mm. The rising line is the stirrups: a cut along the crack severs z·cot θ/s of them, so flattening the crack from 45° to cot θ = 2.5 takes the 315 kN they carry to 787 — 2.5 times as much from the same steel. The falling line is the concrete strut, whose stress is V(cot θ + tan θ)/b_w z and therefore least at 45°. They do not cross in this range, so the stirrups govern throughout and the angle is a free choice.
Fig. 1 Two capacities against the crack angle, for a web 350 mm wide with a lever arm of 630 mm. The rising line is the stirrups: a cut along the crack severs z·cot θ/s of them, so flattening the crack from 45° to cot θ = 2.5 takes the 315 kN they carry to 787 — two and a half times as much from exactly the same steel. The falling line is the concrete strut, least stressed at 45°. They do not cross in this range, so the stirrups govern throughout and the angle is a free choice.

Two and a half times the capacity from the same reinforcement, decided by a number the designer writes down. That is either a scandal or a theorem, and it is a theorem.

It is worth being clear at the outset about what is and is not being claimed. Nothing here says a member gets stronger when a designer writes a smaller angle on a drawing. It says that a member detailed for a flat truss — with the stirrups the flat truss asks for, and the longer flexural bars it also asks for — carries the load that model computes. The choice is a choice about what to build, not about what to assume; and the reason it feels like the second is that on a drawing the two look identical.

Where the freedom comes from

The lower-bound theorem says that any state of stress in equilibrium with the load and nowhere exceeding the material’s strength is safe. It does not say the structure will adopt that state; it says the structure will not collapse below the load that state carries.

A truss at 45° and a truss at cot θ = 2.5 are two such states for the same member. Both satisfy equilibrium everywhere, both are drawn on the same cracked web, and each asks for a different arrangement of reinforcement. Provide what either asks for and the member carries the load — which is the whole content of the theorem and is why the angle can be chosen rather than found.

What the theorem does not give is any preference. Choosing the angle is a design decision made outside the mechanics, exactly as choosing a lever arm in a strut-and-tie model is.

Every one of these models is safe, and they disagree by a factor of two. The tie force in a strut-and-tie model of the same region, against the lever arm the model assumes, as a fraction of the depth. It runs from 3000 kN at a lever arm of 40% of the depth down to 1263 kN at 95%, and every model in the shaded band is in equilibrium with the same load. The lower-bound theorem says all of them are safe if the structure is built to carry what they ask for, so choosing one is not a calculation — it is a decision about where the reinforcement goes and how much the concrete has to be trusted. The band is where the strut angle stays between 25° and 65°, outside which the model stops resembling anything the concrete will do.
Fig. 2 The same freedom in a D-region: the tie force against the lever arm the model assumes, running from 3,000 kN at 40 per cent of the depth to 1,263 kN at 95 per cent. Every model in the band is in equilibrium with the same load and every one is safe if built to carry what it asks. The variable-angle truss is this picture applied to a web instead of to a support region.

Which free body produced the number

The free body is a cut parallel to the cracks, and taking it that way rather than vertically is the entire trick.

The cut that severs the stirrups is the cut that counts them. A cracked web drawn as the truss it has become: a tension chord along the bottom, a compression chord along the top, concrete struts between the cracks and stirrups crossing them. The free body is a cut parallel to the cracks, which over a lever arm of 630 mm severs z·cot θ/s = 7.2 stirrups at cot θ = 2. Every one of them is at yield, so the shear is their number times their strength, and flattening the crack raises the count rather than the strength. The same cut passes through the bottom chord, which is where the chord force the moment diagram does not contain comes from.
Fig. 3 The cracked web drawn as the truss it has become, with the cut that counts the stirrups. Over a lever arm of 630 mm at cot θ = 2 the cut severs z·cot θ/s = 7.2 stirrups, every one of them at yield, so the shear is their number times their strength: 629 kN from 1,149 mm² per metre. The same cut passes through the bottom chord.

A cut along the crack crosses no concrete in tension, because the crack is already there. What it crosses is a set of stirrups and the two chords. Vertical equilibrium of the piece then reads

V=AswsfywzcotθV = \frac{A_{sw}}{s}\, f_{yw}\, z \cot\theta

and the only thing θ\theta does in that expression is decide how many stirrups the cut catches. Flattening the crack lengthens the cut, the cut catches more stirrups, and each of them is still at the same yield strength. Nothing about the steel has improved; the geometry of the free body has changed.

That is worth stating plainly because the usual phrasing — “a flatter truss gives more capacity” — sounds like a property of the member. It is a property of the cut, and the cut is a choice about which free body to draw.

What stops it

If flattening the truss only ever caught more stirrups, the right angle would be as flat as possible. The concrete strut is what ends the argument.

The strut carries the shear along its own line, so its force is V/sinθV/\sin\theta over a face of width zcosθz\cos\theta, giving a stress of V(cotθ+tanθ)/bwzV(\cot\theta + \tan\theta)/b_w z. That expression is symmetric about 45° and rises either side, so the web is least stressed at exactly the angle that needs the most steel.

Flattening the truss saves stirrups and crushes the web. Two capacities against the angle of the cracks, for a web 350 mm wide with a lever arm of 630 mm. The rising line is the stirrups: a cut along the crack severs z·cot θ/s of them, so flattening the crack from 45° to cot θ = 2.5 takes the 491 kN they carry to 1229 — 2.5 times as much from the same steel. The falling line is the concrete strut, whose stress is V(cot θ + tan θ)/b_w z and therefore least at 45°. They cross at cot θ = 2.10, and 1031 kN is the most this section will carry however it is reinforced.
Fig. 4 The same section carrying 900 kN with heavier stirrups. Now the two curves cross, at cot θ = 2.10, and the crossing is the section’s ceiling: 1,031 kN is the most this web will carry however much steel is put in it. Past the crossing the strut governs and adding stirrups changes nothing.

So the angle is free inside a window, and the window has a width the concrete sets. Codes give the range as cot θ between 1 and 2.5 — 45° down to 21.8° — and the upper end is a limit on belief rather than on arithmetic: a crack much flatter than 21.8° is not one anybody has watched a member form.

There is a second consequence of the crossing that is easy to miss. Below it the member is under-reinforced in shear and its failure is a stirrup yielding, which is ductile and visible. Above it the failure is web crushing, which is neither. The angle is a free choice and the ceiling is not, and a design that pushes the angle to save steel is walking toward the failure mode with no warning in it.

The force the moment diagram does not contain

The cut along the crack passes through the tension chord as well as through the stirrups, and what it finds there is the part of this subject that catches people out.

The bar is longer than the moment diagram says. The force in the tension chord along a 9 m span. The lower curve is M/z, which is what a bending calculation gives and which is zero at the support. The upper one adds the V·cot θ/2 the truss puts there: the cut along the crack passes through the chord as well as the stirrups, and it carries half the shear as chord tension. At the support that is 750 kN where bending says nothing at all, and everywhere else it is the same curve moved 788 mm toward the support. A bar cut off where the moment diagram says it may be is a bar that is too short.
Fig. 5 The force in the tension chord along a 9 m span. The lower curve is M/z, which is what a bending calculation gives and which is zero at the support. The upper one adds the V·cot θ/2 that the truss puts there — 750 kN at a support where bending says nothing at all — and everywhere else it is the same curve moved 788 mm toward the support.

Taking moments about the point where the strut meets the compression chord, the tension chord has to balance not only the applied moment but half the shear times cotθ\cot\theta. So

Fchord=Mz+Vcotθ2F_{\text{chord}} = \frac{M}{z} + \frac{V \cot\theta}{2}

which is the same as the moment diagram shifted horizontally by al=zcotθ/2a_l = z\cot\theta/2. A flatter truss saves stirrups and lengthens every bar, which is the second price of the choice and the one that does not appear on the shear calculation at all.

The tension a bar must carry is the moment from further along. A beam with a diagonal crack at 40 degrees, and the tension the bottom bar is asked for. Beam theory takes the tension at a section from the moment at that section; the truss model does not, because the compression in the diagonal above the crack has to be balanced by tension at the bar's far end. The tension diagram is therefore the moment diagram displaced by a_l = z(cot θ − cot α)/2, which for z = 900 mm and vertical links is 536 mm — 21 bar diameters. The development length starts from there, so a curtailment computed from the moment diagram alone stops the bar 536 mm too early at every point it is cut off.
Fig. 6 The same rule read from the reinforcement’s side: a beam with a diagonal crack at 40°, and the tension the bottom bar is asked for. Beam theory takes the tension from the moment at the section; the truss does not, because the compression above the crack has to be balanced at the bar’s far end. The tension diagram is the moment diagram displaced by 536 mm — 21 bar diameters — and the development length starts from there.

The shift rule that every code carries is exactly this, and it is why a bar may not stop where the moment diagram says it may. The angle chosen for the shear check therefore reaches into the flexural detailing, and choosing cot θ = 2.5 to save links lengthens the tension steel by 788 mm at every curtailment point.

What the choice does to the numbers, in one section

It is worth putting the whole trade on one member rather than describing it.

The section above is 350 mm wide with a lever arm of 630 mm in concrete of 35 N/mm², carrying 600 kN of shear. At 45° it needs 2,190 mm² of stirrups per metre and the strut runs at 4.5 N/mm². At cot θ = 2.5 it needs 876 mm² per metre — 60 per cent less steel — and the strut runs 45 per cent higher, at 6.6 N/mm², still comfortably inside the 12.0 available.

The chord force at the support goes the other way: nothing at 45° from the moment diagram, plus 300 kN of shear-induced tension, against 750 kN at cot θ = 2.5. And the shift applied to every curtailment goes from 315 mm to 788 mm.

So the flat truss saves 1,314 mm² of stirrups per metre and costs 450 kN of chord tension and 473 mm on every bar. Whether that is a good trade depends on the price of a fixed stirrup against the price of a longer bar, and on whether the support region has room for the extra tension steel — which is a congestion question rather than a strength one, and is the reason experienced designers on heavily reinforced members often choose an angle steeper than the code allows them to.

The angle the member picks for itself

Nothing above says what the member actually does, and the two questions are worth separating carefully because they get confused.

The cracks form at the angle the elastic stress field had at the moment the concrete’s tensile strength was reached — near 45° in a web with no axial force, flatter where there is compression, steeper where there is tension. That is a measurement and it does not depend on the design.

The truss is a model of what the member does after cracking, at a load far above the one that cracked it. By then the stirrups have yielded, the concrete has redistributed, and new cracks have formed at angles the original field would not have produced. Tests routinely show final crack patterns flatter than the first cracks and flatter than 45°.

So the design angle is not a prediction of a crack angle and does not need to match one. What it must do is be reachable — the member has to be able to deform its way into the state assumed, which needs the stirrups to yield without the concrete crushing first. That reachability is the real content of the ceiling, and it is why the crossing in the figure above is a hard limit while the code’s cot θ ≤ 2.5 is a convention.

There is a satisfying check on all of this in the strain field. A truss at cot θ = 2.5 implies a particular ratio of stirrup strain to chord strain, and measuring both on a test beam says which truss the member has adopted. The measurements come back close to the assumed angle when the member has been detailed for it and closer to 45° when it has not — the member, in other words, adopts the state it has been given the steel to adopt, which is exactly what the theorem promises and is a slightly unnerving thing to see happen.

Everything above assumes stirrups exist. A slab, a footing and a lightly loaded beam usually have none, and they carry shear anyway.

A strength with no mechanism in it, made of four. The shear a member carries with no links in it, split into the mechanisms that carry it, against the member's effective depth on a logarithmic axis. The three bands are calibrated to Taylor's measured shares at one 350 mm × 500 mm member and are then evaluated everywhere else, so the shape of the total is a prediction. Aggregate interlock is the band that dies: it depends on how tightly the crack faces are held together, crack width grows with member depth, and it falls from 59% of a shallow member's strength to 20% of a deep one's. That decay is the whole of the size effect, and the dashed line is the code's fitted k = 1 + √(200/d), which knows nothing about interlock and falls by a factor of 1.52 where the model falls by 1.96 over the same twentyfold range. Dowel action is why the expression contains the flexural reinforcement ratio, which nothing in a truss analogy would predict.
Fig. 7 The shear a member carries with no links, split into the mechanisms that carry it, against effective depth on a logarithmic axis. Aggregate interlock is the band that dies: crack width grows with member depth, so it falls from 59 per cent of a shallow member’s strength to 20 per cent of a deep one’s. The dashed line is the code’s fitted k = 1 + √(200/d), which knows nothing about interlock and falls by 1.52 where the model falls by 1.96 over the same twentyfold range.

None of that is a truss. The strength of a member with no links is a fitted expression with no mechanism in it, assembled from a compression zone that carries shear, an aggregate interlock across the crack, and a dowel action in the flexural bars — which is why the code’s expression contains the flexural reinforcement ratio, a term nothing in a truss analogy would predict.

The two treatments do not meet smoothly, and that discontinuity is real rather than an artefact. A member with a token amount of stirrups is analysed as a truss and comes out weaker than the same member with none, analysed by the empirical route, because the truss model refuses to count the mechanisms the empirical one is built from. Codes patch the gap with a minimum stirrup requirement and a rule about which expression applies where.

What the choice is actually made on

Given a free angle inside a window, the practical question is what decides it, and three considerations do.

Steel quantity. The flattest admissible truss needs the fewest stirrups, and stirrups are labour-intensive to fix. On a heavily loaded beam this dominates.

Web stress. The flattest truss stresses the web most. On a thin web — a precast beam, a box girder — the ceiling arrives early and the angle is forced back toward 45°.

Chord force. The flattest truss adds most to the tension chord and moves every curtailment. On a member where the flexural steel is already congested at the support, a steep truss is cheaper overall even though it needs more links.

The three do not agree, and none of them is mechanics. They are quantities of material and hours of labour, which is a fair description of what a design decision is — and it is unusual in this collection to find a case where the analysis hands over a family of answers and says explicitly that choosing between them is somebody else’s problem.

The same argument, one field over

The variable angle is a particular case of something general, and recognising the shape makes it easier to trust.

A structure with more ways to carry a load than statics can distinguish between has a family of internal states, all in equilibrium with the same load. In an indeterminate frame that family is spanned by the redundants and the elastic solution picks one member of it by stiffness. In a cracked concrete web the family is spanned by the strut angle, and nothing picks a member of it at all — because the material has already been taken past its elastic range and its stiffness is no longer the thing deciding anything.

That is why the two subjects feel so different despite resting on the same theorem. Elastic analysis answers which state; plastic analysis answers which states are permissible and hands the choice back. The variable-angle truss is the second kind wearing the clothes of the first, which is why it is presented in codes as a formula with a parameter in it rather than as a design freedom, and why the parameter is so often left at its default.

The same reading explains what a minimum stirrup rule is for. It is not a strength requirement — the member has already been shown to carry its shear — but a ductility one: enough steel, distributed closely enough, for the web to be able to reach the state the model assumed. A theorem that grants permission always needs something to make the permission usable, and here it is a bar diameter and a spacing.

Where the model stops

The stirrups are assumed to yield. All of them, simultaneously, over the whole of the cut. A stirrup near the compression chord is much less strained than one near the tension chord, and the assumption is a plastic one that needs the steel to be ductile enough to deliver it.

The concrete carries no tension across the crack. That is conservative and it is the reason the truss model underestimates lightly reinforced members: the aggregate interlock in the figure above is real and the truss deletes it.

The lever arm is taken as constant. It is not, and near a support the compression chord curves down into the node, which changes both the strut angle and the number of stirrups the cut catches.

Nothing here is a serviceability statement. A member designed at cot θ = 2.5 has wider cracks at working load than the same member at 45°, because the stirrup stress is higher for the same shear. The ultimate check is satisfied and the appearance is not the same.

The stirrups are assumed vertical. Inclined links change every expression here — the cut catches them at a different rate and the strut angle interacts with the link angle — and bent-up bars, which were the standard arrangement for half a century, are a different model again.

And the angle is assumed uniform along the member. In a member with varying shear the admissible angle varies too, and the shift applied to the flexural bars uses whichever angle was assumed where — a bookkeeping problem that gets solved by using one angle everywhere.

What a designer should carry away

Three sentences survive the arithmetic.

The angle is a decision and the ceiling is not. Anywhere below the crossing the choice is between quantities of two materials; at the crossing the member has met a limit that no reinforcement removes, and the only remedies are a thicker web, a stronger concrete or a deeper section.

Every angle chosen for shear is also chosen for bending. The shift is not a separate rule bolted on afterwards; it is the same free body read at the chord instead of at the stirrups, and a design that flattens the truss without lengthening the bars has taken half of one model.

And the choice is invisible on the finished structure. Two beams detailed at 45° and at cot θ = 2.5 look identical from the outside, carry the same load, and have different amounts of steel in different places. Nothing a reader of the drawing can see records which model was used, which is a good argument for writing the angle on the drawing — and almost nobody does.

Choosing the angle is choosing a load path, and two neighbours make the same choice at different scales. A strut-and-tie model is this freedom applied to a whole region rather than to a section, with the node as the price; and shear across a crack that is already there is the case where the angle has been chosen by a crack rather than by the designer.

The ladder from here

Later rungs on this anchor: prestressed webs, where the axial compression flattens the cracks before any choice is made and the angle is partly decided by the prestress. Shear in members with inclined chords, where the chord’s own vertical component carries part of the shear and the truss is only responsible for the rest. Punching, which is this argument in two dimensions and around a perimeter rather than across a section. Torsion combined with shear, where the same web carries two shear flows that add on one face and cancel on the other. The minimum stirrup rule, which exists to bridge the discontinuity between the two treatments described here. And the size effect worked properly, since the mechanism that dies with depth is the one no truss contains.

Ritter proposed the 45° truss in 1899 and Mörsch developed it, and for seventy years the angle was fixed at 45° because nobody could justify anything else. The variable angle arrived with plasticity theory in the 1970s — Nielsen, Thürlimann and others — and it is one of the clearest cases in structural engineering of a theorem changing a design rule: nothing new was measured, and a quantity that had been a constant became a decision.

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Aggregate interlockConcrete shearEquilibriumFree bodyLower-bound theoremReinforcementShear forceShift ruleSize effectStirrupStrut-and-tieWeb crushing