The strength with no mechanism in it

Almost every strength in this collection can be traced to a free body. A plastic moment is the couple of two stress blocks. A truss member’s force is a joint equilibrium. A punching capacity is a shear stress on a stated perimeter. Even the ones that are fitted — a buckling curve, a fatigue category — are fitted to a shape that a mechanism produced.

The shear strength of a concrete member with no links in it is not like that. What every code in the world gives is a version of

$$v_c = C\,k\,(100\rho f_{ck})^{1/3}, \qquad k = 1 + \sqrt{200/d} \le 2$$

which is three variables raised to fitted powers, multiplied by a term that depends on the depth. There is no free body in it, no equilibrium statement, and nothing that could be drawn. It is a curve through a scatter of about two thousand tests, and its form was chosen because it fits them.

That is not a criticism. It is a description of what is being fitted, and the interesting question is what.

Which free body produced the number #

There is one, and it is unusual: the free body is bounded by a crack rather than by a plane.

Once a concrete member has cracked in flexure, a diagonal crack propagates from the flexural crack toward the compression face. Take the piece of the member on one side of that crack as the free body, and ask what crosses the crack to carry the shear. Four things do.

The uncracked compression zone above the crack tip is intact concrete, and it carries shear the way any solid section does. Its contribution is proportional to how deep the compression zone is, which is a property of the cracked section and therefore of the reinforcement ratio.

Aggregate interlock acts on the crack faces themselves. A crack through concrete does not go through the aggregate, it goes round it, so the two faces are rough at the scale of the largest stones and they engage when they try to slide past one another. This is shear across a crack that is already there in its natural habitat, and the force it carries depends on how far apart the faces are.

Dowel action is the longitudinal reinforcement crossing the crack and being bent by the relative movement of the two pieces. A bar is a very stiff dowel over a very short length, and it carries a shear that depends on how much steel there is.

Residual tension across a narrow crack is real for the first fraction of a millimetre — concrete does not lose its tensile stress instantly at cracking, it softens over a crack opening of a few hundredths of a millimetre — and it disappears first.

Every one of those four is a statement about a free body. What none of them is, on its own, is a formula — and that is why the design expression has none.

The mechanisms, calibrated once and then left alone #

The figure above is built the way this site prefers: each band has the physics of its own mechanism in it, the three are scaled once so that they add up to the measured strength at a single reference member, and everything else on the plot is a consequence.

The compression zone’s band goes as the neutral axis depth. The dowel band goes as the cube root of the reinforcement ratio. The interlock band goes as the crack faces’ engagement, which falls as the crack gets wider — and the crack gets wider as the member gets deeper, because crack width scales with the spacing of the cracks and the spacing scales with the member.

Calibrate at a 300 by 500 member and run the model up and down the depth axis, and the total does something the calibration never told it to do: it falls. It falls by a factor of about two over a twentyfold range of depth, and the fitted $k$ term in the code falls by about a factor of 1.5 over the same range. Two models built from different information agree that a deeper member is weaker in shear, and they disagree about how much by a quarter.

That is worth reading carefully. The code’s size term was fitted to tests and knows nothing about aggregate interlock. The model here was built from interlock and knows nothing about the tests. They land in the same place because the same thing is happening in both.

Where the size effect comes from, exactly #

Almost nothing in this subject admits to a size effect. A yield stress is a yield stress; a modulus does not care how large the member is; a plastic moment is a stress times a modulus and scales exactly. The bigger one is the weaker one is about the general phenomenon, and about how rarely it survives into a design rule.

Shear without links is the exception, and it is in the rule openly. The reason is now visible: the mechanism whose strength does not scale is the one that depends on a length — the crack width, against the size of an aggregate particle that stays 20 mm across whatever the member does.

Look at the interlock band as the member deepens. The model has not changed and neither have its inputs; only the depth being read off it has.

Now take the marker the other way, to a transfer beam or a pile cap, and read the same three bands again.

The whole of the size effect lives in that term, and that term is the one carrying its load across a gap whose width scales with the structure while the thing bridging it does not.

The consequence for design is unpleasant and correct. A 1,500 mm deep transfer beam with no links has, per unit area, about half the shear strength of a 200 mm slab of the same concrete. It also has seven times the area, so it carries about three and a half times the force — which is why the effect is easy to miss. The number that falls is the number in the check, and the number that rises is the one a designer feels.

The reinforcement ratio, which should not be there #

A truss analogy is a beautiful thing. The beam that becomes a truss is the argument that a cracked concrete beam with stirrups is a truss — concrete struts, steel ties, chords top and bottom — and that its shear capacity follows from cutting that truss and summing forces. Nothing in it mentions the longitudinal steel except as a chord.

The expression for a member with no stirrups has $\rho$ in it, and $\rho$ is the flexural reinforcement. Doubling it from 1.2% to 2.4% raises the shear strength by a quarter. A truss analogy predicts nothing of the sort, and the reason it does not is that there is no truss: with no stirrups there are no ties, so whatever is happening is not that.

Two of the four mechanisms explain it. More longitudinal steel means a deeper compression zone, because the neutral axis of a cracked section descends as the tension steel area rises — so the first band grows. And more steel means more dowel, directly. Both are visible in the same picture, with the depth held and the steel doubled instead.

That is a shear strength answering to a bar that was put there for bending, which is a thing no truss can be drawn to explain.

There is a third and subtler reason, and it belongs with stiffer than its cracked section says: more flexural steel means smaller crack widths at a given moment, and smaller crack widths mean more interlock. The dependence on $\rho$ is therefore a dependence on crack width wearing a different name — which is the same variable the size effect turns on. One physical quantity, appearing twice in the formula under two different symbols.

It is worth noticing how strange a job dowel action is asking a bar to do. A reinforcing bar is designed to be pulled, and it builds its force up along its length through bond over tens of diameters. Dowel action asks the same bar to carry shear across a crack by bending, over a length of a few diameters, with nothing but a cover of concrete beneath it to bear against — which is why the dowel band is the one that goes when the cover splits.

The place where none of this applies #

Everything above assumes the member behaves as a beam: that the load travels along it as shear, in the way what a cut reveals describes, and arrives at the support having been carried by the section.

Below about two and a half depths of shear span, it does not.

A load applied close to a support arches straight to it. The compression follows a strut from the load point to the bearing, the tension is the bottom reinforcement acting as a tie, and the shear span stops being a variable in the problem — the strut’s inclination is, and the capacity is decided by the strut’s crushing strength. That is when there is no section to design, and the capacity rises steeply as the load moves closer to the support rather than falling.

Kani drew the resulting curve in the 1960s and it has been called Kani’s valley ever since: capacity against shear span over depth, with a minimum somewhere near two and a half, arching action above it on the left and beam action climbing back on the right. The valley is a picture of two different mechanisms failing to hand over cleanly, and design rules deal with it by enhancing the shear strength for loads inside about twice the depth — an adjustment that is a mechanism change dressed as a factor.

What a member without links is really relying on #

Put the four mechanisms together and one property of the whole arrangement stands out: none of them has any reserve after it starts to fail.

A member with stirrups that reaches its shear capacity has a truss with yielding ties. It deforms, the crack widens, the stirrup strains, and the load can be redistributed elsewhere in the structure while that happens. A member without stirrups has a crack whose faces are engaged by roughness. When the crack widens past the engagement, the interlock does not yield — it disappears, and it takes the dowel with it, because a bar with no concrete under it splits its cover.

That is why shear failures without links are the ones photographed after collapses. They are sudden, they happen at a load that a bending calculation says is safe, and the member gives no warning because none of the mechanisms carrying the load can deform without losing it.

The contrast with bending is exact and worth stating. After the first yield a section goes on carrying, at very nearly the same moment, through rotations of many times the elastic one — and that plateau is what makes the moment that was moved on purpose possible at all, because redistribution requires somewhere to redistribute from that will hold its force while it happens. Shear without links has no plateau to redistribute across. The four mechanisms are a sum of brittle contributions, and a sum of brittle things is brittle.

It is worth being explicit about what interlock is, because the word makes it sound like friction and it is not. The two crack faces are rough at the scale of the largest aggregate, so sliding one past the other makes them ride up over one another and forces the crack to open. Whatever resists that opening — the flexural steel, an axial compression, a stirrup if there were one — is what presses the faces together and turns the roughness into a shear force. Take away the thing that closes the crack and the roughness carries nothing at all, which is why interlock does not yield when it is exhausted. It goes.

The design response is a minimum area of shear reinforcement in almost every member deeper than a slab, and the reason is not strength — it is that a member with a nominal cage of links has a mechanism with a plateau in it, and one without has a mechanism with a cliff.

The step where the two rules meet #

The two calculations — the fitted expression for a member with no links, and the truss for a member with them — do not join up, and the discontinuity is in the wrong direction.

Take the 350 by 700 member. With no links the fitted expression returns about 150 kN. Now add a nominal cage: 6 mm links, two legs, at 300 mm centres, which is 57 mm² per 300 mm. The truss with $\cot\theta = 2.5$ gives

$$\frac{57}{300}\times 0.9\times 700 \times 435 \times 2.5 = 130\ \text{kN}$$

less than the member had with nothing in it. Adding steel has reduced the calculated capacity by 13 per cent.

Nothing physical has got worse. What has happened is that the truss model discards the concrete term entirely, on the reasoning that a member with links develops the crack pattern the truss assumes and the interlock across those cracks is already counted in the strut. That reasoning is defensible at ordinary link ratios and plainly wrong at a nominal one, where the links are too far apart to organise anything.

The practical resolution is to take the greater of the two, which every code does in one form or another. The interesting part is what the step is evidence of: the two expressions are not two values of one function, they are two different mechanisms, and the region between them belongs to neither.

Where the model stops #

The four mechanisms are not separable in a test. Their shares here come from measurements that instrumented cracks and bars, and those measurements have a spread of ten or twenty per cent on every term. The calibration is a reading of the middle of that spread, not a determination.

The size term is capped. The code’s $k$ reaches 2 at $d = 200$ mm and stops, which says that a member shallower than about 200 mm gets no further benefit. That is a fit to where the data ran out, not a statement about interlock, and the model here does not have such a cap — which is one of the places the two curves separate.

And the size effect has a general form this rule is only one instance of. Below a transitional size a cracked structure fails at a strength and above it at a fracture energy, with everything of practical size sitting on the transition between the two. The shear rule’s $k$ term is a fit to a piece of that transition, made without reference to the underlying law, and it inherits the law’s domain along with its shape.

Nothing here is about axial force. A member in compression cracks later and its shear strength rises; a member in tension cracks earlier and it falls. Both effects are large and both act through the same variable — how open the crack is — which is consistent with everything above and is not drawn.

And shear does not arrive alone. The section that is worst in shear is somewhere near a support and the section worst in moment is near midspan, so the two checks are usually made at different places on the same member and neither sees the other — which is both at once and neither matters as a general proposition, and is true here for a reason specific to this failure: the crack the mechanisms cross is a flexural crack that has turned, so a section with no moment on it has no diagonal crack to fail on either.

Where they do arrive together — a continuous member over an internal support has the largest of both within a metre of one another — each check is very nearly unaffected by the other over most of the range, and the interaction only bites in the corner where both are near their limits.

The generalisation #

The habit worth taking away is to ask, of any empirical strength, what it is an empirical fit to.

An expression with fitted powers is not a confession of ignorance. It is usually a statement that several mechanisms are competing and that no single free body governs across the range — which is a real physical situation, and one that a mechanism-based formula would have to handle by taking the minimum of several expressions and would then get wrong at the crossovers.

The test of whether the fit is understood is whether its shape can be predicted from mechanisms that were calibrated somewhere else. Here it can: the size term falls out of interlock, the reinforcement term falls out of the compression zone and the dowel, and the concrete term falls out of both. That is a good deal more than a curve through a scatter, and it is a good deal less than a derivation.

It also gives the fit a domain, which is the practical value of understanding it. A member with a maximum aggregate size of 4 mm, a lightweight concrete whose cracks go through the stones rather than round them, or a member cracked by something other than flexure has lost the mechanism the size term was fitted to — and the expression, which contains no aggregate size and no crack, will go on returning a number. The free body is a choice is usually advice about how to solve a problem. Here it is advice about how to know when a formula has stopped being about one.

The other habit is scale changes everything applied to a check rather than to a structure. Whenever a strength is quoted as a stress, ask what length is hidden in it. If there is one — an aggregate size, a crack spacing, a fracture process zone, a grain — then the stress is not a material property and the member’s size is a variable in a rule that does not appear to have one.