Structural Engineering & Statics
where the load goes
Internal forces

The strength with no mechanism in it

A concrete member with no links in it carries shear, and the expression that says how much is three variables raised to fitted powers with a size term in front. There is no free body anywhere in it. What it is fitting is a competition between four things that carry shear across a crack, and only one of them explains why a deeper member is worse at it.

Assumes The beam that becomes a truss, What a cut reveals, and why it was there all along and The bigger one is the weaker one.

13 min read 9 figures The load must go somewhereWhich failure arrives firstScale changes everything

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

vc=C k (100ρfck)1/3,k=1+200/d≤2v_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.

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 300 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 62% of a shallow member's strength to 22% 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 2.05 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. 1 The shear a member carries with no links in it, split into the mechanisms that carry it. The bands are calibrated to measured shares at one member and evaluated everywhere else, so the shape of the total is a prediction rather than an input.

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.

The neutral axis is wherever the first moment vanishes. A 300 by 556 section with 1800 mm² of steel at a depth of 500, carrying 150 kNm after the tension side has cracked. The neutral axis is no longer at mid-depth: it has risen to 171.9 mm from the top, which is where the first moment of the compression zone plus the transformed steel vanishes. The compression is 13.1 N/mm² at the top fibre and the steel carries 188 N/mm²; the resulting couple is 339 kN on a lever arm of 443 mm, which multiplies back to the 150 kNm applied. The uncracked section would have had 4827×10⁶ mm⁴ against the cracked 1961×10⁶ — a loss of 59% of the stiffness.
Fig. 2 The section the crack passes through. The compression zone is what is left above the neutral axis, and its depth is what the first of the four mechanisms rides on.

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 kk 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

The one design rule that carries a size effect openly. Shear stress at failure against effective depth, for a member with no links in it. The stress falls from 0.79 N/mm² at 150 mm to 0.50 at 3000 — a factor of 1.59 for the same concrete, the same steel ratio and the same everything. Almost nothing else in this subject admits to a size effect at all: a yield stress is a yield stress and a modulus is a modulus. This does, because the mechanism is a crack, and a crack's width scales with the member while the aggregate that has to bridge it does not. The dotted line is the force, which goes on rising with depth — from 36 kN to 449 — so a deeper member carries more and is worse at it, and only one of those is the number in the check.
Fig. 3 Shear stress at failure against effective depth, with the total force on the same axis. The stress falls and the force rises, and only one of the two is the number in the check.

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. In a 150 mm member it is more than half the strength. In a 1,500 mm member it is a fifth. Nothing else in the picture changes much: the compression zone’s share is roughly constant because it is a fraction of the depth, and the dowel share is roughly constant because the reinforcement ratio is. The whole of the size effect lives in one 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 ρ\rho in it, and ρ\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.

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 = 7.9 stirrups at cot θ = 2.5. 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. 4 What a member with stirrups does instead. The truss is real, the ties are the stirrups, and the capacity follows from cutting it — which is why that calculation has a free body and this essay’s does not.

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 ρ\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.

The bond stress is crowded against the loaded end. A 25 mm bar embedded 735 mm, with the force in it and the bond stress on it plotted along the embedment. Uniform bond — the assumption behind every development length ever tabulated — is a flat stress and a straight line of force. An elastic bond of the same peak strength is neither: the slip is largest where the bar is pulled and dies away over 1/α = 470 mm, so the far end of the bar is doing almost nothing. At the code's own length of 29 diameters the elastic bond is 59 per cent used. The uniform answer is what the bond looks like after it has yielded along the whole length, which is a statement about ductility rather than about strength.
Fig. 5 How a bar’s force is built up along its length. Dowel action asks the same bar to do something quite different — to carry shear across a crack by bending, over a length of a few diameters.

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.

Where a section exists, and where it does not. The same beam divided into the regions the two theories own. Within about one depth of a support, a concentrated load, a corner or an opening, the strain is not linear across the section and every calculation on this site that begins by choosing one is inapplicable — those are the D-regions, marked here. What is left between them is the B-region, where beam theory is exact enough to have been trusted for two centuries. On a beam this deep the D-regions are most of it, which is the practical reason the strut-and-tie model exists at all: 35% of this span is a region a section cannot describe.
Fig. 6 Where a sectional calculation applies and where it does not. Inside a distance of about one depth from a load or a support, plane sections do not stay plane and the member is not a beam.

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.

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.

The coefficient is a slope, and that is why it can exceed one. The crack magnified: two rough faces, drawn as a sawtooth at 45° to the plane. Sliding one over the other cannot happen without lifting it, so a shear displacement forces a separation in fixed proportion — the tangent of that angle, which is the number written down as a coefficient of friction and here is 1.00. The bars crossing the plane are stretched by the separation and clamp the faces back together; they are not carrying the shear, they are supplying the normal force that lets the roughness carry it. A bar that is not anchored on both sides supplies nothing.
Fig. 7 Interlock made explicit: the crack faces ride up over one another as they slide, and the steel crossing them is stretched by the opening. What holds the faces together is what makes the roughness work.

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.

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 kk reaches 2 at d=200d = 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.

A strength that is a property of the specimen. Nominal strength against size for geometrically similar specimens of one material. On the left the specimen is too small for a crack to run and the strength is a plateau — a plastic limit, and the regime laboratory specimens sit in. On the right a crack releases more energy than it consumes as soon as it starts and the strength falls as the inverse square root of size, which is the regime real structures sit in. The turn happens at D₀ = 120 mm. A 150 mm specimen reads 2.80 N/mm² and a 2400 mm member of the same material carries 0.92: the test overestimates the structure by a factor of 3.06.
Fig. 8 The general law the shear rule is one instance of. Below a transitional size a structure fails at a strength and above it at a fracture energy, and everything real sits on the transition.

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.

Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight.
Fig. 9 What happens when they do arrive together. Over most of the range each check is unaffected by the other, and the interaction only bites in the corner.

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.

Where this goes next

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 495 mm severs z·cot θ/s = 8.3 stirrups at cot θ = 2.5. 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. The beam that becomes a truss Once a web has cracked in shear there is no shear stress field in it any more. There are concrete struts, two chords and whatever crosses the cracks, and the angle of those cracks is not a property of the material — it is something the designer chooses, and every quantity in the beam moves when it changes. A strength that is a property of the specimen. Nominal strength against size for geometrically similar specimens of one material. On the left the specimen is too small for a crack to run and the strength is a plateau — a plastic limit, and the regime laboratory specimens sit in. On the right a crack releases more energy than it consumes as soon as it starts and the strength falls as the inverse square root of size, which is the regime real structures sit in. The turn happens at D₀ = 120 mm. A 100 mm specimen reads 3.10 N/mm² and a 1500 mm member of the same material carries 1.14: the test overestimates the structure by a factor of 2.71. The bigger one is the weaker one Two geometrically similar beams of the same concrete should fail at the same nominal stress, because a strength is supposed to be a material property. They do not. The large one fails at less, and the reason is that a crack releases energy in proportion to a volume and consumes it in proportion to an area. The coefficient is a slope, and that is why it can exceed one. The crack magnified: two rough faces, drawn as a sawtooth at 54° to the plane. Sliding one over the other cannot happen without lifting it, so a shear displacement forces a separation in fixed proportion — the tangent of that angle, which is the number written down as a coefficient of friction and here is 1.40. The bars crossing the plane are stretched by the separation and clamp the faces back together; they are not carrying the shear, they are supplying the normal force that lets the roughness carry it. A bar that is not anchored on both sides supplies nothing. Shear across a crack that is already there Every shear calculation in this collection starts from an uncracked solid — a principal stress, a shear flow, a diagonal tension. This one starts after the crack, on a plane with no tensile strength at all, and the coefficient it uses is not a coefficient of friction. It is the slope of the roughness. A truss drawn inside a solid, and solved as one. A deep member 4000 mm between bearings and 2000 mm deep, carrying 1200 kN at mid-span. The model is two struts and one tie, on a lever arm of 1600 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 750 kN and each strut at 960 kN, at 38.7° to the horizontal. Spread over a strut width of 812 mm the compression is 3.0 N/mm² against a limit of 15.8 for concrete cracked across its own strut, and the tie needs 1724 mm² of steel. A beam calculation on the same member would have asked the tie for 702 kN, which is 7% less than the model does. When there is no section to design Beam theory needs a section, and a section needs the strain to be linear across it. Within about a depth of a support, a load, a corner or a hole it is not — and those are the regions structures actually fail in.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

Aggregate interlockBondCrack widthCracked sectionDowel actionEmpirical formulaFailure surfaceFree bodyLoad pathReinforcement ratioShearShear flowShear truss analogySize effectStrut and tie