Internal forces

Two models of one bracket

A corbel can be designed as a plane that has to be clamped or as a truss that has to be drawn, and the two are not approximations of each other. They describe different failures, they ask for steel in different places, and the honest answer is that both are checked because neither bounds the other.

Assumes Shear across a crack that is already there, When there is no section to design and The force that arrives along a length.

A corbel is a bracket a few hundred millimetres deep sticking out of a column, carrying a beam that lands on it. It is small, it is common, it is precast by the thousand, and it has two design methods that do not agree.

The disagreement is not a matter of accuracy. The two methods draw different free bodies, describe different failures and put the steel in different places, and a member designed to either alone has been checked against half of what can happen to it.

The first model: a plane to be clamped

Shear friction takes a plane that is already cracked — the vertical face where the corbel meets the column — and asks what carries shear across it.

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.
Fig. 1 The plane 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 1.40 written down as a coefficient of friction. At 0.8 per cent reinforcement the clamping stress is 4.00 N/mm² and the resistance 5.50.

The bars crossing the plane are not carrying shear. They are stretched by the separation the sliding forces, and the tension they develop clamps the faces back together — so the reinforcement supplies a normal force, and the roughness converts that into shear resistance at the rate μ\mu.

That explains the two things about shear friction that look wrong. The coefficient exceeds one, at 1.4 for concrete cast against roughened concrete, because it is the tangent of an asperity slope rather than a coefficient of sliding friction. And a bar not anchored on both sides is worth nothing, because a bar that pulls out cannot develop the tension that does the clamping.

Nothing happens until something moves

The mechanism has an unusual demand: it needs a displacement before it produces a force.

Nothing is resisted until something has moved. Shear resistance against slip along the crack. At zero slip the faces are not clamped and the resistance is whatever cohesion survives, which for a crack that has already opened is nothing at all. Sliding by a fraction of a millimetre forces the faces apart, the bars stretch, and by 0.55 mm they are at yield and the resistance has reached 5.50 N/mm². This is a mechanism that has to move to work, which is why it is checked at the ultimate limit state and never at the serviceability one.
Fig. 2 Shear resistance against slip along the plane. At zero slip the faces are not clamped and the resistance of an already-cracked plane is nothing at all. Sliding by a fraction of a millimetre forces the faces apart, the bars stretch, and by 0.55 mm they are at yield and the resistance has reached 5.50 N/mm².

Half a millimetre is small and it is not zero, and everything that follows from it is on the serviceability side of the design rather than the strength side. The joint opens, the crack at the column face is wider than a flexural crack would be, and the bearing under the beam has moved down and outward by an amount nothing computes.

A mechanism that must move to work is checked at the ultimate limit state and nowhere else, which is a reasonable convention and leaves a real gap: there is no serviceability limit on the slip, so the appearance of a corbel is governed by a quantity no calculation produces.

Where the steel stops helping

The relationship between reinforcement and resistance is linear and then it is not.

More steel across the crack, until the roughness runs out. Shear resistance of the interface against the reinforcement crossing it. The steel clamps rather than carries, so the resistance is the clamping stress times the interlock coefficient and rises in a straight line — until the asperities crush at 5.50 N/mm², which happens at a reinforcement ratio of 0.79%. Past that the line is flat and every further bar is decoration. The dashed line is what the clamping alone would give if the concrete were unbreakable.
Fig. 3 Shear resistance against the reinforcement crossing the plane. The steel clamps rather than carries, so the resistance is the clamping stress times the interlock coefficient and rises as a straight line — until the asperities crush at 5.50 N/mm², at a reinforcement ratio of 0.79 per cent. Past that the line is flat and every further bar is decoration.

The cap is the concrete’s, not the steel’s. Beyond about 0.8 per cent the faces are being clamped harder than the asperities can survive, they shear off, and what is left is sliding on a smooth plane at a much lower coefficient. More steel across a shear plane is worth exactly nothing past the crushing cap, and the remedy for a corbel that fails shear friction is a deeper corbel or a rougher interface rather than more bars.

That is a useful boundary to have, because the temptation with a small member is always to add reinforcement. Here the arithmetic says where that stops working and gives the ratio at which it does.

The second model: a truss to be drawn

Where there is no section to design, a strut-and-tie model draws the load path explicitly. A corbel has no section anywhere in it — it is a D-region from end to end — so this is the natural treatment.

A truss drawn inside a solid, and solved as one. A deep member 1600 mm between bearings and 800 mm deep, carrying 600 kN at mid-span. The model is two struts and one tie, on a lever arm of 640 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 375 kN and each strut at 480 kN, at 38.7° to the horizontal. Spread over a strut width of 375 mm the compression is 3.2 N/mm² against a limit of 18.1 for concrete cracked across its own strut, and the tie needs 862 mm² of steel. A beam calculation on the same member would have asked the tie for 351 kN, which is 7% less than the model does.
Fig. 4 A short deep member 1,600 mm between bearings and 800 mm deep carrying 600 kN, solved as a truss rather than by a formula. The tie comes back at 375 kN and each strut at 480 kN at 38.7° to the horizontal. Spread over a strut width of 375 mm the compression is 3.2 N/mm² against a limit of 18.1, and the tie needs 862 mm² of steel.

The model is a strut running from the bearing plate diagonally down into the column and a tie running horizontally along the top of the corbel to hold the strut’s thrust. The tie is at the top, in the direction the load is not acting, and the strut passes through the body of the member without crossing the column face at any particular place.

Compare that with the first model. Shear friction wants vertical bars crossing a vertical plane; the truss wants horizontal bars along the top. The two models ask for reinforcement in perpendicular directions, which is the sharpest possible statement that they are not two calculations of one thing.

Which free body produced the number

Two free bodies, and the difference between them is the whole disagreement.

Shear friction’s free body is the corbel, cut on the plane of the column face. What crosses that cut is a shear traction, a normal traction and the bars. Equilibrium along the plane says the shear must be carried by the interlock, and equilibrium across it says the clamping force is the bars’ tension. There is no bending anywhere in that statement, because the cut is parallel to the load.

The truss model’s free body is the corbel cut anywhere at all, with the internal forces taken as the members of an assumed truss. Its equilibrium statement is that the truss closes, and the tie force follows from moments about the node at the bottom of the strut.

The two cuts are at right angles to each other. One asks what crosses a vertical plane and the other asks what balances a moment; neither is a special case of the other, and a member can satisfy one and fail the other. That is the technical content of “both are checked”, and it is not a hedge.

The corbel, in numbers, twice

It is worth doing the same bracket both ways, because the two answers are easier to argue about than the two principles.

Take a corbel 400 mm wide and 800 mm deep at the column face, carrying 600 kN with the bearing 300 mm out from the face.

By shear friction, the plane is 400 × 800 mm and the demand is 600 kN, which is 1.88 N/mm². At the rough interface’s coefficient of 1.4 that needs a clamping stress of 1.34 N/mm², which is 0.27 per cent of the plane’s area at a yield of 500 — about 860 mm² of steel crossing the face, in closed links through the depth of the corbel. The check is comfortable: the crushing cap is 5.50 N/mm² and the demand is a third of it.

By the truss model, the strut runs from the bearing to a node at the bottom of the corbel, the lever arm is about 640 mm, and the tie carries 600×300/640=281600 \times 300 / 640 = 281 kN plus the horizontal load the bearing delivers. At 435 N/mm² that is about 650 mm² of steel, running horizontally along the top and anchored at both ends.

Neither number is larger than the other by much, and that is not the point. One is 860 mm² of vertical steel and the other 650 mm² of horizontal steel, and a corbel built with either alone is missing a load path. What actually goes in is both, plus a rule about distributing the links over the top two-thirds of the depth that comes from tests rather than from either model.

When the interface is not rough

The coefficient is a property of the interface’s preparation, and the preparation changes the structure.

The coefficient is a slope, and that is why it can exceed one. The crack magnified: two rough faces, drawn as a sawtooth at 31° 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 0.60. 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. 5 The same plane with a coefficient of 0.60 — a smooth joint, a trowelled surface, or concrete cast against hardened concrete with no preparation. The asperity slope has fallen from 54° to 31°, and with the same 0.8 per cent of steel the resistance falls from 5.50 to 2.40 N/mm².

A factor of 2.3 on the capacity, from a decision made on site about whether to scabble a surface. That is one of the largest sensitivities to workmanship anywhere in this collection, and it is the reason interface preparation is specified rather than assumed.

There is a second way to help the same plane, and it does not involve steel at all.

Nothing is resisted until something has moved. Shear resistance against slip along the crack. At zero slip the faces are not clamped and the resistance is whatever cohesion survives, which for a crack that has already opened is nothing at all. Sliding by a fraction of a millimetre forces the faces apart, the bars stretch, and by 1.25 mm they are at yield and the resistance has reached 3.60 N/mm². This is a mechanism that has to move to work, which is why it is checked at the ultimate limit state and never at the serviceability one.
Fig. 6 The smooth plane again, this time with an external compression of 2.0 N/mm² acting across it. The clamping no longer has to be manufactured entirely by the bars, so the resistance rises to 3.60 N/mm² and the slip needed to reach it grows to 1.25 mm — because the bars are being asked for less, and reach yield later.

A permanent compression across a shear plane is worth clamping steel, one for one, which is why a corbel under a column is easier than one under a beam and why a construction joint in a wall is easier at the bottom than at the top. The check does not care where the normal force comes from.

What each model misses

Setting the two side by side, the omissions are as informative as the results.

Shear friction has no lever arm in it. It does not know how far the load is from the column face, so it gives the same answer for a bearing at 100 mm and at 300 mm. A corbel with a long projection fails by the tie yielding and shear friction has nothing to say about it.

The truss model has no plane in it. It does not know that the interface between two pours is weaker than monolithic concrete, so it gives the same answer for a corbel cast with the column and one cast afterwards. A corbel with a construction joint at its root fails by sliding on that joint and the truss model has nothing to say about it.

Each is blind to exactly what the other is about, which is why codes require both and why the required reinforcement is the sum of two arrangements rather than the larger of two numbers: horizontal ties along the top for the truss and, in addition, closed links through the depth for the plane.

What holds the two together

There is one requirement both models share, and it is the one most often got wrong.

The bond stress is crowded against the loaded end. A 25 mm bar embedded 906 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 = 500 mm, so the far end of the bar is doing almost nothing. At the code's own length of 36 diameters the elastic bond is 52 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. 7 A 25 mm bar embedded 906 mm, with the force in it and the bond stress on it along the embedment. Uniform bond — the assumption behind every tabulated development length — is a flat stress and a straight line of force; an elastic bond of the same peak dies away over 500 mm, so at the code’s own length the elastic bond is 52 per cent used.

Both models depend on a bar reaching yield somewhere it cannot easily be anchored. The shear-friction bars have to develop on both sides of the plane, in a corbel perhaps 300 mm deep. The truss’s tie has to develop behind the node at the outer face, where there is a bearing plate and the free edge of the member and nothing else.

A bar has a length before it is worth anything, and in a corbel there is no room for it. The standard answers are all geometric rather than analytical: loop the tie back on itself, weld it to a cross-bar, weld it to the bearing plate, or bend it down into the column. The detail is the design, and it is why a corbel drawn without its reinforcement is not a design at all — which is the same lesson the node in a strut-and-tie model teaches, on a member small enough to see the whole of.

The failure that decided the subject

Corbels have a poor record, and the reason is worth knowing because it is not either of the mechanisms above.

The commonest observed failure is a splitting crack running from under the bearing plate outward to the free edge, taking the top corner of the corbel off, with the tie bar still intact and still anchored. That is not a shear-friction failure and it is not a tie yielding. It is the node under the bearing plate running out of concrete on its outer face, in a member where there is very little concrete outside the bar.

It is the reason bearing plates on corbels are set back from the free edge, the reason the tie is looped or welded rather than hooked, and the reason a corbel’s projection is limited relative to its depth. None of those rules comes from either model; all of them come from the geometry of a node close to a free surface, which the node check is about and which the two models above take for granted.

So there are really three checks and one of them is a drawing. Two calculations that disagree, plus a set of dimensional rules that neither produces — and the dimensional rules are the ones that have prevented the most failures.

Why two models survive

It is worth asking why the subject tolerates two methods rather than settling on one, and the answer is a decent illustration of what design codes are for.

Both are lower-bound methods, so both are safe if the structure can carry what they ask for. Neither is an upper bound, so neither predicts the failure load; each predicts a load the member can carry. Two lower bounds do not combine into a better lower bound — the true capacity is at least the larger of them — but two lower bounds on different mechanisms do combine into a better design, because the reinforcement each asks for is reinforcement the other cannot supply.

That is why the requirement is additive. It is not conservatism; it is that the two models are talking about different things, and satisfying both is the only way to have said anything about both.

The same pair, elsewhere

Once the shape of the disagreement is clear it turns up in several other places, and recognising it saves rediscovering it.

A construction joint in a wall has a plane to be clamped and a wall to be designed; the joint check is shear friction and the wall check is flexural, and the reinforcement crossing the joint is the larger of the two demands rather than the sum, because here the two models really are about the same steel in the same direction.

A composite floor’s interface is a horizontal plane carrying a shear flow, clamped by the studs’ own tension and by the slab’s weight. The truss model of the beam and the friction model of the interface are again perpendicular, and again both are required.

A precast beam sitting in a pocket has a bearing to check, a plane at the back of the pocket to clamp, and a truss inside the pocket wall to draw. Three models, three free bodies, and the reinforcement is the union of what they ask for.

The general statement is that a member with no section has as many models as it has plausible failure surfaces, and lower-bound reasoning gives no way of ranking them. That is a genuine limitation of the theorem rather than a gap in the codes: a lower bound establishes that a structure will carry a load, and says nothing about which of several arrangements it will use to do it, or about what happens on a surface the chosen arrangement did not cross.

Where the model stops

The coefficient is a fitted number from push-off tests. Those tests measure a specimen as much as a material — the size of the specimen, the way the crack was formed and the bar arrangement all move the answer — and the values in codes are lower bounds to scattered data rather than measurements of anything.

Load reversal is not covered. Cycling a shear plane grinds the asperities off, so the coefficient degrades with each cycle. That makes shear friction a poor seismic mechanism without confinement, and none of the curves above knows about it.

The steel is assumed to yield. A bar that is not anchored, or is anchored in cracked concrete, supplies whatever it can develop rather than its yield force, and the clamping falls in proportion.

Nothing here computes a crack width. The slip is a displacement along the plane and the separation is a displacement across it, and the second is what a reader sees.

And the truss model’s angle is a choice. Every model in the admissible band is safe, and a steeper strut asks for less tie and more node stress. The 38.7° drawn is one member of a family, and the family’s other members give different amounts of steel in the same place.

Both models are lower bounds and both are licensed by the same theorem, which is why they can disagree and both be right. A strut-and-tie model checked at its nodes is the general version of the first, and the load that has to be lifted is the case where choosing the wrong one of two paths leaves reinforcement doing nothing at all.

The ladder from here

Later rungs on this anchor: shear friction under load reversal, and the degradation that makes it a poor seismic mechanism without confinement. Interface shear in composite floors, where the plane is horizontal, the check is a shear flow rather than a force, and the clamping comes from the studs. Push-off tests, which is where every number here comes from and which measure a property of a specimen as much as of a material. Shear keys, which replace the asperities with a geometry somebody chose and turn a statistical quantity into a drawn one. The dapped end and the half joint, which are the two D-regions with worse failure records than the corbel and the same pair of models. And the case that inverts the whole argument: a joint deliberately made smooth and greased so that shear is not transferred, which is a bearing.

Shear friction was proposed by Birkeland and Birkeland in 1966 as a design expedient for precast connections, with an explicitly empirical coefficient and no mechanism attached. The sawtooth explanation came afterwards and is what turned a fitted line into a model — and it is also what predicted the crushing cap, which the original expression did not have and which is now the part of the check that governs most often.

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 interlockAnchorageBondClampingCrack widthEquilibriumInterface shearLower-bound theoremPrecastReinforcementShear frictionStrut-and-tie