Shear across a crack that is already there

Every shear argument so far in this collection begins with a solid that has not broken. A shear flow accumulates across an uncracked section; a principal stress is computed from a state that still has tension in it; a diagonal crack is the result of a calculation rather than its starting point.

There is a large class of real problems where the crack is where the calculation begins. A construction joint between yesterday’s pour and today’s. The interface between a precast unit and the topping cast on it. The base of a shear wall, where a horizontal joint exists whether anybody drew one or not. A corbel, whose failure plane runs down the face of the column. A crack that has already formed under load and now has to keep transferring shear across itself. In every one of them the plane is a decision somebody made, which puts the whole subject inside the free body as a choice.

On all of them the plane has no tensile strength worth counting, and shear still crosses it. The question is how.

The coefficient is a slope #

The design expression is disarmingly simple:

$$v_n = \mu\,(\rho f_y + \sigma_n)$$

and the trouble it causes is entirely in the symbol $\mu$. Read as a coefficient of friction it is absurd: the values in use are 0.6 for a joint cast against hardened concrete left smooth, 1.0 for one deliberately roughened, and 1.4 for a crack through monolithic concrete — and no dry contact between two solids has a coefficient of 1.4.

Friction as this collection has already met it is a block on a plane: a normal force, a capacity of μN, and a reaction that leans out of the normal by the friction angle. Every coefficient in that argument is a property of two surfaces, and every one of them is below one. Nothing in that family of contact pairs approaches 1.4, and the preloaded steel joint drawn further down this page runs on a coefficient of 0.5.

The resolution is that $\mu$ here is not a property of two surfaces at all. A crack through concrete does not separate along a plane — it runs round the aggregate, leaving both faces covered in projections a few millimetres high. Sliding one face along the other is not sliding; it is climbing. Each face rides up the other’s projections, and the separation that results is the slip times the slope of the projections.

So if the asperities stand at an angle $\theta$ to the plane, a slip $s$ opens the crack by $s\tan\theta$, and the force needed to drive the slip against a clamping stress $\sigma$ is $\sigma\tan\theta$ per unit area. The “coefficient of friction” is $\tan\theta$ and nothing else:

A ground joint has been given a shallow slope. A monolithic crack has a steep one. The exactly-one case at 45° is the reminder that nothing about the number is a boundary — it is where the projections happen to stand at half a right angle.

Which free body produced the number #

Take a block of concrete on one side of the crack, cut so that the crack is one of its faces. Three things act on that face: the shear being transferred, a normal compression from wherever the clamping comes from, and the contact forces at the asperity tips.

Each contact force acts perpendicular to the little inclined surface it sits on, so it leans away from the plane’s normal by the asperity angle. Resolve it: its component along the plane is the shear it carries, its component across the plane is the clamping it consumes, and the ratio between them is $\tan\theta$. Sum over every contact and the same ratio survives, which is why the model can be written with a single coefficient at all.

The clamping does not come from outside. It comes from the reinforcement crossing the crack, and that is the part of the mechanism most often described backwards.

The bars do not carry the shear. They are perpendicular to it. What they do is get longer: the separation forced by the sliding stretches them, and a stretched bar pulls the two faces together with a force equal to its area times its stress. The steel supplies the normal force that lets the roughness carry the shear, and its contribution is therefore proportional to $A_{vf} f_y$ and independent of where along the crack it is placed, what diameter it is, or which way the shear runs.

Two consequences follow immediately, and both are practical.

Anchorage on both sides is everything. A bar that can pull out of either side supplies no clamping at all, so the full development length is needed each way from a crack whose position may not be known to within a bar diameter.

External compression is worth exactly as much as steel. The $\sigma_n$ term is a permanent compression across the joint — the weight above a wall base, a prestress, the reaction from a bearing — and it adds directly. A joint at the bottom of a heavily loaded column has a great deal of shear resistance for free.

Nothing is resisted until something has moved #

The curve starts at the origin, which is the whole character of the mechanism: it has to move before it works. A shear-friction joint at zero slip has zero resistance, and every newton it carries has been bought with a fraction of a millimetre of movement and a corresponding opening of the crack.

That is why this is a strength calculation and never a serviceability one. The crack width at which the bars reach yield is around a millimetre — perfectly acceptable at the ultimate limit state, and unacceptable in a water-retaining structure or an exposed face at working load. A designer relying on shear friction has accepted a visible crack as the price of the mechanism.

It also puts the method firmly in the family of things that need ductility. The bars have to reach yield and stay there while the slip continues, so the property that appears in none of the equations is doing the work again: a joint reinforced with a high-strength, low-elongation bar has the same $\rho f_y$ on paper and much less of it in practice.

The steel analogue is exact and worth putting beside it.

The difference between those two curves is the whole of the distinction. A preloaded bolt arrives with its clamping already in place. A shear-friction bar has to be stretched by the failure it is resisting before it clamps anything.

More steel, until the roughness runs out #

The cap is not an arbitrary code limit. The contact between the two faces is happening at the tips of the projections, over a small fraction of the nominal area, so the local compression there is many times the average. Push the clamping high enough and the projections crush, the crack loses its roughness, and the mechanism disappears — leaving a smooth plane and a coefficient near zero.

The saturation ratio is worth carrying because it is small:

$$\rho_{sat} = \frac{v_{max}}{\mu f_y} = \frac{5.5}{1.4 \times 500} = 0.786\%$$

Below about 0.8% of the joint’s area in steel, every bar added buys $\mu f_y$ per unit area — 700 N/mm² of resistance per unit of reinforcement ratio, which is a great deal. Above it, nothing. A joint that needs more resistance than the cap allows needs a bigger joint, or a shear key, or a different load path; there is no arrangement of reinforcement that gets past it.

What the uncracked calculation would have said #

It is worth asking what the ordinary machinery predicts for the same plane, because the two answers are unrelated.

There are two such routes and both are spent by the time this plane exists. One is a state of stress at a point, read on every plane through it, with a principal tension deciding where the crack will form — a calculation shear friction begins strictly after. The other is shear flow accumulated as VQ/It, which needs a continuous section to accumulate across; a crack interrupts the accumulation, which is why the two calculations cannot be blended and why one of them stops where the other starts.

The uncracked calculation predicts a diagonal tension and the plane on which it acts. Once that plane has cracked, the prediction has been spent: there is no tension across the crack, the principal stress trajectories have rearranged, and the remaining question is entirely about what crosses the plane that now exists.

This is the same discontinuity that a cracked section in bending has. The elastic calculation is right up to the crack and irrelevant after it; the post-crack calculation is a different model with different variables. There the neutral axis moves to wherever the first moment of what is left vanishes, and the calculation that applies afterwards shares no variables with the one that applied before. Here the same break happens on a plane rather than through a depth.

Where the mechanism is actually used #

A corbel is the canonical case and the clearest. A bracket projecting from a column carries a beam reaction near its tip; the failure plane is vertical, down the column face; and the horizontal bars at the top of the corbel cross it. Nothing about a bending calculation on the corbel produces those bars.

The other model for the same member is a truss drawn inside the solid, with a compression strut running into the column and a tie across the top of the bracket. Shear friction and strut-and-tie are two ways of paying for one load path, and a corbel is usually checked by both.

A construction joint in a wall or a slab is the most common and the least noticed. Concrete is placed in lifts, each lift has a joint at its top, and the joint is a plane with no tensile strength across it. Whether it needs checking depends on the shear crossing it, and the reinforcement that happens to be continuous through it is usually enough — which is why this check very often passes without being made.

The interface between precast and in-situ concrete is checked this way as a matter of routine, and it is the case where the surface preparation is a design variable: the difference between a smooth joint at $\mu = 0.6$ and a raked one at 1.0 is a 67% change in resistance, obtained with a rake.

The shear an interface has to carry follows the shear diagram along the span, so a joint between two pours has most to do near the supports and nothing at all at mid-span, which is where the reinforcement crossing it is usually checked. Which of the two routes a shear takes when both are available is the ordinary question of whether the stiffest path takes the load, asked across a plane rather than along a member.

The base of a shear wall is the case where the external compression term earns its place. A wall carrying substantial gravity load has a permanent $\sigma_n$ across its base joint — the same weight that resists its overturning is holding its base joint together, and adding a modest external clamping stress of 1.0 N/mm² takes the resistance from 3.50 to 4.90 — a 40% gain from load that was there anyway.

A tension across the plane spends the clamping #

The $\sigma_n$ term was introduced above as a gift: a permanent compression across the joint adds directly to the resistance, and a wall base gets a great deal of it for nothing. The term has a sign, and the other sign is the commonest way this check is failed in service.

A direct tension across the plane does not merely fail to help. It consumes the clamping the bars were providing, one newton for one newton, because the mechanism cannot tell where the normal force came from. The expression is the same one read backwards:

$$v_n = \mu\left(\rho f_y - \frac{N}{A_{cv}}\right)$$

Take the corbel case. A joint 300 mm wide and 400 mm deep is 120,000 mm², so at 3.50 N/mm² it carries 420 kN of shear. Now suppose the bearing on the corbel delivers a horizontal restraint force of a fifth of the vertical reaction — 84 kN, which is the figure design guidance conventionally assumes when a bearing’s behaviour is not known. That is $\sigma_n = -0.70$ N/mm², so

$$v_n = 1.4\,(2.50 - 0.70) = 2.52\ \text{N/mm}^2$$

a 28% loss of shear capacity from a horizontal force nobody listed as a load. And it is the ordinary condition rather than an unusual one: an elastomeric bearing stiffens as it ages, a beam shrinks and drags its support inward, a bearing that was meant to slide seizes, and a thermal movement that had somewhere to go on the day of construction has nowhere to go in year twenty.

Which is why the standard requirement is to reinforce for the two effects separately and additively — $A_{vf}$ for the shear friction, plus $A_n = N/f_y$ on top of it, in the same layer, crossing the same plane. The steel that carries the tension has been spent and cannot also be clamping.

That is the general shape of the trap, and it is not confined to corbels. A shear-friction bar is being asked for a normal force, and any other demand for a normal force across the same plane is in direct competition with it. Flexural tension at the top of a corbel is one. Restrained shrinkage across a construction joint is another. Prying at a connection above it is a third. None of them appears in $\rho f_y$, all of them subtract from it, and the arithmetic that hides them is that they are perpendicular to the shear being checked — the one direction a shear calculation is not looking in.

A smooth joint is a different structure #

The whole of the difference between the two figures is the slope of the sawtooth. That is a genuinely unusual situation in this subject: a change of more than a factor of two in a resistance, produced by neither material nor dimension nor arrangement, but by the texture of a surface that a specification either did or did not ask for.

It also explains why the saturation ratio moves the other way. A smooth joint saturates at $\rho = 1.83\%$ rather than 0.79%, because it needs more clamping to reach the same crushing limit — so a smooth joint can usefully take more reinforcement than a rough one, and still ends at the same cap.

Where the model stops #

The cohesion term has been left at zero throughout. Design expressions usually carry one, representing the concrete’s own contribution before any slip. It is real for a joint that has never cracked and it is not there for one that has, and since the whole premise here is a plane that has already given up, including it would be claiming a resistance that the argument has just spent.

The dowel action of the bars is ignored. A bar crossing a crack does resist slip directly, by bending as a short cantilever in the concrete either side. It is a real contribution and a small one, and it is deliberately not counted — it needs a much larger slip to develop than the clamping does, so counting both at once counts a resistance twice at two different displacements.

The sawtooth is a caricature. A real crack surface has a distribution of asperity heights and slopes, the contact area grows as the clamping grows, and the effective $\mu$ falls slowly as the crack opens. The single-angle model captures the mechanism and not the softening.

Nothing here is cyclic. Under load reversals the asperities grind each other down, the crack opens progressively — the faces cannot re-seat, because the projections that held them apart are gone — and the resistance degrades cycle by cycle. Shear friction in a seismic joint is a different and much less generous calculation.

And the reinforcement is assumed perpendicular to the crack. Bars at an angle contribute both a clamping component and a direct component, which improves the resistance and complicates the expression, and a bar inclined the wrong way across a crack contributes less than a perpendicular one rather than more.

What the pictures cannot show #

The sawtooth in the hero figure is drawn at a pitch of tens of millimetres so that it can be seen. The real projections on a crack through 20 mm aggregate concrete are of that order in height and utterly irregular in spacing, and the smooth periodic figure suggests a regularity that would make the mechanism far more predictable than it is.

The load-slip curve is drawn against a slip axis running to two millimetres, which is drawn as a visible distance and is about the thickness of the line used for the crack. Every displacement on this page is a fraction of a millimetre.

And no figure here can show the thing that decides whether the calculation applies: whether the crack is where it was assumed to be. Shear friction is checked on a plane a designer nominates, and its whole validity rests on that plane being the worst one — which is an argument made in words, before the arithmetic starts.

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. The corbel, in full, where shear friction and strut-and-tie give different answers and both are checked. Interface shear in composite floors, where the plane is horizontal and the check is a shear flow rather than a force. Push-off tests, which is where every number on this page comes from and which measure a quantity that is a property of a specimen as much as of a material. Shear keys, which replace the asperities with a geometry somebody chose. And the case that inverts the argument: a joint deliberately made smooth and greased, so that shear is not transferred, which is a bearing.