Structural Engineering & Statics
where the load goes
Internal forces

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.

Assumes The shear nobody draws, The worst stress is not where the worst bending is and The property that appears in none of the equations.

21 min read 11 figures The load must go somewhereWhich failure arrives firstThe statics of things that do not move

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, and that is why it can exceed oneThe 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.36shearthe bars clamp, and do not carryasperity slope 54° · tan = 1.40clamping stress 2.50 N/mm² · resistance 3.50 N/mm²sliding without separating is not available to a rough crack
Fig. 1 The crack magnified: two rough faces drawn as a sawtooth at 54.5° to the plane. Sliding one over the other cannot happen without lifting it, so a shear displacement forces a separation in fixed proportion. The bars crossing the plane are stretched by that separation and clamp the faces back together.

The coefficient is a slope

The design expression is disarmingly simple:

vn=μ (ρfy+σn)v_n = \mu\,(\rho f_y + \sigma_n)

and the trouble it causes is entirely in the symbol μ\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.

The reaction lies inside the cone, so the block standsA block of 100 on a plane at 15°, against a coefficient of friction of 0.35. Resolving across and along the plane gives a normal force of 96.6 and a friction demand of 25.9, against a capacity of μN = 33.8 — a ratio of 0.77. Added together the two make one contact reaction leaning 15.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 19.3°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 15.0°, which is why the angle of repose is a material property and the size of a heap of sand is not.15°reaction, leaning 15.0° from the normalthe cone: half-angle arctan μ = 19.3°W = 100demand 25.9 against a capacity of 33.8 — F/μN = 0.77the weight appears nowhere in the cone — only the direction of the reaction is asked about
Fig. 2 Friction as this collection has already met it: a block on a plane, a normal force, a capacity 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.

The resolution is that μ\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 θ\theta to the plane, a slip ss opens the crack by stan⁡θs\tan\theta, and the force needed to drive the slip against a clamping stress σ\sigma is σtan⁡θ\sigma\tan\theta per unit area. The “coefficient of friction” is tan⁡θ\tan\theta and nothing else:

slope of the asperities apparent μ
31° 0.60
35° 0.70
45° 1.00
54.5° 1.40

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⁡θ\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 AvffyA_{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 σn\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

Nothing is resisted until something has movedShear 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 3.50 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.00.20.40.60.811.21.41.61.820123456slip along the crack (mm)shear resistance (N/mm²)the bars yield hereclampingfrom the bars3.50
Fig. 3 Resistance against slip. At zero slip the faces are not clamped and the resistance is whatever cohesion survives, which on a crack that has already opened is nothing. By 0.55 mm of slip the bars have reached yield and the resistance has reached its plateau of 3.50 N/mm².

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 ρfy\rho f_y on paper and much less of it in practice.

The steel analogue is exact and worth putting beside it.

A preloaded joint, before and after it slipsTwo preloaded bolts at 172 kN each, on one friction face at μ = 0.5. The joint carries 172 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 250 kN with the bolts now in shear. Two different mechanisms, one joint.00.511.522.533.544.55050100150200250displacement, mmload, kNfriction 172 kNbearing 250 kNslipthe rising branch is drawn, not solved: it is elastic shear of the plates
Fig. 4 A preloaded bolted joint, whose bolts are also not in shear: the preload clamps the plies and friction carries the load, until the joint slips into bearing and a second mechanism takes over. Same arrangement, a coefficient below one, and a plateau rather than a climb — because the clamping there is applied by tightening rather than developed by moving.

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

More steel across the crack, until the roughness runs outShear 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.0.0%0.2%0.4%0.6%0.8%1.0%1.2%1.4%1.6%1.8%2.0%0123456reinforcement crossing the crack, as a fraction of the areashear resistance (N/mm²)the crushing capsaturates at0.79%
Fig. 5 Resistance against the reinforcement crossing the plane. The clamping is proportional to the steel area and the resistance to the clamping, so the line is straight — until the asperities crush at 5.5 N/mm², which happens at a reinforcement ratio of 0.786%. The dashed line is what the clamping alone would give if the concrete were unbreakable.

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:

ρsat=vmaxμfy=5.51.4×500=0.786%\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 μfy\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.

One point, every plane through it, one circleA point carrying 140 N/mm² across one face, 0 across the other and 45 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 83.2 centred at 70.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are 153.2 and -13.2, on planes 16.4° from the face the 140 acts on; the largest shear on any plane is 83.2, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 160.2.στthe x faceσ₁ = 153.2σ₂ = -13.2τ max 83.2the plane turns by 16.4°, the circle by 32.7°von Mises 160.2 N/mm²
Fig. 6 The uncracked route: a state of stress at a point, every plane through it, and a principal stress that decides where the crack will form. This is the calculation shear friction begins after — its answer has already been used up by the time the plane exists.
Shear stress across a sectionThe distribution of shear stress over an I-section, computed as VQ/It by accumulating the first moment of the area above every height. The peak is 0.41 against a mean of 0.19 — a ratio of 2.15 — and it falls at the neutral axis, where the bending stress is zero.neutral axispeak 0.4stressflow, q = VQ ÷ Imean stress 0.19 — the value a shear divided by an area would givepeak 2.15× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does
Fig. 7 And the other uncracked route: 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.

The neutral axis is wherever the first moment vanishesA 300 by 556 section with 1200 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 192.2 mm from the top, which is where the first moment of the compression zone plus the transformed steel vanishes. The compression is 11.9 N/mm² at the top fibre and the steel carries 287 N/mm²; the resulting couple is 344 kN on a lever arm of 436 mm, which multiplies back to the 150 kNm applied. The uncracked section would have had 5040×10⁶ mm⁴ against the cracked 2415×10⁶ — a loss of 52% of the stiffness.x = 1921200 mm² of steel, n = 15b = 30011.9 N/mm²344 kN in the steelz = 436C = T = 344 kN · C·z = 150.0 kNm = the applied momentcracked I 2415×10⁶ mm⁴ against uncracked 5040×10⁶ — 52% of the stiffness gone
Fig. 8 The bending version of the same move: half the section has given up, the neutral axis has moved 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.

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.

A truss drawn inside a solid, and solved as oneA deep member 4000 mm between bearings and 2000 mm deep, carrying 1 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 1 kN and each strut at 1 kN, at 38.7° to the horizontal. Spread over a strut width of 812 mm the compression is 0.0 N/mm² against a limit of 15.8 for concrete cracked across its own strut, and the tie needs 2 mm² of steel. A beam calculation on the same member would have asked the tie for 1 kN, which is 7% less than the model does.1 kNtie 1 kNstrut 1 kN38.7°z = 1600strut 0.0 N/mm² over 812 mm · limit 15.8bursting across each strut 0 kN · tie steel 2 mm²
Fig. 9 The other model for the same member — a truss drawn inside the solid, with a compression strut into the support and a tie across the top. Shear friction and strut-and-tie are two ways of paying for the same 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 μ=0.6\mu = 0.6 and a raked one at 1.0 is a 67% change in resistance, obtained with a rake.

The connection is busiest where the beam is notThe force per unit length the interface has to carry, along a 4 m span under a uniform load, with connectors of stiffness 200. It is largest at the supports and zero at mid-span, which is the shear diagram and not the moment diagram — so the studs go where the bending stress is smallest and the last thing a designer looks at is where the connection works hardest. The peak here is 46.4 against 60.0 for a fully bonded beam of the same section, the difference being that a partly composite beam does not have the full section's shear flow to carry. The total the connectors on one half of the span must transfer is 54.0 kN.05001000150020002500300035004000-40-2002040along the span (mm)force per unit length at the interfacewhat the connectors carryVQ/I, if it were bonded
Fig. 10 And the composite version of the same plane. 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, exactly as the stiffest path takes the load decides which route a shear takes — which is where the reinforcement crossing it is usually checked.

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 σn\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 smooth joint is a different structure

The coefficient is a slope, and that is why it can exceed oneThe 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.16shearthe bars clamp, and do not carryasperity slope 31° · tan = 0.60clamping stress 2.50 N/mm² · resistance 1.50 N/mm²sliding without separating is not available to a rough crack
Fig. 11 The same crack with a shallow roughness — the joint left as cast against a smooth form, at an asperity slope of 31°. Every part of the mechanism is unchanged and the resistance is 1.50 N/mm² instead of 3.50, because the climb is gentler and less clamping is developed per unit of slip.

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 ρ=1.83%\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 μ\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.

Where this goes next

neutral axispeak 0.4stressflow, q = VQ ÷ Imean stress 0.19 — the value a shear divided by an area would givepeak 2.15× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does The shear nobody draws A stack of loose planks slides at its ends when it is loaded. Glue them and the sliding stops — and whatever the glue is now carrying is a stress that no bending calculation contains. στthe x faceσ₁ = 153.2σ₂ = -13.2τ max 83.2the plane turns by 16.4°, the circle by 32.7°von Mises 160.2 N/mm² The worst stress is not where the worst bending is Every stress this collection has quoted is a stress on a particular plane, and neither the bending stress nor the shear stress is a property of the point. Turn the plane and both change; one pair of numbers does not, and on a short beam it peaks where neither of them does. 00.511.522.533.544.55050100150200displacement, mmload, kNfriction 137 kNbearing 188 kNslipthe rising branch is drawn, not solved: it is elastic shear of the plates The joint that carries nothing until it slips Tighten the bolts hard enough and the plates are clamped together with a force nothing applied. The joint then carries shear by friction, the bolts are in tension and not in shear at all, and the load path has nothing in common with the joint it looks identical to. x = 1371200 mm² of steel, n = 7.5b = 30018.0 N/mm²371 kN in the steelz = 404C = T = 371 kN · C·z = 150.0 kNm = the applied momentcracked I 1139×10⁶ mm⁴ against uncracked 3422×10⁶ — 67% of the stiffness gone When half the section has given up Bending theory puts the neutral axis through the centroid. That is a consequence, not a rule — and when the tension side cracks, the same reasoning moves the axis somewhere else entirely.

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Aggregate interlockAnchorageClamping forceConstruction jointCorbelCrackDilatancyFailure surfaceLimit stateReinforcement ratioShear frictionShear transferSlip