Connections

The failure that is in the concrete

An anchor bolt is a steel component and its capacity is usually decided by something else entirely — a cone of concrete pulled out around it, failing in tension, in a material every other calculation on the project has assumed cannot take tension at all. The exponent in the capacity says so: it is not the square the geometry implies.

Assumes Where the structure meets the ground, and when the bolts start working, A check made on a perimeter, not on a section and The bigger one is the weaker one.

A holding-down bolt is designed like any other steel component: an area, a stress, a division. The number that comes out of that division is almost never the capacity, because the bolt is cast into concrete and the concrete goes first.

That is an unusual situation in this collection, and it inverts the usual reading of a connection that is not a point: the region that matters is not inside the joint but in the material underneath it. Everywhere else a connection’s capacity is decided by the connection — a bolt in shear, a weld throat, a plate in bearing. Here it is decided by the material the connection was attached to, in a mode that material is not usually asked to work in at all.

A base plate, and when the bolts start workingA 500 × 500 mm plate carrying 600 kN and 180 kN·m, so the resultant sits 300 mm from the centre against a kern of 83.33 mm. The plate is in bolts engaged: bearing over 150.88 mm at a peak of 20 N/mm², with the holding-down bolts carrying 154.42 kN. The plate lifts at 50 kN·m and crushes at 126 kN·m, and the bolts are not needed until 150 kN·m.600 kN180 kN·mresultant at e = 300, outside the platemiddle third154.42 kNBolts engagedbearing over 150.88 mm at 20 N/mm² · 100% of 20
Fig. 1 A base plate carrying axial force and moment. Past a moment of 150 kN·m the bolts start working, and here they are asked for 154 kN.

Which free body produced the number

Take a cone of concrete with the anchor’s head at its apex and a free surface at its base, and pull the anchor out of it. The forces on that free body are the anchor’s tension, the self weight, and a tensile stress distributed over the conical surface where the cone parts from the rest of the slab.

The whole of anchorage design is that free body, and the two questions it raises are how big the surface is and what stress it can carry.

The surface’s size follows from an angle. Measured breakout cones stand at about 35° to the concrete face, so a cone from an embedment hefh_{ef} reaches 1.5hef1.5h_{ef} out in every direction and its projected area on the surface is about 9hef29h_{ef}^2. Multiply by a tensile stress and the capacity goes as hef2h_{ef}^2.

It does not. The measured capacity goes as hef1.5h_{ef}^{1.5}, and the missing half-power is the whole of the interesting content.

Where the exponent went

A strength that is a property of the specimenNominal 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 = 80 mm. A 100 mm specimen reads 2.80 N/mm² and a 900 mm member of the same material carries 1.20: the test overestimates the structure by a factor of 2.33.10321003161000316201234size (mm, logarithmic)nominal strength (N/mm²)the specimen: 2.80the structure: 1.20the plastic limitfracture mechanicsthe test overestimates by 2.33× · D₀ = 80 mm
Fig. 2 Nominal strength against size for geometrically similar specimens. A 100 mm specimen reads 2.80 N/mm² and a 900 mm member of the same material carries 1.20 — the test overestimates by a factor of 2.33.

A breakout cone is a fracture surface, and fracture surfaces get weaker as they get bigger. The energy released by a crack running through the concrete grows with a volume and the energy consumed grows with an area, so the nominal stress at failure falls as the inverse square root of the size — which takes an exponent of 2 down to 1.5 exactly.

That is the bigger one being the weaker one appearing as the exponent of a design formula rather than as a curiosity about test specimens. It is one of very few places in structural design where a size effect is written directly into a capacity equation instead of being buried in a factor, and the reason is that the effect here is large: over the range of embedments used in practice it costs a factor of two or more.

The same power appears in the concrete cone’s shear counterpart and in punching, and it always has the same origin. Wherever a capacity comes from a tensile fracture surface, a geometric exponent is reduced by a half.

The perimeter, and its counterpart

The related check most designers already know is punching, and it is worth setting alongside because the two behave differently for a reason.

A check made on a perimeter, not on a sectionOne bay of a flat slab, 7.2 m square, on a 300 × 300 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 3462 mm long against 1200 mm round the column itself. The shaded area inside it delivers no shear across it and is subtracted from the load; everything outside arrives through the perimeter. At 12 kN/m² that is 611 kN across 3462 × 180 mm, a shear stress of 0.980 N/mm² against a resistance of 0.678.2d = 360column7.2 m bayperimeter u₁ = 3462 mmshear to carry V = 611 kNv = 0.980 against 0.678 N/mm²145% of the resistance used
Fig. 3 A control perimeter for punching, drawn 2d from the column face with its corners rounded. It is 3462 mm long against 1200 mm round the column itself.

A check made on a perimeter uses a line times a depth; a breakout check uses an area. That difference makes punching resistance nearly linear in the column size and breakout capacity quadratic in the embedment, and it is the reason the two look like different subjects when they are the same failure surface counted two ways.

The comparison is also a reminder of how much of either perimeter is corner. On the slab above, the rounded corners are most of the control perimeter, and on a breakout cone the projected area is dominated by the parts furthest from the anchor — which is exactly the part an edge or a neighbour takes away.

Edges and neighbours

The cone is a volume, and volumes interfere with things.

An anchor near a free edge has its cone truncated: the concrete that would have carried part of the tension is not there, and the capacity falls in proportion to the projected area lost. Below an edge distance of 1.5hef1.5h_{ef} the reduction is roughly linear, and an anchor at the very edge of a slab has lost about half its cone before anything else is considered.

A group of anchors is the same problem inwards. Two anchors closer together than 3hef3h_{ef} have overlapping cones, and the group’s failure surface is one merged cone rather than two — so the group capacity is the projected area of the merged shape, not twice the area of one. Packing four anchors into a small base plate buys four times the steel and very little more concrete.

That is a genuinely counter-intuitive result and it produces the characteristic detail of anchorage design: base plates that are much larger than the column they carry, with the bolts pushed out to the corners, in an arrangement that looks wasteful and is not.

The width nobody drewA gusset plate with a brace bolted to it over 240 mm, and the width the profession has agreed to pretend is carrying the force. Everything else on this site arrives with a cross-section; a gusset does not, because it is a piece of steel with something attached somewhere in the middle of it and there is no geometry that says how much of it is working. The answer is the **Whitmore section**: assume the force spreads at 26° from the first fastener and take the width it has reached at the last, b_eff = w + 2L·tan26° = 324 mm. That is 3.60 times the width anything is actually attached to, and the rule comes from a 1952 master's thesis. It has since been checked against finite element work and holds to about ten per cent, which is fortunate, because moving the assumed angle by ten degrees moves the answer by 35%. On this plate the check that governs is not the stress the rule was written for: it is the Whitmore section buckles, at 637 kN against 1381.the brace force26°b_eff = 324 mm90 mm3.60× the width anything is actually attached togoverns on the Whitmore section buckles, at 637 kN
Fig. 4 The same difficulty in a plate rather than in concrete. A force attached somewhere in the middle of a piece of steel has no cross-section until somebody assumes a spread angle, and the answer moves 35% for ten degrees of assumption.

The Whitmore construction is the plate version of the cone, and it is worth putting beside it because the epistemics are identical. A spread angle is assumed, an effective width follows, and the whole check rests on a geometric convention that has been checked against measurement and is not derivable. Both are conventions with error bars, dressed as geometry.

The bolt’s own problems come first

Before the concrete has a chance, the load has to reach it, and two things happen on the way.

Prying action in a tee stubA tee stub pulled by its web with 180 kN per bolt. The 20 mm flange is in the mechanism regime, so the prying force at the flange tip is 61.88 kN and the bolt carries 241.88 kN — 1.34 times what was applied. The flange stops prying entirely at 36.18 mm thick, and collapses on its own at 110 kN.180 kN appliedbolt 241.88 kNprying 61.88 kNm = 45n = 40flange 20 mm · mechanismbolt force is 1.34 times the applied load
Fig. 5 Prying action in a tee stub. The flange bends, its tip bears back on the support, and the bolt carries 241.88 kN against the 180 applied — a factor of 1.34.

A flexible base plate bends under the bolt tension, its edge bears back against the concrete, and the bolt carries the applied tension plus the prying force. On the tee drawn that is a factor of 1.34, and on a thinner flange it is more. So the tension arriving at the anchor is larger than the tension in the column, and the concrete cone check has to be done on the larger number.

Block shear: the metal between the holesThree bolts in a 12 mm plate end connection. The shaded block tears out along a shear plane 180 mm long and a tension plane 40 mm long. Shear yields first, and the capacity is the sum of two different strengths on two different planes: 506.04 kN, of which the shear plane carries 70.43%.pullshear plane, 180 mmtension plane, 40 mmcapacity 506.04 kN0.6 fu Anv = 387 kN · 0.6 fy Agv = 356.4 kN · fu Ant = 149.64 kNthe yield value governs the shear plane
Fig. 6 The steel version of a block being pulled out. A shear plane and a tension plane fail together at different strengths, and the shear plane carries 70% of the total.

Shear, which fails a different way

An anchor in shear does not pull a cone out of the surface. It pushes concrete off the edge in front of it, a half-cone on its side, and that failure depends on the edge distance rather than on the embedment.

That gives a bolt group two entirely separate geometries to satisfy — deep enough for tension, far enough from an edge for shear — and the two are not traded against each other. It also means the interaction of tension and shear on an anchor is an interaction of two different concrete failures rather than two components of one steel one.

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 4.00 N/mm² · resistance 5.50 N/mm²sliding without separating is not available to a rough crack
Fig. 7 The mechanism behind a shear transfer across concrete. The bars do not carry the shear; they supply the normal force that lets the roughness carry it, and a bar not anchored on both sides supplies nothing.

That last clause is the connection between this essay and every other anchorage question on a project. Shear across a crack works only because the bars crossing it are anchored, and a bar anchored into a cone that has broken out is not anchored at all. The failures cascade.

Working the arithmetic once

It is worth putting numbers on a base plate, because the ordering of the failures is the whole design and it is not obvious from any of the formulae.

Take four M24 grade 8.8 bolts, one at each corner of a 500 mm square plate, cast 300 mm into a slab. The steel capacity of one bolt in tension is about 250 kN, so the group in tension is 1,000 kN if nothing else intervenes.

The breakout cone from a 300 mm embedment reaches 450 mm in every direction. Four bolts at 400 mm spacing therefore have cones that overlap heavily: the merged projected area is roughly a 1,300 mm square rather than four separate 900 mm circles, which is about 40% of the area four independent cones would give. Multiply by a nominal tensile capacity and the group’s concrete capacity lands in the region of 500 kN.

So the concrete governs by a factor of two, and every kilonewton of extra bolt steel bought for that connection is wasted. Two moves recover it: push the bolts apart, which separates the cones and costs only plate; or embed deeper, which buys capacity as h1.5h^{1.5} and costs only a longer bolt. Deepening from 300 mm to 450 raises the concrete capacity by 1.84, and that is usually the cheapest structural intervention available on a whole project.

Neither move appears in a calculation that treats the anchorage as a bolt check, because neither the spacing nor the embedment is in that calculation at all.

Why the whole subject is about ductility

Everything above is a brittle failure. A concrete cone comes out with no warning, no redistribution and no reserve — the property that appears in none of the equations, absent, and a group of anchors that fails this way fails all at once because the cones are connected.

So the design rule that governs anchorage everywhere is not a capacity at all. It is a hierarchy: make the steel govern. Choose an embedment deep enough, and an edge distance and spacing generous enough, that the concrete’s capacity exceeds the bolt’s yield capacity — and then the connection fails by the bolt stretching, which is visible, gradual and redistributing.

A metal does not care what pressure it is under; nothing else agreesStrength against hydrostatic pressure, for a metal and for a pressure-dependent material. Von Mises's criterion contains only stress *differences*, so squeezing a metal equally in three directions does nothing at all to it and its locus is a cylinder along the hydrostatic axis — the flat line. A granular material's is a cone: its strength rises with pressure at a rate fixed by its friction angle, 38°, and this is the same statement as the confinement argument, where a lateral pressure of a twelfth of the concrete's strength raises that strength by half. The two are not variants of one theory; they disagree about whether a quantity appears at all.00.511.522.53050010001500hydrostatic pressure ÷ f_ystrength (N/mm²)a metala material with frictionMises contains only stress differences, so the pressure cancels out of it exactly
Fig. 8 Why concrete and steel cannot be reasoned about the same way. A metal’s strength does not depend on the pressure it is under and a granular material’s does, so the two are not variants of one theory — they disagree about whether a quantity appears at all.

That is the same capacity-design argument that puts a plastic hinge where somebody wants it and keeps the connection elastic around it. Here the thing being protected is the base material rather than a member, and the arrangement is the part that is meant to be weak with the weak part chosen to be the bolt, and the reason is the same: a failure mode is being chosen, because the alternative chooses itself.

Where the model stops

Cracked concrete is a different material. All the numbers above assume the cone forms in sound concrete. An anchor in a region that is already cracked — which is most of a reinforced concrete member under service load — has a capacity roughly 30% lower, and whether a given anchor is in a cracked region is a question about the member’s bending, not about the anchor.

Reinforcement changes everything. Bars crossing the cone surface hold it together and can be designed to carry the whole tension, at which point the breakout check disappears and is replaced by an anchorage-into-reinforcement problem — which is the force that splits what it pushes on and a strut-and-tie model.

A truss drawn inside a solid, and solved as oneA deep member 4000 mm between bearings and 2000 mm deep, carrying 800 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 500 kN and each strut at 640 kN, at 38.7° to the horizontal. Spread over a strut width of 750 mm the compression is 2.1 N/mm² against a limit of 15.8 for concrete cracked across its own strut, and the tie needs 1149 mm² of steel. A beam calculation on the same member would have asked the tie for 468 kN, which is 7% less than the model does.800 kNtie 500 kNstrut 640 kN38.7°z = 1600strut 2.1 N/mm² over 750 mm · limit 15.8bursting across each strut 160 kN · tie steel 1149 mm²
Fig. 9 The replacement, drawn as what it is. A truss inside a solid, solved by the truss solver, with a tie that needs steel and struts that need width.

And the plate itself may not be stiff enough to share. Everything above assumed the four bolts carry equal shares of the tension, which requires a plate stiff enough to distribute it. A flexible plate loads its nearest bolt first — the same non-uniformity a long bolted joint has along its length, one dimension over.

Time is not in any of it. A bolt cast into concrete is protected by the concrete around it, and the cover over it fails by splitting rather than by any of the mechanisms above.

Cover enters twice, and the strength of the concrete enters onceHow long a 25 mm bar has before the cover over it splits, against the cover, split into the two halves it is always split into. Initiation is the time for the chloride front to reach the bar, which goes as the **square** of the cover — Fick's law and nothing else — and it is 9.1 years at 35 mm and 36.2 at 70. Propagation is the time from there to a split cover, which is short: 0.9 years, because the cover cracks at a section loss of 0.17% and no strength check in this collection would notice a loss that small. The pressure the cover can take grows with the cover too, so cover appears in both terms and the concrete's own tensile strength appears in one of them, linearly. That asymmetry is why every durability clause in every code is about cover and crack width, and hardly at all about strength.204060800102030405060cover (mm)years to a split coverinitiation ∝ c²and propagation35 mm: 9.1 yr70 mm: 36.2 yrthe cover splits at a section loss of 0.17%
Fig. 10 How long a bar has before the cover over it splits. Initiation goes as the square of the cover and propagation is short — the cover cracks at a section loss of 0.17%, which no strength check would notice.

The three checks that have to be made, and their order

A complete anchorage design is a list, and the value of the list is that its items belong to different materials and different mechanisms rather than being refinements of one another.

Steel failure of the anchor. An area times a stress, with the prying force added. Ductile, well characterised, and the one everybody computes.

Concrete cone breakout. A projected area times a tensile stress, reduced for edges, spacing and cracking, with the exponent that carries the size effect. Brittle.

Pull-out and splitting. The head or the bond fails locally without a full cone forming, which is a bearing failure under the head for a cast-in anchor and a bond failure for a bonded one. It governs at shallow embedments, where the cone is small enough that the local mechanism is weaker.

The order matters because the three have different sensitivities to the same decision. Deepening the anchor helps the second and the third and does nothing for the first. Moving it from an edge helps the second and nothing else. Making it thicker helps only the first — and if the first was not governing, it helps nothing at all.

That is why anchorage design is one of the few places in this subject where the answer to “it does not work” is almost never “make it bigger”.

What the picture cannot show

The cone is drawn as a cone and it is not one. Real breakout surfaces are irregular, follow the aggregate, and are truncated by whatever reinforcement, ducts and services happen to lie in the way. The 35° angle is a fitted average of a scattered population.

Nor does any figure show the installation. A cast-in anchor’s capacity depends on it being where the drawing says, at the depth the drawing says, in concrete that was properly compacted around its head — and the commonest failure of an anchorage is not any mechanism in this essay but a bolt that was 40 mm shallower than intended.

The generalisation

The habit worth carrying is to ask which material a connection’s capacity belongs to.

A bolted splice’s capacity belongs to the steel. A welded joint’s belongs to the weld metal and the parent plate. An anchorage’s belongs to the concrete, and the concrete is being asked for tension — which is the one thing the rest of the project has agreed it does not have.

That inversion is the reason anchorage is so consistently underestimated. Every instinct built up designing steelwork is about the connected parts, and here the connected part is a passenger: the anchor is a way of addressing a volume of concrete, and its capacity is a property of that volume, its edges, its neighbours, its cracks and its size. Nothing about the bolt appears in the answer at all until the design has been arranged so that it does.

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.

AnchorageBase plateBreakoutBrittlenessCapacity designCoverDuctilityEdge distanceLoad pathPryingPunching shearShear frictionSize effectStrut and tieTensile strength