Structural Engineering & Statics
where the load goes
Internal forces

A check made on a perimeter, not on a section

Every shear check in this collection is made on a plane cut through a member. A slab sitting on a column has no such plane, because the shear leaves in every direction at once — so the check is made on a closed line, and a line grows with the column while the load grows with the square of the bay.

Assumes The shear nobody draws, The load a beam is given is a decision and The slab that spans both ways.

22 min read 11 figures The load must go somewhereScale changes everythingWhich failure arrives first

A beam is checked for shear on a plane cut through it. The plane has a width and a depth, the shear crossing it is a force, and the two divide. The shear nobody draws is on such a plane, and so is every other shear argument in this collection so far, and it is so natural that the shape itself goes unnoticed.

A flat slab sitting directly on a column has no such plane. The load arrives at the column from every direction at once, and any plane drawn through the slab has more slab on both sides of it. Choosing the free body is the whole of the method, and here the choice that works is not a plane at all. What fails is not a section but a surface: a cone or a pyramid of concrete that pushes down through the slab and takes a plug of it with the column. The check that follows is made on a closed line drawn round the column, and that single change of geometry is the whole subject.

A check made on a perimeter, not on a sectionOne bay of a flat slab, 7.2 m square, on a 400 × 400 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 4427 mm long against 1600 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 604 kN across 4427 × 225 mm, a shear stress of 0.606 N/mm² against a resistance of 0.658.2d = 450column7.2 m bayperimeter u₁ = 4427 mmshear to carry V = 604 kNv = 0.606 against 0.658 N/mm²92% of the resistance used
Fig. 1 One bay of a flat slab on a 400 mm square column. The heavy closed line is the control perimeter, standing two effective depths from the column face with its corners rounded at that radius. The shaded area inside delivers no shear across it and is subtracted from the load; everything outside arrives through the line.

Why there is no section

The reason is worth stating in the language of free bodies rather than of codes, because it is the only reason.

A shear force is what a cut reveals: draw a plane through a member, take the piece on one side, and whatever is needed to keep that piece still is the internal force on the plane. For a beam the choice of plane is free but its direction is not — a beam has one axis, and the interesting cuts are perpendicular to it.

A flat slab has two axes and no preference between them. Cut it on a plane through the column in one direction and the free body is half a bay of slab, held up by half a column; the shear on that plane is real and it is not the thing that fails. Cut it in the other direction and the answer is the same by symmetry. Neither cut has isolated the column.

The free body that does isolate it is closed: a surface running right round the column, from the top of the slab to the bottom. Everything outside that surface has to deliver its load across it, and there is no plane anywhere in the statement.

Load, shear and moment — a simple spanThe applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.4 per unit lengthshear16.0moment32.0 at x = 4.00the moment peaks exactly where the shear passes through zero
Fig. 2 For contrast, the shear a beam is checked for: a single scalar along a single axis, at its largest where the moment is smallest. There is one diagram because there is one direction to travel in, and a designer looks at one number per station.
Shear stress across a sectionThe distribution of shear stress over a tall rectangle, computed as VQ/It by accumulating the first moment of the area above every height. The peak is 0.29 against a mean of 0.19 — a ratio of 1.50 — and it falls at the neutral axis, where the bending stress is zero.neutral axispeak 0.3stressflow, q = VQ ÷ Imean stress 0.19 — the value a shear divided by an area would givepeak 1.50× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does
Fig. 3 And the section it is checked on, with the shear stress distributed over it as VQ/It. A beam’s shear check has a width and a depth and a distribution across them; nothing in this figure has an analogue at a slab-column junction, where the equivalent quantity is spread round a perimeter rather than over a face.

Which free body produced the number

The perimeter is drawn at some distance from the column face — two effective depths in the code these numbers come from, half of one in others, and the distance is a convention rather than a discovery. What matters is that the same distance is used to draw the line and to calibrate the stress it is compared against, so the pair is consistent even though neither half is fundamental.

Take the free body to be the whole bay outside that perimeter. The load on it is whatever arrives from the slab it covers, which is the bay area less the small piece inside the line. The only route out is across the perimeter. So

vEd=β VEdu1 dv_{Ed} = \frac{\beta\,V_{Ed}}{u_1\,d}

with u1u_1 the length of the closed line and dd the effective depth of the slab. That denominator is a length times a depth, which is an area, so the quantity is a stress — but it is not a stress anywhere in the slab. It is a force divided by a surface, and its only job is to be compared with a number obtained by dividing other forces by other surfaces in tests.

For the bay drawn: the perimeter is 4,427 mm, of which 1,600 mm is column face; the load inside the perimeter is 1.52 m² of an 51.8 m² bay; the shear to carry is 604 kN; and the shear stress is 0.606 N/mm² against a resistance of 0.658. Ninety-two per cent used, which is the condition a great many real slabs are in.

Where a beam's load comes fromA 7.2 × 7.2 m panel carrying 12 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 13.0 m² each and the short beams a triangle of 13.0 m²; the four areas sum to 51.8 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 43.20 kN/m at midspan.13.0 m²21.60 kN/m13.0 m²21.60 kN/m13.0 m²21.60 kN/m13.0 m²21.60 kN/m7.2 m7.2 m51.8 m² divided, 51.8 m² of panel — the division closes
Fig. 4 Where the 604 kN comes from — the same tributary argument that gives a beam its load, applied to a column instead. For a square bay on a regular grid the column’s area is the full bay, which is the arrangement that makes a flat slab worth building and the arrangement that makes punching hard.

The bay wins, and it wins as a square

Here is the asymmetry that decides everything.

The resistance is a property of the column and the slab. It contains a perimeter, which is four column faces plus a fixed amount of corner, and a depth. Nothing in it knows how far away the next column is.

The demand is a property of the grid. It is a pressure times an area, and the area is the bay squared.

So the two quantities are not merely different, they scale differently, and a bay large enough will defeat any column.

The perimeter grows with the column and the load grows with the bayShear demand divided by shear resistance on the control perimeter of a 400 × 400 mm column in a 225 mm slab, against the bay it supports. The resistance is fixed — it is a property of the perimeter and the slab — while the load rises with the square of the bay, so the curve is a parabola and it crosses one at 7.49 m. Nothing about this check is a stress in a section: it is a force divided by a length of 4427 mm times a depth.45678910111200.511.522.5bay, square (m)demand ÷ resistancethe limit7.49 mon this column
Fig. 5 Demand divided by resistance, against the bay. The resistance is a horizontal line on this plot in every sense — it does not move as the bay changes — and the demand is a parabola through the origin. They cross at 7.49 m, which is the largest square bay this column and this slab will carry without help.

Read the numbers off it: a 6 m bay is at 63% of the resistance, a 7.2 m bay at 92%, an 8 m bay at 114% and a 10 m bay at 180%. The slab has done nothing wrong between the first and the last; the grid has changed.

This is why punching is the failure that decides flat-slab construction rather than a detail within it. A designer choosing a column grid is choosing, before any member has been sized, whether the slab-column junction will work. And the choice is unforgiving in one direction: a bay 40% larger doubles the shear it has to deliver through a perimeter that has not moved.

The scale argument is the same one that separates a beetle from a bridge. Weight grows as a volume and strength as an area; here demand grows as an area and resistance as a length. The exponent gap is one instead of one, and the consequence is identical — there is a size past which the arrangement stops working and no amount of care within it helps.

Depth is in the answer more than once

The obvious repair is a thicker slab, and it is the right one, for a reason worth taking apart.

The effective depth appears three times in the resistance. It multiplies the stress, because the denominator is u1du_1 d. It sets the radius of the corner arcs, because the perimeter stands at two depths from the face, so a thicker slab has a longer perimeter. And it appears inversely inside the empirical size-effect factor k=1+200/dk = 1 + \sqrt{200/d}, which says that thick slabs are less good per unit area than thin ones — a real effect, measured, and one of the few places in this subject where an absolute dimension rather than a ratio decides an answer.

Depth is in the answer twice, and the size effect takes some of it backPunching resistance against the effective depth of the slab, everything else held. The depth enters three times over — the perimeter stands 2d from the face and so grows with it, the resistance is a stress times that perimeter times the depth, and the empirical size-effect factor shrinks as the slab gets thicker. Fitted over this range the resistance goes as d to the power 1.49, which is neither the square the first two terms suggest nor the linear dependence a shear check on a beam would give.1502002503003504004505000500100015002000effective depth (mm)punching resistance (kN)fitted power1.49not 2, andnot 1
Fig. 6 Resistance against effective depth, everything else held. Two of the three appearances of the depth push the resistance up and the third pulls it down; the fitted exponent over this range is 1.49, which is neither the square the first two suggest nor the linear dependence a beam’s shear check would have.

An exponent of 1.49 is a strong dependence. Going from 225 to 300 mm of effective depth — a third more concrete in the slab, and a third more weight on every column and foundation in the building — multiplies the punching resistance by 1.51. Nothing else available moves the answer that far.

For comparison, the concrete strength enters as a cube root. Doubling the characteristic strength from 30 to 60 N/mm² multiplies the resistance by 1.26. Going from 30 to 50 buys 19%, which is less than 25 mm of slab, and the 25 mm is cheaper. The same ordering holds for the property that appears in none of the equations: a stronger concrete is a more brittle one, so the material handle makes the failure worse in the one respect that is not being counted. Punching is a geometry problem wearing a material’s units, and the material is the least effective handle on it.

Most of the perimeter is not the column

The remaining handle is the column, and this is where the arithmetic surprises.

Most of the control perimeter is not the columnPunching resistance against the size of the square column, everything else held. The gain is far weaker than linear — fitted at the power 0.43 — because the control perimeter is the four faces plus two full quarter-circles at each corner, and those corner arcs are 2827 mm long whatever the column does. At the column drawn they are 64% of the whole perimeter.200400600800100012001400020040060080010001200column side (mm)punching resistance (kN)fitted power0.43the cornersdo not shrink
Fig. 7 Resistance against the size of a square column. The gain is far weaker than linear — fitted at the 0.43 power — because the perimeter is four straight faces plus four quarter-circle corners, and the corners are the same length whatever the column does.

The perimeter of a rounded rectangle is 2(c1+c2)+2πa2(c_1 + c_2) + 2\pi a, where aa is the distance to the perimeter. The second term is a complete circle of radius aa, assembled out of four corners, and it does not contain the column at all. At the slab drawn here aa is 450 mm, that circle is 2,827 mm, and it is 64% of the whole perimeter.

The limit is worth stating because it sounds absurd and is exactly right: a column shrunk to a point still has a control perimeter, of length 2πa2\pi a, and therefore still has a punching resistance. That resistance is 2π⋅2d⋅d⋅vRd2\pi \cdot 2d \cdot d \cdot v_{Rd}, which is proportional to d2d^2 and contains no column dimension whatever. A slab does not lose its ability to resist punching as the support gets small; it loses only the part of the perimeter that the support contributed.

Turned round: doubling the column from 400 to 800 mm takes the perimeter from 4,427 to 6,027 mm, a gain of 36%. Doubling it again, to 1,600 mm, buys another 53%. Column sizes are limited by architecture long before this stops being true, which is why the practical version of the move is a column head — a local thickening of the slab or a flare on the column, which increases the perimeter and the depth together.

A check made on a perimeter, not on a sectionOne bay of a flat slab, 7.2 m square, on a 1200 × 1200 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 7627 mm long against 4800 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 571 kN across 7627 × 225 mm, a shear stress of 0.333 N/mm² against a resistance of 0.658.2d = 450column7.2 m bayperimeter u₁ = 7627 mmshear to carry V = 571 kNv = 0.333 against 0.658 N/mm²51% of the resistance used
Fig. 8 The same bay with a 1,200 mm head under the column. The perimeter has gone from 4,427 to 7,627 mm and the utilisation from 92% to 51% — bought with a piece of concrete that is not a structural member in any other sense and appears on no analysis model.

The moment nobody applied to the check

Everything above assumes the shear runs round the perimeter evenly. It does not, whenever the column also transfers a moment to the slab — which is at every edge column, every corner column, and every interior column in a frame carrying lateral load or unequal spans.

A moment transferred into a slab is carried partly by bending across a width of slab and partly by shear running round the perimeter unevenly: more on one side, less on the other. The check absorbs this with a factor β\beta multiplying the whole demand, computed from the eccentricity M/VM/V and from the perimeter’s own first moment about its centre — so a long thin column is worse than a square one of the same area, because its perimeter is worse at resisting a twist.

For the slab drawn, an eccentricity of 150 mm — a modest unbalanced moment — puts β\beta at 1.075 and the utilisation from 92% to 99%. An eccentricity of 400 mm puts it at 1.199 and the slab fails. Nothing about the vertical load changed.

This matters because the eccentricity is often the part of the problem that arrives last. A slab designed on a regular grid under gravity is checked with β=1\beta = 1; the lateral system is designed later; and the moment the column then transfers into the slab was not in the check.

What is actually happening in there

The stress vEdv_{Ed} is a bookkeeping device and the perimeter is a convention. The mechanism is worth naming separately.

Round a column the slab is in a genuinely three-dimensional state: radial compression driving into the column, circumferential tension going round it, and an inclined crack that starts at the column face and runs out and up through the depth. The concrete above that crack is a conical strut carrying load into the column, and it fails by crushing, by loss of aggregate interlock across the crack, or by the top reinforcement tearing out — usually some of each. None of the three is the ductile redistribution that makes a bending failure forgiving, which is the other half of why the check is treated as a hard limit.

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 nearest thing this collection has already drawn: a truss put inside a solid, with the compression drawn as a strut and the reinforcement as a tie. A punching cone is the same idea rotated about the column — a conical strut instead of a pair of them, and a ring of tension instead of a straight tie.

The reason the empirical check survives despite that mechanism being well understood is instructive. A strut-and-tie model of a punching cone can be built and it is sensitive to details a designer does not control: the exact crack angle, the tensile strength across it, the anchorage of every top bar. The empirical expression trades that fidelity for the two variables that actually decide the answer — the perimeter and the depth — and calibrates the rest against several hundred tests.

Reinforcing a surface

Shear reinforcement in a slab is unlike shear reinforcement in a beam, and the difference follows from the geometry again. A beam’s links cross one plane. A slab’s shear reinforcement has to cross a surface, so it is arranged in rings or on radial lines round the column, and every ring further out crosses a longer surface and needs more steel to keep the same effect.

Whatever crosses that surface works the way shear across a crack that is already there works: the bars are not carrying the shear, they are clamping the crack so that the roughness can.

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. 10 Why more steel stops helping. The clamping a bar provides is proportional to its area, and the interlock it enables is proportional to the clamping — until the asperities crush, at which point the line goes flat. Punching shear has its own version of this cap, and it is checked separately at the column face itself.

That cap is the second check every punching calculation carries and the first one people forget. At the column face the stress is 1.68 N/mm² against a crushing limit of 5.28, so the slab drawn here has room. A slab with heavy shear reinforcement can be pushed until that limit governs, and past it no reinforcement of any kind is worth adding: the concrete inside the perimeter has been asked to deliver more than it can carry in compression.

Where the model stops

Nothing here is a two-way slab calculation. The slab’s bending is a separate problem, solved by the strip argument that a square panel spans both ways and a long one does not, and it delivers a load to the column that the punching check then consumes. The two calculations share only that number.

A two-way slab is a one-way slab as soon as it is not squareThe share of the load carried by the strips spanning the short way, against the ratio of the sides. The two families of strips cross at the centre and must deflect equally there, and a strip's deflection goes as the fourth power of its span — so at a ratio of 1.33 the short strips already take 76% and at 2 they take 94%. The panel drawn here is 7.2 × 7.2 m, a ratio of 1.00, and its short strips take 50.0%. Two-way action is worth having at a ratio of one and worth almost nothing by two.11.522.530.40.50.60.70.80.91long span ÷ short spanshare taken by the short strips7.2 × 7.2 m: 50.0%by 2 : 1 it is a one-way slab
Fig. 11 The bending half of the same slab, at the ratio of sides this bay has. A square panel shares its load equally between the two directions, which is why a flat slab on a square grid is efficient in bending — and why so much load arrives at the column, which is what makes it hard in shear. A column that stops and hands its load sideways — a transfer structure — concentrates it further still.

The perimeter is a convention and it is not the failure surface. The real crack does not follow the drawn line, its angle varies with the reinforcement ratio, and the two effective depths are a calibration constant. Codes differ: some draw at half a depth and compare against a larger stress. Both arrive at similar answers for slabs inside the test range and diverge outside it.

The size-effect factor is empirical and capped. k=1+200/dk = 1 + \sqrt{200/d} reaches its cap of 2 at d=200d = 200 mm and keeps falling above it, and there is no mechanism in the expression — it is a curve fitted to tests, and the tests on very thick slabs are few. A transfer slab two metres deep is outside the evidence.

The reinforcement ratio is a flexural quantity being used as a shear one. It enters through a cube root because the top steel controls the crack width across the failure surface, but the value used is an average over a width, and the steel is not uniformly distributed — it is concentrated over the column, which is where the surface is.

And nothing here is about what happens after. A punching failure at one column drops that column’s load onto its neighbours instantly, and those neighbours are already at 92% of their own resistance. It is the most direct route to progressive collapse that a building frame contains, which is why continuous bottom reinforcement through the column — steel that does nothing in the elastic design and everything afterwards — is the standard defence.

What the pictures cannot show

The plan view draws the perimeter as a line on a flat sheet, which is exactly what it is not. The surface it stands for goes down through the slab at an angle nobody has measured, and the figure has no way to show a solid failing on a cone.

Nor can the sweeps show that the resistance they plot is a fitted expression, not a derivation. Every curve here is smooth and confident and the scatter behind it is considerable — the tests these constants come from have a coefficient of variation of well over ten per cent, and the design expression sits low in that scatter deliberately.

And the utilisation of 92% quoted throughout is a number about one bay of one slab, drawn to be near enough to the limit that the sweeps are legible. Reading it as a comment on flat slabs generally would be reading a chosen example as a population.

The ladder from here

Later rungs on this anchor: the edge and corner columns, where the perimeter loses a third or two-thirds of its length and the moment transfer is unavoidable. Shear reinforcement in slabs — links, shear ladders, studs, and why the outermost perimeter has to be checked as well as the innermost. Punching in foundations, where the load is applied upward over a large area and the shear inside the perimeter is subtracted rather than ignored. The flat-slab-with-drops and the waffle slab, which are the same argument answered with geometry. Openings near columns, which cut the perimeter and are the commonest way a punching check is quietly invalidated on site. And the historical case: the flat slab was patented in 1906 with no punching check at all, and the first ones were load-tested to destruction because nothing else would have settled the question.

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. 11.522.530.40.50.60.70.80.91long span ÷ short spanshare taken by the short strips6 × 8 m: 76.0%by 2 : 1 it is a one-way slab The slab that spans both ways A panel supported on four sides sends its load in two directions at once, and the share is decided by a fourth power — so a panel a third longer than it is wide has already stopped being a two-way slab in any useful sense. What it does at collapse is a different calculation with a different answer. 15.0 m²9.38 kN/m15.0 m²9.38 kN/m9.0 m²7.50 kN/m9.0 m²7.50 kN/m8 m6 m48.0 m² divided, 48.0 m² of panel — the division closes The load a beam is given is a decision Every beam calculation on this site so far has started with a load per metre, handed over as though it were a property of the beam. It is not. It is the answer to a prior question nobody draws, and two defensible answers to it differ by sixty per cent on the same floor. 1200 kNtie 750 kNstrut 960 kN38.7°z = 1600strut 3.0 N/mm² over 812 mm · limit 15.8bursting across each strut 240 kN · tie steel 1724 mm² 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.

What links here

Every essay whose body links to this one.

The objects this essay names

Each one links to every other essay that touches it.

Aggregate interlockClamping forceColumn headControl perimeterEffective depthFailure surfaceFlat slabProgressive collapsePunching shearShear frictionShear reinforcementShear stressSize effectStrut and tieTributary area