Half the studs, and most of the beam

Stack two planks and bend them, and each one bends about its own middle: the pair has twice one plank’s stiffness. Glue them and the pair has one neutral axis in the middle of the whole depth, four times the second moment and half the stress. Nothing was added but a restraint on sliding, and that restraint is worth a factor of four.

Every composite floor in the world is built on that arithmetic and none of them is glued. What holds the concrete slab to the steel beam beneath it is a row of headed studs welded through the decking, and a stud is not a restraint on sliding. It is a spring. The slab slides on the steel by a fraction of a millimetre, and the beam sits somewhere between two beams and one.

The question this essay is about is where between, and the answer is not where the intuition of “half the studs, half the benefit” puts it.

Which free body produced the number #

Cut the composite beam at a station and draw the two layers separately.

Both are bent to the same curvature $\kappa$, because they are attached to each other and deflect together. Each carries a moment of its own — $EI_1\kappa$ and $EI_2\kappa$ — and the pair carries an axial force couple as well: compression $N$ in the slab, tension $N$ in the steel, separated by the distance $d$ between the two centroids. The total moment at the cut is

$$M = EI_0\,\kappa + N d, \qquad EI_0 = EI_1 + EI_2$$

That is one equation in two unknowns, and the second comes from the interface. The slip $s$ between the layers grows along the beam at a rate equal to the difference in strain at the interface,

$$\frac{ds}{dx} = \frac{N}{EA^*} + \kappa d, \qquad \frac{1}{EA^*} = \frac{1}{EA_1} + \frac{1}{EA_2}$$

and the connectors turn that slip into a shear flow, $dN/dx = Ks$. Differentiate, substitute, and the whole problem collapses to one linear equation:

$$N'' - \alpha^2 N = -\alpha^2 \bar{N}(x), \qquad \alpha^2 = \frac{K\,EI_\infty}{EA^*\,EI_0}$$

where $\bar{N}$ is the axial force full interaction would have produced and $EI_\infty = EI_0 + EA^*d^2$ is the fully composite stiffness. Newmark wrote it down in 1951 and it has not needed improving.

The boundary condition is the part worth pausing on: $N = 0$ at both ends of the beam, because nothing anchors the slab beyond the last stud. The slab arrives at the support carrying no axial force at all, which means the shear flow is largest exactly where the bending stress is smallest.

Solve that equation for a whole family of connections rather than one, and every answer it can give collapses onto a single curve. The only quantity that decides where a beam sits on it is $\alpha L$, which is why a count of studs is the wrong thing to reason about.

The curve is steep in the wrong place #

The shape of that curve is the finding, and it is almost the opposite of what “partial interaction” suggests.

An ordinary composite secondary beam — a 457 mm section under a 120 mm slab, studs at about 150 mm centres — has $\alpha L$ near sixteen and is 96 per cent composite in deflection. Halve the connection and it is 93 per cent. Halve it again and it is 87. Cutting it by a factor of fifteen, which is a stud every two metres, still leaves the beam 63 per cent composite, and reaching the halfway point between two beams and one takes a factor of about twenty-four.

The reason is in the definition of $\alpha$. The connection stiffness enters under a square root, and it is compared against a group with $EA^*d^2$ in it — the composite action’s own stiffness contribution, which is large because $d$ is the depth of the beam and it is squared. A connection has to be very soft indeed before it is soft relative to that.

So the practical statement is a comfortable one and a slightly deflating one: an ordinary floor is fully composite to within a few per cent, and no reasonable amount of extra connection will improve it. The studs in a real building are governed by strength, not by stiffness — there have to be enough of them to transfer the total axial force at the ultimate limit state — and once there are, the interaction question is settled.

The connectors are not asked for equal shares #

The deflection answer is comfortable. The connector answer is not, and it is where the elastic solution earns its place.

The shear flow at the interface is $dN/dx$, and $N$ follows the moment diagram: largest curvature at mid-span, largest rate of change at the ends. Under a uniform load the flow is very nearly linear, peaking at the supports and vanishing at mid-span. On the beam here it reaches 282 newtons per millimetre at the end and averages 156 over the half-span. The end connectors are being asked for 1.8 times the mean.

Studs are nevertheless installed at a uniform spacing, in every composite floor ever built.

The justification is not that the calculation is wrong. It is that a stud is ductile. A headed stud reaches its capacity at a slip of well under a millimetre and then goes on deforming for five or six millimetres more without losing much load, so a stud that has reached its limit sheds the excess to its neighbours and the group ends up sharing equally. Uniform spacing is a lower-bound design that depends on ductility to become true.

That is exactly the argument a cracked web makes about its stirrups — where the assumption that every stirrup crossed by the crack is at yield is also a plastic redistribution, also depends on ductility, and is also the reason the flattest permitted crack angle is capped. The same sentence appears in three places on this site with three different pieces of steel in it.

The slip itself #

The slip on this beam is 0.47 mm at the support and zero at mid-span. It is small enough to be invisible and large enough to matter twice.

It matters at the interface, because that half-millimetre is what the studs are deforming through and what their own capacity is quoted against. And it matters for the strains: a beam with slip has two neutral axes rather than one, and the strain diagram has a step in it at the interface. Plane sections stay plane within each layer and not across the pair, which is the assumption composite design quietly replaces and rarely states.

That is the assumption underneath every bending calculation in this collection, and it is the one partial interaction breaks. A fully composite section has one linear strain diagram across the whole depth; with slip there are two, one per layer, and the discontinuity between them at the interface is the slip strain — the derivative of the half-millimetre the layers have moved past each other.

Two things called the same word #

There are two quantities in composite design that both get called “the degree of shear connection”, and confusing them is the commonest way to get this subject wrong.

The first is the one this essay has been computing: the degree of interaction, a stiffness question, defined by where the beam’s deflection sits between the two-beam and one-beam limits. It is governed by $\alpha L$, it is in the high nineties for anything ordinary, and it is a serviceability quantity.

The second is the degree of shear connection, a strength question: the total capacity of the studs provided divided by the axial force full composite action would need at the ultimate limit state. A beam with half that force provided has a shear connection of 0.5, and its moment capacity falls along a line between the bare steel section’s and the fully composite one’s — not because anything slipped elastically, but because the studs cannot deliver more compression into the slab than they can carry.

The two answer different questions and move differently. A beam at 0.5 shear connection is typically still above 0.9 interaction, so it is nearly as stiff as a fully composite beam and appreciably weaker. A designer who reads the stiffness number and concludes the strength is fine has made the specific mistake this distinction exists to prevent.

How much of the slab is in the beam #

Every term with a subscript 2 in it needs an area for the slab, and a slab has no natural edge. What is used instead is an effective width, and it is a shear-lag quantity rather than a geometric one.

The compression is delivered into the slab through a line of studs along the beam’s axis, and it spreads sideways from there at a finite rate. A metre out from the line the slab is carrying less than it does directly above it, and far enough out it is carrying nothing at all. The conventional allowance is $L/8$ each side of the beam, capped by half the distance to the next beam — so a 12 m span gets 3.0 m of slab whatever the beams’ spacing, and beams at 4 m centres have a quarter of their slab discounted.

The interesting part is how little that costs the stiffness. The composite term is $EA^*d^2$, and $A^*$ is a series combination, so it is governed by the smaller of the two areas — which is always the steel. Cutting the transformed slab area from 68,600 mm² to 51,400 changes $A^*$ from 8,313 to 7,990: 3.9 per cent, on a 25 per cent change in the slab.

So the effective width, which looks like the crudest assumption in the whole calculation, is one of the least consequential for deflection. Where it is consequential is everywhere else. The same compression is being carried by less concrete, so the slab’s stress rises in proportion. And at the ultimate limit state, where the moment capacity depends on the depth of the compression block, a narrower slab needs a deeper one — which lowers the lever arm and reduces the capacity directly, with none of the series-combination cushioning that protected the stiffness.

It also varies along the span, and in the direction the shear flow does not want. Near a support the axial force has only just begun to be introduced, so it has spread over very little width; at mid-span it has had half the beam to spread across. The single number used in design is a mid-span value applied everywhere, which is generous exactly where the connectors are working hardest.

The failure that has no stud in it #

The studs deliver the compression into the slab at one line, and the slab has to get it out to its effective width. That transfer crosses a vertical plane through the slab running parallel to the beam, on each side of the stud row — and that plane is a shear plane like any other.

It has no stud on it. Nothing crosses it but the slab’s own transverse reinforcement: the mesh, plus whatever bottom bars run across the beam. If that steel is insufficient, the slab splits longitudinally along the line of studs, the compression cannot spread, and the composite action is lost — with every stud still welded, intact, and nowhere near its capacity.

The check is shear across a plane that will crack: the longitudinal force per unit length that has to cross the plane is the same shear flow the studs are carrying, minus whatever the plane on the other side takes, and the resistance is the transverse steel crossing it times its yield stress, with the concrete’s own contribution. On a beam carrying 282 N/mm at its end, a plane taking half of that needs its own reinforcement whether or not the slab was going to have any.

Two things make it easy to miss. The reinforcement in question is in the slab, drawn on the slab’s drawing, sized by the slab designer for the slab’s own bending — so nobody looking at the beam sees it. And the failure it prevents is one the beam calculation cannot represent, because the beam calculation has already assumed the slab acts over its effective width and has no way to ask how the force got there.

It is the same shape of omission as a shear connector that carries no uplift: a mechanism perpendicular to the one being computed, invisible to the computation, and capable of removing the whole benefit at once.

What breaks the comfortable answer #

Three things move a real beam off the flat part of the curve, and all three are common.

A propped construction that is unpropped. If the steel beam carries the wet concrete on its own and the composite section only exists afterwards, the deflection has two parts computed on two different sections, and the first is a bare steel beam at $EI_1$ alone. That is a much larger effect than any amount of partial interaction, and it is a sequence problem rather than a stiffness one.

Deliberate partial shear connection. Where the studs are limited by the decking’s ribs rather than chosen, a designer may accept fewer than full interaction requires and take a reduced moment capacity. That is a strength decision, and its stiffness consequence is the curve above — usually still in the nineties.

A long-term slab. The concrete’s modulus falls by a factor of two or three under sustained load, which changes $EA_2$, $EI_2$ and therefore $\alpha$. The interaction degree barely moves; the composite stiffness itself falls a great deal. The beam gets softer for a reason that has nothing to do with its connection.

Everything above treats the slab’s stiffness as a number, and it is a modulus that is not a constant: under sustained load it is a function of time, and a composite beam’s long-term deflection is dominated by that rather than by anything happening at its interface.

Why it was ever in doubt #

Composite action between steel and concrete was being used before there was any way to calculate what it was worth. The early floors relied on the bond between the concrete and the top flange, plus whatever the encasement contributed, and the bond is not a design quantity: it is present until it is not, and there is no warning.

Two things settled it. The shear connector — a bar bent into a spiral, later a hooked channel, finally the headed stud fired through the decking — turned an unquantifiable bond into a countable component with a load and a slip that could be tested. And Newmark’s differential equation turned a row of springs into a closed-form answer, which is what made it possible to say how much connection is enough rather than more.

The interesting part of the history is that the equation arrived and then largely stopped being needed. It answered the question so decisively — the flat top of that curve — that the design of a composite floor moved to the strength question, where the interesting variation is. The elastic partial-interaction solution is now most useful for the two cases that are not floors: timber-concrete composites, where the connection really is soft, and any assessment of an old structure whose connectors are not what anyone would provide today.

Where the model stops #

The connection is linear. $q = Ks$ with one $K$ is a description of a stud below about half its capacity. At service load that is roughly true and at ultimate it is not true at all — the studs near the ends are past their elastic range, the stiffness there is lower, and the real slip distribution is flatter than the exponential-and-parabola drawn here.

The two layers stay in contact. Newmark’s equations assume no separation, which studs with heads are there to provide. Uplift at the interface is real near the ends of a beam with a heavy point load, and nothing here can see it.

And it is one span, simply supported. In a continuous composite beam the slab is in tension over the support, cracked, and contributing almost nothing to $EA_2$ — so $\alpha$ collapses in exactly the region where the shear flow is largest. The hogging region of a continuous composite beam is a different problem wearing the same name.

What the pictures cannot show #

The slip is drawn as a curve on an axis. In the beam it is half a millimetre distributed along twelve metres, entirely invisible, and detectable only by the fact that the beam has deflected 36.7 mm instead of 33.9.

Nor can any of these figures show what actually decides how many studs a floor gets, which is neither stiffness nor strength but the pitch of the decking’s ribs. A stud goes where a rib lets it go, and the elegant continuous $K$ of the equation is a row of discrete lumps at 150 or 300 mm centres, chosen by a rolled profile.

The assumption the figure rests on #

The connection stiffness $K$ is taken as 600 N/mm of slip per millimetre of beam — a stud of perhaps 90 kN/mm at 150 mm centres, smeared into a continuum. Both halves of that are assumptions. The stud stiffness is a fitted number with a scatter of a factor of two across tests, and the smearing replaces a row of springs with a continuous one, which is right when the spacing is small compared with the length over which the shear flow changes and is exactly wrong near a support, where the flow changes fastest and there may be four studs in the whole region.

None of that changes the total the connection has to move. The shear flow the studs are carrying is $VQ/I$ on the fully composite section, and the truss and the stress field agree about it: the connectors transfer exactly the change in the slab’s axial force, and $VQ/I$ is that change written as a rate. Partial interaction reduces the flow slightly and moves none of it.

Between two beams and one there is a factor of 3.24 in deflection on the section drawn, and everything on this page has been about where in that interval a real floor lands. For anything built normally it lands within a few per cent of the stiff end, and the studs that put it there were chosen for a strength calculation that never asked the question.

The ladder from here #

Later rungs on this anchor: the hogging region of a continuous composite beam, where the slab cracks and the interaction collapses where the shear flow is greatest. Partial shear connection at the ultimate limit state, where the question is a moment capacity rather than a deflection and the answer is a straight line between two points. The stud group near a large point load, where smearing fails and the discrete spacing has to be respected. Composite columns, where the same slip argument runs vertically and there is no bending to drive it. And timber-concrete floors, where the connection is a screw or a notch, $\alpha L$ is genuinely small, and everything comfortable about this essay stops being true.