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.

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. To get the beam down to the halfway point between two beams and one, the connection has to be cut by a factor of about fifteen, which is a stud every two metres.

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.

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.

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.

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.

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.