Two beams, or one beam four times as stiff

Take two identical planks, lay one on the other, and bend the pair. Each plank bends about its own centreline, each carries half the load, and the two faces at the interface slide past one another — visibly, at the ends, by a few millimetres. Now glue them. The pair has one neutral axis, four times the second moment of area, a quarter of the deflection and half the stress. Nothing was added except a restraint on slip, which is not a force, not a material and not a section property, and which is worth a factor of four.

The factor is exactly four and it is worth seeing where it comes from before anything else.

Why four, exactly #

Each plank alone: $I_1 = bh^3/12$. Two of them, each about its own axis: $I_0 = 2bh^3/12$.

The pair bonded, of overall depth $2h$: $I_\infty = b(2h)^3/12 = 8bh^3/12$.

$$\frac{I_\infty}{I_0} = \frac{8}{2} = 4$$

with no dimensions surviving. The stiffness ratio for two equal layers is always four, for any material, any breadth, any depth. Three equal layers give nine; $n$ equal layers give $n^2$, which is the same $d^3$ argument that decides everything about a section’s shape counted a different way round.

The stress ratio is two rather than four, because the section modulus is $I/c$ and bonding doubles $c$ as well: $8/2$ over $2/1$ is exactly 2. So bonding halves the stress and quarters the deflection, which is the arithmetic behind every composite floor, every glued laminated beam, every plywood web and every sandwich panel there has ever been.

The claim that no dimension survives is the kind that is worth watching the solver make rather than taking on the algebra. Halve the planks and run the same experiment.

Every absolute number in that figure differs from the one before it, by a factor of eight in second moment and of eight in deflection, and the ratio is four in both. That is what a dimensionless result looks like when it is computed rather than asserted: the solver was given a different beam and returned the same four. What did change is how far the connectors get: 94% of the way between the two bounds here against 89% for the deeper pair, because the connection stiffness was held at 200 while the beam it serves was made shallower.

Which free body produced the number #

Cut the bonded pair on a horizontal plane at the interface, over a length $dx$, and take the piece above the cut.

Two things act on it: the direct stresses on its two vertical faces, which differ because the moment differs between $x$ and $x+dx$; and whatever the interface supplies. The difference in direct force across the length is

$$dF = \frac{dM}{I}\int y\,dA = \frac{V\,Q}{I}\,dx$$

so the interface must carry $q = VQ/I$ per unit length, with $Q$ the first moment of the area above the cut about the neutral axis of the whole. That is the shear flow this collection has already drawn, read for a purpose it is not usually read for.

For the planks above: $Q = b h \cdot h/2 = 200 \times 150 \times 75$, $I = 450\times10^6$, and at the support where $V = 12$ kN the flow is 60 N per millimetre of length. That is the force the glue line, the nails or the studs have to deliver, and it is a force per unit length rather than a force.

The interface plane is not a plane where anything is happening to the material. It is a plane where a difference is happening — the plank above wants to be shorter than the plank below, and the joint’s whole job is to stop it.

Where the connection actually works #

This is the finding most often got wrong, and the reason is that everything else about a beam points the other way. The moment is largest at mid-span; the stress is largest at mid-span; the deflection is largest at mid-span; a designer checking a beam is looking at mid-span. The connection carries nothing there.

It follows directly: $q = VQ/I$ and $V = dM/dx$, so the interface force is proportional to the rate of change of the moment. A uniformly loaded simply supported beam has $V = 0$ at the middle, and the studs there are along for the ride.

The practical consequences run both ways. Uniform stud spacing along a beam is a fabrication convenience rather than a structural requirement, and it puts a large surplus at mid-span. And a beam that is fine everywhere else can delaminate at its ends, which is why timber beams split along their length at the supports rather than in the middle, and why a hole cut through a web is worst where the shear is.

The total is the integral, and it is the number a designer really needs: 54.0 kN transferred across one shear span here, which is the sum of the force in all the studs on one half of the beam. Divide by a stud’s capacity and the answer is a count.

The middle ground, solved rather than tabulated #

Real connections are neither absent nor rigid. Studs deform, nails bear into wood, glue creeps; the two layers slip a little, the section is not quite composite, and the answer is somewhere between the two bounds.

Newmark’s equation puts the problem in one line. With $N$ the axial force the interface has transferred by station $x$, $k$ the connector stiffness per unit length and $d$ the distance between the two layers’ centroids:

$$N'' - \alpha^2 N = -\frac{k\,d}{EI_0}\,M(x), \qquad \alpha^2 = k\left(\frac{1}{EA^*} + \frac{d^2}{EI_0}\right)$$

where $1/A^* = 1/A_1 + 1/A_2$. It is a linear equation with a particular integral proportional to the moment and a hyperbolic complementary function, and the solver behind these figures writes the solution out rather than looking anything up. The check it makes is that $k \to \infty$ returns the curvature $M/EI_\infty$ exactly — which is not built in anywhere and is the reason the middle of the range can be believed.

The shape of that curve is the practical content of the subject. It says that a modest connection buys most of what a perfect one would, and that chasing the last tenth costs an order of magnitude. Which is why the concept of partial shear connection exists at all in design: a composite floor beam is routinely built with fewer studs than full interaction needs, on the explicit understanding that 80% of the connection buys 95% of the beam.

The connector stiffness $k$ is not the honest variable, though, because a given $k$ means different things on different beams — which is what the two plank pairs showed by reaching 89 and 94 per cent from the same connectors. Newmark’s equation says what the honest variable is: the single dimensionless group $\alpha L$, and drawing the interaction against that instead collapses every beam onto one line.

One curve, and every partly composite beam there has ever been is a point on it. A nailed joist, a glued laminate, a studded floor beam and a bolted flitch plate differ in material, in scale and in the connector’s whole nature, and each of them reduces to a place on that axis. The curve is steep where a real design sits, which is the practical form of the same warning as before — halving the number of studs does not halve anything, and doubling them does not double it either.

The other thing the interface transmits #

A real connector’s load-slip curve has three parts: an initial stiffness, a plateau where it yields, and a large slip capacity beyond that before anything fractures. The partial-interaction analysis above uses only the first of the three, which is why it is a serviceability calculation and not a strength one.

The stiffness $k$ used above is the initial slope. At ultimate load a stud is well past that, on the plateau, and the analysis that applies is a plastic one: all the studs in a shear span reach their capacity, and the compressive force delivered to the slab is however many studs there are times whatever each of them can carry.

Redistribution needs ductility to be real, and a stud has a great deal of it — which is what licenses the plastic treatment.

So a composite beam is analysed by two entirely different methods at its two limit states — an elastic partial-interaction calculation for deflection, and a plastic redistribution for strength — and the two disagree about which studs matter. The elastic one says the studs near the support carry most of the force. The plastic one says every stud carries the same, and that is what makes uniform spacing legitimate at the ultimate limit state even though the elastic distribution is triangular.

The gap between the two is worth measuring rather than describing, and on a real floor beam it is not small.

A factor of 1.81 between the worst stud and the average is exactly the size of discrepancy that a plastic argument has to be good for. It is also the reason a brittle connector — a bolt in a slotted hole, an adhesive, a shear key that splits — cannot be designed this way at all: it has no mechanism for shedding, so the end connector reaches its capacity at 55 per cent of the load the uniform assumption predicts, and the failure then unzips along the beam.

Where the section comes from in the first place #

The two-plank example is deliberately made of one material, so that the factor of four is uncontaminated. A steel-and-concrete beam is a section made of two materials, and the composite second moment has to be computed on the transformed section — after which every argument on this page applies unchanged, with $Q$ and $I$ taken from the transformed shape.

There is one asymmetry worth naming. In a composite floor beam the neutral axis of the composite section usually sits in or near the slab, which means the whole steel section is in tension and the concrete carries the entire compression. That is the arrangement both materials are best at, and it is the reason composite construction is not merely a stiffness trick. It also puts the whole slab in compression, which a material with no useful tensile strength is grateful for.

Stated in the vocabulary of sections rather than of interfaces, composite action is the operation that moves a beam from one shape to another without adding any material. Four shapes of identical area can span a factor of forty in second moment, and bonding two layers is a way of buying that move with a row of studs.

Four is the best case, and a real floor beam does not get it #

The factor of four requires the two layers to be identical, and almost nothing built is. The general result is worth having because it says how much of the four a given pair can hope for.

For two layers whose centroids are a distance $d$ apart,

$$I_\infty = I_1 + I_2 + A^*d^2, \qquad \frac{1}{A^*} = \frac{1}{A_1} + \frac{1}{A_2}$$

so the whole of the composite gain is the term $A^*d^2$, and $A^*$ — the harmonic mean of the two areas, halved — is maximised when the two areas are equal. Two identical planks give $A^* = A/2$ and $A^*d^2 = 3I_0$, hence the four. Make one layer much smaller than the other and $A^*$ collapses toward the smaller area, because a series pair is governed by its weaker member exactly as two springs in series are.

Put a real composite floor beam through it. A 457 × 191 × 74 steel section, $I = 333\times10^6$ mm⁴ and $A = 9{,}460$ mm², under a 130 mm slab of 2.5 m effective width transformed at a modular ratio of 7 — so 357 mm of equivalent steel, $A = 46{,}410$ mm² and $I = 65\times10^6$ mm⁴ — with the two centroids 293.5 mm apart:

$$A^* = \frac{9460 \times 46410}{55870} = 7{,}858\ \text{mm}^2, \qquad A^*d^2 = 677\times10^6\ \text{mm}^4$$

against $I_0 = 398\times10^6$. The gain over the two acting separately is 2.70, not 4. Against the bare steel beam — which is the comparison a designer actually makes, since the slab was not a beam before — it is $1075/333 = 3.23$.

That 3.23 is the number quoted in every composite design guide as “about three times stiffer”, and it is now traceable rather than remembered. It is short of four for one reason: the slab and the beam are not the same size, and the series term punishes the mismatch. Which also says where the remaining benefit is — thickening the slab helps very little once $A_2 \gg A_1$, and lifting the slab further from the steel, by a deeper deck or a haunch, helps a great deal, because $d$ is squared and $A^*$ is already stuck.

The connector also has to hold the layers together #

Everything above treats the interface as resisting slip. It has a second job that no shear-flow calculation reveals, and a connector that does only the first has a known failure mode.

The layers try to separate. A discrete connector transfers its force at a point, so between connectors the two layers bear on each other unevenly and pry apart; a slab on a beam with a load applied between studs lifts locally; and any curvature mismatch — a slab shrinking, a haunch of uneven depth, a beam cambered against a flat slab — opens the interface directly. None of that appears in $q = VQ/I$, which is a statement about forces in the plane of the interface and is silent about forces across it.

The design answer is in the connector’s shape rather than in its capacity. A headed stud has a head because the head is a tension anchor: the usual requirement is that a connector resist a tensile force of about a tenth of its shear resistance, with a head at least 1.5 diameters across and 0.4 diameters deep. A stud without its head is a shear connector in a calculation and a dowel in a slab.

The same requirement is what rules out the interfaces that rely on friction or adhesion alone. A slab cast on an unpainted steel flange bonds to it and carries a genuine shear flow until the first separation, after which it carries nothing anywhere — because bond has no uplift capacity and its failure is unzipping rather than local. That is why composite action is never claimed from bond, and why the two-plank experiment at the top of this essay does not work if the planks are merely clamped: friction resists slip in proportion to a normal force that separation is busy removing.

The same trick, four ways #

The factor is $n^2$ for $n$ equal layers, and the whole of engineering has found ways to collect it.

The cube is easiest to see with the section assembled strip by strip: each strip contributes its own area times the square of its distance from the axis, so material moved outward is rewarded quadratically. Bonding two layers is the operation that lets the far strips count at all — unbonded, each plank’s strips are measured from its own axis and the outermost of them is only half a plank from it.

Glued laminated timber takes the argument to its limit: dozens of thin laminations, glued over their whole area, giving a member whose second moment is that of the solid section and whose material can be graded lamination by lamination. The glue line is the connection and it is continuous, so $k$ is effectively infinite and the beam sits at the top of the interaction curve.

Plywood and cross-laminated timber do the same in two directions at once, and pay for it: the cross-plies carry almost nothing in bending and their thickness still counts toward the depth, so the composite gain is real and the efficiency is not what a solid section’s would be.

Nailed and screwed built-up joists sit low on the interaction curve, at $k$ of the order of tens, and are worth naming because the honest answer for them is often “not very composite at all” — two 47 mm joists nailed together are appreciably stiffer than two loose ones and a long way short of one 94 mm joist.

And a steel-concrete floor beam is the case the whole apparatus was built for, where the connection is a row of welded studs and the design question is how many.

One complication separates the last of those from the others. Composite action does not exist until the concrete has set, so the wet slab, the formwork and the steel’s own weight are carried by the bare steel beam and the composite section is only ever asked for what arrives afterwards. The section that carries the later loads is not the section that carried the earlier ones, superposition has to be applied stage by stage, and an unpropped beam therefore has a locked-in stress state that no analysis of the finished structure contains.

Where the model stops #

Everything above is elastic and uncracked. A concrete slab in the tension zone over a support cracks, its contribution disappears, and a continuous composite beam has a different section in its hogging regions from its sagging ones — which makes the whole member a stepped beam whose steps move as it loads.

Slip is assumed uniform through the depth of the connector. A real stud is a cantilever in a concrete slab with a haunch, a deck profile and a bearing zone at its base, and its stiffness is empirical rather than derived.

Shrinkage and creep are absent. A concrete slab shrinks against a steel beam that does not, and the restraint puts real force through the interface with no load applied — an imposed deformation rather than a load, and one that has to be added to everything here.

And the two bounds are bounds on stiffness, not on strength. A beam with no connection at all still carries load — as two beams — and a beam with full connection carries about twice as much. The strength ratio is two and the stiffness ratio is four, and confusing them is the most common error in this subject.

Nothing here is a construction sequence. A propped composite beam and an unpropped one have the same final section and different histories, and the histories decide the stresses.

What the pictures cannot show #

The slip in the hero figure is drawn as an offset of the two curves’ ends by a visible amount. The real slip at the end of those planks under that load is a fraction of a millimetre, and drawing it to scale would show two lines touching. Every millimetre in that figure is an exaggeration and the caption’s ratio is the honest reading.

Nor can the pictures show that the interface force is a flow rather than a set of forces. The drawing has to put something at the interface, and whatever it puts there looks like a connector at a place — whereas the quantity $q$ is continuous, and the connectors are a discretisation of it that the analysis then smears back out.

And the interaction curve is drawn against a connector stiffness with no units on the axis worth quoting, because $k$ has the awkward dimensions of force per length per length. What a reader wants is a stud spacing, and converting one to the other needs the connector’s own stiffness — a number that comes from a push-out test and from nowhere else.

The ladder from here #

Later rungs on this anchor: partial shear connection at the ultimate limit state, and the interaction curve between degree of connection and moment capacity. The stud itself — its load-slip behaviour, the effect of the deck profile it sits in, and why a stud in a rib is worth a fraction of one in a solid slab. Continuous composite beams, where the cracked hogging region turns the whole member into a variable-stiffness problem. Composite columns, where the interface carries load introduction rather than flexure. Timber-concrete composite floors, where the connection is notched rather than welded and the slip modulus is a design variable. Sandwich construction, where the two faces are joined by a core so weak that shear deflection dominates. And the historical case: composite action was present in every riveted plate girder and every filler-joist floor for fifty years before anybody counted it, which made a great many old structures considerably stronger than their own calculations.