The slip that peaks where the moment turns

The same studs, moved to where the slip is took a simply supported composite beam with half the studs full interaction needs and crowded them toward its supports. Nothing about the beam’s strength changed, and the slip its end studs had to give at 95 per cent of the plastic moment halved, because the slab’s force built up sooner and the sections near the supports acted more nearly as one. The studs were moved to where the slip was, and in a simply supported beam the slip is at the supports.

That essay ended by asking about the beam that most composite floors actually are: continuous over its supports. A continuous span has its slab in compression over the sagging part of the span and cracked in tension over the support, with a point of contraflexure between them. Whether studs graded toward the points of contraflexure rather than the supports would do what the graded studs did there — and whether the hogging region, where the slab is only its reinforcement and the studs seem to carry little, can give studs up to the span — is the question.

A continuous span, and its two kinds of section #

The beam is the earlier essays’ section: a 457 × 191 × 67 universal beam under a 130 mm slab, now continuous over two equal 12 m spans and carrying 30 kN/m. Where the moment sags, the slab works whole, 2.5 m wide; where it hogs, the slab is cracked and only its reinforcement acts — 1 per cent of a 2 m width of slab, 2,600 mm² of bars at its mid-depth. Each span carries 40 studs of 19 mm, each taken as a spring of 100 kN per millimetre of slip.

The count is Newmark’s: the slab and the steel each bend about their own centroids, the studs carry the difference in their strains as slip, and the slab’s force $N$ obeys

$$N'' - \frac{k}{EA_p}\,N = -\frac{k\,r}{EI_0}\,M(x),$$

where $k$ is the studs’ stiffness per unit length, $r$ the lever arm between the slab’s centroid and the steel’s, $EI_0$ the two components’ own bending stiffness and $EA_p$ a combined axial stiffness. It is the same equation the connection is busiest where the beam is not used, solved here with the slab’s properties switching where the moment changes sign, and with the support moment — the beam’s one redundant — found from the condition that the beam is level over its middle support. It is a linear count: the studs are springs and the steel elastic, which is a beam in service rather than at its plastic moment.

With the slab cracked over the support the beam is softer there than in the span, and the support takes 448 kN·m of hogging moment against the 540 that a uniform beam would give it; the span sags 339 kN·m at most. Contraflexure is 9.5 m from the end support, four fifths of the way along.

How composite the beam is shows in its deflection. With the studs rigid, the span deflects 22.2 mm at its largest; with no studs at all, the steel and slab bending separately, 50.5 mm; with the forty studs, 26.8. The connection gives the continuous beam most of what two beams or one found full interaction worth, and the slip that costs the rest is what this essay follows. Rigid studs would also have put 549 kN into the bars over the support instead of 344, and 734 kN into the slab under the sagging peak instead of 632: the studs’ flexibility costs the hogging section a larger share than the sagging one, a first sign of where it matters.

Two peaks, and neither is over the middle support #

The slip starts at its largest at the end support, 0.68 mm, as in a simply supported beam. It falls through nothing under the largest sagging moment, 4.8 m along, where the slab’s force stops growing. Then it grows again the other way, and peaks at 0.62 mm almost exactly at the point of contraflexure. Past that point it falls, and over the middle support it is nothing.

The slip of a continuous composite beam has two peaks of nearly equal size, at the end support and at contraflexure. The span between them is a sagging parabola, zero at both ends, and the slab’s force has to build up from nothing at one end and fall back to nothing at the other. Studs at the two ends of that parabola are doing the same job in opposite directions.

The second peak is at contraflexure, not over the middle support, and that needs explaining, because over the middle support the shear is largest of all: the support takes 217 kN from each span against the end support’s 143. The earlier essay drew the shear flow as following the shear, which in a beam whose studs are rigid it does: the slab’s force follows the moment, and its rate of change follows the shear. But the moment has a kink over the middle support, where the reaction reverses the shear, and a slab force connected through studs that slip cannot follow a kink. By symmetry the slab’s force over the support is at its largest and not changing, so the slip there is nothing, and it takes about a metre of beam for the slip to rise from nothing to what the shear asks. Over that metre the studs carry less than the full-interaction picture says, and the hogging side’s peak is left at contraflexure.

Graded to the supports, the studs miss the second peak #

A simply supported beam’s studs are graded toward its ends. Do the same to a continuous span — more studs per metre near both of its supports, fewer in the middle — and the grading meets the first peak at the end support, but the other end of the grading is over the middle support, where the slip is nothing, and the second peak at contraflexure gets no more studs than before.

Graded to the supports, the end peak falls but the contraflexure peak hardly changes, so the largest slip falls from 0.68 to 0.62 mm and stops there: it is the second peak now. Graded to the slip — studs in proportion to how much the evenly spaced studs slipped, which puts them at the end support and around contraflexure and thins them under the sagging peak and over the middle support — both peaks fall together, and the largest slip is 0.49 mm.

The difference grows with the grading. Grading toward the supports is exhausted at 0.3: by then the end peak has fallen below the contraflexure peak, and no further crowding at the supports touches the latter. Grading toward the slip keeps paying, down to 0.43 mm at 0.9 — a cut of more than a third, from the same forty studs. In practice the grading toward the slip is easy to describe: studs in pairs over the outer quarter of the span from the end support, in pairs again for a couple of metres either side of the point of contraflexure, and singly everywhere else, including over the middle support.

The hogging region cannot give its studs away #

The hogging region seems to be where studs could be saved. The slab there is cracked, so it is only its bars; the composite action it supplies is a fraction of the span’s; and the design rules for continuous composite beams count the hogging region’s studs only for what the bars need to develop. The obvious economy is to thin them there and put the saving into the sagging span, where the slip seems to be.

The figure says the opposite. Halve the stud density beyond contraflexure, moving the studs into the rest of the span, and the largest slip rises from 0.68 to 1.04 mm; thin it to a quarter and it rises to 1.69, two and a half times as much. The reason is the first finding: the hogging region begins at the point of contraflexure, and that is where the second slip peak is. The studs taken from the hogging region are taken first from the place where the slip is largest on that side of the span.

The bars over the support lose their force too, from 344 kN to 251 at half the density and 118 at a quarter. The bars act with the steel only through the studs between them and contraflexure, and with fewer studs they are carried by the slab less and the steel more. What the hogging region’s studs buy is not their own strength but the reinforcement’s, and without them the composite section over the support is nearly the steel beam alone.

There is a small saving, at the very start of the curve: moving a tenth of the hogging studs into the span lowers the largest slip slightly, from 0.68 to 0.67 mm. That is the grading toward the slip again, seen from the other side — the studs nearest the middle support, where the slip is nothing, can go. The ones nearest contraflexure cannot.

What the rule’s count of studs agrees with #

The design rules for continuous composite beams count studs between critical sections: between the end support and the point of largest sagging moment the studs must carry the slab’s compression there, and between that point and the middle support they must carry that compression away and the bars’ tension in. In this beam’s elastic state the slab carries 632 kN under the sagging peak, 4.8 m from the end, and the bars 344 kN over the support, 7.2 m further on. The first length must transfer 632 kN, 132 kN a metre; the second, $632 + 344 = 976$ kN, 136 kN a metre.

So the rule’s averages say the studs should be spaced almost evenly, and evenly spaced studs are what the beam was given. The rule counts the right total on each side of the sagging peak and says nothing about where on that side the studs go. The two slip peaks are inside those lengths — one at the end of the first, one two-thirds of the way along the second — and the grading that halves them moves studs within each length without changing either total. That is also why the rule could not have found the second peak: a count between critical sections is a count of forces, and a slip peak is a statement about how fast the force changes, which the count averages away. The slip the plastic moment asks of the studs found the same blindness in a simply supported beam’s degree of connection, which counts studs and not slip.

Fewer studs over the support move moment into the span #

The bars that stop working take stiffness from the section over the support, and a continuous beam sends its moment to where it is stiff. Thinning the hogging studs to a quarter takes 50 kN·m of hogging moment off the support and puts 20 kN·m more sagging moment into the span: the moment that was shed has to land, and it lands on the sagging section the studs were moved to relieve. The span’s slab force rises with its moment, and its studs carry more.

So the economy undoes itself twice over. The studs moved out of the hogging region leave the contraflexure peak unattended, and the stiffness they took with them moves moment into the span they were moved to.

It undoes itself a third time in service, where the beam spends its life. The bars over the support are there to control the slab’s cracks as much as to carry moment, and a bar controls a crack only in proportion to the force it can pick up from the concrete either side. Bars that the studs have stopped connecting to the steel carry less of the hogging, so the slab’s own strain over the support is set more by the steel beam’s curvature beneath it and less by the bars within it. A floor finish laid over a continuous composite beam sees that as wider cracks over every support, which is the usual complaint about such floors, and the stud layout is one of the things that decides it.

At every span, two peaks #

The two peaks keep their proportions at every span. The contraflexure peak is 0.87 to 0.91 of the end peak from 6 m to 20 m, and contraflexure stays at about four fifths of the span. Both grow faster than the span, because the shear at their ends grows with the span and the slab force the studs must deliver grows with its square.

The two peaks by hand #

The span between the end support and contraflexure is a sagging parabola. With the support’s 448 kN·m of hogging, the end support takes $wL/2 - M_B/L = 180 - 37.3 = 142.7$ kN, and the moment $142.7x - 15x^2$ is zero again at $x = 9.51$ m, where the shear is $142.7 - 30 \times 9.51 = -142.7$ kN. The two ends of a parabola have equal and opposite shears, and the slip, which follows the shear, peaks equally and oppositely at both. That is a statement about statics, not about studs: it holds whatever the support moment, which only moves contraflexure. One support too many found that a redundant support’s reaction is decided by stiffness; where contraflexure falls is decided by the same stiffness, and the second slip peak goes with it.

How large it is follows from Newmark’s equation. Where the studs are stiff compared with the span — here $\alpha L = 9.7$, $\alpha = \sqrt{k/EA_p}$ — the slab’s force follows the moment and its rate of change follows the shear: the slip is $s \approx (r\,EA_p/EI_0\,k)\,V$. At the free end the slab’s force must also start from nothing, which costs a correction of $w/\alpha$ in the shear, 37 kN here:

$$s(0) \approx \frac{r\,EA_p}{EI_0\,k}\left(V_A - \frac{w}{\alpha}\right) = \frac{291.7 \times 5.07 \times 10^8}{6.93 \times 10^{13} \times 333}\,(142.7 - 37)\times 10^3 = 0.68\ \text{mm}.$$

At contraflexure the slab is still whole on the sagging side and only its bars on the other, and the slip’s peak sits between the two values the two sections give, a little short of the end’s.

What the linear count leaves out #

The studs are springs and the steel elastic. The earlier essays’ fibre sections, built as in the section that yields from the outside in, and Ollgaard’s stud curve carried the beam to its plastic moment; this count stops at service. At the plastic moment the studs at both peaks are on the flat part of their curves and the slip grows faster than the load, but the two peaks are set by the two ends of the sagging parabola at any load, so the finding that there are two survives the nonlinearity. How large each is near collapse does not.

The cracked zone is where the moment hogs. A real slab cracks from its restraint as well as its load, and its tension stiffening keeps some of the concrete working between the cracks, so the hogging section is stiffer than its bars alone and the support takes more of the moment. Both move the second peak’s size and not its place.

The load is uniform and on both spans. A load on one span only moves contraflexure in the loaded span toward the middle support and in the other span puts the whole span in hogging, and a floor has to be designed for both. The second peak then wanders with the load, and a grading that follows it has to cover the band it wanders over.

The studs carry only slip. A stud near contraflexure under a cracked slab sits in concrete that is cracked at the stud’s root, and its stiffness and strength there are less than in solid concrete — exactly where this essay puts the most demand.

Still open: the slab that is only partly cracked over the support #

The hogging section here is the steel beam and its bars, with nothing between the cracks. A real slab over a support is cracked at intervals, and between the cracks the concrete still carries tension and hands it to the bars through bond — the tension stiffening that the stiffness a cracked section keeps measured in a slab. Bond between the cracks and studs between the slab and the steel are two connections in series, and a stud near contraflexure is transferring force into a slab that is itself only partly transferring it into its bars. Whether the studs nearest contraflexure, given a slab that is stiffer than its bars, see a larger peak or a smaller one — and whether the bond’s own slip at each crack is where the hogging region’s ductility actually comes from — is the question the bars’ partner asks of the studs.