The same studs, moved to where the slip is

The slip the plastic moment asks of the studs followed a composite beam’s slip along its length and found the reason EN 1994-1-1 has a minimum degree of shear connection: below it, the stress blocks promise a moment that the end studs can deliver only by slipping further than a stud is trusted to. That calculation spaced every beam’s studs evenly, as most beams’ studs are, and it ended on the obvious question that spacing raises. The studs near mid-span slip little and the studs near the supports slip a great deal. A beam that put its studs where the slip is might spend its connection better.

So the experiment is simple to state. Keep the beam, the slab, the stud and the number of studs. Move the studs toward the supports. Ask what happens to the slip the end studs must give, and to the least degree of connection that keeps it inside 6 mm.

The same beam, the same nineteen studs #

The beam is the earlier essay’s: a 457 × 191 × 67 UB in S355 under a 130 mm slab of C30/37, 2.5 m effective width, over 12 m, carrying a uniform load. Its studs are 19 mm headed studs with a resistance of 80 kN each, following Ollgaard’s load–slip curve, and its sections are modelled fibre by fibre, as a section that yields from the outside in is. Full interaction needs 38 studs each side of mid-span; this beam has 19, half connection, and its plastic moment by stress blocks is 850 kN·m, 0.85 of the 1,001 that full connection would give.

The grading is the simplest that has a free parameter. The studs’ density along the half-span falls linearly from the support to mid-span, $\rho(x) \propto 1 + g\,(1 - 2x/(L/2))$, so that $g = 0$ is even spacing and $g = 0.9$ puts nineteen times as many studs per metre at the support as at mid-span. The number of studs is the same for every $g$. A practical grading — studs in pairs over the outer part of each half-span and singly over the inner — is about $g = 0.3$.

Because every stud reaches its capacity at the plastic moment, and the slab’s force at mid-span is their sum, grading changes nothing about the beam’s strength. It changes only how the slab’s force builds up along the half-span, and so how far the studs must slip to deliver it.

Half the slip #

The slip curves keep the shape the earlier essay found — gentle while the steel is elastic, steep once its bottom flange yields — but each grading lowers the whole curve. At 95 per cent of the plastic moment the evenly spaced end studs must slip 8.8 mm. Graded by 0.3, the practical pairs-and-singles layout, they slip 7.1; by 0.6, 5.7; by 0.9, 4.4. The same studs, moved, halve the slip.

Read the other way, the curves say how far each layout gets before its end studs reach 6 mm. Evenly spaced, the beam reaches 88 per cent of its plastic moment; graded by 0.9, 98 per cent. The stress blocks’ promise is the same for both, and the graded beam can keep it.

Why moving the studs moves the slip #

The slip at a beam’s end is an integral: the sum, from mid-span to the support, of the jump in strain across the interface between slab and steel. That jump is what the earlier essay’s two neutral axes are separated by, and it depends on how much of the section’s moment the slab’s force carries — the question two beams, or one beam four times as stiff first asked of an elastic composite beam. A section whose slab is already carrying a large force is nearly one composite section, its slab and steel strain nearly alike, and its slip strain is small. A section whose slab carries little is two members bending side by side, and its slip strain is large.

Evenly spaced studs build the slab’s force up in proportion to the distance from the support: a quarter of the way along, a quarter of the force. The elastic shear flow asks for something quite different — the connection is busiest where the beam is not, at the supports — and a graded layout is a step back toward that diagram. So every section near the support is two members, nearly, and those are the sections that contribute most to the slip at the support, because they are where the integral ends. Graded studs give the slab 42 per cent of its force a quarter of the way along and 72 per cent halfway. The sections near the support become composite sooner, their slip strain falls, and the integral that is the end studs’ slip falls with it.

That is the reason the earlier essay’s suggestion works, and it is a different reason from the one it gave. It is not that the studs near mid-span are idle. On Ollgaard’s curve a stud reaches three quarters of its capacity at a millimetre of slip and nine tenths at two, so evenly spaced studs three quarters of the way to mid-span are already at 95 per cent of their capacity. It is that the studs near the support, by arriving sooner, change the sections the slip is summed over.

The least connection the slip allows #

The earlier essay derived the Eurocode’s minimum degree of connection by asking, for each span, the least connection whose evenly spaced end studs reach 95 per cent of the plastic moment within 6 mm. Ask the same question of graded studs and the answer falls steadily with the grading: 0.69 evenly spaced at 12 m, 0.56 graded by 0.3, 0.48 by 0.6, 0.40 by 0.9. In studs, from 27 a half-span to 22, 19 and 16.

The minimum degree of connection is a property of the spacing as much as of the span. The Eurocode’s line at 12 m, 0.61, sits a little below the even spacing’s 0.69 and well above every graded layout’s answer. A beam whose strength needs only half connection — 19 studs a half-span — is pushed up to 27 by the slip if its studs are spaced evenly, and stays at 19 if they are graded by 0.6. The eight studs the slip demands are the eight that grading saves.

EN 1994-1-1 allows ductile studs to be spaced uniformly between critical sections, on the reasoning that a ductile stud will redistribute its share however it is placed. That is true of the strength and it is why the stress blocks need no grading. The figure says it is not true of the slip, and the slip is what the minimum degree of connection was for.

At every span #

Across spans the two layouts part steadily. At 6 m both are near the bottom of the search, 0.24 and 0.20, because a short beam’s end studs never need much slip. From 8 m up the graded line runs about 0.3 below the even one, and at 20 m it is 0.37 below: 0.51 against 0.88. The even line follows the Eurocode’s straight line within 0.08 from 8 m up, as the earlier essay found; the graded line runs nearly parallel to it, at about two thirds of its height.

So the benefit grows with the span, which is what the slip’s nature as an integral predicts. A longer beam has more half-span for the slip to accumulate over, and more of it near the supports, where grading makes the sections composite sooner.

Under a load the studs were not graded for #

A layout graded for one load is a bet on that load. The question the earlier essay put was whether the gain survives a load that is not uniform, so that the slip is not where the design put the studs.

For the two other symmetric loads a beam commonly carries, it does. A central point load asks little of any stud: the shear is the same all along each half-span, the slab’s force builds up evenly whatever the studs do, and both layouts reach 99 per cent of the plastic moment within 6 mm. Two loads at the third points ask the outer thirds for all the shear and the middle third for none, which is a sharper version of what a uniform load asks, and the graded layout’s advantage is larger than for the uniform load it was graded for: 0.95 of the plastic moment against 0.87.

The loads that would defeat the grading are the ones that put shear where the graded layout has thinned its studs: a heavy load near mid-span on one side only, a moving load whose shear passes through the middle of the span, a beam continuous over its supports whose shear is largest at a point of contraflexure inside the span. None is drawn here, and each is a reason to keep a minimum density of studs at mid-span — which the grading by 0.9 already does, with a tenth of the average density there, and which a practical layout of pairs and singles does generously.

The Eurocode’s line, read again #

The earlier essay found that EN 1994-1-1’s minimum degree of connection for ductile studs is, to within 0.08, what a 6 mm slip limit requires of evenly spaced studs at 95 per cent of the plastic moment. Johnson and Molenkamp fitted it while the Eurocode was being drafted, with the studs spaced evenly, because that is what the rules allowed and what most beams have. Read against the graded line, the rule is not wrong; it is an answer to a narrower question than its wording suggests. It says what connection evenly spaced ductile studs need. A designer who grades the studs is entitled, by the same slip argument, to less — and is not entitled to it by the rule, which gives the same minimum whatever the spacing.

That is the ordinary situation of a fitted rule, and it has a history. Half the studs and most of the beam found that even spacing is itself a plastic assumption: elastically the end connectors are asked for 1.8 times the mean shear, and spreading the studs evenly relies on their ductility to share that out. The minimum degree of connection is where the assumption is paid for, and grading is a way of paying less. The rule is safe for a graded beam, because grading only helps; it leaves the help unclaimed unless someone computes it. For a floor of identical beams that computation is made once, and the saving is made a hundred times.

In a profiled deck #

Most composite floors are cast on profiled steel decking, and there the studs can only go where the deck’s ribs are — one, two, or none in each rib, at the rib pitch. Grading in a deck is therefore coarse: pairs in the ribs over the outer part of each half-span, singles over the inner part, perhaps empty ribs near mid-span. A layout of pairs over the outer third and singles over the rest is close to a grading of 0.3, and at 12 m that grading takes the least connection the slip allows from 0.69 to 0.56 — five studs a half-span.

A deck also lowers each stud’s resistance and can make its load–slip curve less ductile, and studs in pairs in one rib are reduced further than singles. So a pair is not quite two studs’ worth, and the grading a deck permits is partly paid for in each stud’s resistance. That is a calculation with the deck’s own reduction factors in it, and it does not change the direction of the result: the slip is decided near the supports, and the studs that are there decide it.

What the grading costs #

The grading costs nothing in studs, by construction, and nothing in strength. It costs in two other places. The studs near the support are closer together, and a stud’s resistance in a solid slab is reduced when studs are crowded: EN 1994-1-1 sets a least longitudinal spacing of five stud diameters, 95 mm for these studs, which a layout graded by 0.9 approaches at the support. And it costs a decision: an evenly spaced layout is drawn with one dimension, a graded one with several, and the fabricator places them by hand or by programme either way.

Against that, the beam either needs fewer studs for the same moment — eight fewer a half-span at 12 m, if its strength needs only half connection — or reaches more of its plastic moment with the studs it has, as a beam that has yielded uses its ductility in one place to keep its strength in another. On a floor of a hundred beams that is a large number of welds, and the earlier essay’s price list said where they are spent: on the last few per cent of the moment, which is exactly what the slip limit stands between.

The plastic moment and the slab force, by hand #

The plastic moment at half connection is the earlier essay’s: a slab force of 1,502 kN, a block 23.6 mm deep, the steel’s plastic neutral axis 11 mm into its top flange, and 850 kN·m in all — the same for every grading, because it depends only on the slab force at mid-span.

The slab’s force a quarter of the way to mid-span, with every stud at its capacity, is the studs between the support and that point. Evenly spaced, that is a quarter of nineteen. Graded by 0.9, the density at a fraction $u$ of the half-span is $1 + 0.9(1 - 2u)$ times the average, and its integral from 0 to a quarter is $0.25 + 0.9 \times (0.25 - 0.0625) = 0.42$: 42 per cent of the studs, as the figure shows. Halfway, $0.5 + 0.9 \times (0.5 - 0.25) = 0.72$.

The slip itself is not a hand calculation, for the reason the earlier essay gave: it is an integral of a curvature that, near the plastic moment, depends on how far into its yield plateau the steel has been taken. But its direction can be checked by hand. A section with a larger slab force carries more of its moment as a couple between slab and steel and less by bending each, so its curvature is smaller, and the jump in strain across its interface — curvature times the distance between the parts’ centroids, less the force over their combined axial stiffness — is smaller for both reasons.

Smeared studs on the earlier beam #

The studs are smeared. The shear flow along the interface is taken as continuous, with a density that varies; real studs are points, and a graded layout’s sparse studs near mid-span are far apart. Where the spacing approaches the slab’s depth the smeared model is a poor description, and EN 1994-1-1’s maximum spacing — six times the slab’s depth or 800 mm — is the limit beyond which it does not apply.

The rest is the earlier essay’s. A solid slab, Ollgaard’s curve, a steel without strain hardening, a simply supported beam. Each was named there with its direction, and none changes the comparison between the two layouts, which share all of them.

The deck ribs that fix the spacing #

They cannot show a stud’s weld, or the profiled deck most composite floors are cast on, in whose ribs studs can only go where the ribs are — which fixes the spacing at the rib pitch and allows grading only by putting one, two or none in each rib. They cannot show the beam’s deflection, which grading reduces a little by making the beam stiffer near its supports, nor its behaviour after the end studs reach their capacity, which a graded layout delays.

Still open: the beam continuous over its supports #

A continuous composite beam has its slip in the sagging span and its slab cracked in the hogging region over the supports, with a point of contraflexure between them where the shear is largest and the slab’s force changes sign. Whether studs graded toward the points of contraflexure rather than the supports reduce the sagging span’s slip as these did, and whether the hogging region — where the slab’s force is only its reinforcement’s, and the studs carry far less — can give up studs to the sagging span without losing the connection it needs for its own reinforcement to act with the steel, is the question the simply supported beam leaves.