The slip the plastic moment asks of the studs

A steel beam and the slab above it are two beams or one depending on what joins them, and the joining is done by studs welded to the top flange and cast into the concrete. With enough of them the pair bends as a single section, the slab doing nearly all the compressing and the steel nearly all the tensioning, and the beam is twice as strong as the steel alone. With fewer, the studs slip and the two parts bend partly separately. In service that partial interaction is a stiffness question, governed by Newmark’s equation, and a beam with half its studs is over ninety per cent composite in stiffness because the studs near the supports, where the shear flow is largest, do most of the work.

At the ultimate limit state it is a different question, and the answer is written in EN 1994-1-1 as a curve and a line. Both rest on an assumption about the studs that is easy to state and rarely checked, and this essay checks it.

Stress blocks with half the studs #

The beam is a 457 × 191 × 67 universal beam in S355, simply supported, under a solid slab 130 mm deep and 2.5 m wide in C30/37. The steel can develop a tension of 3,004 kN; the slab could take more than twice that in compression, so with full connection the plastic neutral axis is in the slab and the studs must deliver the steel’s whole 3,004 kN between the support and mid-span. At 80 kN a stud, that is 38 studs each side.

With fewer studs, the slab can be given only what they deliver. The degree of shear connection η is that fraction, and the plastic moment follows from stress blocks: a concrete block of depth $\eta N_{c,f}/(0.85 f_{ck} b)$ at the top of the slab, and a steel section that must carry $\eta N_{c,f}$ more tension than compression, which it does by finding its own plastic neutral axis somewhere below the interface.

The curve bows well above the straight line between the steel alone and full connection. Half the studs gives 85 per cent of the moment, not 76; three tenths gives 76 per cent, not 66. The reason is visible in what each stud is buying. The first studs move the slab’s compression from nothing to something with a long lever arm — the slab is at the top of the section, half a metre above the steel’s centroid — while the steel’s own couple barely changes; the last studs move the slab force from almost full to full while the steel’s neutral axis climbs out of the top flange into the slab, and the lever arm they buy is short. Each stud is worth less than the one before.

EN 1994-1-1 allows the curve for ductile studs and the straight line for any connector. The difference between them — the shaded band — is what ductility buys, and ductility here means one thing: the stud can slip a long way while still carrying its full load.

Two neutral axes and a jump between them #

What the stress blocks assume about the slip is visible in the section itself.

The slab has a neutral axis 23.6 mm below its top, and the steel has its own, 11 mm below the interface. Between them, at the interface itself, the slab’s bottom fibre is stretched — it is below the slab’s neutral axis — while the steel’s top fibre is squeezed — it is above the steel’s. Two fibres a hair apart, one in tension and one in compression: the strain jumps at the interface, and the jump is the rate at which the slab and the steel slide past each other along the beam, the slip strain. A fully composite section has one neutral axis and no jump, which is the plane-sections assumption holding across the interface. Every partially connected section has two, and the studs are what is allowing the jump.

The slip itself is the integral of that jump along the beam — the same relation the elastic analysis of the interface uses, now with the section yielding — from mid-span, where symmetry holds the two parts together, to the support, where it is largest. That integral is what the stress-block calculation never computes, and it needs a curvature at every section.

Following the slip along the beam #

The beam is analysed as it would be tested. The section at each point along the half-span is a set of fibres — concrete with no tension and a parabola-rectangle in compression, steel elastic up to yield and plastic after, as in a section that yields from the outside in — in two parts sharing one curvature, with the slab carrying the force the studs have so far passed it and the steel carrying the equal and opposite tension. The studs follow Ollgaard’s load–slip curve, $q = P(1 - e^{-0.709 s})^{0.4}$ with $s$ in millimetres, which rises steeply and approaches the stud’s capacity at a few millimetres of slip. Starting at the support with no slab force and some slip, the slab force is built up stud by stud and the slip run down by the slip strain, and the slip at the support is adjusted until the slip at mid-span is exactly zero.

The curves start gently and finish steeply. Taking the beam to half its plastic moment asks almost nothing of the studs, a millimetre or two, because the steel is still elastic and the curvature small. The last tenth is where the slip is spent: the steel’s bottom flange yields, its curvature grows quickly for a small increase of moment, and the slip strain — which is proportional to the curvature times the distance between the two neutral axes — grows with it. On a steel with no strain hardening the curvature on the plastic plateau grows without limit as the moment approaches its plastic value, so the slip demand does too. The rigid-plastic curve in the first figure is a promise about the end of a curve that runs off the top of this one.

Read against the stud’s 6 mm, the beam with half its studs reaches the plastic curve’s 0.85 of full connection only as far as about 87 per cent of that moment before its end studs have given everything they are trusted to give. The beam with 70 per cent connection reaches about 95 per cent. Full connection never comes close.

Where the slip happens #

The slip is not shared evenly, and that is what makes the end studs the ones that matter.

Every curve is greatest at the support and falls to nothing at mid-span, and with less connection the whole curve is higher. A stud near the support has been asked to slide past everything the beam has done between it and the middle. The studs near mid-span slip a fraction of a millimetre, are on the rising part of their curves and carry less than their capacity — which is why the slab force at mid-span, under a uniform spacing, never quite reaches the η times full connection the stress blocks assumed until the end studs have slipped far enough to bring the middle ones up.

That is also the reason design rules allow studs to be bunched towards the supports for some loadings and require them to be spread for others. Spreading them evenly puts the same number of studs where the slip is small as where it is large.

The span, which the slip integrates over #

The slip at the support is an integral along the half-span, so a longer beam asks more of its end studs even if every section along it is in the same state.

The slip grows nearly in proportion to the span — a gentler law than the fourth power that deflection follows, but the same kind of statement: a check that holds on a short beam is not a property of the beam’s section. A 6 m beam with half its studs needs 5 mm at the end; the same beam at 18 m needs 13. A degree of connection that is ductile enough on a short span is not ductile enough on a long one, and the reason is not that long spans are more heavily loaded — every curve here is drawn at the same share of each beam’s own plastic moment — but that the end stud of a long beam is at the end of a longer integral.

The Eurocode’s line, derived #

Put the two requirements together — reach 95 per cent of the plastic moment, and do not ask any stud for more than 6 mm — and the least degree of shear connection for each span follows.

The two lines run together. From 8 m to 20 m they differ by at most 0.06, and they rise at the same rate, about 0.06 of the connection for every two metres of span. EN 1994’s rule — a linear function of the span with a floor of 0.4 — is the fitted form of exactly this kind of calculation, made by Johnson and Molenkamp while Eurocode 4 was being drafted, with a stud curve, a slip capacity and a beam taken to failure. The minimum degree of shear connection is a ductility requirement wearing the clothes of a strength one: a beam with fewer studs than the minimum is not too weak by the stress blocks; its stress blocks are a promise its studs cannot keep.

Below 8 m the lines part. The slip calculation would allow a 6 m beam to reach its plastic curve with a fifth of its connection; EN 1994 asks for 0.43. The floor is there for reasons the slip does not capture — a beam with very few studs has long gaps between them, and the assumption that their forces can be smeared into a continuous shear flow fails — and it is the rule being cautious where the arithmetic is least reliable.

What the minimum buys and what it costs #

The two figures before this one can be read as a price list. At 12 m the Eurocode’s minimum degree of connection is 0.61, and the slip calculation’s is 0.67. Full connection needs 38 studs each side of mid-span and gives 1,001 kN·m; 0.67 needs 26 and gives 0.91 of that; 0.5 needs 19 and gives 0.85.

So the last twelve studs a half-span — a third of the full set — buy the last nine per cent of the moment, and the seven before them buy only six per cent more. The stress-block curve is so flat near the top that most of a composite beam’s connection is spent on its last few per cent of strength, which is why designers stop well short of full connection whenever the slab is stiff and the beam is not near its capacity. The minimum is what stops them stopping too soon: below 0.67 at 12 m the moment the curve promises is real only if the end studs can slip further than they are trusted to.

There is a second way to read the same list, and it decides how much the straight line costs. At η = 0.5 the line gives 0.76 against the curve’s 0.85: the designer who uses the line because the studs are not certified ductile loses nine per cent of the moment for the same studs, or needs about 0.69 instead of 0.5 — seven more studs a half-span — to get the same moment back. The band between the curve and the line is worth about a fifth of the connection, and it is earned entirely by the stud’s ability to keep carrying its load while it slips.

Six millimetres, and where it comes from #

The 6 mm is not a property of a stud so much as a property of a test. A push-out specimen — a short length of steel beam with a slab cast against each flange, pushed through — is loaded until it fails, and its slip capacity is read where its load has fallen back to the characteristic resistance on the descending branch. EN 1994-1-1 takes 90 per cent of the measured value, and calls a connector ductile if that characteristic slip is at least 6 mm. A headed stud of 19 mm in solid normal-weight concrete usually achieves it; a stud in a thin profiled deck, or one welded through a galvanised sheet with a poor weld, sometimes does not.

That makes the minimum degree of connection a promise built on two tests that never meet. The push-out test measures what a stud can give; the beam test measures what a beam asks for; the line in the last figure is where the two agree. Either can move. A stiffer, stronger concrete makes the stud’s curve rise faster and its slip capacity smaller; a longer or more slender beam asks for more. A beam that has yielded has used its ductility in one place to keep its strength in another, and a composite beam with partial connection is the same bargain made between two materials instead of within one.

The numbers, by hand #

The stress-block moment at η = 0.5 can be checked on the back of a drawing. The slab force is $0.5 \times 3{,}004 = 1{,}502$ kN; its block is $1{,}502{,}000/(25.5 \times 2{,}500) = 23.6$ mm deep, so its centroid is 118 mm above the interface. The steel must carry 1,502 kN of net tension: its full plastic tension is 3,004 kN, so 751 kN of it must be compression, which a 189.9 × 12.7 mm top flange at 355 N/mm² supplies in its top $751{,}000/(355 \times 189.9) = 11$ mm. The moment about the interface is the slab’s $1{,}502 \times 0.118 = 177$ kN·m plus the steel’s own plastic moment with that neutral axis, 673 kN·m, together 850 kN·m — 0.85 of the 1,001 that full connection gives.

The slip is not a back-of-a-drawing quantity, and the reason is in the second figure. It depends on the jump between two neutral axes at every section, which depends on the curvature, which near the plastic moment depends on how far into the plateau the steel has been taken. That is why the stress blocks can be taught in an afternoon and the minimum degree of connection had to be found by computation and written as a fitted line.

A solid slab, ductile studs and a steel that does not harden #

The calculation makes four choices worth naming.

The slab is solid. Most composite floors are cast on profiled steel decking, which reduces each stud’s capacity, can make its load–slip curve less ductile, and changes the slab’s depth of concrete above the ribs. EN 1994’s rule for decking adds conditions to the minimum for exactly that reason.

The studs follow one curve. Ollgaard’s expression is an average of push-out tests; a real stud’s curve depends on the concrete, the weld and whether the slab is cracked over it, and the 6 mm is a characteristic capacity, below which most studs fail later.

The steel does not harden. A real S355 steel gains strength past a strain of one or two per cent, which bounds the curvature on its plateau and keeps the slip finite at the plastic moment. Without hardening the slip demand at full plastic moment is infinite, which is why every figure here stops short of it and reads the slip at a stated fraction of it instead. The fraction chosen, 95 per cent, is a judgement; nine tenths would move the derived line down, and 98 per cent up.

The beam is simply supported under a uniform load. A point load near a support concentrates the shear flow where the studs are fewest, and a continuous beam puts a hogging region with a cracked slab next to the supports, where the slip calculation has to be made with the slab in tension and only its reinforcement working. Both change the answer; the uniform simply supported beam is what the Eurocode’s line was fitted to.

A stud that fractures, and the sag that comes with the slip #

The figures cannot show a stud failing. Every curve here is a demand: the slip the beam asks for. A stud asked for more than it has fractures at its weld, and the stud next to it then carries more and fails sooner, so a beam whose end studs have reached their capacity unzips from the ends inward rather than settling gently. The 6 mm line is the point at which the calculation stops trusting the stud, not a point at which anything visible happens.

They also cannot show the deflection that goes with the slip. A beam taken to 95 per cent of its plastic moment with half its studs is deflecting several times as much as it does at its design service load, and a beam with so little connection that its studs approach their limit has also lost a measurable share of its stiffness at the load where serviceability is checked. The minimum degree of connection protects the ultimate state; the deflection check, made separately, is what protects the floor.

Still open: the beam with its studs where the slip is #

Every beam here has its studs evenly spaced, so the studs near mid-span slip little and carry less than their capacity while the end studs approach their limit. A beam designed with its studs placed where the slip is — more of them near the supports, fewer at mid-span — would spend its connection where it is used, and its end studs would each slip less for the same total. Whether such a distribution can reach the plastic curve with fewer studs in all, and whether the gain survives a load that is not uniform, so that the slip is not where the design put the studs, is the question the uniform spacing leaves.