The beam that pays the fire back in sag

The tension a fire leaves behind held a steel beam at its ends by the structure around it, a tenth as stiff axially as the beam itself, and took it through a compartment fire that heats and goes out. The beam expanded against its restraint, was pushed into compression, yielded short because its strength was falling while its expansion was not, and when the fire went out the expansion came back and the yield did not. A 457 × 191 × 67 universal beam that peaked at 1,150 kN of compression ended the fire pulling 1,670 kN on its connections — more than it had ever pushed, arriving hours after the fire was out, on connections designed to carry shear.

That beam was kept straight. It had to be, because it carried no load: it was a bar between two springs, and the only thing it could do with its length was yield. The essay ended on the beam that a floor is actually made of, which carries its load as it heats, sags as it softens, and is held at its ends while it sags — a catenary while still hot. Whether the hot catenary’s tension and the cooling’s add in the connection, or whether the sag on the way up relieves enough of the compression that less is left to take back on the way down, is the question.

A beam that can sag has a third way to find length #

The beam is the same section, now spanning 9 m and carrying 20 kN/m — 0.39 of its cold plastic moment, a usual load ratio for a floor beam in a fire, where the load is the fire combination and not the ultimate one. It is held at its ends by springs a tenth as stiff as itself, and heated, as before, as a bare steel lump in a fire of 200 MJ per square metre of enclosing surface that takes it to 940 °C.

The model adds one thing to the straight beam: the beam may sag, in a half-sine of amplitude δ. A sagging beam’s arc is longer than its chord by $\pi^2\delta^2/4L$, so a sag is a way of accommodating length; and with an axial force $P$ in the beam, the sag puts a moment $P\delta$ at mid-span that the beam’s bending has to make up to the load’s $wL^2/8$:

$$\frac{wL^2}{8} = M + P\,\delta.$$

Compression makes $P\delta$ add to the load’s moment, which is how a squeezed beam bows; tension makes it carry part of the load, which is a catenary. The beam’s bending yields at its plastic moment at temperature, reduced by the axial force it is carrying — the hinge after the first yield followed in a cold beam — and its axial force yields at its squash load at temperature; past either it keeps the plastic sag or the plastic strain it yielded by. The straight beam of the earlier essay is this model with no load, and the model reproduces it exactly.

Squeezed until it bends #

On the way up the two beams start on the same line: the restraint holds them, their expansion is refused, and the thrust grows with the temperature. The loaded beam leaves the line at 501 °C, with 1,025 kN of thrust in it, and its thrust falls by half in a few degrees. What happened is not buckling in the elastic sense — the beam’s Euler load at that temperature is four times the thrust — but a hinge. The thrust, acting through the beam’s small sag, has added to the load’s moment while the steel’s strength has fallen to 0.78 of its cold value and the thrust itself has eaten more than a third of the moment the section can carry. The mid-span section reaches its reduced plastic moment, and from there the only way for the beam to carry its load and its thrust together is to sag.

Once it sags, it no longer has to be squashed. The sag takes up the length the restraint refuses, and the thrust falls instead of the beam yielding short. The loaded beam finds its length sideways. By 640 °C the thrust has gone; by 700 °C the beam is in tension; at 827 °C it is pulling on its ends with 293 kN — its whole yield in tension at that temperature, because the hot steel’s strength is all a catenary can use.

Then the fire goes out. The straight beam, which shortened plastically on the way up, pulls on its connections all the way down and arrives at 1,669 kN. The loaded beam cools from a catenary that already had its ends pulled in; its tension grows a little as its strength comes back, and it ends the fire at 613 kN.

The sag the fire leaves #

The sag arrives in a step at the hinge and then grows as the beam softens, to 1.28 m at the height of the fire under 20 kN/m — a seventh of the span, the kind of deflection that unprotected steel floor beams have shown in fire tests and that no design check expects. The temperature is a shape found a section’s temperature deciding its strength; here it decides the beam’s geometry too. It hardly recovers on cooling. The beam has bent plastically, and the 1.13 m it keeps is the fire’s own record of how much length it took.

Those numbers are a measure of what the straight beam was being asked to do. A sag of 1.13 m over 9 m holds $\pi^2 \times 1.13^2/(4 \times 9) = 0.35$ m of length in its curve. The steel’s thermal contraction from 940 °C back to ambient is about 0.12 m over the same span. The sagged beam has three times the length its cooling takes back; the straight beam had to find all of its shortening in plastic strain and then in the restraint’s spring.

Even a light load does it. At 5 kN/m, a tenth of the plastic moment, the beam bows at 585 °C and keeps 452 mm. The straight beam is not a lightly loaded beam; it is a beam that has been forbidden to bend, which no floor beam is.

The more it carries, the less it pulls #

The cold tension at the end of the fire falls with the load, from about 1,000 kN at light loads to 613 at 0.39 and 477 at 0.54. Every one of them is well below the straight beam’s 1,669, and the heavier the beam is loaded the more it has sagged and the less it has to be pulled for. At light loads the curve rises a little before it falls: a beam that hinges late, at 585 °C, has already yielded short over part of its heating, and keeps that shortening as well as its sag.

The hot tension goes the other way. A beam that sags more hangs more steeply and carries more of its load as a catenary, and a catenary’s tension is what its sag cannot supply: 41 kN at a tenth of the plastic moment, 293 at 0.39, 397 at 0.54. Above 0.56 the beam cannot find a sag up to a fifth of its span at which the hot steel’s tension and its remaining moment carry the load, and it runs away near the fire’s peak. That is the beam’s real fire limit in this model: not the temperature at which its plastic moment falls to its load, which the hour that is really a temperature found, but the temperature at which the catenary that took over runs out too.

So the loaded beam trades the cold tension for a hot one. It does not add them. The sag that relieves the cooling is the same sag that makes the catenary.

Held hard or softly, it pulls the same #

The earlier essay’s central result was that the cooling tension is the plastic shortening times the restraint: a stiffer surround makes the beam yield shorter and then pulls harder on what it shortened, until at a fifth of the beam’s stiffness the beam ends the fire yielding in tension. For the loaded beam that result disappears. Its end tension is between 497 and 628 kN at every restraint from 2 to 60 per cent of its own stiffness.

The reason is that the restraint no longer decides where the length goes. A straight beam between two springs has nowhere else to put its expansion, and the springs’ stiffness sets how much of it becomes plastic strain. A beam that sags puts it into the sag, by an amount set by its load and its hinge; a stiffer surround only makes it hinge a little earlier. The cold tension that remains is set by the sagged beam’s own geometry and strength, not by the frame. The force the connections must survive in cooling is a property of the beam’s load, not of the building around it. That is the more useful result for a designer, because a beam’s load is known and the axial stiffness of the frame around it, with its connections’ slip and its columns’ bending, almost never is.

The hot catenary pulls on a hot connection #

A cooling tension arrives when the connections are cold and strong. The hot catenary arrives when they are hot. A connection at the end of a beam heats more slowly than the beam’s span, because it has more steel in it and less of it exposed, and fire design takes its temperature as a share of the beam’s. Its bolts lose strength faster than the steel does: by 600 °C a bolt has 0.22 of its cold strength, and by 700 °C a tenth.

A connection is not the pin the beam diagram draws — it has a size and parts, and its bolts are the part that weakens fastest. Give the connections a tension resistance of 600 kN cold — a fifth of the beam’s own yield, a generous tying resistance for a beam-to-column connection. The straight beam pulls on them with 2.78 times that once it is cold, and nowhere else. The loaded beam pulls with 1.02 times it cold, which a connection can nearly carry. But while it is hot, with its connections heating 0.8 as much as the beam, it asks for 3.79 times what the connection has: its catenary’s 228 kN at 870 °C, on bolts at about 700 °C with a tenth of their strength. With connections heating 0.6 as much as the beam the hot demand is 0.9 and the worst is 1.05, on cooling.

Whether a loaded beam’s connections are most at risk in the fire or after it depends on how hot the connections get, which is the one temperature a fire design usually estimates least carefully. For the straight beam the question never arose: it had no hot tension. For the beam that floors are made of, the connection’s own temperature decides which of two failures comes first, and between 0.6 and 0.8 of the beam’s temperature rise the answer changes.

The hot demand also has a simple form. While the beam is yielding as a catenary its tension is its own hot yield, $k_{y,T}A f_y$, so the ratio of demand to a connection whose cold resistance is $F_c$ is $k_{y,T}Af_y/(k_{b,T_c}F_c)$: the beam’s strength factor at its temperature over the bolts’ at theirs, times how much stronger the beam is than its connection. A connection a fifth as strong as the beam needs bolts at least five times as strong, in proportion, as the beam’s steel at the same moment — which at 870 °C means bolts below about 550 °C.

A longer fire, a deeper sag #

The straight beam pulls harder after a longer fire, because a hotter beam yields shorter. The sagging beam pulls less, because a hotter beam sags deeper. In the shortest fire drawn, which peaks at 827 °C, the two are nearly the same, 1,185 kN against 1,212: the beam hinged and sagged half a metre, but it had also yielded short before it hinged, and the shortening dominates. By 300 MJ/m² the sag is 1.57 m and the end tension 463 kN, a quarter of the straight beam’s. At 400 MJ/m² the beam runs away.

The two beams’ answers diverge in the direction that matters least for the connections and most for the floor. A severe fire leaves a sagging beam a modest pull on its connections and a floor that is a metre and a half out of level — usable for nothing until it is replaced.

When the straight beam is the right model #

None of this makes the earlier essay’s straight beam wrong; it makes it the model of a particular member. A beam that cannot sag — a tie, a bracing member, a short beam whose load is small beside its stiffness, a beam held from bowing by the floor it supports through stiff composite action — has nowhere to put its expansion but its own length, and the plastic shortening times the restraint is exactly what it pulls with afterwards. The fire that goes out found that the cooling is where such members are asked the most.

The floor beam is the other kind. Its load is the thing that bends it out of the way, and its hinge temperature, about 500 °C under a typical fire load ratio, is below the 628 °C at which the straight beam reached its largest thrust. The beam leaves the straight beam’s path before the straight beam has done most of its yielding. The question to ask of a member in a restrained fire bay is therefore not how stiff its restraint is but whether its load or its geometry will let it bend before its thrust has squashed it. A member that will bend gives back length as sag; one that will not gives it back as tension.

The hinge by hand #

At 501 °C EN 1993-1-2 gives the steel 0.78 of its yield strength and 0.60 of its stiffness. The section’s plastic moment at temperature is $0.78 \times 355 \times 1.471 \times 10^6 = 407$ kN·m and its squash load $0.78 \times 355 \times 8{,}550 = 2{,}370$ kN. Under 1,025 kN of thrust the reduced plastic moment is $407 \times 1.18 \times (1 - 1{,}025/2{,}370) = 272$ kN·m.

The load’s moment is $20 \times 9^2/8 = 202.5$ kN·m. The beam’s elastic sag just before the hinge is 58 mm — doubled from its cold 26 mm by its falling stiffness and by the thrust — and the thrust’s moment through it is $1{,}025 \times 0.058 = 59$ kN·m. Together, 262 kN·m against a capacity of 272, and the next few degrees close the gap. The Euler load at the same temperature, $\pi^2 \times 0.60 \times 210{,}000 \times 2.938 \times 10^8/9{,}000^2 = 4{,}510$ kN, is four times the thrust: the beam does not buckle; it yields in bending because its thrust has taken its moment capacity.

What the one mode leaves out #

The beam is one half-sine. A real beam’s hinge is local, at mid-span, and the rest of the beam stays straighter than a sine; its sag for a given hinge rotation is a little less and its arc a little shorter. The one mode reproduces the beam’s cold sag to within 3 per cent and catches the hinge, the catenary and the residual sag; it would not catch a beam that hinges near its ends, as one with stiff end connections does.

The steel is at one temperature. A beam under a slab is hotter at its bottom flange than its top, and the gradient bows it downward thermally before any load does. That gradient adds sag on the way up and takes it back on the way down, and would make the hinge arrive sooner.

There is no slab. A floor beam’s slab is the larger part of the floor’s catenary, as the force nobody put in the model found of a slab’s membrane in a cold floor: once the beams sag, the slab hangs between them in two directions and carries its own load as a membrane. That is the subject of a different calculation, and it is the reason real floors in real fires have survived deflections that would have run away a bare beam.

The connections are springs that do not fail. A connection that fractures at the hot catenary’s peak leaves the beam a cantilever at one end, and the load it was carrying as a catenary falls back onto bending it has no strength for.

Still open: the floor whose slab hangs as the beams sag #

Every beam here hangs alone. In a floor the beams sag together, and the slab spanning between them, held at its edges by beams that are themselves sagging, can become a membrane carrying the floor’s load in tension across the bay — the mechanism that let steel-framed floors survive the full-scale fire tests that unprotected beams alone should have failed. The cable that pays the factor back followed a single member’s catenary after a column was lost; a floor in fire is the same geometry arriving by heat. Whether the slab’s membrane takes over before the beams’ hot catenaries run out — and whether the force it hands to the edge beams’ connections is larger than the hot catenary’s, at a moment when those connections are hotter still — is what the slab over the beams asks of the beam alone.