The joint that has to keep turning

The joint that was chosen found the stiffness at which a beam’s support and span moments come out equal, and noted in passing that the joint then has to be strong enough to carry what its stiffness delivers. Most semi-continuous joints are not. An end plate that makes a joint stiff enough to matter usually has a moment resistance well below the plastic moment of the beam it is bolted to, which is what the words partial strength mean. A partial-strength joint yields, and a joint that yields has to go on rotating.

Statics fixes the collapse load #

The beam is a 457×191×67 universal beam spanning 9 m: flexural rigidity $61{,}700\ \mathrm{kN{\cdot}m^2}$, plastic moment 522 kN·m. Its joints resist 261 kN·m, half of that.

Whatever the joints do on the way, the beam can only collapse one way: a hinge at each end and one at mid-span, three hinges, a mechanism. At that moment the end moments are the joints’ resistance and the mid-span moment is the beam’s plastic moment. Statics says the end moment and the span moment of a uniformly loaded beam always add to $wL^2/8$, so the collapse load is

$$w_u = \frac{8(M_p + M_j)}{L^2}$$

which for this beam is 77.3 kN/m. The collapse load of a structure is fixed by its mechanism, and the mechanism has the hinges’ strengths in it and nothing else. The joint’s stiffness is not in the collapse load. On pinned joints the same beam collapses at $8M_p/L^2$, 51.6 kN/m; on rigid full-strength ones at $16M_p/L^2$, 103.1 kN/m. Half-strength joints put it half-way between, whether they are stiff or soft.

Stiffness decides the order #

Stiffness decides what happens before collapse, and the first figure shows it.

While everything is elastic, the joints carry a fraction $f = 1/(1 + 2EI/S_jL)$ of the fixed-end moment $wL^2/12$. At 6 EI/L that fraction is three quarters, and the end moment climbs faster than the span moment. It reaches the joints’ resistance first, at 51.6 kN/m. From there the joints hold 261 kN·m and turn freely, and everything the load adds goes to the span, whose moment climbs until it reaches 522 kN·m at 77.3 kN/m.

A softer joint would change the order. Below a certain stiffness the span yields first, and then the extra load goes to the joints instead, which reach their resistance exactly as the collapse load arrives. Either way the collapse load is the same. What differs is which hinge forms first — and the first hinge is the one that has to keep turning.

The rotation a joint must supply #

Between first yield and collapse, the joints are holding a constant moment while the load goes on rising, so the beam’s ends rotate as the ends of any beam with fixed end moments do: by $L^3/24EI$ for every kilonewton per metre added. The plastic rotation each joint must supply before the span hinge forms is therefore

$$\theta_p = (w_u - w_1)\,\frac{L^3}{24EI}$$

For the beam in the first figure that is $(77.3 - 51.6) \times 729/(24 \times 61{,}700)$, 12.7 milliradians. By a coincidence of these proportions it equals the rotation a fixed-ended beam’s end hinges need, $M_pL/6EI$, which is the textbook figure for a full-strength rigid joint.

The whole calculation for this beam fits in a few lines, and it is worth having in one place. The joints’ stiffness is $6EI/L$, 41,130 kN·m/rad, so $f = 1/(1 + 2/6) = 0.75$. The end moment is $0.75 \times wL^2/12$, which reaches 261 kN·m when $w = 12 \times 261/(0.75 \times 81)$, 51.6 kN/m. By then each joint has turned elastically through $261/41{,}130$, 6.3 milliradians. The collapse load is $8 \times 783/81$, 77.3 kN/m. The load still to come after the joints yield is 25.7 kN/m, and each kilonewton per metre of it turns the beam’s ends by $729/(24 \times 61{,}700)$, 0.49 milliradians, so the joints must turn a further 12.7 at constant moment and 19.0 in all.

Why the ends turn as though the joints were pins #

The factor that turns a load increment into a rotation deserves a sentence of justification, because it is what makes the demand computable at all.

Once the joints have yielded they carry a constant moment. Adding load does not change that moment, so the increment of load is carried by the beam exactly as it would be by a simply supported beam: the constant end moments are already in equilibrium with the load that produced them, and the new load acts on a beam whose ends offer no further resistance to turning. The area of the new load’s M/EI diagram is the change of slope along the beam, and half of it is the rotation at each end: $\tfrac{1}{2} \times \tfrac{2}{3} \times L \times \Delta w L^2/8EI$, which is $\Delta w L^3/24EI$. After the joints yield, the beam’s ends turn as though the joints were pins, while the joints go on holding their moment. That is what a plastic hinge is, and it is why the demand depends on the beam’s flexibility and on nothing about the joint except when it yielded.

What the plastic theorems take on trust #

A plastic analysis of this beam is two lines long: a mechanism with hinges of 261, 522 and 261 kN·m, and a collapse load of 77.3 kN/m. The upper-bound theorem guarantees that no mechanism gives a lower load, and the lower-bound theorem that a moment distribution in equilibrium and nowhere above its resistance cannot be carried at more. Both are satisfied, and neither says anything about the order in which the hinges form.

Both theorems rest on an assumption that is not in either of them: every hinge can go on rotating at its full moment for as long as the others take to form. A beam’s own plastic hinges usually can, if the section is compact. A joint is a different object, assembled from plates, bolts and welds, and whether it can depends on which of those parts yields first. The rotation demand is the price of the plastic theorems’ guarantee, stated as a number, and a design that uses the collapse load without checking it has bought the guarantee without paying for it. That is why a code permits plastic global analysis of a frame with partial-strength joints only where the joints at the hinge locations can be shown to have enough rotation capacity.

That demand is a rotation at constant moment, and it is supplied by something in the joint deforming plastically: an end plate bending in yield lines, a column flange, a web panel in shear. A joint is a set of springs in series, and the one that yields first is the one that must be able to keep yielding. If it is a bolt instead, which breaks rather than yields, the joint has no rotation to give.

A rotation of a few hundredths of a radian sounds small until it is turned into the movement that has to happen somewhere in the steel. An end-plate joint rotates about a point near the beam’s compression flange, so the tension flange moves away from the column by the rotation times the beam’s depth. The 457 mm beam here, turning 12.7 milliradians plastically, opens a gap of about 5.5 mm between its end plate and the column face at the tension flange; at the 21.4 milliradians of the stiff quarter-strength joint it is about 9 mm. That opening has to come from the end plate bending away from the column round its bolts, from the column flange doing the same, or from the bolts stretching. A thin plate can fold that far along its yield lines. A bolt of the usual kind stretches by a small fraction of that before it breaks, so a joint whose bolts are its weakest component has nowhere to find 9 mm, and a joint that is to supply rotation has to be detailed so that the plate yields before the bolts do.

A weak joint on a stiff connection #

The demand grows in the direction nobody’s intuition points.

A stiff joint picks up moment quickly and a weak one reaches its limit early, so a joint that is both yields almost at once: at 20.9 kN/m, a third of the way to collapse. Everything from there to 64.4 kN/m is plastic rotation, 21.4 milliradians of it, supplied at a constant 131 kN·m.

That is more than the 12.7 milliradians a rigid full-strength joint needs. A connection detailed as stiff but not strong asks more of its ductility than one that is both.

A joint with 15 milliradians of plastic rotation capacity does not get there. It fractures at 51.3 kN/m. The beam then has no end moments at all, and its span must carry the whole of $wL^2/8$ — 519 kN·m, three kilonewton-metres short of its plastic moment. It survives, but only just, and only because this beam happened to have that margin. The designer who counted on 64.4 kN/m got 51.3, and the joints that were meant to help have given the beam nothing at all.

Stronger joints need less #

Holding the stiffness at the balanced 6 EI/L and changing the joint’s strength shows the two quantities moving in opposite directions.

At the balanced stiffness the demand is exactly $(L/3EI)(M_p - M_j)$: a straight line to zero at full strength. A joint as strong as the beam, at the stiffness that divides the moments equally, yields at the same load as the span, and neither hinge has to wait for the other. The balanced joint of the essay before this one is the joint that needs no ductility at all — provided it is as strong as the beam. Built at partial strength, the same stiffness asks the joints to rotate in proportion to how much strength they lack.

The collapse load runs the other way, rising linearly with the joint’s strength until the joint is as strong as the beam. Beyond that the hinge forms in the beam beside the joint, and extra joint strength buys nothing.

The stiffness below which nothing rotates #

Varying the stiffness for joints of several strengths gives the whole picture, and a formula for where each curve starts.

The joint and the span yield together when the elastic end fraction is $f^* = 1.5M_j/(M_p + M_j)$, which is a stiffness of $2f^*/(1 - f^*)$ times $EI/L$. Below that stiffness the span yields first and the joint never has to rotate plastically; above it the joint yields first and the demand rises with every increase in stiffness. For a joint as strong as the beam the threshold is 6 EI/L, the balanced joint again. For a quarter-strength joint it is 0.86 EI/L, which is barely above the boundary at which a joint counts as pinned.

So the classification by stiffness gives no warning. At the rigid boundary, the quarter-strength joint needs 19.8 milliradians and the full-strength joint 3.2: the classification calls them both rigid, and one needs six times the ductility of the other. It is the strength, read against the stiffness, that says how much a joint has to turn.

What ductility buys #

The practical question is the other way round: given what a joint can supply, how much of the collapse load does the beam reach?

A joint with no plastic rotation at all is a joint whose strength can only be used up to first yield, and for a stiff weak joint that is a third of what the beam could carry. Every milliradian of capacity buys load in a straight line until the demand is met, and then nothing more. On the softest joint 15 milliradians is more than enough. On the stiffest it is not, and the shortfall costs a fifth of the collapse load.

Codes deal with this by granting rotation capacity to joint details whose weakest component is ductile — an end plate thin enough, relative to its bolts, that the plate folds along its yield lines before a bolt breaks — and by requiring a calculation, or a test, for anything else. What the figures add is which joints most need that grant to be true: the stiff ones that are much weaker than their beams.

The stiffness bought for deflection, and spent at service load #

There is a second reason a stiff, weak joint is a poor arrangement, and it arrives long before collapse.

Joints are made stiff in the first place to reduce deflection. On pins, the 9 m beam carrying 40 kN/m — a little over half its collapse load on quarter-strength joints, and a plausible service load — deflects $5wL^4/384EI$, 55.4 mm, span over 162. Elastic end moments reduce that by $ML^2/8EI$, and with the joints carrying a fraction $f$ of the fixed-end moment the deflection becomes $(5 - 4f)/5$ of the pinned value. Joints of 6 EI/L bring it to 22.2 mm; joints of 25 EI/L, with $f = 0.926$, to 14.4 mm, span over 626.

But at 40 kN/m the elastic end moment on those stiff joints would be 250 kN·m, and the quarter-strength joints resist 131. They have already yielded under the service load. Their end moment is stuck at 131 kN·m, which takes 21.4 mm off the pinned deflection rather than 41, and the beam deflects 34.0 mm — more than twice the elastic estimate, span over 265 instead of span over 626 — with a permanent rotation left in the joints when the load comes off. The stiffness that was specified for serviceability is gone before the beam is in service, and a deflection calculated with it is a deflection the beam will not have.

So the stiff weak joint fails both ways: it needs the most ductility at collapse and it delivers the least stiffness in use. The combination that avoids both is the one the essay before this one arrived at from the other direction: a joint whose strength matches what its stiffness delivers, so that it yields with the span rather than before it.

Where the model stops #

The joints are elastic–perfectly plastic. A real moment–rotation curve bends over gradually, so a joint starts to soften well below its resistance and there is no single first-yield load. The demand computed here is the demand past an idealised knee, and the real joint begins using its capacity earlier.

The beam is elastic until its hinge forms. Plasticity spreads along the span before the hinge is complete, which lengthens the process and adds rotation, and the span hinge needs rotation capacity of its own, which a class 1 section has and a slender one may not.

One span, symmetric, uniformly loaded. A continuous beam with pattern loading yields its joints in a sequence that depends on which spans are loaded, and a support moment that falls short on one side is made up on the other.

The joint carries its shear throughout. The collapse load assumes the connection still transfers the beam’s end reaction after it has rotated plastically, and a detail that relies on its bolts for both actions has to be checked for the combination.

Still open: the column that receives the moment #

A joint that delivers its resistance to the beam delivers it to the column too, and a column designed for axial load now has a moment at every floor where its joints have yielded. That is the first question left open. After it: the unbraced frame, where the joints’ rotation demand is set by sway as well as gravity and accumulates up the height; the continuous beam under alternating patterns of load, where a joint can yield one way and then the other and the question becomes shakedown; and the cooling phase of a fire, where a beam that expanded and yielded pulls on its joints as it shortens, and the rotation a joint supplied on the way up is not the one it is asked for on the way down.