The joint that was chosen

The redistribution nobody chose is about what a real joint does to a beam that was designed as though the joint were something else: a connection drawn as a pin has a stiffness, it takes a share of the fixed-end moment, and the analysis that assumed it away is describing a different beam.

The obvious response is to make the joint match the assumption — stiffen it until it is rigid, or soften it until it is not. There is a third response, and it is better than either.

The beam is sized by the larger of its two moments. The larger of two numbers that add to a constant is smallest when they are equal. So there is a best joint, it is neither of the two idealisations, and finding it is arithmetic.

Which free body produced the number #

The free body is the beam, cut at both ends, with a rotational spring at each.

Crossing each cut is a moment $M_s$, and it is whatever the spring’s stiffness and the beam’s end rotation agree on. Along the beam is the uniform load. Statics then gives the mid-span moment directly:

$$M_{span} = \frac{wL^2}{8} - M_s$$

and that relation is the whole of the argument. The sum is fixed by statics and the split is fixed by stiffness. No amount of joint design changes the sum; a joint decides only where the beam is asked to carry it.

Writing $M_s$ as a fraction $f$ of the fixed-end moment $wL^2/12$ and setting the two equal:

$$f\frac{wL^2}{12} = \frac{wL^2}{8} - f\frac{wL^2}{12} \quad\Longrightarrow\quad f = 0.75$$

Everything cancels. The span, the load, the beam’s stiffness and the joint’s are all gone, and what is left is a pure number: a joint that delivers three quarters of the fixed-end moment balances any uniformly loaded beam.

The design moment there is $0.75 \times wL^2/12 = wL^2/16$, which is half the simple-span value and two thirds of the fully continuous one.

What that is worth #

Half the moment is not half the beam, and it is worth being precise about what it buys.

Section modulus is proportional to moment, so half the moment is half the modulus — but a beam’s modulus goes roughly as its mass to the power of 1.4, so half the modulus is about 40 per cent less steel. On a floor of secondary beams that is a real saving, and it is bought with an end plate and four bolts rather than with a heavier section — the connection that has to make a moment cross a gap doing it deliberately and only partly.

The asymmetry of the curve matters as much as its minimum. Undershooting the balanced stiffness is much worse than overshooting it. A joint at half the balanced stiffness leaves a span moment close to the simple-span value; a joint at twice it leaves a support moment only a little above the balanced one. So a design that aims at the balance point and misses it upward has lost very little, which is what makes the approach usable with a joint whose stiffness is known to perhaps 30 per cent.

Two spans and two joints #

The single beam is the clean case and the arrangement it belongs to is a floor, so it is worth asking what the same argument does to a run of them.

Two facts survive the change of beam and one does not.

The fraction survives. It is 0.75 for every uniformly loaded beam, because it came from an equation with nothing in it.

The saving survives. wL²/16 against wL²/8 is a factor of two on any beam.

The stiffness does not. The joint that delivers 75 per cent of the fixed-end moment on a 6 m beam is not the joint that does it on a 9 m one, because what matters is the ratio of the joint’s stiffness to the beam’s $EI/L$. A standard detail used on a floor of beams of different spans lands at a different point of the curve on each of them — stiffer relative to the long beams and softer relative to the short ones.

That is the practical difficulty with the whole approach and it is worth stating plainly. A joint detail is a fixed object and the beams it serves are not, so a floor detailed with one connection has a range of $f$ across it. The flatness of the curve near the optimum is what makes that tolerable, and it is why the sensible target is a little above 0.75 rather than at it — the long beams then fall to the flat side rather than the steep one.

The classification, and what it is for #

Every code classifies a joint as nominally pinned, semi-rigid or rigid, with boundaries at $0.5EI/L$ and $8EI/L$ for a braced frame. The rung below points out that those are lines drawn across a continuous curve.

What they are for is worth stating, because it is not what it looks like. A classification is a licence to ignore something. A joint classified as rigid may be modelled as rigid, which means the frame analysis does not need the joint’s stiffness in it; one classified as pinned may be modelled as a pin, for the same reason. The boundaries are placed where the error from ignoring the joint’s real stiffness is acceptable — five per cent on the frame’s resistance, in the European rules, which is the sort of number a classification boundary is always secretly made of.

The balanced joint is in neither class. It is squarely semi-rigid, and semi-rigid means the joint’s stiffness has to be in the analysis — which is the cost of the 40 per cent, and the reason semi-continuous design is rare in practice despite being obviously worth doing.

The strength, which is a second requirement #

Stiffness decides the split; strength decides whether the joint can deliver it, and the two are independent properties of the same connection.

A joint at the balanced stiffness has to carry $wL^2/16$, which on the beam here is 68 kN·m. An end plate connection of the right stiffness might have a resistance well above or well below that, because stiffness comes from the plate’s bending and the bolts’ extension while strength comes from whichever component runs out first.

Three cases follow, and only one of them is a design.

Strength above the balanced moment. The joint delivers what the stiffness calculation says and the beam is designed for $wL^2/16$ everywhere. This is the intended case.

Strength below it. The joint reaches its capacity, yields, and sheds moment to the span — so the span moment rises above $wL^2/16$ and the design has to be made on the joint’s resistance rather than on its stiffness. That is a partial-strength design and it is perfectly legitimate, provided the joint has the rotation capacity to do the shedding.

Strength below it and no rotation capacity. The joint reaches its capacity and fails. That is the arrangement the three T-stub regimes are about, and it is why a partial-strength joint has to be detailed so that its weakest component is a ductile one — a plate in bending rather than a bolt in tension.

So a semi-continuous design is three checks rather than one: the stiffness that sets the split, the strength that delivers it, and the ductility that covers the case where it does not.

Why it is rare #

Semi-continuous design has been available and codified for thirty years, the arithmetic is a page, and the saving is 40 per cent of a floor’s secondary steel. It is not much built, and the reasons are worth setting out because none of them is that it does not work.

The analysis has to contain the joint. A frame model with rotational springs at every beam end is a different model from one with pins, and it has to be built, checked and maintained through design changes. The classification exists precisely to allow the springs to be left out.

The detail has to be computed. A joint’s stiffness comes from the component method, which is a calculation per connection type rather than a table lookup, and it is sensitive to things a fabricator may change — a bolt row moved, a stiffener omitted.

And the responsibility is divided. Beams are designed by one party and connections are very often designed by another, later, to a set of end reactions. A design that depends on the connection’s stiffness cannot be handed over that way, and the industry’s contractual arrangement is older than the method.

So it is a case where the mechanics is settled, the benefit is real and the obstacle is entirely in how the work is organised. That is not unusual and it is worth recognising, because it explains why a method can be in a code for decades and remain unused.

The number that keeps appearing #

Three moments have turned up on this page — $wL^2/8$, $wL^2/12$ and $wL^2/16$ — and the relationship between them is worth pausing on because it recurs throughout the subject.

They are the simply supported moment, the fixed-end moment, and the balanced one. Their ratios are 6 : 4 : 3, and none of the three is a property of any beam: they are consequences of a parabola’s area, a fixed end’s rotation being zero, and a split being equal.

Every continuous beam problem is a redistribution between the first two of them, and the third is the arrangement that treats the split as a variable. The moment over the support is the same trade made by a genuinely continuous beam, where the split is fixed by the members rather than chosen; a plastic design pushes it further, to $wL^2/16$ at both locations with hinges at each; and a designer choosing a joint stiffness is buying the same point elastically.

That coincidence — the plastic collapse mechanism of a fixed-ended beam gives $wL^2/16$ at the supports and the same at mid-span — is exact and is not an accident. The balanced elastic design and the plastic collapse mechanism arrive at the same distribution, because both are asking for the two moments to be equal. One gets there by choosing a stiffness and the other by letting the steel choose, and the second is what a plastic hinge is for.

One consequence of the balance point deserves stating separately because it changes what a connection is for. A joint chosen at f = 0.75 is carrying a moment the beam would otherwise have carried at mid-span, so it is a structural member with a design action rather than a device for transferring a reaction. That is a different relationship between the beam designer and the connection designer from the one a simple design has, and it is the reason the arrangement needs the two to be the same person or to be working from the same model.

Where the model stops #

One span, one load case, uniformly loaded. A continuous beam over several spans has a different balance, pattern loading moves both moments in ways that do not cancel, and a point load changes the arithmetic that gave 0.75 — the sum is no longer $wL^2/8$ and the fixed-end value is no longer $wL^2/12$.

The joint is linear. A real moment-rotation curve is nonlinear from a low fraction of its capacity, so the stiffness at working load is not the stiffness at the ultimate limit state — which is why codes use the initial stiffness for serviceability and a reduced one, typically a third, for strength.

Serviceability is not in it. The balanced design has a stiffer beam-end condition than a pinned one, so its deflections are smaller — but the deflection depends on the joint’s stiffness at service load rather than on the moment split, and a beam sized by deflection gets less from this than one sized by strength.

Shear is not in it. The joint still has to carry the beam’s end reaction, and a connection detailed for moment is usually more than adequate for shear — but a partial-depth end plate is not, and the arrangement that gives the stiffness has to give the shear capacity too.

And the column has been ignored. A joint that delivers 68 kN·m to the beam delivers 68 kN·m to the column, which then has a moment it would not otherwise have had. On an internal column with beams both sides they largely cancel; on an external one they do not, and part of the beam’s saving is spent there.

What the pictures cannot show #

The connection itself. Every curve here has the joint’s rotational stiffness on its axis, and nothing on the page says what arrangement of plate, bolts and welds produces a given value — that is the component method’s question, and it answers it as several springs in series whose softest member governs.

The practical consequence is that a designer cannot choose a stiffness directly. They choose a detail, compute its stiffness, and find out where on these curves it lands — which is why the flatness of the curve near the optimum matters so much, and why the design is usually done by picking a standard detail and checking which side of the balance it falls on.

They also cannot show what the joint costs. An end plate with four bolts and a stiffened column flange is several times the fabrication of a fin plate, and the 40 per cent of beam steel it saves has to pay for that. On a long-span secondary beam it does; on a short one it does not, and the crossover is a commercial number rather than a structural one — which is the honest answer to why the arrangement is uncommon.

The assumption the figure rests on #

That the beam’s own stiffness is fixed while the joint’s varies.

It is not, in the design that follows from all this. Balancing the moments makes the beam lighter, a lighter beam has a smaller $EI$, and the classification boundaries and the balanced stiffness are both proportional to $EI/L$ — so the target moves when the beam it was computed for changes.

The iteration converges quickly and in a helpful direction: a lighter beam needs a softer joint to achieve the same split, because the fixed-end moment it attracts is smaller. So a design that sizes the beam at the balance point and then finds the joint stiffness has overshot slightly, which the flat side of the curve absorbs.

That self-correction is worth noticing as a property rather than a convenience. A structure whose stiffnesses are all being chosen at once has no single optimum to converge on, only a family of consistent answers — the same situation an indeterminate structure’s forces are always in, where what a member carries depends on what it was made.

What to carry away #

Three sentences, and the second is the one that is genuinely surprising.

Statics fixes the sum of a beam’s span and support moments and the joint fixes only the split, so the design moment is minimised where the two are equal.

That happens at three quarters of the fixed-end moment, for every uniformly loaded beam, with no property of the beam or the load in it — and the design moment there is wL²/16, half the simple-span value.

And the joint that does it is squarely semi-rigid, which means the classification that exists to let a designer ignore joint stiffness is exactly what has to be given up to get the saving. A joint is neither pinned nor rigid is an observation about real connections; this is the same fact used on purpose.

The ladder from here #

Later rungs on this anchor: the same balance on a continuous beam over several spans, where pattern loading decides it. Partial-strength design proper, where the joint’s resistance rather than its stiffness sets the split and the rotation demand has to be computed. The moment-rotation curve’s nonlinearity, and the two stiffnesses a code uses for the two limit states. The column side of the joint, where the moment delivered has to be carried by a member designed for axial load. And the economics: what a semi-continuous frame costs in fabrication against what it saves in steel, which is the calculation that decides whether any of this is built.