Two hinges let in and nothing lost

The moment over the support set out what continuity buys: a span that shares its moment with its neighbours takes a fifth less at its worst section than a simply supported one, and more in a long run of spans. It mentioned in a single paragraph that a Gerber hinge gives up the redundancy while keeping most of the benefit, and it moved on. Where to put the supports said the same in a sentence, and any structure will carry the unit load used the hinge as a device. So there is a standing claim in those essays, that a hinge inserted at the point of contraflexure changes nothing but determinacy, and it has not been tested.

Heinrich Gerber built the first such bridge over the Main at Hassfurt in 1866. He had a reason that the claim does not cover. Continuous iron beams of that date were at the mercy of their foundations, in the way the support that moved sets out, and a pier settling by a few centimetres could change a beam’s moments by more than its load did. A hinge gets rid of that. The claim to be tested is therefore that the hinges cost almost nothing and the settlement problem goes away. Half of this turns out to be better than advertised, and the other half is something the hinged beam can’t get back.

A hinge is an equation, and two of them finish the beam #

Three equal spans on four supports have four reactions and two equations of statics, so two degrees of indeterminacy. Each hinge is a further equation, the moment at that point being zero. Two hinges finish the problem. Everything in the beam is then decided by statics and nothing by stiffness. In the layout used here, both hinges sit in the centre span, a distance $a$ in from its supports. The middle part is a simple beam of length $L - 2a$ hanging between them, and each end span is a beam with an overhang carrying half of the suspended part’s load on the tip of its arm.

Counting the unknowns gave the count. What matters here is that each of those hinges is a choice, so the value of $a$ decides the moment diagram.

The two diagrams lie close to each other, which is the first surprise, as one of them needed the beam’s stiffness to find it and the other could have been written down with a pencil. With $a = L/4$ the arm is 1.75 m and the suspended part is 3.5 m. Under $w = 35$ kN/m the hanging part hands 61 kN to the tip of each arm, so the support moment is $61 \times 1.75 + 35 \times 1.75^2/2 = 161$ kN·m. The suspended span’s own sag is $35 \times 3.5^2/8 = 54$ kN·m. The end span, with a reaction reduced by its overhang, reaches 141 kN·m. All of those can be checked on the back of the drawing, and the elastic solution with its tenth of $wL^2$ over the support, $0.1 \times 35 \times 49 = 172$ kN·m, isn’t far away.

The position, and the two rules that name it #

Where should the hinge go? Two answers are in circulation, and each has a clear reasoning behind it.

The first answer is the point of contraflexure of a fully fixed span, at $a = 0.2113L$. Put a hinge where the moment is already zero and the diagram can’t change, so the beam has the moments of a fixed-ended one, $wL^2/12$ at the support and $wL^2/24$ at the middle. That is the claim in the earlier essays. It is true of an interior span in a long chain and of no other, since in three spans the end span was never fixed.

The second is the balance. Make the hogging over the support equal the sagging in the suspended span:

$$\frac{w a (L - a)}{2} = \frac{w (L - 2a)^2}{8},$$

which gives $a = 0.1716L$ and a moment of $0.0858wL^2$ both ways, against $wL^2/8$ for a simple span. In a chain of many spans that is the best possible: about 31 per cent less moment than a row of simple spans. This is the number every account of a Gerber beam quotes.

In a three-span beam, both answers overlook the end span. The figure shows it as the third line, and at the balance point it is the largest of the three, at $0.092wL^2$. It falls slowly as the hinge moves towards the middle of the span, since a longer arm pulls down harder on the end span, and it crosses the rising hogging line further along. So under a full load, the best position in this beam is near $0.22L$, further in than the balance and just past the contraflexure, with both moments at $0.086wL^2$. That is the sort of thing that follows from having only three spans, and it’s not a defect of the method. It is an example of what can go wrong when a rule comes with a span count attached and is used without one.

The moving load is where the argument turns #

So far the load has been on every span at once, and that is the load case on which continuity looks its best. Real imposed load comes and goes, and the envelope is not a structure has described the penalty a continuous beam pays for that: load one span and leave the next empty, and the support moment at the empty side falls, so the loaded span’s sag rises. Load two adjacent spans and leave the third, and the moment over the support between them rises past the full-load figure.

The envelope is the result that the essay’s title is claiming. On a dead load of 20 kN/m and an imposed load of 15, the continuous beam reaches 184 kN·m, over the support, and the hinged beam with hinges at a quarter of the span reaches 171 kN·m, in the end span. So the hinged beam needs a smaller section than the continuous one, by 7 per cent at this position and by 8 per cent at its best position, 0.27 of a span, with 169 kN·m.

The reason comes straight from the influence lines. In a continuous beam, every span’s load reaches every section. It has long tails, and that’s what makes the pattern penalty possible, as an influence line is a deflected shape shows. In the hinged beam, a load on the opposite end span reaches nothing beyond its own support. Influence lines in a determinate beam are straight and they stop at the hinges. So a pattern has less to work with, as there are no distant spans able to add to the moment at a given section. The continuous beam’s great virtue is that every span helps every other, and its fault under a moving load is exactly the same fact.

That doesn’t always go the hinges’ way. The result depends on the ratio of imposed to dead load, and it is close. On the 20 kN/m of dead load alone the two reach 92 and 98 kN·m. With imposed load equal to the dead load, 201 and 212. With twice as much imposed as dead, 323 and 327, so the hinged beam’s lead shrinks as the live load grows, and a beam dominated by moving load would see it vanish. What counts is that a hinged beam can match a continuous one at all, and it has been widely assumed that it could not.

A comparison on elastic moments also isn’t a fair one for the continuous beam, and to be fair to it requires a remark. A ductile continuous section can redistribute its peak moment, so 184 kN·m is not its true limit, and at collapse it holds a further reserve that a determinate beam doesn’t have. That belongs to the last section of this argument.

What Gerber bought #

Now the settlement.

The flat line is the point of the whole device. It is exactly flat, by a theorem and not by a number: a determinate structure has no self-stress state, and a support movement on a statically determinate beam moves it as a rigid body with no strain, so there is no force at all that could arise. The continuous beam gains $3.6EI\delta/L^2$ at the settling support, which for these numbers is 73 kN·m per 10 mm. A 15 mm settlement has added 42 per cent to the moment; 40 mm has more than doubled it.

There is a feature of the continuous line that a designer should take note of. It is proportional to $EI$. A stiffer beam gets a larger moment from one settlement. So if the settlement is the issue, stiffening the continuous beam is the wrong way of solving it, and a stronger section gets a higher share of the same trouble. This was what Gerber faced: for a span long enough to need a deep iron truss, $EI$ was huge and the settlement moments enormous. The same mechanism applies to temperature gradients, creep, lack of fit and other imposed deformations, which in a determinate beam produce only movement, and never stress. The redundancy only the sun can find and the settlement that matters is the difference are the corresponding accounts of what is being escaped.

It costs little in stiffness, too. A hinge looks like it should cost a great deal.

A greatest deflection of 6.1 mm against 5.8 is close to no cost at all. It has a local visual price, since the line kinks at each hinge, and in a bridge deck that kink meets every wheel. Expansion joints in the running surface are required at hinges, and those are the maintenance item with the worst record in bridge engineering. Still, the expected penalty on the overall deflection under a full load doesn’t appear here, because each span’s load only has a short way to go to a support, and a chain of short determinate pieces can be as stiff as a continuous one.

The same determinacy shows in construction, where it pays directly. A hinged beam’s anchor spans can be built first, as balanced cantilevers, with no falsework in the river, and the suspended span is lifted in afterwards onto seats already at the right level. Every member’s force at every stage is a matter of statics, so nothing needs to be jacked or adjusted, and a hundred and fifty years ago that was worth a good deal more than a small saving in steel. The Forth Bridge is the argument at full scale, in which the suspended spans are short links between enormous cantilevers.

So far the balance sheet favours the hinges in every column but one.

What has been spent #

Take the same two beams and lose a section of the middle of the centre span, by an impact, a fire, a corroded bearing. Under an accidental combination, the full dead load and 30 per cent of the imposed one, the continuous beam has become two cantilevers, 3.5 m long. It hogs to 150 kN·m, and its section is good for 184, so it stands, deflected and alarming, with nobody hurt. The hinged beam loses a hinge or a seat. Its arms now carry only their own load, at 38 kN·m, and they are completely safe. Its suspended 3.5 m has no way to reach the ground, and it falls.

This is not a shortfall of strength. It is the loss of an alternative path. A determinate truss has no robustness at all priced this for a truss, and the same counting argument gives the same conclusion: a determinate structure is one in which every member and every joint is necessary, by definition. That is what determinate means, and it’s the same sentence that gave the settlement result. Immunity to settlement and vulnerability to a lost hinge are one property read two ways. With no self-stress state there are no locked-in forces and no alternative paths, and a structure can’t have one of these without the other.

The plastic reserve says the same in a less dramatic setting. A continuous end span with the moment reaching its capacity forms a hinge and carries on, and it collapses only at about 11.66 $M_p/L^2$, in this beam 1.25 times the envelope’s load. A hinged beam’s first plastic hinge is a mechanism at once. So its design load and its collapse load are the same number, and the comparison of sections above was between a section with a reserve and one without.

The real weakness is also concentrated at the hinge itself, since a Gerber hinge is usually a half-joint: a dapped end, seated on a nib, with a bearing in a place that can’t be inspected and a deck joint directly above it, leaking. The load that has to be lifted looked at how the forces go round such a corner, and that detail has the worst record of any in concrete bridges. The de la Concorde overpass at Laval in Quebec, whose suspended centre span sat on cantilevered ends, dropped that span in 2006 and killed five people. The inquiry found a shear failure in a cantilever that had grown, over thirty-six years, around poorly placed reinforcement near the seat, with water getting in. Half-joints have since been a focus of inspection in several countries, and few new ones are designed. The moment diagrams above are all still right. They say nothing about a hidden corner that a structure relies on entirely.

So modern bridge decks are usually continuous with their foundations made good, and the settlement is handled by analysis, by jacking, or by choosing where the structure is allowed to move. But the idea of a Gerber beam has not gone away. It survives where determinacy is worth its price: purlin runs, where the sleeve at a lapped joint is a cheap hinge-like detail and there’s no fatigue problem, and sites of expected large ground movement, such as above mining, where no analysis can rescue a continuous structure and the failure of one bay is of little consequence.

What a uniform beam on stiff supports leaves out #

The section is uniform. With a free choice of hinge, one would also taper the section to the diagram. The arms of a real Gerber girder are deep over the supports and shallow at the hinge, which moves the stiffness and, for the continuous beam only, the moments as well.

The live load is on whole spans. A full envelope has shorter patterns on parts of a span, including a load on the suspended part alone or on the arm alone, which can worsen the hinge’s shear. The sections’ moments are close to this, but the hinge’s shear and its seat are designed by a patch of load.

A hinge is a perfect hinge. It may seize, and a seized half-joint or corroded pin develops moment and is then not a hinge at all, in which case the beam quietly becomes continuous at that point, with a section that was never given the steel for it. The hinge barely moves, and it seizes gives the corresponding case for an arch crown.

The settlement is one support, and vertical. A pier that tilts or slides moves a hinged beam as a rigid body too, but a large enough movement unseats it. Gerber beams have collapsed from the shortage of seat length. A determinate structure doesn’t care about force under movement, and it cares very much about whether its parts are still touching.

Lost section means a lost section. The comparison of the two beams has the loss at one place. Lose an end span’s support, and the hinged and continuous beams are both in very serious trouble, so the difference is limited to some failures rather than all.

Still open: how much redundancy a chain needs #

Between a continuous beam and a Gerber beam there’s a range. A long viaduct might be divided into continuous groups of three or four spans with a hinge between each group, as many modern ones are, with both the benefit of immunity for its long-range movement and a local reserve within each group. How many spans in a group gets the settlement moments down to where they don’t matter, and how much of the alternative path a group of a given size still has, is a two-sided optimum. It could be read as a count of how many supports can settle against how many sections can be lost, and neither of the two beams here has anything to say about where that balance falls.