The train that is worse than its heaviest axle

An influence line answers a precise question: as a single unit load walks across a span, how does one chosen quantity at one chosen station vary? The answer is a diagram, and the diagram’s peak says where to put the load to make that quantity as large as it can be.

The question a bridge asks is different, and the difference is not a refinement. A vehicle is not one load. It is a set of axles at fixed spacings, all of which move together, and the arrangement doing the most damage is decided by all of them at once. Putting the heaviest axle at the peak of the influence line is a reasonable guess and it is essentially never right, because while that axle sits at the peak the others are standing wherever the spacings put them — and sliding the group along can trade a small loss under the heavy axle for two larger gains under the others.

The peak is 0.12 metres off midspan. That is a small distance and it is not the point; the point is that it is not zero, and that nothing in the single-load influence line predicts where it goes.

The envelope is a different object from the diagram #

A bending-moment diagram is a picture of one loading. It is a function of position along the beam, for a load case that is fixed.

An envelope is a picture of many loadings at once: at each station, the largest value that station ever experiences over the whole set of load positions considered. No single arrangement of the train produces the envelope; it is an upper contour assembled from different arrangements at different stations, and a beam designed to it is designed for a load case that never occurs. The same is true of the envelope a continuous beam gets from pattern loading, which is assembled from load cases that are each individually real and never simultaneous.

That is not a flaw. It is the correct thing to design for, because the beam has to survive every arrangement, and it also means the envelope cannot be read as a moment diagram. It does not satisfy the differential relations: its slope is not the shear of anything, and its second derivative is not a load. It is a maximum over a family, and maxima over families are not the objects they are maxima of.

Where the maximum actually is #

The absolute maximum moment in a simply supported span under a group of loads obeys a rule discovered before it was proved, and it is worth stating before it is derived because it sounds arbitrary and is not.

The absolute maximum moment occurs under one particular axle, when the midspan bisects the distance between that axle and the resultant of the whole group on the span.

The axle in question is the one nearest the resultant. In the figure above, the three axles are 120, 120 and 80 at 0, 4 and 9 metres from the front, so the resultant sits at $(0 \times 120 + 4 \times 120 + 9 \times 80)/320 = 3.75$ metres from the front axle. The nearest axle is the second, at 4 metres — a quarter of a metre from the resultant. Midspan bisecting that quarter-metre puts the axle at $10 - 0.125 = 9.875$ metres, and that is where the maximum is.

The derivation is a small piece of calculus and worth having. With a resultant $W$ at distance $e$ from the axle of interest, and the axle at $x$ from the left support, the moment under that axle is

$$M(x) = \frac{W}{L}(x)(L - x - e) - \sum_{\text{axles left of } x} P_i a_i$$

The sum is a constant while no axle crosses the support or the station, so differentiating gives a maximum at $x = (L - e)/2$, which is the bisection rule written out. Everything the rule says is contained in the fact that the first term is a parabola in $x$ whose vertex has been shifted by half the offset.

The check that is not a repetition #

The figures here do not use Barré’s rule to find the maximum. They march the train across the span in two-centimetre steps and evaluate the moment under every axle at every position, which is an exhaustive search over the whole family, and the rule is then drawn on top as an independent statement about where the answer should have been.

The two agree to the last digit printed, across trains of different weights and spacings, and reduce correctly to $PL/4$ for a single axle. That is a real check rather than two versions of one calculation: one route is a nineteenth-century geometrical construction and the other is brute force, and they share no algebra.

Getting the check to be a check took some care. The first version evaluated the search only at a grid of stations along the span, and the true maximum falls between grid points — so the search came out slightly below the rule, which looked like the rule being approximate and was the search being coarse. The moment under a group of point loads is piecewise linear with its peaks under the axles, so evaluating under each axle rather than at fixed stations finds the exact answer, and only then can the two be compared.

The sign of the offset also had to be got right, and getting it wrong was not obvious: it puts the construction the same small distance on the other side of midspan, producing a moment fractionally below the true maximum — a discrepancy easy to attribute to discretisation.

What one arrangement looks like #

It is worth putting one member of the family beside the contour built from all of them, because the two are drawn in the same style and mean different things.

Two things are visible here that the envelope suppresses. The moment diagram has straight segments between the axles, because there is no load between them, and it has a kink under each — the shear jumps by the axle load, and the moment’s slope is the shear. And the peak is not at a nice location: it is under an axle, wherever that axle happens to be.

Which free body produced the number #

For the first train in its critical position, the lead axle is at 13.875 metres from the left support, which puts the three axles at 13.875, 9.875 and 4.875.

The free body is the whole beam. Moments about the right support give the left reaction: $R_1 \times 20 = 120(20 - 13.875) + 120(20 - 9.875) + 80(20 - 4.875) = 735 + 1215 + 1210 = 3160$, so $R_1 = 158.0$.

Now cut the beam just to the left of the middle axle, at 9.875, and take the left-hand piece as a second free body. On it are the left reaction and the one axle at 4.875:

$$M = 158.0 \times 9.875 - 80 \times (9.875 - 4.875) = 1560.25 - 400 = 1160.25$$

which is the number the figure prints. The same total weight of 320 as a single load would give an envelope peaking at $WL/4 = 1600$ — 38% more — which is the quantitative version of the observation that spreading a load out is worth something.

The assumption every figure here rests on is that the axles are point loads at fixed spacings on a simply supported span, applied slowly. All three parts of that matter. Real wheels distribute their load over a contact patch and through a deck, which rounds the peaks; real vehicles have suspension, so the axle loads vary as the vehicle moves; and a vehicle crossing at speed applies more than its static weight, which is what a dynamic amplification factor is for.

Why the answer is not at midspan, in words #

The reason the maximum wanders off midspan is worth having intuitively, because the algebra hides it.

For a single load the beam is symmetric and the load is symmetric, so the answer is symmetric: midspan. For a group, the beam is still symmetric but the group is not — its resultant is somewhere other than the axle being examined. The reaction that drives the moment at a station is set by where the whole group sits, and the moment at the station is reduced by whichever axles are between the station and the support. Those two effects are optimised at slightly different positions, and the compromise lands halfway between them, which is precisely what the bisection rule says.

The offset is half the distance from the axle to the resultant, so it is small when the group is compact or nearly symmetric and large when one heavy axle is out on its own. For a long articulated vehicle on a short span, only some of the axles are on the bridge at all, and the resultant of the ones on the bridge is what matters — which changes as the vehicle moves, and is why the search sweeps the lead axle from before the span to past the end of it.

The other question the same machinery answers #

Where the load stands is one question. Where the supports stand is another, and it is the same optimisation seen from the other side.

The structural similarity is exact: in both cases a quantity is minimised over a continuous family of configurations, and the answer is at the point where two competing effects balance rather than where either is individually best. Where to put the supports is the designer’s version of the question and the moving load is the traffic’s version, and a bridge deck is subject to both at once.

Where the model stops #

One span. Everything here is a simply supported beam. On a continuous beam the influence lines have negative regions, so the worst arrangement puts load in some spans and deliberately leaves others empty — pattern loading — and the search is over subsets as well as positions.

One quantity. The envelope drawn is for bending moment. The shear nobody draws has its own envelope with a different critical arrangement. That one is generally worst with the group as close to a support as it will go, and the two envelopes are rarely produced by the same position.

Elastic and small-displacement. The envelope superposes and compares load cases freely, which requires the response to be linear in the load. Once anything yields — a plastic hinge forming under a heavy axle, say — the arrangements stop being independent and the order in which they arrive begins to matter.

Static. A vehicle crossing a bridge excites it. The dynamic amplification depends on the ratio of the crossing time to the bridge’s natural period, on the roughness of the deck, and on the vehicle’s own suspension frequencies — and for short spans it is not a small correction.

One vehicle. Real load models are more than one vehicle, plus a distributed lane load, plus the arrangement across the width of the deck. The multi-lane problem adds a transverse distribution question the entire two-dimensional analysis has assumed away.

The envelope has a limitation it cannot escape, and it is the one already named: it is not a moment diagram, and it is drawn in the same style as one. A reader who integrates it, differentiates it, or looks for the point of contraflexure on it will find nothing meaningful, because those operations belong to a single load case and the envelope is not one. There is no drawing convention that distinguishes the two, and captions have to carry the whole burden.

The generalisation #

Optimising over a family of load positions, rather than analysing one, is a genuinely different kind of engineering question, and the influence line is the device that converts one into the other: it turns “which arrangement is worst” into “where is this function largest”, which is a question a diagram can answer.

Müller-Breslau’s principle is what makes the conversion cheap even for redundant structures — release the quantity of interest, impose a unit displacement, and the deflected shape is the influence line — and it is a consequence of reciprocity rather than of anything about moving loads. That is a striking piece of economy: a theorem about the symmetry of a stiffness matrix answers a question about lorries.

The generalisation past bridges is to any structure whose worst case is a placement rather than a magnitude. Crane gantries, storage racking, floor systems under partition layouts that are not yet known, and stadium seating under crowds that move — all of them are envelope problems, and all of them share the property that the design case is a contour no single event produces. The same shape of reasoning decides where a column is worst braced and where the worst pattern of floor loading sits, and in each the enumeration is over arrangements rather than magnitudes.

Barré published the construction in 1859 for railway bridges, at a time when the axle spacings of a locomotive were the dominant design variable and the calculation was done by hand for every new engine. It has survived intact because it is exact for point loads on a simple span, and because the alternative — the exhaustive search the figures here perform — was not available to anyone until roughly a century after it was needed.

The ladder from here #

Later rungs on this anchor: influence lines for continuous beams, where the negative regions make pattern loading a search over subsets. Influence surfaces for slabs and decks, where the load can be anywhere in two dimensions. The shear envelope and why it disagrees with the moment envelope about everything. Dynamic amplification, and the ratio that decides it. Fatigue spectra, where what matters is not the worst arrangement but the histogram of all of them. And the codified load model, which is a fictitious vehicle constructed so that its envelope covers the envelope of the real traffic — a design object that exists only as an upper contour, and is the ultimate expression of everything in this essay.