Structural Engineering & Statics
where the load goes
Concept

Influence line — where it appears

The value of one effect at one station, plotted as a unit load walks the whole structure. It answers where to put a load rather than what a load does, and its negative regions are where adding load reduces the effect.

Named by 10 essays across 5 fields — each of them below, with the objects they name alongside it.

Influence line for the bending moment at x = 3. The bending moment at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 3.00, giving 2.100.

The worst place to stand

A bridge is not designed for a load. It is designed for a load that moves, and for every station along it there is a different position of that load that does the most damage.

internal-forces · Influence line The worst position is not the obvious one. Three axles totalling 320 units, marched across a span of 20 in steps of 0.02. The envelope is the largest moment each station ever sees; its peak is 1160.2 at 9.88 along the span, which is 0.12 off midspan and occurs under the axle nearest the resultant rather than under the heaviest one. Barré's construction, which places midspan halfway between that axle and the resultant, independently gives 1160.3 at 9.88. The dashed curve is the envelope the same total weight would produce as one load rather than three: its peak is 1600.0, which is 38% more — spreading a load out is worth something.

The train that is worse than its heaviest axle

An influence line says where to stand one load. A vehicle is several loads at fixed spacings, and the worst arrangement never puts the heaviest one at the peak.

internal-forces · Influence line Maxwell's reciprocal theorem. A load at one point and the deflection it causes at another, against the same load moved to the second point and the deflection read at the first. Both integrals return 93.3335, and neither calculation was told about the other. The two deflected shapes are entirely different; the two readings are identical.

The theorem that swaps the question round

Push here and measure there; push there and measure here. The two readings are identical, for every elastic structure, whatever its shape — and that fact turns an influence line into something a model can be asked for directly.

deflection · Reciprocity The envelope is not a state of the structure. Every arrangement of the imposed load on three spans — 8 of them, since each span is loaded or not — drawn faintly, with the greatest sagging and greatest hogging at each station drawn over them. Each faint curve is a real state of equilibrium and satisfies the free-moment identity exactly: mid-span ordinate minus the mean of the end moments is wL²/8, to 3e-16 of it. The envelope satisfies it nowhere, missing by up to 23% — because it is assembled from different load cases at different stations and no arrangement of load produces it. seven of the 8 arrangements are needed to build it; the rest never govern anywhere.

The envelope is not a structure

A continuous beam whose imposed load may sit on any span has eight load cases, and every one of them is a genuine state of equilibrium. The curve the design is made against is not one of them — it is assembled from different cases at different stations, and it fails the identity all eight satisfy exactly.

equilibrium · Load arrangement The worst speed is not the fastest one. Peak deck acceleration against train speed, for a 20 m span at 6.25 Hz under 10 axles 18 m apart. The spikes are not a numerical artefact and they are not about how heavy the axles are: a regularly spaced train is a forcing function with a frequency v/d, and where a multiple of it lands on the bridge's own frequency each coach arrives in step with the motion the last one left. The arithmetic is v = d·f₁/k, which puts peaks at 405, 203, 135, 101 km/h — all of them operating speeds. What fails first is the acceleration rather than any stress: ballast loses its interlock at about 3.5 m/s², and strength does not appear in the equation at all. Here the limit is first passed at 376 km/h.

The train that arrives in time with itself

A single load crossing a span is a mild problem. A train is not one load — its axles are regularly spaced, so the forcing has a frequency of its own, and where a multiple of it lands on the bridge's frequency each coach arrives exactly in step with the motion the last one left behind.

dynamics · Moving load resonance Two answers added, and the answer to the two together, drawn on top of each other. A 8 m beam under a 60 kN point load at mid-span (152.38 mm), under 12 kN/m of uniform load (152.38 mm), and under both at once (304.76 mm). The sum of the first two is 304.76 mm, and the residual between it and the third is zero — not small, zero, to the last bit of the arithmetic. That exactness is not a numerical accident: the governing equation is linear in the load, so the response is a linear operator applied to it, and a linear operator distributes over addition by definition. Every calculation that adds one load case to another is standing on that one line.

The addition everything else rests on

Influence lines add, the unit-load method adds, moment distribution adds, load combinations add, and a stiffness matrix is linear by construction. All of it stands on one sentence with three hypotheses in it — and when they fail, two of the failures point in opposite directions.

deflection · Superposition Influence line for the shear force at x = 10.5. The shear force at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 10.56, giving 0.560.

Two diagonals, one of which is absent

A truss diagonal is sized for the shear in its panel, and near mid-span that shear changes sign depending on where the load stands. A member that can only pull cannot carry the reversed case, so the panel gets a second diagonal — and at any instant one of the pair is not there.

structures · Truss Maxwell's reciprocal theorem. A load at one point and the deflection it causes at another, against the same load moved to the second point and the deflection read at the first. Both integrals return 307.5006, and neither calculation was told about the other. The two deflected shapes are entirely different; the two readings are identical.

An influence line is a deflected shape

Finding where a load has to stand to be worst means solving the structure once for every position it could stand in. Reciprocity says the answer is a single deflected shape — release the quantity being asked about, move it by a unit, and the shape the structure takes is the influence line.

deflection · Reciprocity Two lines and a triangle: a three-hinged arch drawn. A three-hinged arch of 20.0 m span, springings at (0.0, 0.0) and (20.0, 0.0) m and the crown hinge at (10.0, 5.0), under 100.0 kN at 5.0 m. The right half carries no load, so it is a two-force member and its reaction lies along the line from its springing through the crown hinge. That line meets the load's line at K, 7.50 m up, and the left reaction must pass through K too. The triangle of the load and the two reaction directions gives the reactions as 90.1 kN at A and 55.9 kN at B, with a horizontal thrust of 50.0 kN — the values four equilibrium equations return, to 7e-15 kN. The shape of the rib entered nowhere.

The arch that is only its three hinges

A three-hinged arch's reactions come from three points and nothing else, so a parabola, a circle and a portal frame on the same hinges push on their abutments identically. Move a load across and the point where the reactions cross runs along two straight lines through the crown. That is the arch's influence line, drawn with a straightedge — and friction in the hinges it was built around blurs it.

equilibrium · Graphic statics Putting a funicular through three points. Four loads — 40.0 kN at 3.0 m, 60.0 kN at 7.0 m, 30.0 kN at 12.0 m, 50.0 kN at 16.0 m — and three points the polygon must pass through: A and B at the springings and C, 5.0 m above their chord at 10.0 m. A trial pole, dashed, draws a polygon from A that ends 6.39 m below B. The ray through the trial pole parallel to its own closing line cuts the load line at Q, 95.0 kN from the top, which is the left reaction of a simple beam on A and B and does not depend on the pole at all. Every pole whose polygon passes through A and B lies on the line through Q parallel to A B; the one whose polygon also reaches C is 98.0 kN from the load line, which is the moment at C of that simple beam, 490.0 kN·m, divided by C's height above the chord. That is the three-hinged arch's thrust — the four equilibrium equations give 98.0 kN.

The polygon that runs out of freedom

A funicular polygon for given loads has exactly three freedoms, so it can be made to pass through three chosen points and no more. Three points is a three-hinged arch, and the drawing solves it. Take the crown hinge away and one freedom is left over. The drawing then offers a whole family of thrust lines and cannot say which one the arch uses — the rib's stiffness decides, and stiffness is not on the paper.

equilibrium · Graphic statics

Named alongside it

The objects these essays reach for when they reach for this one.

Load arrangementPattern loadingSuperpositionFree bodyIndeterminacyArchAxle trainBending momentDeterminacyEquilibriumFlexibilityGraphic statics

All concepts