Structural Engineering & Statics
where the load goes
Concept

Load arrangement — where it appears

The pattern in which a variable load is placed on a structure, chosen to maximise one action rather than to represent a state. A continuous member has to be checked for several arrangements, and the envelope of their results is in equilibrium with none of them.

Named by 14 essays across 6 fields — each of them below, with the objects they name alongside it.

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 tie is a redundancy, so its stiffness decides the thrust. Thrust and rib bending for a 60 m tied arch of 0.15 rise ratio, against the stiffness of its tie. Cut the tie and the structure is a curved simply supported beam, so the tie force is the one redundant and the force method gives it: with a rigid tie the answer is 2227 kN, within 1.0 per cent of the funicular wL²/8f, and the shortfall is the arch's own axial shortening. A real tie stretches 68 mm and returns 2166 kN — 2.7 per cent of the flexibility is the tie — and whatever thrust the arch does not get, it carries as bending: 760 kNm at 30 m. A tenth of the tie stiffness is not a tenth of the problem; it is a different structure.

The thrust that never reaches the ground

Every arch on this site has ended at the same sentence — the foundation is where an arch is really decided. A tie changes the sentence without changing the arithmetic: the horizontal force is still there, still the same size, and it now closes on itself through a bar at deck level.

structures · Tied arch The snow that left the roof is standing against the wall. A roof in section with a 1.0 m obstruction at its downwind end, drawn with the vertical scale exaggerated 4 times because a metre of snow on twenty metres of roof is thinner than the line it would be drawn with. The balanced layer is 0.48 kN/m² everywhere; the wedge against the wall holds the snow that 30% of the 20 m upwind gave up, so its area is fixed by conservation of mass rather than chosen. That makes it 0.85 m deep and 3.4 m long, with a peak load of 2.60 kN/m² — 5.4 times the balanced value, and still only 27% of the snow on the roof. An intensity several times the design load, made of a quantity nobody would notice had moved.

The load that arrives where the wind stops

Snow is the one load a structure is given rather than subjected to. What falls is spread evenly over a whole region; what a member carries is whatever the wind left above it, and the wind piles it against whatever gets in the way. The heaviest patch on a roof is usually a quarter of the snow on it.

equilibrium · Snow drift A Pratt truss of 8 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing.

The tie that spends an afternoon as a strut

A tension member is chosen by its area and nothing else. A compression member is chosen by how that area is arranged. So a member whose force reverses under some load case is not merely being asked for the same number with the other sign — it is being designed against a different variable, and the same steel can carry seventy times more or less depending on a shape nobody chose for that purpose.

stability · Load reversal Two of these move and the third cannot. The first span of a 3-span beam under 5 kN/m, with the stiffness of the middle span swept over a factor of 25. The support moment and the mid-span moment both move — that is what redundancy does, and it is the whole reason a continuous beam has to be analysed rather than read off. Their combination does not: the mid-span ordinate plus the average of the two end moments is 30.6 kNm at every point on this axis, which is wL²/8 for this span and this load and contains nothing else. The largest departure anywhere on the sweep is 2.3e-16 of the value, which is the arithmetic of the stiffness solution rather than a property of the beam. Continuity buys a distribution and not a capacity, and this is the line that says so.

Two of these move and the third cannot

Cut one span of a continuous beam free and add up the forces on it. What comes out is that the mid-span moment plus the average of the two end moments equals the free bending moment of that span, with nothing else in it — no stiffness, no support settlement, no analysis at all. Continuity moves moment about. It does not reduce the total, and it never has.

internal-forces · Static moment The prop load is decided by the digging, not by the hole. Prop forces in a 12 m excavation propped at 3 levels, with the force each prop reaches at any stage of the sequence drawn thick and the force the finished arrangement gives it drawn thin. The pale dots are the individual stages. The middle prop reaches 214 kN while the dig is at 9.5 m and finishes at 88 — a factor of 2.44 between the two, and the larger one is not in the final analysis anywhere. Terzaghi and Peck's apparent pressure diagram, a rectangle of 49.4 kN/m², reproduces the total of the staged maxima to 2% — which is what it is: an envelope of measured prop loads, back-figured into a pressure, and a shape nothing on a wall is ever loaded with.

Every prop has its own worst day

A braced excavation has no finished state worth analysing. It is dug in stages, a level of props goes in at each stage, and a prop's force is largely fixed the moment it is installed — so the force to design it for is the largest it sees during a sequence that appears on no calculation sheet, and which for the middle prop here is nearly two and a half times what the finished arrangement gives.

structures · Propped excavation 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 Cross the hangers and the chords stop bending. The same tied arch, the same sixteen hangers, the same load on half the span — hung vertically and hung as a network. Vertical hangers make the two chords a Vierendeel frame, which has no truss action at all, so a partial load is carried by bending: 3316 kNm in the tie and 6234 in the arch. Inclined hangers can carry the shear between the chords axially, and the same load gives 686 and 831 — factors of 4.8 and 7.5. The thrust is identical in both, because that is decided by the span and the rise and nothing else.

Cross the hangers and the bending goes

A tied arch with vertical hangers is a Vierendeel frame with a curved top chord — it has no truss action at all, so a load on half the span is carried by bending. Incline the hangers so they cross and the same two chords become a truss.

structures · Network arch Every section is hogged and sagged before the bridge exists. The bending moment envelope of a launched deck, section by section along its own length, taken over every position of the launch. In service each section has one sign; during the launch 100 per cent of them see both, because each passes over every pier and through every span on its way out. The worst launch moment is 42568 kNm against 40500 in service, and with no launching nose at all it would be 185977. That is why a launched bridge is a constant-depth box with symmetric flanges: the design case is not a load, it is a history.

Every section was somewhere else

A bridge pushed out over its piers subjects each of its cross-sections to a history rather than to a load case. Every one passes over every support and through every span, so the design envelope is the envelope of envelopes — and no in-service condition produces it.

structures · Launched bridge 3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 28.1 to 18.0, and a hogging moment of 22.5 appears over the supports where there was none.

The chord is a continuous beam

A truss analysis reports one number per member, and for a chord running through eight panels that number is an axial force with no bending in it. The chord is a single piece of steel spanning between its own nodes, and anything landing between them bends it.

structures · Truss 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 The same load, two diagrams, both in equilibrium. One span of a pair of 9 m spans under 30 kN/m, drawn twice. The elastic solution puts 304 kNm over the support and 171 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 213 and 207: the section the beam needs falls from 304 kNm to 213, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 304 kNm for either — and the second is legitimate for that reason alone. What it costs is 13.0 milliradians of rotation at the support, which the section has to be able to deliver.

The moment that was shed has to land

Redistribution takes a moment off a beam's support and pays for it with rotation. On a beam that is the whole story. In a frame the support is a column, the shed moment does not vanish, and it arrives at a member whose section was chosen from the diagram it has just left.

internal-forces · Moment redistribution 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 Settling down or walking away, cycle by cycle. The total plastic hinge rotation of the beam after each cycle of loading — span 1, both, span 2, neither — with the midspan load at 0.98, 1.02, 1.05, 1.10 times the shakedown load of 126.3 kN. At 0.98 it stops at 0.59 mrad. At 1.02 it grows 4.57 mrad a cycle. At 1.05 it grows 11.43 mrad a cycle. At 1.10 it grows 22.86 mrad a cycle. Nothing collapses in any single cycle; above the shakedown load the beam walks.

The load it can carry once

A two-span beam whose loads come and go span by span collapses at 150 kN under any one arrangement, and walks at 127. Between the two it can carry every arrangement once and none of them forever: each cycle leaves a few more milliradians of rotation at the support and a midspan fifteen millimetres lower. Melan's theorem finds the limit as the last residual moment line that fits, Koiter's as a mechanism no single load state can drive, and a cycle-by-cycle calculation walks exactly where both say it will.

materials · Shakedown

Named alongside it

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

ContinuityFree bodyMoment redistributionBending momentEnvelopeInfluence lineEquilibriumFree momentIndeterminacyLower-bound theoremPlastic hingeStiffness

All concepts