Structural Engineering & Statics
where the load goes
Concept

Continuity — where it appears

The carrying of moment across a support instead of stopping at it, which lowers the peak and costs a stiffness calculation. It drops the mid-span moment by a third and creates a hogging moment where there was none, and it makes the structure's forces depend on the stiffnesses of every span.

Named by 24 essays across 6 fields — each of them below, with the objects they name alongside it.

Where to put the supports. Peak sagging and hogging moment for a uniformly loaded beam, against how far the supports are moved in from the ends. The best arrangement is where the two curves cross, and it is nowhere near the ends.

Where to put the supports, which is not at the ends

Moving the supports of a uniformly loaded beam inward by about a fifth of its length halves the worst bending moment. The load has not changed and nor has the beam.

internal-forces · Support layout 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 30.6 to 19.6, and a hogging moment of 24.5 appears over the supports where there was none.

The moment over the support, and what it buys

Run a beam over its supports instead of stopping at each one, and the mid-span moment falls by a third while a new moment appears where there was none. Nothing was added but continuity.

internal-forces · Continuity 3 continuous spans against 3 simple ones. The bending moment in a continuous beam whose support 1 has settled by 0.01. Three curves: the moment the load causes, the moment the settlement causes on its own — dashed, peaking at 73.5, and in equilibrium with no applied load at all — and their sum, which is what the beam carries, peaking at 73.5 against 24.5 without the settlement. The settlement field is proportional to EI: a stiffer beam is punished harder for the same movement, which is the opposite of every intuition load-carrying gives.

The support that moved

A redundant structure knows things statics cannot see. Settle one support by ten millimetres and a complete set of bending moments appears — in equilibrium with no load at all, and larger for a stiffer beam.

deflection · Indeterminacy The answer arrives in instalments. The hogging moment at support 1 of a three-span beam, cycle by cycle. It starts at the fixed-end moment of 53.3 kNm — the value with every joint clamped — and settles at 64.0 kNm against an exact 64.0. The error falls by about a factor of four per cycle: 14.00, 3.50, 1.95, 0.59 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything.

Solved by passing it around

An indeterminate structure needs simultaneous equations, and for thirty years engineers solved them without writing any down. Clamp every joint, release one, share out what is left over, pass half of it along, and repeat — and the answer walks in, three figures correct after four cycles.

deflection · Moment distribution The props decide where the stress ends up. Bottom-fibre stress in the steel of a 12 m composite beam carrying 12 kN/m of wet concrete and 18 kN/m afterwards. Unpropped, the bare steel takes the first stage alone and reaches 292 MPa; propped, the finished composite section takes everything and reaches 186 MPa — a ratio of 1.57. 62% of the unpropped beam's final stress was locked in before the slab was structural at all. The deflections differ by 1.73 times for the same reason, and no drawing of the finished beam distinguishes the two.

The structure that was never complete

Every analysis in this collection is of a finished structure loaded once. Real ones are built in pieces, and each piece carries whatever was present at the moment it became structural — so the stress in a member depends on when it arrived, which appears nowhere on any drawing.

structures · Construction sequence The moment does not stop at the end of the beam. A portal frame of 8 m by 4 m with fixed bases, carrying 20 kN/m on the beam. The bending moment is drawn on the tension side of every member, and it runs round the corner without a break: 65.2 kNm arrives at the end of the beam and 65.2 kNm leaves down the column, which is the same number, since joint rotational equilibrium is one of the equations the frame solve satisfied. Midspan carries 94.8 kNm, and the two add to 160.0 — the 160.0 kNm of a simply supported span, to 0.0e+0 kNm. The corner takes 61% of the wL²/12 a fully built-in beam would have carried, because the columns are springs rather than walls: the beam-to-column stiffness ratio is 1.27. The beam's moment crosses zero 0.92 m from the corner and the column's 1.33 m above its base.

The moment that goes round the corner

At a rigid knee the bending moment does not stop at the end of the beam. It turns and runs down the column, and in the same instant the beam's shear becomes the column's axial force — while the block of steel that has to carry the turn appears on no member diagram anywhere.

internal-forces · Corner moment A reaction with no load, and the moment it bends the beam with. The prestress moments in a 2-span beam. The primary moment is −P·e, the tendon acting on its own section, and it reaches 540 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 306 kNm — 57% of the primary, with the same sign, so it does not cancel anything. It comes from the middle support refusing to let the beam lift: 51.0 kN pressing down there and 25.5 kN lifting at each end, a reaction set that sums to -2e-13 because nothing external was applied. Its diagram is straight between supports to 2.0e-13% of its own peak, which it has to be: reactions are point forces and a point force puts no curvature in a span.

The prestress that pushes back

On a simply supported beam a tendon is an internal matter and changes no reaction. Put the same beam on three supports and the tendon lifts it off the middle one, the support refuses, and the force it takes to hold the beam down is a reaction produced with no load applied at all.

internal-forces · Secondary prestress The same load, two diagrams, both in equilibrium. One span of a pair of 7 m spans under 5 kN/m, drawn twice. The elastic solution puts 31 kNm over the support and 17 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 21 and 21: the section the beam needs falls from 31 kNm to 21, 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 31 kNm for either — and the second is legitimate for that reason alone. What it costs is 1.0 milliradians of rotation at the support, which the section has to be able to deliver.

The moment that was moved on purpose

The elastic analysis of a continuous beam gives one set of moments. It is not the only set the beam is allowed to have, and taking a smaller one at the support is legal, cheaper, and paid for in a rotation that has to be delivered before the design exists.

internal-forces · Moment redistribution 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 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 The bearing is one length and the web is loaded over another. A load applied over a stiff bearing of 200 mm on the flange of a girder with a 1200 × 8 mm web. The flange bends under it and the yield lines that form spread the load along the web over 659 mm — 3.3 times the bearing, and 70% of the yield resistance is that spread rather than the bearing. The effective length is not a decision anybody made: it is what the flange's own bending stiffness against the web's own strength works out to.

The support that is not a point

A reaction is drawn as a single arrow because the equilibrium equations only need its total. Underneath the arrow is a bearing of some width, delivering a pressure over that width, and almost everything a designer would like to know about the region near a support is a consequence of the width the arrow does not have.

internal-forces · Support width The end bolts do the work and the middle ones very nearly nothing. A lap of 10 bolts at 75 mm pitch transferring 800 kN between two plates, with the force each bolt actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind. The end bolts carry 1.16 of their nominal share and the middle ones 0.89. The reason is not in the bolts: at the leading end the first plate is carrying everything and the second nothing, so the two strain at different rates and the slip between them is largest there. In the middle they strain alike, there is almost no slip, and a bolt with no slip across it transfers almost no force. The mean over the worst is 0.861, and the end bolt has to slip 1.16 mm before the rest catch up.

The joint that has to be as good as the member

A splice exists because members come in lengths and structures do not. It has to deliver the same force, at the same stiffness, in the same distribution across the section, through a discontinuity — and each of those three requirements is met by a different feature of the detail, with the third one usually left to look after itself.

connections · Splice 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 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 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 A reaction with no load, and the moment it bends the beam with. The prestress moments in a 2-span beam. The primary moment is −P·e, the tendon acting on its own section, and it reaches 1440 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 720 kNm — 50% of the primary, with the same sign, so it does not cancel anything. It comes from the middle support refusing to let the beam lift: 102.9 kN pressing down there and 51.4 kN lifting at each end, a reaction set that sums to 0e+0 because nothing external was applied. Its diagram is straight between supports to 3.6e-13% of its own peak, which it has to be: reactions are point forces and a point force puts no curvature in a span.

The tendon that can be moved

Lift a continuous beam's tendon at its interior support without changing its drape and nothing about the beam's total moment changes. The primary falls, the secondary rises by exactly as much, and the pressure line stays where it was — which turns a parasitic effect into a quantity a designer can place.

internal-forces · Secondary prestress The lag follows the shear, so it is worst at the supports. Effective width along the span of a simply supported beam under a uniform load, summed over 25 odd harmonics. Each harmonic has its own half-wavelength L/n and its own, smaller, effective width — 0.815 for the first, 0.400 for the third, 0.269 for the fifth — and near a support the short harmonics carry a bigger share of what little moment there is. So the working fraction is 0.603 at the support against 0.839 at mid-span, a difference of 23.6 percentage points on the same flange. The single sinusoid's answer, 0.815, is drawn as the flat line, and it is only right at mid-span. This is the behaviour a code reproduces by shortening L_e near a support, and the reason it does is that shear lag is driven by w‴, which is the shear force.

The flange works least where the shear is largest

Shear lag is driven by the shear force rather than by the moment, so the effective width of a wide flange is not a property of the beam. It is a function of position along it, worst at the supports, and a single number quoted for a whole span is right at mid-span and nowhere else.

sections · Effective width

Why it converges, and how fast

Moment distribution is an iteration, and iterations do not always converge. This one always does, at a rate the beam's own proportions fix — about a factor of four per cycle on a regular beam and considerably worse on an irregular one, which is where the method's reputation for two cycles being enough comes from and where it stops being true.

deflection · Moment distribution

The joint that was chosen

A joint's stiffness decides how a beam's moment divides between its span and its supports, and the two add to a constant. So there is a stiffness at which they are equal, the beam is sized by the smaller of two numbers rather than the larger of one, and the design moment is half what a simple connection leaves behind.

connections · Joint classification

Told what the far end is doing

Moment distribution discovers, cycle by cycle, that the pinned end of a beam carries no moment — a fact known before any arithmetic started. Telling it instead changes one stiffness from 4EI/L to 3EI/L and the work from thirty numbers to eight, for the identical answer. Cutting the beam on its own axis of symmetry gets it in two.

deflection · Moment distribution

The table that cannot be read halfway

Kani's method converges at exactly the rate moment distribution does, sweep for sweep and digit for digit, because it is the same iteration. What it changes is what is written in the boxes — rotations rather than moments — and that buys a shorter table that repairs its own mistakes and cannot be stopped early.

deflection · Moment distribution

The joint that has to keep turning

A beam on partial-strength joints collapses at 8(Mp + Mj)/L² however stiff the joints are. Stiffness decides only which yields first, and a joint that yields first has to go on rotating at full moment until the span catches up. A weak joint on a stiff connection has the most turning to do — more than a rigid full-strength one.

connections · Joint classification

The column given more than its rectangle

The tributary rule draws a rectangle round each column and hands it whatever stands inside. A floor is continuous over its columns, and a continuous beam does not give each support the load above it — the first interior one takes a quarter more and the end ones a quarter less. On a grid the two directions multiply, and two columns on the same floor differ by a factor of nearly three.

equilibrium · Tributary area

Named alongside it

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

IndeterminacyFree bodyMoment redistributionStiffnessLoad arrangementBending momentMoment distributionCarry-overDistribution factorDuctilityEquilibriumFixed-end moment

All concepts