Concept

Compatibility — where it appears

The requirement that a structure's parts still fit together after they have deformed, which supplies the equations equilibrium is short of. It is what an indeterminate structure needs in addition to equilibrium, and it is the reason such a structure's forces contain a stiffness while a determinate one's do not.

Named by 42 essays across 8 fields — each of them below, with the objects they name alongside it.

Counting unknowns against equations. Three frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint: fewer unknowns than equations is a mechanism, equal is solvable by statics, more needs stiffness.

Counting the unknowns, and finding out whether statics can answer

Two equations per joint, one unknown per member, one per restraint. Subtract, and the sign of the answer says whether the structure is a mechanism, solvable, or beyond what equilibrium alone can settle.

equilibrium · Determinacy
Where the peak moment goes when a restraint is added. The same uniformly loaded beam with three sets of restraints, and the bending moment in each. Adding restraint moves moment from mid-span to the supports and lowers the peak — but only the first case can be solved by statics.

One support too many, and what it costs to know

Add a redundant restraint and the load has two routes to the ground. Equilibrium cannot say how it splits, and the answer turns out to depend on stiffness — which is a different kind of question.

deflection · Indeterminacy
The deflection at x = 4, by virtual work. Three diagrams: the moment from the real load, the moment from a unit load placed where the answer is wanted, and their product. The area under the third, divided by EI, is the deflection — 213.33 here. No standard case was consulted, so the method works for any load pattern at all.

One deflection, without solving everything

To find how far one point of a structure moves, put an imaginary force of one unit there, multiply two moment diagrams together, and integrate. The answer arrives without ever solving for the deflected shape.

deflection · Virtual work
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
Two beams tied together, and the deeper one takes 89% of the load. Two simply supported beams of 6 m, one twice as deep as the other, tied together at midspan so that they have to move as one. A load of 100 kN stands on the tie. Point stiffness is 48EI/L³, so the deeper beam is 8 times as stiff — depth cubed, nothing else — and the load divides in that ratio: 11.1 kN into the shallow beam and 88.9 kN into the deep one, 11% against 89%. Both midspan points move 5.00 mm, which is the whole of the argument: the geometry of the load never entered it. The deflection is drawn 78 times full size — the real sag is 5.00 mm on a 6 m span, about 1 in 1200.

The stiffest path takes the load

When two members share a force the split can be argued about. When they share a displacement it cannot — stiffness settles it, and nothing about the load or the plan drawing gets a vote. The consequence is that stiffening a lightly loaded member raises its stress, and the way to unload something is to soften it.

internal-forces · Load-sharing
The ground pushes back hardest where the beam has gone down furthest. A strip 16.4 m long and 1 m wide on ground of subgrade modulus 50 × 10³ kN/m³, carrying 1000 kN at its centre. The beam settles 3.90 mm under the load and the ground pushes back in proportion — the arrows are k times the settlement above them, peaking at 195 kN per metre — so the pressure diagram is the settlement bowl and not an assumed distribution. The characteristic length 1/β is 2.56 m: the bowl crosses zero at 6.04 m, which is 3π/4 of it, and beyond that the arrows reverse because the beam has lifted off. By 8.05 m — one π/β — the disturbance is 4.3% of what it was, which is why the moment 641 kNm and the peak pressure 195 kN/m contain no length at all. The settlement is drawn 217 times full size — the real bowl is 3.90 mm deep over 16.4 m, about 1 in 4207 — and at true scale the beam would be a straight line.

The beam that sits on the ground

Every other beam in this collection is held at points. A footing is held everywhere, by something that pushes back in proportion to how far it is pushed — and that single change hands the structure a length it did not choose. Two or three of those lengths from the column, nothing knows the load happened.

internal-forces · Elastic foundation
The further it deflects, the harder it pulls back. Total load against midspan sag for a 30 m cable of 1000 mm² prestressed to 500 kN, carrying 5 kN/m. The cubic H³ − T₀H² − w²L²EA/24 = 0 was bisected at every point of the curve, so the sag at the full 150 kN is 0.740 m rather than the 1.125 m the flat-cable formula WL/8T₀ gives — the straight dashed line, which is the tangent to this curve at the origin and nothing more. Its slope is the initial stiffness 8T₀/L = 133.3 kN/m; at the marked point the tangent has reached 341.2 kN/m, 2.56 times as stiff, and the horizontal component of the tension has risen from 500 kN to 760 kN. Nothing about the steel changed. The geometry got better at the job.

The stiffness that comes from the shape

A cable has no bending stiffness whatever, and it still holds up a roof. What resists the load is the change of its own geometry, so its stiffness is a function of the tension already in it — and prestress buys stiffness that no change of material could.

structures · Cable stiffness
The hoops change their mind at an angle no proportion chose. The two membrane forces of a spherical dome of radius 30 m under 3 kN/m² of surface, taken from the crown to a base at 60°. The free body for the meridional force is the cap above a cone of half-angle φ, and vertical equilibrium of it gives N_φ = −wR/(1 + cos φ) directly: -45.0 kN/m at the crown falling to -60.0 at the base, compression everywhere. Equilibrium normal to the surface then gives the hoop force, which starts at -45.0 kN/m and reaches 15.0 — it changes sign, and the angle at which it does was found here by bisecting N_θ rather than quoted: 51.827292°. Setting N_θ = 0 gives cos²φ + cos φ − 1 = 0, so cos φ is (√5 − 1)/2, the reciprocal of the golden ratio — an identity this site's solver gate checks against the bisection to nine decimals rather than asserting, because it is too pretty to be believed on sight. Below that parallel the hoops are in tension, which masonry has none of, and that is where every old dome is cracked.

The surface that carries by being curved

A flat plate spends its thickness on a lever arm of a few millimetres. Curve the same sheet and the load is carried in the surface itself, at a thirtieth of the thickness — and the entire bill is presented at the edges.

structures · Shell action
The depth is decided by how far it moves, not by what it can carry. A column carrying 6000 kN landing 3 m into a 12 m transfer member. The free body is the member itself, cut under the column: M = P·a(L − a)/L = 13500 kNm, with 4500 kN of shear on one side of the cut and 1500 on the other. At an allowable stress that moment asks for 1.94 m of depth — the dashed outline — and keeping the settlement it causes inside the floors' own bending asks for 2.55 m, which is the member drawn solid. 31% more depth is bought by nothing the strength calculation can see. The depth grows as √(P·a), so four times the load is exactly twice the depth, and depth in a transfer member is a storey nobody occupies.

The column that stops

A load path that runs straight to the ground costs almost nothing. Interrupting one costs depth in proportion to the square root of the load times the distance it is moved — and the interruption's own deflection becomes the settlement of everything standing on it.

structures · Transfer structure
The same restraint, twice, with opposite signs. A 4 m strip of 200 mm slab whose ends cannot move apart, against deflection measured in its own thicknesses. The flat line is what a yield-line calculation gives, which is what the same strip would carry if its ends were free: 30.0 per unit width. The rising branch is compressive membrane action — the deflected strip is forced into an arch — and it peaks at 116.6, which is 3.89 times the yield-line load, at a deflection of 0.24 of the thickness. Past that the arch runs out of depth and the load falls back to the flexural one; past a deflection of one thickness there is no arch left and the reinforcement starts carrying the strip as a cable. It gets back to the arch's load at 2.17 thicknesses, which is one part in 9 of the span — a sag nobody would design for and exactly what a floor does instead of falling.

The force nobody put in the model

A slab strip whose ends cannot move apart is not the strip in the yield-line calculation. Deflecting shortens the chord between its ends, the ends do not come in, and the strip is forced into an arch — worth four times the load it was designed for, at a movement nobody would see.

internal-forces · Membrane action
The plan a straight beam does not have. A beam of radius 12 m turning through 60°, seen from above, with bending drawn outward from the axis in one colour and torsion in the other. The load is vertical and uniform and nothing is applied off the axis. Bending reaches 446 and torsion 155; the two peaks are in different places, which is why the section has to be chosen for a combination rather than for either.

Bending that arrives as twist

A straight beam under a vertical load carries no torsion unless something applies one. A beam whose axis curves on plan carries torsion everywhere, from the same load, with nothing applied off the axis — and it cannot be simply supported at all.

internal-forces · Curved in plan
A couple applied to the core, and two columns to make it. A 20-storey core with one outrigger at 59% of its height. The arm is stiff in bending and the perimeter columns are stiff in tension and compression, so between them they resist the core's rotation at that level — a couple of 35283 kNm here, carried as a 294 kN pair in the columns at 120 m centres. The compatibility is one equation: the core's rotation at that level, less what the couple takes back out of it, equals the rotation the arm and its columns allow. The top drift falls from 360 mm to 74, which is 80% of it, and the base moment from 73500 to 38217 kNm. The deflected shape is drawn hugely exaggerated: the real top drift is about one five-hundredth of the height.

The arm that makes the columns work

The perimeter columns of a tall building are already there, already carrying gravity, and already the furthest thing from the centre. They take almost none of the overturning, because a floor slab transmits shear and not moment — and one storey-deep arm at the right height changes that by nearly a half.

structures · Outrigger
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
Between two beams and one, and much nearer one. How composite a beam is, against the one dimensionless group that decides it: αL, where α² = K·EI∞/(EA*·EI₀). At αL = 0 the layers slide freely and the beam is two beams; past about 20 the connection is stiff enough that the last per cent is unbuyable. The beam drawn sits at αL = 15.8 and is 96% composite, deflecting 36.7 mm against 33.9 for full interaction and 110 for none. The curve is steep where a real design sits, which is why halving the number of studs does not halve anything.

Half the studs, and most of the beam

Bonding two layers together quadruples the stiffness of the pair. A real connection is a row of studs that deform, so the layers slide a little and the beam sits between the two answers — but not halfway between, and the number that decides where is a single dimensionless group.

internal-forces · Partial interaction
One coefficient, and nothing else in it. The deflected shapes of one beam under four load cases, each scaled so that its mid-span deflection is the same, with the tangent at the left-hand support drawn on each. The end rotation is that deflection times a coefficient that depends only on the shape of the load: 3.20 for a uniform load, 3.00 for a load at mid-span, 2.99 for a triangular load, 3.60 for a load on half the span. Every material property, every second moment and the span itself cancel out of the ratio θL/δ, so a beam at any deflection limit has an end rotation that is known before anything about it is: at L/360 it is 8.89 milliradians, or 0.51 of a degree.

The angle nobody limits

Every serviceability rule in this collection limits a displacement. What a bearing, a joint and a cladding gap actually have to accommodate is an angle — and the angle is locked to the displacement by a coefficient that contains no material, no section and no span.

deflection · End rotation
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 2229 kN, within 0.9 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 2168 kN — 2.7 per cent of the flexibility is the tie — and whatever thrust the arch does not get, it carries as bending: 739 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 end bolts do the work and the middle ones very nearly nothing. A lap of 8 bolts at 70 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 1.09 of their nominal share and the middle ones 0.94. 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 0.918, and the end bolt has to slip 1.36 mm before the rest catch up.

The bolts that do not share

Every bolted connection in this collection has divided a force by a number of bolts. That is right for a short joint and wrong for a long one, and the reason has nothing to do with the bolts — it is that the plates they join are elastic, and stretch by different amounts at different points along the lap.

connections · Long-joint
Almost all of it is exactly zero. The stiffness matrix of a 2-bay, 3-storey plane frame: 36 freedoms, of which 16.2 per cent of the 36² entries are non-zero. The zeros are not small numbers; they are absences. A member reaches only the two nodes at its ends, so it can contribute nothing to any row belonging to a node it does not touch, and every such entry is zero exactly rather than nearly. The non-zeros therefore sit in a band of width 13 about the diagonal. Before the supports are applied the matrix is singular, and its null space has exactly three dimensions — the three rigid-body motions a plane frame has with respect to the ground, which is the same statement nullVector makes about a truss that is a mechanism, arrived at from the other end.

The matrix that replaced the hand methods

Moment distribution passes moments round a frame until they stop moving. Virtual work computes one deflection at a time. Both are exact and both stop scaling in the low tens of members. What replaced them adds no physics at all — the whole of the invention is the bookkeeping.

deflection · Stiffness method

Choose what to take away

The other machine for a redundant structure works by removing restraints until what is left can be solved by statics, then putting back exactly enough force to close the gaps that opened. Which restraints are removed does not change the answer at all, and changes the arithmetic completely — one choice gives a tridiagonal matrix a person can solve on paper, and another gives a full one.

deflection · Force method

The stiffness the load takes away

Buckling is usually taught as an event — a critical load, a bifurcation, a mode. Written as a matrix it stops being an event at all. A compressive load subtracts a stiffness from the structure, the subtraction grows with the load, and the critical load is simply where what is left reaches zero.

stability · Geometric stiffness

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 moment the beam left behind

A beam reaction is drawn arriving on a column's centreline. It arrives on a cleat a hundred millimetres out from the face, and the difference is a couple that goes into the column and has to be shared between the lengths above and below it. Nothing about it appears in a frame model whose members meet at nodes.

internal-forces · Column eccentricity

The curvature nobody applied

Concrete shrinks as it dries, by about half a millimetre in every metre. In a symmetrically reinforced member that is a shortening and nothing else. In a member with more steel in one face than the other — which is every beam and every slab — the steel holds one side back and the section bends, with no load on it at all.

deflection · Shrinkage curvature

The answer that depends on how it was divided

Every computed answer in this collection came out of a structure chopped into pieces — elements, strips, stations, trial positions. The chopping is invisible in the result and it is not neutral: some divisions give the exact answer, some give one that is always too stiff, and one of them changes nothing but the cost of getting there.

deflection · Discretisation

The check that cannot see the error

Every analysis prints a global equilibrium residual, and it is the first thing anybody looks at. It catches a lost restraint and a load entered in the wrong unit immediately. It is structurally incapable of catching a member whose stiffness is wrong by a factor of ten, because the wrong answer is still in equilibrium with the same loads.

equilibrium · Equilibrium check

The analysis that assumes the answer

A rigid frame is indeterminate, so statics cannot finish it. The hand methods finish it anyway, by assuming where the bending moment is zero and treating those points as hinges. That is not a shortcut around the analysis — it is a different kind of answer, exact in equilibrium and wrong in compatibility, and knowing which half is which is what makes the error a bound rather than a mystery.

internal-forces · Portal method

The length a structure was never given

A disturbance applied at one place dies out over a distance, and the distance is not something anybody chose. A beam forgets a badly applied load over its own depth. A beam on the ground forgets a point load over the fourth root of its stiffness against the soil's. A shell forgets a held edge over the square root of the radius times the thickness — a geometric mean of two lengths three orders of magnitude apart, which is neither of them and is not near either.

internal-forces · Edge disturbance

Built to the wrong length

A redundant structure's members do not have independent lengths. Choose all but one and geometry decides the last, so a member made a different length has to be pulled or pushed into place — and the force required stays in the structure for as long as the structure does. Nothing has been applied to it, there is no load case and no factor, and the members are carrying real force.

connections · Fit-up

The ground that hangs on instead of holding up

A pile is driven through fill that has not finished settling. The fill goes down past the shaft, the friction along that length turns round, and the pile is now carrying the soil rather than the other way about. The worst force is not at the head and not at the toe, and nothing at that depth can be seen.

internal-forces · Downdrag

Two cells, one equation, and a web with nothing in it

Bredt's formula answers a single closed cell because a single closed cell has one unknown and one equation. Put a web down the middle and there are two unknowns and still one equation — and the answer, when the missing statement is supplied, is that the new web carries exactly nothing.

sections · Multicell torsion

The coincidence reinforced concrete stands on

Steel expands at twelve microstrain per degree and concrete at ten. Nobody chose either number, they are not equal, and the seventeen per cent between them is the smallest mismatch of any pair of materials engineering bonds together — which is the reason the most-used structural material on earth does not tear itself apart every summer.

materials · Thermal compatibility

The tension that was left out

A suspension bridge's deck sits on a cable pulling hard along it, and a member with a large tension in it is stiffened by that tension. Leaving the term out of the deck's own equilibrium is what elastic theory does, and on a long span it asks for fourteen times the girder.

structures · Stiffening girder

What the second arm is worth

One outrigger at its best height removes five sixths of a tall core's drift, which sounds like the end of the argument. A second removes half of what is left, a third half of that, and each of them costs a storey of the most valuable floor area in the building — so the question is not where to put an outrigger but how many the arithmetic still justifies.

structures · Outrigger

The drawing that is right except for a rotation

Williot's construction finds every joint of a truss from its members' changes of length alone, in one drawing — and puts the roller six thousand units off its support. The error is one rigid rotation, Mohr's diagram takes it away, and a drawing started from the member symmetry holds still never makes it.

deflection · Truss deflection

The shear that moves the moments

In a statically determinate beam, shear deformation adds movement and changes no force. In a redundant one the forces come from how the beam deforms, so a stubby beam soft in shear carries less moment at its wall and more in its span — and the carry-over factor, the one half every hand method passes along, falls to nothing at φ = 2 and then changes sign.

deflection · Shear deflection

The fixed-end moment is a column stress

A fixed end forbids a change of slope and a deviation, so on a member fixed at both ends Mohr's two theorems both have the answer nothing. Written with 1/EI as a width, those two conditions are the P/A + My/I of a short column under an eccentric load — the end moments are its edge stresses, and a haunched member's are read from where its elastic centre has moved.

deflection · Moment-area

The elastic centre is not on the frame

Cross's analogy turns a member fixed at both ends into a short column and its end moments into edge stresses. A closed frame's analogous column is the frame's own outline, its elastic centre is a point hanging in mid-air inside it, and the horizontal thrust a gravity load produces is that section's bending stress about the axis through that point.

deflection · Moment-area

Any structure will carry the unit load

Virtual work has two readings and each is free exactly where the other is bound. A unit load needs only something to stand on in equilibrium, so the deflection of a beam statics cannot solve comes out of a cantilever statics can. A virtual displacement needs only to fit together, so a reaction comes out of pushing a mechanism — and on a redundant beam the unknown cancels out of the equation and nothing is found at all.

deflection · Virtual work

The other area under the curve

Castigliano's theorem says a deflection is the derivative of the strain energy with respect to the load, and it is true only while the material is linear. Past that, the right energy is the area on the other side of the stress–strain curve. On two aluminium bars at their proof stress the strain energy gives a deflection four times too large, and on a redundant truss minimising it picks a set of forces in perfect equilibrium that no deformed shape can produce.

deflection · Strain energy

The load that is not a load

Settle one foot of a portal frame by ten millimetres and the frame develops moments with nothing applied to it anywhere. In the analogous column the case is simpler than a load case, not harder — the section carries no direct stress at all, and the whole answer is one bending stress. And it scales the wrong way: the moments are proportional to EI, so the stiffer the frame, the more a settlement costs it.

deflection · Moment-area

Three, and what three is a property of

The column analogy works because a closed ring cut once has three redundants and a plane section has three stress resultants. That match is the whole method, and it is topological rather than geometric — a portal, a pitch, a step, a splay and a polygonised arch are all one ring and all exact. Add a second bay and the frame has six redundants with nothing to be a drawing of, and the outer ring on its own is out by 190 per cent.

deflection · Moment-area

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.

StiffnessIndeterminacyBending momentEquilibriumFree bodyLoad pathVirtual workForce methodStiffness methodDeflectionFlexural rigidityFunicular

All concepts