Structural Engineering & Statics
where the load goes
Concept

Stiffness method — where it appears

Writing one equation per freedom and solving for displacements, with member stiffnesses scattered into a matrix indexed by the freedoms they touch. It adds no physics to the hand methods it replaced; the whole of the invention is the bookkeeping, which is why a machine can do it.

Named by 12 essays across 4 fields — each of them below, with the objects they name alongside it.

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 the null vector of the equilibrium matrix 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 Two different structures released, and one bending moment diagram. The bending moment in a continuous beam of 8, 10, 8 m under 12 kN/m, solved twice by the force method with different redundants. The first release puts a hinge over each interior support, so the released structure is a row of simple spans and the redundants are moments. The second removes each interior support, so the released structure is one simple span of the whole length and the redundants are reactions. The two released structures have nothing in common — different shapes, different deflections, different everything — and the diagrams they produce lie on top of each other to 9e-15 of the peak moment. Which restraints are released is a choice about the arithmetic and not about the structure, which is a fact worth trusting: it means a hand calculation can pick whichever release makes the sums easiest and be sure of the answer.

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 check that everything adds up, and the error it cannot see. Four versions of the same 3-bay, 4-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 8% and 32%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 24% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure.

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 hinge is at mid-height in exactly no storey. The height of the point of contraflexure in each column of a 4-storey, 3-bay frame under lateral load, as a fraction of the storey height, against the portal method's assumption that it is at the middle. The exact solution is a plane-frame stiffness analysis of the same frame. In the bottom storey the zero sits at 0.609 of the height, because a fixed base is stiffer than the joint above it and takes more of the column's moment; in the top storey it sits at 0.359, because there is no column above to share that joint. The average over the whole frame is 0.475, which is why the assumption survives — it is right on average and wrong everywhere. The worst error in the storey shear share is 25%, and the column shears still add to the storey shear to 0e+0 of it, because the method is exact statics applied to an assumed structure.

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 same curve, computed twice and from opposite ends. A cantilever of 6 m, with its deflected shape drawn from a double integration of M/EI and the bending moment of its conjugate beam drawn on top of it. The conjugate is the same span carrying M/EI as a load, with its supports transformed — a real fixed end becomes a free end and a real free end becomes a fixed one, because a fixed end has no slope and no deflection and the conjugate therefore needs no shear and no moment there. The two curves agree to 4.7e-6 of the largest deflection, which is the trapezium rule and not the method. The largest deflection is 36.00 mm against the closed form's 36.00. The reaction of the conjugate beam is -9.000 milliradians, which is the real beam's rotation at that support — so the whole of a slope calculation is one reaction.

The beam whose moment is a deflection

A beam's bending moment is the second integral of its load. Its deflection is the second integral of M/EI. They are the same problem, so a deflection can be found by loading a fictitious beam with M/EI and asking a statics question — and the only thing to remember is the supports, which are not remembered but derived, one boundary condition at a time.

deflection · Conjugate beam 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 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 98.0 kNm — the value with every joint clamped — and settles at 156.9 kNm against an exact 156.9. The error falls by about a factor of four per cycle: 21.03, 5.92, 1.54, 0.60 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything.

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 Two load sets with the same resultant and different work. The two ways of putting a uniform load of 10 kN/m onto a beam element 6.00 m long. Both put 30.0 kN at each node, so both have the same resultant and the same moment about any point — they are equivalent for a rigid body. The consistent set adds a couple of 30.0 kN·m at each end, in opposite senses, which is what makes it do the same virtual work over the element's shape functions as the real load does. The couples cancel in the resultant, which is exactly why the resultant cannot see them, and they are the whole difference between an exact answer and one that is a third out.

Equivalent in work, not in resultant

Two force systems with the same resultant and the same moment about every point are interchangeable — for a rigid body. A finite element is not a rigid body, and substituting one for the other on a beam element leaves the tip of a cantilever a third too low with no warning of any kind.

equilibrium · Force couple Every number in the table is a rotation, and none of them is a moment. The rotation contributions of a three-span beam of 8, 10, 8 m under 24 kN/m, sweep by sweep. There are six of them, one per member end, and not one is a bending moment: the moment is assembled at the end from M = FEM + 2m′ + m′ of the far end, and until that is done the table holds quantities that mean nothing on their own. That is the trade. A moment distribution stopped after two cycles hands over moments that are wrong by a known amount and can be used; this table stopped after two sweeps hands over nothing that can be read at all — and it gets there in four sweeps against Cross's own count on the same beam, writing six numbers a sweep rather than one per distributed member end plus a carry-over.

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 whole deflected shape, found from the members' changes of length. A Pratt truss of eight panels at a depth of 1, carrying 15 kN at each top joint, at EA = 1, drawn as built and in its deflected shape, with every movement magnified the same number of times. The shape comes from Williot's construction and Mohr's correction: every joint's movement from the members' extensions alone, less the rigid rotation the supports forbid. The mid-span joint L4 moves down 2019.41 units, and the dots at every joint are a stiffness solution sharing none of that arithmetic, agreeing to 1e-14 of the largest movement. One construction gives all sixteen joints at once.

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 Soft in shear, the wall gives up moment to the span. The bending moment along a 300 × 600 rectangle fixed at its left end and propped at its right, under a uniform load, per unit load and span, at span-to-depth ratios of 4, 2, 1, and for the same beam treated as rigid in shear, dashed. Rigid in shear the wall carries 0.125 wL² and the prop 0.375 of the load. Counting the shear it deforms by, the wall moment falls to 0.119 wL² at 4, 0.105 wL² at 2, 0.070 wL² at 1, and the prop's share rises to 0.381, 0.395, 0.430. The load has not changed and the beam is no weaker: its forces have moved, because in a redundant beam they come from how it deforms.

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

Named alongside it

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

CompatibilityEquilibriumFree bodyIndeterminacyContinuityDeflectionMoment distributionVirtual workConvergenceDegrees of freedomDeterminacyFixed-end moment

All concepts