Equivalent in work, not in resultant

Moving a force establishes the rule that makes most of statics possible: a force can be moved anywhere as long as a couple is added to pay for it, and two systems with the same resultant and the same moment about any point are interchangeable.

That rule has a condition attached, and the condition is usually invisible because it is nearly always satisfied. The two systems are interchangeable for a rigid body. They are not interchangeable for anything that deforms, and every structural model is made of things that deform.

The couples are the difference between an exact answer and one that is a third out.

Which free body produced the number #

The free body is one beam element, cut at its two nodes, and the question is what set of nodal forces stands in for the load along it.

The requirement is not equilibrium. Both sets satisfy equilibrium, which is the trap. The requirement is that the substitute does the same work as the real load over any displacement the element can adopt, because the stiffness method is a statement about energy: the nodal forces are the derivatives of the strain energy with respect to the nodal displacements, and a load enters through the work it does.

An element’s displacements are not arbitrary. They are restricted to what its shape functions can produce — for a cubic beam element, the four Hermite polynomials that give unit displacement or unit rotation at one node with the other three zero. So the condition is four equations:

$$F_i = \int_0^h w(x)\,N_i(x)\,dx$$

one for each nodal freedom, with $N_i$ the shape function belonging to it. For a uniform load on a cubic element that integrates to $wh/2$ at each translation and $\pm wh^2/12$ at each rotation.

The rotational freedoms are the ones the resultant argument has no answer for. A resultant is a force and a moment about a point; the work a load does through a nodal rotation is a different question, and lumping the load onto the translations answers it with zero.

The couples are fixed-end moments #

There is a way of reading the consistent vector that makes it familiar rather than novel, and it is worth having because it connects this page to a century of hand analysis.

$wh^2/12$ at each end of a uniformly loaded member, in opposite senses, is the fixed-end moment — the moment a beam built in at both ends develops under a uniform load, tabulated in every book of hand methods since Hardy Cross.

That is not a coincidence and it is not an analogy. The consistent load vector is, by construction, the set of nodal forces that would hold the element’s nodes still under the applied load — which is exactly what a fixed-end reaction is. Moment distribution begins by locking every joint, computing the fixed-end moments, and then releasing them; the stiffness method assembles the same fixed-end forces into a load vector and releases them all at once by solving — which is what the matrix replaced the hand methods with rather than what it replaced them by.

Two methods, one table. Anybody who has done moment distribution has computed consistent nodal loads by hand, and the reason the connection is easy to miss is that one of them calls the quantity a moment and the other calls it a load.

It also gives the quickest way to remember the vector: whatever a table of fixed-end moments says for that load and that member is the rotational part of its consistent load vector, with the sign convention adjusted. A triangular load, a point load off centre, a partial UDL — all of them are tabulated, and all of the tables are load vectors.

What it costs #

The cost is easiest to see on the case where the exact answer is known.

A third. And the reason it is a third rather than a rounding error is worth stating: lumping the load moves half of it to the tip, where it does far more work than it does distributed along the span. A cantilever under a tip load of $wL/2$ deflects $wL^4/6EI$; under the real uniform load it deflects $wL^4/8EI$; and the ratio of those is exactly 4/3. It is the same comparison a unit load makes between two load positions, with the arithmetic run for a reason nobody intended.

The exactness on the other curve is the more surprising half. A finite element solution is normally an approximation that improves with refinement, and here one element gives the exact nodal deflection of a cantilever under a uniform load. The reason is that the element’s shape functions are cubics, the exact deflected shape is a quartic, and a quartic’s nodal values are reproducible by a cubic with the right nodal forces — the error is in the shape between the nodes, not at them.

That is a general property worth knowing: for a load whose exact solution differs from the element’s own space by something that vanishes at the nodes, the consistent load vector gives exact nodal answers on any mesh.

Why nothing gives it away #

The most useful thing about this failure is its signature, and the signature is that there is none.

The reactions are right. Both load sets have the same resultant, so the support reactions and the fixed-end moment are identical and correct — 180 kN·m on the cantilever here, from either set. A model checked by summing its reactions against the applied load passes.

Equilibrium is satisfied everywhere. The nodal forces balance the element forces at every node, because that is what the solve enforces. An equilibrium check finds nothing.

And the deflected shape looks plausible. It is smooth, it has the right sign, it is largest where it should be, and it is a third too large.

So the error survives every check a modeller normally makes, and it shows up only as a comparison against a hand calculation — which is exactly the comparison that stops being made once the model is trusted. The check that cannot see the error is the general form of it and this is the finite-element instance — a model that satisfies every check it is given and answers a different question.

The sign change is worth noticing. The lumped error is not conservative in any consistent direction: it overstates a cantilever’s deflection and understates a simple span’s, because what it does is move load toward the nodes and whether that helps depends on where the nodes are.

Where it actually bites #

Modern software builds its own load vectors and gets this right, so the question is where a designer still meets it — and there are four places.

A load applied at nodes by hand. Putting a pressure onto a model by computing tributary areas and applying point loads at the nodes is exactly the lumped substitution, done deliberately. On a fine mesh it is harmless; on the four-element beam somebody meshed to save time it is not.

A mass matrix. The same distinction exists for mass — consistent against lumped — and there the lumped version is more accurate for some quantities and is used deliberately, because a diagonal mass matrix makes explicit dynamics possible. The trade is different and the mathematics is the same.

An element without the freedoms. A truss element has no rotational freedoms, so there is no consistent vector for a distributed load on it at all — the load has to be lumped, and the element’s own bending is not being modelled anyway. Putting a distributed load on a truss member in a model is a request the element cannot honour and usually a modelling error one level up.

A hand-built spreadsheet model, where every load vector is typed. Those still exist for the small repetitive jobs — a run of purlins, a rack, a set of standard frames — and they are the models least likely to be checked against anything.

And the recovery of internal forces. Even with the right load vector, the element’s internal moment diagram is the one its shape functions give, which for a uniform load is a straight line between the nodal values rather than the parabola the beam actually has. Most software adds the exact element solution back afterwards; software that does not shows a moment diagram missing $wh^2/8$ at mid-element.

What it is worth on a real model #

The size of it on a job is worth a paragraph, because the answer is “usually nothing, occasionally a great deal”, and knowing which is the useful part.

A floor grillage meshed at one element per beam, with the slab load applied as point loads at the beam intersections, is the lumped case at its worst — and the deflections it returns are the ones a serviceability check is made against. On a 6 m secondary beam that is a third; on the same beam meshed into four it is 2 per cent.

A transfer structure modelled with a few elements per member, checked for deflection, is where the error is largest and where the consequences are worst, because a transfer beam’s deflection is what everything above it inherits.

And a global model of a whole frame, meshed at one element per member with member loads applied properly by the software, has none of this at all — the loads are consistent, the nodal answers are exact, and the only approximation left is the one between the nodes — which is the shear and moment picture rather than the deflection.

So the exposure is a modelling habit rather than a piece of software. It is present wherever somebody has converted a load into point forces by hand, which is a thing done under time pressure, on the models that are built quickly, which are the ones nobody checks against a hand calculation.

The general principle, which is older than the method #

Underneath all of this is a statement about what “equivalent” means, and it predates finite elements by two centuries.

Two force systems are equivalent with respect to a class of displacements. For a rigid body that class is translations and rotations of the whole body, and equivalence in resultant and moment is exactly equivalence in work over that class — which is why the rigid-body rule is a work argument in disguise.

Enlarge the class and the equivalence has to be re-earned. An element that can bend admits displacements a rigid body cannot, and a substitution that ignores them is equivalent over a smaller class than the one the problem uses.

Which gives a rule that transfers well past this page: whenever a load is simplified, ask what motions the simplification is equivalent over, and whether the structure has any others. The unit load method is the same statement used constructively — a virtual force system chosen to extract one displacement — and the load vector here is that machinery running in the opposite direction.

Where the model stops #

Beam elements and one load case. The same argument runs for plates, shells and solids, where the integrals are over areas and the shape functions are bilinear or quadratic — and where the consistent vector for a uniform pressure on an eight-node element has negative corner forces, which look wrong and are right.

The element is straight and prismatic. A tapered element’s shape functions are not the Hermite cubics, so its consistent vector is not $wh^2/12$, and software that uses a prismatic vector on a tapered element is making a smaller version of this same error.

Nothing here is nonlinear. Once the stiffness depends on the displacement, the load vector is re-formed at every step and its consistency matters at every one.

And the exactness is a property of this load on this element. A load that varies along the element in a way its shape functions cannot follow gives nodal answers that are consistent and not exact, and the difference between the two words is the whole reason meshes are refined at all.

What the pictures cannot show #

The moment diagram inside the element, which is where the rest of the answer is. With consistent loads the nodal displacements are exact and the internal distribution is still an approximation: the element reports a linear moment where the beam has a parabola, and the difference at mid-element is $wh^2/8$ — on the one-element cantilever, the whole of the mid-span moment.

That is the practical reason to mesh a beam more finely than the exactness argument suggests. The nodal answers are exact and the picture between them is not, and a designer reading a moment diagram off a coarse model is reading the element’s interpolation rather than the structure’s behaviour. That is the same distinction between a computed value and a sampled one that every discretised model carries.

The assumption the figure rests on #

That the load is known well enough for the distinction to matter.

A uniform load of 10 kN/m is a modelling idealisation of a floor build-up, a partition allowance and an imposed load, each of which is known to perhaps ten per cent. Against that, a 33 per cent error is large and a 2 per cent one is not — so the practical rule is not “always use consistent loads” but “know which error is which”.

What makes this one worth a page anyway is that it is free to remove. A load vector costs nothing to get right, it is right on every mesh, and it removes an error that is otherwise paid for in elements. That is an unusual combination — most improvements in accuracy cost something — and it is the reason every solver written since about 1965 does it without being asked.

The history, which is a lucky choice #

Consistent load vectors arrived with the finite element method’s variational formulation in the early 1960s, and the reason they are universal now is that the formulation makes them unavoidable rather than optional.

The earlier structural matrix methods — the ones that grew out of aircraft analysis in the 1950s — assembled stiffness from member equations and put loads on by hand, and the fixed-end moment tables did the job. What the variational route added was a derivation: minimise the total potential energy over the space the shape functions span, and the load term falls out as an integral of the load against those functions. There is no choice to make.

That is worth noticing as a property of a good formulation. A method derived from a principle produces its own conventions, and the ones it produces are the right ones for reasons the method can state. The lumped alternative survives only where a different principle recommends it — the diagonal mass matrix in explicit dynamics — and in that case it is a deliberate approximation with its own error analysis rather than a shortcut.

The historical curiosity is that the answer was already tabulated. Fixed-end moments were in every handbook by 1935, computed for moment distribution, and they are the rotational half of every consistent load vector the finite element method has ever needed. The method did not discover them; it explained why they were the right numbers to have written down.

The ladder from here #

Later rungs on this anchor: consistent loads for plate and shell elements, where the corner forces of a uniform pressure change sign and look like a mistake. The consistent mass matrix and the lumped one, and why the trade goes the other way in dynamics. Reduced integration and the load vectors that go with it. Element internal force recovery, and the difference between a nodal answer and a field. And the deeper question this belongs to: what a finite element solution actually converges to, which is the minimum of a functional over a restricted space rather than the solution of an equation — and which is why the answer depends on how it was divided in a way that has nothing to do with arithmetic error.