Everything adds to nothing, and that is the whole of statics

A bridge that is standing still is not doing anything. That is the entire content of statics, and it is worth stating in the negative because the whole subject is built on what is absent: no acceleration, no rotation, nothing changing.

Nothing changing means two sums come to zero. The forces on any piece of the structure add to nothing, and the moments about any point add to nothing. Everything else — reactions, member forces, shear, moment, the size of a beam — is those two sentences applied to a well-chosen piece.

Two statements, and why one is not enough #

The force equation is the obvious one. Add up everything pushing and pulling, in each direction, and the total must vanish:

$$\sum F_x = 0, \qquad \sum F_y = 0.$$

The moment equation is the one that does the work:

$$\sum M = 0 \ \text{about any point.}$$

It is easy to think the second is a refinement of the first. It is not — it is independent, and without it almost nothing can be solved.

Consider two equal and opposite forces applied at different places on a beam. The force sums are satisfied exactly: up equals down, left equals right. The beam nonetheless spins, because the two forces form a couple, and a couple has a moment with no resultant force at all. A force equation cannot see it. Only the moment equation can.

That is the reason a statics problem in a plane has three equations rather than two, and it is the reason a plane structure needs three restraints rather than two. The third restraint exists to stop rotation, and a structure that has two is a turnstile.

Choosing where to take moments #

The moment equation holds about any point, and that freedom is the most useful thing in the subject.

Taking moments about a support kills that support’s reaction, because a force through a point has no moment about it. One equation, one unknown, no simultaneous solving. Then the force equation gives the other reaction in a line.

For the beam above: taking moments about the left-hand support removes the left reaction entirely, leaving the loads and the right reaction. The right reaction follows immediately; the vertical force sum then gives the left. Two equations, taken in the right order, and no algebra beyond dividing.

Choosing badly gives two equations in two unknowns and the same answer after more work. Choosing well is the whole of technique, and it is worth noticing that the technique is a choice about where to stand, not about the structure.

The three marks are the freedom stated as a picture. A reader who suspects the choice of centre is doing something to the answer can watch it not doing anything, which is a harder thing to believe from an algebraic identity than from three sums that all read zero on one drawing. The choice changes the labour and never the result.

The cantilever makes the independence of the two equations concrete. Its single support has to provide a force and a moment, and the moment is not implied by the force — a support that could push but not resist rotation would let the beam swing down like a gate.

Where the resultant of a spread load acts #

A uniform load is not a force at a point, and treating it as one is the commonest simplification in the subject.

For the purposes of the two equations, a distributed load can be replaced by a single force equal to its total, placed at the centroid of the loaded region. That is exact, not approximate, as far as the external equilibrium is concerned — the reactions come out identical.

It is emphatically not exact for anything internal. The shear and moment along the beam are completely different for a uniform load than for the equivalent point load, and using the point-load substitution to compute them is a real error rather than a small one. A load spread over a span produces a parabolic moment diagram; the same load concentrated at mid-span produces a triangular one with a peak twice as high.

Symmetry makes the check easy to read here and the substitution easy to trust. Move the load onto half the span and the two reactions separate, and the resultant moves with the centroid of whatever region is now loaded; the arithmetic is the same and only the numbers change.

The rule is worth stating as a boundary: replacing a distributed load by its resultant is legitimate outside the free body and illegitimate inside it.

Reading the equations backwards #

The two statements are usually used to find unknown reactions from known loads. Run them the other way and they say something stronger: they constrain what a structure can be.

Two equilibrium equations are available at each joint, and one unknown arrives with each member and each restraint. Three frames differing by a single member therefore fall into three different worlds: too few unknowns and the structure is a mechanism, exactly enough and it is solvable, too many and it is something statics cannot finish. Counting unknowns against equations decides whether the two sentences are sufficient. A frame with fewer unknowns than equations cannot satisfy them all and moves; a frame with more has a family of solutions and needs to know how stiff its members are before it can pick one.

The middle case is where statics lives, and it is narrower than it looks. Most real structures are in the third category, and the fact that they are solved anyway is a matter of adding stiffness to the analysis — a whole extra body of theory that exists because two sentences ran out.

The forces that are not there #

The most useful habit in the subject is asking what a support can and cannot do.

A roller can push perpendicular to its surface and nothing else. A pin can push in any direction but cannot resist rotation. A built-in end can do all three. Each is an idealisation of a real detail, and the idealisation is a statement about which forces are permitted to appear in the equations.

Choosing the wrong idealisation is the most consequential error available. Treating a connection as pinned when it actually resists rotation puts moment where the analysis says there is none — and that is where cracks appear. Treating it as fixed when it is not underestimates the moment at mid-span. Neither error shows up in the arithmetic, because the arithmetic is correct for the structure that was described.

The pair is the force equation as a drawing, and for a century that drawing was how structures were solved: a drawing board, a scale rule, and a polygon that had to close. The gap in the second figure is not an error of drafting. It is the resultant, measurable with the same scale rule, and it points in the direction the joint would accelerate.

What the moment equation measures #

A moment is a force times a distance, and the distance is the perpendicular one from the point to the line of action.

The same beam and the same load, with nothing altered but where it sits, makes the lever arm the only variable in the picture. That is the cleanest way to see what the moment equation is measuring, because everything the force equation can see has been held fixed.

The linearity is worth dwelling on because it is the source of most of the leverage in structural design. Doubling a force doubles its moment; doubling its distance does the same. Since distances are usually cheaper to change than forces, nearly every efficient structure is one that has arranged for a large lever arm — the depth of a truss, the depth of a beam, the spread of a foundation.

The same picture also explains why the units are awkward. A moment has dimensions of force times length and no direct physical analogue: nothing about a beam is “40 kilonewton-metres” in the way it is “3 metres”. The quantity is a bookkeeping device for rotation, and it is real in exactly the sense that its absence keeps buildings up.

Three moment equations, or one, and the check that costs a line #

“Moments about any point” sounds like an infinite supply of equations. It is not, and the exact statement is worth having because it produces a free check on every calculation.

For a plane body there are three independent conditions and no more. Writing the moment equation about a second point adds nothing new — it is a linear combination of the first moment equation and the two force equations, which is Varignon’s theorem in its practical form. Any fourth equation is guaranteed to be satisfied automatically if the first three are.

What is genuinely free is the choice of which three. The standard set is $\Sigma F_x$, $\Sigma F_y$, $\Sigma M$. An equally valid set is $\Sigma F_x$ together with moments about two points, provided the line joining those points is not perpendicular to the $x$ axis. A third valid set is moments about three points, provided the three are not collinear. Those provisos are not pedantry: choose collinear points and the three equations become dependent, the system is singular, and the arithmetic produces either nonsense or a division by zero, which is the same trap as a badly arranged set of restraints in a different costume.

The practical value is the check. Solve for the reactions using moments about one support and the vertical force sum; then take moments about the other support and see whether the answer closes. It cannot fail unless something is wrong, and it costs one line.

For the beam in the first figure — loads of $12$ at three metres and $8$ at six, on an eight-metre span — moments about the left support give

$$R_2 = \frac{12\times3 + 8\times6}{8} = \frac{84}{8} = 10.5,$$

and the vertical sum gives $R_1 = 20 - 10.5 = 9.5$. Now the check, taking moments about the right support:

$$R_1 = \frac{12\times5 + 8\times2}{8} = \frac{76}{8} = 9.5. \checkmark$$

Two routes, one answer. Every beam figure on this site runs exactly that comparison inside the solver before it draws anything, and refuses to draw if the two routes disagree.

The size of the disagreement is worth reading rather than merely noticing, because the common mistakes each leave a signature. A closure error equal to one of the loads means that load was entered on one side and forgotten on the other. An error equal to a load times a distance means it is in the right sum and the wrong place. An answer exactly twice what it should be means something was counted in both routes. And an error that is a few per cent of nothing in particular is usually arithmetic rather than modelling, which is the one case where re-doing the sum is the right response. The reason is a rule worth adopting: an assertion that has never rejected anything proves nothing, and a check that is guaranteed to pass when the work is right and to fail when it is not is the cheapest quality control available in the subject.

Necessary, and — once — sufficient #

Equilibrium is a necessary condition. A structure that does not satisfy it is moving, and there is no arrangement of material that rescues it. The converse fails: a force system that satisfies equilibrium says nothing about whether the material can supply the forces in it, and a beam in perfect equilibrium at ten times its capacity is in equilibrium right up until it is not.

There is, however, one circumstance in which equilibrium alone is enough to prove a structure safe, and it is one of the most useful results in the subject.

The lower-bound theorem of plasticity says: if any distribution of internal forces can be found that is in equilibrium with the applied loads and nowhere exceeds the capacity of the material, then the structure will not collapse under those loads. The distribution does not have to be the real one. It does not have to be compatible with any deformation. It only has to exist.

The theorem is worth pausing on, because it rescues something that appeared lost. Earlier in this essay, statics ran out at the redundant structure: infinitely many force systems satisfy equilibrium and the equations cannot say which is real. The lower-bound theorem says that for a ductile material, it does not matter which is real. Pick any of them, check it against capacity, and if it passes, the structure is safe — because a ductile structure loaded beyond the elastic distribution will redistribute toward whatever equilibrium state it can find, and a state has been shown to exist.

Two familiar techniques are this theorem in use. The thrust line drawn inside a masonry arch is a statement that a force system exists in equilibrium and within the material, which is a proof of safety without any claim to be what the arch is actually doing — and it is why an arch that has cracked and settled is not thereby unsafe. Strut-and-tie modelling in reinforced concrete does the same for regions where beam theory fails: invent a truss inside the concrete, check the struts against the concrete and the ties against the steel, and if it works, the region works.

The price is ductility, and it is not negotiable. The theorem requires the material to be able to deform enough to reach the assumed state without fracturing on the way. For steel it holds comfortably. For masonry in compression it holds well enough that a millennium of buildings testifies to it. For a brittle material, or a connection that fails suddenly, it does not hold at all, and the reassurance it offers is withdrawn exactly where reassurance is most wanted.

What survives when the model changes #

There is one more reason the two sentences are worth taking seriously, and it has to do with how much of structural engineering they outlive.

Every model in the subject makes assumptions that later models discard. Beam theory assumes plane sections stay plane; elastic analysis assumes stress is proportional to strain; a first-order analysis assumes the geometry does not change; a pin-jointed truss assumes joints carry no moment. Each of these is true in a range and false outside it, and each has been superseded for some class of problem by something more careful.

Equilibrium is not on that list. Every one of those models satisfies it, and so does every model that will replace them. An elastic analysis and a plastic one give different distributions of moment along a beam, and both give the same total reaction. A cracked concrete section and an uncracked one have different stiffnesses, different neutral axis positions and different stress distributions, and the sum of the forces on the cut face is identical in each. A nonlinear finite-element analysis run overnight on a million elements is producing, among other things, an answer that satisfies the same two sentences on every element in it.

This is the practical reason a structural engineer checks a computer output by hand. Almost nothing about a large analysis can be verified by inspection: the stiffnesses cannot be checked, the element formulation cannot be checked, the mesh cannot be checked. The total vertical reaction, however, must equal the total applied load, and it takes ten seconds. A model that fails that check is wrong regardless of how sophisticated it is, and a great many modelling errors — a load applied twice, a support in the wrong place, a unit misread, a member left unconnected — announce themselves there and nowhere else.

Where the model stops #

The two equations are exact for a rigid body. Real structures are not rigid, and four consequences follow.

Deflections change the geometry. The equations are written on the undeformed shape, which is a first-order approximation. Where deflections are large enough to move the lines of action, the load starts amplifying itself and equilibrium has to be written on the deformed shape instead.

Statics cannot see stiffness. For a determinate structure that does not matter. For anything else, the distribution of load depends on relative stiffness, which the equations do not contain.

Nothing here mentions material. The reactions on a steel beam and a timber beam of the same span under the same load are identical, and so are their bending moments. Everything about whether either survives is a separate question.

Support idealisations are approximations. As above, and the most consequential ones in practice.

There is also a limit that belongs to the drawing. Every figure here shows a plane structure with forces in that plane. Real structures are three-dimensional, with six equations rather than three, and forces that leave the page — wind on a face, the twist of an eccentric beam, the out-of-plane restraint that stops a truss falling over sideways. A plane analysis assumes something else is looking after that direction, and occasionally nothing is.

The ladder from here #

Later rungs on this anchor: the free body as a deliberate choice of boundary. Distributed loads and where their resultants act. Couples and pure moments. The method of sections. Three-dimensional equilibrium and its six equations. Determinacy counting done properly, including internal releases. Virtual work as an alternative to summing forces. And the point at which equilibrium alone stops being enough, which is the entrance to everything else in the subject.

The two equations were written down by Archimedes for levers, extended by Simon Stevin to the inclined plane, and given their modern form by Varignon around 1725. Nothing has been added to them since.