The equation that is not new, and the three that are

Three equations hold a plane body still: the forces sum to nothing horizontally, the forces sum to nothing vertically, and the moments sum to nothing. That is the whole of statics in the plane, and everything else follows from it.

It is also a claim about counting, and counting is where it goes wrong. Three is not a limit on how many equations can be written — infinitely many can be written, one for every point in the plane a moment might be taken about. Three is the number of them that carry information. Every other equation that can be written about the same free body is a combination of these three, which means it is satisfied automatically the moment they are, and asking it produces the number zero with no effort and no news.

The distinction matters because the whole apparatus of determinacy rests on comparing a count of unknowns with a count of equations, and a count of equations that includes duplicates gives the wrong answer in the direction that feels like success.

What a rigid body can be told to do #

A rigid body in a plane has three ways to move: it can slide in one direction, slide in a direction at right angles to the first, and rotate. Any motion it might make is some combination of those three, which is what it means to say the body has three degrees of freedom.

Equilibrium is the statement that none of those three motions is happening, and that is why there are three equations. They are not three arbitrary conditions that happen to be useful; they are one condition per way of moving. A body with a fourth way to move would need a fourth equation, which is exactly what happens in three dimensions — six freedoms, six equations, and the two extra moments and one extra force that a cut in three dimensions exposes.

This is the reason no fourth equation exists in the plane. There is nothing left for it to prevent.

The second moment centre #

The most common way to write a fourth equation is to take moments about a second point. It feels like new information, because the lever arms are all different and the arithmetic is genuinely different work.

It is not new information, and the reason is a piece of geometry old enough to be named after a seventeenth-century Jesuit. Varignon’s theorem says the moment of a set of forces about any point equals the moment of their resultant about that point. Take moments about $A$, then about $B$: both sums are the moment of the same resultant, evaluated at two places. If the resultant is zero — which is what the two force equations already assert — then its moment is zero everywhere, and the second sum was guaranteed before it was written.

Formally: for a body with resultant force $\mathbf{R}$ and moment $M_A$ about a point $A$, the moment about any other point $B$ is

$$M_B = M_A + \mathbf{R} \times \mathbf{r}_{AB}$$

If $\mathbf{R} = 0$ and $M_A = 0$, then $M_B = 0$ for every $B$ in the plane. There is no choice of $B$ that recovers anything.

The practical consequence is one every engineer meets: a second moment equation is an excellent check and a useless equation. It cannot find an unknown that the first three did not, and it will catch an arithmetic slip immediately, because a slip breaks the guarantee. Used as a check it is worth the thirty seconds. Used as a source of a fourth unknown it produces $0 = 0$, and the identity is a message rather than a failure.

When three moment equations are legitimate #

There is a genuine alternative set, and it is worth stating precisely because the imprecise version of it circulates widely.

Instead of $\Sigma H = 0$, $\Sigma V = 0$, $\Sigma M_A = 0$, a plane body can be analysed with $\Sigma M_A = 0$, $\Sigma M_B = 0$, $\Sigma M_C = 0$ — three moment equations and no force equations at all. This set is valid, and it is independent, provided $A$, $B$ and $C$ are not collinear.

The condition is not a technicality. Suppose the three points lie on a line, and suppose the body is not in equilibrium but is acted on by a single resultant force whose line of action is that same line. That resultant has zero moment about every point on its own line of action, so all three equations are satisfied by a body that is being pushed along the line. Three equations, all satisfied, and the body accelerates. Collinear centres cannot detect a resultant along their own line, so they cannot be a complete set.

Move any one of the three off the line and the escape closes: a force along the old line now has a lever arm about the displaced point, and the third equation catches it.

There is a two-moment set as well, similarly conditioned: $\Sigma M_A = 0$, $\Sigma M_B = 0$, and $\Sigma F = 0$ in any direction not perpendicular to $AB$. The perpendicular case fails for the same reason — a resultant along $AB$ has no moment about either point and no component in the direction being summed.

The figure is the caveat drawn rather than stated. Every moment centre this family can mark lies on the beam’s own axis, so no placement of them ever escapes collinearity, and a reader looking for the legitimate three-moment set has to look at a body whose forces are not all parallel.

The pattern behind all three sets is the same. Any three conditions will do, so long as between them they can detect all three of the body’s freedoms.

Which free body produced the number #

The figure above is a beam with two downward point loads of 12 and 8 at 3 and 6 along a span of 8, on a pin at the left and a roller at the right.

The free body is the whole beam, cut away from the pin and the roller, with the two reactions drawn as the forces those supports were applying. Moments about the left support: $12 \times 3 + 8 \times 6 - R_2 \times 8 = 0$, so $R_2 = (36 + 48)/8 = 10.5$. Vertical forces: $R_1 = 20 - 10.5 = 9.5$.

Now take moments about the right-hand support instead, as the generator does at the third marked point: $R_1 \times 8 - 12 \times 5 - 8 \times 2 = 9.5 \times 8 - 60 - 16 = 76 - 76 = 0$. And about the load at 3: $-9.5 \times 3 + 8 \times 3 - 10.5 \times 5 + 12 \times 0 = -28.5 + 24 - 52.5 + 57 = 0$, taking the reaction at the roller with its own lever arm.

Three zeros, one fact. The figure prints each residual rather than asserting it, which is the only reason the claim is checkable at all: a description that said “and all three come to zero” would be equally convincing if the generator had silently drawn a different beam.

The equation that was never there #

For this particular free body the situation is worse than redundancy, and it is worth being blunt about because it is the case that turns up in practice most often.

Every force on that beam is vertical. The moment of a vertical force about a point does not depend on the point’s height — only on the horizontal distance between them — so moving a moment centre anywhere in the plane, not merely along the beam, returns the same equation. The three-moment-centre set is unavailable for a body loaded this way no matter where the points are put, because all of them are effectively collinear with respect to the only forces present.

And the horizontal equation, $\Sigma H = 0$, reads $0 = 0$ before anything is computed. There are no horizontal forces to sum.

So a simply supported beam under vertical load does not have three equations. It has two, and it has two unknown reactions, and the reason such a beam is determinate is that those numbers match — not because three is bigger than two by a margin nobody checked. The moment the loading acquires a horizontal component, or a support acquires a horizontal restraint, the third equation switches on and the third unknown appears with it.

Three forces meeting at a point is what the moment equation looks like when it has something to say. On a beam under vertical loads all the lines of action are parallel — they meet at infinity, the concurrency condition is trivially satisfied, and the geometry is telling the same story as the algebra.

The fourth equation that is real #

There is one way to get a fourth equation, and it is not a trick of where the moments are taken. It is a different free body.

A three-pinned arch has four unknown reaction components — two at each springing — and the whole structure offers three equations. It is nonetheless determinate, and the extra equation comes from cutting the structure at the crown hinge and writing the moment equation for one half alone. A hinge transmits no moment, so the moment of everything on the left half about the crown pin is zero, and that is a statement the whole-structure equations do not contain.

The distinction is exactly the one this essay is about. Taking moments about the crown for the whole arch is a second moment centre and gives nothing. Taking moments about the crown for half the arch is a new free body, with its own three equations, and one of them contains the internal actions at the cut. The pencil mark is in the same place on the page; what changed is which body the sum is being taken over.

This is the general mechanism by which every determinate structure with more than three reactions is solved. The method of sections is the same move applied to a truss, the propped cantilever is the same move refused — there is no hinge to cut at, so the fourth equation has to come from compatibility instead of statics. A condition equation is a new body, and a new body is three more equations at the price of the unknowns exposed on its cut.

The counting error this prevents #

An essay about not being able to write a fourth equation would be pedantry if the mistake stayed on paper. It does not.

The standard determinacy count for a pin-jointed frame is $m + r$ against $2j$: one equation per joint per direction. For a rigid frame the count is $3m + r$ against $3j$ plus the releases. Both counts assume every equation is independent, and both are necessary rather than sufficient for exactly the reason this essay is about: an equation that duplicates another leaves the structure with fewer real constraints than the arithmetic promises.

The failure mode is specific. A structure whose count says determinate and whose equations are not independent is a mechanism, and it will be analysed happily by hand — because hand analysis walks joint by joint and never assembles the whole system to notice the dependency.

Two frames differing by one member, one determinate and one a mechanism, are told apart by the count without difficulty, and that is the reason the count is trusted. The case it does not handle looks exactly like the determinate one, which is the subject of the next rung on this ladder.

The one thing the three equations give away for free #

Everything above is about equations that add nothing. There is a property of the same three that adds a great deal, and it is worth setting beside them because it comes from the same place: they are linear in both the loads and the unknowns.

Linearity means the solution of a sum is the sum of the solutions. Solve the beam for the 12 kN load alone and get one pair of reactions; solve it for the 8 kN load alone and get another; add the two pairs and the result is the answer for both loads together, exactly. Nothing was approximated. Superposition is not a technique laid on top of statics — it is a property the three equations already have, and it is the reason a designer can hold a library of standard cases and build any loading out of them.

The consequences run through the whole of practice. Influence lines are superposition used to place a moving load. Load combinations — dead plus imposed plus a fraction of wind, each with its own partial factor — are superposition with coefficients, and a factored combination is analysed by scaling and adding results rather than by re-solving. The decomposition of an awkward distributed load into a rectangle and a triangle is superposition applied to the load rather than to the answer.

The limits are exactly where the linearity goes. A structure whose lever arms change appreciably under load has equations written on a geometry that depends on the answer, so the second-order case is not superposable: the sway under wind and gravity together is more than the sum of the two taken separately, which is the whole point of the amplification. A material past yield is not superposable either, for the same reason in a different variable. Both failures are announcements that the equations being solved are no longer the linear three this essay has been counting.

Where the model stops #

Rigid bodies only. Everything here is a statement about a body that does not deform, or whose deformation is small enough that the equations may be written on the undeformed geometry. When the deformation is large enough to change the lever arms, the equations become nonlinear and the tidy count of three stops being the whole story — which is what second-order analysis is.

Three equations is the count for the whole body. Cut the body into pieces and each piece brings three of its own, along with the internal forces on the cut faces as new unknowns. This is why the method of sections works: it does not create information, it exchanges one body’s three equations for two bodies’ six, at the price of three unknowns on the cut. The arithmetic is always conserved.

A support condition is not an equation. It is an unknown. The commonest confusion in the whole subject is between the three equations, which are a property of the plane, and the reactions, which are a property of the supports. Adding a support never adds an equation.

The figures have a limitation they cannot escape either. A moment centre is drawn as a small circle on the page, which makes it look like a physical thing — a pin, a hinge, somewhere the body is attached. It is none of those. It is a place a sum is being taken about, chosen for arithmetic convenience, and the body neither knows nor cares where it is. Every free-body diagram in every textbook has this problem, and the honest correction is to remember that the circle is a pencil mark rather than a piece of hardware.

The generalisation, and where it came from #

The count of independent equilibrium equations equals the number of degrees of freedom of the body, and that statement is more general than statics. It is the finite-dimensional case of the virtual-work principle: a body is in equilibrium if the work done by the applied forces vanishes for every admissible virtual motion, and the number of independent equations is the dimension of the space of those motions. The unit-load method is the same principle used in the other direction, with a fictitious force chosen to extract a real displacement.

Seen that way, “three in the plane, six in space” stops being a fact to memorise. It is a dimension count, and it comes out right for a body on a frictionless plane, for a mechanism with one freedom, and for a continuum with infinitely many — where the equilibrium equations become the differential equations of elasticity and the count becomes an argument about function spaces.

The same dimension count is what survives when the body stops being still. A body in steady motion is not in equilibrium, but d’Alembert’s move — carry the mass times acceleration across the equals sign and call it a force — hands the three sums a problem they can answer without gaining or losing a single equation. The count of three is a count of freedoms, and a body that is accelerating has exactly as many freedoms as one that is not.

Varignon published his theorem in 1687 as a proposition about the composition of forces, in a tradition running back to Stevin’s parallelogram, and it took until Poinsot’s work on couples in 1803 for the moment of a system about an arbitrary point to be handled cleanly enough that the redundancy of the second moment equation was obvious rather than surprising. The intervening century is full of otherwise careful calculations that solve for one unknown twice and treat the agreement as a result.

The ladder from here #

Later rungs on this anchor: the six equations of the three-dimensional body and the two moments the plane case throws away. Equilibrium as a variational statement rather than a set of sums. The equations of a body in steady motion, where the sums come to mass times acceleration and every result here survives with one term added. Equilibrium of a deformable continuum, where the three sums become three differential equations. And the case where the equations are all satisfied and the structure still moves, which is the next rung and is a fact about rank rather than about counting.

The immediate sequel is the count that does not see it: what happens when a structure has exactly as many unknowns as equations, and the equations turn out not to be independent after all.