Structural Engineering & Statics
where the load goes
Equilibrium

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

A structure that stays put obeys two statements — the forces on it sum to zero, and so do the moments. Every number in the subject comes out of those two sentences.
10 min read 6 figures The statics of things that do not moveThe load must go somewhere

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.

A beam, its loads and its reactionsA free body: the beam cut away from everything it touches, with the forces that were doing the touching drawn on it. The reactions are computed from the loads, so their arrows are to scale relative to each other.1289.510.5ΣM about one support gives the other reaction; ΣF then gives the first
Fig. 1 A beam, its loads and the reactions the supports must supply. The reaction arrows are computed from the loads rather than drawn at plausible lengths, so moving a load toward one end lengthens that end’s arrow and shortens the other.

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:

∑Fx=0,∑Fy=0.\sum F_x = 0, \qquad \sum F_y = 0.

The moment equation is the one that does the work:

∑M=0 about any point.\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.

A cantilever and its fixingA free body: the beam cut away from everything it touches, with the forces that were doing the touching drawn on it. The reactions are computed from the loads, so their arrows are to scale relative to each other.3 per unit length24.0fixing moment 96.0ΣF = 0 and ΣM = 0 give both the force and the moment at the wall
Fig. 2 A cantilever under a uniform load. There is only one support, so it must supply everything: a vertical force to balance the load and a moment to balance the load’s moment about the wall. Both are computed here.

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.

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.

Counting unknowns against equationsThree 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.m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case
Fig. 3 Three frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint: too few unknowns is a mechanism, exactly enough is solvable, and too many is a structure 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.

A closed force polygonThe forces on a joint, laid tip to tail. Equilibrium is the statement that the polygon closes, and the gap when it does not is the out-of-balance force, to scale.load 60strut 84.9tie 60starts and ends here
Fig. 4 The forces on a joint laid tip to tail. Equilibrium is the statement that the polygon closes, which is the force equation drawn rather than written.
A force polygon that does not closeThe forces on a joint, laid tip to tail. Equilibrium is the statement that the polygon closes, and the gap when it does not is the out-of-balance force, to scale.load 60strut 84.9tie 60out of balance: 19.2
Fig. 5 The same construction with one force too small. The gap that opens is the out-of-balance force, to scale, and something has to supply it or the joint moves.

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 moment is the force times the distanceOne force applied at five distances from a pivot, with the moment it produces drawn as a bar. The force never changes; only the arm does, and the moment follows it exactly.pivotthe same force of 20, moved along the levermoment about the pivot120 × 1 = 20220 × 2 = 40320 × 3 = 60420 × 4 = 80520 × 5 = 100
Fig. 6 One force at five distances from a pivot, with the moment it produces drawn as a bar. The force never changes; only the lever arm does, and the moment follows it exactly.

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.

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.

Where this goes next

2012.57.5shear 7.5moment 22.5the cut, at x = 5nothing was applied here — the internal forces are what the left-hand piece needs The free body is a choice, and choosing it well is the whole skill Cutting a structure open is not a step in the method. It is the method — and where the cut is made decides whether the answer takes one line or twenty. m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case 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. 2012.57.5shear 7.5moment 22.5the cut, at x = 5nothing was applied here — the internal forces are what the left-hand piece needs What a cut reveals, and why it was there all along Cut a beam anywhere and two quantities appear on the face — a shear force and a bending moment. Nothing was applied there. They are what the material was already doing.