Answering one question without solving the rest

A truss bridge has perhaps sixty members. Somewhere in it is the one that governs the design, and it is almost always a chord near mid-span. Solving the whole frame to find that one number is a great deal of work to answer a question about one bar.

There is a technique that answers it in a line. Cut the truss clean through, keep one side, and take moments about a point chosen so that everything unwanted disappears.

The cut has to sever exactly three #

The rule that makes the method work is arithmetic. A plane free body has three equations available, so a cut that exposes three unknowns is solvable and a cut that exposes four is not.

For a parallel-chord truss that means a vertical cut through a panel, severing the top chord, the bottom chord and the diagonal between them. Cut anywhere else and the count goes wrong: through a joint and four or five members are severed at once; through two panels and six.

The count is the same one that decides whether the whole structure can be solved at all, applied to a piece rather than to the whole. Three unknowns, three equations, one answer — and the fact that the rest of the truss has been discarded costs nothing, because the piece being kept is in equilibrium on its own.

That last point is worth dwelling on, because it is the whole justification and it is easy to skip. The left-hand piece is not an approximation to the truss. It is a body that is genuinely not moving, and therefore genuinely obeys the two sums, and the forces on its cut faces are genuinely whatever the removed half was supplying. Choosing where to draw the boundary is the only decision being made.

Where to take moments, which is the whole technique #

Three unknowns and three equations can always be solved. Solving them simultaneously is tedious, and the method of sections exists to avoid it.

A force through a point has no moment about that point. So if moments are taken about the joint where two of the three cut members meet, those two contribute nothing at all, and the third stands alone:

$$F \times d = \Sigma(\text{moments of everything else on the free body}).$$

One equation, one unknown, no elimination. And the right-hand side is a quantity that has nothing to do with the truss — it is the bending moment of an equivalent beam at that station, computed from the loads and the reactions.

The moment centre has to be found rather than assumed, and this is the one place the method reliably catches people out. In a Pratt truss the diagonals reverse their lean at mid-span, so the joint at which the top chord and the diagonal meet is at one end of the panel in the left half of the truss and at the other end in the right half. Using the wrong one gives an answer that is wrong by a whole panel’s worth of moment — a plausible number, of the right order, in the right units, and incorrect.

The generator behind the figures on this page computes the moment centre from the members it actually severed, then does the one-line calculation and compares it with the whole-truss elimination. If the two disagree it refuses to draw. That check exists because the wrong moment centre produces a figure that looks entirely convincing.

The three answers, and the three different questions #

One cut yields three numbers, and each comes from standing somewhere different.

The top chord comes from moments about the bottom joint on the far side of the panel. Both the bottom chord and the diagonal pass through it.

The bottom chord comes from moments about the top joint where the top chord and the diagonal meet.

The diagonal needs no moment centre at all. Both chords are horizontal, so the vertical force sum on the free body contains only the diagonal, and the diagonal’s vertical component is exactly the shear passing the cut.

That third case is the most revealing, because it says what the diagonals are for. The chords carry the moment; the diagonals carry the shear. A diagonal’s force is the panel shear divided by the sine of its inclination, so the diagonals near the supports work hardest and the ones at mid-span carry almost nothing — the exact opposite of the chords, which peak in the middle.

Reading a truss that way makes its shape explicable rather than decorative. The chords are heaviest at mid-span and the web heaviest at the ends, which is why a bridge truss is often deepest in the middle and why the end diagonals are the largest members in many frames.

The check the three answers owe each other #

Three unknowns were found by three separate one-line calculations, each using a different equation. That leaves something over, and what is left over is a test.

Each of the three answers used one equation: two moment equations about two different points, and the vertical force sum. A plane free body has exactly three independent conditions, so all three have now been spent — but the horizontal force sum is a legitimate combination of them, and it has not been written down. Writing it down cannot produce new information and can produce a contradiction.

On a parallel-chord truss the horizontal sum contains the two chord forces and the horizontal component of the diagonal, and nothing else, since the loads and reactions are vertical. So the three answers have to satisfy

$$F_{\text{top}} + F_{\text{bot}} + F_{\text{diag}}\cos\theta = 0.$$

For the cut in the hero figure the three come back as $-52.94$, $+47.06$ and $7.72$, with the diagonal inclined at $\cos\theta = 0.762$. The sum is $-52.94 + 47.06 + 5.88$, which is zero to the last digit the solver carries.

A check of that kind is worth more than its arithmetic. It is guaranteed to pass when the work is right, so passing tells nothing; the value is entirely in what happens when it fails. A mistaken moment centre, a lever arm read as the panel width instead of the truss depth, a load left off the free body, a reaction with the wrong sign — every one of those breaks the horizontal sum, and by an amount that names its own size. And it fails silently in no other way: each of the three answers on its own remains perfectly plausible.

The general shape of this recurs across the subject. Solving a determinate structure uses up exactly as many equations as there are unknowns, and any further equilibrium statement that can be written is then a free consistency test. The shear diagram that must return to zero is the same idea along a beam; the force polygon that must close is the same idea drawn rather than written. Where an equation is left over, it should be used, because an assertion that has never rejected anything proves nothing.

What the equivalent beam is doing there #

The right-hand side of the moment equation deserves a second look, because a quantity from a completely different structure has turned up in a truss calculation.

The moment of the loads and reactions about a point, taken on the piece to one side of a cut, is the bending moment of a beam of the same span carrying the same loads. Nothing about the truss entered it. So the chord force is

$$F = \frac{M_{\text{beam}}}{d},$$

which is the couple argument arriving as a consequence rather than as an assumption. For a parallel-chord truss the two are exactly equal, not approximately.

The correspondence has a practical use beyond elegance. A designer wanting the chord force at any station can read the beam’s moment diagram, divide by the depth, and have the answer without touching the frame at all. It also explains what changes when the chords are not parallel: the lever arm in the denominator becomes the perpendicular distance from the moment centre to the chord’s line of action, which for a curved top chord varies along the span — and a truss shaped so that this distance follows the moment diagram has chords of constant force, which is what a bowstring truss is.

Where the method is genuinely better #

The method of joints solves everything. The method of sections solves one thing. That sounds like a limitation and it is usually an advantage.

Checking rather than computing. An engineer given a computer output for a truss can verify one chord force in thirty seconds with a section cut. Verifying the whole frame would mean repeating the analysis. The single-member answer is exactly what a check needs.

The member that governs. Design is decided by the worst member, and for a parallel-chord truss under gravity load that member is known before any analysis: the chord nearest mid-span. Cutting there answers the design question directly.

Frames the joint method struggles with. The method of joints requires a joint with no more than two unknowns to start from, and some geometries have none. A section cut has no such requirement.

Structures that are not trusses at all. The count is what matters, not the frame type. A multi-storey building frame analysed by hand is cut horizontally through a whole storey, with an assumption supplied about how the shear divides between the columns; that is the portal method, and it is a section cut with one extra sentence added to make a redundant frame answerable. The same manoeuvre appears in arch analysis, in shear-wall assessment, and anywhere else a large structure has to yield one number quickly.

Non-vertical cuts. Nothing requires the cut to be a straight vertical line. It may be stepped, or inclined, or wrapped around a sub-assembly, so long as it severs three unknowns. A cut around a single joint is the method of joints; the two techniques are the same argument at two scales, and the family of cuts in between is available whenever it happens to be convenient.

What it costs, and where it stops #

The savings are real and the technique buys them by giving something up.

One member at a time. For a full analysis the method of joints or a matrix solution is faster, because each section cut repeats the work of computing the reactions and the moment. Answering ten questions means ten cuts.

No deflections. The method is equilibrium and nothing else, so it says nothing about how far anything moves. A member sized by stiffness is not sized by this calculation.

Determinate frames only. A cut through three members of a redundant truss gives one equation relating three unknowns that are not determined by equilibrium. The method returns a relationship rather than a number, and no choice of moment centre rescues it.

Nothing about capacity. The answer is a force. Whether the member can carry it is a separate question, and for a compression chord it depends on the distance between its restraints rather than on the number the cut produced.

The habit it teaches #

Underneath the technique is a disposition worth having independently of trusses.

The instinct when facing a large structure is to solve it. The method of sections says: decide what is actually wanted, then draw the boundary that isolates it, then choose where to stand so that everything else drops out. Two of those three steps are about arranging the question rather than answering it.

That instinct transfers. Enlarging the free body to make a difficult force internal is the same move in reverse — pushing the boundary out until an unknown disappears rather than pulling it in until an unknown is exposed. Both are decisions about where the cut goes, and in both the arithmetic afterwards is mechanical.

It also explains a persistent feature of hand structural analysis, which is that the experienced version looks nothing like the taught version. The taught version solves systems. The experienced version cuts once, in the right place, takes moments about the right point, and writes one line — and it looks like intuition when it is entirely a choice about geometry.

Where the model stops #

Pin joints. As with everything about trusses, the cut faces are assumed to carry axial force only. Real joints are rigid and transmit some moment, so a real cut face has three quantities on it rather than one, and the extra two are ignored on the argument that they are small and self-limiting.

Loads at panel points. A load applied between joints puts bending into the chord, which the section cut does not report. The axial force it returns is correct and incomplete.

Straight members and small deflections. The moment arms are measured on the undeformed geometry, so a frame that has swayed has lever arms the calculation does not know about.

Three unknowns exactly. A cut through four members is not a harder version of the same problem; it is an unsolvable one. Some frames — a K-truss, a subdivided panel — have no three-member cut in some regions, and the method has to be applied twice with an intermediate result, or abandoned.

The figures share a limitation worth stating plainly. The removed half of the truss is drawn faintly, which makes it look as though it is still contributing something. It is not — it has been replaced entirely by the three forces on the cut faces, and if those three are correct then the ghosted members could be anything at all. What is drawn faintly is a reminder of what the structure looks like, and it is not part of the calculation.

The ladder from here #

Later rungs on this anchor: choosing the moment centre when the chords are not parallel. Cuts that are not vertical. The method applied to three-hinged frames. Sub-assembly cuts and free bodies around whole storeys. The portal method and the cantilever method, which are section cuts with an assumption added to make a redundant frame tractable. Sections in space trusses, where the count becomes six. And the relationship between a section cut and the equilibrium matrix, which is the same operation in two notations.

The technique is usually credited to Ritter in 1862, who published it as a way of avoiding the labour of graphical analysis for a single member. It has survived every change in how structures are analysed since, because the question it answers — what is the force in that one member — has never stopped being the question that gets asked.