Which member moved the roof

A loaded beam sags because it curves. Every fibre above the neutral axis shortens, every fibre below it lengthens, and the section rotates a little more at each station along the span; the deflected shape is the accumulation of those rotations. A truss does none of that. Its members are straight bars carrying axial force only, they are still straight after the load arrives, and there is no curvature anywhere in the structure to integrate.

The roof comes down all the same. The only thing that has changed is that each bar is a little longer or a little shorter than it was, and the joints have moved to the positions those new lengths allow. A truss deflection is therefore not an integral at all. It is a sum with one term per member, and every term belongs to an object that can be bought, welded and paid for separately.

The widths in that drawing are shares of a movement, and nothing else. It is worth putting the same frame beside the picture a truss is usually shown in, which is coloured by force.

The sum, and what each term is #

For a pin-jointed frame the unit-load method reduces to

$$\delta = \sum_{\text{members}} \frac{F f L}{EA},$$

with $F$ the force in a member under the real load and $f$ the force in the same member under a single unit load placed where the answer is wanted.

The term is easier to read split in two. The quantity $FL/EA$ is a length: it is how much that member actually got longer or shorter under the real load. Call it the member’s extension. Then $f$ is the fraction of that extension which arrives at the joint being measured — the leverage the member has over that particular movement. So

$$\delta = \sum f \times (\text{extension}),$$

and the deflection of a roof is a weighted sum of twenty-one small length changes.

That is the whole argument of this essay, and everything below is a consequence of it. The sum is attributable. A beam’s deflection can be broken up by place — this third of the span contributes that much — but not by thing, because a beam is one continuous object. A truss’s deflection breaks up into parts that have part numbers.

Two free bodies, and only one of them is asked a question #

The two force sets in the sum come from two separate analyses of the same frame, and naming them separately is the point of the method.

The first free body is the whole truss under its real load. Five panel loads of 10 kN at the top nodes, a pin at the left support and a roller at the right, so vertical equilibrium of the whole gives 25.0 kN at each end. Cutting one joint out of the frame and writing $\Sigma H = 0$ and $\Sigma V = 0$ for it gives two member forces, and walking joint to joint gives the rest.

Cutting a whole section rather than a joint gets a chord force in one line. A vertical cut through the panel between $x = 2$ and $x = 3$, with moments taken about the top node at $x = 2$, gives $25 \times 2 - 10 \times 1 = 40$ against a lever arm of $0.85$, so the bottom chord there carries $47.1$. The same cut, moments taken about the bottom node at mid-span, gives $25 \times 3 - 10 \times (1+2) = 45$ over $0.85$, and the top chord at mid-span carries $-52.9$. Answering one question without solving the rest is exactly what this method is for, and the two numbers it returns are the two largest terms in the sum.

The second free body is the same truss under a unit load at the point in question, and nothing else on it. No panel loads, no self-weight, one downward force of one unit at the bottom node at mid-span. Its reactions are 0.5 each, its mid-span moment is $0.5 \times 3 = 1.5$, and the top chord force it requires is $1.5/0.85 = 1.765$.

This second analysis is the only place in the whole calculation where the question “deflection where?” is asked. The real analysis knows the load and nothing about the question; the unit analysis knows the question and nothing about the load. Ask for the deflection of a different joint, or in a different direction, and only the second one changes.

Multiplying: the top chord at mid-span contributes $52.9 \times 1.765 \times 1/1 = 93.4$, which against a total of $631.06$ is the $14.80\%$ printed on the hero figure. The bottom chord at mid-span contributes $47.1 \times 1.176 = 55.4$, or $8.77\%$. The picture is the arithmetic.

Two different kinds of nothing #

The sum has twenty-one terms and three of them are zero, for two entirely different reasons — and telling the two apart is the most useful thing the method does.

The familiar zero. Take the bottom node one panel in from the left support as a free body. Three members meet there: two bottom chords, collinear, and one vertical. No load is applied at that node, so $\Sigma V = 0$ has exactly one term in it and the vertical carries nothing. The same argument holds at the node one panel in from the right. Those two members have $F = 0$, so their terms vanish however long they are and whatever they are made of — and, because the zero has to be exact, they are also the cheapest available check that the statics is right.

The stranger zero. Take the top node at mid-span as a free body. Three members meet there too: two top chords, collinear and horizontal, and the vertical hanging below. Under the real load a 10 kN panel load lands on that node, so $\Sigma V = 0$ gives the vertical exactly $-10.0$ kN — the whole panel load, one of the harder-working members in the frame. Under the unit load nothing at all is applied at that node, the same three members give the same equation with the load term removed, and $f = 0.000$.

The product is zero because one factor is. A member can be carrying the full applied load and be worth precisely nothing to stiffen, and no total, however carefully computed, can ever say so. It takes a per-member sum, and this is the finding a per-member sum exists to make.

Being idle is not a property of the member, though. It is a property of the pair, member and question: the mid-span vertical is idle with respect to the movement of the node beneath it. It remains the load path for that panel’s entire share of the roof, and taking it out is not an option — the frame that survives losing a member is a different question from the one asked here. The zero says something narrower and more actionable: money spent making that member stiffer buys no reduction in that deflection.

The truss with no verticals at all #

If the five verticals of a Pratt truss are between them 2.0% of the movement, a frame without any is worth looking at.

This is a different frame rather than the Pratt with its verticals deleted — different node positions, twenty-three members instead of twenty-one — so the comparison is between two designs and not between one design and its own subset. Even so, the headline is that a truss without the entire member class that contributed 2.0% is, if anything, slightly stiffer: 596.97 against 631.06.

The composition has tidied itself in the process. With no verticals to hold, the diagonals do all the web’s work and take exactly 20.0%, and the two chords split the remaining four fifths evenly at 40.0% each. Three quarters of the movement was in the chords of the Pratt truss and four fifths is in the chords here, which is the same finding twice.

Depth changes the proportions, not just the answer #

Depth is the cheapest strength there is, and the reason is a lever arm: a chord force is the bending moment divided by the depth.

Both $F$ and $f$ in a chord’s term carry that reciprocal, since the unit load produces a moment diagram of its own that the same lever arm has to resist. So a chord’s contribution falls roughly as $1/d^2$. A web member’s does not: a diagonal gets longer as the truss deepens, and its force is set by the shear rather than by the moment, so its term falls much more slowly. The total is a mixture of two different behaviours, and sweeping the depth separates them.

The fitted exponent is the evidence. A pure chord structure would come out near $-2$ and a pure web structure near $0$; $-1.49$ is a mixture whose proportions are changing as the sweep runs, which is exactly what the lower panel shows. A deep truss is not simply a stiffer truss. It is a differently constituted one, and past some depth stiffening the chords stops being the thing to do.

That is not a small effect at ordinary proportions, and the ranking flips well inside the range a designer would consider.

Two of the essay’s own claims have just changed places. At a depth of 0.85 the ranking says stiffen the top chord at mid-span; at a depth of 1.5, the same truss, the same load and the same question say stiffen the end diagonals. The list is not a fact about trusses. It is a fact about this truss, at this depth, for this question — which is precisely why it has to be computed rather than remembered.

The chords take over with span #

The other sweep runs the opposite way and is much steeper.

An exponent of 3.65 sits just below the span to the fourth that a beam of fixed section and proportional load obeys, and the shortfall has a cause rather than being noise: the chord terms are heading for a fourth power and the web terms are not, so while the mixture is still shifting the fitted slope lies between the two.

The 0.608 to 0.952 curve is the computed form of a piece of received wisdom — the chords grow with the span — and it says something the wisdom does not. At a short span the web is nearly 40% of the movement, so a long-span roof and a short one are not the same design problem scaled. The long one is designed at its chords and merely detailed at its web; the short one is not.

There is a second essay in this phase asking the same shape of question about a different object. The flange that is not all there asks which part of a wide flange is actually carrying stress, and answers with a distribution across the width rather than with a yes. This essay asks which member is actually carrying the roof’s position, and answers with a distribution across the member list. Both replace a boolean with a ranking, and in both the useful content is in the tail.

The beam that cannot be taken apart this way #

The contrast with a beam is worth drawing explicitly, because it is what makes the truss case special rather than merely convenient.

That beam’s movement does decompose — into a bending term and a shear term, and by position along the span through the product integral. But neither decomposition hands anyone a shopping list. There is no component of that beam that owns the 0.0049 mm; the shear deformation is distributed through the same steel that is doing the bending, and stiffening it means changing the whole section.

A truss’s parts list is genuine. The 14.80% belongs to one bar, and a different bar can be ordered. This is why the method is worth more on a truss than on a beam even though the theorem behind it is identical, and it is the reason a truss ranking survived as a hand calculation long after most hand calculations stopped being done.

Where the model stops #

Every member has the same EA. The whole sum was computed at EA = 1, so the shares reported are shares of a frame that is uniformly stiff, and $\delta$ simply scales as $1/EA$. Give each member its real area and the ranking moves — and since the ranking is what would be used to choose those areas, the honest reading is that the figures show the first iteration of a loop, not its answer.

The joints are pins. Every real truss joint is a plate with bolts or welds through it, and it restrains rotation. Real joints add stiffness the sum does not include, so the computed deflection is an over-estimate; they also add secondary bending the sum cannot see at all. The pin assumption is conservative for the number and silent about the stresses.

The frame must be statically determinate. The solver refuses a redundant one rather than approximating it, and the refusal is correct: a least-squares force set for a frame with one support too many is not the distribution the members carry, so a ranking built from it would be a ranking of nothing. Redundant trusses need a compatibility method, and the unit-load sum then reappears inside it as the way the flexibility coefficients are computed.

Geometry is not updated. $F$, $f$ and $L$ are all taken on the undeformed frame, so this is a first-order calculation like everything else on this site. For a shallow truss under a load large enough to matter, second-order effects add to the movement and the sum does not know.

The load arrives only at panel points. The load a member is given is a decision, and putting all of it on the top nodes is the decision made here. A purlin landing mid-panel bends the top chord locally, which is a deflection with no term in this sum at all.

What the drawings cannot show. The member widths in the contribution figures are shares of a movement — not forces, not areas, not stresses. A hairline in the hero figure may be a member in heavy compression, and the mid-span vertical is drawn as one. The figures also show a single question: every share on them is a share of the mid-span bottom-chord deflection, and asking for the movement of a top node, or a horizontal movement, would redraw all of them. There is no such thing as the contribution of a member in general.

The ladder from here #

Later rungs on this anchor: the Williot diagram, which found these same joint displacements graphically from the member extensions in the 1870s and is the geometry behind the sum. Negative terms — a member whose force reverses between the real case and the unit case makes the structure stiffer at that point, and it is rare and real. Deflection under a lack-of-fit rather than a load, where a member built slightly short is an extension imposed by hand. Camber, which is the deliberate use of exactly that. Thermal movement of a truss, where the extension of each member is $\alpha \Delta T L$ and the same $f$ weights it. The unit-load sum inside the force method, where the release is a member force and the compatibility condition is a relative movement. Optimal member sizing, where the ranking is fed back into the areas until it stops moving, and the result is the fully-stressed design that turns out not to be the lightest one. Deflection of a three-dimensional truss and a space frame, where each joint has three freedoms and a great deal more of the frame is idle than anyone expects. And the same sum written for a cable, whose stiffness is not in its material at all — where the geometry changes enough that $f$ has to be recomputed as the load goes on, and the linear sum stops being available.