The drawing that is right except for a rotation

The sum that ranks a truss’s members by what they contribute to its deflection answers one question at a time. Its unit load sits at one joint and points one way, and every virtual member force $f$ belongs to that question and to no other. The eight-panel Pratt truss carrying 15 kN at each top joint moves 2,019.41 units at mid-span, at EA = 1, and the sum that says so says nothing whatever about the joint beside it.

The deflected shape is a different object. It is sixteen joints, each moving in two directions, and whoever is checking a roof for ponding, a bridge deck for its drainage fall or a truss for the clearance beneath it needs all of it. Done by unit loads it is up to thirty-two virtual analyses, each with its own twenty-nine member forces. Williot, in 1877, did it in one drawing.

A truss moves only because its members change length #

A statically determinate truss has no way to move other than by its members getting longer and shorter. Its joints are pins and its members are straight bars, and a triangle of bars cannot fold: for every set of three lengths there is exactly one triangle. Build the truss from triangles and the same is true of the whole. Give every member its new length and the new shape is fixed, apart from where the whole thing sits in the plane.

So the input to a deflection calculation is a list of extensions, one per member, $e = FL/EA$, and nothing else. Not the loads, which have already done their work by producing the forces. Not a unit load. Not an energy. On this truss the list is short enough to read: the bottom chord lengthens by 52.50 in each end panel and by 112.50 in the two middle ones; the top chord shortens by 90.00 near the ends and 120.00 in the middle; the end posts shorten by 105.00; the diagonals lengthen by 75.00, 45.00 and 15.00 working inward; and the verticals shorten by 37.50, 22.50 and 15.00, except the two beside the supports, which carry nothing and do not change at all.

What turns a list of lengths into a drawing is one fact about small movements. When a bar AB turns through a small angle about A, the end B moves at right angles to the bar — which is the direction of motion about any point of rotation, to first order. When the bar lengthens by $e$, B moves by $e$ along it. Any small movement of B relative to A is a mixture of the two, so the only thing the member fixes is the component along itself:

$$(\mathbf{d}_B - \mathbf{d}_A)\cdot\hat{\mathbf{u}}_{AB} = e_{AB}$$

One equation per member, and a line in the plane of possible movements of B: everything on it is consistent with the member’s extension, and the member has no opinion about where on the line B lies. Two members meeting at a joint give two lines, and two lines in a plane cross at one point. That is the whole of Williot’s construction.

Every joint from two joints already found #

The drawing is made in a space of displacements rather than positions. A point in it is a movement, not a place, and the truss’s own shape never appears.

It needs somewhere to start. One joint is declared to be still, and its movement is the origin of the drawing, called the pole. One member through that joint is declared not to turn, so its far end moves only by the member’s extension, along the member. From then on the construction repeats one step. Find a joint connected by members to two joints already placed. From the point for each of those, lay off the connecting member’s extension parallel to the member — away from the joint already found if the member lengthens, toward it if it shortens. From the end of each, draw the perpendicular. The two perpendiculars cross at the new joint’s movement.

On the Pratt truss the order takes fifteen steps for fifteen joints, and the count is worth noticing. The first step uses one member and each of the fourteen after it uses two, so the construction uses twenty-nine members, each of them exactly once. That is $2j - 3$ for sixteen joints, which is the member count of a statically determinate plane truss. The construction is not merely consistent with determinacy; it is determinacy, read as geometry. A truss with a thirtieth member would present, at some step, a third line that need not pass through the crossing of the other two, and the drawing would have to decide which member to disbelieve.

The drawing that results is tall and thin, and its most striking feature is the one that is wrong.

The roller has left its support #

The roller at L8 comes out 6,227.94 units above the pole. It cannot be there. The roller holds L8 at the height of L0, which is the only thing a roller does. And the mid-span joint L4, on a truss that visibly sags, sits 1,094.56 units above the pin in the drawing, as though the middle of the truss had risen.

Nothing in the construction consulted a support. It used the pin only as a place to start, and it used an assumption the truss does not honour: that the bottom chord in the end panel stays level. It does not. The truss sags, the first panel’s chord slopes down from the pin toward mid-span, and holding it level has swung the whole truss upward about L0 by exactly that slope.

The error can only be a rotation, and the argument is short. Every member’s extension is honoured at every step, so the drawn movements and the true ones both satisfy all twenty-nine member equations. Their difference therefore changes no member’s length. A movement that changes no member’s length in a rigid truss is a movement of the truss as a rigid body. In the plane that means two translations and one rotation, and nothing else. The pole fixed the translations when it put L0 at zero, which is where the pin really holds it. What is left is a single small rotation about L0, and the roller decides it: turn everything until L8 is back at the height of the pin.

That rotation carries L8, eight panels from the pin, through 6,227.94 units, which is 778.49 per panel. 778.49 is exactly the true downward movement of L1, and it has to be. L0–L1 was held level, so the drawing’s L1 is wrong by precisely its real drop over one panel’s length. The rotation in the drawing is 3.08 times the movement the whole calculation exists to find.

Mohr’s truss, turned through a right angle #

The correction was Mohr’s, a decade after Williot, and it is drawn rather than computed. A small rotation about L0 moves each joint at right angles to the line joining it to L0, by the angle times that distance. Plotted in the space of displacements, those movements form a figure with the same proportions as the truss itself, turned through a right angle and scaled down by the angle. Draw that figure with its pin at the pole and every joint has two points in the diagram: its image on Mohr’s truss and its Williot point. The joint’s true movement is the arrow from the first to the second.

It is the same Mohr who turned a beam’s deflection into a bending-moment diagram on a fictitious beam, and the two are one programme. Graphical statics had already made force a matter of drawing; Cremona’s reciprocal figure gave every member force of a truss in one diagram. The Williot–Mohr diagram does for movement what that figure does for force, and for a determinate truss the resemblance is closer than analogy. Cremona’s figure finds member forces joint by joint in an order where each joint has at most two unknowns. Williot’s finds joint movements joint by joint in an order where each joint has two unknowns. The table of numbers that says which members meet at which joints is the same table read in two directions: the equilibrium equations read it by joints and the compatibility equations by members, and each is the transpose of the other. On a simple truss the order that lets one construction proceed, reversed, lets the other proceed too.

The arrows say several things at a glance that no single sum says. L8’s arrow is horizontal and 615.00 long: the roller slides. L4’s points 2,019.41 down and 307.50 across, because the bottom chord’s left half has lengthened by exactly that much and the pin holds its left end. And each arrow is the short difference between two points that are thousands of units apart.

How large the correction is depends on where the drawing starts #

That last observation is a practical defect of the drawing as a drawing. The answer at L4 is the difference between a Williot point and a Mohr image, both some three thousand units from the pole. A pencil working at a scale that fits 6,258 units of diagram onto a sheet is making its answer out of two much larger measurements, and whatever error it makes in either is carried whole into a result a third their size.

But the correction was never forced by the truss. It came from the choice of starting member, and any member will do, since every choice produces a drawing that is right up to some rigid motion. So the size of that motion can be compared across choices.

The ranking has a plain reason. The correction is the rotation the held member actually undergoes, carried out to the far end of the truss. The members near the supports turn most: the end post and the first chord panel follow the slope of the deflected shape where it is steepest. The members near mid-span turn least, because the deflected shape is flattest there. And one member does not turn at all.

Start where symmetry holds still #

The truss is symmetric and so is its load, so its deflected shape is a mirror image of itself about the central vertical. A vertical line that is its own mirror image stays vertical. L4–U4 shortens by 15.00, and it does not turn.

Now the drawing’s only mistake is where it put the pole. It declared L4 still, and L4 in fact moves 2,019.41 down and 307.50 across. The pinned support therefore comes out at exactly that movement reversed, and so does the roller, apart from its horizontal slide. Every point in the drawing is out by the mid-span joint’s own movement and by nothing else. Mohr’s truss has shrunk to a single point. The drawing is the deflected shape relative to mid-span, symmetric about its own centre line as the truss is, with pairs of joints L1 and L7, L2 and L6, sitting level with each other on either side.

The accuracy problem goes with the rotation. Every arrow is now one fixed shift, and the diagram’s scale is set by the truss’s own movements rather than by three times them. This is the ordinary advice of the drawing-office textbooks, and the chart above is the measurement behind it: start in the middle, from a member the loading keeps from turning.

A Warren truss makes the advice slightly less obvious to follow, since it has no vertical at all.

The member symmetry holds still is the central top-chord member, which is its own mirror image lying horizontally. Held level from U4, the construction takes sixteen steps for sixteen joints and uses all thirty-one members once each, again $2j - 3$. Started from the pin instead, the Warren’s correction is 3.13 times its answer, a shade worse than the Pratt’s; started from the middle, it is zero. The lesson does not depend on the arrangement of the web. It depends only on whether some member lies on the axis of symmetry, and a symmetric truss with an even number of panels always has one.

One construction against seven unit loads #

The claim at the start was that one drawing replaces a family of analyses, and it can be checked joint by joint.

Seven unit loads, seven sets of virtual forces, 203 products $FfL/EA$, for the vertical movements of one chord. The horizontal movements are seven more analyses and the top chord’s are fourteen more. The construction gave all of them, from twenty-nine extensions and fifteen crossings, and it never computed a virtual force at all.

The comparison is not a contest the unit-load method loses; the two answer different questions. A unit-load calculation earns its place when one movement is wanted, and its terms are what rank the members. It cannot rank anything without a question, and the construction cannot rank anything at all: the drawing contains no record of which member produced which part of which movement. A stiffness solution gives every movement at once as the construction does, by assembling and solving all the compatibility and equilibrium equations together. For a determinate truss Williot’s construction solves the same compatibility equations one joint at a time, in an order where each step has exactly two unknowns. It is back-substitution, drawn.

What the drawing shows that a sum hides #

The whole shape carries relationships that are invisible one joint at a time, and several can be read straight off the numbers.

The roller’s slide is the bottom chord’s extension, summed. L8 moves 615.00 horizontally, and the bottom chord’s eight extensions — 52.50, 52.50, 90.00, 112.50, 112.50, 90.00, 52.50, 52.50 — add to 615.00. The bottom chord is a straight row of members between the two supports. To first order a joint’s vertical movement does not change its horizontal distance from its neighbours, so the horizontal movements accumulate along the chord member by member, and only extensions enter.

The top chord’s joints move the other way relative to each other. U1 moves 630.00 to the right and U7 moves 15.00 to the left. The difference, 645.00, is the top chord’s total shortening, summed the same way along its six members. A top-chord joint near the roller moves back toward the pin while the bottom-chord joint beneath it moves away.

The mid-span vertical’s shortening is the whole difference between its ends. U4 moves 2,034.41 down and L4 moves 2,019.41, and the 15.00 between them is L4–U4 shortening under its panel load. That is the member the per-member sum found contributes nothing to L4’s movement. Both statements are true together. Its shortening reaches U4 in full and reaches L4 not at all, and a drawing of the whole shape shows exactly where each member’s change of length goes.

And the verticals beside the supports make L1 and U1 move together. L1–U1 carries nothing, so its length does not change, and the two joints drop by the same 778.49 while moving horizontally by different amounts.

A list of extensions from anything #

The construction never asks why a member changed length. It takes a list of extensions and returns the movements that fit them, and a member force was only ever one way of producing the list.

That makes it the natural instrument for movements no load causes. A member heated or cooled has an extension $\alpha\,\Delta T\,L$. A member fabricated a few millimetres short has an extension of minus a few millimetres. A truss built to the wrong shape on purpose is a truss whose bottom chord is given a chosen set of shortenings, so that the drawing of those alone is the camber. In each case the drawing is identical in method, and only the list differs.

It also gives a check that owes nothing to any analysis of forces. Give every member the same strain and the truss must scale about its pin without changing shape: every joint moves outward from L0 in proportion to its distance from it, and the roller slides with no rotation needed. The construction returns exactly that, because a uniform strain of a determinate truss is a similarity transformation, and a similar figure has the same angles.

Where the construction stops #

Small movements only. The perpendicular is the tangent to the arc a rotating member’s end really follows, and the error is of second order in the rotation. For a truss moving a few hundredths of its panel length it is invisible; for a truss whose geometry changes enough to change its forces, like a cable whose stiffness comes from its shape, the drawing is the wrong tool for the same reason the linear sum is.

Determinate trusses, or trusses whose forces are already known. The list of extensions needs member forces, and a redundant truss’s forces depend on its movements. The force method’s release breaks the circle: remove a redundant member, draw the released truss, read the gap across the removed member, and find the force that closes it. The drawing works on the released truss, and the extensions of the real one come afterwards.

An order must exist. Each step needs a joint joined to two joints already placed by members that are not in line. A simple truss, built by adding one joint and two bars at a time, always has such an order, and the order is the one it was built in. A compound truss of the kind the method of joints also stalls on, and every complex truss, can have none: at some point every remaining joint has at most one placed neighbour, and the construction stops. The usual way in is to swap one member for a substitute that makes the truss simple, draw that, and correct for the swap afterwards, which is the force method again in graphical form.

Pins and axial strain. The members are bars, so everything the sum omits the drawing omits too: bending in continuous chords, slip in bolted joints, and secondary moments at rigid connections.

And a drawing’s accuracy is its scale. A correction three times the answer is a real loss on paper and none at all in arithmetic, which is why the start in the middle mattered to the people who drew these and does not matter to a program reproducing them.

Still open: the truss whose extensions are chosen rather than computed #

The construction separates the geometry of a truss from the cause of its movements, and several further questions follow from that separation. A truss under temperature alone — the sunlit top chord of a long roof truss, whose extensions are imposed while every member force stays zero, so the drawing is the whole answer. Lack of fit in a redundant truss, where the Williot diagram of the released truss gives the gap and the self-stress closing it follows. Camber specified as chord shortenings, drawn directly, with the per-member shares from the sum saying which members it is cheapest to shorten. The influence line for a member force, which is itself a deflected shape: remove the member, impose a unit extension across the gap, and the Williot diagram of the resulting mechanism is the influence line in one drawing. And the kinematic reading stated in full, in which Cremona’s figure and Williot’s are the equilibrium and compatibility matrices of one truss drawn as each other’s transposes, and the question is whether a truss that defeats one construction always defeats the other.

The unit-load sum and the Williot diagram were published thirteen years apart and are one theorem. The sum weights each extension by a virtual force and adds; the drawing carries each extension to the joints it moves and draws. Virtual work proves the first; the geometry of small rotations proves the second; and both rest on the fact that a determinate truss has exactly as many members as it has independent ways to move.