Structural Engineering & Statics
where the load goes
Concept

Truss — where it appears

An assembly of straight members meeting at joints that carry no moment, in which every member is a two-force member with one number in it. Its joints carry no moment by assumption and a great deal in reality, and the difference is a secondary stress that fatigue rather than strength notices.

Named by 19 essays across 3 fields — each of them below, with the objects they name alongside it.

The force runs along the chord whatever route the member takes. A member pinned at two points 10 m apart, loaded only at those two points, and bowed 1.2 m off the line between them. Equilibrium leaves the two end forces no choice: equal, opposite, and along the chord. So at every section the axial force is P cos α, the shear is P sin α, and the bending moment is the force times the perpendicular offset — 240 kNm at the crown for 200 kN at 1.2 m, which is a multiplication rather than an analysis. Straighten the member and the moment diagram is identically zero; that is the case a truss member is in, and the reason it carries one number.

The member with only one direction

If a body is in equilibrium under forces applied at exactly two points, those forces are equal, opposite and along the line joining the points. It is three conclusions from two equations, it is the shortest real theorem in statics, and nearly everything that follows depends on it without saying so.

equilibrium · Two force member A derivative taken with a ruler, and the step that makes it worst. Castigliano's theorem says the deflection is ∂U/∂P, and the derivative here is taken numerically — two solves at ±dQ and a central difference. Against the unit-load answer of 1.720635e-2 it agrees to 1.6e-13, which for a linear structure it must: ∂N/∂P is exactly the force a unit load produces, so the two expressions are the same sum written twice. The error against step size is the classic pair of straight lines — truncation falling as the step shrinks, round-off rising as the difference of two nearly equal energies loses its digits — meeting near dQ = 1.2e+1. For a linear structure the truncation term is exactly zero, so what is drawn here is round-off alone.

The deflection that is a derivative

A structure's strain energy is one number. Differentiate it with respect to a load and out comes the displacement under that load — and the trick that makes it a method rather than an identity is that the load does not have to be there.

deflection · Strain energy Three ways to move the same column, and they are not close. The same 2000 kN moved 3 m across 14 m, built three ways and drawn to one scale. The deep beam is 1.14 m of concrete, 24.1 tonnes, and settles 60.6 mm in the long term. The storey-deep truss takes the same moment as a couple at 3.6 m centres, so its chords carry M/h and it weighs 2.6 tonnes — a fifth of the beam — while settling 15.8 mm, and it does not creep. The wall is 72.8 tonnes and hardly moves at all, 1.71 mm, of which 36% is shear rather than bending — which is what a member as deep as it is long always does, and is why beam theory does not describe one. A wall as a deep beam is the stiffest of the three by a factor of 35.4.

The same span, four ways

A beam, a truss, an arch and a cable can all cross the same gap under the same load, and the choice between them is usually described as a matter of judgement or of taste. It is neither. Each carries the load by a different mechanism, each mechanism has a different exponent, and an exponent decides the ordering at every span rather than at some spans.

structures · Form selection The corner columns take what the middle ones did not. Axial stress in the columns across one flange face of a 36 by 36 m framed tube, at the base. Plane sections says the flat line: every column on the face at the same distance from the neutral axis, therefore at the same stress. The solved distribution is the curve — 78.5 N/mm² at the corner against 13.4 in the middle, a ratio of 5.88. The middle columns lag because the only route the axial force has into them is the in-plane shear of the spandrel frame, bay by bay from the corner. The face is carrying its resultant on an effective width of 45 per cent, and the tube deflects as though its second moment were 64 per cent of the gross.

The columns that lean

A framed tube carries its wind load by bending the spandrel beams between its columns, and it does it badly — the corner columns take nearly six times what the middle ones do. Tilt the columns instead, so the perimeter is triangulated, and the same shear is carried axially. The concentration falls to 1.21 and the tube recovers most of the stiffness the plan said it had.

structures · Diagrid A closed force polygon. The 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.

One drawing solves the whole truss

The method of joints solves a truss one joint at a time, and each solution is thrown away as soon as the next begins. Drawn instead of computed, the joints share their edges — every member's force appears once in a single figure, and the figure's own closure is the check.

structures · Truss A Pratt truss of 8 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing.

Halving the panel buys a shorter strut

Subdividing a truss into more panels of the same span and depth barely changes the chord forces, because the couple that carries the moment has not moved. What it changes is the length of every compression member, and a buckling capacity goes as the inverse square of a length.

structures · Truss 3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 28.1 to 18.0, and a hogging moment of 22.5 appears over the supports where there was none.

The chord is a continuous beam

A truss analysis reports one number per member, and for a chord running through eight panels that number is an axial force with no bending in it. The chord is a single piece of steel spanning between its own nodes, and anything landing between them bends it.

structures · Truss Influence line for the shear force at x = 10.5. The shear force at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 10.56, giving 0.560.

Two diagonals, one of which is absent

A truss diagonal is sized for the shear in its panel, and near mid-span that shear changes sign depending on where the load stands. A member that can only pull cannot carry the reversed case, so the panel gets a second diagonal — and at any instant one of the pair is not there.

structures · Truss Every member's share of the movement, and they are not the members expected. A Pratt truss of eight panels at a depth of 1, carrying 15 kN at each top node, with the movement of the bottom chord at mid-span attributed member by member. The unit-load sum δ = ΣF·f·L/EA gives 2019.41 at EA = 1: 49.4% from six top chords, 30.8% from eight bottom chords, 16.8% from eight diagonals, 3.0% from seven verticals. The single worst member is a top chord at mid-span at 11.9% of the whole. Each member is drawn at the width of its own share. The same deflection from a stiffness solution that shares none of this arithmetic is 2019.41, a relative residual of 1.4e-14.

The member that is not worth stiffening

A truss's deflection is a sum of one term per member, and a term is zero whenever either force in its product is. A vertical carrying the whole of a panel load can contribute nothing at all to the movement — which a total can never show and a per-member sum shows nothing else.

deflection · Truss deflection Removing each member in turn. Every member of a 8-panel pratt truss removed one at a time, with the worst demand on the survivors plotted against the member removed. Four of the 35 leave a mechanism — the bars drawn to the top of the frame — and for those there is no redistribution to compute, because there is no structure left. The rest redistribute, and the worst of them asks a survivor for 2.04 times what it carried before. A single number for robustness does not exist: it depends on which member goes.

A determinate truss has no robustness at all

Remove any one of a Warren truss's thirty-one members and what is left is a mechanism. Not weakened — gone, with no set of forces that holds the load in any position. Robustness is not a property a structure has by degree; it is bought by adding members that carry nothing until something else stops carrying, and a truss without them has none of it to measure.

structures · Robustness The whole deflected shape, found from the members' changes of length. A Pratt truss of eight panels at a depth of 1, carrying 15 kN at each top joint, at EA = 1, drawn as built and in its deflected shape, with every movement magnified the same number of times. The shape comes from Williot's construction and Mohr's correction: every joint's movement from the members' extensions alone, less the rigid rotation the supports forbid. The mid-span joint L4 moves down 2019.41 units, and the dots at every joint are a stiffness solution sharing none of that arithmetic, agreeing to 1e-14 of the largest movement. One construction gives all sixteen joints at once.

The drawing that is right except for a rotation

Williot's construction finds every joint of a truss from its members' changes of length alone, in one drawing — and puts the roller six thousand units off its support. The error is one rigid rotation, Mohr's diagram takes it away, and a drawing started from the member symmetry holds still never makes it.

deflection · Truss deflection The section, once the joint has supplied the fourth equation. Everything above a cut through panel 3. Four severed members; three equations. Moments about the mid-joint of the horizontal remove both diagonals and leave both legs. Moments about the point where the legs' lines meet, 36.0 m up, remove both legs and leave both diagonals. With the joint's result that the diagonals are equal and opposite, that second equation has one unknown: 20.83 kN in each diagonal. The legs follow: 26.1 T and 86.3 C.

The cut that needs a joint first

The method of sections works because a cut through three members leaves three unknowns and a point about which two of them have no moment. A K-braced tower has no such cut anywhere: every section severs two legs and two diagonals. One joint in the middle of a horizontal supplies the missing equation, and only in that order does each step have one unknown — after which the diagonals turn out to be carrying not the shear but the moment about the point where the legs would meet.

equilibrium · Method of sections A roof truss and its reciprocal figure. Left, a pitched roof truss of 8 m under three loads of 10.0 kN, its 13 members drawn in the colour of their force — 5 in tension, 6 in compression and 2 carrying nothing — with a letter on every region outside it between one external force and the next and a number on every cell inside it. Right, the force diagram: every lettered or numbered space is a point, every member is the line between the two spaces it separates, drawn parallel to the member and as long as its force, and every joint of the frame is a closed polygon. The eight joints and eleven spaces of the frame have become eight polygons and eight distinct points — fewer points than spaces, because some spaces land on one point, as the two either side of a member carrying nothing always do. Force times length adds to 190.0 kN·m over the tension members and 250.0 over the compression members, and the difference, −60.0, is fixed by the loads and where they act, whatever frame carries them. Each point was placed by crossing one member, and the 8 crossings not used to place anything all close to within 5e-15 kN.

Every space a point, every joint a polygon

A truss's force diagram is a second drawing of the truss in which the joints have become polygons and the spaces between members have become points. Maxwell showed in 1864 that the exchange runs both ways, so a designer can draw the forces first and ask what shape carries them. His theorem also says which frames have such a diagram at all, and the answer is a surprise, because it is about polyhedra.

equilibrium · Graphic statics Two pencil lines, and where they cross. Two lines drawn with a pencil 0.2 mm wide are two bands, drawn here much wider than a pencil so the shape can be seen, and they cross not at a point but in a parallelogram. At 60° apart the parallelogram's long diagonal is 2.0 pencil widths — 0.40 mm; at 12° apart the parallelogram's long diagonal is 9.6 pencil widths — 1.91 mm. The crossing's uncertainty along the bisector is the width divided by twice the sine of half the angle, so it grows without limit as the lines turn parallel, and a construction that finds a point by crossing two lines inherits it.

How wrong a drawing is

A pencil line is a band, and two bands cross in a parallelogram that grows as they turn parallel. The accuracy of a graphical construction is therefore a property of the angles it makes, not of the hand that made it — and the worst case is the shallow arch, the structure the method was most used on. Measured properly, the drawing's error there is the size of the builder's, and the check draughtsmen relied on cannot see it.

equilibrium · Graphic statics One volume of steel, divided three ways. A Pratt truss of eight panels at a depth of 1, carrying 10 kN at each top joint, drawn three times with every member as wide as its area, the total volume the same in each. With equal areas the mid-span deflection is 43503; fully stressed, with area in proportion to force, 32702; with area in proportion to the square root of the product of each member's real and virtual forces — the division that makes mid-span as stiff as this steel can make it — 31318. The fully stressed truss is 4 per cent short of the stiffest; the equal-area truss is 39 per cent short. No member is allowed less than 10 per cent of the equal area; the deflections are in units of load × length / (E × volume).

The truss that is stiff by accident

Give a truss a fixed volume of steel and ask how to divide it among the members. Sized for strength — every member at the same stress — its mid-span deflection comes within four per cent of the stiffest that steel can make, although stiffness was never asked about. The reason is an inequality, and the same inequality says where the accident stops: at the quarter point the strength design is sixty-nine per cent short of the best.

deflection · Truss deflection The strength design, and the least steel that is stiffer. A Pratt truss of eight panels at a depth of 1, carrying 10 kN at each top joint, each member drawn as wide as its section from a catalogue whose sections step by 25 per cent in area, the smallest 10 per cent of the largest member's need. Above, every member at the smallest section that carries its force. Below, the least steel that makes mid-span 1.50 times as stiff with every member still strong enough — the discrete optimum — with the 24 members it made larger drawn in the second colour: six of six top chord members, eight of eight bottom chord members, six of eight diagonals, four of seven verticals. The optimum uses 44 per cent more steel than the strength design, and it puts it where a member's real and virtual forces are both large; members either force leaves small are not touched.

The calculus answer, rounded, is the worst one

A real truss is built from a catalogue, and every member is rounded up to the next section. That rounding costs its stiffness almost nothing: the few per cent that separate the strength design from the stiffest survive it. What does cost is the next step. When a deflection limit governs, the obvious move — take the continuous optimum and round it up — needs more steel than any other way of stiffening the truss, and the exact discrete answer is within one per cent of a bound no catalogue can beat.

deflection · Truss deflection The forces the girder can hold with nothing on it. The one self-stress state of a K-braced girder of eight 3 m panels, 3 m deep: member forces in equilibrium at every joint with no load and no reaction. It lives entirely in the two middle panels — twelve members: the four diagonals meeting at the middle joint of the central vertical in tension, and the four central chord members and the half-verticals at the outer side of each middle panel in compression, at 0.89 and 0.45 of the diagonal force. It is symmetric about mid-span, so a symmetric load can call on it as freely as any other: symmetry does not supply the missing equation. Every other member of the girder is idle in it, drawn faint.

The redundancy only the sun can find

A K-braced girder that is symmetric about mid-span has one member more than statics can resolve, and it is at the centre, where two K's meet at one joint. Symmetry does not remove it, because the forces it allows are symmetric themselves. Loads barely touch it — a section calculation that ignores it is exact for a load at almost any joint. What finds it is a millimetre of misfit or a sunlit top chord, which put forces into the middle of the girder that no load calculation contains.

equilibrium · Method of sections Fully stressed, with a choice of diagonals. Two trusses sized so that every member is at the allowable stress under the full load, each member drawn as wide as its area, tension and compression in two colours. Above, an eight-panel Pratt truss as deep as a panel is long, with a counter-diagonal in each of its six interior panels, loaded by 10 at every bottom joint, found by resizing and reanalysing until nothing changes; below, the same truss without its counters. The counters in the two middle panels have shrunk to nothing (dotted) and the four nearer the supports have stayed, working in compression beside the diagonals in tension. The truss that kept them needs 1,200 units of steel against the Pratt's 1,220, and deflects 24.0 at mid-span against 26.0.

The truss whose forces follow its sections

In a determinate truss each member's force is fixed before its section is chosen, so sizing for strength and sizing for stiffness can be done in either order. Put a counter-diagonal in every panel and they cannot. The fully stressed design becomes an iteration that starves some counters to nothing and keeps others, lands on a different truss from every start, and — whichever it lands on — weighs the same and deflects the same. Then enlarge one group of members to stiffen it, and a vertical nobody touched is overloaded by 58 per cent.

deflection · Truss deflection

The factor of two belongs to one mode

A member that fails suddenly hands its force to the structure around it all at once, and the convention is to double the static answer: a load applied suddenly to a spring overshoots to twice its static deflection. A truss is not one spring. Take a diagonal out of a counter-braced truss in an instant and some members swing to three times their change of force, one swings the wrong way first, and a bottom chord whose force does not change at all passes through half as much again as it carries — because every mode overshoots by two, at its own time, and a member is a sum of modes.

structures · Robustness

Named alongside it

The objects these essays reach for when they reach for this one.

DeterminacyVirtual workDeflectionFree bodyFunicularGraphic staticsChord forceEquilibriumForce polygonFully stressed designIndeterminacyLoad path

All concepts