Structural Engineering & Statics
where the load goes
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A Pratt truss of 6 panels

Rendered here at the parameters it defaults to, with every essay that calls it — which is the same list as the blast radius of changing it.
A Pratt truss of 6 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.

A Pratt truss of 6 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.

16 essays call truss. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this generator changes every one of these figures.

The same beam, cut at x = 5. A beam separated at one station. On the exposed face a shear force and a bending moment appear, equal and opposite on the two pieces, with values obtained by summing the forces on whichever piece is easier. Equilibrium

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.

 Counting unknowns against equations. Three 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. Equilibrium

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.

 The count is necessary and not sufficient. Two pin-jointed frames, each satisfying m + r = 2j exactly. One of them folds anyway, because the equations are not independent; the ghosted outline is the motion that costs no member any change of length, drawn at an exaggeration of 0.55 of the span. Equilibrium

The count that does not see it

A frame can have exactly as many unknowns as equations and fold up anyway. The count asks whether there are enough equations; it never asks whether they are different from one another.

 A Pratt truss of 6 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing. Structural form

The triangle that cannot fold, and everything built out of it

A square of pinned bars is a mechanism. A triangle is not, and that single fact is the reason trusses exist and the reason they look the way they do.

 Chord force against truss depth. The force in a truss chord for a fixed bending moment, against the depth of the truss. The relationship is a reciprocal: the chords form a couple whose lever arm is the depth, so a shallow truss pays for it steeply. Structural form

Depth is the cheapest strength there is

Doubling the depth of a truss halves its chord forces without adding a gram of material to the chords. Nothing else in structural design is that cheap, and almost every structure has already spent it.

 A 8 mm plate, and the width it can be. The elastic critical stress of a plate in compression against its width, with the yield stress drawn across it. Below 370 mm the plate reaches yield before it buckles; above it the plate ripples first, and the fraction of the width still carrying load falls away — at 700 mm only 47 per cent of it is still working. Stability

The plate that ripples, and the width that is left

A wide thin plate in compression buckles at a stress that has nothing to do with the strength of the material. It then goes on carrying load — the middle drops out, and the edges work harder.

 Every member's share of the movement, and they are not the members expected. A Pratt truss of six panels at a depth of 0.85, carrying 10 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 631.06 at EA = 1: 47.2% from four top chords, 28.5% from six bottom chords, 22.3% from six diagonals, 2.0% from five verticals. The single worst member is a top chord at mid-span at 14.8% 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 631.06, a relative residual of 3.6e-15. Deflection

Which member moved the roof

A beam sags because it curves. A truss has no curvature anywhere — it comes down because every one of its members changes length, and the sum of those changes, weighted member by member, is a ranking that names which ones are worth stiffening. Usually not the ones a designer worries about.

 A buckled panel is a truss that nobody drew. A 1000 × 1000 panel of 6 mm web, at d/t = 167. It buckles in shear at 63.8 N/mm², which is 383 kN — and it then carries 696 kN, 1.82 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 22.5° with a membrane stress of 252 N/mm² over a width of 541 mm, and it pulls on the flange at 221.3 N per millimetre of its length. A web that never buckled at all would have reached 953 kN, so the panel ends at 73% of a stocky web's capacity on a fraction of its steel. Stability

The panel that carries more after it has failed

Everywhere else in this field a critical load is where the argument ends. A thin web is the exception — it buckles visibly, in waves anybody can see, and then goes on to carry nearly twice as much again by turning itself into a truss nobody drew.

 The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 12 m by 1500 mm, under 100 kN at mid-span. There is no diagonal in it, so each panel's 50 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 25.0 kNm, and it adds to an axial force of 200 kN from the global moment at the same point. The girder deflects 6.20 mm against 3.18 mm for the same members triangulated — 1.95 times — and 68% of that movement is chord bending that a diagonal would have removed entirely. Structural form

The truss with no diagonals

A rectangle without a diagonal is a mechanism, so a Vierendeel girder makes its corners rigid instead. The shear a diagonal would have carried as axial force now travels as bending in the chords, and bending is a far more expensive way to move a force.

 A frame with no plane to be drawn in. The tetrahedron, solved: three equations at every free joint, one axial force in every member. Members drawn heavy carry more; tension and compression are separated by the sign that came back from the solve rather than by inspection. The count is m + r = 12 against 3j = 12, which makes it exactly determinate, and joint equilibrium closes to 1.8e-15. Structural form

Three equations at every joint

A plane truss is determinate when m + r = 2j. A space frame needs 3j, and that one changed digit is why a cube of twelve bars is six mechanisms short while looking perfectly solid — and why every three-dimensional frame ever built is made of triangles in several planes at once.

 The cut that severs the stirrups is the cut that counts them. A cracked web drawn as the truss it has become: a tension chord along the bottom, a compression chord along the top, concrete struts between the cracks and stirrups crossing them. The free body is a cut parallel to the cracks, which over a lever arm of 495 mm severs z·cot θ/s = 8.3 stirrups at cot θ = 2.5. Every one of them is at yield, so the shear is their number times their strength, and flattening the crack raises the count rather than the strength. The same cut passes through the bottom chord, which is where the chord force the moment diagram does not contain comes from. Internal forces

The beam that becomes a truss

Once a web has cracked in shear there is no shear stress field in it any more. There are concrete struts, two chords and whatever crosses the cracks, and the angle of those cracks is not a property of the material — it is something the designer chooses, and every quantity in the beam moves when it changes.

 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. Stability

The tie that spends an afternoon as a strut

A tension member is chosen by its area and nothing else. A compression member is chosen by how that area is arranged. So a member whose force reverses under some load case is not merely being asked for the same number with the other sign — it is being designed against a different variable, and the same steel can carry seventy times more or less depending on a shape nobody chose for that purpose.

 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. Structural form

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.

 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. Structural form

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.

 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. Structural form

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.

 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. Structural form

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.

The library, page 6 of 7 — where truss sits