Structural Engineering & Statics
where the load goes
Generator

The truss generator

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 panelsA 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.tensioncompression2 carrying nothing

14 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.

2012.57.5shear 7.5moment 22.5the cut, at x = 5nothing was applied here — the internal forces are what the left-hand piece needs 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.

 the loadroller: vertical onlypin: any directionall three lines meet here Equilibrium

Three forces must meet at a point, and a drawing can find it

A body held by exactly three forces has their lines of action concurrent. That is a theorem, it is enough to solve for direction and magnitude, and for a century it was done with a straightedge.

 m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case 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.

 one panel braced twice, the next not at allm 9 + r 3 = 2j 12 · rank 11a mechanismthe same count, properly arrangedm 9 + r 3 = 2j 12 · rank 12stands up 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.

 tensioncompression2 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.

 0.511.52050100150200250300depth of the truss2501671251007150the same moment, resisted by a longer lever arm 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.

 20H 10.0 M 22.2H 10.0 M 22.2the two base shears add to the applied 20 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre Structural form

The frame that leans, and what stops it

A rectangle of pinned bars folds flat. Make the corners rigid instead of adding a diagonal and it does not — which buys an unobstructed opening and costs bending in every member of it.

 worst secondary bending: 24.3% of the axial stress, in the member markedaxial force there 16.7 kN · end moment 0.20 kNm · slenderness of the member 18the same members, the same loads, the same solver — only the releases differ Structural form

The joint that is not a pin

Every truss on this site is analysed as though its joints were frictionless pins. Almost none are. The bending that follows is called secondary, which is a claim about size — and the claim is checkable.

 the station being watched, x = 3unit load, at its worst position2.100shaded: where a spread load must stand to make this quantity worstthe horizontal axis is where the load is, not where the beam is cut Internal forces

The worst place to stand

A bridge is not designed for a load. It is designed for a load that moves, and for every station along it there is a different position of that load that does the most damage.

 span ÷ depth = 8plane sections holdspan ÷ depth = 4plane sections holdspan ÷ depth = 2off by 19%span ÷ depth = 1off by 31%the assumption is the theory — everything else is arithmetic on top of it Sections and stress

Plane sections stay plane, and what the assumption costs

Beam theory rests on one sentence about geometry. It is very nearly true for a slender member, wrong for a deep one, and everything in the subject that fails does so where it stops holding.

 K = 0.5both ends fixedK = 0.7one fixed, one pinnedK = 1both ends pinnedK = 2fixed at the base, free at the topsame column, same section, four ways of holding the endsthe load at which each buckles goes as 1 ÷ K² — a factor of sixteen across this row Stability

The ends decide the length that matters

Four columns of identical height and section, buckling at loads sixteen times apart. Nothing differs but what is holding the two ends.

 1002003004005006007000200400600plate width (mm)slender beyond 370 mmyieldcritical stress — inverse square in the widthwhat the plate actually delivers, over its full width 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.

 simply supportedstatics alonesag 32.0propped at one endneeds stiffnesssag 18.0hog 32.0built in at both endsneeds stiffnesssag 10.7hog 21.3the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment Deflection

One support too many, and what it costs to know

Add a redundant restraint and the load has two routes to the ground. Equilibrium cannot say how it splits, and the answer turns out to depend on stiffness — which is a different kind of question.

 real Mpeak 32.0a unit load, here and nowhere elseunit mM × marea ÷ EI = 213.33the unit load is the only place the question 'deflection where?' is asked Deflection

One deflection, without solving everything

To find how far one point of a structure moves, put an imaginary force of one unit there, multiply two moment diagrams together, and integrate. The answer arrives without ever solving for the deflected shape.

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