Structural Engineering & Statics
where the load goes
Generator

The shear-moment 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.
Load, shear and moment — a simple spanThe applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.4 per unit lengthshear16.0moment32.0 at x = 4.00the moment peaks exactly where the shear passes through zero

24 essays call shear-moment. 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.

 25.0the cut — three members, three unknowns47.06-52.947.72moments about here kill two of the threesolved without touching any of the other 18 members Equilibrium

Answering one question without solving the rest

A truss of fifty members can be interrogated about one of them. Cut through three, take moments about the point where two of them meet, and the third falls out in a single line.

 resultant 24.0at x = 5.33, the centroid of the areamomentspread: 24.6replaced: 42.7reactions agree exactly (8.00 and 8.00); the peak moment does not Equilibrium

The load that is spread out, and the force that replaces it

A distributed load can be swapped for a single force at its centroid. The reactions come out identical and the bending moment does not, and knowing which side of the cut the swap is legitimate on is most of the skill.

 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.

 cable: pure tensionreflected herearch: pure compression Structural form

The shape that carries itself, and the arch that is its reflection

Hang a chain and it takes the one shape that carries its load in pure tension. Turn the shape upside down and it carries the same load in pure compression. That is what an arch is.

 10148126H = 27.9, the same at every stationeach vertex is a load; each slope is the running vertical sum ÷ H Structural form

The polygon that finds the shape

A hanging string under five loads has no smooth curve in it — it has five vertices and six straight segments, and every slope in it is a running sum divided by one number.

 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.

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

What a cut reveals, and why it was there all along

Cut a beam anywhere and two quantities appear on the face — a shear force and a bending moment. Nothing was applied there. They are what the material was already doing.

 4 per unit lengthshear16.0moment32.0 at x = 4.00the moment peaks exactly where the shear passes through zero Internal forces

The diagram is an integral, and that is why it can be drawn by eye

Load, shear and moment are one function and its two integrals. Once that is seen, the diagrams stop being things to calculate and become things to sketch.

 00.050.10.150.20.250.30.3501020304050overhang, as a fraction of the spanbest at 21% — peak 8.8sagging, mid-spanhogging, over the supportmoving the supports in by a third of the way halves the worst moment Internal forces

Where to put the supports, which is not at the ends

Moving the supports of a uniformly loaded beam inward by about a fifth of its length halves the worst bending moment. The load has not changed and nor has the beam.

 sagging hinge at 4.69hinge at the fixed endlowest upper bound: 7.29every hinge position gives an upper bound on the collapse loadassumed position of the sagging hingecoefficient 11.66 Mp ÷ L² Internal forces

After the first yield, which is not the end

A steel beam whose extreme fibre has reached yield has not failed. It has started forming a hinge, and collapse waits until there are enough hinges to make a mechanism.

 moment19.6 sagging24.5 hogging30.6 if the spans were simplereactions 14.0 38.5 38.5 14.0 — the inner supports carry far more than a sharethe continuous case needed stiffness; the comparison did not Internal forces

The moment over the support, and what it buys

Run a beam over its supports instead of stopping at each one, and the mid-span moment falls by a third while a new moment appears where there was none. Nothing was added but continuity.

 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.

 12012080resultantmidspan05101520020040060080010001200station along the spangreatest moment ever seen there1160.2 at 9.88Barré: 1160.3 at 9.88 Internal forces

The train that is worse than its heaviest axle

An influence line says where to stand one load. A vehicle is several loads at fixed spacings, and the worst arrangement never puts the heaviest one at the peak.

 neutral axiscompressiontensionI = 29.97 × 10⁶Z = 299.7 × 10³peak stress 200.2σ = M y ÷ I, at every height Sections and stress

Bending is a pair of forces, pushing and pulling

A bending moment is not a mysterious twisting. It is a push near the top of a section and a pull near the bottom, separated by a lever arm — a couple, made out of stress.

 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.

 neutral axispeak 0.4stressflow, q = VQ ÷ Imean stress 0.19 — the value a shear divided by an area would givepeak 2.15× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does Sections and stress

The shear nobody draws

A stack of loose planks slides at its ends when it is loaded. Glue them and the sliding stops — and whatever the glue is now carrying is a stress that no bending calculation contains.

 00.20.40.60.80246810applied load ÷ buckling load1.1×1.3×1.4×2.0×3.3×10.0×first-order analysis says the answer is always 1×one over one minus the ratio Stability

The load that makes itself worse

A structure that has leaned carries its weight off the axis, which makes it lean further. The amplification is one over one minus the load ratio, and it runs away long before the buckling load.

 2000400060008000100001200002004006008001000distance between lateral restraintsthey cross at 3803the plastic capacity of the sectionelastic critical momentSt Venant torsion alone — what is left at long lengthswarping dominates here Stability

The beam that fails sideways

A deep narrow beam bending in its strong plane can, at a moment well below its capacity, swing out of that plane and twist. The failure has nothing to do with how much it can carry and everything to do with what is holding it.

 0.60.811.21.41.61.8200.511.5span, relative to the firstthey cross heredeflection runs out at 1.40strength runs out at 1.54the limitstrengthdeflection Deflection

Stiffness is not strength, and usually it is the one that governs

A beam can be nowhere near failure and still be unusable, because it has moved too far. For most long-span members that limit arrives first, and stronger steel does not help at all.

 11.522.533.54050100150200250300span, relative to the first1×5×16×81×256×moment: the squareload: the first powerdeflection: the fourth Deflection

Span to the fourth, which is why spans are short

Doubling a span multiplies its deflection by sixteen. No other relationship in ordinary structural work is that steep, and it is the reason long spans are always a different kind of structure.

 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.

 20 at 2δ at B = 93.33320 at 6δ at A = 93.334the shapes have nothing in commonand the two readings agree to 1e-14which is why an influence line can be measured by pushing the structure where it is easy to push Deflection

The theorem that swaps the question round

Push here and measure there; push there and measure here. The two readings are identical, for every elastic structure, whatever its shape — and that fact turns an influence line into something a model can be asked for directly.

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