Structural Engineering & Statics
where the load goes
How this site is made

The figure library

Every picture here is generated from code at build time. This page lists the generators, each rendered at its defaults.

No figure on this site is a drawing that was made once and saved. Each one is a function: it takes parameters and returns SVG, so the same generator produces the 27° incline and the 5° incline without either being redrawn.

That is the reason the collection can keep growing without the illustrations drifting apart. A generator is written once, checked once, and every essay that calls it inherits the same line weights, the same colour roles, and the same behaviour in dark mode. There are 24 of them so far.

bending-stress

Bending is a push and a pullA section carrying a bending moment, with the stress at every height computed as the moment times the distance from the neutral axis divided by the second moment of area. It is compression above and tension below, and zero exactly at the neutral axis.neutral axiscompressiontensionI = 29.97 × 10⁶Z = 299.7 × 10³peak stress 200.2σ = M y ÷ I, at every height

cable-arch

The cable and the arch are the same curveThe shape a cable takes under a uniform load is a parabola, and it carries that load in pure tension. Reflected, the identical curve carries the same load in pure compression, which is what an arch is.cable: pure tensionarch: pure compression

cut-diagram

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

deflected-shape

The deflected shape is the moment, integrated twiceA loaded beam with its deflected shape above and its bending-moment diagram below. The shape was obtained by integrating the moment twice and fitting the constants to the supports; the vertical scale is exaggerated enormously.the largest movement, at x = 4.00momentdrawn at roughly three hundred times the real deflection —a beam at its serviceability limit moves about a three-hundredth of its span

determinacy

Counting unknowns against equationsThree 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.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

effective-length

The ends decide the length that mattersFour columns of identical height and section, buckling under four sets of end conditions. The effective length factor is the fraction of the column that behaves like a pin-ended one, and the buckling load goes as its inverse square.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

euler-buckling

The column curveFailure load against slenderness, as a fraction of the squash load. A stocky column crushes; a slender one buckles at the Euler load; the crossover is where the two curves meet, and real columns fall below both near it.5010015020000.20.40.60.811.2slenderness (effective length ÷ radius of gyration)they cross at λ = 75squashingEuler bucklingreal columns, which are neither

force-polygon

A closed force polygonThe 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.load 60strut 84.9tie 60starts and ends here

free-body

A beam, its loads and its reactionsA free body: the beam cut away from everything it touches, with the forces that were doing the touching drawn on it. The reactions are computed from the loads, so their arrows are to scale relative to each other.1289.510.5ΣM about one support gives the other reaction; ΣF then gives the first

joint-cut

Joint 0 of the truss, cut outOne joint of the truss with every force acting on it. Two equations — the horizontal and vertical sums — are enough for a joint with no more than two unknown member forces, which is the whole method.HV29.4-38.6reaction 0.0reaction 25.0ΣH = 0 and ΣV = 0, and nothing else is needed

moment-arm

The moment is the force times the distanceOne force applied at five distances from a pivot, with the moment it produces drawn as a bar. The force never changes; only the arm does, and the moment follows it exactly.pivotthe same force of 20, moved along the levermoment about the pivot120 × 1 = 20220 × 2 = 40320 × 3 = 60420 × 4 = 80520 × 5 = 100

p-delta

The load that makes itself worseThe amplification of a deflection against the ratio of applied load to buckling load. A structure at half its buckling load deflects twice as far as first-order analysis predicts, and the curve runs away well before the load is reached.00.20.40.60.80246810applied load ÷ buckling load1.3×1.7×2.5×5.0×10.0×first-order analysis says the answer is always 1×one over one minus the ratio

parallel-axis

Moving the flanges apartThe second moment of area of an I-section against its depth, with the flange and web areas held constant. The growth is close to quadratic, because the parallel-axis term dominates everything the flanges contribute about their own centres.1001502002503000M10M20M30M40M50M60Moverall depth1.0×2.6×4.9×7.9×13.8×21.3×same steel, moved apart

redundant-beam

One support too manyThe same uniformly loaded beam with three sets of restraints, and the bending moment in each. Adding restraint moves moment from mid-span to the supports and lowers the peak — but only the first case can be solved by statics.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

second-moment

Every strip counts by the square of its distanceA rectangular section divided into equal strips, with each strip's contribution to the second moment of area drawn beside it. The strips are identical in size; only their distance from the neutral axis differs.neutral axiscontribution of each striptotal I = 79.86 × 10⁶the outer strips do almost all of the work

section-compare

The same material, four waysFour cross-sections of identical area, so identical weight and cost, with the second moment of area computed from each profile's own geometry. Only the arrangement differs, and the stiffest is many times the flattest.the same, laid flatI = 0.06 × 10⁶1.0× the firstsquareI = 0.75 × 10⁶13.3× the firsttall rectangleI = 10.00 × 10⁶177.8× the firstI-sectionI = 24.29 × 10⁶431.8× the firstevery section here has an area of 3000 — only the shape differsthe bar is the second moment of area, to scale

shear-moment

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

slenderness-cost

Length costs more than it looksThe same column section at four lengths, with the buckling capacity of each drawn as a bar. Capacity falls as the inverse square of the length, so a column three times as long carries a ninth as much.1× the length100% of the capacity1.5× the length44% of the capacity2× the length25% of the capacity3× the length11% of the capacityidentical section, identical material, identical end conditions

span-fourth

Deflection goes as the fourth power of the spanDeflection against span for a constant load intensity and section, with a straight line for comparison. Doubling the span multiplies the deflection by sixteen, while the bending moment only quadruples.11.522.533.54050100150200250300span, relative to the first1×5×16×81×256×moment: the squareload: the first powerdeflection: the fourth

support-position

Where to put the supportsPeak sagging and hogging moment for a uniformly loaded beam, against how far the supports are moved in from the ends. The best arrangement is where the two curves cross, and it is nowhere near the ends.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

three-force

Three forces must meet at a pointA body held by two supports and one load. The reaction at the roller is vertical and the load's direction is given, so their lines of action fix a meeting point — and the pin reaction has to point at it.the loadroller: vertical onlypin: any directionall three lines meet here

truss

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

truss-depth

Chord force against truss depthThe 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.0.511.52050100150200250300depth of the truss2501671251007150the same moment, resisted by a longer lever arm

which-governs

Which limit arrives firstUtilisation of the strength limit and of the deflection limit, against span. Strength grows as the square of the span and deflection as the fourth power, so the two cross — and past the crossing a beam is sized by how far it moves rather than by what it can carry.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