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Ladders

A field says what an essay is about. A ladder says what else there is to say about it — the distinct arguments that stand against one idea, from the one that introduces it to the one that assumes all the others.
cable: pure tensionreflected herearch: pure compression

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.

2 rungs · structures
crown hinge — no moment here, by constructionH = 23.4H = 23.427.027.0thrust line and axis coincide — the definition of funicular

The hinge put in on purpose

An arch with two pinned feet cannot be solved by statics. Add a third hinge at the crown — deliberately weakening it — and the whole structure falls out of one moment equation.

1 rung · structures
neutral axiscompressiontensionI = 29.97 × 10⁶Z = 299.7 × 10³peak stress 200.2σ = M y ÷ I, at every height

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.

1 rung · sections
5010015020000.20.40.60.811.2slenderness (effective length ÷ radius of gyration)they cross at λ = 75squashingEuler bucklingreal columns, which are neither

Strong enough and still falls over

A column can fail at a fraction of the load its material could carry, by going sideways. Buckling is a failure of stability rather than of strength, and it is decided by geometry.

1 rung · stability
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

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.

1 rung · internal-forces
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

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.

1 rung · equilibrium
4 per unit lengthshear16.0moment32.0 at x = 4.00the moment peaks exactly where the shear passes through zero

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.

1 rung · internal-forces
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

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.

1 rung · equilibrium
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

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.

1 rung · stability
1289.510.5ΣM about one support gives the other reaction; ΣF then gives the first

Everything adds to nothing, and that is the whole of statics

A structure that stays put obeys two statements — the forces on it sum to zero, and so do the moments. Every number in the subject comes out of those two sentences.

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

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.

1 rung · equilibrium
the loadroller: vertical onlypin: any directionall three lines meet here

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.

1 rung · equilibrium
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

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.

1 rung · deflection
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

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.

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

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.

1 rung · internal-forces
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

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.

1 rung · stability
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

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.

1 rung · equilibrium
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

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.

1 rung · sections
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

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.

1 rung · internal-forces
1002003004005006007000200400600plate width (mm)slender beyond 370 mmyieldcritical stress — inverse square in the widthwhat the plate actually delivers, over its full width

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.

1 rung · stability
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

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.

1 rung · structures
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

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.

1 rung · deflection
neutral axiscontribution of each striptotal I = 79.86 × 10⁶the outer strips do almost all of the work

The material far from the middle does nearly all the work

A strip of steel contributes to bending stiffness in proportion to the square of its distance from the centre. Move the same steel outward and the section gets stiffer for nothing.

1 rung · sections
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

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.

1 rung · stability
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

The same steel in a different shape, and a factor of forty

Four sections of identical area, identical weight and identical cost. The stiffest is dozens of times the stiffest of the flattest, and the only thing that changed was the arrangement.

1 rung · sections
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

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.

1 rung · deflection
web centrelineshear centree = 31.7no twisttwiststhe flange flows are equal, opposite, and separated — which is a coupleand nothing about the section's 20.19 × 10⁶ second moment predicts it

The point that is not in the section

A channel loaded down its web twists. To stop it, the load must be applied through a point outside the steel entirely — in the air beside the section, where nothing can be attached.

1 rung · sections
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

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.

1 rung · sections
11.522.533.54050100150200250300span, relative to the first16×81×256×moment: the squareload: the first powerdeflection: the fourth

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.

1 rung · deflection
0.511.52050100150200250300depth of the truss2501671251007150the same moment, resisted by a longer lever arm

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.

1 rung · structures
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

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.

1 rung · internal-forces
tensioncompression2 carrying nothing

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.

1 rung · structures
real Mpeak 32.0a unit load, here and nowhere elseunit mM × marea ÷ EI = 197.50the unit load is the only place the question 'deflection where?' is asked

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.

1 rung · deflection

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