Structural Engineering & Statics
where the load goes
Concept

Second moment — where it appears

The integral of area times the square of its distance from an axis, which is what makes depth the cheapest strength there is. It goes as the cube of depth for a rectangle, so a member twice as deep is eight times as stiff for twice the material.

Named by 25 essays across 7 fields — each of them below, with the objects they name alongside it.

Loaded straight down, and moving sideways. An equal angle with a moment applied about the horizontal axis. Its principal axes lie at 45.0° to the drawn ones, so the neutral axis runs at -30.6° rather than horizontally, and the section moves 59% as far sideways as it moves down. The product of inertia that causes it is -1.066 × 10⁶ mm⁴, and it is zero for every section drawn in this field until now.

Loaded straight down, and it moves sideways

Every section drawn here so far had an axis of symmetry, and that symmetry has been doing silent work. Take it away and a vertical load produces a neutral axis that is not horizontal, a deflection that is not vertical, and on one ordinary section a sideways movement larger than the downward one.

sections · Principal axes The middle third, computed. The kern of a 400 × 600 mm rectangle, computed by asking, for every direction, how far the resultant can move before the far face would be pulled. It reaches ±100.0 mm vertically and ±66.7 mm horizontally, which are h/6 and b/6 exactly, and the region between is a rhombus rather than the ellipse the two numbers suggest — a resultant on a diagonal has less room than either axis allows.

The middle third

A material that cannot be pulled imposes a condition on where the load may land, and the condition is a region rather than a point. For a rectangle it is the famous middle third; for every other section it is a shape nobody quotes, and one ordinary section's is nearly twice as generous as the rule allows.

sections · Kern What is left after the first fibre yields, which is a property of shape. The shape factor — plastic modulus over elastic — for six sections, computed by finding each one's equal-area axis and summing ±f_y over it. The numbers contain no dimension, no stress and no material: a rectangle is exactly 3/2 whatever its size, a diamond exactly 2, a circle 16/3π. The spread is the argument. An I-section keeps only 13 per cent in reserve past first yield, because nearly all its material is already at the extreme fibre and there is nothing further in to recruit; a diamond keeps 100 per cent, because most of its material is near the middle and doing very little elastically. So the section shapes that are best at elastic bending are the ones with the least left afterwards, which is exactly backwards from the way the reserve is usually described.

What is left after the first fibre yields

The elastic section modulus stops at the moment the outermost fibre reaches yield. Nothing else in the section has, so it goes on taking load — and how much more it takes turns out to be a property of the shape alone, with no dimension, no stress and no material anywhere in the answer.

sections · Shape factor One slit, and the torsional stiffness falls by a factor of hundreds. A 200 by 200 box of 8 mm wall, drawn closed and then slit along its length. Closed, the torque runs round the wall as a shear flow and the torsion constant is 5.66×10⁷ mm⁴; slit, the loop is broken and only each wall's own thickness resists, giving 1.31×10⁵ mm⁴. The ratio is 432 to one, so the same torque twists the slit section 432 times as far and raises a peak shear stress 36 times as high. Nothing about the material changed.

The slit that costs a factor of six hundred

Bending stiffness cares where the material is, and changes by a factor of two or three between sensible sections of the same area. Torsional stiffness cares whether the material forms a closed loop, and the penalty for not doing so is an order of magnitude squared.

sections · Torsional constant The cheapest way out of being round. A ring under uniform external pressure, drawn in its first four buckling modes with the pressure each one needs underneath it, in N/mm². The pressure has no direction: it stays normal to the wall wherever the wall goes, so it does work on any change of shape that reduces the enclosed area, and the ring buckles into whichever shape is cheapest. Bare, that is the oval — n = 2 at 3EI/R³ — and the modes rise as n² − 1, so three lobes cost 2.67 times as much. Nothing in the drawing prefers any orientation, which is the point — a column has an axis to buckle about and a ring has none.

The pressure that needs no direction

Every buckling problem in this collection has had a load with a direction — a column pushed along its axis, a plate along its edge, an arch by what is on it. A buried pipe has none. The pressure is the same everywhere, it stays normal to the wall as the wall moves, and it does work on any change of shape that reduces the area inside.

stability · Ring buckling Two cantilevers, or one wall, and the beams decide which. The deflected shape of a coupled pair of 6 m walls, drawn against the two limits it lies between. Release the coupling beams entirely and the pair is two independent cantilevers, deflecting 111 mm. Make them rigid and it is one composite wall of the full width, deflecting 16 mm — 6.8 times stiffer, because the lever arm between the wall centroids is 8.40 m and everything inside either wall is smaller than that. Real beams of 600 × 350 mm over a 2.4 m opening land at 23 mm and carry 63% of the base overturning as an axial couple rather than as wall bending. The degree of coupling never reaches one, because a beam of finite depth cannot suppress the walls' curvature entirely.

Two walls that agreed to be one

A pair of shear walls with a row of doors between them is the commonest lateral system there is, and it has two readings that differ by a factor of seven. What decides which one applies is a beam 600 mm deep over a 2.4 m opening — and most of the overturning ends up as an axial couple that no bending diagram contains.

internal-forces · Wall coupling A fourth power, and then a cliff. The factor of safety against rolling, against beam length, for one section hung from a roll axis 0.9 m above its centre of gravity. Nothing about the section changes along this axis. z̄ goes as the fourth power of the length — 0.236 m at 30 m becomes 0.747 m at 40 — and the factor of safety is proportional to (y_r − z̄), so it does not decline gently: it falls away and then stops existing. The working factor of 1.5 is lost at about 41 m, and past 42 m there is no hook height at all at which this beam hangs stably. Which is why long girders are lifted with the picks moved inboard, or with the beam braced, or not in one piece.

Hung from above and still unstable

A rigid body hanging from a point above its centre of gravity is a pendulum and cannot fall over. A beam is not rigid, and tilting it puts a component of its own weight sideways — which bows it, which moves its centre of gravity further out. Past a length there is no hook height at which it hangs stably at all, and the length arrives as a fourth power.

stability · Lift stability Three lines through one point, and three different winners. Modulus against density on logarithmic axes, with a guide line for each of three indices drawn through mild steel. A performance index E^(1/n)/ρ is a straight line of slope n on these axes, so ranking materials by it means sliding the line up and to the left and seeing what it leaves behind. The three lines have three different orders, which is why a table of properties cannot answer the question on its own: for a tie the answer is carbon fibre, for a beam it is timber at 4.28 times steel, and for a plate timber wins by more still. The construction is Ashby's; the arithmetic on it is this site's.

The ranking belongs to the load case

Every table of material properties ever printed ranks by strength, and every one of them answers a question nobody asked. What a structure wants is the least mass for a stated performance, and the combination of properties that gives it changes with the shape of the member — so timber beats steel four to one as a beam and loses to it as a tie, without either material changing.

materials · Material index The answer is continuous and the catalogue is not. Capacity bought against capacity required, over a real rolled series. The straight line is what a continuous section would give — exactly the moment asked for, and nothing can be bought on it. The staircase is what a catalogue gives: each tread is one section, each riser is the step to the next, and the vertical gap between the two is steel that is paid for and does nothing. The steps in this series run from 23% to 59% in plastic modulus, so the average waste is 15.0% and the worst is 46% — just above a riser, where the section below has been missed by a kilonewton-metre. The 1200 kNm marked buys a 686×254×125 at 1418 kNm, which is 85% utilised. Two things follow that a continuous treatment cannot see: the sensitivity of a design to an assumption is zero over most of a tread and enormous at a riser, and an optimisation that returns three significant figures is answering a question with twelve answers in it.

The answer is continuous and the catalogue is not

Every optimisation in this subject returns a number with three significant figures in it, and nothing with three significant figures can be bought. What can be bought is a rolled series whose steps are a quarter to a half apart, so the member that goes on the drawing is on average a tenth stronger than the one that was calculated and can be a third stronger for no reason at all.

sections · Available sections A section has two areas and the tables give one of them. Peak shear stress divided by the mean, for four sections of exactly the same gross area and depth. The mean is V/A and is the number a first calculation uses; the peak is what the material actually sees, and the ratio between them is a property of shape alone. A rectangle's is 1.5 — the parabola's peak over its average — and it is one of the few numbers in this subject that is exactly derivable and universally ignored. An I-section's is near 1.98, and the reason is on the second bar: 97% of the shear is inside a web that is 56% of the area. So the flanges carry the moment and almost none of the shear, and the web carries the shear and almost none of the moment — which is why a shear check on an I-section uses the web area and a moment check uses the whole section, and why the two checks are about two different pieces of steel.

The section has two areas

A shear force divided by the area of the section is not the shear stress anywhere in it. A rectangle's peak is exactly one and a half times that number and an I-section's web carries nearly all of the shear over a fifth of the area, which is why a moment check and a shear check on the same member are checks on two different pieces of steel.

sections · Shear area The group is not weaker; it is very much softer. A 3 × 3 pile cap on the left, with each pile's share of 9.0 MN and 4.5 MNm in meganewtons — N/n plus M·y/Σy², the same three terms in the same order as a bolt group under an eccentric load and a section under biaxial bending. The corner piles take 1.25 times the average and a pile added at the centroid would change that by nothing at all, because it adds to neither second moment. On the right is the effect a bolt group cannot have: the piles share ground, so the stress bulbs overlap and the group settles 3.9 times as much as a single pile at the same load per pile, rising to 14.2 for 144 of them. The capacity check everyone makes — block failure against the sum of the piles — comes out at 4.54 here and does not govern at all. The check nobody tabulates is the one that does.

Nine piles, and four times the settlement

A pile cap divides its load between its piles by the same three terms a bolt group uses and a section under biaxial bending uses. What a bolt group does not have is neighbours it shares ground with — and the group effect that matters is not the strength check everybody makes, but a stiffness effect nobody tabulates.

structures · Pile group The lining that carries less for being weaker. Bending moment and hoop thrust in a circular lining, against the lining's own bending stiffness, both as fractions of the free-ring values. The ground arrives already stressed — 500 kPa vertically and 300 horizontally at K₀ = 0.6 — and the difference between them tries to squash the hole into an ellipse. A lining stiff enough to refuse absolutely collects the whole distortion pressure, p₂R²/3 = 300 kNm/m; one flexible enough to go with the ground collects nothing, because there is no curvature change left to resist. The thrust is the flat line: it comes from the mean stress rather than the difference, so it does not move at all. Putting 8 joints in this ring drops the moment to 34% of the solid one and leaves the thrust exactly where it was, which is why a segmental lining is jointed and why the intuition carried over from a beam is inverted here.

The lining that is stronger for being weaker

A tunnel lining is not loaded. The ground arrives already stressed and the hole wants to squash into an ellipse; the lining's only job is to refuse, and how much moment it collects depends entirely on how hard it refuses. Make it stiffer and it takes more. Make it flexible — put joints in it, make it thin — and it takes almost none, while the hoop thrust it carries does not move at all.

structures · Tunnel ring The average is not the answer, and it is unsafe. Critical load of a pinned column whose middle third has been given a different stiffness, against the whole-column Euler load, with the two numbers a hand check reaches for beside it. The eigenvalue is taken from K − P·Kg over 24 elements, so nothing here is a formula for a stepped column — it is the same computation the uniform case gets. At a middle third of 0.50 times the rest the true load is 0.612 of Euler's, the arithmetic average says 0.832 and the weakest segment says 0.496. The average is high by 36% and it is high on the unsafe side, because the third of the column it is averaging over is the third where the mode has all its curvature. The weakest-segment answer is safe everywhere and wasteful by about as much.

An average stiffness is not a safe stiffness

Euler's load belongs to a column of one EI. Give the same column two, and the temptation is to average them — which is wrong, and wrong in the unsafe direction by a quarter. Buckling weights stiffness by the square of the curvature of the mode, so the middle of a pinned column decides everything and the ends decide almost nothing.

stability · Stepped column The tube flattens because of the bending, and then cannot carry it. Moment against curvature for a long tube of radius 300 mm and wall 4 mm. Compression on one face and tension on the other are both directed along a curved line, so each produces an inward transverse pressure and the circle is squashed into an oval by the bending it is carrying. That reduces the second moment, so the curve bends over and reaches a limit point — no bifurcation, no imperfection, nothing to be sensitive to. It arrives at an ovalisation of exactly 2/9 for every tube of every size in every material, at 1018 kNm, where the secant stiffness has fallen to 67 per cent of the undeformed value and the tangent stiffness is zero. The relaxed path reproduces the closed form to 0.004 per cent.

The tube that flattens itself

Bend a tube and the compression on one face and the tension on the other are both running along a curve, so both push inward. The circle becomes an oval, the second moment falls, and the moment–curvature curve turns over at a limit point that needs no imperfection, no bifurcation and nothing to be sensitive to.

stability · Brazier buckling The section that is checked is not the section that was chosen. A 457 mm beam coped 50 mm deep over 120 mm to frame into a girder. What is left is a tee with a section modulus of 3.836e+5 mm³ against the whole section's 1.438e+6 — 27 per cent. The moment at the end of the cope is the reaction on a lever arm of 130 mm: 23.4 kNm, giving 61 N/mm² and a flexural utilisation of 0.17. The web now has a free edge along the cope, so its buckling coefficient collapses from 4 to 0.425 — a factor of 9.4 — and the re-entrant corner has a stress concentration of 5.5 on a 10 mm radius.

The section that is checked is not the one chosen

A beam framing into a girder has its top flange cut away so the two can sit at the same level. What is left is a tee with a quarter of the section modulus, a web with a free edge, and a re-entrant corner — and the beam was selected on a table entry that describes none of it.

connections · Coped beam The best design is where two failures arrive together. A fixed area of steel rolled into tubes of every proportion, with the three things that can end each one. Euler's load goes as r² because I = A r²/2; the local buckling stress goes as 1/r² because the wall thins as the tube grows; squashing does not care. The capacity is the lowest of the three, so it has a maximum — and the maximum is exactly where the two buckling curves cross, at r/t = 129 and 2364 kN, which the closed form r*, the fourth root of αAL² over π³β√3, reproduces to 0.52 per cent. That is the general result and it is not about tubes: the optimum of a minimum of a rising and a falling curve is always their intersection, so optimising a design against two failure modes puts both of them at the design point — which is the one configuration imperfections hurt most.

The best design is the most sensitive one

Take a fixed area of steel and roll it into a tube. Euler's load rises with the radius and local buckling falls with it, so the capacity has a maximum — and the maximum is exactly where the two failure modes arrive together, which is the one configuration imperfections hurt most.

stability · Wall optimum The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 21 m by 6000 mm, under 300 kN at mid-span. There is no diagonal in it, so each panel's 150 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, 131.3 kNm, and it adds to an axial force of 263 kN from the global moment at the same point. The girder deflects 109.09 mm against 6.53 mm for the same members triangulated — 16.70 times — and 98% of that movement is chord bending that a diagonal would have removed entirely.

The frame is a girder stood on end

Every unbraced building frame is a Vierendeel girder turned through ninety degrees, and the identification is not an analogy — it is the same equations with the axes swapped. Which means the frame inherits results that read as absurd for a building — more bays is stiffer, a wider building is not, and doubling one section property halves the sway.

structures · Vierendeel The distortion runs the length of the span, and a diaphragm stops it. Longitudinal stress at a corner of the box from distortional warping, along a 60 m span carrying 45 N/mm at 2.0 m off the axis. With no interior diaphragm it peaks at 73 N/mm², which is 83 per cent of the bending stress the girder was designed for. Two interior diaphragms take it to 18. The governing length is Winkler's: the distortion decays over 23.6 m, so a diaphragm helps its neighbours only if it is closer than that, and past it the spacing stops mattering.

One diaphragm is nearly none

A box girder's distortion decays over a length the section decides, and on a sixty-metre span that length is twenty-four metres. So a single diaphragm at midspan sits further from each end than the distortion can reach and removes a third of the problem; three diaphragms remove nine tenths. The spacing rule is not span over five — it is a property of the plates.

sections · Box distortion

The eccentricity at right angles to the drawing

A bracket's load stands off the plane of its welds as well as being offset within it, and the second eccentricity produces a completely different object — bending about an axis through the group rather than torsion about a point in it. The weld line has no compression zone to argue about, so its neutral axis is its own centroid, and the whole length works.

connections · Weld group

The fixed-end moment is a column stress

A fixed end forbids a change of slope and a deviation, so on a member fixed at both ends Mohr's two theorems both have the answer nothing. Written with 1/EI as a width, those two conditions are the P/A + My/I of a short column under an eccentric load — the end moments are its edge stresses, and a haunched member's are read from where its elastic centre has moved.

deflection · Moment-area

The elastic centre is not on the frame

Cross's analogy turns a member fixed at both ends into a short column and its end moments into edge stresses. A closed frame's analogous column is the frame's own outline, its elastic centre is a point hanging in mid-air inside it, and the horizontal thrust a gravity load produces is that section's bending stress about the axis through that point.

deflection · Moment-area

The centre that hangs in the air

Cross's analogy turns a closed frame into a short column and reads its redundant moments off as that column's stresses. Nothing in it requires the frame to be straight. Run round a fixed arch, the analogous column's centroid is a point two thirds of the way up the rise, on no part of the arch at all, and the bending stress about it is the arch's horizontal thrust. For a parabola under a uniform load the answer it returns is a diagram of nothing, which is exactly right.

deflection · Moment-area

The axes that have to be turned first

Make one column of a portal shorter than the other and the analogous column stops being symmetric about a vertical line. It acquires a product of inertia, its principal axes tilt seven and a half degrees, and the thrust and the redundant shear stop being two separate divisions. Using the elastic centre and nothing else — which is what the symmetric construction looks like from outside — reports 17.0 kN·m at the left foot where the frame carries 27.8.

deflection · Moment-area

The load that is not a load

Settle one foot of a portal frame by ten millimetres and the frame develops moments with nothing applied to it anywhere. In the analogous column the case is simpler than a load case, not harder — the section carries no direct stress at all, and the whole answer is one bending stress. And it scales the wrong way: the moments are proportional to EI, so the stiffer the frame, the more a settlement costs it.

deflection · Moment-area

Three, and what three is a property of

The column analogy works because a closed ring cut once has three redundants and a plane section has three stress resultants. That match is the whole method, and it is topological rather than geometric — a portal, a pitch, a step, a splay and a polygonised arch are all one ring and all exact. Add a second bay and the frame has six redundants with nothing to be a drawing of, and the outer ring on its own is out by 190 per cent.

deflection · Moment-area

Named alongside it

The objects these essays reach for when they reach for this one.

StiffnessFlexural rigidityIndeterminacyColumn analogyCompatibilityCritical loadElastic centreFree bodySection modulusSection shapeSuperpositionBuckling

All concepts