Structural Engineering & Statics
where the load goes
Concept

Plastic hinge — where it appears

A cross-section that has yielded through its depth and rotates at nearly constant moment, redistributing load rather than failing. It has to deliver a rotation as well as a moment, and how much it can deliver is decided by the section's plate slenderness or by its confinement.

Named by 30 essays across 9 fields — each of them below, with the objects they name alongside it.

The collapse mechanism of a propped cantilever. A collapse mechanism, with the hinge position found by searching rather than quoted. Every position gives an upper bound on the collapse load; the lowest is 7.29, at a hinge 58.6 per cent along, which is a coefficient of 11.657 times Mp over the square of the span.

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.

internal-forces · Plastic hinge Two materials pulled until they stop. Two stress-strain curves — mild steel, cast iron — plotted to a strain of 2.0%. One of them has a plateau, so the stress at which yielding starts is something the specimen does rather than something anyone chooses. No offset construction is drawn.

The property that appears in none of the equations

Ductility is in no design formula on this site. Every method on this site depends on it — and a brittle structure does not merely fail early, it makes the analysis wrong.

materials · Ductility Prying action in a tee stub. A tee stub pulled by its web with 100 kN per bolt. The 20 mm flange is in the one-hinge regime, so the prying force at the flange tip is 50.63 kN and the bolt carries 150.63 kN — 1.51 times what was applied. The flange stops prying entirely at 26.97 mm thick, and collapses on its own at 110 kN.

The force the bolt never saw applied

Pull a tee stub with a hundred kilonewtons and its bolt carries a hundred and fifty. The extra comes from the flange bending and pressing its own edge against the thing it is bolted to, and no free body of the connection as a point contains it.

connections · Prying A ground motion, on a structure of 1.00 s period. Displacement against time for a single-degree-of-freedom structure of natural period 1.00 s and 5.0% damping, under a ground motion of 3.5 m/s² peak. The elastic peak is 74.47 mm, and the same frame given a 4th of that strength peaks at 69.87 mm and comes to rest 11.82 mm from where it started.

The earthquake asks for a displacement

A structure a quarter as strong as the elastic demand does not deflect four times as far. It deflects almost exactly as far, yields on the way, and survives — which is why no ordinary building is designed for the force an earthquake would apply if it stayed elastic.

dynamics · Ductility demand A two-way slab is a one-way slab as soon as it is not square. The share of the load carried by the strips spanning the short way, against the ratio of the sides. The two families of strips cross at the centre and must deflect equally there, and a strip's deflection goes as the fourth power of its span — so at a ratio of 1.33 the short strips already take 76% and at 2 they take 94%. The panel drawn here is 6 × 8 m, a ratio of 1.33, and its short strips take 76.0%. Two-way action is worth having at a ratio of one and worth almost nothing by two.

The slab that spans both ways

A panel supported on four sides sends its load in two directions at once, and the share is decided by a fourth power — so a panel a third longer than it is wide has already stopped being a two-way slab in any useful sense. What it does at collapse is a different calculation with a different answer.

structures · Two-way spanning Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight.

Both at once, and neither matters until it does

A section carrying shear has less moment capacity, and the reduction is the web's share of the plastic modulus times one minus the root of one minus the shear ratio squared. On a rolled beam that share is a quarter, so half the shear capacity costs three and a half per cent — and then the last tenth costs more than the first eight.

sections · Shear moment interaction The same load, two diagrams, both in equilibrium. One span of a pair of 7 m spans under 5 kN/m, drawn twice. The elastic solution puts 31 kNm over the support and 17 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 21 and 21: the section the beam needs falls from 31 kNm to 21, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 31 kNm for either — and the second is legitimate for that reason alone. What it costs is 1.0 milliradians of rotation at the support, which the section has to be able to deliver.

The moment that was moved on purpose

The elastic analysis of a continuous beam gives one set of moments. It is not the only set the beam is allowed to have, and taking a smaller one at the support is legal, cheaper, and paid for in a rotation that has to be delivered before the design exists.

internal-forces · Moment redistribution Every path to the ground goes through the link. A braced bay 8 m by 4 m whose two diagonals stop 800 mm apart instead of meeting. The storey shear reaches the ground through the diagonals, and the vertical components they deliver to the beam have to pass through the segment between them: the link carries 47% of the applied shear as a shear force, at a lever arm short enough that its ends reach 0 kNm while the rest of the beam carries 0. The deflected shape drawn is the solved one, magnified — the real drift under this load is 0.008 mm. Everything outside the link is designed to stay elastic while the link is yielding, which is what makes the mechanism a choice rather than a hope.

The part that is meant to be weak

A braced frame is stiff and has nowhere to yield. A moment frame yields everywhere and is soft. Move the two diagonals a metre apart along the beam and the whole storey shear has to pass through the segment between them — which keeps most of the stiffness and puts every yielding in one member the designer chose.

structures · Eccentric brace The same concrete, held sideways. Two stress-strain curves for one concrete. The lower is a cylinder test: it peaks at 30 N/mm² near a strain of 0.002 and has nothing left by 0.0035, because it fails by splitting apart sideways. The upper is the same material inside a 12 mm hoop at 100 mm centres, which cannot stop it expanding but can make the expansion stretch steel: the lateral pressure of 2.48 N/mm² — 8% of the strength it is multiplying — takes the peak to 44.4 and the ultimate strain to 0.028. The strength gain is 1.48 times and the strain gain 8.0; the area under the curve, which is the toughness, goes up by 11. It is the third number the confinement is provided for.

Squeezed sideways into a different material

Concrete in a cylinder test fails by splitting apart sideways under a load pushing it down. Put a hoop round it and the splitting has to stretch steel — and a lateral pressure of a twelfth of the strength raises the strength by half and the ultimate strain by eight.

materials · Confinement The two theorems close on the answer from opposite sides. A pinned-base portal frame under 200 kN at mid-span and 80 kN at the eaves, with a plastic moment of 200 kNm. The three horizontal lines are mechanisms — work equations, each an upper bound: the beam mechanism gives 1.000, sway 1.250 and the combined one 0.714. The curve is the lower bound: for each value of the one redundant, the largest load factor whose moment field stays inside the plastic moment everywhere. Its peak is 0.714, and it touches the lowest mechanism exactly — so the collapse load is known rather than bracketed. Using the beam mechanism instead would have claimed 1.40 times the real capacity, and the sway one 1.75 times, both of them on the wrong side.

Two ways of being wrong

Plastic analysis has two theorems and they point in opposite directions. Any equilibrium field that nowhere exceeds the plastic moment gives a load at or below the collapse load; any mechanism gives one at or above it. Only one of those errors is safe.

equilibrium · Bound theorems 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 The point the rafter turns about, which is off the frame. A pitched portal of 8 m span and 4.0 m to the eaves, with a 1.5 m rise, collapsing. Each rigid part of the mechanism rotates about some point: the left column about its base hinge, the right about its own. The rafter between them does neither, and its centre is found by one rule — two bodies joined at a hinge share that hinge, so the second body's centre lies on the line through the first body's centre and the hinge, extended. Two hinges give two lines and they cross at (8.0, 11.0) metres, which is 5.5 m above the ridge and outside any drawing of the frame itself. From there the whole collapse is two ratios of lengths and no trigonometry: the load factor is 1.339. Flatten the roof and the centre descends; make the two lines parallel and it goes to infinity, which is the statement that the rafter translates instead of turning.

The point the mechanism turns about

A collapsing frame is a chain of rigid pieces, and every piece is rotating about some point. Find those points and the whole collapse load reads off two ratios of lengths, with no trigonometry anywhere — and for a pitched roof the point in question is well above the top of the drawing.

equilibrium · Instantaneous centre Six ways for one dowel to fail, and the capacity is the smallest. Johansen's single-shear mechanisms for a 12 mm dowel through 40 and 40 mm members, each drawn as the shape it is: the dowel straight and the timber crushing, the dowel rotating rigidly, one plastic hinge, then two. The capacity under each is that mechanism's own, and the joint's strength is the smallest — 5.02 kN by mode c, which is the dowel rotates rigidly and both members crush. That is the kinematic theorem of plasticity: every mechanism gives an upper bound and the true collapse is the lowest of them. The crushed timber is shaded, and the circles are plastic hinges in the steel.

The smallest of six failures

Everything else in this collection that fails does so in one way at a time. A dowel through timber does not — the wood can crush while the steel stays straight, or one plastic hinge can form in it, or two — and the capacity is the smallest of the six, which is the kinematic theorem of plasticity applied to a joint rather than to a frame.

connections · Dowel yield How fast the strain arrived, which a quoted strength does not record. The dynamic increase factor on strength against strain rate, over eight decades. Steel follows Cowper and Symonds' fit, whose constant D = 40.4 s⁻¹ is not an arbitrary parameter — it is the rate at which the material is exactly twice as strong. Concrete in tension follows the model code's two-branch curve and is steeper. The four marked regimes are the argument: a testing machine works at about 10⁻⁴ per second, an earthquake at 5 × 10⁻³, a vehicle impact at a half, a blast at a hundred, and the enhancement across them runs 1.00, 1.08, 1.32, 2.04. So this is a correction that is either negligible or decisive with very little in between, which is why no seismic code carries it and every blast code does. What does not rise is the modulus, which is a lattice property, and the ultimate strength rises only a third as much — so the ultimate-to-yield ratio closes from 1.56 to 1.23 and the material has less warning left in it than it started with.

The steel that is stronger in a millisecond

Every strength quoted anywhere in this collection was measured at about a ten-thousandth of a strain per second, because that is what a testing machine does, and nothing on a drawing says so. Load the same steel a million times faster and its yield stress rises by a third.

materials · Strain rate What the shape of a load in time is worth, for two load shapes. The peak displacement as a multiple of the static deflection, against the load's duration divided by the structure's natural period, for two load shapes: a load that rises linearly, then stays; a rectangular pulse, then nothing. The lines are closed forms and eight dots are the peak of a complete time integration of an oscillator of 0.300 s period under that load, agreeing with the line to within 0.19% everywhere.

The load that is over before it has moved

A blast delivers an enormous pressure for a few milliseconds. Everything else in this field asks what force a structure can carry; a load that has come and gone before the structure has travelled any distance is not asking that question, and the answer turns out to depend on the mass and the ductility with the strength barely in it.

dynamics · Blast The two braces balance until one of them buckles. An inverted-V brace after the compression member has gone. While both braces are elastic they carry equal and opposite forces and their vertical components cancel on the beam above, which is why the beam in a chevron bay is usually sized for gravity alone. The compression brace buckles at 445 kN and then sheds most of what it was carrying — 30% is left here — while the tension brace goes on to yield at 1065. The difference between the two vertical components is 659 kN, applied at the middle of the span with no help from either brace, and it asks the beam for 1317 kNm against the 200 kNm the gravity load asks for — 6.6 times as much. The beam drawn does not: 1517 kNm against a capacity of 731. The force is not a load case anybody applies; it is what the frame leaves behind on its way to the state it will actually be in.

The force the brace leaves behind

Two braces meeting under a beam carry the storey shear as a tension and a compression whose vertical components cancel, so the beam above sees nothing. They cancel only while both braces are elastic. Once the compression brace buckles it sheds most of its force, the tension brace goes on to yield, and the difference is a point load at midspan that nobody applied.

structures · Chevron brace A transverse load with nothing applied. Web slenderness against web thickness, with the limit the flange's own curvature sets. A flange carrying 6213 kN and curved to a radius of 592 m needs 10.5 N per millimetre of radial force to stay on its curve, and the only thing available to supply it is the web. Nothing has been applied to the girder: the load comes from the deflected shape, which is why a straight beam has none of it and a beam at a plastic hinge has a great deal. Setting the radial force against the web's own plate-buckling resistance gives, in four lines, h_w/t_w ≤ k·(E/f_yf)·√(A_w/A_fc) — the form the codes use, arrived at without them. The constants differ: an elastic flange strain gives k = 1.34 and the rule uses 0.3, a factor of 4.5, and the gap is the curvature assumed. k goes as the inverse square root of the flange strain, so 0.3 is a flange strained to 3.4% — which is what a plastic hinge does to it. The rule is not conservative; it is written about a different beam.

The web that is crushed from inside

A plate girder's compression flange is curved by the beam's own deflection, and a curved force needs a transverse load to stay on its curve. The only thing available to supply it is the web. So a deep girder can buckle its web vertically with nothing applied to it at all, and the rule that prevents it is the only clause in the codes about a load no load case contains.

stability · Flange induced The hinge is at mid-height in exactly no storey. The height of the point of contraflexure in each column of a 4-storey, 3-bay frame under lateral load, as a fraction of the storey height, against the portal method's assumption that it is at the middle. The exact solution is a plane-frame stiffness analysis of the same frame. In the bottom storey the zero sits at 0.609 of the height, because a fixed base is stiffer than the joint above it and takes more of the column's moment; in the top storey it sits at 0.359, because there is no column above to share that joint. The average over the whole frame is 0.475, which is why the assumption survives — it is right on average and wrong everywhere. The worst error in the storey shear share is 25%, and the column shears still add to the storey shear to 0e+0 of it, because the method is exact statics applied to an assumed structure.

The analysis that assumes the answer

A rigid frame is indeterminate, so statics cannot finish it. The hand methods finish it anyway, by assuming where the bending moment is zero and treating those points as hinges. That is not a shortcut around the analysis — it is a different kind of answer, exact in equilibrium and wrong in compatibility, and knowing which half is which is what makes the error a bound rather than a mystery.

internal-forces · Portal method

Two of these move and the third cannot

Cut one span of a continuous beam free and add up the forces on it. What comes out is that the mid-span moment plus the average of the two end moments equals the free bending moment of that span, with nothing else in it — no stiffness, no support settlement, no analysis at all. Continuity moves moment about. It does not reduce the total, and it never has.

internal-forces · Static moment

The axis that moves when the section yields

An elastic section bends about its centroid. A fully plastic one bends about the axis that halves its area, and for anything symmetric those are the same line — which is why the distinction is almost never met. For a tee they are a fifth of the depth apart, and three things follow that the elastic calculation gives no warning of.

sections · Equal-area axis

The moment that was shed has to land

Redistribution takes a moment off a beam's support and pays for it with rotation. On a beam that is the whole story. In a frame the support is a column, the shed moment does not vanish, and it arrives at a member whose section was chosen from the diagram it has just left.

internal-forces · Moment redistribution

The map with three regions

A structure carrying a constant load and a cycling temperature has three possible fates and only one of them is a collapse. It can stay elastic, it can yield once and then stop, or it can gain a little more deformation every cycle for ever — and the third has no failure load at all.

materials · Shakedown

The wide side goes inside

A crane hook's section is a trapezoid with its broad face towards the centre of curvature, and that is not a casting convenience. Turn the same section round — same area, same depth, same moment — and the stress at the fibre that breaks rises by thirty-nine per cent. The shape is doing two things at once, and only one of them is in a straight beam's arithmetic.

sections · Curved beam

The tension has to pull on something

A buckled web carries its shear on a diagonal band of membrane tension, and the band pulls sideways on the flanges and stiffeners that bound it. That pull is the design output nobody plots — it runs from 72 to 603 newtons per millimetre across ordinary panel proportions, it is largest exactly where the panel is most efficient, and at the end of the girder there is nothing beyond to take it.

stability · Tension field

Four corners and a mechanism

The check on a hole in a web adds an axial stress to a local bending stress and compares the sum with the yield stress at one corner. What ends the opening is four hinges arriving together — and the gap between the two is a factor that varies along the span from two and a half to exactly one.

sections · Web opening

The section that governs is inside the haunch

A tapered cantilever has its worst section somewhere along it because the moment grows linearly and the modulus quadratically. A haunched rafter has the same competition with a step in it, and the step is where the check lands — at neither end of the member, at a station no formula names.

sections · Tapered member

The pattern that stopped describing the building

A pushover analysis pushes with a load pattern chosen to resemble the first mode, and by the time the structure has done anything worth analysing it no longer has that mode. Recomputing the pattern as the frame softens changes the base shear by nine per cent and the drift at the fourth floor by a factor of four.

dynamics · Pushover

Any structure will carry the unit load

Virtual work has two readings and each is free exactly where the other is bound. A unit load needs only something to stand on in equilibrium, so the deflection of a beam statics cannot solve comes out of a cantilever statics can. A virtual displacement needs only to fit together, so a reaction comes out of pushing a mechanism — and on a redundant beam the unknown cancels out of the equation and nothing is found at all.

deflection · Virtual work

The load it can carry once

A two-span beam whose loads come and go span by span collapses at 150 kN under any one arrangement, and walks at 127. Between the two it can carry every arrangement once and none of them forever: each cycle leaves a few more milliradians of rotation at the support and a midspan fifteen millimetres lower. Melan's theorem finds the limit as the last residual moment line that fits, Koiter's as a mechanism no single load state can drive, and a cycle-by-cycle calculation walks exactly where both say it will.

materials · Shakedown

The hinge that forms in the other flange

An end plate bolted to a column flange is two tee stubs sharing one bolt, and the force at their tips is one force pressing on both. Checked separately, as the component method checks them, the end plate carries 109 kN a bolt and the column flange 104. Together they carry 92 — by a mechanism with one hinge in each flange that neither tee stub has on its own.

connections · Prying

Named alongside it

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

DuctilityFree bodyLoad pathCollapse mechanismEquilibriumLower-bound theoremMoment redistributionRotation capacityPlastic momentBound theoremsCapacity designHysteresis

All concepts