Structural Engineering & Statics
where the load goes
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The statics of things that do not move

Every result here is obtained by imagining a motion that does not happen and insisting the sums cancel. Nothing in the subject is measured directly.
A beam, its loads and its reactions. A 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. Equilibrium

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.

 The same beam, cut at x = 5. A 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. 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.

 Three forces must meet at a point. A 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. Equilibrium

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.

 Counting unknowns against equations. Three 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. Equilibrium

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.

 A truss cut through panel 3. The truss severed through one panel, with everything to the right removed and the three cut member forces drawn on the exposed faces. Taking moments about the marked joint removes two of the three unknowns, so one equation gives the third: -52.94. 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.

 A triangular load and the force that replaces it. A triangular distributed load with its resultant computed by integration: an area of 24.0 acting at 5.33 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is 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.

 A beam, its loads and its reactions. A 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. The moment of everything on the body is summed about three marked points, and every sum comes to zero. Nothing here is a new equation: with only vertical forces on the body, the moment about a point does not depend on that point's height, so every centre anywhere on the plane returns the same equation — and the third equilibrium equation, the horizontal sum, reads nothing equals nothing. Equilibrium

The equation that is not new, and the three that are

A plane free body yields exactly three independent equations. Most attempts at a fourth are one of the first three wearing different clothes — and on a beam under vertical load, one of the three is already saying nothing.

 The count is necessary and not sufficient. Two pin-jointed frames, each satisfying m + r = 2j exactly. One of them folds anyway, because the equations are not independent; the ghosted outline is the motion that costs no member any change of length, drawn at an exaggeration of 0.55 of the span. Equilibrium

The count that does not see it

A frame can have exactly as many unknowns as equations and fold up anyway. The count asks whether there are enough equations; it never asks whether they are different from one another.

 The joints are not pins, and this is what that costs. A 4-panel Pratt truss solved twice on the same stiffness matrix: once with a moment release at every member end, which is the pin-jointed idealisation, and once with the joints continuous, which is what welding them produces. The axial forces are the same to within a per cent; the bending the second solution adds is worst in member 0, where the bending stress reaches 24.3% of the axial stress. Members are shaded by that ratio. Structural form

The joint that is not a pin

Every truss on this site is analysed as though its joints were frictionless pins. Almost none are. The bending that follows is called secondary, which is a claim about size — and the claim is checkable.

 The same beam, cut at x = 5. A 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. 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.

 Where plane sections stop staying plane. Strain across a cut face at four span-to-depth ratios, with the straight line the theory assumes drawn faintly behind. For a slender beam the two coincide; for a beam as deep as its span the real distribution is nothing like a straight line, and beam theory has no claim on 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.

 A column that was never straight. Load against lateral deflection at mid-height, for a column starting with an initial bow of 0.002. There is no critical value to reach: the deflection grows from the first increment, slowly at first and then without bound as the ratio approaches 1.00 — which is the Euler load, and which the column therefore never attains. The perfect column, drawn for comparison, sits on the vertical axis until it arrives there and then has no answer at all. Stability

The column that was never straight

Euler's load is the load at which a perfectly straight column becomes indifferent to being bent. No column is perfectly straight, so no column ever reaches it — and the load it never reaches can still be measured.

 A brace is a stiffness requirement, not a strength one. Critical load against brace stiffness for a pinned column braced at mid-height. The curve climbs from the unbraced Euler load of 9.87EI/L² and flattens at 39.48EI/L², which is the Euler load of the braced segment — past that the column buckles in a shape the brace does not obstruct, and further stiffness buys nothing. The knee is at about 159EI/L³. A stiffness of 60EI/L³ is marked, reaching 21.75EI/L². Stability

The brace that need not be strong

A brace holding a column at mid-height carries almost no force. What it has to be is stiff — and the stiffness required is exact, large, and reached at a knee past which more buys nothing at all.

 The deflection at x = 4, by virtual work. Three diagrams: the moment from the real load, the moment from a unit load placed where the answer is wanted, and their product. The area under the third, divided by EI, is the deflection — 213.33 here. No standard case was consulted, so the method works for any load pattern at all. 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.

 3 continuous spans against 3 simple ones. The bending moment in a continuous beam whose support 1 has settled by 0.01. Three curves: the moment the load causes, the moment the settlement causes on its own — dashed, peaking at 73.5, and in equilibrium with no applied load at all — and their sum, which is what the beam carries, peaking at 73.5 against 24.5 without the settlement. The settlement field is proportional to EI: a stiffer beam is punished harder for the same movement, which is the opposite of every intuition load-carrying gives. Deflection

The support that moved

A redundant structure knows things statics cannot see. Settle one support by ten millimetres and a complete set of bending moments appears — in equilibrium with no load at all, and larger for a stiffer beam.

 Three materials pulled until they stop. Three stress-strain curves — mild steel, high-strength steel, aluminium alloy — 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. The 0.2% offset construction is drawn on the high-strength steel: a line of slope E from a strain of 0.002, cutting the curve at 460 N/mm². Materials

The stress at which nothing in particular happens

One material in six has a yield point that a specimen actually does something at. For all the others the yield stress is a construction — a line drawn at an arbitrary offset — and every strength calculation for those materials depends on it.

 What it costs to reach the plastic moment, for two shapes. Moment against curvature for two cross-sections of identical area (3000 mm²) and identical depth (200 mm), in mild steel, each divided by its own first-yield moment and its own first-yield curvature. The rectangle has a shape factor of 1.50 and reaches 98% of its plastic moment at 4.1 times the curvature at first yield; The I-section has a shape factor of 1.09 and reaches 98% of its plastic moment at 1.1 times the curvature at first yield. The dashed lines are the rigid-plastic moments, computed from the equal-area axis rather than read off the curves, and no curve reaches its own. Materials

The section that yields from the outside in

A rectangle has half again as much moment in reserve past first yield as its elastic capacity suggests, and an I-section has a seventh. Read as a ranking that gets it backwards — the reserve is bought with curvature, and the rectangle pays four times as much of it.

 A I-section at 80% of its plastic moment. The same I-section drawn three ways: the shape, the strain across its depth, and the stress that strain produces in mild steel. The strain diagram is a straight line, because plane sections stay plane whatever the material is doing. The stress diagram is not: 0% of the area has yielded, working inward from both faces, and the neutral axis sits at 70.9 mm against a centroid at 100.0 mm. The compression resultant is 322.5 kN and the tension resultant 322.5 kN, on a lever arm of 180.1 mm, which multiplies back to the 58.1 kNm the section is carrying. A rolled residual stress pattern of ±30% of yield is locked in before any load arrives. Materials

The stress that was there before the load

A rolled steel section leaves the mill carrying eighty N/mm² of stress with nothing applied to it, in a pattern that sums to no force and no moment. It is invisible to every calculation and it is the knee in every column curve.

 What is left when the load comes off. Mild steel taken to a strain of 0.60% and then unloaded to zero stress, at which point the strain has not returned to zero: 0.469% of it is permanent. Materials

What is left when the load comes off

Unload a section that has yielded and it does not return to nothing. It returns to a self-equilibrating stress field it did not have before, a permanent set, and an elastic range wider than the one it started with.

 Which of them stops moving. Three load cases on the same rectangle, each a constant moment plus a temperature profile cycled from nothing to a peak and back, over sixteen cycles. At 30% of the plastic moment with a 20°C profile it never yields at all; At 60% of the plastic moment with a 120°C profile it shakes down; At 85% of the plastic moment with a 200°C profile it ratchets, at 1.9% of the first-yield curvature per cycle. The ratcheting case never collapses and never returns: it simply arrives somewhere further round every cycle, which is a serviceability failure that no collapse calculation contains. Materials

The structure that settles down, and the one that walks

A load that is safe applied once may not be safe applied ten thousand times. Nothing about that is fatigue — the structure never breaks, it simply arrives somewhere slightly further round every cycle, until it has arrived somewhere unusable.

 The deflection that arrives years late. The multiplier on a concrete member's deflection under a sustained load, against time. The elastic deflection arrives on the day the load does and is the 1.0 at the left. After a year it has been multiplied by 3.00, after five years by 3.29, and it approaches 3.38. Nothing has been added to the load and nothing about the strength has changed: this is a serviceability failure arriving on a structure that passed every strength check on the day it was built. Materials

The deflection that arrives three years late

A concrete beam that passes every check on the day it is built goes on deflecting for a decade, and ends up three times where it started. Nothing about the load changed, and nothing about the strength was ever in question.

 The stress that leaks away. A restrained shrinkage strain of 300 microstrain in concrete of modulus 32000 N/mm². Ignoring creep it produces 9.60 N/mm², which is above the tensile strength of 3.5 and predicts that every restrained concrete member ever cast has cracked. Counting creep by the superposition integral leaves 2.32 N/mm² after 27 years, and the one-line age-adjusted shortcut at the usually quoted ageing coefficient of 0.8 leaves 3.31. The two disagree — this creep function implies an ageing coefficient of 1.32, not 0.8 — and both are below the tensile strength, so the conclusion turns on counting creep at all rather than on how it is counted. Materials

The strain that was imposed, and the stress that leaked away

Multiply a restrained shrinkage strain by the modulus and the answer is three times the tensile strength — which predicts that every restrained concrete member ever cast has cracked. Most have not, and the reason is that the material creeps while it is being stressed.

 The crack length at which the strength stops mattering. Failure stress against crack length for a toughness of 100 MPa√m, with two steel grades drawn. The falling curve is fracture — Kc divided by Y times the root of pi a — and it does not know what the yield stress is. The horizontal lines are the grades. At 275 N/mm² the two cross at a crack 33.5 mm long; At 460 N/mm² the two cross at a crack 12.0 mm long. The stronger grade's transition is the shorter one — raising the yield stress does not raise the strength of a cracked member, it only shortens the crack that takes it away. Materials

The flaw that sets the strength

A member with a crack twenty millimetres long fails at its yield stress. Make the steel stronger and the crack that does it gets shorter, so the same flaw that was harmless in the weaker grade decides the stronger one.

 A general force system is a screw, not a force. Two forces that are neither parallel, nor concurrent, nor coplanar. Their resultant is (30, 0, -60) and no point in space reduces them to it alone: the moment left on the central axis is (14.40, 0.00, -28.80), which is parallel to the resultant and has a pitch of 0.4800. The simplest object equivalent to the system is therefore a force and a couple about the same line — a wrench — and a planar drawing can never produce one, because a coplanar system has its moment perpendicular to its resultant by construction. Equilibrium

Moving a force, and what it costs

Every free body on this site begins by putting a force somewhere convenient. That move is free along the force's own line, costs a couple across it, and in three dimensions leaves behind something no choice of point can remove.

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