Structural Engineering & Statics
where the load goes
Concept

Moment redistribution — where it appears

The migration of bending away from a section that has yielded, toward one that still has capacity to take it. The collapse load does not know the elastic distribution, so redistribution costs nothing in strength and is paid for in a rotation the section has to deliver.

Named by 17 essays across 6 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 3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 30.6 to 19.6, and a hogging moment of 24.5 appears over the supports where there was none.

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.

internal-forces · Continuity Where the peak moment goes when a restraint is added. The 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.

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.

deflection · Indeterminacy 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.

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.

deflection · Indeterminacy 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 Moment against rotation, for three real joints. Three connections on one plot, with the classification boundaries for a beam of EI/L = 14000 drawn as rays through the origin. web cleats is pinned, flush end plate is semi-rigid, extended end plate is semi-rigid. The boundaries are multiples of EI/L, so the same joint is rigid on a short stiff beam and semi-rigid on a long slender one.

Neither pinned nor rigid, which is every real connection

Frame analysis offers two options for a joint and reality supplies a continuum between them. Worse, the boundaries are not properties of the connection at all — the same end plate is rigid on a short stiff beam and semi-rigid on a long slender one.

connections · Joint stiffness What the joint does to the beam. End moment as a fraction of the fixed-end value wL²/12, against the joint's rotational stiffness, for a beam of EI/L = 14000. At the rigid boundary of 112000 kN·m/rad the joint delivers 80% of it and at the pinned boundary 20%. Everything between the two lines is a redistribution nobody chose and every analysis assumed away.

The redistribution nobody chose

A beam designed as simply supported, on connections that are not pins, has end moments the analysis never predicted and a mid-span moment smaller than it was sized for. Usually that is safe. It is never intentional, and there is one direction in which it is not safe at all.

connections · Joint classification 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 The envelope is not a state of the structure. Every arrangement of the imposed load on three spans — 8 of them, since each span is loaded or not — drawn faintly, with the greatest sagging and greatest hogging at each station drawn over them. Each faint curve is a real state of equilibrium and satisfies the free-moment identity exactly: mid-span ordinate minus the mean of the end moments is wL²/8, to 3e-16 of it. The envelope satisfies it nowhere, missing by up to 23% — because it is assembled from different load cases at different stations and no arrangement of load produces it. seven of the 8 arrangements are needed to build it; the rest never govern anywhere.

The envelope is not a structure

A continuous beam whose imposed load may sit on any span has eight load cases, and every one of them is a genuine state of equilibrium. The curve the design is made against is not one of them — it is assembled from different cases at different stations, and it fails the identity all eight satisfy exactly.

equilibrium · Load arrangement 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 Two of these move and the third cannot. The first span of a 3-span beam under 5 kN/m, with the stiffness of the middle span swept over a factor of 25. The support moment and the mid-span moment both move — that is what redundancy does, and it is the whole reason a continuous beam has to be analysed rather than read off. Their combination does not: the mid-span ordinate plus the average of the two end moments is 30.6 kNm at every point on this axis, which is wL²/8 for this span and this load and contains nothing else. The largest departure anywhere on the sweep is 2.3e-16 of the value, which is the arithmetic of the stiffness solution rather than a property of the beam. Continuity buys a distribution and not a capacity, and this is the line that says so.

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 3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 28.1 to 18.0, and a hogging moment of 22.5 appears over the supports where there was none.

The chord is a continuous beam

A truss analysis reports one number per member, and for a chord running through eight panels that number is an axial force with no bending in it. The chord is a single piece of steel spanning between its own nodes, and anything landing between them bends it.

structures · Truss The same load, two diagrams, both in equilibrium. One span of a pair of 9 m spans under 30 kN/m, drawn twice. The elastic solution puts 304 kNm over the support and 171 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 213 and 207: the section the beam needs falls from 304 kNm to 213, 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 304 kNm for either — and the second is legitimate for that reason alone. What it costs is 13.0 milliradians of rotation at the support, which the section has to be able to deliver.

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 A reaction with no load, and the moment it bends the beam with. The prestress moments in a 2-span beam. The primary moment is −P·e, the tendon acting on its own section, and it reaches 1440 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 720 kNm — 50% of the primary, with the same sign, so it does not cancel anything. It comes from the middle support refusing to let the beam lift: 102.9 kN pressing down there and 51.4 kN lifting at each end, a reaction set that sums to 0e+0 because nothing external was applied. Its diagram is straight between supports to 3.6e-13% of its own peak, which it has to be: reactions are point forces and a point force puts no curvature in a span.

The tendon that can be moved

Lift a continuous beam's tendon at its interior support without changing its drape and nothing about the beam's total moment changes. The primary falls, the secondary rises by exactly as much, and the pressure line stays where it was — which turns a parasitic effect into a quantity a designer can place.

internal-forces · Secondary prestress The studs are evenly spaced and the demand is not. The force per unit length the shear connection carries along half of a 12 m composite beam, from Newmark's solution. It is largest at the support — 282 N/mm — falls to nothing at mid-span, and averages 156: the end studs are asked for 1.81 times the mean. Studs are nevertheless placed at a uniform spacing, and the justification is the one the variable-angle truss uses for its stirrups — a ductile connector sheds what it cannot carry to its neighbours, so the uniform distribution is a plastic redistribution and not a description of the elastic state.

The connection is busiest where the beam is not

A composite beam's studs are spaced evenly along it and the demand on them is not even at all. It peaks at the supports, where the bending stress is nothing, and falls to zero at mid-span, where the section is working hardest — so the connection is designed from a diagram nobody looks at.

internal-forces · Composite action The joint yields first and the span takes the rest. The end and span moments of a beam of span 9 m and flexural rigidity 61,700 kN·m², with a plastic moment of 522 kN·m, on joints of rotational stiffness 41,133 kN·m/rad, 6.00 EI/L, and resistance 261 kN·m, as a uniform load rises to collapse. While elastic the joints carry 0.75 of the fixed-end moment. The joints reach their resistance first, at 51.6 kN/m, and from there every further increment of load goes to the span, which reaches 522 kN·m at 77.3 kN/m. The collapse load is 8(Mp + Mj)/L² = 77.3 kN/m, between the 51.6 at which the beam would collapse on pins and the 103.1 it would reach on rigid full-strength ends, and it contains the joint's strength and not its stiffness.

The joint that has to keep turning

A beam on partial-strength joints collapses at 8(Mp + Mj)/L² however stiff the joints are. Stiffness decides only which yields first, and a joint that yields first has to go on rotating at full moment until the span catches up. A weak joint on a stiff connection has the most turning to do — more than a rigid full-strength one.

connections · Joint classification Settling down or walking away, cycle by cycle. The total plastic hinge rotation of the beam after each cycle of loading — span 1, both, span 2, neither — with the midspan load at 0.98, 1.02, 1.05, 1.10 times the shakedown load of 126.3 kN. At 0.98 it stops at 0.59 mrad. At 1.02 it grows 4.57 mrad a cycle. At 1.05 it grows 11.43 mrad a cycle. At 1.10 it grows 22.86 mrad a cycle. Nothing collapses in any single cycle; above the shakedown load the beam walks.

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

Named alongside it

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

ContinuityPlastic hingeDuctilityIndeterminacyEquilibriumFree bodyLoad arrangementLower-bound theoremStiffnessRotation capacityBending momentCollapse mechanism

All concepts