Structural Engineering & Statics
where the load goes
Generator

Where the peak moment goes when a restraint is added

Rendered here at the parameters it defaults to, with every essay that calls it — which is the same list as the blast radius of changing it.
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.

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.

13 essays call redundant-beam. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this generator changes every one of these figures.

Where to put the supports. Peak sagging and hogging moment for a uniformly loaded beam, against how far the supports are moved in from the ends. The best arrangement is where the two curves cross, and it is nowhere near the ends. Internal forces

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.

 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. Internal forces

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.

 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. Deflection

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.

 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.

 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 area is the rotation, and its first moment is the movement. A 6 m cantilever under a tip load of 10, with the M/EI diagram beneath it. The shaded area is 180.00, which by the first theorem is the change of slope along the whole member. Its centroid is at 2.000 m, and the first moment about the tip is 720.00 — which by the second theorem is the deviation from the tangent, and for a cantilever that tangent is horizontal, so it is the deflection itself. Integrating the curvature twice instead gives 720.00. Deflection

The area of a diagram is a rotation

A deflection is the double integral of a bending moment, and the two constants of integration are the whole difficulty. Mohr's theorems replace them with two pictures — an area, and where that area's centre of gravity sits.

 Two differences up the same building, peaking in different places. Differential shortening between a perimeter column and the core of a 40-storey building, plotted up the height. The part driven by load peaks at level 20 — exactly half way up, because a floor near the top has almost nothing built above it to shorten what is beneath, and a floor near the bottom has almost nothing beneath it to shorten. The part driven by shrinkage does not care what is above it at all and accumulates all the way to the roof. Their sum is worst at level 40, at 43 mm, which across a 9 m bay is a floor out of level by one in 208. Deflection

The columns are shorter than the core

Every column in a tall building gets shorter as the building is built on top of it, and the core beside it gets shorter by a different amount. The floors between them tilt by the difference — and the difference is largest exactly half way up, because a floor near the top has almost nothing built above it and a floor near the bottom has almost nothing beneath it.

 The building does not care how far it went down; it cares how much it tilted. Five footings on soil that is 35% as stiff under one of them, carrying 60 kN/m. They settle between 12 and 54 mm, and the number that matters is neither of those: it is the angular distortion between neighbours, 4.32 per thousand, or one in 231 — against a limit of one in 500 for cracking in finishes, which this does not. A building that went down half a metre uniformly would be undamaged and would need a new front step; this one has moved a twentieth as far and has cracked. Deflection

The settlement that matters is the difference

A building that goes down half a metre uniformly is undamaged and needs a new front step. One that goes down a twentieth as far, unevenly, has cracked. The superstructure can even the difference out — and the only way it can do so is by carrying the difference itself, as a force.

 Half the beam does nearly all of the deflecting. The virtual-work integrand M·m/EI along the member, normalised to its own peak, with the running share of the answer beside it. The integrand is a density: it says how much of the deflection each millimetre of the beam produced. For this case the half nearest the root supplies 87.5 per cent of it, and the rest of the member supplies the remainder. Stiffening the busy 50 per cent by 1.5 times takes the deflection down by 29.2 per cent; the same material spent on the quiet end takes it down by 4.2 — a factor of 7.0 for the same steel. The map of what is contributing is not the map of where the moment is largest, and the second is the one that gets drawn. Deflection

Where a deflection comes from

The unit-load method gives a deflection as an integral, and this collection has treated that integral as a number to evaluate. It is not a number. It is a density, and it says which millimetres of the member produced the answer — which is not the same map as where the moment is largest.

 Almost all of it is exactly zero. The stiffness matrix of a 2-bay, 3-storey plane frame: 36 freedoms, of which 16.2 per cent of the 36² entries are non-zero. The zeros are not small numbers; they are absences. A member reaches only the two nodes at its ends, so it can contribute nothing to any row belonging to a node it does not touch, and every such entry is zero exactly rather than nearly. The non-zeros therefore sit in a band of width 13 about the diagonal. Before the supports are applied the matrix is singular, and its null space has exactly three dimensions — the three rigid-body motions a plane frame has with respect to the ground, which is the same statement nullVector makes about a truss that is a mechanism, arrived at from the other end. Deflection

The matrix that replaced the hand methods

Moment distribution passes moments round a frame until they stop moving. Virtual work computes one deflection at a time. Both are exact and both stop scaling in the low tens of members. What replaced them adds no physics at all — the whole of the invention is the bookkeeping.

 Two different structures released, and one bending moment diagram. The bending moment in a continuous beam of 8, 10, 8 m under 12 kN/m, solved twice by the force method with different redundants. The first release puts a hinge over each interior support, so the released structure is a row of simple spans and the redundants are moments. The second removes each interior support, so the released structure is one simple span of the whole length and the redundants are reactions. The two released structures have nothing in common — different shapes, different deflections, different everything — and the diagrams they produce lie on top of each other to 9e-15 of the peak moment. Which restraints are released is a choice about the arithmetic and not about the structure, which is a fact worth trusting: it means a hand calculation can pick whichever release makes the sums easiest and be sure of the answer. Deflection

Choose what to take away

The other machine for a redundant structure works by removing restraints until what is left can be solved by statics, then putting back exactly enough force to close the gaps that opened. Which restraints are removed does not change the answer at all, and changes the arithmetic completely — one choice gives a tridiagonal matrix a person can solve on paper, and another gives a full one.

 The same curve, computed twice and from opposite ends. A cantilever of 6 m, with its deflected shape drawn from a double integration of M/EI and the bending moment of its conjugate beam drawn on top of it. The conjugate is the same span carrying M/EI as a load, with its supports transformed — a real fixed end becomes a free end and a real free end becomes a fixed one, because a fixed end has no slope and no deflection and the conjugate therefore needs no shear and no moment there. The two curves agree to 4.7e-6 of the largest deflection, which is the trapezium rule and not the method. The largest deflection is 36.00 mm against the closed form's 36.00. The reaction of the conjugate beam is -9.000 milliradians, which is the real beam's rotation at that support — so the whole of a slope calculation is one reaction. Deflection

The beam whose moment is a deflection

A beam's bending moment is the second integral of its load. Its deflection is the second integral of M/EI. They are the same problem, so a deflection can be found by loading a fictitious beam with M/EI and asking a statics question — and the only thing to remember is the supports, which are not remembered but derived, one boundary condition at a time.

The library, page 4 of 7 — where redundant-beam sits