Structural Engineering & Statics
where the load goes
Theme

One support too many

Indeterminacy: more restraints than equations. It buys robustness, costs a stiffness calculation, and makes a structure sensitive to things statics cannot see.
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 three-pinned arch, rise 2.6 on span 9. A three-pinned arch under a uniform load. One moment equation about the crown hinge gives a horizontal thrust of 23.37, with no stiffness and no assumption about the section. The thrust line lands on the axis everywhere, so there is no bending anywhere in the arch. Structural form

The hinge put in on purpose

An arch with two pinned feet cannot be solved by statics. Add a third hinge at the crown — deliberately weakening it — and the whole structure falls out of one moment equation.

 A portal frame swaying under 20 kN. A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 10.0 and 10.0 and add to the applied 20; the peak moment is 22.2. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face. Structural form

The frame that leans, and what stops it

A rectangle of pinned bars folds flat. Make the corners rigid instead of adding a diagonal and it does not — which buys an unobstructed opening and costs bending in every member of 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. 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.

 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. Internal forces

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.

 The ends decide the length that matters. Four columns of identical height and section, buckling under four sets of end conditions. The effective length factor is the fraction of the column that behaves like a pin-ended one, and the buckling load goes as its inverse square. Stability

The ends decide the length that matters

Four columns of identical height and section, buckling at loads sixteen times apart. Nothing differs but what is holding the two ends.

 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.

 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.

 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. Materials

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.

 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 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.

 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. Connections

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.

 A joint is springs in series. The five components of an end-plate joint, with each bar the flexibility it contributes. The column flange in bending is 39.62% of the total on its own, and doubling its stiffness raises the joint's by a factor of 1.25 — while doubling the stiffest component buys 1.05. The joint's rotational stiffness is 25227.71 kN·m per radian. Connections

A joint made of springs in series

A connection's stiffness is not a property of the connection. It is the series combination of the flexibilities of everything the force passes through — so the softest component decides, and stiffening any of the others changes almost nothing.

 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. Connections

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.

 Three modes of a five-storey frame. The first three mode shapes of a five-storey shear frame, from the eigenvalue problem rather than sketched. Mode 1 has a period of 0.6 s, no node and carries 88.0% of the mass; Mode 2 has a period of 0.21 s, one node and carries 8.7% of the mass; Mode 3 has a period of 0.13 s, two nodes and carries 2.4% of the mass. The nth mode crosses the axis n−1 times, which is a theorem rather than a drawing convention. Dynamics

A structure has more than one period

One mass on one spring has one period. A building with eight floors has eight, each with a shape of its own, and the second one bends the building into a curve nobody drew. They are not harmonics, they do not interfere, and each behaves as though the others were not there.

 Three legs, three equations, one answer. A rigid top on three legs carrying 120 kN at (0.3, 0.2) m. The three equilibrium equations available — one vertical and two moments — leave three unknowns, so the system is exactly determinate and the reactions are 56.8, 13.6, 49.6 kN. Move the load anywhere and the answer moves with it; nothing about the legs' stiffness enters. Equilibrium

Six equations, and the drawing shows three

Every essay so far has taken place on a piece of paper, where equilibrium is three equations and a structure is a diagram. The real object has six, the extra three are the ones nobody writes down, and the difference between three legs and four is not a matter of degree.

 It moves, or it pushes. Never both, and never neither. A 30 m steel member 30 °C warmer than it was built, in three conditions. Free, it grows 10.8 mm and carries nothing. Held, it moves nothing and carries 75.6 MPa in compression — which is E·α·ΔT and contains neither the length nor the area of the member, so the identical stress arises in a two-metre strut. Held by a spring it does some of each: 3.2 mm of movement and 53.2 MPa, and the split is decided by the spring rather than by the member. Deflection

The movement nobody applied

A temperature change is the only load in this collection that a structure can decline. Let it move and it produces a movement with no stress; hold it and it produces a stress with no movement — and that stress contains no length, no area and no second moment, so a bracket and a bridge girder carry exactly the same one.

 The answer arrives in instalments. The hogging moment at support 1 of a three-span beam, cycle by cycle. It starts at the fixed-end moment of 53.3 kNm — the value with every joint clamped — and settles at 64.0 kNm against an exact 64.0. The error falls by about a factor of four per cycle: 14.00, 3.50, 1.95, 0.59 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything. Deflection

Solved by passing it around

An indeterminate structure needs simultaneous equations, and for thirty years engineers solved them without writing any down. Clamp every joint, release one, share out what is left over, pass half of it along, and repeat — and the answer walks in, three figures correct after four cycles.

 Take that one away and the load finds another route. A 6-panel pratt truss under 20 kN at each top node, before and after member 2 is removed. The load redistributes. The worst-affected survivor now carries 2.03 times what it did, and four members that carried nothing before are now working. Whether that is survival depends on how much spare capacity was there, which is a different question from whether the frame was strong enough. Structural form

The structure that survives losing a member

Every check in this collection asks what a structure carries. None of them asks what is left when part of it is gone — and two frames with the same members, the same weight and the same factor of safety can answer that question completely differently.

 Two shapes that are the wrong way up for each other. Deflected shapes of a 20-storey building under a uniform wind, drawn to the same scale. The wall alone bends: its shape is flattest at the base and steepest at the top, reaching 146 mm. The frame alone shears: it is steepest at the base where the storey shear is largest, reaching 140 mm. Tied together at every floor they reach 58 mm — less than a quarter of either, and less than the 72 mm two springs in parallel would give, because each is stiff exactly where the other is not. Structural form

How a tall building stands still

A shear wall bends and a framed tube shears, and the two deflected shapes are the wrong way up for each other. Tie them together at every floor and the pair is stiffer than the sum of their stiffnesses — because near the base the wall holds the frame back and near the top the frame holds the wall.

 The reaction lies inside the cone, so the block stands. A block of 100 on a plane at 15°, against a coefficient of friction of 0.35. Resolving across and along the plane gives a normal force of 96.6 and a friction demand of 25.9, against a capacity of μN = 33.8 — a ratio of 0.77. Added together the two make one contact reaction leaning 15.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 19.3°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 15.0°, which is why the angle of repose is a material property and the size of a heap of sand is not. Equilibrium

The force that is whatever it needs to be

Every other force in statics has a value the equations produce. Friction has an inequality instead, so it takes whatever value equilibrium demands and the bound only ever says no — which means a problem with friction in it has a range of answers rather than one.

 The moment does not stop at the end of the beam. A portal frame of 8 m by 4 m with fixed bases, carrying 20 kN/m on the beam. The bending moment is drawn on the tension side of every member, and it runs round the corner without a break: 65.2 kNm arrives at the end of the beam and 65.2 kNm leaves down the column, which is the same number, since joint rotational equilibrium is one of the equations the frame solve satisfied. Midspan carries 94.8 kNm, and the two add to 160.0 — the 160.0 kNm of a simply supported span, to 0.0e+0 kNm. The corner takes 61% of the wL²/12 a fully built-in beam would have carried, because the columns are springs rather than walls: the beam-to-column stiffness ratio is 1.27. The beam's moment crosses zero 0.92 m from the corner and the column's 1.33 m above its base. Internal forces

The moment that goes round the corner

At a rigid knee the bending moment does not stop at the end of the beam. It turns and runs down the column, and in the same instant the beam's shear becomes the column's axial force — while the block of steel that has to carry the turn appears on no member diagram anywhere.

 Two beams tied together, and the deeper one takes 89% of the load. Two simply supported beams of 6 m, one twice as deep as the other, tied together at midspan so that they have to move as one. A load of 100 kN stands on the tie. Point stiffness is 48EI/L³, so the deeper beam is 8 times as stiff — depth cubed, nothing else — and the load divides in that ratio: 11.1 kN into the shallow beam and 88.9 kN into the deep one, 11% against 89%. Both midspan points move 5.00 mm, which is the whole of the argument: the geometry of the load never entered it. The deflection is drawn 78 times full size — the real sag is 5.00 mm on a 6 m span, about 1 in 1200. Internal forces

The stiffest path takes the load

When two members share a force the split can be argued about. When they share a displacement it cannot — stiffness settles it, and nothing about the load or the plan drawing gets a vote. The consequence is that stiffening a lightly loaded member raises its stress, and the way to unload something is to soften it.

 The ground pushes back hardest where the beam has gone down furthest. A strip 16.4 m long and 1 m wide on ground of subgrade modulus 50 × 10³ kN/m³, carrying 1000 kN at its centre. The beam settles 3.90 mm under the load and the ground pushes back in proportion — the arrows are k times the settlement above them, peaking at 195 kN per metre — so the pressure diagram is the settlement bowl and not an assumed distribution. The characteristic length 1/β is 2.56 m: the bowl crosses zero at 6.04 m, which is 3π/4 of it, and beyond that the arrows reverse because the beam has lifted off. By 8.05 m — one π/β — the disturbance is 4.3% of what it was, which is why the moment 641 kNm and the peak pressure 195 kN/m contain no length at all. The settlement is drawn 217 times full size — the real bowl is 3.90 mm deep over 16.4 m, about 1 in 4207 — and at true scale the beam would be a straight line. Internal forces

The beam that sits on the ground

Every other beam in this collection is held at points. A footing is held everywhere, by something that pushes back in proportion to how far it is pushed — and that single change hands the structure a length it did not choose. Two or three of those lengths from the column, nothing knows the load happened.

All themes