Structural Engineering & Statics
where the load goes
Concept

Stiffness attracts load — where it appears

The rule that a stiff element takes force in proportion to its stiffness, whether or not that is where the strength happens to be. It is the reason a strengthening scheme can make matters worse, and the reason an unintended stiff element such as a masonry infill takes force nobody assigned it.

Named by 7 essays across 5 fields — each of them below, with the objects they name alongside 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.

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 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.

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.

internal-forces · Load-sharing The same strain, two moduli, and a width multiplied to say so. A timber section with a steel plate in it, carrying 20.0 kNm. Plane sections stay plane, so the strain at a height is the same in both materials; Hooke's law then puts the stresses in the ratio of the moduli, which here is 19.09. Multiplying the stiffer material's WIDTH by that ratio gives a fictitious section of one material with the same neutral axis and the same forces — 595.2×10⁶ mm⁴ of it, against 351.0 for the same shape with the moduli ignored. The steel plate is 3.8% of the area and carries 43% of the moment, at 96 N/mm² against the timber's 5.0. The transform is not an approximation: it is compatibility and Hooke's law written down.

A section made of two materials, one of them pretended away

Multiplying a material's width by the ratio of the moduli produces a fictitious section of one material with the right neutral axis and the right forces. It is not a trick — it is compatibility and Hooke's law written down — and it says a stiff material takes what its modulus asks for.

sections · Transformed section One of these two curves is a stiffness and the other is a statement of statics. The torque a spandrel beam carries, against how much of its torsional stiffness is left. The rising curve is compatibility torsion — a floor beam framing into the side of the spandrel, which shares its fixed-end moment of 197 kNm between the spandrel's torsional stiffness and its own flexural one. Uncracked, the spandrel takes 51% of it, or 100 kNm; at a quarter of that stiffness it takes 21%, or 41 kNm, and the floor beam picks up what was shed. The flat line is equilibrium torsion — a canopy cantilevering 2.2 m off the same spandrel, whose 116 kNm is fixed by statics and contains no stiffness at all. The first can be designed away by accepting a rotation. The second cannot be designed away by anything.

The torque that should not be shed

A compatibility torque can be let go, because the load has somewhere else to go. What the rule does not say is what it costs the somewhere else — a twenty per cent rise in a floor beam's midspan moment, a crack width nobody limits, and a rotation the spandrel has to actually deliver. There is a size of torque past which shedding is the wrong answer, and no code states it.

internal-forces · Compatibility torsion Removing each member in turn. Every member of a 8-panel pratt truss removed one at a time, with the worst demand on the survivors plotted against the member removed. Four of the 35 leave a mechanism — the bars drawn to the top of the frame — and for those there is no redistribution to compute, because there is no structure left. The rest redistribute, and the worst of them asks a survivor for 2.04 times what it carried before. A single number for robustness does not exist: it depends on which member goes.

A determinate truss has no robustness at all

Remove any one of a Warren truss's thirty-one members and what is left is a mechanism. Not weakened — gone, with no set of forces that holds the load in any position. Robustness is not a property a structure has by degree; it is bought by adding members that carry nothing until something else stops carrying, and a truss without them has none of it to measure.

structures · Robustness Three fasteners, three completely different clocks. What each of three shear fasteners carries against how far the joint has moved. A 400 kN fillet weld is linear to 0.4 mm and then gone — it is stiff and it is not ductile. A 380 kN bolt in a hole 2 mm larger than itself carries nothing until the hole closes and then rises over several millimetres of hole elongation. A 250 kN preloaded bolt is at its slip resistance in under a tenth of a millimetre and holds it until it slips. Two of these in one joint are at the same displacement, so the one that gets there first carries the load — and the weld gets there 23 times sooner than the bearing bolt does.

Two fasteners that never arrive together

Every steel code forbids adding a weld's capacity to a bolt's in one shear joint, and states it as a rule rather than deriving it. It is derivable. Two fasteners in parallel are at the same displacement rather than the same force, and a weld has ruptured at four tenths of a millimetre while a bolt in a standard hole has not yet touched the side of it.

connections · Weld group The bolt force is the larger of two lines. What an M20 bolt in a 25 mm tee flange actually carries, against the tension applied to the flange. Preloaded to 171 kN it starts there and climbs at Φ = 0.185 — the bolt's own stiffness over the bolt's plus the clamped plates', 857 against 3781 kN/mm — so 18 per cent of every kilonewton applied reaches it and the rest is unloading the contact. At 138 kN the contact runs out and the line joins the one an ordinary bolt has followed from the start, climbing at 2.12. The two lines meet, so the strength is the same either way; what differs is the slope, by a factor of 11.5. The flange's own mechanism is at 172 kN, comfortably past the crossing.

The bolt that was already stretched

Prying is a lever that needs the flange to lift before it can act, and a preloaded bolt does not let it. The bolt carries a fifth of every kilonewton applied until the plates part, and everything after that is the ordinary calculation — which is why preload changes the fatigue answer by a factor of a thousand and the strength answer by nothing at all.

connections · Prying

Named alongside it

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

DuctilityCompatibilityIndeterminacyLimit analysisLoad pathRedistributionBearingBolt groupBolt tensionCentroidClamping forceCompatibility torsion

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