Generator

Two beams, or one beam four times as stiff — the interface shear generator

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.

Two beams, or one beam four times as stiff. Two 200 × 150 planks spanning 4 m under 6 per millimetre. Loose, they have 112.5×10⁶ mm⁴ between them and deflect 16.2 mm, with the two faces at the interface sliding past one another. Bonded, the pair has 450.0×10⁶ — exactly 4 times as much, because doubling a depth cubes — and deflects 4.0 mm at half the extreme-fibre stress. Nothing was added but a restraint on slip. With connectors of stiffness 200 the same beam deflects 5.4 mm, which is 89% of the way from one bound to the other.

7 essays call interface-shear. 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.

Internal forces

Two beams, or one beam four times as stiff

Stack two planks and they bend as two beams whose faces slide past one another. Bond the faces and the pair has one neutral axis, four times the second moment and half the stress. Nothing was added but a restraint on slip.

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.

Sections and stress

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.

Between two beams and one, and much nearer one. How composite a beam is, against the one dimensionless group that decides it: αL, where α² = K·EI∞/(EA*·EI₀). At αL = 0 the layers slide freely and the beam is two beams; past about 20 the connection is stiff enough that the last per cent is unbuyable. The beam drawn sits at αL = 15.8 and is 96% composite, deflecting 36.7 mm against 33.9 for full interaction and 110 for none. The curve is steep where a real design sits, which is why halving the number of studs does not halve anything.

Internal forces

Half the studs, and most of the beam

Bonding two layers together quadruples the stiffness of the pair. A real connection is a row of studs that deform, so the layers slide a little and the beam sits between the two answers — but not halfway between, and the number that decides where is a single dimensionless group.

Two faces, a couple, and a core that does none of it. A sandwich section 61.4 mm deep: two 0.7 mm faces separated by 60 mm of core. The bending is carried as a couple between the faces — 41 N/mm² of tension in one and compression in the other, over a lever arm of 60.7 mm — and the core carries a shear stress of 0.047 N/mm² and nothing else. The parallel-axis term is 98.8% of the section's second moment; the faces' own bending about their own centroids is 0.004% of it, and the core's is 1.2%. Separated by nothing at all the same two faces would be 2.3e+4 times less stiff.

Sections and stress

Two skins and the space between them

A sandwich panel is a section made of a material that carries the bending and a material that carries none of it. The parallel-axis term is not a correction here — it is 98.8 per cent of the second moment — and the shear deflection is not a correction either.

The worst force in a pile is not at the top of it. A 0.6 m pile 24 m long through ground that is settling, carrying 800 kN at its head. Above the neutral plane the soil moves down past the shaft and the friction acts downward, so the axial force grows with depth; below it the friction acts upward in the ordinary way and the force falls again to the 300 kN the base takes. The maximum is 1282.87 kN at 13.78 m — 1.60 times the load applied, and it is at a depth where nothing is applied, nothing is connected and nothing can be inspected. A pile section chosen for the head load is under-sized by that factor over the middle third of its length.

Internal forces

The ground that hangs on instead of holding up

A pile is driven through fill that has not finished settling. The fill goes down past the shaft, the friction along that length turns round, and the pile is now carrying the soil rather than the other way about. The worst force is not at the head and not at the toe, and nothing at that depth can be seen.

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.

Internal forces

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.

A short timber beam is a shear problem, and a steel one never is. Utilisation of the bending and shear checks on one beam, against span-to-depth. The two cross where the ratio equals f_m ÷ f_v exactly — no load, no width and no span survives the cancellation — which for this timber is 6.7 and for steel is 1.73. So a timber beam shallower than about six times its depth is governed by shear parallel to the grain, and a steel beam would have to be shorter than twice its own depth before the same thing happened, which is not a beam. The third check is bearing across the grain, which does not move with the span at all: on the beam drawn it is at 0.40, and it is the one that governs.

Sections and stress

The shear that decides a timber beam

A steel beam is never governed by shear, because its bending strength is only 1.73 times its shear strength and no beam is that short. Timber's ratio is 6.7 along the grain and 23 across it, so shear governs at proportions people build every day.

The library, page 2 of 7 — where interface-shear sits