Structural Engineering & Statics
where the load goes
Concept

Truss analogy — where it appears

The model in which a cracked concrete beam is treated as a truss of concrete struts, steel ties and chords. It turns a shear calculation into an equilibrium statement about members, and the angle chosen for the struts is a design decision rather than a measured property.

Named by 5 essays across 2 fields — each of them below, with the objects they name alongside it.

The cut that severs the stirrups is the cut that counts them. A cracked web drawn as the truss it has become: a tension chord along the bottom, a compression chord along the top, concrete struts between the cracks and stirrups crossing them. The free body is a cut parallel to the cracks, which over a lever arm of 495 mm severs z·cot θ/s = 8.3 stirrups at cot θ = 2.5. Every one of them is at yield, so the shear is their number times their strength, and flattening the crack raises the count rather than the strength. The same cut passes through the bottom chord, which is where the chord force the moment diagram does not contain comes from.

The beam that becomes a truss

Once a web has cracked in shear there is no shear stress field in it any more. There are concrete struts, two chords and whatever crosses the cracks, and the angle of those cracks is not a property of the material — it is something the designer chooses, and every quantity in the beam moves when it changes.

internal-forces · Shear truss analogy The same beam, the same load, and one of them has to lift it. Two lower-bound models of one beam, differing only in which chord the 600 kN is applied to. The chord forces are identical in both — the moment diagram does not know where the load arrived — and so are the struts. What is not identical is the vertical at the load: nothing in the top-loaded model, and the whole 600 kN in the hung one, because a bottom-chord panel point touches no strut and the load has nowhere to go but up. That tie is 1200 mm² of steel against the 96 mm² the shear calculation asks for over the same length — a factor of 12.5, in the same place, and additional to it.

The load that has to be lifted

Every shear calculation on this site assumes the load arrives on the top of the beam and walks down a diagonal to the support. Hang the same load from the soffit and the diagonal cannot start — the load has to be carried up to the compression zone first, by a tie no sectional calculation contains.

internal-forces · Indirect support The bond stress is crowded against the loaded end. A 20 mm bar embedded 806 mm, with the force in it and the bond stress on it plotted along the embedment. Uniform bond — the assumption behind every development length ever tabulated — is a flat stress and a straight line of force. An elastic bond of the same peak strength is neither: the slip is largest where the bar is pulled and dies away over 1/α = 471 mm, so the far end of the bar is doing almost nothing. At the design rule's length of 40 diameters the elastic bond is 55 per cent used. The uniform answer is what the bond looks like after it has yielded along the whole length, which is a statement about ductility rather than about strength.

The force that arrives along a length

A bolt takes its force at a hole and a weld along a line. A reinforcing bar has no such place — it is a smooth cylinder in a hole of its own shape, and the only thing stopping it sliding out is a stress smeared over its surface. So the force in it is not a number, it is a function of position.

internal-forces · Bond The chords take the shear the web is credited with. A cantilever of 6 m tapering from 400 to 1200 mm, under a 120 kN tip load, with the shear divided between the web and the two inclined chords. The chord force is M/z and it is not horizontal, so its vertical component is real: V_web = V − (M/z)·dz/dx, which is Résal's result of 1899. At the root the web is left with 33 per cent of the applied shear — exactly d₀/d₁ for a straight taper, with no length, load or material in it. Turn the same member round and the figure becomes 300 per cent: the chords stop helping and start adding, at the section where the moment is largest as well. And the limit is worth having — a member whose depth is proportional to its moment leaves the web nothing to do at all, which is the triangular cantilever every crane jib is.

The shear the chords take

Every shear check in this collection has assumed the two chords of a beam are parallel, so that the whole of the shear crosses the web. Taper the member and that stops being true — and the sign of the correction is decided by which end the haunch is at.

internal-forces · Inclined chord Where a bar may stop, and how far past there it goes anyway. The tension the bottom steel must carry along a 9 m beam, and the resistance of the bars actually present, drawn as a staircase. The demand is the moment diagram divided by the lever arm and then SHIFTED 270 mm toward midspan, because the truss inside the beam delivers its shear diagonally — the two constructions agree to 0.36 per cent, which is the second-order term and nothing else. Each curtailed layer then runs a further 1150 mm to develop, so the outer layer stops at 245 mm rather than the 1665 mm the moment diagram allows. The tail is 1420 mm at each end — 16 per cent of the span — and it is what turns a 20 per cent saving into 2.2.

Where a bar may stop

The moment diagram falls away from midspan, so the steel midspan needs is not needed everywhere, and curtailing it saves real money. Then two things get in the way, and between them they take nine tenths of what the moment diagram promised.

sections · Curtailment

Named alongside it

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

EquilibriumFree bodyLever armStrut-and-tieAnchorageDetailingDevelopment lengthLoad pathReinforcementShearShift ruleBond

All concepts