Structural Engineering & Statics
where the load goes
Concept

Shear — where it appears

The internal force acting across a cut rather than along it, equal to everything the free body on one side of that cut carries transversely. Its diagram is the derivative of the moment diagram and the integral of the load, so its zeros locate the moment's peaks.

Named by 7 essays across 4 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 Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight.

Both at once, and neither matters until it does

A section carrying shear has less moment capacity, and the reduction is the web's share of the plastic modulus times one minus the root of one minus the shear ratio squared. On a rolled beam that share is a quarter, so half the shear capacity costs three and a half per cent — and then the last tenth costs more than the first eight.

sections · Shear moment interaction 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 Two cantilevers, or one wall, and the beams decide which. The deflected shape of a coupled pair of 6 m walls, drawn against the two limits it lies between. Release the coupling beams entirely and the pair is two independent cantilevers, deflecting 111 mm. Make them rigid and it is one composite wall of the full width, deflecting 16 mm — 6.8 times stiffer, because the lever arm between the wall centroids is 8.40 m and everything inside either wall is smaller than that. Real beams of 600 × 350 mm over a 2.4 m opening land at 23 mm and carry 63% of the base overturning as an axial couple rather than as wall bending. The degree of coupling never reaches one, because a beam of finite depth cannot suppress the walls' curvature entirely.

Two walls that agreed to be one

A pair of shear walls with a row of doors between them is the commonest lateral system there is, and it has two readings that differ by a factor of seven. What decides which one applies is a beam 600 mm deep over a 2.4 m opening — and most of the overturning ends up as an axial couple that no bending diagram contains.

internal-forces · Wall coupling A strength with no mechanism in it, made of four. The shear a member carries with no links in it, split into the mechanisms that carry it, against the member's effective depth on a logarithmic axis. The three bands are calibrated to Taylor's measured shares at one 300 mm × 500 mm member and are then evaluated everywhere else, so the shape of the total is a prediction. Aggregate interlock is the band that dies: it depends on how tightly the crack faces are held together, crack width grows with member depth, and it falls from 62% of a shallow member's strength to 22% of a deep one's. That decay is the whole of the size effect, and the dashed line is the design code's fitted k = 1 + √(200/d), which knows nothing about interlock and falls by a factor of 1.52 where the model falls by 2.05 over the same twentyfold range. Dowel action is why the expression contains the flexural reinforcement ratio, which nothing in a truss analogy would predict.

The strength with no mechanism in it

A concrete member with no links in it carries shear, and the expression that says how much is three variables raised to fitted powers with a size term in front. There is no free body anywhere in it. What it is fitting is a competition between four things that carry shear across a crack, and only one of them explains why a deeper member is worse at it.

internal-forces · Concrete shear Two identical pipes, and one carries three times the other. Load per metre on a buried conduit against the depth of cover, in trench widths, with the weight of the prism of soil directly above it drawn between them. A conduit laid in a narrow trench is stiffer than nothing and softer than the sides: the backfill settles relative to the undisturbed ground, the friction on the trench walls acts upward, and the conduit gets 64% of the prism. Lay the same conduit on the ground and build an embankment over it and it is now stiffer than the fill beside it, the interior prism settles less, the friction acts downward, and it gets 172% — a factor of 2.71 between two pipes with nothing different but which way the ground moved. The equation is Janssen's, the same one a silo wall obeys, with a trench for a silo; both curves start on the prism line, because with no depth there is no shear to redistribute anything. This is why a flexible pipe is buried rather than a rigid one: making the conduit weaker moves it down the page.

The pipe decides what the soil weighs

A buried conduit is not loaded by the soil above it. It is loaded by whatever share of that soil the relative movement leaves it — and which way the shear on the sides of the prism acts depends on whether the conduit settles more or less than the ground beside it. Two identical pipes under identical fill, one carrying two thirds of the prism and one carrying nearly twice it.

equilibrium · Soil arching Cross the hangers and the chords stop bending. The same tied arch, the same sixteen hangers, the same load on half the span — hung vertically and hung as a network. Vertical hangers make the two chords a Vierendeel frame, which has no truss action at all, so a partial load is carried by bending: 3316 kNm in the tie and 6234 in the arch. Inclined hangers can carry the shear between the chords axially, and the same load gives 686 and 831 — factors of 4.8 and 7.5. The thrust is identical in both, because that is decided by the span and the rise and nothing else.

Cross the hangers and the bending goes

A tied arch with vertical hangers is a Vierendeel frame with a curved top chord — it has no truss action at all, so a load on half the span is carried by bending. Incline the hangers so they cross and the same two chords become a truss.

structures · Network arch

Named alongside it

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

Free bodyStrut-and-tieCracked sectionEquilibriumLever armLoad pathStiffnessTruss analogyAggregate interlockAxial forceBending stressBond

All concepts