Structural Engineering & Statics
where the load goes
Concept

Shear reinforcement — where it appears

Steel carrying the tension diagonal of a concrete member's internal truss, usually links crossing the potential shear crack. Its area follows from the shear force, the strut angle assumed and the lever arm, and it is separate from any steel needed to suspend a load into the compression zone.

Named by 3 essays across one field — each of them below, with the objects they name alongside it.

A check made on a perimeter, not on a section. One bay of a flat slab, 7.2 m square, on a 400 × 400 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 4427 mm long against 1600 mm round the column itself. The shaded area inside it delivers no shear across it and is subtracted from the load; everything outside arrives through the perimeter. At 12 kN/m² that is 604 kN across 4427 × 225 mm, a shear stress of 0.606 N/mm² against a resistance of 0.658.

A check made on a perimeter, not on a section

Every shear check in this collection is made on a plane cut through a member. A slab sitting on a column has no such plane, because the shear leaves in every direction at once — so the check is made on a closed line, and a line grows with the column while the load grows with the square of the bay.

internal-forces · Punching shear 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 studs reach out to where the concrete is enough. One column of a 260 mm flat slab (d = 225 mm) carrying 12 kN/m² on 400 × 400 mm internal columns, the moment it hands the column putting the shear 300 mm off centre, on a 10.0 m bay, to scale. The inner check, on the control perimeter 2d out (solid), needs reinforcement: 12 rails of studs, 6 perimeters of them from 0.5d at 0.75d spacing, 1,415 mm² on each perimeter. The outer perimeter (dashed), where the concrete alone carries the shear, stands 5.4d from the face — 9,173 mm long against the control perimeter's 4,427 — and the last studs must be within 1.5d inside it, 3.9d from the face.

The studs that send the check outward

Shear studs round a column fix a failing punching check, and they turn one check into three. The studs carry the control perimeter; past them the concrete alone must carry a perimeter long enough to need no help, and that outer perimeter stands further from the column the larger the load, so the studs follow it out. At a 10 m bay they reach four effective depths from the face, and the steel they need grows as the shear to the power 3.5. The third check, at the column face, is the one no stud reaches, and it ends the series at 11.8 m.

internal-forces · Punching shear

Named alongside it

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

Strut-and-tieControl perimeterEffective depthFlat slabPunching shearAggregate interlockBrittle failureClamping forceColumn headCompression strutDetailingEccentricity

All concepts