Structural Engineering & Statics
where the load goes
Concept

Aggregate interlock — where it appears

The mechanical engagement of the projections on a cracked concrete surface, which is what carries shear across a plane with no tensile strength left. It falls away as the crack widens, which is why it is credited only where the crack is held narrow by reinforcement crossing it.

Named by 5 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 coefficient is a slope, and that is why it can exceed one. The crack magnified: two rough faces, drawn as a sawtooth at 54° to the plane. Sliding one over the other cannot happen without lifting it, so a shear displacement forces a separation in fixed proportion — the tangent of that angle, which is the number written down as a coefficient of friction and here is 1.40. The bars crossing the plane are stretched by the separation and clamp the faces back together; they are not carrying the shear, they are supplying the normal force that lets the roughness carry it. A bar that is not anchored on both sides supplies nothing.

Shear across a crack that is already there

Every shear calculation in this collection starts from an uncracked solid — a principal stress, a shear flow, a diagonal tension. This one starts after the crack, on a plane with no tensile strength at all, and the coefficient it uses is not a coefficient of friction. It is the slope of the roughness.

internal-forces · Shear friction 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 Flattening the truss saves stirrups and crushes the web. Two capacities against the angle of the cracks, for a web 350 mm wide with a lever arm of 630 mm. The rising line is the stirrups: a cut along the crack severs z·cot θ/s of them, so flattening the crack from 45° to cot θ = 2.5 takes the 315 kN they carry to 787 — 2.5 times as much from the same steel. The falling line is the concrete strut, whose stress is V(cot θ + tan θ)/b_w z and therefore least at 45°. They do not cross in this range, so the stirrups govern throughout and the angle is a free choice.

The angle is a choice, not a property

The truss inside a cracked concrete web has a strut angle, and nothing measures it. The designer picks it, the stirrup requirement falls as it flattens, the web stress rises, and every choice in between is a different structure that carries the same load.

internal-forces · Concrete shear More steel across the crack, until the roughness runs out. Shear resistance of the interface against the reinforcement crossing it. The steel clamps rather than carries, so the resistance is the clamping stress times the interlock coefficient and rises in a straight line — until the asperities crush at 5.50 N/mm², which happens at a reinforcement ratio of 0.79%. Past that the line is flat and every further bar is decoration. The dashed line is what the clamping alone would give if the concrete were unbreakable.

Two models of one bracket

A corbel can be designed as a plane that has to be clamped or as a truss that has to be drawn, and the two are not approximations of each other. They describe different failures, they ask for steel in different places, and the honest answer is that both are checked because neither bounds the other.

internal-forces · Shear friction

Named alongside it

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

Strut-and-tieFailure surfaceShear frictionSize effectAnchorageBondClamping forceCrack widthEquilibriumFree bodyLower-bound theoremReinforcement

All concepts