Structural Engineering & Statics
where the load goes
Concept

Node — where it appears

A point in a strut-and-tie model where struts and ties meet and their forces are resolved. It is the part of such a model that is actually checked, because the concrete there carries the resultant over an area the model itself has to define.

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

Four ways to make a cell resist being racked, and one that is not one. One cell of a grid shell under the membrane shear it has to carry, by the four mechanisms available for carrying it, with the racking each produces over a 30 m span under 1.2 kN/m² of asymmetric load. A serviceability limit of span/250 is 120 mm. Four pin-jointed bars in a quadrilateral have no in-plane shear stiffness whatever — the cell folds, and the answer is not a large deflection but a mechanism. Rigid nodes carry the shear by bending the members over a cell, which smears to 12EI/s³ and comes to 0.35% of what a continuous sheet of the same stretching stiffness gives: 1205 mm, ten times the limit. One diagonal per cell, or a third member direction, carries it axially instead and lands within a factor of two of the sheet. That is the whole difference between a grid shell and a row of arches.

A shell only if the grid takes shear

A curved surface carries load in its own plane at a fraction of the material a flat one needs, and every gridshell ever built is an attempt to buy that with members instead of with a surface. The attempt succeeds or fails on one property nobody draws — whether four bars meeting at a corner can resist being racked — and a pinned quadrilateral grid cannot resist it at all.

structures · Gridshell A truss drawn inside a solid, and solved as one. A deep member 5000 mm between bearings and 2500 mm deep, carrying 2400 kN at mid-span. The model is two struts and one tie, on a lever arm of 2000 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 1500 kN and each strut at 1921 kN, at 38.7° to the horizontal. Spread over a strut width of 1031 mm the compression is 3.7 N/mm² against a limit of 18.1 for concrete cracked across its own strut, and the tie needs 3448 mm² of steel. A beam calculation on the same member would have asked the tie for 1404 kN, which is 7% less than the model does.

The node is the part that is checked

A strut-and-tie model is safe if every part of it can carry what the model asks. The struts are wide and lightly stressed, the ties are steel and easy, and the whole of the difficulty collects at the points where they meet — in a volume of concrete a few hundred millimetres across that no drawing shows.

internal-forces · Strut-and-tie

Named alongside it

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

AnchorageBearing stressBondBracingBucklingConcrete strengthD-regionEquilibriumEquivalent continuumGridshellInternal forcesLattice

All concepts