Structural Engineering & Statics
where the load goes
Concept

Tie — where it appears

A member carrying tension, provided to close a force that would otherwise have to be delivered to a support. An arch's tie carries the whole thrust, its stiffness decides how much thrust the arch develops, and whatever thrust is lost to its stretching arrives as bending in the rib.

Named by 5 essays across 3 fields — each of them below, with the objects they name alongside it.

The force runs along the chord whatever route the member takes. A member pinned at two points 10 m apart, loaded only at those two points, and bowed 1.2 m off the line between them. Equilibrium leaves the two end forces no choice: equal, opposite, and along the chord. So at every section the axial force is P cos α, the shear is P sin α, and the bending moment is the force times the perpendicular offset — 240 kNm at the crown for 200 kN at 1.2 m, which is a multiplication rather than an analysis. Straighten the member and the moment diagram is identically zero; that is the case a truss member is in, and the reason it carries one number.

The member with only one direction

If a body is in equilibrium under forces applied at exactly two points, those forces are equal, opposite and along the line joining the points. It is three conclusions from two equations, it is the shortest real theorem in statics, and nearly everything that follows depends on it without saying so.

equilibrium · Two force member The tie is a redundancy, so its stiffness decides the thrust. Thrust and rib bending for a 60 m tied arch of 0.15 rise ratio, against the stiffness of its tie. Cut the tie and the structure is a curved simply supported beam, so the tie force is the one redundant and the force method gives it: with a rigid tie the answer is 2227 kN, within 1.0 per cent of the funicular wL²/8f, and the shortfall is the arch's own axial shortening. A real tie stretches 68 mm and returns 2166 kN — 2.7 per cent of the flexibility is the tie — and whatever thrust the arch does not get, it carries as bending: 760 kNm at 30 m. A tenth of the tie stiffness is not a tenth of the problem; it is a different structure.

The thrust that never reaches the ground

Every arch on this site has ended at the same sentence — the foundation is where an arch is really decided. A tie changes the sentence without changing the arithmetic: the horizontal force is still there, still the same size, and it now closes on itself through a bar at deck level.

structures · Tied arch 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 An enhanced strength that is the strength of a tie. Bearing strength as a multiple of the design cylinder strength, against how far the load is allowed to spread, with the bursting tension the spread creates on the same axis. The enhancement is √(A₂/A₁) and it reaches 2.80 for the 250 mm pad on a 700 mm block drawn — 47.6 N/mm² against a design strength of 17.0. There is no material property in that statement beyond the one being enhanced, and the reason is on the second curve: a load that spreads does so along inclined struts, a pair of inclined struts has a horizontal component, and that component is 16.1% of the load. It has to be tied. 1099 mm² of steel is what the enhancement actually is, and the cap of three is not a property of concrete — it is the angle past which nobody believes the strut.

Three times as strong under a smaller pad

Press a small plate onto a large block of concrete and it will carry three times the stress a cylinder of the same concrete fails at. The enhancement is a ratio of areas with no material property in it, which should be a warning: what has actually been measured is not the concrete's strength but the strength of a tie holding it together.

internal-forces · Bearing stress 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.

EquilibriumFree bodyLower-bound theoremStrut-and-tieBearing stressFunicularLoad pathStrutAnchorageAnchorage zoneArchAxial shortening

All concepts