Structural Engineering & Statics
where the load goes
Concept

Bearing stress — where it appears

The compressive stress where one component presses on another over a contact area, rather than through a section. It is checked on an assumed area whenever the load spreads — under a base plate, at a bolt hole, at the toe of a masonry wall — and the area is usually the assumed part.

Named by 8 essays across 5 fields — each of them below, with the objects they name alongside it.

The force spreads, and the spreading needs a tie. The end block behind an anchorage of 1200 kN on a 200 mm plate, in a section 700 mm deep. Half the force enters at the quarter point of the plate and leaves at the quarter point of the section, so a strut between the two rises 125 mm and needs a transverse tie to turn it. Placing the tie 0.5 depths from the face makes that tie force 214 kN — and at exactly half a depth this reproduces Guyon's 0.25P(1 − a/h) to the digit, which makes that famous coefficient a lever arm somebody chose rather than a property of concrete. The bearing stress under the plate is 20.0 N/mm² against 5.7 once the force has spread.

The force that splits what it pushes on

A prestressing tendon delivers its whole force through a plate a fraction of the section deep. One depth further along the stress is uniform, and the spreading in between requires a transverse tension nobody applied — the force that splits end blocks, and the only number in the design that no equilibrium equation on the member can see.

internal-forces · Anchorage zone The line, and the stone it has to stay inside. A masonry pier 9 m high, 1.6 m thick at the top and battered 12% on its outer face, taking a thrust of 40 kN per metre of run at 25° to the horizontal. The line drawn through it is the locus of the resultant on each horizontal cut: everything above the cut is the free body, and the resultant's position is the moment divided by the vertical force. The dashed pair is the middle third, inside which no tension is implied anywhere on the joint. The line stays inside the stone throughout and reaches the base at 0.503 m from the centre, against a half-width of 1.34 m — but outside the middle third, so part of the base joint is open and the toe is carrying a triangle. Nothing about the strength of the masonry appears anywhere in this figure, and that is the point.

The weight that makes it safer

Every load in this collection makes a structure worse. A pinnacle does not. A masonry pier fails when the line of compression leaves the stonework, and adding weight at the top rotates that line back towards the vertical without adding anything the pier cannot carry — so the stone is not being strengthened, it is being aimed.

structures · Buttress A straight line, and the comfortable case is already two thirds down it. The capacity of a 215 mm masonry wall as a fraction of its squash load, against the eccentricity of the resultant. Nothing in this figure is a buckling calculation. A material that cannot be pulled bears on a strip of width 3(t/2 − e) under a triangular stress block, so the capacity is exactly 1.5f(t − 2e) — a straight line, zero when the resultant reaches the face, and already at 67% at the edge of the kern. The middle third is treated everywhere as the comfortable case; a wall loaded there has given away a third of its capacity before slenderness has been mentioned. The lower line is the same wall with slenderness in it, which enters as an ADDITIONAL eccentricity of 17.4 mm rather than as a reduced stress — h_ef²/2400t, for h_ef = 3000 mm. Euler's load for this wall is 9.6 times what the eccentricity rule allows, which is why no masonry calculation contains it.

It does not buckle, it runs out of width

Every stability failure in this collection is a member that could have carried tension deciding to go sideways instead. Masonry cannot carry tension, and its failure under an eccentric load is not a bifurcation at all — the bearing area simply shrinks until it runs out. The capacity is exactly linear in the eccentricity, Euler's load is ten times anything allowed, and no material property appears until the very end.

stability · Wall slenderness 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 A short timber beam is a shear problem, and a steel one never is. Utilisation of the bending and shear checks on one beam, against span-to-depth. The two cross where the ratio equals f_m ÷ f_v exactly — no load, no width and no span survives the cancellation — which for this timber is 6.7 and for steel is 1.73. So a timber beam shallower than about six times its depth is governed by shear parallel to the grain, and a steel beam would have to be shorter than twice its own depth before the same thing happened, which is not a beam. The third check is bearing across the grain, which does not move with the span at all: on the beam drawn it is at 0.40, and it is the one that governs.

The shear that decides a timber beam

A steel beam is never governed by shear, because its bending strength is only 1.73 times its shear strength and no beam is that short. Timber's ratio is 6.7 along the grain and 23 across it, so shear governs at proportions people build every day.

sections · Shear flow The part of a base plate that is delivering anything. A plan of the plate: the 260 × 260 column in the middle, and the shaded region it can reach — its own outline grown by 43 mm, which is t√(f_y/3f_jd) for a 20 mm plate on grout at 20 N/mm². That is 78176 mm² of a 200000 mm² plate, or 39 per cent of it. The corners are outside it and are carrying nothing: a plate cantilevers from the column's perimeter and runs out of bending capacity at c, so making it larger in plan beyond that changes nothing at all, and making it thicker changes everything.

The plate that is only as big as it is thick

A base plate's design is presented as a bearing calculation: an area, a bearing strength, and a check that the pressure fits. The area in that calculation is the plate, and the plate cannot deliver it — a 20 mm plate on a 500 square reaches 43 mm past the column and the corners carry nothing at all.

connections · Base plate The compression is under the flange, not at the edge of the plate. A 500 × 400 mm base plate, 20 mm thick, under a 260 mm column carrying 300 kN and 120 kN·m, with holding-down bolts 60 mm from the tension edge. A plate this thick can deliver bearing only 43 mm past the compressed flange, so the compression sits under that flange, 124 mm from the column's centre, and the lever arm to the bolts is 314 mm: the bolts carry 264 kN and the flange zone 564 kN. The dashed block is the rigid-plate answer, 54 mm deep at the plate's far edge with its resultant 223 mm out, which needs only 128 kN in the bolts. Both satisfy equilibrium; only one is a plate that can carry the pressure where it is drawn.

The compression that stays under the flange

Put a moment on a base plate and a pressure-block calculation pushes the compression out to the plate's edge, where the lever arm to the holding-down bolts is longest. A plate that is not rigid cannot put it there. The compression stays within a few tens of millimetres of the compressed flange, the lever arm shrinks, and the bolts carry twice what the pressure block said.

connections · Base plate

Named alongside it

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

EquilibriumFree bodyLower-bound theoremEccentricityStrut-and-tieAnchorage zoneBase plateBurstingColumnConcrete strengthD-regionEffective area

All concepts