Structural Engineering & Statics
where the load goes
Concept

Thrust line — where it appears

The locus of points through which an arch's internal compression passes, which the arch stands as long as the line stays inside the masonry. Its position is not unique — a masonry arch has a family of admissible lines — and the arch stands if any one of them fits inside the material.

Named by 14 essays across 6 fields — each of them below, with the objects they name alongside it.

The cable and the arch are the same curve. The shape a cable takes under a uniform load is a parabola, and it carries that load in pure tension. Reflected, the identical curve carries the same load in pure compression, which is what an arch is.

The shape that carries itself, and the arch that is its reflection

Hang a chain and it takes the one shape that carries its load in pure tension. Turn the shape upside down and it carries the same load in pure compression. That is what an arch is.

structures · Funicular The funicular polygon for five loads. The shape a string takes under 5 point loads, with a vertex at every load and a constant horizontal component of 27.9 throughout. The end segments carry the most — 38.6 against 28.0 in the flattest one — because they are steepest.

The polygon that finds the shape

A hanging string under five loads has no smooth curve in it — it has five vertices and six straight segments, and every slope in it is a running sum divided by one number.

structures · Funicular A three-pinned arch, rise 2.6 on span 9. A three-pinned arch under a uniform load. One moment equation about the crown hinge gives a horizontal thrust of 23.37, with no stiffness and no assumption about the section. The thrust line lands on the axis everywhere, so there is no bending anywhere in the arch.

The hinge put in on purpose

An arch with two pinned feet cannot be solved by statics. Add a third hinge at the crown — deliberately weakening it — and the whole structure falls out of one moment equation.

structures · Arch A line of thrust, and the masonry it has to stay inside. An arch ring of 9% of the span in thickness, rising 28% of the span, under its own weight as a uniform load. Any horizontal thrust between 3.85 and 5.23 puts a line of compression entirely inside the masonry, so the arch stands — and which of them it actually takes is not decided by statics. The two extremes are drawn: the minimum-thrust line, which rides high at the crown and low at the haunches, and the maximum-thrust line, which does the opposite.

The line that must stay inside

A masonry arch does not stand because its shape is right. It stands because some line of compression can be drawn inside the stonework — any one will do, and there are infinitely many to choose from.

structures · Arch The middle third, computed. The kern of a 400 × 600 mm rectangle, computed by asking, for every direction, how far the resultant can move before the far face would be pulled. It reaches ±100.0 mm vertically and ±66.7 mm horizontally, which are h/6 and b/6 exactly, and the region between is a rhombus rather than the ellipse the two numbers suggest — a resultant on a diagonal has less room than either axis allows.

The middle third

A material that cannot be pulled imposes a condition on where the load may land, and the condition is a region rather than a point. For a rectangle it is the famous middle third; for every other section it is a shape nobody quotes, and one ordinary section's is nearly twice as generous as the rule allows.

sections · Kern 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 two theorems close on the answer from opposite sides. A pinned-base portal frame under 200 kN at mid-span and 80 kN at the eaves, with a plastic moment of 200 kNm. The three horizontal lines are mechanisms — work equations, each an upper bound: the beam mechanism gives 1.000, sway 1.250 and the combined one 0.714. The curve is the lower bound: for each value of the one redundant, the largest load factor whose moment field stays inside the plastic moment everywhere. Its peak is 0.714, and it touches the lowest mechanism exactly — so the collapse load is known rather than bracketed. Using the beam mechanism instead would have claimed 1.40 times the real capacity, and the sway one 1.75 times, both of them on the wrong side.

Two ways of being wrong

Plastic analysis has two theorems and they point in opposite directions. Any equilibrium field that nowhere exceeds the plastic moment gives a load at or below the collapse load; any mechanism gives one at or above it. Only one of those errors is safe.

equilibrium · Bound theorems 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 The middle third, computed. The kern of a 300 × 600 mm rectangle, computed by asking, for every direction, how far the resultant can move before the far face would be pulled. It reaches ±100.0 mm vertically and ±50.0 mm horizontally, which are h/6 and b/6 exactly, and the region between is a rhombus rather than the ellipse the two numbers suggest — a resultant on a diagonal has less room than either axis allows.

The strength thrown away on purpose

Masonry, concrete and soil are all analysed as though they had no tensile strength whatever. Each of them has some. The decision to set it to zero is the single most consequential modelling assumption in the subject, it is safe for one kind of check and unsafe for another, and almost nothing that uses it says which.

materials · No tension The tendon is a load, pointing the other way. A 14 m beam with a parabolic tendon dropping 260 mm to midspan, stressed to 1440 kN after losses. Its curvature pushes the beam up along its whole length with an intensity of 8Pe/L² = 15.28 kN/m, against an applied 17.63 kN/m — so 2.34 kN/m is left to bend anything, and the beam carries 57.4 kNm where an unstressed one carries 432 kNm. What the section then feels is 6.40 MPa of uniform compression and very little else.

The load that comes from changing direction

A force that travels in a straight line asks nothing of anything. Bend its path and it asks for a transverse load of F over R along every millimetre of the curve, and that load is real, is nowhere on the load schedule, and is the same statement behind a prestressing tendon, a hoop force, an arch thrust and a web that buckles with nothing applied to it.

internal-forces · Deviation force A fixed arch, and the section it is a drawing of. A parabolic arch of 30.0 m span and 6.0 m rise, fixed at both springings, carrying 20.0 kN/m over the span. On the right the analogous column: the arch's own centreline drawn as a section of width ds/EI, so it is narrow where the arch is stiff. Its area is 32.99 and its elastic centre sits 3.86 m above the springings — 2.14 m below the crown and on no part of the arch at all, which is the point of it. The released moments loaded onto that section give a direct stress of 3054.0 kN·m and a bending stress whose gradient is the horizontal thrust, 375.0 kN.

The centre that hangs in the air

Cross's analogy turns a closed frame into a short column and reads its redundant moments off as that column's stresses. Nothing in it requires the frame to be straight. Run round a fixed arch, the analogous column's centroid is a point two thirds of the way up the rise, on no part of the arch at all, and the bending stress about it is the arch's horizontal thrust. For a parabola under a uniform load the answer it returns is a diagram of nothing, which is exactly right.

deflection · Moment-area Two lines and a triangle: a three-hinged arch drawn. A three-hinged arch of 20.0 m span, springings at (0.0, 0.0) and (20.0, 0.0) m and the crown hinge at (10.0, 5.0), under 100.0 kN at 5.0 m. The right half carries no load, so it is a two-force member and its reaction lies along the line from its springing through the crown hinge. That line meets the load's line at K, 7.50 m up, and the left reaction must pass through K too. The triangle of the load and the two reaction directions gives the reactions as 90.1 kN at A and 55.9 kN at B, with a horizontal thrust of 50.0 kN — the values four equilibrium equations return, to 7e-15 kN. The shape of the rib entered nowhere.

The arch that is only its three hinges

A three-hinged arch's reactions come from three points and nothing else, so a parabola, a circle and a portal frame on the same hinges push on their abutments identically. Move a load across and the point where the reactions cross runs along two straight lines through the crown. That is the arch's influence line, drawn with a straightedge — and friction in the hinges it was built around blurs it.

equilibrium · Graphic statics Putting a funicular through three points. Four loads — 40.0 kN at 3.0 m, 60.0 kN at 7.0 m, 30.0 kN at 12.0 m, 50.0 kN at 16.0 m — and three points the polygon must pass through: A and B at the springings and C, 5.0 m above their chord at 10.0 m. A trial pole, dashed, draws a polygon from A that ends 6.39 m below B. The ray through the trial pole parallel to its own closing line cuts the load line at Q, 95.0 kN from the top, which is the left reaction of a simple beam on A and B and does not depend on the pole at all. Every pole whose polygon passes through A and B lies on the line through Q parallel to A B; the one whose polygon also reaches C is 98.0 kN from the load line, which is the moment at C of that simple beam, 490.0 kN·m, divided by C's height above the chord. That is the three-hinged arch's thrust — the four equilibrium equations give 98.0 kN.

The polygon that runs out of freedom

A funicular polygon for given loads has exactly three freedoms, so it can be made to pass through three chosen points and no more. Three points is a three-hinged arch, and the drawing solves it. Take the crown hinge away and one freedom is left over. The drawing then offers a whole family of thrust lines and cannot say which one the arch uses — the rib's stiffness decides, and stiffness is not on the paper.

equilibrium · Graphic statics Two pencil lines, and where they cross. Two lines drawn with a pencil 0.2 mm wide are two bands, drawn here much wider than a pencil so the shape can be seen, and they cross not at a point but in a parallelogram. At 60° apart the parallelogram's long diagonal is 2.0 pencil widths — 0.40 mm; at 12° apart the parallelogram's long diagonal is 9.6 pencil widths — 1.91 mm. The crossing's uncertainty along the bisector is the width divided by twice the sine of half the angle, so it grows without limit as the lines turn parallel, and a construction that finds a point by crossing two lines inherits it.

How wrong a drawing is

A pencil line is a band, and two bands cross in a parallelogram that grows as they turn parallel. The accuracy of a graphical construction is therefore a property of the angles it makes, not of the hand that made it — and the worst case is the shallow arch, the structure the method was most used on. Measured properly, the drawing's error there is the size of the builder's, and the check draughtsmen relied on cannot see it.

equilibrium · Graphic statics

Named alongside it

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

FunicularArchEccentricityEquilibriumHorizontal thrustFree bodyGraphic staticsKernBearing pressureDeterminacyForm-findingIndeterminacy

All concepts