Structural Engineering & Statics
where the load goes
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Drawing as calculation

Force polygons, funicular shapes and Cremona diagrams solved real structures for a century. The drawing was not an illustration of the answer — it was the answer.
Three forces must meet at a point. A body held by two supports and one load. The reaction at the roller is vertical and the load's direction is given, so their lines of action fix a meeting point — and the pin reaction has to point at it. Equilibrium

Three forces must meet at a point, and a drawing can find it

A body held by exactly three forces has their lines of action concurrent. That is a theorem, it is enough to solve for direction and magnitude, and for a century it was done with a straightedge.

 A beam, its loads and its reactions. A free body: the beam cut away from everything it touches, with the forces that were doing the touching drawn on it. The reactions are computed from the loads, so their arrows are to scale relative to each other. The moment of everything on the body is summed about three marked points, and every sum comes to zero. Nothing here is a new equation: with only vertical forces on the body, the moment about a point does not depend on that point's height, so every centre anywhere on the plane returns the same equation — and the third equilibrium equation, the horizontal sum, reads nothing equals nothing. Equilibrium

The equation that is not new, and the three that are

A plane free body yields exactly three independent equations. Most attempts at a fourth are one of the first three wearing different clothes — and on a beam under vertical load, one of the three is already saying nothing.

 The count is necessary and not sufficient. Two pin-jointed frames, each satisfying m + r = 2j exactly. One of them folds anyway, because the equations are not independent; the ghosted outline is the motion that costs no member any change of length, drawn at an exaggeration of 0.55 of the span. Equilibrium

The count that does not see it

A frame can have exactly as many unknowns as equations and fold up anyway. The count asks whether there are enough equations; it never asks whether they are different from one another.

 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. Structural form

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.

 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. Structural form

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.

 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. Structural form

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.

 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. Structural form

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.

 Load, shear and moment — a simple span. The applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear. Internal forces

The diagram is an integral, and that is why it can be drawn by eye

Load, shear and moment are one function and its two integrals. Once that is seen, the diagrams stop being things to calculate and become things to sketch.

 Influence line for the bending moment at x = 3. The bending moment at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 3.00, giving 2.100. Internal forces

The worst place to stand

A bridge is not designed for a load. It is designed for a load that moves, and for every station along it there is a different position of that load that does the most damage.

 The worst position is not the obvious one. Three axles totalling 320 units, marched across a span of 20 in steps of 0.02. The envelope is the largest moment each station ever sees; its peak is 1160.2 at 9.88 along the span, which is 0.12 off midspan and occurs under the axle nearest the resultant rather than under the heaviest one. Barré's construction, which places midspan halfway between that axle and the resultant, independently gives 1160.3 at 9.88. The dashed curve is the envelope the same total weight would produce as one load rather than three: its peak is 1600.0, which is 38% more — spreading a load out is worth something. Internal forces

The train that is worse than its heaviest axle

An influence line says where to stand one load. A vehicle is several loads at fixed spacings, and the worst arrangement never puts the heaviest one at the peak.

 A column that was never straight. Load against lateral deflection at mid-height, for a column starting with an initial bow of 0.002. There is no critical value to reach: the deflection grows from the first increment, slowly at first and then without bound as the ratio approaches 1.00 — which is the Euler load, and which the column therefore never attains. The perfect column, drawn for comparison, sits on the vertical axis until it arrives there and then has no answer at all. Stability

The column that was never straight

Euler's load is the load at which a perfectly straight column becomes indifferent to being bent. No column is perfectly straight, so no column ever reaches it — and the load it never reaches can still be measured.

 The deflection at x = 4, by virtual work. Three diagrams: the moment from the real load, the moment from a unit load placed where the answer is wanted, and their product. The area under the third, divided by EI, is the deflection — 213.33 here. No standard case was consulted, so the method works for any load pattern at all. Deflection

One deflection, without solving everything

To find how far one point of a structure moves, put an imaginary force of one unit there, multiply two moment diagrams together, and integrate. The answer arrives without ever solving for the deflected shape.

 Maxwell's reciprocal theorem. A load at one point and the deflection it causes at another, against the same load moved to the second point and the deflection read at the first. Both integrals return 93.3335, and neither calculation was told about the other. The two deflected shapes are entirely different; the two readings are identical. Deflection

The theorem that swaps the question round

Push here and measure there; push there and measure here. The two readings are identical, for every elastic structure, whatever its shape — and that fact turns an influence line into something a model can be asked for directly.

 Three times the stress, and it does not matter how big the hole is. The hoop stress around a circular hole in a wide plate pulled at 100 N/mm², from Kirsch's exact solution. At the sides of the hole it is 3.0 times the applied stress — 300 N/mm² — and the factor is the same for a hole of any radius, because the radius cancels. At the top and bottom of the hole it is -1.0 times the applied stress, which is compression in a plate that nothing is pushing. The disturbance dies quickly: the stress is within 5% of the applied value by 3.5 hole radii, which is Saint-Venant's principle with a number on it. Materials

The hole that multiplies the stress by three

The stress at the side of a hole is three times the applied stress whatever the hole's size, and at the top and bottom of the same hole it is minus one times it — compression in a plate that nothing is pushing.

 A bolt group under an eccentric load. A 3 by 2 bolt group carrying 100 kN at 150 mm from its centroid, with the resultant force on each bolt drawn to scale, by the elastic vector method. The load is shared equally and the torque is not, so the worst bolt carries 50.37 kN against 16.67 kN of direct shear alone — 3.02 times as much. Connections

The bolt that carries more than its share

Six bolts, one hundred kilonewtons, and a worst bolt carrying fifty. The load is shared equally and the torque is not, and the second one is invisible on any drawing where the connection is a point.

 The net section, and the path the tear takes. A 200 mm plate with two holes staggered by 50 mm at a gauge of 60 mm. The straight path through one hole leaves 178 mm; the diagonal path through both leaves 166.42 mm after the s²/4g correction adds 10.42 mm back. The shorter of the two decides, at 83.21% of the gross section. Connections

The tear that goes diagonally, and the correction that has no derivation

Stagger the holes so that no straight line crosses more than one and the plate does not get its strength back. The tear runs at an angle instead, and the arithmetic that makes it come out right is a century-old piece of curve-fitting nobody has improved on.

 Throat stress round a fillet weld group. A c shape weld group carrying 100 kN at 150 mm from its centroid. The peak throat stress is 0.88 kN per mm of throat, at (79.5, -100); the worst point at maximum radius from the centroid carries 0.88. Checking by radius is right here, and points at identical radius differ by a factor of 1. Connections

The corner that is not the worst point

Check the point furthest from the centroid. It is the standard rule for a weld group under an eccentric load, it is exactly right for some shapes, and for others it misses the peak by sixteen per cent — or picks one of four points it cannot tell apart whose stresses differ by two thirds.

 A base plate, and when the bolts start working. A 500 × 400 mm plate carrying 600 kN and 90 kN·m, so the resultant sits 150 mm from the centre against a kern of 83.33 mm. The plate is in partial contact: bearing over 300 mm at a peak of 10 N/mm², with the holding-down bolts carrying 0 kN. The plate lifts at 50 kN·m and crushes at 120 kN·m, and the bolts are not needed until 150 kN·m. Connections

Where the structure meets the ground, and when the bolts start working

Push a base plate off its middle third and it lifts off the foundation. The holding-down bolts then carry exactly nothing, and go on carrying nothing until the plate has crushed the concrete underneath it.

 Where a beam's load comes from. A 8 × 6 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 15.0 m² each and the short beams a triangle of 9.0 m²; the four areas sum to 48.0 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 15.00 kN/m at midspan. Equilibrium

The load a beam is given is a decision

Every beam calculation so far has started with a load per metre, handed over as though it were a property of the beam. It is not. It is the answer to a prior question nobody draws, and two defensible answers to it differ by sixty per cent on the same floor.

 Three legs, three equations, one answer. A rigid top on three legs carrying 120 kN at (0.3, 0.2) m. The three equilibrium equations available — one vertical and two moments — leave three unknowns, so the system is exactly determinate and the reactions are 56.8, 13.6, 49.6 kN. Move the load anywhere and the answer moves with it; nothing about the legs' stiffness enters. Equilibrium

Six equations, and the drawing shows three

Every essay so far has taken place on a piece of paper, where equilibrium is three equations and a structure is a diagram. The real object has six, the extra three are the ones nobody writes down, and the difference between three legs and four is not a matter of degree.

 Loaded straight down, and moving sideways. An equal angle with a moment applied about the horizontal axis. Its principal axes lie at 45.0° to the drawn ones, so the neutral axis runs at -30.6° rather than horizontally, and the section moves 59% as far sideways as it moves down. The product of inertia that causes it is -1.066 × 10⁶ mm⁴, and it is zero for every section drawn in this field until now. Sections and stress

Loaded straight down, and it moves sideways

Every section this collection has drawn had an axis of symmetry, and that symmetry has been doing silent work. Take it away and a vertical load produces a neutral axis that is not horizontal, a deflection that is not vertical, and on one ordinary section a sideways movement larger than the downward one.

 The answer arrives in instalments. The hogging moment at support 1 of a three-span beam, cycle by cycle. It starts at the fixed-end moment of 53.3 kNm — the value with every joint clamped — and settles at 64.0 kNm against an exact 64.0. The error falls by about a factor of four per cycle: 14.00, 3.50, 1.95, 0.59 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything. Deflection

Solved by passing it around

An indeterminate structure needs simultaneous equations, and for thirty years engineers solved them without writing any down. Clamp every joint, release one, share out what is left over, pass half of it along, and repeat — and the answer walks in, three figures correct after four cycles.

 The props decide where the stress ends up. Bottom-fibre stress in the steel of a 12 m composite beam carrying 12 kN/m of wet concrete and 18 kN/m afterwards. Unpropped, the bare steel takes the first stage alone and reaches 292 MPa; propped, the finished composite section takes everything and reaches 186 MPa — a ratio of 1.57. 62% of the unpropped beam's final stress was locked in before the slab was structural at all. The deflections differ by 1.73 times for the same reason, and no drawing of the finished beam distinguishes the two. Structural form

The structure that was never complete

Every analysis in this collection is of a finished structure loaded once. Real ones are built in pieces, and each piece carries whatever was present at the moment it became structural — so the stress in a member depends on when it arrived, which appears nowhere on any drawing.

 The reaction lies inside the cone, so the block stands. A block of 100 on a plane at 15°, against a coefficient of friction of 0.35. Resolving across and along the plane gives a normal force of 96.6 and a friction demand of 25.9, against a capacity of μN = 33.8 — a ratio of 0.77. Added together the two make one contact reaction leaning 15.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 19.3°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 15.0°, which is why the angle of repose is a material property and the size of a heap of sand is not. Equilibrium

The force that is whatever it needs to be

Every other force in statics has a value the equations produce. Friction has an inequality instead, so it takes whatever value equilibrium demands and the bound only ever says no — which means a problem with friction in it has a range of answers rather than one.

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