Structural Engineering & Statics
where the load goes
Concept

Resultant — where it appears

The single force a system of forces adds to, which exists as a force alone only when the system is coplanar, concurrent or parallel. A general spatial force system reduces to a force and a couple rather than to a force, which is why a wrench is the general case and a single resultant is a special one.

Named by 12 essays across 3 fields — each of them below, with the objects they name alongside it.

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.

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.

equilibrium · Graphic statics A triangular load and the force that replaces it. A triangular distributed load with its resultant computed by integration: an area of 24.0 acting at 5.33 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.

The load that is spread out, and the force that replaces it

A distributed load can be swapped for a single force at its centroid. The reactions come out identical and the bending moment does not, and knowing which side of the cut the swap is legitimate on is most of the skill.

equilibrium · Distributed load 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.

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.

internal-forces · Influence line Five loads behind one wall, and the water is the biggest. The horizontal pressure on a 6 m wall retaining soil at 18 kN/m³ with a friction angle of 30°, a surcharge of 10 kPa and the water table 2 m down, drawn once as the profile the wall feels and then once per term. The terms are surcharge 20.0 kN/m at 3.00 m, soil above water 12.0 kN/m at 4.67 m, soil at the water table 48.0 kN/m at 2.00 m, submerged soil 27.2 kN/m at 1.33 m, water 78.5 kN/m at 1.33 m, and they sum to 185.7 kN/m — matched to 7e-8 by integrating the drawn profile numerically. The largest single term is the water, at 78.5 kN/m: water has no shear strength, so its coefficient is exactly one where the soil's is 0.333, and it acts on top of the soil's effective stress rather than instead of it. The combined resultant sits at 1.90 m above the base, 0.317 of the height rather than the third point at 2.00 m that a pure triangle would give.

The load that depends on what carries it

Every other load in this collection is a number the structure is given. Retained soil is not — it pushes with a fraction of its own weight, and the fraction is decided by how far the wall moves. Six millimetres of retreat on a six-metre wall takes a third off the load, and being held still puts it back.

equilibrium · Lateral pressure A general force system is a screw, not a force. Two forces that are neither parallel, nor concurrent, nor coplanar. Their resultant is (30, 0, -60) and no point in space reduces them to it alone: the moment left on the central axis is (14.40, 0.00, -28.80), which is parallel to the resultant and has a pitch of 0.4800. The simplest object equivalent to the system is therefore a force and a couple about the same line — a wrench — and a planar drawing can never produce one, because a coplanar system has its moment perpendicular to its resultant by construction.

Moving a force, and what it costs

Every free body on this site begins by putting a force somewhere convenient. That move is free along the force's own line, costs a couple across it, and in three dimensions leaves behind something no choice of point can remove.

equilibrium · Force couple Wrong in shape, right in two integrals. The compression zone of a C30 section with its neutral axis 150 mm down, drawn twice. The curved outline is the real parabolic-rectangular stress distribution — the material's own law read off the linear strain profile plane sections supplies. The rectangle over it is what every design office uses instead: intensity η f_cd = 16.5 MPa over a depth λx = 125 mm. The two shapes are visibly different and give the same answer, because a bending calculation asks a stress distribution only two questions — how much compression there is, and where its resultant acts. Both are 619 kN at 62.4 mm from the face. The factors are α = 0.8095 and β = 0.4160, and λ = 2β follows from wanting the same centroid. A triangle and a full rectangle match neither integral and are nowhere near.

Deliberately the wrong shape

Concrete in compression follows a curve, and no design office has ever integrated it. Every code in the world replaces it with a rectangle of reduced depth and reduced intensity, and the answer is right to a fraction of a per cent — not because the shapes are similar, which they visibly are not, but because a bending calculation only ever asks a stress distribution two questions.

sections · Stress block The wind pushes on one face and pulls on three. A 30 × 20 m building in plan, with the measured pressure coefficient on each face and the arrows drawn in the direction the pressure acts. Only the windward face is pushed; the other three are sucked, and the side faces are sucked hardest of all at c_p = -0.7. The horizontal resultant is 842 kN at a velocity pressure of 0.9 kPa, and the arithmetic of it is the whole point: the leeward suction pulls the building downwind, so it ADDS, supplying 38% of the answer, while the two side faces cancel each other exactly and supply none of it. The coefficients are wind tunnel data; what is computed is the free body they are applied to.

Most of it is suction

A wind load is drawn as arrows pressing on the windward face, which is where about three fifths of it comes from. The rest is a pull on the back. The two side faces carry the largest suctions on the building and contribute nothing at all to the answer — and the inside of the building, which nobody draws, decides whether the roof stays on.

equilibrium · Wind pressure Every pressure points at the pin, so the water lifts nothing. A radial gate of radius 8 m holding 6 m of water, with its pivot 6 m above the sill. The pressure on a curved surface cannot be obtained by multiplying anything by anything, so it is integrated round the arc: the horizontal component comes to 176.6 kN/m and the vertical to 110.5. Both are recoverable without any integral at all — the horizontal is the pressure force on the surface's own vertical projection, γH²/2 = 176.6, and the vertical is the weight of the water standing above it, 110.5. They agree to 0.000 per cent. And because every pressure is normal to a circle, every one of them passes through the centre: the moment of the whole 208 kN/m about the pivot is -3.4e-15 kNm, against 353 for a flat gate on the same hinge.

Every pressure points at the pin

Pressure acts normal to a surface, so on a curved one every element pushes in a different direction and no multiplication gives the resultant. Two free bodies recover it without an integral — and on a circular surface a third observation makes the whole force disappear from the equation a hoist has to satisfy.

equilibrium · Curved surface pressure Two load sets with the same resultant and different work. The two ways of putting a uniform load of 10 kN/m onto a beam element 6.00 m long. Both put 30.0 kN at each node, so both have the same resultant and the same moment about any point — they are equivalent for a rigid body. The consistent set adds a couple of 30.0 kN·m at each end, in opposite senses, which is what makes it do the same virtual work over the element's shape functions as the real load does. The couples cancel in the resultant, which is exactly why the resultant cannot see them, and they are the whole difference between an exact answer and one that is a third out.

Equivalent in work, not in resultant

Two force systems with the same resultant and the same moment about every point are interchangeable — for a rigid body. A finite element is not a rigid body, and substituting one for the other on a beam element leaves the tip of a cantilever a third too low with no warning of any kind.

equilibrium · Force couple Both checks pass, and the contact lets go. A contact pressed together by 1000, with μ = 0.4. Every tangential force the contact can supply lies inside a disc of radius μN = 400.0, because the friction law bounds the length of the force and not its components. The contact is asked for 300.0 one way and 300.0 the other way — 75% and 75% of the radius taken one at a time — and 424.3 together, 106% of it. Each one-direction check passes and the force does not fit: the square those checks describe reaches √2 times further at its corners than the contact can.

Seventy-five per cent each way

A contact asked for friction in two directions at once can supply a force of a certain length pointing any way it likes, so its limit is a disc and not a square. Two checks made one direction at a time, each passing at seventy-five per cent, describe a contact that has already let go.

equilibrium · Friction Four forces pair off, and the line joining the pairs carries both resultants. A beam 8 m long held by the vertical link at the left end, the strut at 6 m, the horizontal link at the right end, under a load of 10 kN at 3 m inclined at −60.0°. Pairing the load with line A: the two cross at P, (0.00, 5.20) m, and lines B and C cross at Q, (6.00, 0.00). The resultant of the first pair passes through P and that of the second through Q, and since they balance each other both lie on PQ, dashed. The force polygon on the right is the load, then A, B and C, closing where it began, with PQ's direction as the diagonal that splits it into two triangles. The forces are A 4.33 kN along its line, B 6.12 kN along its line, C 0.67 kN against its line, as the three equations of equilibrium give them.

The line that pairs four forces

Three forces in equilibrium meet at a point; four need not. But they pair off. The resultant of two passes through the point where their lines cross, the resultant of the other two through theirs, and the two resultants must share the line joining those points. Culmann's line turns a four-force body into two triangles — and the method of sections into a drawing.

equilibrium · Graphic statics The pole, the strings, and the resultant of any number of forces. Five downward loads on a span of 10 m — 30 kN at 1.5 m, 20 kN at 3.5 m, 45 kN at 5.0 m, 25 kN at 7.0 m, 35 kN at 8.5 m — adding to 155.0 kN. On the right, the loads laid end to end down one line, with a pole 60.0 kN to the left of it and a ray drawn to every division between them. On the left, the funicular polygon: each segment parallel to the ray of the loads it has passed, so the shape is the one a string carrying these loads would hang in. The first and last strings are extended until they cross, at 5.24 m, and that crossing is where the 155.0 kN resultant acts — the same station the moment sum Σ P x / Σ P gives, 5.24 m, reached with no pole in it at all.

The pole decides the drawing, not the answer

Five forces will not pair off the way four do. They need a point that is nowhere on the structure — chosen freely, by whoever is holding the pencil — and the string of lines it generates. Every choice draws a different polygon and finds the same resultant, and the shape it draws turns out to be the beam's bending moment diagram.

equilibrium · Graphic statics

Named alongside it

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

Free body diagramEquilibriumLine of actionCentroidForce polygonFree bodyFunicularGraphic staticsLever armLoad pathBending momentConcurrency

All concepts