Structural Engineering & Statics
where the load goes
Concept

Free body diagram — where it appears

A chosen piece of a structure drawn with every force the rest of the world applies to it, which is where a number is allowed to come from. It is the only device in the subject that turns an assertion into a computation, and every number this collection quotes is traceable to one.

Named by 13 essays across 3 fields — each of them below, with the objects they name alongside it.

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.

Everything adds to nothing, and that is the whole of statics

A structure that stays put obeys two statements — the forces on it sum to zero, and so do the moments. Every number in the subject comes out of those two sentences.

equilibrium · Equilibrium The same beam, cut at x = 5. A beam separated at one station. On the exposed face a shear force and a bending moment appear, equal and opposite on the two pieces, with values obtained by summing the forces on whichever piece is easier.

The free body is a choice, and choosing it well is the whole skill

Cutting a structure open is not a step in the method. It is the method — and where the cut is made decides whether the answer takes one line or twenty.

equilibrium · Free body 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 truss cut through panel 3. The truss severed through one panel, with everything to the right removed and the three cut member forces drawn on the exposed faces. Taking moments about the marked joint removes two of the three unknowns, so one equation gives the third: -52.94.

Answering one question without solving the rest

A truss of fifty members can be interrogated about one of them. Cut through three, take moments about the point where two of them meet, and the third falls out in a single line.

equilibrium · Method of sections 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 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.

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.

equilibrium · Equilibrium The same beam, cut at x = 5. A beam separated at one station. On the exposed face a shear force and a bending moment appear, equal and opposite on the two pieces, with values obtained by summing the forces on whichever piece is easier.

What a cut reveals, and why it was there all along

Cut a beam anywhere and two quantities appear on the face — a shear force and a bending moment. Nothing was applied there. They are what the material was already doing.

internal-forces · Internal forces 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.

The connection is not a point, and every diagram so far says it is

Every free body drawn here has joined its members at points. Real structures fail at the joints far more often than in the members, and the reason is that a joint is exactly the region the theory behind every other page explicitly excludes.

connections · Connection design The moment does not stop at the end of the beam. A portal frame of 8 m by 4 m with fixed bases, carrying 20 kN/m on the beam. The bending moment is drawn on the tension side of every member, and it runs round the corner without a break: 65.2 kNm arrives at the end of the beam and 65.2 kNm leaves down the column, which is the same number, since joint rotational equilibrium is one of the equations the frame solve satisfied. Midspan carries 94.8 kNm, and the two add to 160.0 — the 160.0 kNm of a simply supported span, to 0.0e+0 kNm. The corner takes 61% of the wL²/12 a fully built-in beam would have carried, because the columns are springs rather than walls: the beam-to-column stiffness ratio is 1.27. The beam's moment crosses zero 0.92 m from the corner and the column's 1.33 m above its base.

The moment that goes round the corner

At a rigid knee the bending moment does not stop at the end of the beam. It turns and runs down the column, and in the same instant the beam's shear becomes the column's axial force — while the block of steel that has to carry the turn appears on no member diagram anywhere.

internal-forces · Corner moment 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 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 A pin's force does not pass through its centre. Left, a pin of radius 0.15 m in its hole, with a coefficient of friction of 0.15. The reaction at the contact is inclined by φ = 8.5° to the radius through it, because the friction it can develop is that fraction of the force pressing the surfaces together, and the perpendicular distance from the pin's centre to that inclined line is R sin φ = 0.022 m. Every position the contact can take gives a line tangent to the same circle, shaded. Right, a link 0.48 m long pinned at both ends, at the same scale — 3.2 pin radii, which is a stubby linkage rather than a structural tie, drawn that way because at the thirty radii an ordinary tie has, the two circles are smaller than the pencil: its force is a common tangent to the two circles — the two solid lines for the two senses of rotation, the two dashed ones for the senses in which its ends turn oppositely — and the dashed centre line every construction in this collection draws is none of them. A pin of this size carrying 5.0 MN delivers a couple of 111.3 kN·m to whatever it is pinned to, which is the same offset read as a moment rather than as a distance.

The pin that is not a point

Every line of action drawn so far passes exactly through a pin's centre, which is true of a frictionless pin and of nothing else. A real one carries its force tangent to a small circle instead, so a link's line is a band, a construction's answer is a range, and a support drawn as a hinge hands a couple of a hundred kilonewton-metres to whatever it is pinned to.

equilibrium · Graphic statics

Named alongside it

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

Line of actionResultantGraphic staticsMoment equilibriumConcurrencyForce polygonFunicularLoad pathMethod of sectionsMoment centreRigid bodySaint-Venant's principle

All concepts