Structural Engineering & Statics
where the load goes
Series

Graphic statics — the series

4 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · equilibrium
  2. 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.

    part 2 · equilibrium
  3. 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.

    part 3 · equilibrium
  4. 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.

    part 4 · equilibrium

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