The line that pairs four forces

Three forces in equilibrium must meet at a point, and the point is what makes a drawing enough to solve them: intersect two lines of action, pass the third through the crossing, and close a triangle. A fourth force breaks that. Four forces in equilibrium need not share any point, and a body held by four is the common case — a beam on three supports under one load, a truss cut through three members with the loads beside the cut.

Karl Culmann’s Die graphische Statik of 1866 solved it with one more line.

A beam on three links #

The beam in the figure is held by three links. Each is a member with only one direction: pinned at both ends and loaded only at its pins, so the force it applies to the beam lies along the link, and only its size is unknown. The vertical link at the left end can push or pull vertically. The strut at 6 m acts along a line at 135°. The horizontal link at the right end acts along the beam’s own axis.

That is four forces on one body: a 10 kN load whose line and size are both known, and three forces whose lines are known and whose sizes are not. Three unknowns, three equations of equilibrium for a body in a plane, and so a determinate problem. But no three of the four lines meet at a point, so the three-force construction has nothing to start from.

Why the resultants share a line #

Split the four forces into two pairs: the load with the vertical link, and the strut with the horizontal link. Any two forces whose lines cross have a resultant that passes through the crossing, because each force has no moment about a point on its own line, so neither has any moment about the crossing, and neither does their sum.

The load’s line and the vertical link’s line cross at P, at (0, 5.20) m, above the left end of the beam. The strut’s line and the horizontal link’s line cross at Q, at (6, 0), which is the strut’s own foot, because the horizontal link acts along the beam’s axis. So the resultant of the first pair passes through P, and the resultant of the second pair passes through Q.

The body is in equilibrium, so those two resultants balance each other: they are equal, opposite and — the part that matters — along the same line. A single line passes through both P and Q, and it is PQ. The direction of both resultants is now known before either is computed, and that is Culmann’s line.

It is the three-force theorem used twice. The first pair and the resultant of the second, reversed, are three forces meeting at P; the second pair and the resultant of the first, reversed, are three forces meeting at Q.

Two triangles #

The construction closes the force polygon in two steps.

Draw the load to scale. From its tip, draw a line parallel to the vertical link; from its tail, a line parallel to PQ. They cross, and the triangle they make gives the vertical link’s force and the size of the resultant along PQ. Then, on that resultant, draw lines parallel to the strut and to the horizontal link, and the second triangle gives their sizes.

The four forces laid end to end — load, vertical link, strut, horizontal link — close on the starting point, and PQ’s direction is the diagonal that divides the quadrilateral into the two triangles. On this beam the vertical link carries 4.33 kN upward, the strut 6.12 kN along its line, and the horizontal link 0.67 kN against the direction its line was drawn in, which is all the horizontal thrust the strut’s inclination and the load’s leave unbalanced.

Those are the numbers the three equations of equilibrium give, solved directly as a system with no Culmann line in it. The drawing and the algebra agree to the last figure because they are the same three conditions, arranged differently: the construction uses the two force triangles as two equations each, and the fact that both resultants share PQ as the moment equation.

Which free body produced the number #

The free body is the whole beam, and the construction never cuts it.

That is worth saying because the pairing looks like a cut. It is not. It is a grouping of the forces on one free body into two sub-resultants, chosen so that each group’s resultant has a known point on its line. Nothing about the beam’s internal forces enters, and the construction gives only the three link forces — the external forces on the chosen body.

The choice of pairing is free, and it is a choice of convenience. Culmann’s construction needs P and Q to exist, which fails only when a pair of lines is parallel, and it is easier to draw when P and Q land on the sheet and not far outside it.

Three lines, one answer #

Every one of the three unknowns can be paired with the load, and each pairing has its own Culmann line.

The three lines cross the sheet in completely different places, one steep and two shallow, and they are all correct. Every Culmann line of a body gives the same forces, because each is a statement of the same equilibrium, and a drafter who wants to check a construction draws a second one on another pairing. If the two answers disagree, one of them has a pencil error in it; if they agree, both are almost certainly right, because an error would have to repeat itself on a different line to survive.

Culmann taught at the polytechnic in Zurich, and his construction spread with his students into the design offices of the railway age. One of them, Maurice Koechlin, went on to work out the structure of the tower in Paris that carries Gustave Eiffel’s name, and the iron lattice of that tower was sized by exactly this kind of drawing — pairs of forces, crossings on the sheet, and triangles closed with a straightedge.

That is the practical content of graphic statics and the reason it lasted a century: a drawing that checks itself is worth more than a calculation that has to be done again. It was dropped, in the 1960s, because a calculator made the algebra cheaper than a straightedge — not because anybody found the construction wrong.

Culmann’s own use: a section through a truss #

Culmann’s main application was not a beam on links. It was a truss cut through three members, and the result is a graphical form of the method of sections.

The piece of truss left of the cut has, as external forces, the support reaction, the loads on its top joints, and the three cut members. The reaction and the loads are known and can be added into a single resultant: 25 kN up at the support and 10 kN down at each of two top joints make 5 kN up, acting at the point where their moments balance, which is 6 m to the left of the support. The resultant of the loads on a free body need not lie on the body, and here it lies well outside the truss. What remains is one known force and three unknowns along known lines, exactly the beam problem again.

Paired with the bottom chord, whose line is horizontal, the resultant crosses it at P, (−6, 0). The top chord and the diagonal meet at their shared joint, the top joint at (2, 1), and that is Q. Two triangles on PQ give the three member forces, and they are the forces the joint equations give.

The construction is short enough to do by hand, and doing it once shows every number arriving. PQ runs from (−6, 0) to (2, 1), so it rises one unit in eight, and its direction is $(8, 1)/\sqrt{65}$. The first triangle holds the 5 kN resultant, the bottom chord’s force along the horizontal, and a force of size $r$ along PQ. Their vertical components must cancel, so $r/\sqrt{65} = -5$ and $r = -40.3$ kN; their horizontal components must cancel too, so the bottom chord carries $-8r/\sqrt{65}$, which is 40 kN, pulling away from the cut: tension. The second triangle has that same force along PQ, reversed, as its closing side, split between the top chord along the horizontal and the diagonal along its 45° line. The diagonal must supply the 5 kN of vertical component alone, so it carries $5\sqrt{2}$, 7.07 kN in tension, and the top chord takes what is left of the horizontal, 45 kN pushing toward the cut: compression.

Read against the beam the truss replaces, those are two familiar quantities. The bottom chord’s 40 kN is the bending moment at the section through Q, 40 kN·m, carried as a couple over the 1 m depth; the diagonal’s vertical component is the panel’s shear, 5 kN. The Culmann construction is laying out the moment and the shear of the equivalent beam, in the geometry of the truss, without ever writing either down.

Q is not an accident. Ritter’s method of sections finds the bottom chord’s force by taking moments about the joint where the other two cut members meet, because both their forces pass through it and drop out. That joint is Q. Ritter’s moment centre is Culmann’s second crossing: taking moments about Q gives the bottom chord in one line — 25 kN × 2 m less 10 kN × 1 m is 40 kN·m, over a lever arm of 1 m, 40 kN — and drawing the Culmann line through Q gives it in two triangles. The two methods were published four years apart and are one idea, written once as arithmetic and once as a drawing.

A second cut #

Moving the cut moves the resultant, and the construction follows.

Nearer the support, less of the load has been taken off the reaction, so the resultant is larger and lies closer to the truss. The diagonal carries three times as much, because it carries the panel shear, and the chords less, because the moment is smaller near the support.

One cut would defeat the construction. In the middle panel of a symmetric truss with an odd number of panels and a symmetric load, the forces left of the cut add to no resultant at all, only a couple — equal and opposite vertical forces a distance apart — and then there is no known line to pair with a member. The chord forces come from the couple directly, and the diagonal carries nothing, which is the method of sections’ answer too, reached without a construction because there is nothing to construct.

When three lines nearly meet #

The determinacy count — three unknowns, three equations — is satisfied by every arrangement of the three links. It says nothing about whether the answer is reasonable.

The vertical and horizontal links cross at the left end of the beam. Taking moments about that point, their forces drop out and only the load and the strut are left: the strut must supply the load’s moment about the point, 25.98 kN·m, on whatever lever arm its line has. As the strut’s line approaches the crossing, the arm shrinks and the strut’s force grows as its inverse. At 4.24 m the strut carries 6.12 kN; at 0.04 m, 735 kN. The other two links grow with it to balance its components.

In between, the vertical link’s force passes through nothing and changes sense near 2.15 m, and the horizontal link’s near 3.77 m, which is where their curves dip on the logarithmic axis. At zero distance the three lines meet at a point, and a load that misses that point has a moment about it that none of the three can balance. The beam can rotate about the crossing, which is the point a mechanism turns about.

The same arithmetic is a design rule. Choosing the angle and position of a strut or a brace is choosing its lever arm about the point where the other supports’ lines cross, and the force it attracts is the load’s moment about that point divided by the arm. A brace set so that its line passes near that crossing looks economical on the drawing — short, neat, out of the way — and carries many times the load it was drawn for, along with the other supports that balance it.

A support arrangement whose lines nearly meet is nearly a mechanism, and the count cannot tell. The count that does not see it is the same failure in a truss: the right number of members arranged so that some of them are nearly in line. In Culmann’s drawing the warning is visible: the crossing Q moves toward the load’s line, the Culmann line swings until it is nearly parallel to one of the forces, and the triangles become long and thin, with sides meeting at shallow angles that no pencil can locate. The drawing is ill-conditioned exactly when the structure is.

What a drawing gave that algebra did not #

Solved as algebra, the four-force body is three simultaneous equations in three unknowns — easy with a calculator, tedious by hand, and silent about whether the numbers are sensible. Solved by Culmann’s construction it is one straight line and two triangles, each of which can be checked by eye, and the geometry of the construction shows which force is large and why.

Two things in particular are visible in the drawing and invisible in the algebra. The magnitudes’ dependence on geometry, which shows as the length of the triangles’ sides against the load’s, so a construction whose triangles are long and thin announces a force much larger than the load. And the role of each support, which shows as where the crossings fall: a support whose line passes close to the load’s has little to do, and one whose line passes far from every crossing is the one carrying the moment.

Where the model stops #

The lines of action are known exactly. A link carries force along the line through its pin centres only if its pins are frictionless. A real pin has friction, and the force can pass anywhere within a small circle round each pin centre, so the line is known only to within a band — and a construction that depends on a crossing far from the body amplifies that uncertainty.

The body is rigid. The links stretch and the beam bends, and the lines of action move slightly as they do. For an ordinary structure the movement is negligible; for a mechanism near the concurrent position it is not, because a small movement changes a small lever arm by a large fraction.

Four forces, in a plane. Five or more forces on one body are handled by pairing repeatedly or by the funicular polygon, and a body in space has six equations and needs a construction in three dimensions that graphic statics never made convenient.

Still open: five forces, and the pin that is not a point #

Four forces pair once. Five or more need either a sequence of Culmann lines, each combining a known resultant with one more force, or the funicular polygon, which finds the resultant of any number of forces with a pole and a string of lines, and which is where graphic statics becomes a method for shape as well as force. Bow’s notation, which names the spaces between forces rather than the forces, and makes the force polygon and the space diagram read as each other’s duals. The three-hinged arch, whose two reactions each pass through a hinge and so are found by pairing the load with the arch’s own geometry. And the pin with friction, whose line of action is not a line but a band tangent to a small circle round each pin, which turns every construction on this page into a range.