The member with only one direction

Take any body. Apply forces to it at exactly two points and nowhere else. Insist that it is in equilibrium.

Three things follow, and they follow from $\sum \mathbf{F} = 0$ and $\sum \mathbf{M} = 0$ and nothing else. The two forces are equal in magnitude, because the forces have to cancel. They are opposite in direction, for the same reason. And they are along the line joining the two points, because a moment taken about either point has only the other force in it, so that force must have no lever arm about the first — which fixes its line.

It is the shortest theorem in statics with any real content, and it is doing the work under an enormous amount of this collection. A truss member carries one number because it is a two-force member. A strut’s line of action is known before any analysis begins for the same reason. A funicular polygon exists because each of its segments is one.

Which free body produced the number #

The member itself, cut free of everything. The whole force of the theorem is in that boundary rather than in any arithmetic done afterwards: a cut that crosses two load points and nothing else reduces three unknowns at each end to a single magnitude on a line already known.

Two forces act, $\mathbf{F}_A$ at $A$ and $\mathbf{F}_B$ at $B$. Force equilibrium gives $\mathbf{F}_B = -\mathbf{F}_A$ immediately. Moment equilibrium about $A$ gives

$$\mathbf{r}_{AB} \times \mathbf{F}_B = \mathbf{0}$$

and a cross product vanishes only when the two vectors are parallel. So $\mathbf{F}_B$ is along $AB$.

Notice what the derivation did not use. It did not use the shape of the body, its material, its cross-section, whether it is straight, or whether it is one piece. A bent bar, a curved rib, a chain, a whole substructure — anything at all, provided forces reach it at two points and only two.

That generality is the theorem’s value and it is also where the trouble is.

The theorem says nothing about the member being straight #

This is the part that gets forgotten, and it is the reason the page exists.

The forces run along the chord between the pins. The material may take any route it likes between them. So at any section of a curved two-force member, resolving the chord force into components along and across the local tangent:

$$N = P\cos\alpha, \qquad V = P\sin\alpha, \qquad M = P\,y$$

where $\alpha$ is the angle between the tangent and the chord and $y$ is the perpendicular distance of the section from the line of action.

A curved two-force member is in bending everywhere, and the moment is the axial force times the offset.

For a 10 m member bowed 1.2 m off the chord carrying 200 kN, the crown moment is $200 \times 1.2 = 240$ kNm. On a 200 by 300 rectangle that is 80 N/mm² of bending against 3.3 of axial — the bending stress is 24 times the axial one, in a member that a truss analysis would have reported as carrying 200 kN and nothing else.

Nothing in that diagram is a property of the curve. The theorem fixed the line of action, the section’s distance from it is read off the drawing, and $M = Py$ multiplies the two — so two members with the same pins, the same force and the same offset at midspan carry the same moment there however differently they travel between the ends.

Six times the axial stress for one section depth of bow #

The ratio above is worth having in a form that travels. Bending stress over axial stress is

$$\frac{M/Z}{P/A} = \frac{Py/Z}{P/A} = \frac{y}{Z/A}$$

and for a rectangle $Z/A = d/6$, so the ratio is $6y/d$ exactly. A member bowed by one section depth carries six times as much bending stress as axial. Two depths, twelve. Three, eighteen.

That is the single most useful number on this page and it is why straightness tolerances exist. A column out of straight by $L/500$ over 5 m is 10 mm off; on a 200 mm deep section that is a tenth of a depth, so 0.6 of the axial stress in bending — which is the whole content of the imperfection allowance in every column curve in the world, arrived at from the two-force theorem rather than from a code. A perfectly straight strut is a two-force member with $y = 0$ everywhere and a real one never was, so the moment that follows is exactly the $Py$ above with a deflection in place of a fabrication tolerance.

The sign the theorem does not give, and why it decides everything #

The theorem returns a magnitude and a line. It does not return a sign, and the sign is the difference between two members that behave nothing alike.

Take the bowed member above — 10 m, 1.2 m off the chord, 200 kN — and put it in tension instead of compression. The arithmetic of the previous section is unchanged: the moment at the crown is still $Py = 240$ kNm, still 24 times the axial stress on that section, and the theorem is as true as it ever was. What differs is what happens next.

In tension the force pulls the ends apart along the chord, and the material between them is off the chord. So the load acts to reduce $y$: the member straightens, the offset falls, and the moment falls with it. The bowed tie is its own remedy, and if it is slender enough to bend appreciably it will pull nearly straight and end up carrying close to pure axial force whatever it looked like when it was fabricated. A chain is the limiting case — no bending stiffness at all, so $y$ goes to whatever the load demands and the funicular shape is simply the shape in which $y$ has finished moving.

In compression the same geometry has the opposite sign. The force pushes the ends toward each other, the material off the chord is pushed further off, $y$ grows, the moment grows with it, and the growth feeds itself. That is the whole of the load that makes itself worse, and the two-force theorem is where it starts: $M = Py$ is not a correction to a strut’s design, it is the strut’s design, with the only question being how much $y$ has grown by the time equilibrium is reached.

So the $6y/d$ ratio of the previous section is read differently in the two cases. For a tie it is an upper bound reached at zero load and released as the load arrives — which is why a straightness tolerance on a tie is a fabrication convenience and a straightness tolerance on a strut is a strength requirement. For a strut it is a starting value that the load multiplies, and the multiplier is the amplification factor rather than one. The same theorem, the same offset, the same arithmetic, and one of the two members repairs itself while the other destroys itself.

Where the theorem is being used without being named #

A truss. Every member is pinned at two joints and loaded nowhere else, so each carries one axial force. That is the entire justification for the method of joints: the unknowns are one number per member rather than three, and the count that decides determinacy is built on it. Load a truss member between its joints and it stops being one — which is why a purlin landing at mid-panel is an event rather than a detail.

A funicular polygon. Each segment of a hanging chain between two loads is a two-force member, so its direction is the direction of the force in it. That is why a funicular can be drawn: the polygon of forces gives the directions, and the directions give the shape.

An arch’s thrust line. The line of thrust is the locus of points through which the resultant passes, and the arch is a two-force member between any two hinges. A three-pinned arch has each half as a genuine two-force member between the crown pin and the springing pin, which is why its thrust is determinate — the hinge is put in on purpose to buy exactly that. Give the member thickness and the same statement becomes a line that must stay inside it: a masonry arch is not shaped to the funicular, it only has to have one somewhere within its own section, and the offset $y$ is then measured from that line rather than from a chord.

A pin-ended strut. Its buckling problem is the two-force theorem plus a deflection: the force is along the chord, the material has moved off it, and $M = Py$ with $y$ growing.

What the four have in common is that the theorem is nowhere stated in any of them. It is spent at the moment somebody writes one unknown against a member, draws a polygon whose sides are parallel to a structure, or calls a thrust determinate, and none of those moves announces which theorem paid for it.

The count that the theorem makes possible #

Determinacy counting on this site has always started with “one unknown per member”, and that phrase is the theorem in disguise.

A general plane member has three internal forces at any section — axial, shear and moment. A two-force member has one. So a plane frame of $b$ members and $j$ joints has $3b$ unknowns if the members are rigidly connected and $b$ if they are two-force members, and the whole difference between a truss count and a frame count is that substitution. The count itself is written for members with one unknown each, so every entry in it is the two-force theorem being spent.

That is why the pin-jointed idealisation is so useful and so persistent. It is not that real trusses have pins — most of them are welded or bolted rigidly — it is that the count, the analysis and the intuition all become an order of magnitude simpler when every member has one number, and the error from pretending is small provided the members are slender and the loads arrive at the joints.

What the pretending costs is secondary bending. Rigid joints put moments into truss members that the pin-jointed count has no term for, and the size of them depends on the members’ slenderness rather than on the joints — so a joint that is not a pin is a small error in a slender truss and a large one in a stocky truss, which is a frame wearing a truss’s clothes.

What breaks it #

One thing, and only one: a force applied anywhere other than the two points.

Put a uniform transverse load of 10 kN/m on the 10 m member above, alongside its 200 kN of chord force, and the end forces immediately tilt off the chord by $\arctan(wL/2P) = 14$ degrees. The member is now a three-force member at best, and its moment diagram is the two-force parabola plus the transverse load’s own.

The two contributions do not have to add. A tied arch’s rib carries a thrust that produces $Py$ hogging and a load that produces sagging, and the whole art of shaping an arch is choosing $y$ so that the two cancel — which is what a funicular shape is, defined the other way round.

There is a version of this that is easy to walk into on a drawing, because the offending force is one nobody thinks of as a force. Frame a secondary member into a truss diagonal at mid-length — a tie holding a service run, a light strut bracing something out of plane, a hanger for a walkway — and the diagonal has three load points, not two. It has stopped being a two-force member, and it stopped the moment the connection was drawn rather than the moment any load arrived.

The usual defence is that the secondary member carries nothing much. That is a statement about a load case rather than about the structure, and the analysis has already been changed regardless: the diagonal’s force is no longer along its own chord, so its direction is no longer known, so the joint equations at both its ends no longer close in the member forces alone. Whether the resulting bending is small is a separate question with its own answer, and it is usually yes. Whether the member is still a two-force member is not a matter of degree, and the answer is no.

The practical rule that comes out of it is to treat any attachment to a member between its two nodes as a modelling decision requiring a sentence, rather than as a detail. The theorem’s hypothesis is not “the member is straight” or “the joints are pins”; it is “forces reach this body at exactly two points”, and that hypothesis is falsified by a drawing, not by a load.

Two more subtleties are worth naming because they are the usual real-world escapes.

Self-weight. A member always has some, so no real member is strictly a two-force member. Whether it matters is a ratio: the transverse moment $wL^2/8$ against the axial force times the tolerance on the offset. For a light truss diagonal it is negligible; for a heavy inclined strut it is not.

Friction at the pins. A real pin transmits a small moment, so the line of action is not exactly through the pin centre. It is a small effect and it is the reason a pinned strut’s effective length is quoted with hedging.

The lever arm is the whole of what the theorem removes. A two-force member has no lever arm to any section on its own line of action, so moving a force — the operation that trades a shifted line for a couple — has nothing to trade here, and every departure from the chord is a lever arm being handed back.

Reading the theorem backwards #

There is a design move hidden in all of this, and it is one of the most powerful ones in the subject.

The theorem says: given two load points, the force is along the chord. Read it the other way — given a set of loads, find the shape for which the member is in pure axial force — and the answer is the funicular. That is what a hanging chain finds by itself, what a cable does under any load, and what an arch does when it is built to the inverted shape.

The design move is to choose $y$ so that $Py$ cancels whatever the transverse loads are doing. A parabolic arch under a uniform load is exactly that: the offset from the chord at every station is $M(x)/H$, so the two-force moment $Hy$ is $M(x)$ and cancels it, leaving pure compression.

Which means the two-force theorem is not merely a simplification for analysing trusses. It is a statement of what a structure has to look like if it is to carry its load without bending — and the whole family of funicular structures, from a suspension bridge to a masonry vault, is the set of shapes that satisfy it.

Where the model stops #

It is a statement about forces, not about stresses. A two-force member has a known line of action and says nothing about how the force is distributed across the section at the ends. Near the pin the distribution is whatever the connection makes it, and Saint-Venant’s principle needs a member length or so to sort it out.

It assumes the two points are points. A real connection has dimensions, and a bolt group or a welded end transmits its force over a region — so the “line joining the two points” is a line joining two centroids, and if the connection is eccentric the theorem is being applied to the wrong points.

And it says nothing about stability. A two-force member in compression obeys the theorem exactly right up until it buckles, at which moment $y$ starts growing and the moment with it. The theorem is about equilibrium and buckling is about which equilibrium. $M = Py$ with $y$ fixed is this page; $M = Py$ with $y$ depending on $M$ is second-order analysis, and the two differ by a feedback loop rather than by a mechanism.

What the pictures cannot show #

The bowed member is drawn with its offset exaggerated. A real strut out of straight by $L/500$ would be indistinguishable from a straight line at the scale of any figure here, and the moment it carries would still be 60% of its axial stress.

Nor can the drawings show that the theorem is about a body rather than a bar. The two-force member in a real structure is often a whole substructure — a truss between two pins, a frame hung at two points — and the conclusion applies to it identically.

The assumption the figure rests on #

Every figure here assumes the pins are frictionless and the member is loaded nowhere else. Both are approximations, and the honest statement is that the theorem is exact for an idealisation and useful for reality in proportion to how nearly the idealisation holds. It is the assumption that is doing the work, and it is the assumption that has to be checked — not by looking at the connection detail, but by asking what else is touching the member.

The history, which is older than statics #

The theorem is usually credited to nobody, which is fair — it is too short to have needed inventing. But its consequences were being used long before anybody wrote $\sum \mathbf{M} = 0$.

Hooke’s anagram of 1675, unscrambled after his death as ut pendet continuum flexile, sic stabit contiguum rigidum inversum — as hangs the flexible line, so but inverted stands the rigid arch — is the two-force theorem applied to a chain and then turned over. He had no algebra for it and did not need any: a hanging chain demonstrates the result physically, because a chain has no bending stiffness and therefore cannot be anything but a series of two-force members.

Poleni used exactly that in 1748 to test the dome of St Peter’s, hanging a chain loaded to represent the masonry and checking that the inverted shape fitted inside the dome’s section. That is a structural analysis with no equations in it, and it is the two-force theorem doing every bit of the work.

The line from there to the graphic statics of the nineteenth century, and from there to the force polygons on this site, is unbroken. What made drawing into calculation was a theorem that says a member’s direction is its force’s direction.

A polygon of forces makes the point in one sentence. It closes because the forces balance, which is ordinary equilibrium and would be true of any three concurrent forces; its sides are parallel to the members because each member is a two-force member, which is this theorem and nothing else. Take the second half away and the two drawings — the structure and the force diagram — have no correspondence left between them, and graphic statics stops being a method.

The ladder from here #

Later rungs on this anchor: the three-force member developed properly, and why concurrency is one equation weaker. The two-force member in three dimensions, where the same argument gives the same line and the counting changes. Self-weight as the standard violation, and the threshold at which a strut has to be designed as a beam-column. The funicular defined as the shape that makes a loaded member two-force, which turns the theorem into a design tool. Buckling as the theorem plus a growing $y$. And the connection question this page keeps deferring: what a “pin” has to look like before the line of action is where the drawing says it is.