Moving a force, and what it costs

A force is drawn as an arrow, and an arrow has a tail. Almost every calculation in statics begins by moving that tail somewhere more convenient — to the centroid of a bolt group, to the centreline of a column, to the point about which moments are being taken — and the move is made so automatically that the price of it is rarely written down. There is a price. It is exactly one couple, it is the reason a bracket’s bolts work harder than the load suggests, and in three dimensions it leaves behind a residue that no amount of further moving can get rid of.

The rule is short enough to state in one line and is worth stating carefully, because two of its three clauses are usually dropped. A force may be moved anywhere on its own line of action for nothing; moved off that line by a distance $e$, it must be accompanied by a couple of magnitude $Pe$; and a couple has no point of application at all. The first clause is transmissibility, the second is the whole of connection design, and the third is the one that makes the arithmetic work.

The two systems no equation can tell apart #

Two force systems are statically equivalent when they have the same resultant force and the same resultant moment about one point — and if they agree about one point they agree about every point, because changing the reduction point changes both moments by the same $\mathbf{c} \times \mathbf{R}$.

That is a strong statement and a narrow one. It says that the six equilibrium equations cannot distinguish the two panels of the figure above. It does not say the two panels are the same loading, and the whole of this collection’s business is with the quantity that does distinguish them.

So the honest form of the rule is: the reactions do not know, and the interior does. Choosing the free body is the skill, and the reason it is a skill is precisely that a substitution which is invisible from outside the body can be catastrophic inside it.

Which free body produced the number #

Take the body of the first figure, cut it on a vertical plane through the centreline, and sum moments about a point on that cut.

In the left-hand panel the applied force sits at $e = 180$ mm on one side of the cut. If the cut is taken on the other side, the force does not appear on the free body at all: the internal moment on the cut is whatever the support reactions require, and nothing else. In the right-hand panel the force has been moved onto the centreline and a couple of $60 \times 0.180 = 10.8$ kNm added. Now a cut on either side sees the couple, because a couple is a free vector with no point of application — it belongs to the body, not to a place on it.

That is the substitution’s failure mode, exactly located. It has not changed a reaction; it has moved a moment across a cut. Which is another way of stating Saint-Venant’s principle: the two loadings differ, and the difference dies away over a distance of the order of the body’s own depth, so at a distance the substitution is harmless and near the point of application it is not.

The couple is what the lever arm buys #

A couple is two equal and opposite forces a distance apart, and the arithmetic that makes it useful is that its moment is the same about every point — $P \cdot d$, whatever the point, with no lever arm to measure from anywhere.

There is a temptation to treat the couple as bookkeeping — a term that appears because a force was moved and disappears when it is moved back. It does not disappear. It reappears as a real demand on a real detail, and the clearest place to watch it happen is a bracket.

The two halves travel differently and that is the finding. The direct force is shared by counting: four bolts, a quarter each. The couple is shared by lever arm: each bolt’s contribution is proportional to its own distance from the centroid, so the group resists the couple with $\sum r^2$ — 45,003 mm² here — exactly as a section resists bending with its second moment of area.

Sweeping the eccentricity shows what kind of quantity this is. At $e = 0$ the couple’s share of the worst bolt is zero and the group is carrying 25 kN a bolt. At 50 mm it is already 32%; at 125 mm, 54%; at 250 mm, 70%; at 500 mm, 82.5%. The share rises fast at first and then saturates, because the direct component is fixed while the torsional one grows without limit — so for any bracket worth calling a bracket, the connection is being designed for the move rather than for the load.

The part that cannot be moved away #

Everything so far is planar and everything so far has been a matter of choosing where to put the arrow. Reducing about a different point changes the couple by $\mathbf{c} \times \mathbf{R}$, and a cross product is perpendicular to $\mathbf{R}$. So there is a component of the moment that no choice of point can touch: the component along the resultant.

$$p = \frac{\mathbf{M}\cdot\mathbf{R}}{|\mathbf{R}|^2}, \qquad \mathbf{M}_\parallel = p\,\mathbf{R}$$

$p$ has the units of a length and is called the pitch. It is an invariant of the force system: not of the drawing, not of the reduction point, not of the coordinate axes.

The perpendicular part can always be driven to zero, and the point at which it vanishes is the central axis:

$$\mathbf{c} = \frac{\mathbf{R} \times \mathbf{M}}{|\mathbf{R}|^2}$$

which is derived rather than quoted: it is the shift for which $\mathbf{M} + \mathbf{c}\times\mathbf{R}$ comes out parallel to $\mathbf{R}$. For the two forces of the figure, that axis passes through $(1.6,\ 0.6,\ 0.8)$.

Three systems that always reduce to a single force #

Before the general case, the special ones, because they are the reason the general case is so easily missed. A force system reduces to a single force whenever $\mathbf{M}_\parallel$ vanishes, and it vanishes automatically in three common arrangements.

Concurrent. If every force passes through one point, the moment about that point is zero, so the pitch is zero. That is the reduction three-force problems are solved by: with three forces on a body and no couple, the lines of action must meet, because two of them meet somewhere and the third must pass through the same point or the sums cannot cancel.

Parallel. If every force points the same way, both $\mathbf{R}$ and every $\mathbf{r}\times\mathbf{P}$ are constrained: the moment is perpendicular to the common direction, so again $p = 0$. The single force is at the centroid of the load, and replacing a spread load by its resultant is that reduction used every day.

Coplanar. If every force lies in one plane, every moment is normal to that plane and the resultant lies in it, so $\mathbf{M}\cdot\mathbf{R} = 0$ identically. Three arbitrary forces in a plane — $(3,-4)$, $(-1,6)$ and $(2,1)$ at three unrelated points — reduce to a resultant of $(4,3)$ on a line through $(-1.86,\ 2.48)$, and the computed pitch comes out as exactly zero.

Every drawing on this site is coplanar, which is why the general result never appears in it and why it comes as a surprise when it does.

What a general system reduces to #

Take two forces that are neither concurrent, nor parallel, nor coplanar: 10 units pressing down at $(2,0,0)$, and 5 units pushing sideways at $(0,3,0)$. The resultant is $(5, 0, -10)$, of magnitude 11.1803. The moment about the origin is $(0, 20, -15)$. The pitch is $\mathbf{M}\cdot\mathbf{R}/|\mathbf{R}|^2 = 150/125 = 1.2$, which is not zero, so no point in space reduces this pair to a force alone.

The simplest object equivalent to a general force system is therefore a force and a collinear couple — a wrench, or in the older language a screw. Poinsot published the result in 1804, and the name is not a metaphor: pushing and twisting about the same line is exactly what a screwdriver does, and the pitch has the units of a length for exactly the reason a screw’s pitch does.

Two consequences worth keeping. A general spatial system has no line of action. Asking “where does the resultant act?” is a question with no answer, and only the accident of coplanarity makes it seem like a reasonable one. And the pitch is a genuine invariant, in the same family as the bimoment and locked-in residual stress: quantities that survive every operation the equilibrium equations can perform and are therefore invisible to them.

The substitution that changes the interior #

The rule’s first clause — a force may be slid along its own line for nothing — deserves the sharpest statement of its limits, because it is the one that gets used without thinking.

Sliding a force along its line is free for a rigid body, which is a body with no interior worth asking about. For every body this collection is interested in, it is a change of loading whose effects are confined to the neighbourhood of the move — and “confined to the neighbourhood” is not “absent”. A column loaded through its centroid and the same column loaded at its face with a compensating moment applied to the beam are statically equivalent and are not the same detail; the first has a uniform stress block and the second has a stress concentration and a length of column over which the two have not yet reconciled.

This is why a moment applied to a beam needs its point named. The couple that pays for a translation is a free vector to the equilibrium equations and it is not free to the material: somewhere, two forces a distance apart are actually being applied, and where they are applied decides the local stress field entirely.

Where the model stops #

Rigid bodies do not exist. Every statement about equivalence above is a statement about the six equations, and the six equations are blind to compatibility. Two statically equivalent loadings on a redundant structure can produce genuinely different reactions, because the reactions in a redundant structure depend on stiffness, and stiffness depends on where the load actually is. The wrench and the pitch survive; the reaction distribution does not.

A couple’s “no point of application” is an idealisation of a pair. Any real couple is delivered by forces on an area, and the area matters as soon as the question is local. A moment connection delivers its couple through two flange forces on two identifiable patches of steel, which is the whole reason the connection field exists.

Nothing here is dynamic. A force system on a body that is accelerating is not in equilibrium at all, and the reduction above is a statement about the system of forces, not about what the body does with it. The extra term is the mass times the acceleration, and it is not a load anybody applied.

The pitch is invariant, not conserved. Adding a force to the system changes it. It is a property of a given set of forces, not a quantity that persists as loads come and go.

What the pictures cannot show #

The wrench figure draws a ring about the central axis to stand for the couple, and a ring is not what a couple looks like — a couple has no shape, no size and no location, and the ring is a symbol whose radius was chosen to be legible. Any other radius would have been equally correct, which is precisely the property being illustrated and is exactly what the drawing cannot convey.

The bracket figure draws two arrows at each bolt and adds them. Real bolts do not carry two forces; they carry one, and the decomposition into a direct part and a torsional part is a way of computing that one, not a description of anything happening in the steel. The decomposition is also elastic, and the same group analysed by its instantaneous centre gives a different and larger answer, because the elastic method assumes a rotation about the centroid that a group at failure does not have.

And the invariant figure plots a moment against distance along a line that a reader cannot see. The line is not a feature of the structure; it is a locus derived from the force system, and the whole of the figure’s content is that one of the two curves in it is flat.

The ladder from here #

Later rungs on this anchor: the reduction of a distributed loading over a surface to a wrench, and why a wind pressure field has a centre of pressure only when it is planar. The null system and the reciprocal screw, which is where this argument meets kinematics — a body free to move along a screw is in equilibrium under any wrench reciprocal to it, and the whole of screw theory is that pairing. Statically equivalent load systems in finite element practice, where a nodal load set is chosen for equivalence in the work sense rather than the resultant sense, and gives different answers for that reason. The centre of gravity as a special case of the parallel reduction, and the conditions under which it exists at all in a non-uniform field. And the moment about an axis rather than a point, which is the operation every torsion calculation on this site quietly performs.

Poinsot’s own route to the wrench came from a question about the composition of rotations, not about statics; he was looking for the object that plays the part of the resultant in the kinematic problem and found that the two questions have the same answer. Statics inherited it, drew everything in a plane for the next two centuries, and mostly forgot it.