Moments and couples

A force does two things to a body: it tends to move it along, and it tends to turn it. So far we have only dealt with the first, because all the forces passed through one point. Real structures are loaded at many different points, and the turning effect, called the moment, is often what governs the design. A cantilever balcony fails by rotating off its support, a crane tips over its outriggers, a bolt is tightened by a twisting moment, and a beam bends because of the moments along it.

This chapter defines the moment of a force in two and three dimensions, introduces Varignon's theorem and the cross product, finds moments about an axis, and then treats couples: pairs of forces that produce pure turning. It finishes with equivalent systems, the tool that lets you replace a messy set of loads with a single force and couple. Those ideas are what make rigid-body equilibrium, in the next lesson, possible.

The moment of a force

The moment of a force about a point O measures its tendency to rotate a body about that point. Its magnitude is the force times the perpendicular distance from O to the force's line of action:

The distance \(d\), called the moment arm, must be perpendicular to the line of action. Moments are measured in N·m or kN·m. In two dimensions the moment has a sense, counterclockwise or clockwise, and we usually take counterclockwise as positive. In vector terms the moment points perpendicular to the plane by the right-hand rule: curl the fingers of your right hand in the direction of rotation and your thumb gives the direction of the moment vector. Counterclockwise in the x-y plane is +z.

Two consequences follow straight from \(M = Fd\). A force whose line of action passes through O has no moment about O, however large it is. And a force can slide along its line of action without changing its moment, because the perpendicular distance does not change. That is the principle of transmissibility from Lesson 1 again.

Varignon's theorem

Finding the perpendicular distance can take awkward geometry. Varignon's theorem offers a shortcut: the moment of a force about a point equals the sum of the moments of its components about that point. Split the force into x and y components at any point on its line of action, and add their moments.

If the force acts at a point \((x, y)\) relative to O, the x-component has moment arm \(y\) and the y-component has moment arm \(x\). With counterclockwise positive:

This formula handles signs automatically, provided you use signed coordinates and signed components. It is the 2D version of the cross product you will meet shortly.

Moments in practice: tipping and stability

Many design checks compare two moments about the same point: one that tends to overturn a body and one that resists. Mobile cranes, retaining walls, signs in wind, and loaded shelving are all checked this way. The body tips about its edge when the overturning moment exceeds the restoring moment from its own weight.

Notice how the safe load depends on the radius. Doubling the outreach to 12 m halves the tipping load. Crane load charts reflect exactly this, which is why booming out with a suspended load is one of the most common causes of crane overturns.

Moments in three dimensions: the cross product

In 3D the perpendicular distance is hard to see, and the moment has a direction that matters. The general definition uses the vector cross product. If \(\mathbf{r}\) is a position vector from O to any point on the line of action of \(\mathbf{F}\), then

Expanding the determinant:

The magnitude is \(rF\sin\theta\), which is \(F\) times the perpendicular distance, so it agrees with \(M = Fd\). The direction is perpendicular to both \(\mathbf{r}\) and \(\mathbf{F}\), by the right-hand rule. Order matters: \(\mathbf{r}\times\mathbf{F} = -\mathbf{F}\times\mathbf{r}\). Always put \(\mathbf{r}\) first. In 2D, with \(r_z = F_z = 0\), only the k-component survives, and it is exactly \(xF_y - yF_x\).

Moment about an axis

Often you need the turning effect about a particular line, not a point: a door about its hinges, a shaft about its bearings, a hatch about its hinge line. The moment about an axis \(a\) through O, with unit vector \(\mathbf{u}_a\), is the component of \(\mathbf{M}_O\) along that axis:

This is called the scalar triple product. A force parallel to an axis, or one whose line of action crosses it, has no moment about it. You cannot open a door by pushing toward the hinges or by pulling straight up.

Couples

A couple is two forces of equal size, opposite direction and parallel lines of action, separated by a distance \(d\). The forces cancel, so a couple produces no net force, only turning. Its moment is

and remarkably, it is the same about every point. To see why, take moments about any point O at distance \(a\) from one force: \(F(a + d) - Fa = Fd\). The \(a\) drops out. So a couple moment is a free vector: it can be moved anywhere on the body without changing its effect on the body as a whole.

Couples are everywhere. Two hands on a steering wheel, the twist applied by a tap wrench, the torque from an electric motor on its shaft, and the end moment that a welded connection delivers to a beam are all couples. Different force pairs give the same couple if \(Fd\) is the same: 100 N on a 0.4 m wheel equals 200 N on a 0.2 m wheel.

Equivalent systems

Two force systems are equivalent if they have the same resultant force and the same resultant moment about any point. Equivalent systems have the same effect on a rigid body's equilibrium (though not on its internal stresses). This lets you replace a complicated set of loads with something simpler.

Moving a force

A force can be moved to a parallel line of action through another point, provided you add a couple equal to the moment of the force about the new point. Physically: add equal and opposite forces at the new point. One of them, with the original force, forms a couple; the other is the moved force.

That is why structural engineers care so much about eccentricity. A column designed only for the axial load would be unsafe if the load is applied off-centre, because the hidden couple adds bending stress.

Reducing a system to a single resultant

Any system of forces in a plane can be reduced to a single force \(\mathbf{R} = \sum\mathbf{F}\) acting at O plus a couple \(M_{RO} = \sum M_O\). If \(\mathbf{R}\) is not zero, you can go one step further and shift \(\mathbf{R}\) sideways until its own moment equals \(M_{RO}\). The result is a single force with no couple, located at perpendicular distance \(d = M_{RO}/R\) from O.

The same idea gives the resultant of a distributed load. A uniform load \(w\) (in kN/m) over a length \(L\) is equivalent to a single force \(wL\) at the middle of the loaded length. A triangular load is equivalent to its area, \(\tfrac{1}{2}wL\), at the centroid of the triangle, one third of the way from its tall end. You will use both constantly in the next lesson.