Moments and Couples

Introduction

When a force acts on a rigid body in a direction that does not pass through its axis or point of rotation, it tends to cause the body to rotate. This rotational effect is known as a moment. Moments and force couple systems are essential for analyzing how forces produce rotation and how their combined effects influence the equilibrium of a structure.

2D Moments

For planar problems, moments are calculated using perpendicular distances (moment arms). Moments are similar to torque in physics, as both have units of force times distance. Angled forces must be resolved into components before solving.

$M_{O} = \Sigma Fd$

3D Moments (Vector Method)

Moments in 3D are computed using the cross product: $\overrightarrow{M} = \overrightarrow{r} \times \overrightarrow{F}$, where $\overrightarrow{r}$ is the position vector from point to force and $\overrightarrow{F}$ is the force vector. In determinant form:

Moment About an Axis

To find the moment about a specific axis, take the dot product of the axis unit vector and $\overrightarrow{r} \times \overrightarrow{F}$:

$M_{axis} = {\overrightarrow{u}}_{axis} \bullet (\overrightarrow{r} \times \overrightarrow{F})$

In determinant form: $M_{axis} = det$

Force Couple Systems

Occasionally, forces will cancel each other out but still result in a moment about a central point. The type of system is known as a force couple system. This entails a set of two equal but opposite forces set at equal distances from a given point. Force couple systems are represented as a moment about the given point with a magnitude twice what one of the forces would result in.

$M_{O} = \Sigma Fd$, where $F$ is one of the coupled forces and $d$ is the distance between the couple

Resultant Force & Moment

A system of forces can be reduced to $\Sigma F = F_{R}$ and $\Sigma M_{A} = M_{R}$, replacing a system with a single resultant force and a couple moment at a point.

Example 1:
Find the Resultant Moment about Point A

1) Resolve forces into x and y component magnitudes:

	$F_{1}$	$F_{2}$	$F_{3}$
x	$(250\ N)sin30{^\circ}$	$(300\ N)cos60{^\circ}$	$\frac{3}{5}(500\ N)$
y	$(250\ N)cos30{^\circ}$	$(300\ N)sin60{^\circ}$	$\frac{4}{5}(500\ N)$

2) Multiply each force by the perpendicular distance of that force to point A. $F_{1,\ x}$ and $F_{2,\ x}$ have no moment arms, so they are excluded from the equation.

Assuming CW to be positive, $\Sigma M_{A} = F_{1,\ y}(2\ m) + F_{2,\ y}(5\ m) - F_{3,\ x}(4\ m) + F_{3,\ y}(5\ m)$

Substituting and solving: $\Sigma M_{A}\ = 2532\ N \cdot m$

Example 2:
Find the Moment of Force about Axis AC

1) Find position vector for axis AC and from $F$ to the axis:

$\overrightarrow{AC} = C - A = 4\widehat{i} + 3\widehat{j} + 0\widehat{k}$

${\overrightarrow{r}}_{F \rightarrow AC} = 0\widehat{i} + 0\widehat{j} + 2\widehat{k}$

2) Solve for AC axis unit vector:

$|u_{AC}| = \sqrt{4^2 + 3^2 + 0^2} = 5\ ft$

${\overrightarrow{u}}_{AC} = \frac{4}{5}\widehat{i} + \frac{3}{5}\widehat{j} + 0\widehat{k}$

3) Solve $M_{AC} = {\overrightarrow{u}}_{AC} \bullet {(\overrightarrow{r}}_{F \rightarrow AC} \times F)$

To express as cartesian vector, multiply by the axis unit vector:

$(14.4\ lb \cdot ft)({\overrightarrow{u}}_{AC}) = (11.5\widehat{i} + 8.64\widehat{j})\ lb \cdot ft$