4 Force System Resultants

Learn how to reduce force systems to equivalent forces and moments, shift their reference points, and identify important planar and three-dimensional cases.

Equivalent force-couple systems

A force system can be replaced, for its external effect on a rigid body, by an equivalent force and moment at a chosen reference point. The replacement preserves the system’s total force and total moment, but it does not necessarily preserve the detailed distribution of loading.

Finding the and moment

For forces $\mathbf{F}_i$ applied at position vectors $\mathbf{r}_i$ measured from point $O$ , and applied couples $\mathbf{C}_j$ , the at $O$ is defined by

\mathbf{R}=\sum_i\mathbf{F}_i

and

\mathbf{M}_O=\sum_i(\mathbf{r}_i\times\mathbf{F}_i)+\sum_j\mathbf{C}_j.

The cross product gives each force’s moment about $O$ . Each applied couple is included directly because its moment is a free vector, so its contribution is the same about any reference point.

For a two-dimensional system, take counterclockwise moments as positive. A force applied at $(x,y)$ , with components $(F_x,F_y)$ , contributes

M_O=xF_y-yF_x.

Changing the reference point

To move an equivalent system from point $O$ to point $P$ , let $\mathbf{r}_{OP}$ be the position vector from $O$ to $P$ . The remains unchanged, while the moment at $P$ is

\mathbf{M}_P=\mathbf{M}_O-(\mathbf{r}_{OP}\times\mathbf{R}).

Changing the reference point changes the moment assigned to the system, but the force and moment together still represent the same physical loading.

Reducing a general force system

Use a consistent reference point and track force components, positions, signs, and units when reducing a general force system.

Choose a reference point, such as a connection, support, or convenient coordinate origin.
Resolve each force into coordinate components if needed, and record its position relative to the reference point.
Add the force components to obtain the $\mathbf{R}$ .
Add the moments of all forces about the reference point and all applied couples to obtain the $\mathbf{M}_O$ .
Represent the original loading by $\mathbf{R}$ and $\mathbf{M}_O$ acting at that point, and check units, signs, and directions.

Worked example in two dimensions

At $O$ , a $4\,\mathrm{kN}$ force acts in the $+x$ direction at $(0,2\,\mathrm{m})$ ; a $3\,\mathrm{kN}$ force acts downward at $(3\,\mathrm{m},0)$ ; and a $2\,\mathrm{kN\cdot m}$ counterclockwise couple is applied.

The is

\mathbf{R}=(4,-3)\,\mathrm{kN}.

Taking counterclockwise moments as positive, the total moment about $O$ is

M_O=(0\times 0-2\times 4)+(3\times(-3)-0\times 0)+2=-15\,\mathrm{kN\cdot m}.

Thus, the equivalent system at $O$ consists of a $4\,\mathrm{kN}$ force in the $+x$ direction, a $3\,\mathrm{kN}$ force downward, and a $15\,\mathrm{kN\cdot m}$ clockwise couple. The total includes both the force moments and the applied couple.

Shifting a planar force’s line of action

In a planar system with a nonzero , the force and couple can be represented by a single force whose line of action is shifted to produce the same moment. For the example above, place the at $(5\,\mathrm{m},0)$ . Its moment about $O$ is

5(-3)-0(4)=-15\,\mathrm{kN\cdot m},

which reproduces the moment of the .

Special cases and three-dimensional systems

The and moment identify several useful reductions:

If $\mathbf{R}=0$ and $\mathbf{M}_O\ne 0$ , the system reduces to a .
If $\mathbf{R}\ne 0$ and the moment is zero at a point, the system reduces to the acting through that point.
If $\mathbf{R}=0$ and $\mathbf{M}_O=0$ , the system has zero net force and moment and is in equilibrium as a rigid-body loading system.
In three dimensions, a general force-couple system cannot always be replaced by one force alone. It can be reduced to a : a force and a moment parallel to that force, after shifting the force’s line of action to remove the perpendicular part of the moment.

What equivalence preserves

preserves the net force and moment on a rigid body, not the detailed distribution of loading. Equivalent systems are therefore useful for analyzing overall external effects, but may not preserve local stresses or internal forces.

The central procedure is to sum all forces and to sum the moments of those forces and all applied couples about a chosen point. This gives an . In planar cases with a nonzero , the couple can generally be eliminated by shifting the force’s line of action; in three dimensions, a residual moment parallel to the may remain.