General Distributed Load
with Load Intensity of f(x)
(units force/distance)

In many static problems, applied loads are given as distributed force loads. This is similar to stacking sand bags on a beam so that the load is distributed across the beam instead of at one location (point load). To help make the problem easier to solve, it is convenient to convert the distributed load into equivalent point loads.

If the loading on the object is a set of parallel discrete forces, the resultant force is simply the sum of all the forces, or

F_{R} = ΣF_{i}

In the previous section, three scalar equations were derived that determine
the position r of a force F_{R} that
represents the force resultant to a system
of discrete forces. If all forces are parallel
to
the y direction then the three scalar equations simplify
to

Σ(zF) = z' ΣF

Σ(xF) = x' ΣF

Here x and z are the Cartesian coordinates of the individual forces,
and x' and z' are the coordinates of the force resultant F_{R}.

Rearranging these two equations gives

If the forces are further restricted so that they
all lie in the x-y plane, then z' = 0 and only the second equation applies.

If instead of a system of point loads, consider a continuous distributed force
f(x) that acts in the x-y plane and is parallel to the y axis, then through
calculus the second equation (x') above becomes

The force resultant is simply the force magnitude F_{R} given
by

The force magnitude F_{R} is located a distance x' from the
origin.

Uniform and Triangular Line Loads

Resultant Force and Location for
Two Distributed Loading Types

In the case of a uniform line load as shown, it is unnecessary to perform the integrations because the force resultant is always the value of the distributed load multiplied by the distance over which it acts. The location of the force resultant is always the center point (centroid) of the distributed load.

For a triangular line load, it can be shown that the force resultant is one half of the peak value of the distributed load multiplied by the distance over which it acts. The location of the force resultant is two-thirds of the distance from the vertex to the peak value of the load.

A load of the type shown is said to be a composite load since it can be generated with a combination of simpler loads such as a uniform and a triangular line load.

Loads of this nature can be converted to force resultants by splitting the load into its composite parts, solving for the force resultant of each part, and then combining the forces into a force resultant for the entire load.

As an example, the diagram at the left can be splitting to a triangular
and rectangular distributed load. The two separate loadings can be combined
as,