What is a Distributed Load?
Typically when a body experiences a force it is not in the form of a point load, rather the load is distributed across and area of the object. For instance, the weight of a non-symmetric object is not distributed evenly across its area due to the non-symmetric distribution of mass. Fortunately, for the purpose of static analysis, distributed loads can be expressed as a mathematically equivalent point load, simplifying the solution process.
Equivalent Point Load
Distributed loads can be expressed as a singular equivalent point load applied at the center of mass for the distributed load. In 2D the center of mass (\overline{x}, \overline{y}) is defined by:
\displaystyle \overline{x}=\dfrac{1}{A}\int xdA
\displaystyle \overline{y}=\dfrac{1}{A}\int ydA
The magnitude of the equivalent point load for a given force distribution in the x direction, w(x) (N/m), is simply the sum of all infinitesimal forces.
\displaystyle F=\int w(x)dx
For a distributed force in the shape of a right triangle:
Triangle has a base of b, slope of m, therefore resulting in a height of h=mb.
Force distribution is w(x)=mx
Point of equivalent force:
\displaystyle \overline{x}=\dfrac{1}{A}\int xdA
\displaystyle \overline{x}=\dfrac{2}{mb^2}\int mx^2dx
\displaystyle \overline{x}=\dfrac{2mb^3}{3mb^2}
\displaystyle \overline{x}=\dfrac{2}{3}b
Which is the centroid of the given distributed load.
Magnitude of equivalent force:
\displaystyle F=\int mxdx
\displaystyle F=\dfrac{mb^2}{2}
Which is the “area” of the distributed load.