# Equivalent Point Load

An equivalent point load is a **single point force** that will
have the **same effect** on a body as the original loading condition,
which is usually a distributed force. The equivalent point load should
always cause the same linear acceleration and angular acceleration as
the original force it is equivalent to (or cause the same reaction
forces if the body is constrained). Finding the equivalent point load
for a distributed force
often helps simplify the analysis of a system by removing the integrals
from the equations of equilibrium or equations of motion in later
analysis.

## Finding the Equivalent Point Load

When finding the equivalent point load we need to find the magnitude, direction, and point of application of a single force that is equivalent to the distributed force we are given. In this course we will only deal with distributed forces with a uniform direction, in which case the direction of the equivalent point load will match the uniform direction of the distributed force. This leaves the magnitude and the point of application to be found. There are two options available to find these values:

- We can find the magnitude and the point of application of the
equivalent point load
**via integration**of the force functions. - We can use the
**area/volume and the centroid/center of volume**of the area or volume under the force function.

The first method is more flexible, allowing us to find the equivalent point load for any force function that we can make a mathematical formula for (assuming we have the skill in calculus to integrate that function). The second method is usually faster, assuming that we can look up the values for the area or volume under the force curve and the values for the centroid or center of volume for the area under the curve.

### Using Integration in 2D Surface Force Problems:

Finding the equivalent point load via integration always begins by
determining the mathematical formula that is the **force function**.
The force function mathematically relates the magnitude of the force (F)
to the position (x). In this case the force is acting along a single
line, so the position can be entirely determined by knowing the x
coordinate, but in later problems we may also need to relate the
magnitude of the force to the y and z coordinates. In our example to the
left, we can relate magnitude of the force to the position by stating
that the magnitude of the force at any point in Newtons per meter is
equal to the x position in meters plus one.

The magnitude of the equivalent point load will be equal to the area under the force function. This will be the integral of the force function over it's entire length (in this case from x = 0 to x = 2).

Now that we have the of the magnitude of the equivalent point load
such that it matches the magnitude of the original force, we need to
adjust the position (xeq) such that it would cause the same **
moment** as the original distributed force. The moment of the
distributed force will be the integral of the force function (F(x))
times the moment arm about the origin (x). The moment of the equivalent
point load will be equal to the magnitude of the equivalent point load
that we just found times the moment arm for the equivalent point load (xeq).
If we set these two things equal to one another and then solve for the
position of the equivalent point load (xeq) we are left with the
following equation.

Now that we have the magnitude, direction, and position of the equivalent point load, we can draw the point load in on our original diagram. This point force can be used in place of the distributed force in further analysis.

### Using the Area and Centroid in 2D Surface Force Problems:

As an alternative to using integration, we can use the area under the force curve and the centroid of the area under the force curve to find the equivalent point load's magnitude and point of application respectively.

The **magnitude** (Feq) of the equivalent point load
will be equal to the **area under the force function**. We can find
this area using calculus, but there are often easier geometry based ways
of finding the area under the area.

The equivalent point load will also **travel through centroid
of the of the area under the force function**. This allows us to
find the value for xeq. The centroid for many common shapes can be
looked up in tables, and the parallel axis theorem can be used to
determine the centroid of more complex shapes (see the centroid page for
more details).

### Using Integration in 3D Surface Force Problems:

With surface force in a 3D problem, the force is distributed over a surface, rather than along a single line. To find the magnitude of the equivalent point load we will again start by finding the mathematical equation for the force function. Because the force is distributed over an area rather than just a line, the magnitude of the force may be related to both the x and the y coordinate, rather than just the x coordinate as before.

The magnitude of the equivalent point load (Feq) will be equal to the volume under the force curve. To calculate this value we will integrate the force function over the area that the force is applied to. To integrate this function F(x,y) in terms of the area, we will furthermore need to break the integral down and integrate over x and then integrate over y.

Once we solve for the magnitude of the equivalent point load, we can then solve position of the equivalent point load. Since the force is spread over a surface, we will need to calculate both the x (xeq) and the y (yeq) coordinates for the position. The process for solving for these values is similar to what was done with only an x value, except we change the moment arm value to match the equivalent point load coordinate we are looking for.

In each of the equations above, we will need to expand out the area integral into x and y integrals (as we did for Feq) in order to be able to solve them.

### Using Volume and Center of Volume in 3D Surface Force Problems:

Just as in the 2D problems, there are some available shortcuts to
finding the equivalent point load in 3D surface force problems. For a
force spread over an area, the **magnitude** (Feq) of the
equivalent point load will be equal to the **volume under the
force function**. The equivalent point load will also **
travel through the center of volume of the volume under the force
function**. This should allow you to determine both xeq and yeq.

The center of volume for a shape will be the same as the center of mass for a shape if the shape is assumed to have uniform density. It should be possible to look these values up for common shapes in a table. Again the parallel axis theorem can be used to find the center of volume for more complex shapes (See the Center of Mass page for more details).

### Using Integration in Body Force Problems:

When we jump to body forces, the magnitude of our force will vary can vary with x, y, and z coordinates. This means that our force function can include all of these variables (F(x,y,z)). To find the magnitude of the equivalent point load we integrate over the volume, breaking the volume integral down into x, y, and then z integrals.

To find the point of application of the equivalent point load, we will need to find all three coordinate positions. To do this we will expand out the equations we used with two coordinates to include the third coordinate (zeq).

Again you can expand the volume integrals out here like you have done for the magnitude of the equivalent point load.