So far, we have been dealing with concentrated,
or point forces.
We depict a concentrated force using an arrow.
This arrow represents the magnitude and the
direction of the force, and we apply the force
at a specific point.
In real life, forces are not always concentrated
and not applied at a point.
Consider this bookcase.
Here the books on the shelf create a distributed
force along the shelf.
In many such instances, loads are distributed
along lines or over surfaces or over volumes.
We call these forces as distributed forces.
Look at this example.
Sandbags are placed on the beam creating a
distributed load over the beam’s entire
length.
The wind load on the highway signboard acts
on the entire surface.
Fluid pressure acts on the entire submerged
area.
Wind loads, water loads, people sitting in
a classroom and cars on a bridge are all examples
of distributed loads.
A distributed force is at any point is characterized
by its intensity and the direction.
So, there are two quantities, when we deal
with a distribution: a loading curve and the
intensity of the force.
The intensity of a force acting on a line
is defined by force per length unit.
This is expressed in Newton per meter in SI
the system, or pounds per foot in the US system.
Now the question is how do we deal with distributed
forces when solving problems.
As you can imagine, it is convenient to replace
the distributed load with an equivalent concentrated
force because we know very well how to handle
concentrated forces.
Fortunately, this is relatively an easy two-step
process.
First, to convert the distributed load to
an equivalent concentrated load, we determine
the total force.
The total force is the area under the loading
curve.
It is easy to find the area when we deal with
a uniform distribution like the one shown
here.
It is a rectangle.
Finding the area of a rectangle is something
we all know very well.
Here the area is defined by multiplying the
length of the beam and the intensity of the
force.
Second, we need to figure out the location,
where to apply this force.
It turns out the force must pass through the
centroid of the area under the curve.
That’s it.
Let us look at this example, where we have
a uniformly distributed load on the 2-meter
long beam.
The intensity of the force is 200 Newtons
per meter.
The area of under the loading curve is the
total force, and it is equal to 200 x 2 = 400
Newton.
The centroid of this area is right in the
middle of the rectangle, so the force is applied
at distance of 1 m.
Simple!
Isn’t it?
Let’s solve a few problems and get more
practice in the next video.
