Let's first look at this example:
for this pile of sand, we know that its total weight is distributed throughout the pile. 
You can tell it is made up of the individual weight of each sand grain, and even for an individual sand grain, its weight is still distributed 
through its volume, and is made up by the differential weights of each tiny particle inside.
But
it will be convenient if we can use a concentrated force, W, to replace the distributed weight. 
This force W must provide equivalent effect as the distributed weight. That is to say, 
W not only needs to equal to the resultant force of the distributed load, 
but it must be properly positioned as well in order to have the same moment with respect to any point as the distributed load. 
From experience, we know this force W needs to be placed at the weight center of the body.
In this video, we are going to learn how to use a concentrated force to represent a simple distributed load. 
Let’s look at this piece of board, resting on the ground.
It has a length of L and width of b, 
and its surface is defined as the x-y plane. 
There is a distributed load acting on this board. 
Along the x direction, the pressure is a function of position,
represented as p as a function of x. 
We know from physics that pressure is force over area,
and it has a unit of Newton per meter squared, 
or Pascal in SI unit system, 
and pound per inch squared,
or psi in US customary unit system. 
Along the y axis direction, 
the pressure follows a uniform distribution, 
and that is why it is called a simple distributed load.
This means the pressure profile is the same across the width of the board.
Therefore, 
we can summarize the pressure across the width of the board at a particular position.
Let w be p multiplied by b, and w is known as the load intensity,
which is force over length, 
and it has unit of Newton per meter in SI unit, and pound per inch in US customary unit system. 
Now we can reduce this into a two dimensional problem. 
w(x), the load intensity function, which appears just like the pressure function, 
now represents the distributed load acting on the board. 
Since w is force over length, the total force over a very small length dx at an arbitrary position x 
is dF equals to w times dx.
And we can also calculate the moment of dF about any point, 
but for convenience’s sake, let’s say point O, 
and dMo 
will be x times dF that equals to w x dx. 
Now to find the resultant force of the distributed load, we can integrate dF across the length of the board, and from calculus we know that 
integration of a function equals to the area under the curve. 
Also we can integrate dM to find the total moment caused by this distributed load to point O. 
Note that the negative sign again indicates clockwise rotational effect. 
Now, since we want to use the resultant force F_R to replace the distributed load, this resultant force must be placed at a particular 
location, say x bar, so that F_R creates the same moment about point O as the original distributed load. 
Therefore, M_RO the resultant moment, equals to negative x bar times F_R,
and x bar equals to the magnitude of M_Ro divided by F_R. 
Later we will learn that this force passes through the centroid, point C, of the area. 
Note that you can summarize the total moment of the distributed load about any other arbitrary point, 
but your calculation will still place the resultant force at the same location. 
Let’s look at two special cases. First, when the load intensity has a uniform distribution across 
the length of the member, in other words, w is a constant value, w_0. 
The resultant force equals to the area under the line, w_0 times L, 
and it locates at the center of the rectangle,
with x bar being half of L. 
Another situation is when the load intensity function w(x) follows a linear distribution, 
varying from w_0 to 0 across the length of the board. 
The resultant force equals to the area of the triangle, 
half of w_0 times L, 
and it locates at point C, the centroid, that is 1/3 distance from the base of the triangle, 
with x bar being 1/3 of L. 
In this example we need to replace the applied distributed load on this member with a single force acting on the AB segment.
Here we have a uniform distributed load that takes the shape of a rectangle,
and a linear distributed load that takes the shape of a triangle,
and another linear distributed load, again taking the shape of a triangle.
Because the shapes are regular we can easily determine their resultant forces by calculating the area under the curve. For the first 
uniform distributed load, its resultant force equals to the area of the rectangle,
80 times the length 1.8 m, and that is 144 newton. The location is right at the center of the rectangle.
For the next linear distributed load the resultant force equals to the area of the triangle,
which is ½ times
the height times the base, and that is 72 newton.
The location is at 1/3 location from the base.
And for the next linear distributed load, the same, the resultant force equals to the area of the triangle,
and again the location is at the 1/3 location from the base.
And now the distributed loads have been replaced by their respective concentrated forces, their locations have been specified, 
we can easily determine the total resultant force along the x direction,
along the y direction, and also the total moment calculated about point A.
The force components can be moved to a location x so that they create the same moment about point A.
Based on that information we can calculate x to be the magnitude of the moment divided by the magnitude of the vertical force component,
to be 1.27 m. We can also calculate the resultant force,
and now we have successfully replaced the original distributed load with a single equivalent force, with its location specified.
