Let’s talk about simple distributed loads. As a very general and simple example, imagine you’re
in your room and you see your book shelf. Here, we have our very expensive textbooks all lined up
nicely. If we were to figure out a moment created about a point on this shelf or a force at a
certain point, it’d be hard to do since the weight created by the books is distributed across the
whole shelf. Hence the name, a distributed load. So what we need to do is be able to represent a
distributed load as a resultant force. Removing the books, we see a graph which represents the
load. We can represent this load with an equation, we will label it W. To find the resultant force,
all we need to do is find the area under the graph, in other words, integrate it.
The lower bound is where the distributed load starts from, and the upper bound is where it ends.
Finding just the resultant force isn’t enough, we also need to know where to place it at. To do
that, we use this equation. The x bar tells us the geometric center of the area under the distributed
load, which is where we place our resultant force. Here, x represents the variable in our equation,
this part is the equation of our curve, and the bottom is the same integral we used before,
in other words, it’s the area under the graph. Now If the distributed load is simple, for example,
made of rectangles and triangles, you don’t have to integrate, instead, simply find the
area under the graph by multiplying length by height, or using the appropriate equation for an
area of a triangle. We then place the individual resultant forces at the centroid of each piece.
For a rectangle, that’s simply the midpoint. For a right-angle triangle,
it’s 1/3 the length of the base. If the question asks for the resultant force at
a certain point, for example, at point A, then we just add up all the individual resultant forces.
The moments about that point are also found by multiplying each individual resultant force
by the perpendicular distance. Sometimes however, we need to place our resultant force at a location
so that it accounts for the moment as well. For example, let’s assume our individual resultant
forces created a total moment of 20 N.m about point A and our resultant force was 4 n.
Then all we need to do is place our resultant force 5 m away from point A,
which would then account for the total moment since it’s force times perpendicular distance.
You will understand this much better with examples so let’s get started. The first 2 will be simple
shapes, and the last 2 will be with integrals. Let’s take a look at this problem where we have
to find an equivalent resultant force, and find it’s location measured from point O. Notice how
the 4kn/m load is a square, and the 6kn/m load is a triangle. Since these are basic shapes,
we don’t need to integrate to find our answer. We just need to figure out the area of each load.
Let’s start with the square shaped load. All we do is multiply the length by the load. We place
this force at the midpoint of its length, which is 1 m away from point O. Next, we look at the
triangle piece. So remember, we find the area of a triangle by multiplying ½ times base times height.
To place this force, we need to figure out the centroid location, which would be 1/3 the length
of the base, so that’s 0.5 m. The total distance from point O would then be 2 m plus 0.5, giving us
2.5 m. The total resultant force for both pieces is simply the addition of the two values we found.
Now we need a place to put this resultant force. Since we aren’t going to show the resultant
force at point O, we actually need to place it somewhere that accounts for each individual moment
created by the individual resultant forces. To do that, we need to sum the moments about
point O. Let’s assume clockwise moments to be positive. So for the rectangle, we have force F1,
which is the force we found for the rectangular piece, times the distance from point O. Next,
we have the triangle resultant force, and the distance from point O is 2.5 m. The total moment
created by the two forces would be the addition of these 2 values. So again, the 2 individual
resultant forces would create a total moment of 19.25 kNm about point O. Since we are going to
show a single resultant force, we have to make sure our resultant force is placed at a certain
location that would give us the same exact moment as did the individual resultant forces.
So where can we place it? Well, let’s assume that we place the resultant force a distance of d
away from point O. To find it, all we need to do is equal the resultant force
times d to the total moment. This value we just found tells us that if we place our resultant
force 1.54 m away from point O, it’ll create the same moment as did the other 2 forces. So now,
we represented the distributed load as a single resultant force which accounts for all the forces.
Let’s take a look at this problem where we have to find an equivalent resultant force
and a couple moment about point A. So unlike the previous question, we are going to show both the
resultant force and the moment at point A. Now if we look at this distributed load, we can
see that it’s actually made up of a triangular piece and a rectangular piece. Let’s start with
the triangular piece. This part is 1kN/m, and the length is 1.8 m. The location of this force would
be at 1/3 the length of the base. Next, we have the bottom rectangular piece. The location of
this force would be at the midpoint of the length. Next, we have the area of this rectangle. Again,
it would be placed at the midpoint of the length. Now to find the equivalent
resultant force of all of these forces, we just need to add it all up. But notice how
the 2 forces here is at an angle of 60 degrees. That means we need to break them into components.
For force 2, it only has a y component. Now let’s add up the x components and the y components.
We just found the components of our resultant force, so we have to find the magnitude now. Just
the magnitude isn’t enough, we also need to show an angle. So lets draw out the components of our
resultant force and use the inverse of tan to find it. We found our resultant force, so now we need
to find an equivalent couple moment about point A. We will assume clockwise moments to be positive.
Let’s start with the triangle force. The perpendicular distance from point A is 0.6m. Next,
we have the moment for the rectangular piece. Lastly, we have the moment created at the top.
Note the perpendicular distance for this force, is the cosine of the slanted piece plus the 0.6 m.
The resultant moment is the addition of all of these values. With this,
we can replace all the forces with a single resultant force and couple moment at point A.
Now we are going to look at an example where the distributed load is shown to us with an equation.
In a problem like this, the resultant force is the integral of the equation given to us.
The lower bound would be where the distributed force starts, which is at 0m,
and the upper bound is where it ends, and that’s 3 m. Solving the integral gives us the resultant
force, and since the distributed force was downwards, our resultant force is also downwards.
Now need to find where our resultant force is going to be placed, so we have to use
the equation I mentioned before. On top, we have x times the equation given for the distributed load.
The upper and lower bounds are the same. On the bottom, we have the same integral as before, which
is the resultant force. Solving gives us 2.4 m. So this value we got is measured from the w axis,
in other words, the left side, and that’s where we place our resultant force. The question asks us to
find the moment about point O, which means we need the distance from point O to the resultant force,
which is 0.6 m. Now we can find the moment by multiplying the distance by the resultant force.
If our bar was able to move freely, then the resultant force would create
a counter clockwise movement, so our moment is counter clockwise. This tells us the moment
created about point O from the distributed load. Let’s take a look at one last example where we
need to represent the distributed load as a single resultant force. So as with the previous question,
we start off with an integral to find the resultant force. The lower bound starts at 0
m and the upper bound is 6 m. Solving gives us the resultant force which is downwards
since our distributed load was also downwards. Next, we need to find the location where we
place our resultant force. For that, we use this equation. Solving gives us our answer.
This answer tells us where to place our resultant force which would represent the distributed load.
That should cover the types of problems you will face when it comes to simple distributed
loads. I hope this video helped and if it did, please consider sharing it with your friends and
classmates, they too might find it helpful. Thanks for watching and best of luck with your studies.