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>> Hi, this is Module 21 of An Introduction to Engineering Mechanics.
So let's take a look at the forest here, what I've called the forest before.
This is the overview of the course, as opposed to the trees as we go through each
lesson. We've gotten quite a ways along in the
course. We've, we've done the representation of
forces in 2 and 3D. We've went through the syst-, the part of
the course on particle equilibrium, we've learned about the concept of moments and
how to follow the moment of force about a point in a line or an act a line or an
axis, we've also discuss couples we're in the section about equilibrium equations
and equivalent systems, we're working on resultants and, where they act looking at
distributed forces finding centroids and then method we all use last time in what
I'm going to do again today is another example.
Who have using the method of composite parts.
Because it's a really important skill for engineers to know.
And then we'll go into the last block of instruction, starting with the next
module. So, the learning outcome for today is to
solve another problem using the method of composite parts to find a resultant force
and couple on a, on a, on a body. And so here, here's the problem we're
going to work with we want to, we have a, a, cantilever beam here, it's fixed at the
left hand end, it's like a diving board if you will, here's the beam itself and I've
got some distances on there and I've got some sort of a load, distributed load,
starting with a 100 pounds per foot up here, wrapping down to 60 pounds per foot,
leveling off, and then going down to zero at the end.
And so, we want to find an equivalent resultant force, and a couple at point o
over here on the left. And so, what I'd first like you to do for
practice is to break this load into. Using the techniques that we learned last
module, breaking it into standard shapes, and finding the resultant force for those
standard shapes, and the location of those resultant force for each of those standard
shapes. And so take, take some time to do that and
then come on back. Okay.
Just as a hint, in case you couldn't get started, what I want to do is I want to
show you the standard shapes that you should break this in to.
So, if we draw a line here, and a line here, and a line here.
Those are the standard shapes you should work with.
You should work with this triangle and this rectangle, this rectangle and this
triangle. And so, if you didn't finish up now is a
good time to go ahead and finish and find the result in forces and their locations
for each of these standard shapes. And then come on back.
Okay, these are the results you should have come up with.
For the upper triangle here, its magnitude is the area for a triangle one half the
base times the height or 100 pounds. Its location is 1 3rd from the left-hand
side, or 2 3rds from the right-hand side, so it's 1.67 feet from point o, 1 3rd of
five feet. This rectangle has an area of 300 pounds.
It's resultant force is acting in the middle of this rectangular section, which
is. 2.5 feet.
This rectangle is a, a base of 60 and a height of 4, for 240 pounds.
Its location is halfway along this four feet section so it's 5 plus 2, or 7 feet
from the left, left hand edge. And finally the rectangle here is one half
base times height, or 120 pounds, going from the left, you go 5 feet, 4 feet.
And then one-third of four feet. And so, if that's the result you got,
you're in pretty good shape for breaking the system into composite parts, and
finding those resultants, and their locations.
So now what I want to do is I want to come up with a single equivalent force and
couple at point o. So this will be system two.
So I'll draw my cantilever beam again. And we know from earlier lessons, when we
found equivalent systems where we work with equivalent systems that the sum of
the forces on system 1 has to be equal to the sum of the forces on system 2, and so
in this case. We sum the forces on system 1.
That's going to be, and I'll, I'll call down positive, so I'm going to have 100
pounds, plus 300 pounds, plus 240 pounds, plus 120 pounds equals the Total sum of
the forces on system, two. And so that ends up equaling, 760 pounds,
and so that's one of our answers. That's the result in force.
At point O, and so that would act down over here at point O on system 2 and will
be equal to 760 pounds. Okay, lets now keep on going and find or
satisfy the condition for sum of the moments being equal on both system 1 and
system 2 So let's start out by drawing system 21 here again.
What we've done so far, this is system two And this is point O.
We found that the result in force at point o is 760 pounds down.
And now we have to satisfy the other condition, which is the sum of the moments
About a common point on both systems, so this is point o in this case, have to be
equal. And so let's go ahead and sum moments
about o on system one. And so we've get, 100 is and the, and the
distance to that perpendicular distance to that line of action of that force is 1.67,
so I've got 100 times 1.67 and then I've got 300 times 2.5 and then I've got plus
240. Perpendicular distance is 5 plus 2, or 7.
And plus 120 times the perpendicular distance to the line of action at that
force, which is 5, plus 4 is 9 plus 1.33, is 10.33.
And, all of that has to equal the moment about o on our second system.
So, the sum of the moments about O on system 2, if you do that math, is 3,836.
The units are. Foot, pounds.
And these individual resultants cause a clockwise rotation, so the moment of o or
the couple at o has to also be clockwise. And so that's my equivalent couple.
And I can draw it on the figure. Clockwise array, arrow and it's going to
be 3836 foot pounds. And so we've completed the problem, we've
replace the loading shown to the right with an equivalent resultant force and
couple at point O, and I'll see you next module.
Dr. Wayne Whiteman, PESenior Academic Professional
