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Hi and welcome to module 23 of an introduction to engineering mechanics.

Â For today, we're going to learn, or we're going to recall actually the 2 dimensional

Â and 3 dimensional static equilibrium equations that we learned earlier.

Â And then we're going to employ rules for drawing a free body diagram, which we've

Â done before, but we're going to do again. And we're going to sketch a free body

Â diagram for a body. So first of all on the static equilibrium

Â equations as a review. This is from lesson, or module 13.

Â And so if you don't remember this, go ahead and review module 13 again.

Â But we said that we would have a static equilibrium condition if there was a

Â balance of forces acting on a body and a balance of moments acting about any point

Â on the body and that's vectorially. In scalar form we, we had the two

Â dimensional and the three dimensional equations for static equilibrium.

Â And again, if you don't recall those I'd like you go back to module 13 and, and,

Â and review that one more time. So, we're going to learn about a free body

Â diagram, more about a free body diagram. We've used it earlier in the course.

Â We're going to. This is a really important tool.

Â This is a tool that engineers will use throughout their career.

Â It's a sketch which represents a body isolated from its surrounding, and it

Â shows all of the forces and moments on that body.

Â And I refer to it as an equilibrium tool. It's a very, very important graphical tool

Â that you must learn to be a successful engineer.

Â So, here the, the free body diagram procedures I've, I've listed them here but

Â the easiest way to learn how to do the, the procedures are to go through an

Â example problem. So, here is an example, this is a great

Â part of the course because now we get to start looking at.

Â Some real world engineering examples. And so, this is a beam, typical beam.

Â Might be in a building, might be in a mechanical structure, might be in an

Â aerospace structure. Beams are everywhere but, this beam has a

Â couple of constraints on it. On the right-hand side.

Â It has what's called a pin reaction. Here, six inches from the left-hand side

Â it has what's called a roller reaction. And on the right hand, or on the left-hand

Â side here we have a five pound external force that's acting down.

Â And so I have a, a model of this. Engineering system over here.

Â So here, here's my beam. I've got a pin reaction here.

Â And we'll talk more about that pin reaction.

Â I've got a roller reaction here. And then I've got a weight acting down on

Â the, on the left-hand side. Now in this case it's not five pounds.

Â But it's a scale model, so you get the idea.

Â So, that's, that's the physical model. So, we want to go ahead now and draw the

Â free body diagram. And so, I'm going to, first of all

Â identify the body of interest. And in this case that's going to be the

Â beam, okay? And then, I'm going to sketch that body

Â free of constraints or separated from the rest of the world.

Â And so, here's my beam. [inaudible] completed step two.

Â Now I want to apply any external fo-, forces or moments acting on the body.

Â In this case I don't have any external moments acting on the body, but I do have

Â an external force over here on the left-hand side which is five pounds.

Â So I can replace that with a force acting down on the left hand side which is 5

Â pounds. The other external force which is the body

Â force for the beam would be its weight and that would act right in the center here.

Â I'm going to push that down and I'll just call it weight, it would be known, it

Â would be given and so we've completed step 3.

Â The next step is to replace the constraints with forced reactions and

Â moment reactions. And we have two constraints on this body.

Â We have this pen reaction and, or pen constraint and we have this roller

Â constraint. So, let's look at those on my scale model.

Â So, first of all I'm, I'm taking off the five pound weight.

Â I've replaced that with a force and I'm going to look at the pin reaction now.

Â This pin reaction alone, it prevents motion left or right, it promotes,

Â prevents motion up or down, but it doesn't prevent rotational motion.

Â So there's no, what we call moment reaction there, it only constrains motion

Â in two orthogonal directions. See, if I remove the pin I can now move

Â left or right. I can move up and down.

Â In fact, I can move in, in any direction linearly.

Â And so, that pin reaction can be replaced on my free body diagram with two, pin

Â constraint can be replaced on my free body diagram with two force reactions

Â orthogonal to each other. And so let's go ahead and label these

Â points. I'm going to call this point a, and this

Â point b. And, on my diagram here, this'll be point

Â a, and this will be point b. And so on the pin reaction, when I pin

Â restraint when I remove that constraint I replace it with 2 force reactions,

Â orthogonal to each other. I'm going to replace it with b so y in the

Â y direction and b sub x in the x direction.

Â And as long as they are orthogonal 90 degrees to each other.

Â I can draw them in any direction. I could have drawn one this direction and

Â one this direction. They have to just cover the entire 2D

Â space. Okay, the only other constraint we have is

Â this roller constraint, which is at point a, and so let's go see what kind of force

Â reactions we get from a roller constraint. So, now I've, I've removed the five pound

Â force from the left hand side. I've removed the pin constraint.

Â All that's left is the roller constraint. And you'll see that the roller doesn't

Â prevent motion left or right, it doesn't prevent rotation, so there's no moment

Â reaction that stops rotation. The only thing it does is it stops motion

Â down. And an ideal roller would not only stop

Â motion down, it would also stop motion up. You can think of an ideal roller as the,

Â the roller mechanism that's on the top of the closet doors perhaps in your home,

Â where it rolls along the track and it, and it can't go up or down.

Â So there's only one direction that is constrained from motion and so on my free

Â body diagram over here, I would draw just a force reaction in the y direction.

Â I'll go ahead and put a sub y. So that's complete step four.

Â The last step we have to do is add dimensions.

Â That's relatively straight forward. We have six inches from the left to point

Â a, the weight is in the center since the entire beam is 23 inches long.

Â That would be 11 and a half inches to the, to the weight.

Â And then finally we have 17 inches on the right-hand side to the, back to the roller

Â reaction. So this is 17 inches.

Â And so that's how we draw a free body diagram.

Â We've identified the body of interest of being.

Â We've sketched the body free of constraints.

Â We applied external forces. There was no moments in this case.

Â The external forces being the five pound weight, and the weight of the body itself.

Â We replaced the pen constraint and the roller constraint with forced reactions

Â and we added dimensions. And so, that's, that's it.

Â See you next module.

Â