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. >> Hi and welcome to module 28 of An

Â Introduction to Engineering Mechanics. We're getting close to the end of the

Â course so I thought I'd, wear a nice bright red tie to celebrate the conclusion

Â of the course. Here's today's, learning outcomes.

Â First we're going to recall the 3D static equilibrium equations.

Â And then we're going to apply those equilibrium equations to solve for the

Â force reactions and moment reactions acting on a body.

Â And so if you go back to earlier modules, these were the static equilibrium

Â equations in vector form. We had to have a balance of forces and a

Â balance of moments about any point we choose.

Â And I also show the scalar form down there, because in 2D, as we've solved

Â problems in the last few modules, we get three independent equations.

Â Now, in 3D, we're going to have six independent equations.

Â So this is the problem we're going to look at to solve a 3D equilibrium example.

Â This is a real world piping system. In fact, this is a piping system that

Â I've, I've taken a picture of in one of our undergraduate teaching labs here at

Â Georgia Tech. And so I've physically modeled a portion

Â of that piping system here, here on the, the left.

Â And we need to determine the force and moment reactions that this fixed support

Â at the bottom for the physical model of the of, of this piping system or a portion

Â of the piping system. There is a 10 pound force here.

Â And a five pound force here that are both parallel to the, the y-axis, this being

Â the y-axis. And there's an applied moment of 20

Â foot-pounds, applied right here and that's due to a wrench, a wrench kind of turning

Â that thing. And so, my question to you is that's the

Â physical model of this real world system. What do we do first?

Â And you should know by now what you'd do first.

Â Take a minute, think about it, write it down.

Â Okay, what you should first do is, is, is always draw a free body diagram.

Â And so, go ahead and draw that free body diagram for this system.

Â And after you've got that complete I want you to come back and see how well you did.

Â . Okay, here's the good freebody diagram of

Â the system. In this case our body of interest is the

Â portion of the pipe that we're studying. We had the five pound force here.

Â We have a 10 pound force here. We have that 20 pound, foot pound couple.

Â I went ahead and made the assumption that I neglected the, the weights of the pipe.

Â Because they were insignificant as compared to the applied forces and

Â moments. Now, the, the, the big thing that you

Â should have, worked on, is the force and moment reactions at the fixed support here

Â down at the bottom. And so let's look over here.

Â This is the bottom portion of the pipe. This pipe would be firmly attached to, to

Â the ground, okay? And so, it doesn't allow, in fact, let me

Â call up my coordinate system here. It would not allow any motion in the x or

Â the y or the z directions. So we're going to have force reactions in

Â the x, y and z directions. It doesn't allow any moment about the

Â x-axis. It doesn't allow any moment about the

Â y-axis. And it doesn't allow any moment about the,

Â the z-axis as well. So it, it constrains that.

Â And so we're going to have all three force reactions and all, all three moment

Â reactions acting at the bottom of, of this pipe and we're going to call that point A.

Â Okay, so that's what I've drawn here on my, on my free body diagram.

Â I have my three force reactions and we're going to call this point A.

Â And three moment reactions, and I can draw the moment reactions with a double arrow

Â pointed in the positive directions. And I'm using the right hand rule.

Â Let's say that this 10 pound force is applied at a point B and this 5 pound

Â force is applied at point C. And so there you have a good free body

Â diagram. So what do we do next?

Â In fact, you should think about that and, and you can even get started and go as far

Â as you can. And then when you can't get any further

Â come on back and see how you're doing. .

Â Okay, now we want to determine the force and moment reactions at the fixed support.

Â And so what we have to do is, apply the equations of equilibrium.

Â And so we have two, we're going to work vectorally in 3D, rather than scalar form

Â like we did in 2D. And so the first equilibrium equation

Â we're going to work with is the sum of the forces, factorally equals 0.

Â And so let's look and see what we have here.

Â So we've got the A x force reaction down at the support.

Â It's in the positive x direction, so that's A x i.

Â And they we have the A y force reaction, so it's plus A y J.

Â And then we have the A z force reaction, plus A z K.

Â I'm going to go ahead and label those points again here.

Â So this is point A, this is point B, this is point C.

Â What other forces do we have acting on this system?

Â Well we have this ten pound force, and we said it was acting in the y direction.

Â That's positive y direction, so we have plus 10 J.

Â And we also had this five pound force acting along the y-axis, but it's in the

Â negative y direction. And so this is going to be minus 5J equals

Â 0. And so what we can do at this point is we

Â can equate components. We can say that the I components all equal

Â 0, the J components all equal 0 and the K components all equal 0, so we're going to

Â equate components of that equation. And so we get A x in the I direction, and

Â there's nothing else in the I direction, so Ax must be equal to 0 in this case.

Â And A y in the j direction, so we have A y let's see.

Â That's in the K direction, plus 10 in the j direction, minus 5 in the j direction

Â equals 0. And so, that leads us to A y equals minus

Â 5. And so what that's telling me is, I've

Â drawn A y in the positive y direction. Since I got a negative value, A y must be

Â in the negative j direction. So A y, vectorally, is equal to minus 5j

Â pounds. And so now we've got 2 of those force

Â reactions. A x, we said was equal to 0.

Â A y now is minus 5j pounds. And finally, we equate the K components.

Â X at I component, J comp, K component, A z.

Â There is no other K component so that's also equal to 0.

Â And so now we've solved for the three force reactions.

Â All we have to do now is solve for the moment reactions and so we'll do that

Â next. .

Â And for the moment reactions, we use the balance of motion factorially so we have

Â sum of the moments, equals 0. Let's see.

Â So let's look, use these moment reactions here at the base of the, of the pipe.

Â Okay, we have a moment reaction around the x, y and the z axis.

Â So we have moment, in fact let me do my points here again.

Â This is point A, this is point B, this is point C.

Â And we've got the moment A sub x in the I direction plus the moment A sub y in the J

Â direction plus the moment A sub z in the K direction.

Â We have this 20 pound applied moment due to the wrench.

Â And so, if you come over here and you look, this 20 pound forces is going in

Â this direction. And so that means it's poitning down with,

Â by the right hand rule, so it's in the minus k direction.

Â And so we've got minus 20K. And then we have a couple of applied

Â forces that are going to cause a moment about point A.

Â In fact, one thing I should have said when I sum moments was I, I, I intuitively

Â said, okay, I'm going to sum moments about point A.

Â But I really should specify and say that I'm summing moments about point A.

Â Because I could sum moments about B or C. And I could solve the problem that way as

Â well, and you may want to try that. So but let's keep on going here.

Â We've got, minus 20. And then, we had this 10 pound force

Â acting at point B. So, remember, in 3D, think about it, and

Â how do we define the moment in 3D due to a force?

Â Eh, you may want to go back and review earlier modules.

Â But after you've thought about it what you should remember is that you do R cross F,

Â factorially. And so this is going to be plus r from

Â point A to B, crossed with the force itself, which in this case is 10j.

Â And then we have one more force, which is at point C.

Â So we have plus r from A to C, crossed with that force which is minus 5 j.

Â Okay? And all of that has to equal 0.

Â Finally, next we need to put in what r A B and r A C is.

Â R A B is just the position vector from A to B.

Â And we see that we go three units in the z direction, or k direction, so we're

Â putting 3k here. And for AC, we've got, position vector two

Â units. There are 2 feet in the x direction.

Â And 9 plus 1.5 is 10.5 in the K direction. So RAC is 2 i plus 10.5 k.

Â Okay, now we can do those, those cross products.

Â We've got m A x i plus m A y j plus m A z k minus 20K.

Â And then 3K cross 10j K, K cross j is minus i, so that's going to be minus 30i.

Â And then we've got 2i cross minus 5j. I cross j is K, so that's going to be

Â minus 10 K. And then we've got 10.5k crossed with j.

Â K cross j is minus i. But we've got 10.5 times minus 5.

Â So a minus a minus is going to be a plus. And so that's going to be plus 52.5 i

Â equals zero. And then what do we do?

Â Think about it. You should remember from the last slide

Â how, what we do next. What we should do is we should go ahead

Â and equate the components, equate I J and K components.

Â . And so, for the i components, we've got m

Â A x, m A sub x. And we've got minus 30.

Â And plus 52.5 equals 0. So m A x should be minus 22.5, right?

Â If i carry this to the other side that's a negative 52 plus.

Â So that's m A x is minus 22.5. The units are well let's do the units at

Â the end. So we've got that.

Â Okay, and then we're going to equate the y components.

Â So I've got m at point A about the y-axis. What else do we have?

Â We don't have any other moments about the y axis, so that's just going to equal

Â zero. And then finally, we've got, equating the

Â components for the z axis, we've got m about point A around the z axis, minus 20.

Â . Yeah minus 20, and let's see, minus 10

Â equals zero. And so, m A sub z is equal to 30.

Â And so, now we can write our moment about point A, vectorally.

Â And we get a moment about A is equal to, we've got minus 22.5 in the i direction

Â plus 30 in the k direction, and the units are foot pounds.

Â And we've now solved for all of the force reactions and all of the moment reactions

Â for this piping system to keep it in static equilibrium, so that it doesn't

Â fall over, go anywhere. So that's, that's the, the reactions that

Â we need to have this piping system stay in place as shown.

Â And so that's it for today's module.

Â