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. >> Hi and welcome to module 28 of An

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

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

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

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

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

force reactions and moment reactions acting on a body.

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

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

balance of moments about any point we choose.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

. Okay, here's the good freebody diagram of

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

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

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

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

Because they were insignificant as compared to the applied forces and

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

diagram. So what do we do next?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

What other forces do we have acting on this system?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

next. .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Okay? And all of that has to equal 0.

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

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

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

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

units. There are 2 feet in the x direction.

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

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

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

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

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

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

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

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

equals zero. And then what do we do?

Think about it. You should remember from the last slide

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

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

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

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

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

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

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

the end. So we've got that.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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