Hi, welcome to Module 11 of Application in Engineering Mechanics.

Today we're going to apply the method of joints or

pins, and the method of sections to study 3D trusses.

Last time we looked at this tower crane, and we actually did an example.

We modeled it as a 2D structure.

But in reality, it's more of a three, it's actually a 3D structure.

So, we're going to learn how to make our models a

little bit more complex and analyze them in three dimensions.

And so I have a, a small model, or section, of this tower crane.

And this is the type of structure you'll be able

to analyze as a result of this series of modules.

Here's a space truss, very similar to the, the, the tripod.

we have an applied force at point D, down here, that's given.

we have a, a, a pin connection at C,

or what's also called a ball and socket joint.

And the only other nonzero reaction components are at points A and B as shown.

And we want to solve this.

We're going to do the piece of the time, in this module

we're just going to find the reactions at a, b, and c.

So my question to you is, what should you do first?

As always, we're going to draw a free body diagram of the entire structure.

Let's go ahead and do that together.

these forces are already shown at the bottom, and so the only thing we need

to put our forced reactions on is and, and possible moment reactions is at the pin.

Connection at C.

And so let's look more closely at this pin connection at C.

This is, this is a generic ball and socket joint, or a pin joint that I've got here.

what that does is it prevents motion in the X,

in the Y, and in the Z direction, linear motion.

But it doesn't prevent rotation about the X axis, it doesn't rot prevent

rotation about the Y axis and it doesn't rot prevent rotation about the Z axis.

So the only reactions that you get from this

ball and socket or pin joint are forces in the

X, Y and Z direction.

oops, this is Y, FCY. FCX and FCZ.

And so now I have a good free body diagram.

And what do you do next?

Think, take a minute, keep on going if you can,

without coming back, if you can continue to solve the problem.

Okay, what we're going to do is now apply the equations of equilibrium.

And what I'm going to do, is I'm going to

apply the sum of the moments about point C.

to get one of the equations of equilibrium.

So, we'll do that together.

now you're going to have to be more careful now and go back to the things you

learned in my earlier course in introduction to engineering

mechanics by doing this moment equation in three dimensions.

And so.

We're going to have to have none of these force reactions at c are

going to cause a moment, or a tendency to cause a moment about point C.

So, the first we're going to look at is FAX, and so we're going to have r from

c to a as a vector crossed with

FAX. Plus, now FAY is also going to cause

a rotation about point C, so we'll do that in three dimensions, or vectorially.

So we have ARCA crossed with FAY. And

then we have a force here at D, so

we're going to go plus r CD crossed with

F, which is given. And then finally, one more force

that's going to cause rotation about point C.

And that's plus r CB

crossed with FBx.

Equals 0. So rCA is, we go in the X direction 3

feet, in the Z direction minus 12, so that's

going to be 3i minus 12k.

Crossed with FAX. FAX is all in the X direction, so

it's going to be the magnitude of FAX in the i direction.

Plus r CA, again, is 3i minus 12 K,

crossed with FAY, which is all in the Y direction.

So that's FAY in the J. And I have RCD, so

plus RCD to go from R from C to D, walking from tail to

head as we did in my earlier course. We've got 3 in the X

direction, so 3 i. Plus 4j minus

12k crossed with

the given force of 100i minus 100k.

Plus, the last force is from r from C to B.

So that's nothing in the x direction, 4 in the j direction, and minus

12 in the k direction. Crossed with FBX, which is all

in the i direction, equals

zero. Now, if you do those cross products and

the mathematics. What you'll come out with for your

moment equation is 3FAY in the K direction,

plus 12 FAY in the i direction,

minus 12. FAX in the j

direction, plus 300j,

minus 400k, minus 400i,

minus 1200j. Minus

4 FBXk,

minus 12 FBXj.

Now, if you couldn't do those cross products and

come up with that result, again, I think you

need to go back and review, the skills that

we learned in my earlier course, introduction to engineering mechanics.

And so once we have that the question is what we, what do we do now?

And think about that.

'Kay, there's my equation the moment equation that I just came up with.

And what we're going to do, is we need

to solve for these reactions, by matching components.

And so, I've got i components on the left hand side, j components and k components.

I have no components on the right hand side

in the i, j, and k direction because that's zero.

So let's go ahead and match the, the i

components together.

And I get on the left-hand side I have 12

FAY. And I have minus 400

and equals zero. So what that tells

me is FAY, ends up equaling

33.3 or FAY as a vector

is 33.3 in the j direction, pounds.

So, that's one of my reactions that I want to solve for.

Now, I would like you to go ahead and do the j and

k to match the j and k components to come up with other solutions.

And come on back and see how you did. 'Kay, here's the j and k components.

You should be able to go through and check your work.

you've get two equations, two unknowns. You solve for F B x's minus 75, in the i

direction, and FA FAFB x's minus 75, and FA x ends up being 0.

And so here I've drawn those components that we came up with.

I came up with 33.3 together with you, here's the two that you came up with.

And so, our last three reaction forces we need to determine

are the x, y, and z and the way to do that is by using, instead

of a moment equation now, sum of the forces in the x y and z directions.

So, let's do some of the forces in the x together, and we get we

have FC x, for this one.

and then

we're going to have plus FB x. And the only other one we have is on this

given force F, there's 100 points in the i direction, so that's plus 100 equals 0.

I can substitute in for what I found, the value of FB x is being minus 75.

And a result is F cx equals 25. And then

why don't you go ahead and do the similar forces

in the y and z direction. And once you do that you

should get FC y is minus 33.3, FC z equals 100

and so now I also have the reaction forces at C.

when I put them all together, I, I get minus

25 in the i direction, minus 33 in the

j direction, plus 100 k pounds, all acting at C.

And we have now found all the reactions acting on this truss structure,

and we'll come bac Next module to do the rest of the problem.