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This is module eight of applications in engineering mechanics.

Today's learning outcome is to make use of

the method of sections to analyze truss structures.

We've learned the method of joints or pins, but today

we're going to go through and use the method of sections.

So here's our problem.

We're asked to determine the force in member GH here of the truss shown below.

And for the method of sections, we're going

to isolate now more than one joint.

by isolating more than one joint, it'll be a non-concurrent force

system, because all of the lines of actions of the forces won't

go through a single point, and so we'll have three Independent

equations of equilibrium and therefore be allowed to solve for three unknowns.

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obviously we want to cut through the member

of interest and we're only going to cut through it once here.

So we're going to go through member GH to see what the forces in, in it.

We're going to continue to cut through the

truss and isolate a complete portion of the truss.

But we're only going to cut through the members once and we're not

going to cut through the joints and so I'm going to go up here.

And cut around

like this.

Okay and so that will allow me to say that using

the principles we've, we've talked about in earlier lesson, the entire

free-body diagram or the entire body is in equilibrium plus just

this section of the truss also has to be in equilibrium.

And so, we're going to go ahead and draw that free-body diagram.

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And so here is my section

that I'm interested in. I've cut through this member,

G H. I've cut through this member, D H.

And I've cut through this member E

D. And so I am going to assume those

cuts to be in tension, if they end

up being negative then I will known

those numbers were actually in

compression, so I've got F GH,

F DG and F DE. Plus I have my roller reaction down

here, externally applied to this section. And the roller reaction just gives me a

force in the Y direction, which I'll assume to be up.

It's F sub y.

So that's a free body diagram of the section I'm interested in.

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You'll notice in my, my rules down here, I will

only solve for the external reactions if they're within the cut.

And so in this case, we do have the external reaction F sub y, within the cut.

So I'm going to go ahead and solve for it.

And the way we do that is

by looking at a free-body diagram of the entire truss itself.

And so, in addition to this Fy support for

the entire body, we also have a pin reaction

over here the entire truss has a pin reaction over here A sub x and A sub y,

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There is my, entire free body diagram with,

for the entire truss, and my question to you

is, what equation should you use to solve for

Fy which is going to be within this sectional cut?

And so think about that for a second and come on back.

'Kay, what you should have found or stated was let's go

ahead and just use these sum of the moments about A.

By doing that I have only, I only have one unknown, F sub

y, and be able to calculate that reaction, that's within the sectional cut.

So I'd like you to do that on your own, when you've finished and solved for

Fy come back and we'll, we'll do it

together, make sure you got the correct answer.

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Okay here's the, the calculation for F sub y, I chose

counter-clockwise arbitrarily as my sign convention for assembling my equation.

I have this one kilonewton force here.

It's going to cause a clockwise rotation about A,

therefore it's negative in accordance with my sign convention.

So I have minus 1 times its moment arm The perpendicular

distance between the line of action of that force, and the

moment, point about which we're taking the moment, which is three.

You keep doing that for the 1.5 K, kilo-Newton force.

The three Kilo-Newton force, the two Kilo-Newton force in F sub y, when

you solve the equation, you find that F sub y is 3.92 Kilo-Newtons up.

So now we can go back to our sectional cut.

because we've taken care of one

of the unknowns.

And so here's my section free body diagram, again.

That I've drawn on the first slide.

But now, instead of having four unknowns, where we had F sub DE, F

DG, F GH, and R, we now know what this RFY is.

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and so I do also have a a, an exploded or an blown up version of

the cut itself, so you can see where I cut through

the members and what the internal forces within those members look like.

Let me help you in, in visualizing that, the situation.

So the question becomes. We wanted to find the force in GH.

And so here's FGH.

And so what equation should you use to, to, to solve for FGH?

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You should've said, let's go ahead and sum moments abut point D.

we could sum forces in the x direction, but then we

would have the unknown FGH, and the unknown component of FDE.

If we sum forces in the y direction, it wouldn't have helped us with FGH, and so

we can just one equation of equilibrium, by summing moments about point D.

We'll solve directly

for the force in member GH.

And that's because FDG and FDE, the other two unknowns,

have their line of action that go through point D.

And one other thing I'd like you to note is, we can

sum moments anywhere on the body or what we call the extended body.

So it could be off the body.

And so in this case our section is down here.

Point D is off the body but we're perfectly fine in summing moments.

It still has

to be equal to zero for the body to

be in equilibrium or this section to be in equilibrium.

So we'll go ahead and do this together.

If I sum moments about D, I've chosen counter-clockwise as positive.

I've got FGH times its moment arm. and it's going to be negative

in accordance with my sign convention. So I've got minus FGH times 4 and then

I've got the 3 kilo newton force which is also going to

cause a clockwise rotation about G, so that's going to

be minus 3 times its moment arm which is 1.5.

And the only other force that's going to cause a

rotation about point D is the 3.92 force reaction

at at F and so if I look at that, its going to

cause a counter clockwise rotation that's positive

in accordance with my sign conventions, so it's

plus 3.92 times its moment arm which is 4.5

equals 0. If you calculate that math you'll

find FGH equals 3.29. It's positive, I've assumed that this

member is in tension, so therefore since it's positive, it is indeed in tension.

So I get FGH equals 3.29, the

units are kilo-Newtons. And it's

in tension, so I'll put T in, in brackets.