Continuing our discussion of trusses,

in the last section we saw how to analyze simple trusses by the method of joints.

In this section I want to talk about analysis by the method of sections.

Now, the method of sections, we divide the truss into two distinct sections or parts.

Then we replace the cut bars by their external force, by their internal forces,

which then become external forces on the free body diagram of the separate segment.

And the basic principle is that if the structure as a whole is in equilibrium,

which it is Then each of the separate subsections must also be an equilibrium.

So to illustrate this, let's consider the problem that we looked at in

the last segment and let's suppose that we have this simple truss here and

we're interested in finding the force in this central member here.

Member BE.

So, how would we go about doing that if we analyze this by the method of joints?

Well, the first method would be to find the reactions.

In other words the vertical reaction forces,

let's say R1 and R2 here By the usual methods of statics by

applying a free body diagram to the whole truss.

Then we would proceed node by node.

So in this case we would start at node A here where

we note that we have two unknowns.

The forces in the member AF and the force in the member AB.

Then, we go to Node F.

So we already know the force here.

So then we can find the other two unknown forces in FE and FB.

Finally, we go to Node B, this Node right here.

And now we know this force here and this force, so we can finally solve for

the force that we're looking at, the force in BE, in other words,

a fairly lengthy procedure to find the force in that one particular member.

Now let's suppose that we apply the method of sections to this.

So, in the method of sections, I'm going to make a cut through here,

which cuts through the bar whose force we are trying to find, and

then separate it into two sections.

So, heres the free body diagram, of the entire structure, with a cut.

Then I'm going to split those two sections apart Split the truss into two parts,

separated by that cut.

So here it is with the two segments separated, and

here is the free body diagram of the left hand section there.

So now I still show the external reactions here, R1,

and the external load, L, but now the forces in the members.

Have been replaced.

Those internal forces now become external forces to the free body diagram, as shown.

And the basic principal here is that entire subsection must

itself be an equilibrium.

So to apply that, for example.

I can apply equilibrium of that structure And we note, by the way,

that we have three unknowns here, EF, BE, and BC, but, generally speaking,

we have three equilibrium equations, so therefore we can solve it.

It's statically determinant.

But in this case we're only looking for the force in the member BE So

the easiest way to do this is simply sum the forces in the vertical direction.

Sum F y is equal to 0, and the forces that we have are R1, L, and

the vertical component of the force in BE.

And of course there is no vertical component of EF and BC,

because those are horizontal members So, therefore our equation is this,

R1 minus L plus BE sign theta is equal to 0.

So rearranging for the force that we are looking for

BE is L minus R1 over sign theta, we know L and

presumably we've already calculated R1,

therefore we can calculate the force in the member So

therefore this is a much quicker way to find the force in

some particular internal member in certain cases.

Here's an example.

We have a truss here which is 4 meters tall and

consists of five panels horizontally, each one 3 meters long And

it's loaded by two loads of ten kilonewtons and ten kilonewtons as shown.

So we're going to do a couple of questions related to this.

First we want to find the magnitude of the force in the member CG,

which is this one right here.

And these are the possible answers.

So to analyze this, I'm going to cut this and I'll make my cut through

the center here, cutting through the member that we're interested in.

And here is the free-body diagram of that structure.

So, the reaction forces here, I think you can see by symmetry,

that the upward reaction force at the support A is going to be 10 kN.

And there's no horizontal component of force there,

because there are no forces acting on this structure.

So here is our free body diagram.

With our internal forces in the members replaced by external forces,

where as usual, I'm assuming that they're all intention,

in other words pointing away from the node points.

So, we're looking for the force in the member CG.

So the easiest way to get this is to just sum the forces in the vertical

direction here, then I can solve it with just one equation.

So the forces we have are 10 kN here, 10 kN down and the component of CG.

And all of these triangles here are 3-4-5 triangles so

the cosine of the angle there is just 4/5.

So rearranging,

we find rather surprisingly that the force on that member is 0.

So the correct answer is A.

And, just as a side note.

Any member in a truss that has no force in it is called a zero force member.

In other words,

you could remove that member without anything happening to the structure.

Although it may not be a good idea to do so.

Now the next question, the magnitude of the force CD So

now we want to find the force in this member, CD.

So my same curt and my same free-body diagram applies,

and the answer is most nearly which of these.

So in this case, the easiest thing to do,

and there are different ways to do it, but in this case,

the easiest thing Is to take moments about the support A.

So the moments are, and

I'll assume that clockwise is the positive direction for moments.

I'll just assume that from my sign convention.

So the equations we have then are the moment of CD.

Is CD times it's moment arm.

This distance, which is four meters.

Plus, the moment of this force, ten kilonewtons,

ten times it's moment arm, which is six meters.

Then of course, this force HG has no moment because it passes through

the The point A and CG has no moment because it's a zero force member.

It doesn't have any force in it.

So rearranging, we see that CD is minus ten times six over four,

is equal to minus fifteen kila-newtons.

And the minus sign there tells us that member is in compression.

So that is in compression, but the question only asks for

the magnitude of the force not, whether it was intentional compression,

so the magnitude of the force is 15 kiloNewtons, and the correct answer is C.