In the previous video, we talked about solving trusses using the method of joints.
That is very useful when we have to solve for a large number of members in a truss, or for all
the members. But what if you only wanted to find the forces in a few members? Instead of solving
the whole truss, we can use a method called the method of sections. It’s based on the fact that
if the whole truss is in equilibrium, then any segment of the truss is also in equilibrium.
Then we can apply an equation of equilibrium or a moment equation to solve for unknowns. Let’s say
we have a truss like this and we want to figure out the forces in these 3 members. The first step
is to find the reactions at one support. Then we section it off, in other words, we cut the truss
through the members that we are trying to solve. It’s good to try and cut the members we are trying
to find. Once we do that, we forget one part of the truss, most of the time, we forget about the
part where we didn’t find the reactions at. Now here is where things change. From this point on,
the easiest way to think about this problem is to think of all of this as the truss,
and these cut members as external forces. In other words, this is one single object, and
these are the forces that are applied to it. You can assume the direction of these forces, if you
get a negative value, then it’s opposite to your assumption. After that, you can solve for these
forces using equations of equilibrium or moment equations. This might be a bit confusing now, but
after we go through a few examples, hopefully you will have a firm understanding of how to do this.
Just a reminder from the previous video, that if a force comes towards a pin,
then the member is in compression and if a force leaves a pin, it’s in tension.
Let’s take a look at this problem where we need to find the forces in members GH,
BC and BG. Since all of the unknowns we need to find are on the left side, we need to
figure out the reactions at pin A. We can write a moment equation about point E which allows us
to directly solve for the vertical reaction. We can see that there won’t be an x reaction
at pin A since there are no horizontal forces. Now these are the forces we need to figure out,
so the best place to make a cut is right through them like this. We can forget about the right side
and just focus on the left side. Now I am going to assume that the forces in our cut members
would be like this. Here, you have to think of this as being our truss, and these cut members,
as external forces being applied to our truss. So these are internal forces, and these are external
forces. Before we do anything else, let’s quickly figure out these angles. Notice we can draw a big
right angle triangle like this. Using the inverse of tan, we can figure out this angle. Now notice
that this angle would be the same since if we create a right angle triangle like this,
it’s a similar triangle. The bottom angle can
also be found using inverse of tan. Now, let’s start with a moment equation about
point B. That would eliminate forces BC and BG. It also eliminates the 5 kN force being applied since
its line of action goes through point B. Again, note that the force in member AH is an internal
force, so we don’t need to worry about it. It is the same as when we wrote a moment equation
to figure out the reaction at A. We didn’t care about the internal forces, in the same way, when
we write our moment equation about point B, we don’t care about the internal forces in member BH
or AH. This leaves just force HG and the reactions at A. We also need to figure out this height, that
can be found by using the angle we found earlier giving us 1.5 m. Now let’s write our equation and
we will pick clockwise to be positive. So we have the vertical reaction at A times the perpendicular
distance to pin B, then we have the force applied at A times the perpendicular distance,
this force creates a counter clockwise moment, so it’s negative. Lastly, we have the x component of
force HG times the perpendicular distance. The y component does not create a moment
since it’s line of action goes through point B. Let’s solve. We got a positive value, which means
our assumption was right, and that also means this member is in compression since the force is coming
towards pin H. Remember, when a force comes towards the pin, the member is in compression.
Now, we can write a moment equation about pin A. That would eliminate forces HG, and BC since their
lines of action goes through pin A leaving us with just force BG. Again, member BH does not matter to
us since it’s an internal member. Note that only the y component of force BG creates a moment since
the x component’s line of action goes through pin A. Also, the reactions at A are eliminated
since their lines of action goes through pin A. Let’s solve. We got a positive value,
which means our assumption was right, and since this force is leaving pin B, it’ll be in tension.
We figured out 2 of the 3 members, so let’s find the last one. For that, we can write a
moment equation about point H. That eliminates force HG along with the 5 kN force applied.
Note that only the x component of force BG creates a moment. Let’s solve. We got a positive value,
so our assumption was right and since the force is leaving pin B, it’ll be in tension.
We now solved for the required members using just 3 moment equations.
Let’s take a look at this problem where we need to find the force in members BF,
BG and AB and determine whether they are in tension or compression.
Now I am going to cut the truss right here, and forget about the bottom part. The reason is, if
we just focused on the bottom, we’d have to figure out the reactions at the pin and roller. Now I am
going to assume the forces are going to be like this. As before, this part is our truss, and these
are our external forces. First, let’s write a moment equation about point B. Doing so eliminates
the 10kn force being applied at pin B, along with forces BG and BA leaving us with just force FG.
This will help us figure out the unknowns we require. To figure out the force in member BG,
we need to eliminate force BA. We can do that by writing a moment equation about point D.
Note that only the x component of force BG creates a moment about point D.
We know the angle is 45 degrees since it’s a 90 degree angle divided in 2.
Let’s solve. Since we got a positive value, our assumption was right,
and the force is coming towards pin B, which means the member is in compression.
To figure out member AB, we can write a single equilibrium equation for the y axis forces.
Again, we are considering just external forces. So we have the force FG, which we already found,
then the y component of force BG, which we also found, and lastly, the 5 kn applied at point E.
Note that we picked up to be positive, so the 5 kN force and force AB are both negative since they
point downwards. Let’s solve. We got a positive value, so our assumption was right, and since
it’s leaving pin B, the member is in tension. Figuring out member BF is very easy. All we
need to do is isolate joint F, just like we do in the method of joints. We have force EF,
FB and FG. Notice how we have no other horizontal forces, in other words, there is no force being
applied in the x direction. So if you write an equation of equilibrium for the x-axis forces,
the only force we have is BF, which means it’s zero. So that’s a zero force member.
And those are our answers. If we were to solve this using the method of joints,
we’d have to start up here, and solve for all of these members until we got to the very last ones.
Using the method of sections, we solved for the required members using just 4 equations.
Let’s take a look at this example. In this problem, we need to figure out the force in
members DC, HC and HI and determine whether the members are in tension or compression. First,
we need to figure out the reactions at the roller. To do that, let’s write a moment equation about
point A. Let’s solve. Now, these are the members
that we need to figure out, so I am going to cut right here. I am going
to assume these directions for our forces. Again, this is our
truss now, and these are the external forces. To begin, we need to figure out this angle. We
can use inverse of tan to do so. Now, we will write a moment equation
about point H. This eliminates forces HC and HI, leaving us with force DC to figure out.
Note that the 50kN force is also eliminated since it’s line of action goes through pin H.
Let’s solve. Our assumption was right and since the force is coming towards pin D,
that means this member is in compression. Next, we need to find the force in member HI.
We can do that by writing an equation of equilibrium for the y-axis forces. So we have all
the external forces, including the y component of force DC. Let’s solve. We got a positive value, so
our assumption was right, so that means the member is in tension since the force is leaving pin H.
We can do the same to find force HC, this time, an equation for x axis forces. Other than force HC,
the only other x axis force is the x component of force DC.
Lets solve. Notice that we got a negative value, which means our assumption was wrong.
So the force is actually going away from the pin at H. That means that force is in tension.
Let’s take a look at one last example. In this question, we need to figure out the force in
members JI and DE. So knowing that we’re going to be focusing on the right side of the truss,
let’s figure out the vertical reaction at roller G. Let’s write a moment equation about point A.
Solving give us the reaction at G. Now, these are the members we need to figure out.
So I am going to cut the truss like this. Now I will assume the directions of the
forces. So as usual, this is our truss, and these are now external forces applied to the truss.
To solve for force DE, we can write a moment equation about pin I. This eliminates forces
JI and IO, and we can also get rid of force EO since it’s line of action crosses pin
I. We only have 2 forces which create moments. Let’s solve. We got a positive
value so our assumption was right and that means the member is in tension.
To find the force JI, we can write a moment equation about point E. Solving
gives us the force in member JI. This force is in compression since it’s coming towards pin I.
I hope this video gave you a better understanding of how to use the method of sections to solve for
trusses. If this video helped, please consider sharing it with your friends and classmates,
they too might find it helpful. Thanks for watching and best of luck with your studies!