Truss Analysis, Method of Sections
There are two basic techniques for analyzing statically determinate trusses:
The Method of Joints
and The Method of Sections
Generally speaking, the Method of Joints is preferred when the entire truss is to be analyzed,
when all the member forces are to be calculated.
The method of sections, on the other hand,
provides a quick way for calculating the force in a few members only.
We discussed the method of joints in a previous lecture.
In this presentation we're going to cover the method of sections.
This method is based on the principle that when a structure is in the state of equilibrium,
every substructure within it must also be in equilibrium.
For example, given that this truss is in the state of equilibrium,
if we cut it like so,
then the left substructure and the right substructure must be in equilibrium.
Needless to say, such a cut is imaginary,
done on paper for the purpose of analyzing the truss only.
Naturally, the actual truss members need to remain intact if the structure is to carry the intended load.
This method of cutting a structure into parts
is a convenient way for determining some of the unknown member forces,
without needing to analyze the entire truss.
More specifically the method enables us to rather quickly
calculate the force in one or more cut truss members.
In this particular example,
we can determine the force in members BD, BE and CE without needing to analyze the entire structure.
Generally, we may be able to cut a truss in several ways.
The challenge in using the method of sections is to identify an effective way to cut the truss,
and then decide how to use the equilibrium equations to find the desired unknown forces.
Let's shed more light on the subject using three examples.
Example 1: I'm going to use this example to present the steps involved in using the method of sections.
Here is a statically determinate truss subjected to two concentrated loads.
The truss rests on a roller support at point A, and a pin support at point H.
We wish to determine the force in members BD and CE.
First we need to calculate the support reactions.
To do so we're going to write the equilibrium equations for the entire structure.
Here are the equations:
We have three equations and three unknowns.
Solving the equations for the unknown reactions we get:
Ay=0.25 kN
Hy=3.75 kN
Hx=-3 kN
Now, to find the force in member BD, we need to cut the structure through the member.
The most suitable cut here is one that passes through members BD, BE and CE.
Since the cut goes through BD and CE, we can calculate both member forces using the same set of equations.
When we cut a truss into two parts,
we must show the internal axial force in each cut member in the free-body diagram.
Here since we have cut through members BD, BE and CE,
we need to show the axial force for each of these members in both the left and the right substructures.
So for the left sub structure, the forces are drawn like this:
For the right substructure,
they are drawn like this:
For the sake of consistency we assume that all the members are in tension,
therefore the axial force associated with each cut member is drawn
pointing away from the joint connected to the member.
We can determine these unknown member forces by applying the equilibrium equations
to either the left free-body diagram or the right one.
Let's use the left free-body diagram.
Here are the three equilibrium equations:
For the sum of the forces in the x-direction we have:
3+Fbd+Fce+(Fbe Cos 45)=0
For the sum of the forces in the y-direction we have:
0.25-(Fbe Sin 45)=0
For the sum of the moments, say about point B, assuming clockwise direction is positive we have:
(0.25)(5)-Fce(5)=0
Solving these equations for the unknowns we get:
So member CE carries a tensile force of 0.25 kN, and BD carries a compressive force of 3.5 kN.
Example 2:
This example illustrates the convenience of using
the method of sections for finding one member force at a time.
Say we wish to determine the force in member CE.
Here we can determine the unknown force without needing to calculate the support reactions.
Further, we only need to use one of the equilibrium equations,
the moment equation for this purpose.
Let's cut the truss like this:
The free-body diagram for the top substructure looks like this:
The diagram shows three unknown forces: Fce, Fcd, Fbd.
Since we can write three equilibrium equations, then we can calculate all three forces.
But we're interested in calculating Fce only.
Therefore it makes sense to simplify the solution by using one equation only.
The equation of interest must only have Fce as an unknown.
No other variable should appear in the equation,
otherwise we would not be able to use it to solve directly for Fce.
If we examine the free-body diagram closely,
we can see that all the unknown forces, except for Fce, pass through point D.
So if we take some of the moments about D, only one variable appears in the moment equation.
Here is the equation.
We are assuming clockwise rotation is positive.
(5)(4)+(3)(2)+(2)(2)+Fce(2)=0
Solving the equation for Fce we get:
Fce=-15 kN
Therefore member CD carries a compressive force of 15 kN.
See if you can determine the force in members GI and DF on your own.
I will show you how in a few minutes, at the end of the lecture.
Example three:
In this example we examine a case in which no matter how we cut the truss,
we end up with no less than four unknown forces.
So how does the method of sections work in this case?
How do we solve for the unknowns using only three equilibrium equations?
Suppose we have a K-truss, as shown.
We wish to determine the force in each member along the top chord of the truss.
That is, we wish to find the force in members AB, BC, CD and DE.
Let's start by finding the force in member BC.
To find this force, let's cut the truss like this:
Then the free-body diagram for the left substructure looks like this:
Note that the free-body diagram contains four unknown forces: Fbc, Fnc, Fni and Fhi.
But we can only write three equilibrium equations:
This system of equations does not have a unique solution as there are more unknowns than equations,
therefore we cannot calculate Fbc this way.
So is there another way of cutting the structure, so that Fbc can be easily calculated?
The answer is yes.
In fact, there are two other ways of cutting the truss that give us what we need.
We can cut the truss like this,
or like this:
For the first scenario, the free-body diagram for the left truss segment looks like this:
Note that all the unknown forces, except for Fbc, pass through point H.
For the second scenario, the free-body diagram looks like this:
Interestingly, here also all the unknown forces, except for Fbc, pass through H.
We only need one equilibrium equation to find Fbc.
Let's use the moment equilibrium equation written about point H.
We can use either of the two free-body diagrams.
Here is the equation:
(2)(4)+4Fbc=0
or Fbc=-2 kN
To find the force in member AB, we can cut the truss like this:
Here is the free-body diagram for the left substructure.
Writing the moment equilibrium equation about point G we get:
(2)(2)+4Fab=0
Solving this for Fab we get:
Fab=-1 kN
How about Fcd, how do we calculate it?
Given the symmetrical nature of the structure and the applied load,
one can conclude that members CD and BC carry the same force.
That is, member CD also carries a compressive force of 2 kN.
Alternatively, we can employ the method of sections by cutting the truss like this:
Then take the right free-body diagram.
Taking some of the moments about J, assuming counterclockwise is positive, we get:
(2)(4)+4Fcd=0
or Fcd=-2 kN
Also to find the force in member DE,
we can either take advantage of the symmetry of the system,
which tells us Fde=Fab,
or we can use the method of sections by cutting the truss like this:
Here is the free-body diagram for the right substructure:
Here is the moment equilibrium equation written about point K:
Here is the solution:
So just like member AB, DE carries a compressive force of 1 kN.
This example leads us to an important conclusion:
When using the method of sections,
we may be able to find a desired member force using only the moment equilibrium equation.
This generally works if we can find a point in the free body diagram
that all the additional unknown forces pass through.
Before we conclude this session, let's complete the solution for Example 2.
Here is the truss:
We wish to determine the force in members GI and DF.
To find Fgi, we cut the truss like this:
The free-body diagram for the top part of the structure looks like this:
Note that both Ffh and Fgh passed through point H.
Therefore the moment equilibrium equation written about point H,
gives us the force in member GI.
The equation is: (5)(8)+(3)(6)+(2)(2)+Fgi(2)=0
Solving it for the unknown force we get:
Fgi=-31 kN
Finally to find Fdf, cut the truss like this:
Here is the free-body diagram for the top part:
The moment equilibrium equation taken about point G is:
(5)(6)+(3)(4)-Fdf(2)=0
Solving the equation for the unknown, we get:
Fdf=21 kN
That is, member DF carries a tensile force of 21 kN.
Here are a few practice problems:
Using the method of sections, find the force in each marked member in these trusses: