In the previous lecture,
we showed how the moment distribution method
can be used to analyze an indeterminate frame
subjected to a horizontal joint load.
What if, instead of a joint load, the frame is subjected to a member load, like this:
How do we analyze the frame then?
So far, we have examined two scenarios.
In Lecture SA40, we considered a frame subjected to member loads with no side-sway.
And in Lecture SA41, we showed how to analyze a frame with side-sway due to a horizontal joint load.
Here, we are facing a different scenario.
We have a frame that undergoes side-sway due to a member load, not a joint load.
To analyze this frame, we can draw from the previous scenarios.
We will employ the principle of superposition for this problem.
With this principle,
we can break down the behavior of our frame
into the behavior of two simpler, more familiar frames.
But how do we arrive at these two frames?
We know how to analyze our frame if it is restrained against lateral displacement.
So if we place a roller at the top of the frame, like this,
then we can analyze it using the moment distribution method as was discussed in Lecture SA40.
Consequently, we end up with a horizontal reaction force at the top of the frame.
Let's refer to the force magnitude as R.
We know that R is zero in the actual frame.
To satisfy this loading condition,
we take the reaction force at B,
change its direction,
and place it at the top of this frame.
Note how the algebraic sum of these forces satisfies the loading condition in the original frame.
According to the principle of superposition,
we can conclude that each member force,
or support reaction in this frame added to the corresponding force in this frame
gives us the related force in the original frame.
Since we already know how to analyze both of these simpler frames,
we can easily determine the member forces and support reactions in the original frame.
The assumption here is that all three frames are identical
with respect to their geometry and material properties.
Let's examine this solution strategy in more concrete terms.
Suppose we want to analyze the frame shown here using the moment distribution method.
We know that the frame is going to have a lateral displacement, like this:
So we can view it as a superposition of these two frames,
where R in frame two is the magnitude of the horizontal reaction at B in frame one.
Per Lecture SA40, we can analyze frame one using the moment distribution method.
Instead of going through the specifics that you can find in video SA42-A,
I'll just show you the results here:
Note the horizontal reaction at B.
Its magnitude is 11 kN.
This means R equals 11 kN.
Now we can analyze frame 2 using the technique discussed in Lecture SA41.
Here are the results of the analysis:
The details are presented in video SA42-B.
According to the principle of superposition,
the support reactions for the original frame can be obtained
by adding the corresponding support reactions in frames 1 and 2. So we can write the following:
The moment distribution method offers a convenient technique
for analyzing simple indeterminate beams and frames using hand calculations.
See if you can analyze the following frames using this method: