Here is a frame structure that undergoes side-sway due to a horizontal joint load.
How do we analyze such a frame using the moment distribution method?
We know how to calculate the distribution factors for the joints.
Here they are:
If you're not sure how this is done, see Lectures SA38 through SA40.
But how do we determine the fixed-end moments?
We need to know them in order to start the iterative process of balancing the joints.
We know how to calculate fixed-end moments when one or more of the members are subjected to loads.
But here, there are no member loads.
There is only one joint load.
So how should we approach the problem?
Would it be possible to turn the lateral displacement represented with Delta into fixed-end moments?
That is to say, does a bending moment develop at the fixed ends of a member
if it experiences a relative displacement like Delta?
To answer the question, we can use the slope-deflection equations.
Here they are for member AB:
If the member is assumed to be fixed at both ends,
in other words, if theta_A and theta_B are zero,
then the two equations reduce to:
These are the bending moments at the fixed ends of the member, or our fixed-end moments.
Therefore, using the slope-deflection equations,
we can easily turn the lateral displacement of the frame
into fixed end-moments to be used in the moment distribution method.
But we still have a problem.
We don't have a numeric value for Delta.
We cannot have Delta's value until we analyze the frame and determine its support reactions.
But to analyze the frame, we need to know Delta.
To resolve this circular road-block,
we are going to use the guess-and-check technique.
What are we going to guess?
We are going to guess a value for Delta.
This guess will allow us to compute the fixed-end moments,
which in turn lets us start the moment distribution process to analyze the frame.
Since we are guessing at this stage,
we could pick a value for Delta that would give us
a round number for the fixed-end moments,
like negative 10 kN-m.
And since we are not going to use Delta beyond this point,
its value is not really needed here.
All we need to know is that there exists a value for Delta
such that when used in these equations,
we get negative 10 for the fixed-end moments.
In a way, we are assuming a value for the fixed-end moments.
Using the same Delta,
we can also determine the fixed- end moments for member CD.
They are also negative 10 kN-m.
We then set up our moment distribution table like this:
And start the iterative process until the member-end moments are accurately determined.
Now, we apply the equilibrium equations to determine member-end shear forces and support reactions.
Here are the reactions based on the assumed fixed-end moment values.
We know that if these were the actual reaction forces for the frame,
then the equilibrium conditions would have held true.
Let's check and see how good our initial guess was.
We know that the sum of the forces in the X direction must be zero.
That is, the sum of the support reactions in the X direction must be equal to 10 kN.
The horizontal reaction forces at A and D add up to 3.75.
So we are off by a factor of 2.67.
That is, we would have ended up with the correct support reactions
if our initial assumed value for the fixed-end moments was negative 26.7, instead of negative 10.
Does this mean that we need to go back and regenerate the moment distribution table?
No, not really.
We can simply multiply the answers, the member- end moments and support reactions by 2.67,
in order to arrive at the correct values.
So the actual support reactions for the frame are:
We will examine a more general case of frames with side sway in the next lecture.
For now, see if you can analyze the following frame structures using the moment distribution method.