The moment distribution method is an approximate technique
for analyzing indeterminate beams and frames.
In lecture SA37, I presented a mathematical basis for this method.
Here, I am going to demonstrate the ease with which the method
can be used to analyze continuous beams.
Let's start with a revised version of the example problem we considered in the last lecture,
a continuous beam subjected to a concentrated moment.
We wish to analyze it using the moment distribution method.
As the name suggests, this method involves distributing moments.
What does that mean exactly?
It means, given a bending moment at a joint,
we distribute it among the other joints of the structure.
The joints that receive a part of that moment in turn distribute it among the other joints.
This process repeats itself until every joint is balanced with respect to the bending moment.
Therefore, if a joint connects two members, like this,
then any bending moment found at the joint,
first needs to be distributed between the near ends of the two members.
How?
In proportion to the relative stiffness of the members.
The more rigid member absorbs a larger share of the moment.
We measure the stiffness of each member in terms of its E, I and L.
More specifically, we define EI/L as the stiffness coefficient.
For member AB, we get EI/4.
For BC, we get EI/6.
The total stiffness at the joint is the sum of these two coefficients.
Here, since both members have the same EI,
the sum can be written as 5EI/12.
Then, any bending moment present at joint B
is distributed between members AB and BC,
using the following distribution factors.
Member AB absorbs 3/5 of the moment.
Member BC absorbs 2/5 of the moment.
Note that the sum of the distribution factors equals one.
This is true for every joint in every structure, without exception.
We start by drawing our moment distribution table, like this:
We write the distribution factors in the first row of the table.
Since there is only one member connected to the pin support at A,
the distribution factor at the joint equals one.
This means, any bending moment that appears at A
is going to be distributed entirely to the left end of member AB.
This is also true for joint C.
Any bending moment that appears at C goes to the right end of member BC.
At joint B, we write 3/5 in the left column,
and 2/5 in the right column.
These are the distribution factors we computed a minute ago.
The applied bending moment at B is placed at the top of the table, like this:
In determining the sign of the moment we adopt our slope-deflection sign convention.
This means, a counterclockwise moment applied directly to a joint is considered a negative moment.
Hence, we have negative 10 kN-m here.
Now we are ready to proceed with the iterative process of the moment distribution method.
There are two basic steps in this process:
Balancing joint moments and carrying over member-end moments.
In the balancing step, we make sure the sum of the moments at each joint adds up to zero.
In the carry-over step,
we distribute half of any moment added to one end of a member, to its other end.
So, if an outcome of the balancing step is the addition of bending moment M to the table,
then in the carry-over step we place half of M at the other end of the member.
We start the iterative process by balancing all the joints.
Joints A and C are balanced already.
The total moment at each joint is zero.
To balance joint B, we calculate the total moment at the joint.
It is negative 10.
We then balance it by distributing positive 10
to the members connected to the joint in proportion to their distribution factor.
This means we write: 10 times 3/5, or 6 in the left column.
And we write: 10 times 2/5, or 4 in the right column.
Notice that now the sum of the moments at the joint equals zero.
Then, we carry over half of each added moment value to the other side of the member.
This means, half of six carries over to the other end of member AB,
and half of four carries over to the other end of BC.
These carried over moments make joints A and C unbalanced, 72 00:05:29,160 --> 00:05:32,500 the sum of the moments at these joints is no longer zero.
So we need to balance them.
The total bending moment at A is 3 kN-m.
To balance the joint, we add negative 3 to the left column.
The total bending moment at C is 2 kN-m.
We add negative 2 to the right column to balance the joint.
We then carry over half of each recently added moment value to the other side of the member.
This means, we write negative 1.5 in the second column,
and negative 1 in the third column.
Consequently, joint B has to be balanced again.
Sum of the moments at the joint equals negative 2.5.
We add positive 2.5 to the joint using the distribution factors, like this:
We then carry over half of each moment to the other side of the member.
In the next step, we balance joints A and C.
Then, we carry over the moments.
We continue this process until the moment values get close to zero.
We then add up the values in each column to get the member-end moments.
So, the bending moment at the left end of member AB is zero.
The moment at the right end of AB is 6.
The moment at the left end of BC is 4.
And the moment at the right end of BC is zero.
Now we can draw the free-body diagram for each beam segment
and calculate the remaining unknown member forces, like this:
Next, let's consider the case where the beam is subjected to a member load.
Here is a uniformly distributed load applied to segment AB:
How does that change the iterative process of calculating member-end moments?
It does not change the process at all.
The only change that we need to make is to the initial moment distribution table.
We are going to add a row to the table for fixed- end moments associated with the distributed load.
Recalling from the slope-deflection method,
we know that a beam of length L, subjected to a uniformly distributed load of w,
would have wL^2/12 for its fixed-end moments.
In our case, this means the left and right fixed-end moments for AB each has a magnitude of 4 kN-m.
The left moment, since it is counterclockwise, is considered positive.
And the right moment is considered negative.
We entered these values in our table, like this:
These are considered the initial moments at joints A and B
to be balanced once the iterative process begins.
We begin the process by balancing the existing moments at joints A and B.
We add negative 4 to the left column.
The value to be added to the second column equals 4 times 3/5, or 2.4.
The value in the third column equals 4 times 2/5, or 1.6.
Note that the sum of the moments at each joint is now zero.
In the carry-over step, we carry half of negative 4 to the second column,
half of 2.4 to the first column,
and half of 1.6 to the fourth column.
Then we balance all three joints.
And, add the carry-over moments to the table:
We balanced the joints and carry over the moments again.
After a few more iterations, it becomes clear that we have reached the end of the process.
Now, add up the column values to determine the member-end moments.
We then apply the equilibrium equations to each segment to complete the analysis.
Here are the support reactions for the beam: