In this lecture, we are going to illustrate

the use of the moment distribution method for analyzing frame structures with no side-sway. The analysis here is very similar

to the case of continuous beams. First we calculate the distribution

factors and fixed-end moments. Then we set up the moment distribution table and go through the process of balancing the joints

and determining the carry-over moments repeatedly, until a desired accuracy is reached. For our frame structure, we already know

that distribution factors at the supports. If you are not sure how these values

are obtained, please review lecture SA39. For the interior joints, we calculate

the distribution factors this way: Noting that the moment of inertia for the columns

is twice the moment of inertia for the beam, We write the stiffness coefficients as 2EI/4

for the columns, and EI/6 for the beam. This makes joint stiffness at B: 2EI/3 Hence the distribution factors at the joint are: As for joint C, its total stiffness is also 2EI/3. This makes the distribution

factors 3/4 at the column end, and 1/4 at the beam end. Now, let’s calculate the fixed-end moments. Member BC is subjected to a

concentrated load at its midpoint. This results in a fixed-end moment of

6 kN-m at either end of the member. To draw the distribution table

for the frame, flatten it, like this: And place the distribution factors in this manner: Then, write the fixed-end moments. The table is now set up, ready for the iterative moment distribution process. We start by balancing the joints. Initially, only joints B and C need to be balanced. We then carry over half of the balancing moments

to the other side of each member, like this: We balance the joints again, and add the carry-over moments to the table. Let’s balance the joints one more time. Note that past this point any additional iterations

would not significantly alter the moment values, therefore, we mark this as the

end of the iterative process. We can now determine the member-end

moments by adding up the values in each column. Finally, we apply the equilibrium

equations to each member, in order to determine the remaining unknown forces. Here are the support reactions for the frame: Let us examine another case. Here, we have a joint that connects three members. This leads to an important question. How do we set up the moment distribution

table when such a joint is present in the structure? Let’s start by determining the distribution factors. At the fixed supports we have: For joint B we can write: And for joint D we get: The fixed-end moments for member BD are: To set up the moment distribution table

we imagine a flattened frame, like this: However, as you can see, the frame

cannot be flattened completely, which means we cannot set up

a table with contiguous columns. Since the table needs to have

a column for each member end, we need to create a separate pair

of columns for joint C, like this: Then, the distribution factors are placed like this: To add clarity to the table, let’s label its columns. I am going to label this column AB. This means the column contains bending

moment values for the lower end of member AB. And this column, I label BA. It is for bending moment values

at the upper end of member AB. We label the remaining table

columns in a similar manner. Note that for CD, bending moments at the lower

end of the member need to be written here: And the values for the upper

end of the member go here: This is important to keep in mind, since half

of any value that is added to this column, needs to be carried over to this column: Here are the fixed-end moment values: Now we can begin the moment distribution process. Let’s start by balancing joint D. We need a moment of 6 kN-m to balance the joint. This value needs to be distributed among the three

members according to the distribution factors, like this: We then balance joint B. Since all the joints are balanced, we determine the

carry-over moments and add them to the table, like this: Note that for joint D, we carried half of

this balancing moment to this column. We balanced the joints again, then add the carry-over moments to the table. We continue this process until the

moment values become negligible. Once we reach the end of the process, we add up

the values in each column to get the member-end moments. We can then calculate the member-end

shear forces and support reactions. Here are the results:

very good

Greetings thank you very much for the lesson.

TÜRKİYE den selamlar

very thanks

dr u r really providing quality education to us ,thanx for u r lecture, are u gonaa also cover stiffness matrix of structure analysis and what are the other topic u gonna cover

Your videos are the best, very thorough and concepts are clearly explained at a good pace.

i need to know calculation system of multi stories

these videos are the best!

6 can be divided 4plus 2 or 3plus3 or 4.2 plus 1.8, why it is 3.5 plus 1.5, explain these simple things

Sir will you please deliver us the solution of the excersize questions?

Sir if member end moment and fixed end moment are showing opposite sign still why are u taking both the moment in the same sense(clockwise and counterclockwise)?

@07:20, the rightest column 0.82 should be negative.

thank,s alot

Ly

Where are the answer? So that i can check of im wrong or not

the only best place to clear doubts online…thnx a lot

Hi ma'am, may I know how did you get the reactions? Thanks.

How to get 1.93 Kn?

Thanks a lot, Dr. Structure. Your videos are highly informative and easy to understand.

I have a question. If the horizontal member in a frame contains an internal hinge, then how do we proceed with the method? What would be the distribution factors?

Others use 4EI/L for member stiffness factor , but you use only EI/L. Can u pls explain why?

Why are you considering clockwise moment negative?