 Hello, this is Dr. Mahesh Kalyanshiti, Associate Professor, Department of Civil Engineering, Valchand Institute of Technology, Swalapur. In this session, we will discuss about the analysis of indeterminate structure by moment distribution method, and specifically we will focus on beams. The learning outcome will be at the end of this session, students will be able to analyze indeterminate beams using moment distribution method. Let us take an example. Now analyze the beam as shown in figure one by moment distribution method and EI is constant. So we have this typical beam here where A and D ends are fixed and at B and C we have an interior support and this beam is subjected to loading of UDL 20 kN on span BC and a point load of 250 kN at the center of CD span. Now let us see how to analyze this indeterminate beam now using moment distribution method. We have to start with the determination of the stiffnesses. So we have to find the stiffness of all the members and then we have to find the stiffness of joints. So if I consider joint B here, we have two members meeting here BA and BC. So the stiffness of member BA will be 4EI by L since opposite end is fixed here and for BCL so it will be 4EI by L because C is interior support. So we have 4EI by L for both the spans accordingly we substitute the values and the summation of these two stiffnesses will give me a stiffness of joint B. So the stiffness of joint B is nothing but it is a summation of KBA and KBC. Similarly we determine the stiffness for member CB as well as CD now for CB we have again 4EI by L and for CD since it is fixed end we have 4EI by L accordingly the summation of these two stiffnesses gives the stiffness of joint C. Then we have to calculate the distribution factor as we are aware the distribution factor is nothing but it is a ratio of the stiffness of the member to the stiffness of the joint. So if I consider joint B here the distribution factor BA and BC will be the stiffness of the member divided by stiffness of the joint. So denominator is the stiffness of the joint numerator is the stiffness of the member the ratio is 0.5. For C also we can determine the distribution factor CB again you can see here stiffness of CB divided by stiffness of joint C now the denominator here it comes out to be 0.4 and for CD it comes out to be 0.6. Then the next step is to calculate the fixed end moments. Now we know that for AB there is no load therefore fixed end moment for AB span is not present however for BC we have the loading present and it is a typical case of fixed beam subjected to UDL. So we have a standard case and the fixed end moments are WL square by 12 accordingly FEM BC is minus because it is anticlockwise and FEM CB is plus since it is clockwise so the magnitudes are minus 240 and plus 240. Then for CD also we determine the fixed end moment as the standard case is known to us the fixed end moments are PL by 8. Therefore CD is minus PL by 8 that is minus 250 and DC is plus 250. As per our standard cases we get these fixed end moments. Before we proceed further let us go for some review questions. So I request you to take a pause and answer these questions. Addition of distribution factor of all the members meeting at a joint is always equal to four options are there. The second question is the stiffness of member with far end fixed is the four options. So you think over it get the answer and then resume the video. Welcome back. These are the answers for the review questions. The first question the answer is one because the distribution factor summation of distribution factors at a joint is always equal to one and the stiffness of member with far end fixed is four year well that already we discussed. Let us continue with our problem. Now once we get all the basic data then we have to go for the moment distribution process for that we have to prepare one table here where in all the information are provided joints A B C D so four joints are there member. So A B B A at B we have two members at C we have two members and at D we have one member. Then the distribution factor of all the members are written here as we already determined then the fixed end moments also we determined that is to be substituted here. So fixed end moment will be only for BC span so BC and CB it is minus 240 plus 240 and of course it is for CD also since we have a load over CD minus 250 plus 250. Then after this we shall go for the distribution for that we have to balance the joints first. For example if I look at joint B here the BC and BA it is not balanced. So there is an unbalanced moment of plus 240 so that plus 240 is to be distributed here based on the distribution factor 0.5 0.5 so 120 transfers is 120 transfers here. In the same way here at CL so we have unbalanced moment minus 10 and that plus 10 we have to apply out of plus 10.4 fraction is applied here and 0.6 fraction is applied here. This is how we go for the distribution first and once the distribution is over then we go for a carryover. So from B carryover is possible towards A from C carryover is possible towards D from B carryover takes place towards C and from C also the carryover takes place towards B. So these all carryovers will perform here so this minus 120 sorry plus 120 50 percent transfers here then this 120 60 transfers here 4 2 transfers here then from 6 3 is transferred here. In this way we go for the carryover. Now once the carryover is complete then the balancing gets disturbed again. You can see here now this 2 is an additional moment come to joint B. So therefore this 2 is to be balanced therefore minus 2 is to be applied out of that minus 2 50 percent is transferred here 50 percent is transferred here then plus 60 is unbalanced moment. So minus 60 we have to apply out of that minus 60 0.4 percent is transferred here 0.6 percent is transferred here. So this is how we distribute it and after the distribution is over again carryover takes place. In this way it is an iterative process once this cycle first cycle is over we have to go for a second cycle third cycle. So number of cycles we have to perform till we get a very small and fractional moment left you can see at this point 0 1 0 1. So at that level we can stop that process and then we can take a summation of all these moments so that we get the final moments of all the members. So here at the last row you can see the summation is taken and these are the final moments of the members. Once the moments are determined then we shall go for the free body diagrams. So using our final values which obtained in the previous slide we can draw the free body diagram AB. So here this is 62.5 AB and BA is 125.2 kilo Newton meter then BC 1 0 sorry it is 125.2 kilo Newton meter only this moment is 281.5 kilo Newton meter and for CD in the same way we draw the free body diagram. And once we get the free body diagram we can draw the bending moment diagram as well as shear force diagram. So with the help of this bending free body diagram see you can see here this bending moment diagram is constructed so here it is 62.5 moment positive is present here. So here it is 125.2 kilo Newton meter at support B so it is 125.2 since it is hogging it is minus here also we get a hogging moment 281.5 kilo Newton meter here and here the last moment is 234.3 kilo Newton meter which is hogging in nature here. So in this way we draw the bending moment diagram from the free body diagram and this is positive this is negative. In this way we can just give the proper label to this. These are the references which are used for the presentation thank you thank you very much.