 Hello, this is Dr. Mahesh Kalyanshetty, Associate Professor, Department of Civil Engineering, Walchan Institute of Technology, Sualapur. Today, we will discuss about the analysis of interdisciplinary structure by a moment distribution method wherein we will focus on beams with sinking of supports. The learning outcome of this session will be, at the end of this session, students will be able to incorporate the sinking effect in the analysis of indeterminate beams using moment distribution method. Let us take one example and then we will discuss how the sinking effect is to be considered in the analysis. Analyze the beam as shown in figure by a moment distribution method wherein EI is constant and it is also given that support B sinks down by 10 mm and EI value is to be taken as 75,000 kNm2. As we know that, first of all, we need to calculate the stiffnesses of members and joints. Now we have two interior joints here, joint B and C. Let us calculate the stiffness of the joints. For that, we need to calculate the stiffness of the members. So if I consider a joint B here, so we have two members meeting there, BA and BC. So the stiffness of BA will be 4EI by L since the opposite end is fixed and stiffness of BC also will be 4EI by L since it is a continuous support. And the stiffness of the joint B will be summation of the stiffness of the member BA and BC as shown here. In the same way, if I consider joint C, we have two members meeting at C again, that is CB and CD. So stiffness of CB is 4EI by L again because opposite end is continuous and stiffness of CD also is 4EI by L since it is fixed. So the total of these two will give you the stiffness of joint C. In this way, we can calculate the stiffness of joint B and joint C. Then we have to calculate the distribution factors of joint B and C as we already discussed that the distribution factor is nothing but it is a ratio of the stiffness of the member divided by the stiffness of that joint. So let us consider joint B here. At joint B, we have two members BA and BC. So distribution factor BA and distribution factor BC will be equal to the stiffness of that member divided by the total stiffness of the joint. So denominator is nothing but it is a total stiffness of the joint and it works out to be 0.5. Again if I go to joint C, so here also we have two members CB and CD. So the distribution factor CB will be stiffness of the CB divided by the stiffness of the joint C. Here it is a denominator. So it works out to be 0.4 and the distribution factor CD is again the ratio of the stiffness of the CD. This is stiffness of CD divided by the stiffness of joint C which works out to be 0.6. And as already also we discussed that the summation of the distribution factors shall be always equal to 1 at all the joints. So we can see here at joint B both the members have got the distribution factor 0.5 therefore it is 1. At joint C the distribution factor CB is 0.4 and CD is 0.6. So therefore summation works out to be 1. Now after this we have to incorporate the sinking effect also in the analysis. Now many time in the structure the sinking of the support takes place because of the settlement of the supports and this sinking of the support creates an excessive moment in the members. So these members also we shall consider in the moment distribution process. Now as it is shown here if suppose one of the support sinks down by amount say delta then there will be a fixed end moment developed having a magnitude of 6ER delta by L square. So as you can see here if the right hand support sinks down by delta the moment produced will be 6ER delta by L square and the moment produced will be equal to opposite to the rotation of the beam. Now in this picture it is seen that due to sinking the beam is rotating in clockwise direction you can see here this point comes down here. So the rotation of the beam is in clockwise direction therefore the fixed end moments developed will be in anticlockwise direction. So you can see here these are two anticlockwise moments and the magnitude of these moments will be 6ER delta upon L square and the reactions also will be produced accordingly at the supports. So the support reactions produced due to sinking is again 12ER delta upon L cube. So same reactions we observe here. The same thing if suppose the left hand side left hand support sinks down the same phenomenon takes place here. The moment produced will be now in a clockwise direction 6ER delta by L square and the reactions will form an anticlockwise moment. So if the moment is clockwise in nature the reactions will form anticlockwise moment. Same is the case in the first picture also. So therefore whenever the sinking of the support takes place we consider a moment of 6ER delta upon L square at both the supports where delta is the amount of sinking at the support. So let us consider this effect in the given problem. Now the fixed end moments first of all we need to calculate. So span AB since there is no load here actually therefore there is no fixed end moment due to loading. However due to sinking there will be a moment produced just in the previous slide we discussed. Now this kind of settlement will take place here. So therefore the MAB and MBA both are present. So due to loading as I told that since there is no loading present no fixed end moment. However due to sinking there is a fixed end moment produced and that is 6ER delta by L square and as per our sign convention the anticlockwise moment we considered as negative. So 6ER delta by L square and it works out to be minus 31.25 kilo Newton meter. Let us go for the next support BC. So for support for span BC there are two effects. One is the fixed end moment due to loading as well as fixed end moment due to sinking. So due to loading we know that it is WL square by 12. Therefore due to loading the fixed end moment BC is nothing but WL square by 12. So negative sign indicates it is anticlockwise moment and CB is positive since it is clockwise in nature. So it works out to be minus 240 kilo Newton meter and plus 240 kilo Newton meter. And the second effect is the sinking effect as we know that the left support C sorry B is sinking down. Therefore the moment produced is clockwise in nature and that is 6ER delta by L square. Since the moments are clockwise in nature these are taken as positive and it works out to be 31.25 kilo Newton meter. For the total fixed end moment for span BC will be summation of these two that is due to loading as well as due to sinking and this works out to be total FM BC is minus 208.75 and total FM CB will be plus 271.25 kilo Newton meter. Then the last span left is CD. Now for CD there is no sinking effect because sinking is at B only. So CD is free from sinking therefore the fixed end moment is due to loading only and we know that this standard case the left hand side it is PL by 8, right hand side PL by 8. So it is anticlockwise therefore CD is minus 250 and FM DC is plus 250 kilo Newton meter and due to sinking there is no FM. So all these fixed end moments we have to consider for the further process. Before that let us take a pause here and give the answer to this question. First one is due to sinking the FM produced at fixed end is so four options are given and the second question is due to sinking the shear force produced at the fixed end is the four options given. So think over it take a pause when you get the answer resume the video welcome back. So these are the answers for this question come to the most important part of the problem now after getting all the distribution factors fixed end moments all these information is to be filled in the standard table. Now we have a joint A, B, CD so the member AB at B we have BA and BC so it is written here at C we have CBN, CD and at D we have DC. So these are the distribution factors which are observed B is 0.5, 0.5 and at C 0.4 and 0.6 fixed end moments we calculated all those fixed end moments are to be written here accordingly and then we have to go for the moment distribution process. We have to balance the joints now for example if I consider the joint B here so it is unbalanced because we have minus 31.25 and minus 208 so there is a total minus 240 so minus 240 is unbalanced so plus 240 we have to apply out of plus 240 50 percent here 50 percent here it is to be applied so here also at C also if you look at it is unbalanced so find the unbalanced moment and opposite of that unbalanced moment is to be applied and out of that total 0.4 fraction will be transferred here, 0.6 fraction will be transferred here and then the carryover will take place since the A is fixed end so carryover 50 percent is transferred here and then from C also towards D 50 percent is transferred and from B carryover takes place towards C and from C also carryover takes place towards B. So in this way the carryover takes place and we have to go for number of iterations you can see here after the carryover again the balancing get disturbs and that is to be again balanced so in this way the iterations are to be taken so after taking four iterations it is observed that the unbalanced moment left is very very small and therefore we can stop there and after the process is over you can take a summation of all these moments the final moments will be obtained here so the summation of moment AB, BA, BC, CB, CD and DC all are given here once this moment distribution process is over then the free body diagrams are to be determined are drawn so as next job is to draw the bending moment diagram so based on the moments produced here at the supports this is the bending moment diagram we obtain so these are the references which are used for the presentation thank you