 So, this is Dr. Mahesh Kalanshiti, Associate Professor, Department of Civil Engineering, Walshan Institute of Technology, Swalapur. In this session, we will discuss about the analysis of indeterminate structure by moment distribution method. Specifically, we will focus on non-sway frames. The learning outcome of this session will be, at the end of this session, the students will be able to analyze the non-sway frames using moment distribution method. Let us take an example. Analyze the frame as shown in figure by moment distribution method and EI is constant for all the members. So, in this figure, you can see a member AB, BC, CD and CE and we have hinge supports at D and E and fixed support at A. The loading UDL intensity is 45 kilonewton per meter and the point load of 60 kilonewton is present at B. Now, let us discuss how the analysis is to be carried out for this non-sway frame. Now, this is non-sway frame whereas, you can see here the sway of the frame is been restricted by this support, therefore, it is called non-sway frame. Otherwise, if you do not have any restrictions in the lateral direction, the frame will sway. In that case, we need to go for a separate method of sway analysis that we will see in the further sessions. Now, let us begin with the analysis. As usual, first of all, we have to calculate the stiffness of the member and joints. So, we have two interior joints here, B and C which have got the distribution factors. Therefore, the stiffness of B and C is to be calculated. So, at joint B, we know that two members are meeting B, A and B, C and the stiffness of B, A and B, C will be 4A by L because it is opposite and it is fixed here for B, A and for B, C it is interior support, therefore, it is 4A by L and therefore, the stiffness of the joint B is summation that is KBA plus KBC. For joint C, we will calculate CB will have 4A by L and CE will have 3A by L. Now, we can see here the opposite end is hinge, therefore, CE will be 3A by L and also CD will be 3A by L. So, CD is 3A by L, CE is 3A by L, whereas CB is 4A by L. So, the three members are actually meeting at joint C. Therefore, the summation of all these stiffnesses will give me total stiffness of joint C that is summation. Then, we will calculate the distribution factors. So, for joint B, the distribution factor BA is the stiffness of BA divided by the stiffness of joint B which comes out to be 0.545. For BC, again it is actually by this method also we can calculate as already we discussed that the summation of all the distribution factor at any joint is always equal to 1. So, we have only two members meeting at joint B and out of that one member has got 0.545. Therefore, the remaining next member is naturally it is 1 minus 0.545 that is 0.455. Then joint C, again we have the three members CB, it is coming 0.330, CD it is coming 0.298 and CE is coming 0.372. As per our usual notations, these are the distribution factors at joint C. Then the fixed end moments we need to calculate. So, the member AB since we do not have any load on this member it is 0. Now, member BC we have the UDL so we know that it is WL square by 12. So, left side it is minus since it is anticlockwise. So, FEM BC is minus 135 and FEM CB is plus 135 and the member CE again we do not have any load here. Therefore, this has got fixed end moment 0 and member CD also no load is there therefore, the fixed end moment is 0. Let us take a pause and answer this question sway in the frames are produced due to unsymmetrical support conditions, unsymmetrical loading conditions, unsymmetrical geometrical conditions all of the above. Think over it get the answer and resume the video. Welcome back. Now, this is the question asked sway in the frame are produced due to the correct answer is all of the above because due to symmetry of unsymmetry of all these either it may be a support conditions, it may be a loading conditions, it may be a geometrical conditions any one kind of unsymmetry will lead to sway in the frame. Therefore, D is a correct answer. Let us continue with the problem now the most important thing is this moment distribution table. So, already we determine the distribution factor of all the members and the fixed end moments also is calculated. So, this is to be arranged in this table. So, joints A, B, C, D, E the members will write here and the distribution factor member B we calculated as 0.545 and 0.455 and for C we have three members and all the three distribution factors we calculated which are written here. Then the fixed end moments as we discussed that only for BC member the fixed end moment is present rest all it is 0. So, only BC and CB the fixed end moments are present after writing all these things then we shall proceed with the distribution. So, it is minus 135 is unbalanced moment therefore it is to be balanced here as per their distribution factor here also it is 135 unbalanced so minus 135 we need to apply out of minus 135.33 faction is developed here 0.298 fraction is transferred here and 0.372 fraction is transferred here and then the carryover will take place. So, carryover from B to A is possible as well as from B to C also is possible. So, you can see here the carryover is possible so this carryover will take place now this carryover will disturb the joints again and again we will go for the distribution because minus 22.3 is unbalanced moment produced now this is to be balanced therefore plus 22.3 is to be applied as per the distribution factors we distributed here and here also minus 30.7 is to be applied and as per the distribution factor it is to be distributed in this way we will continue with number of iterations and once we get a very fractional value we will stop. So, after 3-4 iterations it is observed that the process is over then the take a summation of all these members so AB, BA, BC, CB, CD and C all the summations we can take it here. So, these are the final moments of the members and these we can represent here with the help of this free body diagram now you can see here the vertical member the members the moment is 44.5 here then 89.1 then here 89.1 as per their sign so if the moment is positive there we shall show the clockwise moment if the moment is anticlockwise we shall show it anticlockwise sorry if the moment is negative we shall show it anticlockwise. In this way as per our sign convention this free body diagrams are shown here now there is no moment at D and E as you can see D is hinged and E also is hinged therefore there is no moment transferred here. So with the help of this free body diagram now we have to draw the bending moment diagram so the bending moment diagram before that we will just look at this free body diagram accordingly we will draw the bending moment diagram now you can see here now one sign convention we follow here if the moments are producing the tension outside we take it as negative if the moments are producing tension inside we take it as a positive moment so here you can see at moment at A the moment is creating tension inside therefore it is a positive moment 44.5 and here you can see this moment is creating a tension outside therefore it is taken as negative negative 89.1 here you can see here same the case here so here this also is negative this also is negative so both are hugging so hugging moments are shown here and in between due to this UDL we get a positive moment so the maximum positive moment we get 101 km and here also for this member CE the negative moment is produced here this is hugging and here also the tension outside therefore it is minus and then it is 0 at D the moment is 0 at E also the moment is 0 in this way we will construct the bending moment diagram these are the references which are used for this presentation thank you thank you very much