 This is Dr. Mahesh Kallanshiti, Associate Professor, Department of Civil Engineering, Valchin Institute of Technology, Solapur. In this session, we'll discuss about analysis of indeterminate structure by moment distribution method, wherein we'll focus on sway frames. The learning outcome will be, at the end of this session, students will be able to analyze sway frames using moment distribution method. Let us take an example, as shown here, frame subjected to a point load, and we have to analyze this frame using moment distribution method. So before we start, we have to first decide whether it is a sway frame or non-sway frame. Now from the loading condition, it is clearly indicated that it is a sway frame because the loading is not symmetric. Therefore, there will be a sway produced in the frame. Therefore, we have to adopt the sway frame analysis for this case. Now we have to follow certain steps here for the analysis of sway frames. These are given here. First of all, the frame shown is first held from the side sway by applying an artificial joint support at sea. So this is our problem actually, but first of all we have to prevent the side sway by assuming one artificial support here, and we have to carry over the moment distribution process. So the moment distribution is applied and by static equilibrium, the restraining force R is to be determined. The moment obtained thus are called as non-sway frame. It means here in this sketch you can see here, we have to apply the restraining force here and then we have to carry over the moment distribution and based on that, we have to calculate the moments produced in the members. So these moments, we call it as a non-sway moments. Later on, once we get this reaction, now after the analysis is over for the restrained case, we can determine how much R is produced here. Now in the second step, we apply the same force here on the frame and we do the moment distribution again. So the third step as it is shown here, the equal but opposite restraining force is then applied to the frame here as shown. The moment in the frame are calculated. The moment obtained thus are called as sway frames. So in this way, we can calculate the non-sway frame and sway frames. The summation of these two gives us the final moments. So this is an logical process of the sway frame. Let us apply these particular steps to our problem. So first of all, as usual, we have to calculate the distribution factors for the member, sorry, for the joint B and C. So as usual at B, we have two members. So the distribution factors calculated as 0.5, 0.5. Then for C also, we have two members. So the distribution factor is again 0.5, 0.5. Since all the members have got the same length and the same conditions, therefore we get all the distribution factors as 0.5. Then we have to calculate the fixed end moment. So from the figure, we have only fixed end moment for a member BC, rest all there is no lateral load. So we have to consider only the fixed end moment for the BC. And again, we have to consider this particular frame as a non sway frame by applying an artificial support here. So this is a restrainment which we have produced here for the sway. For this conditions, now we calculate the fixed end moments BC, since it is a case of eccentric load, therefore these are the fixed end moments produced. Accordingly, we get the fixed end moment BC is minus and minus 10.24 and for CB, we get plus 2.56. After that, we go for that, our conventional moment distribution process wherein we write all the joints here, all the members and the distribution factors are 0.5 as we discussed here and the fixed end moment we got only for the BC member. Therefore BC and CB, these are the fixed end moments that we get. Then we proceed with the balancing of the joint as usual and carryover wherever applicable. So in this way, in this case, from B, carryover is possible towards A and from CL, so carryover is possible towards D. And from B, carryover is towards C and from C, carryover is towards B. So all the carryover is taken care of and the distribution is continued and this iterative process where to continue till we get a very small and negligible amount of moment left. So after the moment distribution process, we got the final moments here in all the members. So these moments will use to calculate the restrained force R. So the equation of equilibrium are applied to the free body diagram of the columns in order to determine AX and DX. So this AX and DX means these are the shears that you can see here. Now these two members, we got the end moments. In the previous slides, we have seen that MAB, MBA, all the moments we calculated. So these moments are shown here and the free body diagram is drawn and from the free body diagram, we can calculate the value of R. Basically here AX we can determine and DX also we can determine by applying simple static equilibrium conditions. So by doing this, we get the R that is a restraining force as 1.73. So actually AX we get 1.73 kilonewton and DX we get 0.81 kilonewton. So by applying the equilibrium conditions, we get the restraining force R 0.92 kilonewton. Before we proceed, let us go for this review questions. I request you to take a pause and answer these review questions. The first question is, if three members are meeting at a joint and the stiffness of the members are K1 is EI, K2 is 2 EI, K3 is 1.5 EI. The distribution factor for the member is, member 1 is, so these are the four options. Then the second one, the distribution factor of the member OC of the frame as shown in figure. So in this figure, the distribution factor of OC you have to calculate. So four options are given here. So the correct answer, these are C for first question and 0.4 for the second question. Let us proceed further. Now, once we get the restraining force, then the same restraining force will apply here. So then equal and opposite values of R equal to 0.92 kilonewton must be applied to the frame at C and the internal moments are to be computed. So now for that, what happens actually, our job is to get the moments produced due to the restraining force. But for this, we need to calculate the sway which is produced because of R, which is bit complex process. Therefore, we adopt the reverse process wherein we assume some arbitrary movements are produced here because of the R. So sway is taking place towards the right side. Therefore, the moments are forming anticlockwise direction and arbitrary moments we assume. And later on we'll use one distribution factor for that and then we'll calculate it. So assume a force of R is equal applied at C causing the frame to deflect as shown. Then as I told that we apply the arbitrary moment 100 kilonewton meter anticlockwise for AB and DC. And then we calculate what is R produced due to this arbitrary moments. Then we go for the distribution process. Now again, you can see the arbitrary moments we write here minus 100. All the moments are anticlockwise in nature. Therefore, it is considered as minus. And then distribution factors already determined. Therefore, these are to be calculated. So based on these all we get the moments here. We have to do the same process. We get the final moments at the end. Now these final moments actually are due to the assumed value. That we go for a next concept what is called proportioning of moment. Now from the equilibrium, the horizontal reaction at A and D are calculated. So here you can see the moments we just draw plot here and corresponding value of reaction R we calculate. Before that we have to calculate AX and DX by static equilibrium conditions. So AX we got 28 and DX we got 28. So again, as a whole, if I apply the equilibrium equation R must be equal to DX plus AX. So that value comes up to be 56 kilonewton. So R dash is 56 kilonewton which creates a moment as shown in free ball diagram. However, we need to the moments corresponding to R dash as 0.92 because we got the restraining force as 0.92 kilonewton but we got the restraining force as 56 kilonewton based on our assumed arbitrary moments. Therefore, the moments caused by R equal to 0.92 kilonewton can be determined by proportioning. And this proportioning can be done using a multiplication factor as 0.92 upon 56. So based on this concept, we can determine the final moments. So final moment is equal to non-sway moment plus the modification factor into moments corresponding to arbitrary fixed end moments. So if I consider AB, so the non-sway moment for AB is 2.88 and this bracket, this 0.92 upon 56 gives me a multiplication factor and minus 80 is the moments corresponding to arbitrary fixed end moment. Therefore, we get MAB as 1.57. Correspondingly, we calculate the moments of remaining members as shown here. Then this is what is the free body diagram from these particular values. So once we get the free body diagram, then based on this, we draw the bending moment diagram with our conventional assumption that tension outside is negative and tension inside is positive. So you can see here, this is a diagram. These are the references which are used for the presentation. Thank you.