 This is Dr. Mahesh Kalyansheti, Associate Professor, Department of Civil Engineering, Valshan Institute of Technology, Swalapur. Today, we will discuss about analysis of indeterminate structure by moment distribution method, where we will focus on only introduction. As it is covered in the earlier sessions, we have many methods for the analysis of indeterminate structures and we are also aware that there are basically two approaches of indeterminate structural analysis. One is called as force method of analysis, another is displacement method of analysis. In the force method of analysis, we have these many methods, whereas in the displacement method, we have these methods. Out of these, today we will focus on moment distribution method, which is the very easy and simplest method among all. So therefore, from this discussion, we may say that the moment distribution method is one of the methods of displacement method of structural analysis. The second distribution method was first introduced by Professor Haradi Kross of Phillionaire University in 1930. Earlier, people used to struggle when to solve such kind of complex kind of problem with the available methods at that time. However, after the introduction of this moment distribution method, the analysis became easy and since this moment distribution method is very simple and quick, therefore, it becomes very popular after 1930. This method provides a convenient means of analyzing statically indeterminate beams and rigid frames. So we can analyze beams as well as rigid frames and the most importantly, it was used when number of redundants are large and when other methods becomes very tedious. So as I told already that when we have large number of redundants, then other methods becomes very, very tedious. However, moment distribution method is very simple in that case. It is very popularly used on them. Now we will look at some important terms which are used in this topic such as the stiffness. So stiffness is nothing but it is a moment required to produce a unit rotation or slope at a simply supported end of a member and it is denoted by a symbol K. Now we have two cases for this stiffness, usually in case of the beams. We have these two cases. One is when both ends are hinged and another is when one end is hinged and other end is fixed. So let us discuss these two cases and we will see how the stiffnesses of the members is to be determined. Say the first one is beam hinged at both ends. So this is how we can look at this in A and B both are hinged and the moment now the moment required to produce a rotation unit rotation at A is called as stiffness and opposite end or the far end is hinged. In that case, we can have this value as 3EI by L. So from the structural analysis concepts we can derive especially the slope deflection method we can apply here and we can derive this equation. So the moment required to produce unit rotation is 3EI by L. Therefore we call this 3EI by L as a stiffness of this particular member which is denoted by K whereas the second case beam hinged at near end and fixed at far end. So this is how we get this problem where you can see the B is now fixed and for this case if I apply the moment to produce a unit rotation here then that moment required will be equal to 4EI by L. So therefore again 4EI by L we call it as a stiffness of this particular member. Therefore we have two cases as I told the first case is when the opposite end is hinged stiffness is 3EI by L and when the opposite end is fixed stiffness is 4EI by L. So this is how we remember these values and we will use further when we solve the problems. Later on let us go for the another term carry over factor. So whenever the moment is applied to one end some fraction of the moment is transferred to the opposite end such as so a moment applied at the near end induces a fixed far end equal to half its magnitude in the same direction. So that we can see here in this particular problem when I apply the moment here half the moment is transferred to the opposite end you can see half M and if the opposite end is fixed then only this transfer takes place otherwise this transfer does not take place and this we call it as carry over factor. So carry over factor is half now whereas in the second case if I look at the opposite end is a pin or hinged therefore no moment is transferred here therefore the carry over factor for this case is 0. So this is how we take the meaning of carry over factor so it depends upon the opposite end if the opposite end is fixed it is half if opposite end is pin it is equal to 0. So I request you to take a pause and answer these MCQ questions first one is moment distribution method is invented by so four scientist names are given you have to identify the correct one the second one the carry over factor in a prismatic beam whose far end is hinged is so four values are given so we just think over it take a pause answer these questions and resume the video welcome back. So this is the first question moment distribution method is invented by Hardy Cross as it has been discussed already in the earlier slides and the second question carry over factor in a prismatic beam whose far end is hinged so if the far end is hinged no carry over is possible therefore it is 0. Let us continue with the another very important concept in this method that is called distribution factor so whenever we apply a joint moment to any of the joint then the member is transferred to different different the moment is transferred to different members as per their distribution factor so the factor by which the applied moment is distributed to the member is known as the distribution factor. So this will discuss with one small case here so suppose the four members are meeting at one particular joint and these member have got different different support conditions so member 1, member 2, member 3 and member 4 and if this joint is subjected to a moment M so you can see some fraction of the moment M is transferred to the member 1 some fraction of the M is transferred to second member third member and fourth member that will discuss how to find it now first of all we have to find the stiffness of the joint so stiffness of the joint is nothing but it is a summation of stiffness itself all the members which are meeting at that point so therefore we have four members meeting at this joint therefore stiffness of all the four members we have to add so that we get a stiffness of the joint then first member stiffness of the first member as we know that there is a hinge end far end is hinge therefore it is 3EI by L the second member far end is fixed therefore it is 4EI by L third member far end is hinge therefore again 3EI by L and fourth member far end is fixed therefore it is 4EI by L in this way we can calculate the stiffnesses of all these members independently and you can take a summation of all these you will get total stiffness of the joint now the distribution factor is nothing but it is a ratio of the stiffness of the member divided by stiffness of the joint so if I consider the first member here the distribution factor for the first member df1 is equal to stiffness of member number 1 divided by stiffness of joint so k1 upon k in the same way m2 is sorry distribution factor for second member is k2 upon k distribution factor for third member is k3 upon k distribution factor for fourth member is k4 upon k in this way we can find out the distribution factors and once you get the distribution factors we can find out the moment transfer to these members such as say first member moment transferred is m1 so m1 will be equal to distribution factor for the first member multiplied by m it means some fraction of this m is transferred to m1 and that fraction we call it as distribution factor in the same way m2 m3 and m4 all these will just continue with the same logic and one more thing that all the distribution factors naturally have to be in fraction and summation of all the distribution factors shall be equal to always 1 so that we have to take care and accordingly we have to find the distribution factor so in the while we discuss the numerical examples there we will learn in detail how the distribution factors are to be determined these are some sign conventions which we follow in this topic such as the support moments so moments we take this sign convention clockwise moment we consider positive anticlockwise moment we consider negative same with the rotations also that is slope clockwise slope we consider positive anticlockwise we consider negative whereas in the settlements the sinking of the support because many time we come across with the sinking of the supports so there we consider this sign convention the settlement will be taken as positive if it rotates the beam as a whole in clockwise direction whereas the settlement will be taken as negative if it rotates the beam as a whole in anticlockwise direction so these are some fixed and moments for the beam so these are all fixed moments we need when we solve the problems some typical cases are presented here say simply the fixed beam subjected to udl so fixed moments is ql square by 12 if the intensity of load is q if it's a point load at the center so these are the fixed and moments if suppose there is a centric point load then these are the fixed and moments if there is a triangular load these are the fixed and moments if suppose a couple is present in between the span then these are the fixed and moments with the directions that you can consider these are the references which are used for this presentation thank you thank you very much