 Hello, everyone. In this course, when we discussed about idealization of a real system, we only talked about single degree of freedom system. Now, single degree of freedom system, although it is the most simplified representation of a dynamic system, it may not always be appropriate. So, we may have to consider multi degree of freedom system, which may be able to represent the behavior of structure more accurately. So, to get into that, first we look at different type of multi degree of freedom system and then like we did for a single degree of freedom system, we will first set up the equation of motion for a multi degree of freedom system and then see how the equation of motion can be solved. So, let us get started with this lecture. Till now, we have studied about single degree of freedom system and then we also studied about continuous, continuous basically systems in which this continuous system was reduced to a single degree of freedom system or more specifically generalized SDF system by using a shape function. Now, we said that this continuous function which was reduced using phi to a generalized S2DF generalized SDOF system. We said that this was an approximate method because we did not exactly know what was the deflected shape phi. However, now what we are going to do, we are going to simplify our system to a multi degree of freedom system and we are going to find out the exact deflected shape using the methods of analysis in the subsequent chapters. And as we have previously discussed, any system in reality, let us say even if you consider a beam, a cantilever beam, something like this. So, in reality, it would have infinite degrees of freedom. Now, it might be appropriate sometimes to reduce it to a single degree of freedom system which would of course provide us limited result in terms of deformation. If we want the variation of internal forces then we analyze it as a continuous system using this method that we have discussed in previous chapter. However, if you want more accurate representation, ok, then we can approximate the same system as multi degree of freedom system and depending upon the number of degrees of freedom we considered, we can get results that are quite accurate, ok. And to demonstrate the application of multi degree of freedom system, we will start with a shear type building, ok. And we discussed in detail what does a shear type building mean in terms of its behavior. So, we said that for a shear type building, ok, the masses are lumped at each floor and we represent the deflected shape using horizontal degrees of freedom at each floor level, ok. So, if it is a 3 degree of 3 storey building, it would have 3 degree of freedom system. And let us say the deflected shape would look like something like this, ok. And I do not know exactly what that deflected shape is, but I would be able to find that out exactly. However, before that what I need to do is set up the equation of motion, ok. So, first let me draw the deflected shape, ok. It would look like something like this here or there is no flexure here. Remember, in a shear type building, there is no flexure deformation in the beam, ok. So, let me just delete this, ok. So, it looks like something like this and this one also looks like something like this, ok. So, the masses are here and we are going to represent the degrees of freedom as u1, u2 and u3, ok. Now, like we have previous leader, we can also assign a damper to it, ok, which would something like let us say this between each ends of a story. So, between the two floors of a storey, ok, the floor and the roof of the storey. And let us say assign some parameters. So, this damper has like C1, C2, C3. Now, we said that at any storey if we consider in a multi-degree of multi-storey building. So, let us say at any storey J, ok, if there is a storey deformation, ok, there would be a storey shear, ok. So, that is why it is called shear type building, because the deformed shape is only represented through shear deformation in the building. So, if there is a relative deformation between two floors or sorry floor in the roof of a storey, then there would be a storey shear that is generated. One would be due to the elastic deformation of the columns that connects the floor and the roof of that storey and the second would be if we are considering damping, ok. So, the first is the stiffness contribution to the storey shear, which is written as KJ, which is the storey stiffness, ok. KJ times the storey drift and storey drift is nothing but deformation at level J, ok, minus deformation at the storey below that, ok. So, Uj minus Uj minus 1. So, this is the stiffness contribution. Similarly, we can also find out the damping contribution to the storey shear as Cj times storey drift velocity, let us call it. So, Cj times Uj, Uj minus 1, ok. This is the velocity. Now, what do we need to do? We already have established the solution. We have already obtained the solution for a single degree of freedom system, which was of this form, right. If you remember, this is M u double dot, C u dot, this K u dot is equal to P of t. Now, for this case, only one degree of freedom u was considered, but now if you look at here, ok, we do not have a single degree of freedom. So, we do not have a single deformation. We have u1, we have u2 and u3, ok. Similarly, forces might be at different levels. So, let us say this is P1, P2 and P3, ok. So, we have basically forces and deformation that are not a single valued function, but it is like a vector of multiple quantities. So, here instead of a single quantity, the deformation u actually is a vector. Let us represent it as a vector. If it is a 3-storey building, let us say it is u1, u2, u3 and if it is a n-storey building, it could be u1, u2 up to un. Similarly, the force P, the applied force can be represented as another vector P1, P2, P3, ok. And how do we write it? It depends on where the forces are being applied or whether the degrees of freedoms are being defined, ok. So, u and P are written like that, ok. And what I want that if somehow the equation of motion for this type of multi-degree freedom system, it can be reduced in this form here, then the job would become much simpler for P, because I have already, I already have the solutions that we have derived for different type of loading and conditions, ok. Now, there would be a key difference as I said between SGF and MDF system and that would be, we now have multiple quantities here. So, now we are not dealing with the single quantity, we would be dealing in vectors to represents the, let us say force vector or the displacement vector and matrices which you would see is used to represent the stiffness matrix, mass matrix and the damping matrix, ok. So, let us see how to derive the equation of motion for a multi-degree freedom system starting with this 3-story building that we have. So, to do that what I am going to do, I am going to cut this building, ok, at each floor as I have shown here and then draw the free body diagram and then write down the equation of motion for each of these 3 stories, ok. So, let us do that, let us see what do we get. So, let us first draw the free body diagram of the topmost flooring, ok. Remember we have a mass m here, there is a force P3 which is being applied, now it is moving in the rightward direction as U3 and the floor below it is moving in the again rightward direction with the U2. So, differential displacement or the story drift is actually U3 minus U2, ok. So, there would be force, stiffness contribution of the force when I cut this story at it would be FS3, ok, which would be equal to k3, ok. So, I am assuming the story stiffness is to be represented as k1, k2, k3, just write it here, ok. And then I have another force, if I have defined the dampers between the story as Fd3 equal to, sorry this, let me just finish this up. So, this is FS3 equal to k3 times U3 minus U2 and Fd3 is C3 times U3 dot minus U2 dot, ok. And because it is moving on the rightward direction, it would have inertial force which is equal to mass of that floor times U3 double dot which is the acceleration of that mass, ok. So, this is the free body diagram of the third story. Now, if we consider for the second story, remember I have forces from above and below both. Now, from the above the forces would be equal and opposite to FS3 and Fd3, ok. So, it would be in the opposite direction, ok. So, I have FS3 in this direction and Fd3 in this direction. And I have this applied force which is equal to P2, ok. This is M2. So, I have the inertial force which is acting opposite to the direction of motion as mass times the acceleration. And then from the story below, again I would have a force FS2 which is equal to the story stiffness times the story drift of that floor, remember the deformation here is U2 and below it is U1 and that would be opposite to the direction of the deformation. So, FS2 is this much and then F2 to U would be C2 times U2 dot minus U1 dot, ok. So, this is for the second story. And then now let us write down the, draw the free body diagram for the first story that we have, ok. I am going to follow the same procedure that I have done for the previous two stories. So, I would have FS2 here equal and opposite to the forces that I have applied on the story above and Fd2, ok. There is force P1 here, ok. And then I have the inertial force which is M1, ok. So, let me write it here again M1 times U1 double dot. And then I have force which is equal to FS1 which is equal to K1 which is the story stiffness times the story drift of the first floor which is U1 at this point. However, at the second the floor is basically the ground. So, it is at has zero velocity. So, I will write this as one K1 times U1 and similarly Fd1 would be C1 times U1 dot, ok. All right. So, I have now three free body diagram. This is for the third story. This is for the second story and this is for the first story, ok. I am starting with the third story because it just makes the job easier for me to explain in terms of forces from the top to bottom. Now, we can go ahead and write down the equilibrium equation for each floor, ok. So, the dynamic equilibrium equation of motion for each floor and then combine them as a single equation, ok, in terms of stiffnesses and matrices. And so, let us see how do we do that. And in terms of writing down the equation, let me first write down the equation of motion for the first story, ok. So, I am going to write down here if you look at here M1 U1 plus K1 U1, ok, plus C1 U1 dot and then I have forces Fs2 and Ft2 which are acting opposite to these forces and the expression for these forces are basically I have already given you Fs2 and Ft2. So, let me write it here. This is U2 minus U1, ok, and then minus C2 U2 dot minus U1 dot and this is equal to the applied force which is P1, ok. So, we can write down in this form, ok. And then let us combine the similar terms together. So, this is M1 U1 double dot plus K1 plus K2 U1 dot minus K2, sorry, there is no dot here, it is just the deformation. So, K2 times U2 plus C1 plus C2 U1 dot minus C2 U2 dot and this is equal to P1. So, this is for the first story, ok. This is for the first story. Now, let us go ahead and write down the similar equation of motion for the second story. So, for the second story again I can write similarly as M2 U2 double dot plus K2 U2 minus U1 plus C2 U2 dot minus U1 dot, ok. And then minus Fs3 and Ft3 which are acting opposite to these forces and the expression for Fs3 and Fd3 are actually given here. So, let me just write that K3 U3 minus U2 minus C3 U3 dot minus U2 dot and this is equal to the applied force P2, ok. So, again I can combine similar terms together as a coefficients of the deformation acceleration and velocity and write this one as, so this one would be minus K2 times U1 plus K2 plus K3 times U2 and minus K3 U3, ok. And then I am going to write the damping term minus C2 times U1 dot plus C2 plus C3 U2 dot minus C3 3 dot equal to P2. So, this was for the second story, ok. Let us write down the equation of motion for third story which in fact is quite simple because there are no forces from above. So, I would write it as M3 U3 dot plus Fs3 and Ft3 which can be written as K3 U3 minus U2, ok, plus Ft3 which is C3 U3 minus U2 dot and this is equal to the applied force which is equal to P3, ok. So, again let me write it as this minus K3 U2 plus K3 U3, ok and this minus C3 U2 dot plus C3 U3 this is equal to P3. So, now what we see we have three equations here 1, 2, 3 this is for the third story here, ok. So, these are three simultaneous equation, ok and you know you might have studied this in your vector algebra how to write these simultaneous equation, ok in terms of a vector and matrices. So, we combine them together, ok and what I am going to do here just for your understanding I am going to simplify it even further and write down only for once just first time here later I am not going to repeat the same thing, ok. So, that you know that how do we write down this in the terms of matrices and vectors, ok. So, to do that I know that for acceleration there are coefficients m1 m2 and m3 and then vectors that we are considering would be acceleration vector, velocity vector and the deformation vectors, ok. Now in this case let us write down the equation 1 like this m1 U1 dot there are no coefficients to U2 acceleration U2 U2 and the acceleration due to U3, ok. So, I am going to write this one as ok just 0 coefficient plus let us now route first the damping term because conventionally we write down the damping term before the stiffness terms. So, in terms of damping if we look at it we have C1 plus C2 times U1, ok and then the coefficient of U2 dot which is this and there is no coefficient of U3 dot. So, I am just going to write it 0 similarly the stiffness terms, ok it would be k1 plus k2 times U1 minus a2 times U2, ok plus 0 times U3 and this is equal to p1. I am going to do exactly the same thing for the equation 2 and equation 3 as well, ok. So, this would be 0 times U1 double dot now I have m2 times U2 double dot. So, I am going to write it like that and then no term for the U3 double dot for this I am going to write this one as plus sorry in this case is it would be minus k2 times U1 then I have k sorry I was writing down the damping term first, ok. So, let me just do that C2 times U1 double dot plus C2 plus C3 times U2 double dot and then C3 times U3 double dot, ok and then I can write down the stiffness terms which are a2 times U1 plus k2 plus k3 times U2 minus k3 times U3 and this is equal to p2. Now, let us write down the third equation this one in the similar form 0 times U1 double dot, ok minus or plus let us say there is no term here. So, U2 double dot plus m3 U3 double dot, ok and then there are no terms here for the U1 dot, ok. So, it would be 0 U1 dot minus C3 U2 dot plus C3 U3 dot. Similarly, the stiffness terms can be written as 0 times U1 plus minus k3 times U2 and plus k3 times U3 and this is equal to p3. So, now if you look at it all of these are multiplication of some matrices and vectors. For example, if you consider this part here and then this part here and then this part here let us compartmentalize these three sets of equation like that. So, this one is nothing, but multiplication of the matrix m1 0 0 0 m2 0 and 0 0 m3 this matrix multiplied with this vector here, ok and I would take the negative terms these terms inside. So, that it is all plus here and then I have the damping matrix which is basically multiplication of this matrix here C2 C2 plus C3 sorry this is a negative sign here because I have taken it inside then minus C3 then 0 minus C3 and then C3 is multiplied with the vector the velocity vector which I have here and then the stiffness terms which can be written as k1 plus k2 minus k2 0 then minus k2 k2 plus k3 minus k3 0 minus k3 and k3. So, this is U1 U2 U3, ok and this is P1 P2 P3, ok. So, if you look at carefully this three sets of simultaneous equation can be written it like this and if you look at carefully this is nothing, but we this is the mass matrix for a multi degree of freedom system and here it is a 3 degree of freedom system for the 3 storey building, ok and this is the acceleration vector plus the damping matrix times the velocity vector and then the stiffness matrix times the deformation vector and this is the applied load vector, ok. So, we have written down our equation of motion again in a similar form that we had we have been using for a single degree of freedom system. How to solve that, ok we will discuss later, but the first step is to actually find out the equation of motion. So, this is the equation of motion which is the first steps towards the solution of a multi degree of freedom system, ok and the throughout this chapter our goal is to develop or determine the equation of motion for multi degree of freedom systems, ok. So, this is the equation of motion, ok. Now, let us come back look at this equation that we have here and then make out some observation here, ok. So, if you look at here what do you see about the mass matrix here, ok the mass matrix is actually a diagonal matrix. So, there are only elements m 1, m 2, m 3 which are along the diagonal and the off diagonal terms are 0, ok. So, the question becomes would it always be like that, ok. So, mass matrix is typically you would obtain is a diagonal matrix as long as it is a. So, mass matrix would always be a diagonal matrix for a lumped mass systems, ok. So, for lumped mass systems where the degree of freedom correspond to the lumped masses then you are going to obtain a diagonal mass matrix, ok. This would be different you might obtain off diagonal terms for a continuous or distributed mass system. Let us say you have a beam in which the mass is distributed along length. In that case if you define degree of freedom let us say u 1 here and u 2 here to represent let us say translational motion like this. So, let us say this is u 1 and this is sorry, this is u 2 and this is u 1. Then you might see that your mass matrix would not be diagonal and you would have term off diagonal term as well. But typically and which happens for a multi-story buildings, shear building if you have a lumped mass system where the degrees of freedom have been defined at each masses then you would get a diagonal mass matrix, ok. Now, let us look at the stiffness matrix here and the damping matrix here. The first observation that I see is that a stiffness and the damping matrix are symmetric matrix, ok. If you look at it we have the diagonal elements k 1 plus k 2, k 2 plus k 3 and k 3. But the off diagonal terms are actually mirror image of each other along that diagonal. So, minus k 2 here, minus k 2 here, minus k 3 here, minus k 3 here and same is the case for the damping matrix. It is also symmetric. Now, as long as our system is linear what we would see that the mass, the stiffness matrix that we get would always be symmetric, ok. And the third observation that we see that if we define, ok, if we define our damping matrix or the dampers in such a way that it represents the same degrees of freedom to which the stiffnesses have been defined then the form of stiffness matrix, sorry the damping matrix would be same as the stiffness matrix, ok. And this is especially true for multi-degree of or multi-storey building in which dampers are defined at each story. Although, you know, we will discuss later that we do not actually get our damping matrix like this because in reality there is no damper, you know, between different story except in the cases where you actually install dampers, but to get the damping matrix we utilize some other methods to get the damping matrix, ok, utilizing the mass and stiffness matrices, ok. But that is for the later discussion, ok. What is important to note here is that if I have a multi-storey building and I have defined the dampers between different stories then the form of the damping matrix would be similar to as the stiffness matrix, ok, all right. So, these are the few of the observations that I have made. Now, once I have found out the equation of motion for multi-storey building, ok, let us consider some different representation. Remember, I have said that for a shear type building here, ok, there is no flexure deformation, all the mass are concentrated at the floor level, ok, and the degrees of freedom are basically horizontal deformation at each level. So, many times you would also see this representation, this lollipop representation being used to define a shear type building and basically u1, u2, ok, to define the degrees of freedom here, ok, and the story stiffness we say is k1, k2, k3. So, you should be aware of these representation in terms of dynamic behavior these basically, this is basically used to represent the same multi-storey building, ok. So, this is p1, p2, p3 with the masses m1, m2, and m3, ok. So, this is also used to represent the same shear type building. Another representation that is we have been like you know considering is the spring mass tamper representation, ok. And again, we can utilize the same method to draw the spring mass representation. Remember, the first story is actually fixed to the ground, ok. So, to draw the spring mass representation, let us start with the first story here, ok. I have the mass m1 which represent the first story mass, let us draw the tamper as well. So, this m1 which is the first story is now connected to the second story, ok. Let us first write down k1 and c1 representing the stiffness stories here, stiffness of the story, the first story here. And then this is connected to the second story through the spring k2 and the tamper c2, ok. This is m2. And again, m2 is connected to third story, ok, c3 and k3 here. And this is m3. And the degrees of freedom are u1, u2, u3, ok, with respect to initial undeformed position or in this case, that would also be the equilibrium position, ok. And the forces that are being applied are p1, p2, p3, ok. So, this is another spring mass tamper representation of the same system. And you know, we can again consider the equilibrium motion and get the similar kind of equation, the equation that we have got here, ok. So, this is the second type of representation that we typically consider, all right, ok. So, now, we have seen that different type of representation, ok. Now, let us consider different components of a multi-degree of freedom system, ok. And we are going to utilize again the same shear type building, ok. Why do we do that? We will see the utility of that representation later. So, components, ok. Basically, we know that and we did it for single degree of freedom system. Any structure would have three components under dynamic load, the stiffness component, then the damping component, ok. And then the third component, which is the mass component. And when we apply external force on that system, then, so let us say this is the system here, ok. So, let me just draw the system, same three-story building, ok. I have applied forces P1 here, P2 here and P3 here. So, this is the combined system, ok. So, let me just draw all the components of it, ok. This overall system is nothing but combined representation of the three components, which is the first one is basically the stiffness component in which we consider the bare frame without mass or damper. So, we have bare frame, ok, in which we apply forces FS3, FS2, FS1 which represents the stiffness forces in the system, ok. So, the bare frame is there. In the second one, we are going to consider the damping component, ok. That would only have the damper and no frame, ok, or mass, ok. So, only the damper would be there. So, let me draw with dashed line here, ok. There is a damper here, which is like this. So, in this case, again I have Fd3, Fd2 and Fd1, ok. So, this is for displacement u, this is for velocity u dot and then the third component is the mass component, where we are not going to consider the frame or the damper, but we are only going to consider the masses at different levels. So, I only have these masses at different level and the forces that are being applied are actually the inertial forces. So, I am going to write it here F i1, F i2 and F i3, ok. And in general, the total applied force is actually the sum of the individual components. So, let me just write down the acceleration here, double dot, ok. So, the P or the vector P here is actually sum of vector Fs, sum of vector Fd and sum of vector F i. Now, we already know that the same expression we had used for single degree of freedom system, here we can write this as P as equal to remember Fs we have found out as the stiffness matrix times the displacement vector, the damping forces Fd we have found out as damping matrix times the velocity vector Fis the mass matrix times the acceleration vector, ok. Alright, we are going to see later and like you are not later but like just in the next section itself that we are going to discuss that this representation is quite useful in other type of method to find out the equation of motion, ok. And let us now talk about that other type of method. So, the method that we are going to discuss it is called influence coefficient method, ok. So, we are going to call it or it is called influence coefficient method, ok. Now, you might have come across this influence coefficient method in your structural mechanics courses or advanced structural mechanics courses, ok. We will see that how to utilize that to get the equation of motion and let me give you background why we need to do that. If you look at this equation of motion, so this can be either obtained using free body diagram and then finding out writing down three simultaneous equation and writing down all these equation. Now, this sometimes become little bit complicated for complex systems, ok. And you will see that this method that influence coefficient method that we are describing many times this provide an useful like you know it provides useful means to obtain the same equation of motion using an alternative method. So, this is not something you know this is just an alternative of finding out this equation of motion, ok. So, utilizing this influence coefficient method we are again going to find out the equation of motion, ok. And that is what we plan to discuss, ok. Remember that we are only considering linear systems. So, for linear systems principle of superposition is also valid, ok. Principle of superposition is valid, ok. So, utilizing this principle of super position and the influence coefficient method we are going to find out the equation of motion, ok. And let us see how do we do that. So, first let us start with the, ok, stiffness matrix. Now, using this influence coefficient method, third basically what do we do? We individually find out the stiffness matrix and damping matrix and then mass matrix, ok. Most of the times what you will see that we only find out the stiffness and the mass matrix and damping matrix we typically find it as using damping ratio that is obtained experimentally and then combining the mass and the stiffness matrix. So, we do not typically like you know go ahead and do that, but we will going to discuss that anyway here, ok. Now, for the stiffness matrix remember we have considered forces like FS1, FS2 and FS3, right. So, we say that, ok. Our goal here is to relate the external forces, FSj, ok or let me just first do it like this. Vector FS is nothing but FS1, FS2, FSj let us say it goes up to FSn. This equal to, ok, k matrix times u1, u2, uj and then u1, right. So, this is how we write down our FS matrix and the stiffness matrix elements are basically here, ok. So, basically our goal is to relate the external force that is FSj, ok to uj and whatever the coefficient that we get is actually constitute the this stiffness matrix, ok. So, the way we apply this influence coefficient method we apply a unit displacement along j, ok. We apply a unit displacement along degree of freedom j, ok, while keeping other displacements equal to 0. So, if I have a 3 degree of freedom system let us say I apply u1 equal to 0 and keep u2 and u3 equal to 0, ok. So, unit displacement along d of j and 0, ok, at all other dofs, ok. Now, we know that typically when we apply any displacement at a degree of freedom then other degree of freedom would also deform that I know. However, if we want to maintain that that I have unit deformation at a degree of freedom j, but it is 0 at all other degrees of freedom I will have to apply forces at each and every degree of freedom to achieve that state of deformation, ok. Now, before going into that and further discussing remember that if we apply a force F on any spring or a stiffness component, ok and it deforms by a deformation let us say u. So, let us say F equal to k u, ok. For unit displacement u equal to 1, ok, this stiffness k is nothing but force that is being applied on the system, ok. This is the concept that we are going to extend to, this is a single degree of freedom we are going to extend to a multi degree of freedom, ok. So, what do we say here that if I have let us say F sj equal to coefficient let us say k ij times uj here if uj equal to 1, ok then k ij is nothing but the force that is being applied on that particular degree of freedom, ok. So, the stiffness coefficient k ij of this stiffness matrix here, ok. k ij is basically the force that is required, ok. So, force required along d u f i due to unit displacement, unit displacement at d u f j, ok. So, if I have let us say in this case if I consider for a three-story building here basically what I am saying if I consider element k 3 1 it is nothing but the force at d u f 3 due to unit displacement along d u f 1, ok. Now, if you look at carefully let me say I apply a unit displacement u 1 equal to 1, ok. So, let us say I have this building, ok and I am applying unit displacement along 1 and keeping others 0. So, the shape would look like something like this, right, ok. So, what this basically says if I apply unit displacement 1 here I need to apply force which is equal to k 1 1 here, k 2 1 here and k 3 1 here, ok and this would these forces would result in a case where u 1 equal to 1 and u 2 equal to u 3 equal to 0. Now, if you look at it k 3 1, k 2 1, k 1 1 are nothing but the first column of the stiffness matrix, ok. Similarly, if I do like this in which I apply unit displacement and node 2 so and 0 everywhere else. So, it would look like something like this again I have to apply force which is k 2 2 here remember the second subscript of k, ok. So, if I have a k ij, ok, j is where the displacement is being applied and i is basically the degree of freedom at which you are considering the force, ok. So, j is d u f at which displacement is applied and i is the other degree of freedom. So, in this case remember 2 is the degree of freedom at which the displacement is applied. So, the force applied at this degree of freedom 2 would be k 2 2 here it would be force applied at degree of freedom 1 due to unit displacement at degree of freedom 2 force applied at degree of freedom 3 due to unit displacement of d u f 2, ok. So, this is the case when u 1 equal to u 3 equal to 0 and u 2 is 1. Similarly, so now this elements if you look at it it forms the second column of the stiffness matrix, ok. Sorry this is 2 2 here. So, let me rewrite it and then this becomes k of 3 1 sorry this third one we still need to find out. So, let us just write down the third case here. In the third case, ok, I have unit displacement and 3 and then 0 displacement everywhere else. So, u 1 equal to u 2 equal to 0 and u 3 equal to 1, ok. So, that I need to apply force at degree of freedom 3 due to unit displacement at degree of freedom 3 force at degree of freedom in this case basically equal to 2 at degree of freedom 3 in this case force at degree of freedom 1 due to unit displacement of degree of freedom 3, ok. So, the third one is basically k 1 3 k 2 3 k 3 3, ok. So, using this procedure we can get our k equivalent. Now, if the storage stiffnesses are given to you k 1 k 2 k 3 and similarly, ok for all these cases remember we are assuming the same storage stiffness. Again, we can consider the equilibrium of each and every floor to find out what is k 3 3 k 2 3 k 1 3 or k i j in terms of k 1 k 2 k 3, ok. So, I will I will I will give you an example, ok. So, let us take example of this first case, ok and let us cut this, ok at 3 level and find out what is the value. So, if I consider the first story which is nothing, but so it would deform something like this here, ok. So, first story something like this where this is unit deformation. Now, to maintain that I am applying a force k 1 1 here. Now, because of deformation there would be forces from below which would be story stiffness k 1 times the story drift which is 1 at this level and 0 at the ground. So, it would be 1 minus 0. Now, remember there would be forces from above as well and it would be in this direction and what would be that force the story stiffness k 2, ok times the deformation of the floor above and subtracted by this deformation here. Now, deformation of the floor above is 0, but the deformation of this floor here is this k 2 as well. So, k 1 1 would be nothing, but k 1, ok plus k 2 here, ok. So, we can write down the equation of motion and get k 1 1 is k 1 plus k 2. Now, if I consider the story above this, so the second story, ok which is let us say something like this is here. On this I am applying force k 2 1 here, there is force from below which in which direction k 2 times, ok whatever the deformation at this point which is 0, ok minus the deformation of the floor below which is 1. So, in this case k 2 1 comes out to be equal to minus k 2 and then I consider the story above which does not have any story drift, but ok. So, basically it is k 3 1 and because it does not have any story drift, would there be any force here? So, f s 3 is actually equal to 0 here because f s 3 is what? f s 3 is k 3 times u 3 minus u 2 both are equal to 0, ok. So, k 3 1 equal to 0. So, let us look at it k 1 1 k 1 plus k 2 k 2 1 minus k 2 k 3 0 and then I still need to find out these two columns which I can do by finding out or repeating the same procedure that I have just described. Now, if you compare the first column the equation that we have derived here look at this k 1 plus k 2 minus k 2 and 0, ok k 1 plus k 2 minus k 2 and 0. So, we got exactly the same column, ok. And you will see if you repeat this procedure here and here you are going to get exactly the same stiffness matrix, ok. So, this let me write it as minus k 2 0 k 2 plus k 3, ok minus k 3 minus k 3 here and this is k 3. So, this is the k matrix. So, using the influence coefficient method, ok we have been able to derive the stiffness matrix, ok like this. Now, you might think at this point that this seems I mean I mean this seems does not it does not seem any bit easier from the previous method that we described. But remember for this case it might not be, but some of the other cases you will see and we will do some examples, ok. This method the influence coefficient methods actually is much simpler that writing down the simultaneous equation and solving and formulating the equation of motion, ok. And this also provides us some insight into the dynamic behavior of the system, ok. So, this is how we get the stiffness matrix, but remember we still need to find out the damping and the mass matrices. Now, damping matrix again we are going to follow the same procedure, ok. We are going to say remember damping force is what f t equal to c times u dot. So, if, ok if u dot equal to 1 then the damping coefficient is basically the force applied on that particular degree of freedom. For multi-degree of freedom system again we are doing to extend that we are going to say c i j, ok. Remember for stiffness matrix we said that k i j for this we are going to say that c i j is nothing but the external force in d u f i due to unit velocity now not the unit deformation unit velocity, ok in d u f j, ok. So, again we are going to follow the same procedure we are going to apply u j u dot j the velocity at the j th degree of freedom as equal to 1 while keeping velocity at other degrees of freedom equal to 0 and then finding out the forces that are needed to maintain that velocity profile and that would give me the column for that particular degree of freedom j, ok. And I am going to repeat this for all the degree of freedom and I will keep on getting this basically the damping matrix, ok. If it is a multi-degree of freedom system, ok. What you can do, you find out the stiffness matrix and if your damper are also connected between the same degrees of freedom you can just replicate that in terms of c 1, c 2, c 3 however that might not always work. So, just be careful if it is not a shear type building or if like in a degrees of freedom corresponds to different velocity terms then your damping matrix would be different, ok. So, c 2, c 2 plus c 3, ok. This is minus c 3, this is 0 minus c 3 and c 3, ok. Now the third term is still remaining that is the getting the mass, ok matrix. So, let us see how do we get that, how do we get the mass matrix. Now to get the mass matrix we are again going to employ the same thing, ok. Remember mass the inertial force is mass times acceleration. So, if we apply unit acceleration then mass is nothing but the inertial force that is being applied, ok. The procedure remains same in this case we are going to say m ij which is the ith column sorry the element at the ith row and the jth column in the mass matrix is nothing but force the external force in dofi due to unit acceleration now, ok. So, unit acceleration in dufj, ok. Now for our case, ok the it is very simple because if you consider this representation here, ok. The acceleration representation here I only have mass, right and if it is a lumped mass system, ok. So, let me just draw it here remember there is no damper here there is no frame here only these masses are here and if I apply a unit acceleration, ok. So, let us say u 1 is equal to 1 and u 2 and u 3 accelerations are 0 then this is f 1 due to unit acceleration at 1 this is a degree of freedom 2 due to unit acceleration at 1 degree of freedom 3 due to unit acceleration at 1, ok. So, in this case if the unit acceleration is only applied at 1 there are no forces in f 2 1 or f 3 1 because these are not connected, right. So, in this case I can directly say if the unit acceleration is only applied at mass m 1 f 1 1 would be or not let us not call it f 1 1 sorry let me rewrite it. Let us call it m 1 1. So, mass influence coefficient m 2 1 and m 3 1, ok. So, let us consider the free body diagram of this which is a force is being applied m 1 1 and the acceleration is 1 here, right and it is not connected to any of the masses. So, what does that mean m 1 1 the force is equal to this is the mass m 1 m 1 times 1 which is equal to m 1. Now, if we consider the this one here, ok the free body diagram the acceleration is equal to 0, ok the force this is applied is m 2 1. So, m 2 1 is equal to whatever the mass here is m 2 times u 2 double dot which is equal to m 2 times 0. So, 0 similarly, m 3 1 is also equal to 0. So, and this is what this would give us a diagonal matrix because the first column that I get as m 1 0 0 similarly I would get 0 m 2 0. So, for lump mass system it would only produce acceleration at that particular degree of freedom at which you are considering the unit acceleration, ok this is your mass matrix, ok and this you would always as I previously mentioned get this for a lumped mass system a diagonal mass matrix. However, if you consider cases, ok where you have the bar let us say distributed mass, ok and in this let us say it is supported by spring. So, that you are representing it the defined position through 2 degree of freedom let us say u 1 and u 2 because in general it can deform I mean it can rotate and it can translate. So, you need 2 degree of freedom to represent the defined position of the system with respect to its initial position. Now, in this case you have degree of freedom u 1 in u 2. So, if you want to get the mass matrix for this what you will do in the first case, ok let us say you consider u 1 equal to 1 and u 2 equal to 0. So, this is the acceleration profile you will get and acceleration at any distance x is would be equal to x by L, ok it increases from 0 to 1. In the second case, ok u 1 is equal to 1 and u 2 is equal to sorry u 1 is equal to 0 and u 2 equal to 1. Again if you consider x from this direction remember I have consider x from this direction you consider x from this direction your acceleration at x would be equal to x by L. Now, let us say in these degree of freedom you have applied force at degree of freedom 1 due to displacement, accelerate unit acceleration at degree of freedom 1, then force at degree of freedom 2 due to unit acceleration at degree of freedom 1. Similarly, force at degree of freedom 1 due to unit acceleration at 2, force at degree of freedom, 2 due to unit acceleration at 2. So, basically your mass matrix is m 1 1, m 1 2, m 2 1, m 2 2. So, this system would give you your first column, this system would give you your second column ok. Now, if you look at carefully and if you consider the equilibrium of this, you can never have m 2 1 equal to 0 or in this case m 1 2 equal to 0 ok. So, here you would have non-zero off diagonal term as well. So, you would have some term here, some term here and some term here. So, this is a distributed mass system and in this case I do not get a diagonal mass matrix ok even if I employ the influence coefficient ok. So, this is one of the example ok and you would see similar other example where the mass matrix is non-diagonal ok. So, using the influence coefficient method, we have been able to derive the mass matrix, then the stiffness matrix or let us say first the damping matrix and then the stiffness matrix. Damping matrix if it is a shear type building, we just replicate the mass stiffness matrix and this would be equal to the applied force vector along the respective degrees of freedom ok. So, this is another way of finding out the equation of motion using influence coefficient method and depending upon the problem, you might observe that one method or the other one it would work better, you have to select that method, but remember one thing in principle it does not matter what method you use, you would still get the same answer. So, it might be that your selection of method might make your job little bit more complicated, however the final answer would still be the same ok. Alright, so with this I would like to conclude this class. In the next class, we are going to see more example and application of these two method ok. Thank you.