 Welcome back everyone. So, in today's class, we are going to learn how to find out response of a multi-degree of freedom system subject to external force. We have already discussed how to find out the free vibration response of a multi-degree of freedom system by decomposing a multi-degree of freedom system into multiple single degrees of freedom system and then combining the response of each single degree of freedom system to get the total response. We are going to do somewhat similar for the force response as well. In this case additionally, we are going to basically represent force as sum of or not the sum of but a vector representation of force at different degrees of freedom and then to represent it as a contribution of force in each mode through the diagonalization method that we have learned. So, let us get started. So, till now we have learned how to set up equation of this form. So, mass matrix times the acceleration vector and then damping matrix times the displacement velocity vector, then stiffness matrix times the displacement vector and now this would be equal to a force vector. Now remember, if we have a multi-degree of freedom system and let me take the example of this which is basically two-storey shear type building we have been discussing till now and this is 2k and 2m and this is k and m. So, let us say these are the two degrees of freedom we are considering and the forces that are being applied on these degrees of freedom are basically p not sin omega t here and there is no force being applied here. So, this is 0. So, then this excitation vector can be written as p not sin omega t and 0. So, now we have to solve this multi-degree of freedom system. Remember there would be n coupled differential equations which are basically coupled through the mass matrix, damping matrix and the stiffness matrix and like we did for the solution of free vibration response of damton and dam system what we are going to do here, we are going to uncouple these n coupled differential equations by multiplying them with modal matrix phi t and writing down my displacement vector as sum of contribution to the displacement in each mode. So, basically u n would be contribution due to first mode let us say it is u 1 plus u 2 and so on and basically this is nothing but summation let us say 1 to n mode shape times the modal coordinate let us say this is q r here. So, what we are going to do we are going to further write this as this is modal matrix times modal vector ok. So, we are going to write our u vector as modal matrix times q here and pre multiply this equation not with phi t, but basically phi t or yeah let us just multiply with phi transpose here transpose of the modal matrix. So, what we are going basically going to get is basically this equation here phi t m again the modal matrix and then we have the modal coordinate ok basically the modal accelerations. Then I would have this quantity here ok and then modal velocities the vector of modal velocities and then similarly for the stiffness matrix I have this here times the modal coordinate q and times phi t into p of t ok. Now, as we know considering the property the modal orthogonality of the mode that if the two different modes are multiplied ok then let us say this is phi n times phi r this is basically equal to 0 if these correspond to different mode and same goes for the stiffness as well and if I have a classical damping matrix the same relationship holds true for the damping matrix as well. So, basically what I get here is diagonal matrices I get here diagonal matrices mass times acceleration vector damping times velocity vector ok and stiffness times this vector and this is equal to vector p t ok all right. So, this quantity here is basically your m this quantity here is c and this quantity is k these are these all three are basically diagonalized mass matrix damping matrix and stiffness matrix ok. So, the element of these matrices basically m 1 m 2 and m n and same as like you know c 1 c 2 c n we have already seen that what do we get that as for example, my m n is basically transpose of the shape vector for the nth mode times the mass matrix times phi n similarly c n is phi n t times the damping matrix times phi n and then k n is basically phi transpose phi n ok and this is basically a new matrix for which the elements are p 1 p 2 p and then so on p n t ok like that where p n t is nothing, but nth mode transpose times the p vector that we had assembled ok. So, now basically what I have been able to do reduce my n degree of multiple freedom of systems to n single degree of freedom systems for which I am going to just show you the spring mass damper representation. So, basically I have this spring here this damper ok. So, this is let us say for the nth mode ok this is k n here this is c n the mass is m n ok and the displacement vector becomes q and t and the force that is being applied on this one is p n t ok. So, basically this is the vector that you or this is the system the single degree of freedom system that we get by modal decomposition of the multiple degree of freedom system. So, what we have been able to do reduce my multiple degree of freedom system which were in terms of degrees displacement at degrees of freedom 1 2 3 as u 1 u 2 u 3 and so on ok. Let us say u n 2 let us say this is multiple degree of freedom system in terms of n single degree of freedom system with coordinates q 1, q 2, q n ok. Now, remember I would have n such single degree of freedom system and these q 1, q 2 they do not correspond to any specific degree of freedom remember the original equations were in terms of degrees of freedom, but here these are modal coordinates in order for you to convert this modal coordinate to corresponding displacement at the degrees of freedom. So, let us say you are considering nth contribution of the nth mode to the total displacements at each degree of freedom you would need to multiply the q n modal coordinate with the basically the shape vector ok or the mode shape here. Once you multiply that it will give you the contribution of that particular mode to the total displacement at each degree of freedom ok. So, all these equations when we solve these equations this single degree of freedom system it gives us the modal coordinate not the displacement at any degree of freedom ok. We will convert it and once we have for each single degree of freedom or for each mode like this we can sum it up ok to get the total displacement at each degree of freedom ok and we can submit over all the modes ok. So, this is how we basically solve our multiple degree of freedom system subject to any external excitation force ok and we will take up one example ok we will take up one example and then see how to actually get the response ok. So, let us take the example that we have been doing so far. So, what do basically we have here is this same two-story shear frame building ok which I am now representing as this lollipop model here. So, this is k this is 2k ok this is u 1 and u 2 here. Now, we have already derived the mass for this as mass matrix as 2m 0 0 m ok I am just going to write down here u 1 u 2 as well. So, that you do not get confused my stiffness matrix as 3k minus k minus k and k ok and on this one the external force that is being applied is p naught sin omega t at degree of freedom 1 ok. So, p naught sin omega t at degree of freedom 1 and there is no force at degree of freedom 2 ok. So, this is my excitation vector. Now, what I need to find out is basically the steady state response at each for each degree of freedom. Steady state let us say displacement response ok at each degree of freedom. Now, before getting into that remember when we had a single degree of freedom system in which the variable the displacement variable was u ok and it was subjected to p naught sin omega t for that the steady state response was. So, let me write down the corresponding SDUF system if you remember when you had an equation like this here and do not confuse between these parameters I am just writing it this is a different single degree of freedom system altogether ok. So, when you had this kind of single degree of freedom system subject to p naught sin omega t excitation the steady state response was ut equal to applied force amplitude divided by the stiffness of the system times 1 divided by 1 omega minus omega n where omega n was the natural frequency of vibration of that single degree of freedom system and then sin omega t ok. So, this was the response ok. So, we are going to utilize this, but now remember we are solving in terms of model coordinate q 1 and q 2 here not q here ok. So, just keep that in mind alright. So, before solving this equation the first step that we need to do is that we need to find out the mass matrix the diagonal mass matrix and diagonal stiffness matrix. So, that I can get the generalized mass m 1 and m 2 for each mode and generalized stiffness k 1 and k 2 for each mode ok. So, that I can write down the equation of this form m 1 q 1 double dot plus k 1 q 1 is equal to p 1 which is the normalized excitation for mode 1. Similarly, I need that so that I should be able to write down this equation as well equal to p 2 ok. So, m 1 m 2 can easily be found out if we write m 1 as phi 1 transpose mass matrix times phi 1 and if you remember we had already derived the mode shape and frequency for this system ok and I am just going to write it here. The frequency for the first mode was k by 2 m and the mode shape was half and 1 ok. For the second mode the frequency was 2 k by m and the second mode shape was minus 1 1. We had normalized with respect to the top degree of freedom which is the second degree of freedom here ok. So, we can substitute those quantities for m 1 m 2 and so on. So, let us say m n is this ok and basically your k n is phi n t times the stiffness matrix times the phi n ok. So, we can substitute it for 1 and 2 and we can get the normalized masses and stiffnesses as m 1 equal to 3 m by 2 ok m 2 equal to 3 m and then k 1 equal to 3 k by 4 and k 2 equal to 6 k ok. You can do that matrix multiplication and you can check that if these are correct or not ok. Once we have found out let us get our the generalized excitation force for each mode. So, first I want to get what is the p 1 t. So, basically phi 1 t times the vector p t which is nothing, but half times p naught sin omega t and 0 which gives me this as p naught sin omega t ok. Similarly, my p 2 t for the second mode is basically mode shape transpose which is minus 1 and 1 here then again the same vector and this gives me minus p naught sin omega t. So, the two equations that we are solving basically becomes this equal to p naught by 2 sin omega t and this minus p naught sin omega t ok. So, these are the two equation that we are trying to solve here ok and we can utilize this general solution to write down. Now, remember what I am going to do here this I am going to write as response modification factor for the first mode as R 1 and for the second mode as R 2 ok. So, that I do not have to keep writing the same ratio again and again ok. So, let us do that. So, q 1 t is basically the magnitude of the applied force which is p naught by 2 divided by the stiffness for that mode which is k 1. Remember not the stiffness of the story this is the k 1 for the stiffness for the generalized modes ok. So, 2 by k 1 and I can substitute k 1 here as 3 k by 4 and this times let us say R 1 times sin omega t and this I would get as 2 p naught by 3 k R 1 sin omega t. Similarly, my q 2 t I can obtain as p naught or minus p naught which is the amplitude of the force divided by k 2 and k 2 is 6 k here ok times R 2 times sin omega t ok. So, this I get as minus p naught by 6 k R 2 sin omega t all right. So, once we get q 1 and q t q 2 ok. Remember the next step is basically find out the combined response before we get into that since we have obtained q 1 t this is the modal coordinate not the displacement of any degree of freedom. To get the displacement of any degree of freedom let us consider u 1 t here which is basically contribution or if it is called u and t contribution of the n th mode to the total displacement response ok which is basically u t here. So, u 1 t is basically obtained as phi n t times q n t ok. So, let us first get what is the contribution of first mode to the total response and this would give me the contribution at each degree of freedom the displacement contribution ok. So, my phi 1 is half 1 times 2 p naught by 3 k R 1 sin omega t. Similarly, my u 2 I can get it as minus 1 1 ok minus p naught by 6 k times R 2 sin omega 2. So, the total displacement response we can get as by summing these up ok summing up the contribution of each mode ok that will give me the total displacement at each degree of freedom ok. So, that I can write it as p naught by 6 k 2 R 1 plus R 2 sin omega t and p naught by 6 k 4 R 1 minus R 2 sin omega t ok. And this basically corresponds to the displacement history at each degree of freedom 1 and 2 ok alright. So, we have seen that if the external excitation force is given we can construct the force vector p t ok and then we can utilize the model analysis procedure to find out the response at each degree of freedom. So, this was the only focus on finding out the response at each degree of freedom that the displacement response ok. Now, it might happen ok the displacement response in addition to displacement response you might also need to find out the forces ok. So, basically if you let us say it could be any type of any multi degree of freedom system, but let us take example of a multi-story building ok and let us say it is the shear type building something like this ok. So, multiple stories are there here and u 1, u 2 the degrees of freedom are like that ok. So, to find out basically the forces or the internal forces in the system subject to any external excitation ok. Let us see how do we do that ok. So, remember first we found out we have found out how to get the total displacement response using total displacement as sum of individual displacement for each mode ok which is nothing, but summation n over all the degrees of freedom phi n and q n ok and this was the modal analysis procedure that we had followed. Now, let us see how to get the element forces ok. So, to get the element forces there are basically two procedures ok and we will discuss each of this procedure one by one. So, in the first procedure we find out the contribution of the nth mode to the total displacement u and t basically ok using this once we find out q and t multiplying this with the mode shape. Once we found out u and t if we know the relationship between the displacement and the internal forces we can utilize those relationship to find out internal forces. For example, let us say this is the jth history and this is j minus 1th story ok. So, if I define my story shear has in this story as vj and let us say I am considering nth mode. So, in the nth mode basically story shear in the jth story would be whatever the story stiffness is kj let us say ok times the story drift ok and I will say that this is for the nth mode. So, it would be kj times ujn minus uj minus 1n ok and these ujn and uj minus 1 we already know from here because remember u and t is what? It is u1, n, u2, n so on ujn ok and so on ok. So, let us say n degree of freedom so for the nth mode ok this is the contribution of the nth mode through the displacement at each degree of freedom. So, we can find out this or we can also write this as phi jn minus phi j minus 1n times q and t. So, this we can also write ok. So, this is when we directly know the relationship between the displacements that we have obtained and the forces in the system using this type of relationship ok. The second relationship the second procedure that we follow ok is basically the equivalent static procedure ok. So, basically this is equivalent static procedure ok and in equivalent static procedure what do we do? We find out what are the equivalent static forces that are acting and then using the static analysis of the structure we get the forces or the storage here or the story movement at different level ok. So, let us say I have a multi-story building like this again ok. So, the equivalent static force is nothing but the stiffness ok times the u and t. Now, this is for the nth mode ok. So, this can be further written as k and this can be written as phi n times q and t and remember from the eigenvalue equation this is nothing but omega n square m times phi n times q and t ok. So, we can utilize this and then we can find out at each degree of freedom what is the, what are the equivalent static forces? Let us say here it is f 1 n f 2 n and similarly at some story here let us say this is f j n and so on ok. So, the equivalent static forces at each degree of freedom in the nth mode. If we have that finding out finding out the story shear at any level is as simple as summing up all these equivalent static forces above that level. So, this would be let us say summation k equal to j 2 n and f k n ok. So, this is summing up this f k n ok. And remember I mean we have derived this like you know given the example of multi-story building, but the same procedure would also be applicable for any other type of multi-degree of freedom system. Only thing is that this equation might differ ok. So, let us say if I have ok a multi-degree of freedom system like this ok. And we can consider different modes of excitation in that also we can apply. Let us say if the nth mode is that we can apply f 1 n and f 2 n like that and equivalently we can find out what are the base shear. For example, in this case the base shear would be ok in the let us say nth mode it would simply be f 1 here if we consider the equilibrium. So, when we apply the equivalent static forces we do not need to know the relationship between the displacement and the forces ok in the particular element ok. We can just consider the equilibrium of the structure and we can find out forces utilizing the applied equivalent static forces ok. And this can be applied for the examples to find out the forces and the internal forces and the moments in the system alright. So, we are going to conclude this lecture here today. In the next lecture we are going to see how to apply these procedures to find out the seismic response of a structure or a multi-degree of freedom system. Thank you.