 Welcome back everyone. So far we have learned how to set up the equation of motion of a multi-degree of freedom system and we also saw that a multi-degree of freedom system is represented by different modes. So the total response is actually contribution of the response of each mode. In today's class what we are going to learn is basically how to get the response of a multi-degree of freedom system in terms of response at each degrees of freedom. So we are talking about basically displacement response and we will extend the same concept to find out other type of response quantities like shear forces and moment. And we are going to see for both type of system undamped and damped system how to actually get the expression for the displacement response. So let us get started. In the last class we discussed basically the frequency and the mode shapes of a multi-degree of freedom system. So we have frequencies and mode shapes. And we said that if multi-degree of freedom system is excited through some excitation in general the response of each degree of freedom would not be harmonic but there exist few characteristic shapes in which if they are provided displacement according to those shapes then the multi-degree of freedom system would have harmonic response at each degree of freedom and it is going to maintain its shape throughout the vibration and those characteristic shapes are actually called mode shapes. And because the response is harmonic each of these modes have their own frequencies which is basically the time taken to complete one cycle of motion. Okay, that is the time period and we can get frequency using the relationship as 2 pi by time period. So we discussed that and we said that the total response of a system would be contributed by each of these mode and let us say if the mode shapes are represented through this vector phi 1 and phi 2 and so on. Then we said that the total response would be combination of each the contribution due to each of these mode shapes. Okay and depending upon how much of those contribution are some modes are going to govern the response and we are going to learn about that later. But right now let us consider that the total response which is nothing but the response at each degree of freedom let us say and so on. Okay can be written as the response due to all mode shapes which are represented through this shape vectors okay and so on. Now we already know the frequency we already know the mode shape. So the problem statement here is given the vector u is it possible to find out the factors that need to be multiplied to these mode shapes so that we can get the modal decomposition or modal expansion we call it okay of this displacement vector here. So basically our objective here is to find out these factors here because once we derive expression to find out these factors then we can find out the total response as a linear combination of contribution due to different mode shapes okay. So let us see how do we do that. So this is basically called this what we have written here is modal expansion okay modal expansion of displacements. This is called modal expansion of the displacement and let us look at that how to get those factors. So let us assume that the vector u is basically okay so I am just going to write these expression that we have written here in terms of summation. So let us say we have n modes here n degrees of freedom okay so we will have phi which is the mode shape here times qr okay and this qr is basically these factors that I am talking about. So our goal is to find out those qr's. Now we can write the same summation in terms of multiplication of modal matrix times a vector q as well okay where vector q contains all these factors as column vector okay. So these are called modal coordinates okay the vector q is basically q1 q2 and so on and these are called modal coordinates okay and these are the factor that are required to find out the total response okay by combining different modes here okay. So let us see how we can find out these individual qr here okay. Now we are going to utilize again the condition of orthogonality of the modes which basically says that if I have two different modes okay then if I take the product with respect to the mass matrix okay the product of the transpose of one mode times the mass matrix times the product of another mode that would be equal to 0 okay as long as those are different mode shapes okay. So what I am going to do here I am going to pre-multiply this expression that we have written okay with phi n to the power t times the mass matrix. So we get here as well on the right hand side inside the summation I will have phi nt times m okay remember this is phi r here k0 to 1 this is phi r times qr. Now as we know due to modal orthogonality when I expand this summation okay when I execute this summation what will happen most of the term will vanish except when r is equal to n sorry r is equal to n that is the only term that is going to survive okay. So knowing that the expression that I will get is actually phi nt times m and now I will substitute r equal to n okay this is n and this is qn here okay. So qn the modal coordinate I can write it as phi n to the power t times m times u and this divided by this expression here. Now if you look at the denominator this is nothing but the diagonal element of the diagonal mass matrix okay so basically what I am saying here this is nothing but in the denominator I can write this as mn which is the nth diagonal element of the diagonal mass matrix okay. So we have found out qn now so we can do that for all the mode shapes and find out these factors multiplicative factors for all the mode shapes and once we know that then we can combine modes using those factors okay this would be more clear after we do one example okay. So let us take an example that we have been doing till now so basically I have okay the same two-story representation of a building in which this here is 2k this is k this is 2m and this is m okay. Now we know that the mode shapes for we had derived the mode shapes and we had got that for this is degree of freedom one and this is 2 so this was half and one we had normalized with respect to the top story or u2. So my phi2 here is minus 1 and 1 okay now the question is that at any time instant let us say okay u of t at any time instant let us say u is given as 11 and what I have to find out basically the modal expansion of the displacement vector so I have to find out the expansion of this in terms of some factor times the first mode shape plus some factor times the second mode shape so basically I have to find out those factors q equal to q1 and q2 okay. So let us see how do we do that all right so my q1 will be nothing but a transpose of phi1 which so let me just first write down anyway this would be phi1 times u here okay this would be phi1 sorry phi1 transpose here phi1 transpose times phi1. Now if you look at this let us substitute those vectors here so this is half 1 and the mass matrix is of course 2 m 0 0 m and here is the displacement vector for which we are doing the modal expansion so this is 1 and 1 here and in the denominator basically I have half 1 then 2 m 0 0 m times 1 1 okay so this I get as 4 by 3 similarly q2 I can find out as by substituting phi2 okay so this will be phi2 transpose m u phi2 m phi2 the value that you you get is basically you will be going to get q2 as minus 1 by 3 okay so you got this factors let us see whether we get the same expression not so basically my u1 or the basically the u vector the displacement vector is 1 and 1 okay so let us use the 4 by 3 times the first motion which is 1 by 2 and 1 and then I have minus 1 by 3 the q2 here which is minus 1 and 1 okay so if you look at here get this as 2 by 3 plus 1 by 3 okay plus 4 by 3 minus 1 by 3 so this we get as 1 1 okay so that expansion is actually correct and if we want to represent this in terms of or like you know using some schematic in terms of the mode shape problem is that or the state the problem statement is initially it is given displacement which is 1 at each degree of freedom so you need displacement at both positions so this is 1 here and this is 1 here okay now this is equal to you look at the first mode okay the mode shape is okay so this is basically 2 by 3 and 1 by 3 sorry 2 by 3 and 4 by 3 so I can just write it like this this is 4 by 3 and this is 2 by 3 here plus sum of the second terms which is plus 1 by 3 and minus 1 by 3 so let me again draw this here okay at the second degree of freedom something like this here and then okay so this is minus 1 by 3 this is 1 by 3 here so this is what we get okay so this is nothing but 4 by 3 times 5 1 here and this is minus 1 by 3 5 2 okay so the total response which is basically you is represented like this so you can see that through the modal expansion okay at any time instead whatever the displacement is given can be expanded in terms of contribution of the first mode and the second mode and so on if it has n modes then there would be n such mode shapes okay and vice versa it can be also true so the idea is that if we have n number of modes they can be combined using these factors q1 q2 and so on to find out the total displacement at all degrees of freedom okay and we are going to see how we are going to get these vectors now remember this q that we have obtained here it is at any time instant okay but in general q is a function of time okay so this is for any time fixed time instead equal to let us say t0 but in general q is basically the function of time which is the time variation okay so at any time it would be distributed or contributed by different modes so let us get into that so now what we are going to do here actually going to solve the equation for free vibration response so remember that our equation of motion is m times u plus k times u is equal to 0 okay we have said that my now we are going to write in terms of as a function of time t okay so this is nothing but r equal to 1 and then the linear combination of linear sum of all the mode shapes so it would be let us say phi r times or let me just write in terms of n here okay so let us say n equal to 1 to n this is phi n times qn here now it is a function of t so we need to find out this qn of t okay and we have previously said that because each mode shapes respond as they respond in a harmonic fashion so the displacement q and t which is the generalized displacement or the time variation of that mode shape can be written as a n cos omega n t plus b n sin omega n t okay where this a n and b n are basically constants that need to be found out using initial condition if this is our equation of motion the initial conditions are typically given at displacement vector at time t equal to 0 and the velocity vector at time equal to 0 and utilizing this we find out a n b n okay and omega n is basically the frequency the circular frequency of the nth mode okay so let us see how do we get the constant a n b n because once we get that then we can find out the overall displacement vector okay if you utilize the initial condition okay we have the expression for you here let us get the so let me just rewrite it okay this is nothing but summation of phi n which is the mode shape times q and t which is a n cos omega n t plus b n sin omega n t okay and we can similarly get the velocity by differentiating with respect to time the same expression here this would be omega n phi n and here it would be minus a n sin omega n t plus b n cos omega n t okay so let us substitute t equal to 0 to get these expressions here so that will give that would give us u 0 as summation n equal to 1 to n phi n times a n and then summation sorry the vector velocity vector u dot 0 has summation n equal to 2n omega n times phi n times b n okay so this is the expression to that we would utilize to find out the constants a n and b n now if you look at a n and b n and let us say my qt is written as this expression here okay a n cos omega n t so this is the expression that the general expression right solution that we had assumed sin omega n t now if we say that let us assume a n equal to q sorry not qt q dot 0 q 0 and b n omega n equal to q dot 0 okay then I can write this expression as q 0 cos omega n t plus q dot 0 divided by omega n sin omega n t remember these are also constants here but why I am writing it like this so that my qt is similar to the expression that we had obtained for single degree of freedom system in terms of u 0 and 0 to 0 but now we obtain here in terms of the modal coordinate okay so I am just making this substitution here so that it is we can correlate it to single degree of freedom system so if that is there then this expression can be written as n equal to this phi n times q 0 and this expression as phi n times omega n q dot sorry no omega n just q dot 0 okay now we can look at these two expressions and q 0 and q dot 0 are nothing but you know the expressions similar to what we had derived here okay so we can utilize the same expression to find out the value of q 0 and q dot 0 okay so basically so then remember this is q n 0 for nth mode okay so just be careful with that all things we are doing for the nth mode this is again q n here q n q n here okay so my q n of 0 becomes phi n t times mass and this vector is basically u 0 now and again in the denominator I have m n of n q dot 0 is phi n of t u dot 0 the m n is here so utilizing these two expressions you can get q n 0 and q dot n 0 and once we have that then the expression for q and t is known sine omega n t and once q and t is known then we can do that for each and every mode and we can find out the total response using this summation that we had derived okay so this is basically n equal to 1 to n phi n times q and t okay so basically we saw that how to find out the response the free vibration response of an undamped system okay and basically what we do we find out the time variation of the generalized coordinate for each mode okay we find out the constants unknown constant using the initial conditions and finally we combine all those modes using the modal coordinate at any time t okay and we can find out the response total response is a function of linear combination of all those mode shapes okay so we can do an example of this and then see okay demonstrate the free vibration response so let us again take the same example okay so this is here 2k and k here this is 2m and m here okay again phi 1 and phi 2 have been given to it now what is given as initial conditions have been given to us so u0 is basically half and 1 remember this is u1 here and this is the second degree of freedom u2 and the initial velocity is also given as 0 0 okay so let us see what do we get as the overall response okay the ut value to do that first I need to find out these factors q1 of t and then q2 of t okay let us see how do we get that so basically q1 of t would be nothing but phi 1 t mass times u0 divided by m1 which let me write it as phi 1 t mass phi 1 here okay and once we substitute the all the values here so this phi t would be basically half 1 then I have the mass matrix 2m 0 0 m and then u0 which is basically half 1 okay and I again have half 1 2m 0 0 m and half 1 okay so we can get this q1 t as this value as okay sorry this is q1 0 here not t as the numerator and denominator is same we get this as one similarly q2 0 we can get as doing the same thing okay just replace phi 1 by phi 2 and this would you will see that after you substitute this parameter this comes out to be 0 in addition to that q1 dot 0 and q2 dot 0 would be equal to 0 because my u dot the initial velocity or degrees of freedom equal to 0 this is equal to 0 so we can go ahead and in our expression we can substitute q1 of t would be basically equal to phi 1 times okay or let me just first write down q1 at t is equal to q0 cos omega 1t plus q dot of 0 divided by omega sin omega 1t okay this is omega 1 let us see okay this is equal to 0 q0 is 1 this I am left with cos omega nt okay q2 of t would be equal to 0 because my q2 of 0 is equal to 0 and q2 dot 0 equal to 0 okay so these two q1 of t and q2 of t we have obtained as these expressions so remember my ut would be nothing but summation n equal to 1 to 2 here phi n times q nt which is basically phi 1 times q1 of t phi 2 times q2 of t which basically second term is equal to 0 so this is I get this I get as half and 1 and times cos omega 1t okay so this is the response that I get for each degree of freedom or it combined the uts 1 by 2 and 1 times cos omega 1t okay so this was the first part let us again find out the response for another case in which okay so I am considering another case and in this case the initial displacement has been given as minus 1 by 2 and 2 okay and q2 dot is actually equal to 0 okay so we can again follow the same procedure okay and we can write down the displacement as 1 by 2 and 1 or let me before writing that remember this is my first degree of freedom this is the second degree of freedom okay I can write this as phi 1 times q1t and phi 2 times of q2 of t so this we get as half times 1 times cos omega 1t plus minus 1 and 1 the second mode shape times cos omega 2 okay so this is the response for the second case now the reason that we did like in the two cases I wanted to show you something you have a look at the response here and the response here what can you tell me about these responses if you look at it and if you the response for the first one this is actually a harmonic motion at each degree of freedom right because it is a cos omega t type okay as a cos function however in the second one I have some function times or some vector times cos omega 1t plus another vector times cos omega 2t so this is not a harmonic variation because I have two frequency and it cannot be written as some constant times some cos or sin variation okay had it been the same angle omega 1 and omega 2 equal to omega then I would I could have done that okay but not in this case so in the first case the response is harmonic whereas in the second case the response is non-harmonic okay one more thing to notice here in the first case I can see only the contribution due to the first mode there is no contribution of the second mode there is no omega 2t term here okay however here you can see that there is contribution of both modes okay so only the contribution of first mode in the first case while in the second case there is contribution of both mode and that can be explained directly if you look at the initial displacement okay for the first case remember the initial displacement was half and one which is also the first mode shape of the given structure and by definition as we have previously discussed if you assign initial displacement to the structure which is according to the one of the characteristic shape then it is going to respond harmonically while maintaining its shape and it is going to vibrate in a particular mode of vibration so there would be no contribution of any other mode so that is why we do not see any contribution of other modes because it is being vibrated with initial displacement which is one of its mode shapes compared that to second one the initial displacement here is not one of its mode shapes so in general when you provide initial displacement we will have contribution of both modes and the response would not be harmonic at each degree of freedom okay and we have just seen example of that okay whatever we had discussed okay all right so I hope this example is clear now let us get into damped free vibration okay so we have discussed undamped free vibration now let us discuss damped free vibration now for the damped free vibration we will now have additional damping here with the damping matrix C here so this is equal to 0 okay now in this case if you look at it here I have this damping matrix okay now the reason I was able to solve the undamped free vibration because my mass matrix and my damping matrix could be diagonalized okay the solution for damped free vibration would depend on the fact that whether my C matrix can be diagonalized or not because if I am able to diagonalize my C matrix or the damping matrix then I can uncouple all the equation of motion for all the different modes so I would have n uncoupled differential equations okay and then I could solve those and combine the responses using the modal expansion factors okay so whether I can solve this analytically or not it would depend on C okay so basically C plays a big role now if C is a classical damping matrix okay so if C is a classical damping matrix if okay so let me write it here C is classical damping matrix if it can be diagonalized okay and how do we diagonalize this okay how do we diagonalize any matrix we multiply with the modal matrix okay and then see whether it can be diagonalized or so let us say it becomes C to C and as C and this is 0 and 0 okay and if it can be diagonalized we call it a classical damping matrix there are non classical damping matrix as well and there are some methods of solution for those as well but that is not within the scope of this course okay so let us consider classical damping matrix where C can be diagonalized and then see if we can get solution to this damped free vibration so what we are going to do here I am going to write my U as the modal matrix times the modal coordinate vector Q here okay of course it is Qt I am dropping the t term here okay so first I am going to substitute this and then I am also going to pre-multiply with the modal matrix transpose so this is the acceleration of the modal coordinate here again phi t times C times U dot which I can write it as Q dot here and then phi t times K times phi times Q now as we know this is diagonal mass matrix this is diagonal damping matrix and this is diagonal stiffness matrix so this can be written as mass times the Q vector okay the diagonal mass matrix times the modal acceleration vector and then this has diagonal damping matrix times velocity vector here and then the diagonal stiffness matrix times the Q vector here okay and because remember all these expressions are in terms of m1 m2 and so on let us say mn here like this and then Q1 Q2 and so on similarly c1 cn Q1 Q2 plus k1 k2 all these are uncoupled n uncoupled differential equation okay and for the nth mode okay for the nth mode I can write it as mn Qn plus cn Qn dot plus kn Qn is equal to zero which is of the form of the free vibration response of a single degree of freedom system so I have basically decomposed my multiple degree of freedom system through the modal decomposition into n single degree of freedom system and then I am going to find out the response for each system which would be basically response for each mode in terms of the modal coordinate Q and t and once I have the Q and t we know that we can combine all the modes using the multiplication of the mode shape times the Q and t using this expression here okay now like what we did for free vibration response of a damped un-damped system we can do the same thing for the damped system as well okay and Q and t can be similar the obtain okay now to do that let me first do this write down the same equation that we have okay by dividing throughout by mn and writing it in this form okay Qn plus 2 zeta n omega n okay times Qn omega n square times Qn equal to zero now if you look at here we have damping for the nth mode so now for a multi-degree of freedom system the way the damping ratio is defined there is a separate damping ratio for each of the mode okay and that damping ratio is nothing but cn divided by c critical which is 2mn omega n okay and my Q and t can be written as n zeta omega n times t here then Qn zero now I would have remember now there is a damping in each of the modes so I have to consider the damped frequency so this is the omega n dt okay plus fn dot zero zeta n omega n times omega n d times sin omega n okay this is similar to the expressions that expression that we had obtained for the damped response of a single degree of freedom system only now we are considering everything for a particular mode so this is the response for the nth mode and we have defined the damped frequency for the nth mode as omega n times 1 minus zeta n square okay and the total response I can write using the expression that I had here has ut equal to summation n to n phi n Q and t where Q and t is basically this expression here okay so this is how we get the response the damped response of a free response of a damped system okay now if we notice in this case typically we do not get the damping matrix I have told you that you know you can get the damping matrix and see if it can be diagonalized or not okay but in practice there is no way we could get the damping matrix by just considering damper element in a structure okay so what we typically do for practical analysis we assume or we determine experimentally the damping ratio for each mode because remember now the damping is defined for each of these modes okay and there are some experimental methods to do that okay or we can assume some values based on the data that we have observed in the past from the testing or the response of other type of structures similar structures