 We have been discussing modeling of frequency response functions and the role played by damping models in doing that exercise, so we are dealing with linear damping models and we broadly classify the damping models as being viscous or structural and within each of these groups we have classical and non-classical models. The classification into viscous and structural is something to do with the way the energy dissipated in a cycle behave as a function of frequency, and the classification into classical and non-classical depends upon whether the undamped normal modes uncouple the equation of motion or not. In a classical damping model the undamped normal modes diagonalize the damping matrix whereas in non-classical damping models that won't happen. So we discussed the problem of finding frequency response function for viscously damped multi-degree freedom system, so if we drive the S degree of freedom, S degree of freedom by a unit harmonic force we have shown in the previous lecture that the matrix of frequency response functions can be evaluated either directly by inverting this matrix, this so-called dynamic stiffness matrix or we can evaluate in terms of undamped normal modes and we get this as a summation. Now we also discussed few issues about truncation of this series, so if we omit the first M1 modes, M1-1 modes and the last M2 plus 1 to the N number of modes this will be the kind of representation and these two terms here and here provide the corrections for omitting the contributions from normal modes which have not been included in this summation, so we discussed this in the previous lecture. Now we will address few points before we move on to discussion on non-viscously damped systems, so let's ask the question how do we determine damping matrix given modal damping ratios in viscously damped systems? So let's consider a N degree of freedom classically and viscously damped system, the governing equation will have this form, F of T is exaltation and X of 0 and X dot of 0 are the specified initial conditions. So we introduce the transformation X equal to phi Z, where phi is the matrix of undamped normal modes which have been normalized so that phi transpose M phi is I and phi transpose K phi is the diagonal matrix of the eigenvalues which are the squares of natural frequencies. Now this leads to set of uncoupled equations in the new coordinate system which is a family of single-grave freedom systems and we discussed how to evaluate the initial conditions without inverting the phi matrix. Now this C bar be the diagonal matrix of the modal bandwidths or these terms 2 eta N omega N, now the question is if I provide C bar how do I get C? You will see later in the course that we may like to integrate these equations directly and in which case I need C matrix whereas the damping model may be in terms of modal damping ratios, so the question we are asking is given the modal damping ratios how do we construct a spatial damping model? This is a modal damping model, I want a spatial damping model, so clearly C bar is phi transpose C phi, so if you have a full square matrix for the modal matrix you can directly evaluate this by inverting, so you need to invert two matrices phi transpose and phi, so in principle this is possible but in practical situations first of all we don't like to invert matrices if we can help avoiding that we would be happy to do so, secondly more importantly we will not be evaluating phi as a square matrix, we will not be evaluating all the modes that are possible in a given discretized system, we may be focusing on only first few modes, in which case phi will be a rectangular matrix, the number of rows will be equal to degree of freedom but number of columns will be equal to the number of modes that you want to include in the analysis, so in which case using this formulary will not be feasible, so now how do we proceed? So what we can do is there is a simple way of overcoming this problem, suppose if I consider the orthogonality relation I as phi transpose M phi, now I can post multiply by phi inverse, so phi inverse will be phi transpose, I am post multiplying phi transpose M phi phi inverse, now phi phi inverse is identity matrix so therefore phi inverse is phi transpose M, similarly if I pre-multiply by phi transpose inverse I will get this which is M phi, so for phi inverse and phi transpose inverse I can use these representations and I can get the damping matrix without inverting the phi matrix or its transpose, so this is a simple way of evaluating the damping matrix. Now another question is what happens if natural frequency is repeat, it is not very unusual if structure is symmetric in a geometric sense the eigenvalues can repeat, for example a square plate if you look at normal modes what is a normal mode in next direction would also be a normal mode in Y direction, suppose I am talking about a square plate suppose all around simply supported if one of the mode shapes in this direction is this the same mode shape will also be possible in the other direction, so for the same eigenvalue there will be two eigenvectors so how do we deal with such situations? So we will return to this later but we will now consider a simple example artificially constructed example where we can see that the eigenvalues repeat, so this system has three masses and a set of springs, so the data is that all springs are having same value K, all masses have the same value M, and all dampers have same matrix C, so now the mass matrix will be this damping matrix, stiffness matrix will be this and damping matrix will be this, so the characteristic equation will be given by the determinant of K-M omega square equal to 0 which is this, if we solve this characteristic equation we see that first natural frequency is K by M and the second and third natural frequency square will be 4K by M, so the second and third eigenvalues repeat. Now how do we handle this in evaluation of eigenvectors and in calculating the response? So let us find the eigenvectors, suppose if omega 1 square is K by M the first eigenvalue then to determine eigenvectors I go back to the original eigenvalue statement of the eigenvalue problem and put omega 1, omega as omega 1 square and I get this equation. Now in this since eigenvector has a free floating constant that means any constant multiple of an eigenvector is also an eigenvector, so there is a non-uniqueness to the extent of a scalar multiplier we can start by assuming that R1 is 1 and it implies by writing this equation that R2 and R3 would also be 1, you can verify that, so because 2 minus 1 minus 1 is 0 so you can see that they satisfy that, so the first eigenvector can be written with this normalization as 1, 1, 1. For the second eigenvalue that is omega 2 square by 4KM I write the equation for the eigenvector we can see we see that the equation is R1 plus R2 plus R3 equal to 0, so there are infinitely many solutions that are possible, but what we do is we select deliberately only two of, only one of them so that we retain the orthogonality property, since this eigenvalue repeats this will be the equation for the next eigenvalue also, so let us do here let R1 be 1 and R2 be 1, so then R3 would become minus 2, so phi 2 the second eigenvector will be of this form. Now let us come to the third eigenvector, the governing equation is same as what it was here because eigenvalue is the same, so R1 plus R2 plus R3 equal to 0, now what I do is I select now the third eigenvector so that it is orthogonal to both first and second eigenvector, so I assume that phi 3 transpose is 1A and minus 1 plus A, so then we can see that R1 plus R2 plus R3 is 0, so A is a free floating constant, so how do we select A? We can select any A as far as this equation is concerned this is true for any A, but what we do is we select A such that phi 2 transpose M phi 3 is 0, so if we impose that condition I get A as 0 and the third eigenvector is this, so if we now consider the model matrix to be 1 1 1, 1 1 minus 2, 1 minus 1 0, then we'll return to this equation with a forcing on the one of the degrees of freedom, and now if I impose this transformation and use the orthogonality relation for the Z1, Z2, Z3 I get these three equations, which are again uncoupled and these are the generalized forces that we have to use, so we can solve this problem and construct back the solution using X equal to phi Z, okay, so that means if eigenvalues repeat we can construct eigenvector so that the required orthogonality property is still possessed by the calculated eigenvectors, so a remark can be made at this stage, if we consider a N degree of freedom system if the Nth natural frequency repeat R number of times where R can be between 2 and N, the eigenvectors associated with the repeated eigenvalues can be selected through a process of orthogonalization to be orthogonal to the structure mass matrix, and hence also the orthogonal to the structure stiffness matrix, okay, so this orthogonalization procedure you have to implement if you find that your system has repeated eigenvalues, so this helps us to uncouple the equation. Now we talked about proportional damping and non-proportional damping models in the context of discretized system, it may be useful to visit the equation for the beam vibration and see what is meant by mass proportional damping and stiffness proportional damping as far as the beam equation is concerned, so it turns out that if you consider the beam equation we can add, we have added now here four terms, this is the one, this is the second, this is third, and this is four, which represents energy dissipation properties. If these energy dissipation properties are absent their traditional equation is dou square by dou X square EI Y double prime plus MY double prime Y double dot equal to F, now this term nu of X into dou cube Y by dou X square dou T is actually damping which depends on strain rate, and it is viscous damping. The second term is strain rate dependent structural damping because you see here in the denominator I have put omega to ensure the required energy dissipation characteristics as a function of frequency, and this C into Y dot is a traditional velocity dependent viscous damping, so this doesn't depend upon strains, it only depends on the displacements. Now this term is the velocity dependent structural damping, so this is how the damping terms appear if I were to write the original partial differential equation. Now if I make now this nu of X to be proportional to the flexural rigidity then we get stiffness proportional viscous damping, if I make G of X to be proportional to EI I get stiffness proportional structural damping, similarly if C is made proportional to mass M of X I get mass proportional viscous damping, and finally if I make H of X to be proportional to M of X then I get velocity dependent structural damping. So when we discretize this equation we get MX double dot plus C X dot plus K X equal to F of T, but it is useful to have this insight at the original PDE level, so we know what we are doing at the original PDE level when we assume alternative models for damping. Now let us now move on to other damping models, so we have finished discussing viscous classical damping systems, systems with viscous and classical damping, now we will move on to structural damping models and begin by discussing classical damping models, so how do we discuss this? So again whenever we are talking about structural damping I want to emphasize that such models are applicable only for steady state response analysis, you cannot do transient analysis, no time domain analysis is applicable. So we consider N degree of freedom system with harmonic forcing, so this equation can be used only for the purpose of calculating frequency domain response, it's not a differential equation with T as independent variable although it is made out to be like that, so while interpreting this equation you should only use this only to find the frequency response function. Now let Rth degree of freedom be driven by an unit harmonic excitation, now we are interested in steady state, suppose in the steady state the response becomes harmonic Q e raise to I omega T, the system is linear it is being driven harmonically and we are considering steady state, so in steady state the system responds at a driving frequency in a harmonic manner, so if I now substitute this into the governing equation I get this as a dynamic stiffness matrix this into Q is yeah, so if you are interested in finding the matrix of receptances or the frequency response function you can invert this matrix directly for every omega, okay, so this is a direct solution to the given problem, so we are interested in finding a solution which employs normal modes and I want to write this as a summation because if I use this approach for every value of omega that I am interested I need to invert to this matrix which can be computationally demanding, so the note is that determination of Q by direct inversion of stiffness, dynamic stiffness matrix for every omega is possible, however this would be computationally intensive and does not provide insights into the behavior of the frequency response, it is character less description. Now let phi be the matrix of undamped normal modes such that it is mass normalized and phi transpose K phi is the diagonal matrix of square of the eigenvalues, now we will make the transformation Q as phi Z, so the usual steps if I now find Q e raise to omega T will be this and this will be if I now substitute into this equation I will get in frequency domain this equation and if I am pre-multiplied by phi transpose I get this, now if I take phi transpose and phi inside I will get minus omega square phi transpose M phi plus phi transpose K phi plus I into phi transpose D phi Z is the generalized force phi transpose F. Now suppose D is classical, by that what I mean the same this phi transpose D phi is diagonal that means undamped normal mode diagonal as a damping matrix, if that is possible then it follows let us call D bar as phi transpose D phi, so this will be phi transpose M phi is I that is how we have normalized, phi transpose K phi is the diagonal matrix of square of eigenvalues which is capital lambda and this I D bar into Z is phi transpose F, so this matrix minus omega square I plus lambda plus I D bar is a diagonal matrix now, right, so its inversion is a simple task, so if I write now this in the scalar form and look at the Kth term I get this as an expression for ZK, so this can be since only one of the ordinates is being driven here this summation collapses to a single term this is phi R, so this is ZK, now if I go back to Q it is phi Z, so if I am interested in QJ we have to sum over all the modes I get this expression, so this quantity is the receptance as we have seen and this is given by this where we are now getting this without inverting a matrix as a summation, okay, now this is symmetric and not Hermitian as I have been pointing out, so this is the receptance function, so this completes the formulation for FRF or classically damped, classically damped and structurally damped systems. Now let us move on to questions about treatment of non-proportional damping, first we can ask the question where do you expect to get non-proportional damped systems, typically whenever we have structures made up of different subsystems with different materials we can expect that the resulting damping matrix will be non-proportional, for example in an industrial building there will be soil in the foundation and there will be RCC superstructure there could be steel piping and metallic equipment and there could be a metallic truss so on and so forth, so the entire structure consisting of soil, superstructure, the secondary systems and the trusses etc. consist of several materials, several subsystems each subsystem made up of a different material, so for this type of systems we can expect that the damping in the system would be not classical, that means undamped normal mode will not uncouple the equation of motion. So there would be other complicated features associated with energy dissipation at joints like you may have welded joints, bolted joints and so on and so forth, so how does energy dissipate at the joints, so this also could contribute to damping being non-proportional. So if we now consider a system made up of NS subsystems, the system damping matrix can be viewed as summation I equal to 1 to NSCI, where NS is number of subsystems with different materials, CI is a contribution to the structural damping matrix from the ith subsystem to the C matrix. Now if I assume that Rayleigh's type of model is applicable for each of the subsystem, I can write C as alpha I Mi plus beta I Ki, right, this ith material, then even for this case if I now compute phi transpose C phi I will get this, on the right hand side I get phi transpose Mi phi plus phi transpose Ki phi, but we know that phi transpose M phi is diagonal, it doesn't mean that phi transpose Mi phi will be diagonal, so this the resulting C matrix would still be non-diagonal. So now this leads to a few questions, one of that is what is the mathematical framework to uncouple the equation of motion in such cases, should we use direct integration or direct analysis in frequency domain by inverting the dynamic stiffness or is there a mode super imposition type of representation possible, then are there any simplification possible so that the damping remains classical and yet at the same time we take into account the fact that the structure is made up of subsystems with different materials, okay, so these two questions we will now consider and try to find answers to these questions. So I will start by first considering viscous damping and we'll focus on non-classical damping, so the topic now is analysis of systems with non-classical viscous damping, so we can start with a preliminary ideas on single degree freedom system, suppose I have an undamped single degree freedom system I know that if you consider free vibration the equation Mx double dot plus Kx equal to 0, and if you assume solution to be of this form this is admissible provided S satisfies this equation where S is plus minus S square root K by M, so it is plus minus I omega, so the solution itself will be of this form, which is harmonic at frequency omega which is square root K by M, this is well known. Now if you assume that system is damping, system has damping and it is underdamped then we again consider this equation and consider these roots to be complex conjugates, this is what happens when damping is less than critical, so in this case the solution will be of this form, okay, X naught and X naught dot are the initial conditions, this is the decaying, this is the term contributes to the decay of the solution, and as you can see eta is sitting here. Now this again I can write it in this form X naught bar exponential e raise to minus eta omega t cos omega t minus theta naught bar, so if you now compare this solution for undamped system and this solution for the damped system we can see that there are couple of things that change, the frequency is now a damped natural frequency it is affected by damping but the change is marginal, but the amplitude is now modulated by an exponentially decaying function, so for undamped systems the characteristic roots are pure imaginary, motion is periodic and motion is sinusoidal at system natural frequency, for a damped system characteristic roots are complex with nonzero real parts and these roots are complex conjugates, motion is a periodic and motion is exponentially decaying sinusoidal at damped natural frequency, okay, this is for single degree freedom system. Now we can ask the question what happens to multi degree freedom systems in presence of damping, do we get a similar features in our response, so let's begin that discussion, we will consider N degree of freedom system let Q be the vector of displacements, so the equation of motion will be MQ double dot plus CQ dot plus KQ is F of T with prescribed initial condition, now for reasons that would become clear soon we augment this equation with an identity that is MQ dot minus MQ dot equal to 0, and this is a given governing equation, so we write now these two equations together in a matrix form as shown here, so you can see here 0 into Q double dot plus MQ dot minus MQ dot into 0 Q is 0, that is your first equation, the second equation is the governing equation of motion, so we have not altered the mathematical nature of the problem, we have rewritten the governing equation in a slightly different way. Now I will call now this vector Q dot Q as the state of the system at time T and denote it by Y, so I call Y as Q dot Q and this 0 F of T I call it as forcing vector, now I will introduce two notations, this matrix I will call as A and this matrix as B, now with these notations we can make few observations, we can see that this A and B matrices are symmetric, okay, A transpose is A, B transpose is B, but they are not positive definite, okay, there are negative terms here and so on and so forth. Now the governing equation therefore can be now be written as AY dot plus BY equal to F of T, so the original set of governing equation in the configuration space was a set of N coupled second order differential equations, in the state space the equations are now there are two N equations and they are all first order equations, so I have now two N coupled first order equations, A and B are now the new structural matrices, they don't have the interpretation of being mass and stiffness and damping and so on and so forth, nor there is any energies associated with A and B matrices, so they have some abstract meaning, and A and B are non-diagonal, they are symmetric, now the question we can ask is can we uncouple this set of first order differential equation? Now this representation as I mentioned is called state space representation and this Y which is Q dot Q is called a system state at time T, now as before what we want to do is we want to introduce a transformation Y equal to TZ, now T is a 2N by 2N matrix, Z is a 2N cross 1 new coordinate system, and our objective is to find T so that upon making this transformation in the Z coordinate system the equations become uncoupled, how to find this capital T? For that we will again start with the free vibration problem AY dot plus BY equal to 0, and now what we will do is we will seek the solution in the form R e raise to alpha T, okay, alpha is complex, it's not pure imaginary as we used to do for un-damped system, it is now a complex number with nonzero real part and nonzero imaginary part, now if I substitute into this the eigenvalue problem that I will get will be BR is equal to minus alpha AR, okay, e raise to alpha T cannot be 0 for all T therefore the term inside the bracket should be 0 that leads to this eigenvalue problem, so this is a generalized eigenvalue problem, it's an algebraic eigenvalue problem, and the characteristic equation is given by determinant of B plus alpha A must be equal to 0, now this will lead to a 2N order polynomial and it will have 2N roots which will be complex valued, now we cannot rank order them because complex number cannot be ordered, so we simply list them as 2N eigenvalues, associated with each of these eigenvalues there will be a complex valued eigenvector. Now we can make one observation if you now consider BR is equal to minus alpha AR and take conjugation on both sides I will get B star R star is equal to minus alpha star A star R star, but A and B are real valued, so B star is same as B, A star is same as A, so that would mean it will be BR star is equal to minus alpha star AR star, so what does it mean? If alpha, R is an eigenpair its conjugate is also an eigenpair, that means eigenvalues and eigenpairs appear as complex conjugate pairs, okay, and the order of this equation is always E1 because I start with the N degree of freedom system and the state space model has 2N degrees of freedom, so this will be always true. Now let us denote the eigenvalues as alpha 1, alpha 2, alpha N, then their conjugates alpha 1 star, alpha 2 star, alpha N star, similarly for the eigenvectors I will write it as R1, R2, RN, and then their conjugates R1 star, R2 star, RN star. Now let us try to understand the nature of an eigenvector, so to do that let's recall that Y is made up of Q dot and Q, suppose now the system is vibrating in its normal mode the solution for displacement alone can be written as phi into E raise to alpha T, now Q dot will be therefore alpha phi exponential alpha T, so if I now write the state vector in the eigenvector when the system is vibrating its normal mode it will be of the form alpha phi phi into E raise to alpha T, because Q dot is this and Q is this, so I can write in this form, so that would mean each of the eigenvector will be of this form, eigenvalue multiplied by phi and phi, okay, so the kth eigenvector is of the form alpha K into phi K and phi K, now what I do is I define a modal matrix capital Psi where I will now assemble these vectors R1, R2, RN and their conjugates in this order, so I have this Psi, now for each of these R1, R2 I will write it as alpha 1, phi 1, phi 1, so alpha 2, phi 2 for R2, and their conjugates will appear in the next set of terms here, so now if I introduce N by N matrix phi with phi 1, phi 2, phi N as columns, and capital Lambda as a diagonal matrix of the N eigenvalues, then we will see that the modal matrix in this problem has this structure, it will be of the form phi lambda phi, phi star lambda star phi star, so it is important to recognize that this structure is present in the modal matrix. Now we can now talk about orthogonal relations, so let us consider Rth and S eigenpairs, so the eigenvalue problem for Rth, eigenvalue will be BRR is minus alpha R ARR, similarly for S I get this equation, so let's name these equations as 1 and 2, I will pre-multiply 1 by RS transpose and 2 by RR transpose, so I get these equations, so first one is RS transpose BRR is minus alpha RRS transpose AR, so the next equation is RR transpose BRS minus alpha S RR transpose ARS, now we will transpose both sides of 4, I will get RS transpose B transpose RS and minus alpha S RS transpose A transpose ARR, now the benefit of writing the original equation in the specific form that we chose now pays dividends here, we know that A and B are symmetric, so A transpose is A, B transpose is B, so if you go back when we wrote this additional equation the way we converted this N second order equations to 2N first order equation there are other ways of doing it I will come to that later, this way of doing ensures that these coefficient matrices are symmetric, okay, so that is helpful to consider orthogonal relation, so now since the A and B are symmetric this equation can be written in this form, so this is equation 5, now if you subtract 3 and 5 I get alpha S minus alpha R is RS transpose ARR equal to 0, so I can demand that RS transpose ARR is 0 whenever R is not equal to S, if alpha R equal to alpha S I can still insist that RS and RR can be chosen so that this relation holds, this is what I showed in the case of a repeated eigenvalue. Now once this is 0 for R not equal to S you can go back to any of these equations for example here, and if right hand side is 0 for S not equal to R, left hand side will also be 0, so automatically I get this, so in the matrix form we can rewrite this, but before that we can normalize the eigenvalues, so what we can do is we can select the normalization constant so that RS transpose ARS is 1 for S equal to 1 to 2I, so in that case what happens? It will be RS transpose ARR will be the chronicle delta, and RS transpose BRR will be minus alpha R into chronicle delta, so now in terms of the modal matrix that we showed well before the orthogonality relation with this normalization in place will be psi transpose A psi is I, and psi transpose B psi is this diagonal matrix, the first block has lambda and the next block is its conjugate, and the half diagonal terms are zeros. Now I can write the Rth eigenvalue in this form minus eta R omega R plus this, if we indeed do that we can see that the Kth coordinate in Rth mode will have this type of representation, so this is quite similar to the free vibration format of the free vibration solution for a single degree freedom system that we saw, so there's a connection that I want you to notice. Now let's look at force response analysis, so I have now AY dot plus BY is F of T with these initial conditions and this is a forcing vector, so I make this substitution, Y is psi Z, and I put it back in the original equation I get this equation, now if I pre-multiply by psi transpose I get this. Now psi transpose F, let us first look at it, psi transpose is this and F is this, so this will be, since there's a zero here the structure of the generalized forcing vector will be this. Now psi transpose A psi is I, and psi transpose B psi is this matrix and this is Z, and this is a forcing vector, so I will get now for each of the Z uncoupled equation, but we can write Z as two vectors U and V, if I write like this you can see that these two equations can be written, these two N equations can be written in terms of a set of two N uncoupled equations as shown here, one for U and one for V. Now let's consider the case when Rth degree of freedom is driven harmonically by E rise to I omega T, okay, now the equation for U and V will be of this form, so F of, now if you write now for F of T the summation, this summation if you expand since only one coordinate has a nonzero value this summation collapses to a single terminal, so in the, as T tends to infinity UK of T will be given by this, and similarly as T tends to infinity VK will be given by this, okay. Now if I put back into the original coordinate system Y will be psi Z, so Q dot Q is this, so Q will be Phi U plus Phi star V, so if I now write QJ in this form we can see that as T tends to infinity I get the expression for the frequency response function in terms of the normal modes and the eigenvalues and their conjugates, so this is the transfer function in terms of the modal description, in the modal domain, again we can see that this function is, this is a receptance which is symmetric, but it is not Hermitian. Now let us consider the other case where Rth degree of freedom system is driven impulsively, this is a viscously damped system therefore I can talk about tire response in time domain also, so I will get now the response, I mean I get the equation for U and V in this form, so the equation for UK will be this, and equation for VK will be this, now you can see here that U and V will form a conjugate, complex conjugate pair, so VK of T will be this, and UK of T is this therefore VK is nothing but UK conjugate, so now if I use those relations we can go back to Y coordinate system Y is psi Z, and for psi I am writing this, and if I use the fact that U and V are conjugate pairs, now the impulse response function will be in terms of this, okay, again this is impulse response function is obtained in terms of the eigenvectors and the eigenvalues, and it is real valued here, okay, although the eigenvalues and eigenvectors are complex value, we can quickly go through a numerical illustration, so let's consider 3 degree freedom system configured as shown here, and these are the numerical values, and if I write the equation of motion for this system I get for the undamped free vibration this is the equation, so we can quickly do the some calculations because it's a simple problem, we will consider two alternative models, one is proportional damping matrix with damping ratios specified for the three modes to be 0.01, 0.03, and 0.05, this is one model, and associated with this will be the C matrix which we can compute following the procedure that I outlined, and the second model is a non-proportional damping matrix model where off diagonal, you see this is almost similar to this except that the off diagonal terms are 0, and the diagonal terms are nearly, you know, of the same order as this, this is 100, 175, 200, now for the undamped system this is a mass, this is stiffness, after we insert the numerical values, we can show that the undamped normal modes this is a modal matrix, and this is a matrix of eigenvalues, you can see that phi transpose M phi is the identity matrix, phi transpose K phi is the lambda matrix which is same as this, squares of the eigenvalues, or the eigenvalues themselves squares of the natural frequencies. Now if I now consider C that is the first case where C is proportional we can see that phi transpose C phi is a diagonal matrix, so the system has these three damping ratios and these three natural frequencies, and these are the complex, you know, roots of the characteristic equation which we get as this, for this problem. Now what I will do now is, even for this case where damping is proportional we can see now if I approach the formulation through A and B matrices, so what happens, will I get the same answer or will I get something different? We should expect to get the same answer, suppose for the MCK damping matrices that we have chosen this will be the A matrix, this will be the B matrix, you can see that now this is 6 by 6, these two are 6 by 6 matrices. Now there are too many numbers here but you need to follow a few details, this psi is the 6 by 6 complex valued eigenvector matrix, so you can see that they appear as conjugate pairs, this is not ordered in the way that I showed, this is output of eigenvalue solver so I am just reporting the way we got the results. Now you can see here this is one eigenvector its conjugate appears here, this is the next eigenvector its conjugate is here, okay, and this is this, so what I mean to say is this is not in this form, okay, doesn't matter. Now you can quickly see that psi transpose A psi is a diagonal matrix, psi transpose B psi is also a diagonal matrix, you must notice that psi transpose A psi is diagonal but it is not an identity matrix, it has not been normalized to be a diagonal matrix, sorry an identity matrix. Similarly psi transpose B psi is a diagonal but diagonal entries are not equal to the eigenvalues, okay, so that you should notice. Now the matrix of eigenvalues are shown here and this has a square matrix and as a column it is shown here, so you can see here that they are appearing as complex conjugates. Now if you now compare this with what result that we got earlier we see that we are getting the same results, for example the first pair of roots here is same as these roots, the next pair is this which is this, third pair is this, this is this, so what does it mean? That means the system is classically damped so it doesn't really matter which approach you take you will get the same roots. Now how do we see the nature of eigenvalue? So in the, if you look at now the eigenvectors for the undamped case we can see that for this eigenvalue all the 3 degrees of freedom are imperfectly in phase, whereas here this is negative this is positive therefore the phase difference is 180 degrees and similarly this is again phase difference is 180 degrees. So in undamped free vibration all points vibrate harmonically either perfectly in phase or perfectly out of phase, there is no other phase difference is possible, okay. The phase differences will occur because of damping but in classically damped system the question that we should ask is what happens to the phase difference? So if we ask that question if you look at this psi matrix the question about phase difference is not, cannot be easily answered, so what we do is we normalize this eigenvector so that I fix the third entry here to be 1, so you recall that the modal matrix is of this form so the lower half of this is on displacements and the upper half is on velocities, so I make the first displacement to be 1, that is a normalization you know I have the freedom to do that because there is an arbitrary scaling factor. Now consequently the eigenvectors will have this appearance, now if I look at angle associated with this if I find the angle of that I get this, so now you focus on the lower half of it I see that the phase angles are either perfect out of phase or perfect in phase, so the way the eigenvalues are organized you can see that the third eigenvalue here is the first eigenvalue in the undamped system so you can see that all points are in perfectly in phase, for this eigenvalue this is perfectly out of phase with the other 2 degrees of it, right, so you follow, so this is because we are using classically damped systems. Now let us go to the non-proportional damping matrix, so the undamped in normal mode is this, now if I simply find phi transpose C phi this won't be a diagonal matrix, that is why I call this is non-classical, first you should convince that we are dealing with a non-classical damping, so this is how we could quickly verify that. Now let us form the A and B matrices, this is again 6 by 6 matrices here and perform the eigenvalue analysis, this is a sine matrix as thrown out by the eigenvalue solver without any adjustment of the normalization, you see psi transpose A psi is again diagonal, psi transpose B psi is diagonal, so there again psi transpose A psi is not an identity matrix and psi transpose B psi although it is diagonal the diagonal entries are not eigenvalues, good. Now this is the eigenvalues written as a square matrix, diagonal matrix I can write it as a column vector and I get these as eigenvalues, okay, now this is a damped system, so now undamped natural frequencies are shown here, see you can see now there is a slight change in this number 10.691 it is 10.7074 etcetera, okay. Now again I normalize the first degree of freedom I will arbitrarily make it as 1, okay, and find the angles now, the moment you find the angles you see something interesting happening, the phase differences are now nearly 0 but slightly changing, this is minus 0.18 degree to 1.75 degree, whereas here it is not perfectly out of phase instead of being 180 degrees it is 178 degrees, okay, so this is how the damping influences the normal modes, that means the all points on the structure will now be executing a decaying harmonic oscillation but different points will not be in perfectly in phase or perfectly out of phase in non-classically damped system, in classically damped systems again in free vibration all points will be executing decaying exponentially decaying harmonic motions but the motions will be either perfectly in phase or perfectly out of phase, but here there is a phase difference you know other than perfect in phase or perfect out of phase there is a slight modification to that, this is how the damping manifest itself in the Eigen vectors, so if you are doing an experiment and if you measure normal modes you will always measure damped normal modes, because in a there is no magic switch to remove damping in an experimental work, in a computational work you can put C equal to 0 and find out the Eigen values very easily but that is a counterpart of that in experimental work is not possible, so you will have to deal with damped normal modes if you are doing an experiment, so whenever you measure normal mode there are methods for that you will see that there will be a slight phase differences in a lightly non-proportional damped system, these phase angles will be clearly close to 0 or 180 degrees. Now I have now discussed non-classical viscous damped system, now I will extend the formulation to structural damping with non-classical damping, so we consider the N degree of freedom system, this is the structural damping term which is the complex part of the stiffness, and again I drive the Rth degree of freedom harmonically, this is the only case that we can consider there is no transient analysis possible here. Now we need to now find a transformation Q equal to TZ so that T transpose MT and T transpose K plus IDT are diagonal, okay. Now this has the form of MQ double dot plus KQ equal to 0 undamped pre-vibration but K is complex valued, so the normal mode should be orthogonal I mean the transformation matrix should be orthogonal to the complex valued stiffness matrix, so that is the novel feature here, so what we do we towards achieving this we consider the eigenvalue problem K plus ID psi is minus SS square M psi, okay. I cannot call it as a free vibration problem here, it's a simply a mathematical statement of an eigenvalue problem because you cannot talk of free vibration for this problem, so the characteristic equation is given by this and we have N complex valued roots and associated eigenvectors, now you should notice now that these eigenvalues will not appear as complex conjugates, they are N complex numbers, similarly eigenvectors are N complex eigenvectors and N need not be a even number, okay. So this all the eigenvectors are assembled in psi matrix and we will normalize that so that psi transpose M psi is psi and psi transpose K plus ID psi is a diagonal matrix, okay. You can prove the orthogonality using the arguments which I have now used several times, you consider R and S pair and you know do go through the few steps of calculation, you can show that these orthogonality relations are true here. Now you come to the force vibration problem, you make the substitution Q is equal to psi Z, substitute back I get this, pre-multiply by psi transpose I get this, now psi transpose M psi is I and psi transpose K plus ID psi is another diagonal matrix, that is a diagonal matrix of complex valued eigenvalues as shown here, so the equation for ZK will be of this form, so I can find amplitude in steady state to be this, so go back to the Q coordinate system I get this and this is my expression for therefore F, this SK is now a complex valued number, SK square, so again we can consider as a simple example, we will have the same M and K and this is a undamped to normal mode matrix and this is a undamped natural frequencies, phi transpose M phi is diagonal, phi transpose K phi is this, so for this data now we will introduce a proportional structural damping model and a non-proportional structural damping model, so let us start with case of proportional damping matrix, so I will take D to be 0.03 times K, so K star will be K plus ID and phi star D phi will be this, okay, this is a mere transposition, it is not a conjugation plus transposition, it is mere transposition, that is also important to note. Now see undamped normal model matrix uncouples the equation of motion, in this case, psi is this, you can see phi transpose D phi where phi is undamped model matrix, it uncouples the damping matrix, so I can take psi to be this, psi transpose M psi is this, psi transpose K plus ID psi is also this, that means even if I find complex valued psi, the uncoupling takes place. So now we will scale the first eigenvector, element of the eigenvector to be 1 again, again you see if I find the angle of psi the phase differences are either 180 degrees or 0, so the points are still perfectly in phase or perfectly out of phase, because this is a proportionally damped system. So now if I find eigenvalues for the damped system the eigenvalues will be of this form, okay, now undamped system if you recall the eigenvalues obtained are this, so if you compare these two for the kind of notations we have used the real part of eigenvalues will be the natural frequencies here, okay, these three terms appear as a three natural frequencies and these are the terms associated with damping, so this is a classical structural damping. Now let's consider a not proportional damping, so again I consider D to be this and K star is K plus ID, quickly we can verify phi transpose D phi is not diagonal, so it's not a classically damped system, so now I can do a complex eigenvalue analysis and I get this as psi matrix, so we can verify psi transpose M psi is this, psi transpose K psi is this, these are diagonal but psi transpose M psi is not identity, this is not, this is again diagonal but diagonal entries are not the eigenvalues, so I will scale now the eigenvector so that the first element is 1, I get this as a scaled psi matrix and if I now compute the angle of this you see now I get a phase difference of about 2 degrees and less than 1 degree here and 0.6 degrees and 1.8 degrees and about 4 degrees and here, so this is a manifestation of the role played by damping on eigenvectors. Now you can look at now the eigenvalues for the damped system, I get this you can see here the real parts are again close to the undamped square of the natural frequencies but they are slightly different because of influence of damping, okay. So now if we can quickly summarize now what we have done is we have evaluated frequency response functions for both viscous and structurally damped systems, by direct calculation for viscously damped system the FRF matrix is given by this, for structurally damped system the FRF matrix is given by this, but if you want a calculation based on modal superposition then you have to find the appropriate modal coordinates for that. So if we talk about viscously damped system with classical damping we have shown that it is given in terms of these undamped eigenvectors and natural frequencies like this. For similarly for structurally damped system with classical damping again with the undamped normal modes we get these solutions, for viscously damped system with non-classical damping we go to that A and B matrices and we get now the frequency response function in terms of complex valued eigenvectors and complex valued eigenvalues, this is how. For structurally damped system with non-classical damping the one that we did last again, now this is summation in terms of complex valued eigenpass. So this is direct calculation, this all these are through modal superposition, so this is what basically I intended to demonstrate. Now I also pose this question, we pose basically two questions when we are talking about non-proportional damped system, the first one was what is the mathematical framework to uncouple the equation of motion, this we have answered now. The other question that is remaining is are there any simplification possible so that a damping remains classical and at the same time we take into account the fact that structure is made up of subsystems with different materials. So this takes us into damping model, material damping model we will consider that in the next lecture, so we will conclude this lecture at this point.