 In the previous lecture we discussed about CISO that is single input single output system in which one excitation is on the single degree freedom system and there is one output and for that we obtain the necessary relationships and in that also we discussed several things like what is the relationship that exist between the power spectral density function of velocity with the power spectral density function of displacement then power spectral density of acceleration with power spectral density function of displacement and so on. Now before we start the multi degree of freedom system in which we will have a number of inputs to the system and there will be a number of outputs from the system and the inputs would be represented in terms of a vector and outputs will be also a vector and as we discussed last time when you have a number of random processes in a particular vector then the process is described with the help of the power spectral density function matrix rather than a single power spectral density function. Let us try to recapitulate some of those fundamental things that we discussed in yesterday's lecture so that we can effectively understand the MDF system subjected to a random excitation vector. First thing that we discussed yesterday was y is equal to A into X where y is a vector X is a vector and it is connected through a matrix A. In that case SYY will be power spectral density function matrix and SXX is the power spectral density function matrix. SYY will be related to SXX with the help of this equation. SXY that is the cross power spectral density function between the output and input that is SXY is will be equal to A into SXX. Now if we consider a multi degree freedom system then we can write down the equation of motion and if Pt is the excitation vector then one can obtain the relationship between the frequency contents of the response vector to the frequency content of the excitation with the help of a FRF that is frequency response function matrix H omega and this H omega is equal to K minus M omega square plus IC omega inverse of that that is what we discussed yesterday. Therefore X omega can be written as X omega equal to H omega multiplied by P omega where P omega is the frequency contents of the force vector Pt. In case of single degree of freedom system this H omega matrix is written as small h omega that is a single quantity which is called also the FRF that is frequency response function of a single degree freedom system. Now SXX and SYY these matrices say for example if it is a vector of size 2 then this will be the nature of the power spectral density function matrix the diagonal terms are the power spectral density functions of X1 and X2 and the op-diagonal terms would represent the cross power spectral density function between X2 and X1 and this will be between X1 and X2. If we have a more number of elements in the vector then obviously we will have X1, X1, X2, X1, X3, X1 so on. This will be the cross power spectral density function between different responses and this relationship also was shown that is X1, X2 cross power spectral density function between X1 and X2 is equal to the complex conjugate of the cross power spectral density function between X2 and X1. In case of the matrix that means if it is not a single power spectral density function if this is a matrix then the relationship is SX1, X2 will be equal to X2, X1 complex conjugate and transpose of that. So that is the difference between the single quantity as the cross power spectral density function and the SX1, XX2 or the cross power spectral density function between X1 and X2. In that case X1 and X2 will be the two vectors. Now the diagonal terms of the power spectral density function matrix are real and they are frequency dependent and generally we do not write omega in front of it but it is understood that for each frequency we have a matrix like this therefore as we go on varying the frequency then we have for each frequency one matrix like this. Now with the help of this we would now go into the multi degree of freedom system and in the previous lecture also we had shown that X omega is equal to H omega into P omega that they are all complex numbers T omega is obtained from first Fourier transform and SXX the power spectral density function of the response is equal to absolute value square of H omega into the power spectral density function of P. Now this can also be written as H omega into SPP into H omega star because this multiplied by this becomes equal to the absolute value square and the cross power spectral density between the response between the force and the response is equal to H omega SPP. Apart from that we also had proved in the previous lecture that the power spectral density function of velocity is equal to omega times the power spectral density function of displacement and power spectral density function of acceleration is equal to omega to the power 4 into the power spectral density function of displacement. Now in case when this X dot and X they are vectors then they are matrices and the power spectral density matrix of velocity is related to the power spectral density function of displacement again through this relationship only thing that these changes to matrices. So we had proved that SXX dot plus SX dot X is equal to 0 that is the cross power spectral density function between the velocity and displacement that if that if it comes into the solution or analysis then we ignore that that means we set that to equal to 0 because of this reason. Similarly if the velocity and acceleration for that if we have a cross power spectral density function or cross power spectral density matrix then we ignore that because of this reason that this plus this that becomes equal to 0. And also we have seen that SXY is equal to SYX star T that is the complex conjugate of this and transpose of that. So with these few things that we described in the previous lecture we come into the response of a multirigrafiton system excited by a force vector Pt. Now for single point excitation we know that the force vector Pt is equal to minus m into i into X double dot G where i is a influence coefficient vector consisting of 1 1 1 or 1 0 1 0 1 0 and so on that you have seen in the case of the deterministic analysis. X omega we can write down to be equal to H omega matrix multiplied by P omega that is what we have just discussed where H omega is the frequency response function matrix for the system multirigrafiton system and P omega is the frequency content vector of the excitation Pt and X omega is the frequency contents of the response. Now if this relationship holds good then from the just discussion that we had from that we can write down the power spectral density function of SXX that is the power spectral density function matrix of X is equal to H omega matrix multiplied by the SPP matrix that is the power spectral density function of the excitation into H omega star t. In the case of the single degree freedom system if you recall we had we could write it as small H omega multiplied by SPP multiplied by small H omega star and H omega star small H omega star and this H omega if you multiply that becomes the absolute value square of H omega square. If H omega is a matrix then we have to make it a transpose of the complex conjugate of the matrix H omega. So then only it will become the absolute value square. So this is a standard relationship that we use for the case of a multiregree of freedom system. What you have to simply do is that you have to find out the FRF that is the frequency response function matrix for the system which is equal to K minus M omega square plus I C omega and inverse of this matrix if you take at every frequency then you get H omega at every frequency and SPP at every frequency is known. So you have to carry out this matrix multiplication and from that you will get the value of the power spectral density function matrix at each value of omega. Now in case of the single point excitation system SPP that turns out to be M i into I T M t into S X double dot G and these follows from this relationship that is P t is equal to minus M i X double dot G and therefore if we wish to write down the value of SPP then it will be M i into S X double dot G multiplied by the I T M t that is A S X A t that what we have discussed just now. So using that one can write down the SPP that is it will be M i into I T M t in place of writing I T M t on this side we have taken it on this side because S X double dot G is a single quantity in the case of the single point excitation. So therefore whether we keep I T M t on this side of S X double dot G or in this side it does not matter. Now with the help of this relationship that we obtain for SPP the S X X that is power spectral density function matrix of the response can be written in this particular form that is H M i then I T M t H star t into X double dot G where X double dot G is a power spectral density function of the ground motion. So it is a single quantity at each frequency and the cross power spectral density function between the ground acceleration and the displacement that will be equal to minus H M i X double dot G and that follows from the equation that again we just examine that is S X S X output input correlation is equal to A into S X. So in place of S A we write down here now H M i so that is that becomes the pre multiplier for the S X double dot G. Now using these relationships a problem is solved and this problem was the problem in which we had two responses at the non support degrees of responses the problem is I think the problem is this was the problem and that is we had u 1 and u 2 as that is placements and we had three ground excitations and in this case all the three ground excitations are assumed to be the same and therefore we had a R matrix that is in place of R matrix we had only a I matrix. So this is the I matrix you can see here the I matrix is equal to I 11 that this is because of the fact that the ground excitations at the three supports are assumed to be the same they are not different therefore the I influence coefficient vector will be equal to 1 and 1. So we know the mass matrix we can construct the C matrix from the relationship that is alpha m plus beta k and alpha and beta can be obtained from the two frequencies that you have obtained. So with the help of that the alpha value had become 0.816 and the beta value is 0.0027 this is the k matrix and from there we can obtain for each frequency the h matrix that is by inverting this matrix and obtain the value of the power spectral density function matrix of the two displacements using the previous equation and the diagonal terms of the power spectral density function matrix that is the PSDFs of the two displacements. So that is what is shown here in the plot you can see that it is the power spectral density function ordinates for displacement u1 and this is the frequency and that is how the PSDF of the displacement u1 varies with frequency and one can see that this is a picking nearly at about 12.5 and the first frequency of the system is 12.25. So therefore we see that there is a the peak of the power spectral density function of displacement u1 is around that frequency. This is the power spectral density function of the displacement u2 these area under the curve of this power spectral density function of u1 and u2 they provide the variance and the square root of variance gives the standard deviation of the displacement and in case of zero mean process the standard deviation is equal to the root mean square value. So the root mean square value of u1 is 0.0154 and the root mean square value of the displacement and 2 or u2 is 0.0078. Now when you consider the multipoint excitation system then the S this is this is Sxx this is not Sx double dot g it is wrongly written over here it will be Sxx that is the power spectral density function of the matrix of the response is equal to now h m in place of r now we have in place of i we have got now r. So h m r x double dot g is a matrix now is not a single quantity because the size of the matrix would depend upon the number of the point of excitations and it is multiplied on the right hand side by rt mt and h star t. So that is the general relationship between the displacement and the excitation displacement is a vector the force is a vector and therefore the this is a matrix of displacement and this is a matrix of the or power spectral density function matrix of excitation. And r is the influence coefficient matrix that we have discussed when we are solving the problem for the multi degree of freedom system subjected to the deterministic ground motions which cause different excitations at different supports because of the time lag Sx double dot g x that is the cross power spectral density between the excitation and the displacement is given by this expression again and here Sx double dot g again is a matrix the size of this Sx double dot g matrix is s into s if s is a number of support and r is a influence coefficient matrix that is obtained from a static analysis of the system that we have seen before is of the size of n into s that is if there are n degrees of freedom and if there are s number of support excitation then the displacements that will be caused at the n non support degrees of freedom that will be represented by a matrix n into s many a time we are not interested in all the response quantities say only a limited number of response quantities may be involved for example in this problem we consider that this is a frame subjected to two ground excitations different ground excitations arising because of the time lag and there are five degrees of freedom in the system that is non support degrees of freedom then h omega matrix will be equal to a 5 by 5 matrix now say we are interested in only the response x 1 and x 4 in that case the we can have a reduced frequency response function matrix which will be called h bar omega equal to 2 by 5 that is we can select the first row and the fourth row and make a modified h omega matrix called h bar omega matrix and its size will be equal to 2 into 5 now s x x now will be a 2 by 2 matrix because we are only interested in the responses x 1 and x 4 so the diagonal terms of these matrix would give you the power spectral density function of 1 and 4 and the off diagonal terms will give the cross power spectral density function between 1 and 4 and 4 and 1 so that is what the s x x matrix will represent and it has the same equation as that of 4.77 that we had just discussed that is h m r s x double dot g then r t m t h star t only in place of h star and h we write now h bar and h bar star where h bar is this modified frequency response function matrix and if we put the sizes of different matrices here we can see that this finally leads to this multiplication finally leads to a 2 by 2 matrix which would be the size of the matrix s x x so in many problems what we do is that we may not be interested in all the displacements or all the response quantities and for that we take a reduced frequency response function matrix and use the same equation that we have stated over here with the help of equation 4.77 now the equations that I had shown before or in the previous slides they have been obtained or derived with the assumption of ergodicity and that we did purposefully because it is easy to understand this relationship by assuming the stationary process to be ergodic because in that case we will be dealing with only one time history and with the help of the Fourier transform of that we can construct all that relationship however all the relationship that we have derived in the previous slide they can be also derived without the assumption of ergodicity that is by assuming the process to be a stationary random process and in that case what we will do is that we will start with the autocorrelation and cross correlation functions of the process or of the two processes and then take a Fourier transform of them and using that would come to all the equations that we have derived before and we have seen before that there exist a relationship between the autocorrelation function and the power spectral density function through a Fourier transform pair. Now let us solve a problem of multi degree or multi point excitation system that is example 4.2 if a time lag of 5 second is introduced between the supports find the PSDF of u1 and u2 that is the problem this problem in this problem we now assume that the x double dot g1 x double dot g2 and x double dot g3 are different and there is a time lag of the 5 second between these excitations. So the time lag between x double dot g1 and x double dot g3 is 10 second x double dot g1 and x double dot g2 is 5 second and x double dot g2 and x double dot g3 is 5 second. So these are the time lags then first we have to construct sx double dot g matrix that will be a 3 by 3 matrix of the ground motion and the diagonals of this would be the power spectral density function of the excitations at this point and at this point and the cross power spectral density functions would represent the cross power spectral density function on between this excitation and this excitation and so on. So this is the sx double dot g matrix a 3 by 3 matrix. So if we have to construct that matrix then let us recapitulate what we discussed in the seismic input part of our study. In the seismic input if you recall we discussed about a coherence function and one of the form one of the empirical equation of the coherence function which is a very simple one is this that is exponential of minus absolute value of rij into omega divided by vs where rij is a distance between the point i and j that is at these two points we have excitations and the distance between these is rij. The vs is the seismic wave velocity so the same earthquake train or ground motion or time history of ground motion that is travelling with a velocity of vs in that case one can replace rij by vs by the phase or the time lag that is rij divided by vs is equal to the specified time lag that is the time that will be required to travel from i to j if the velocity is vs. Now this has been specified that is between the supports the time lag is 5 second so in this we can replace this quantity rij by vs by 5. Now once we have done that then coherence function between the first support and the third support will be equal to exponential of minus this quantity will be 10 second into omega. Similarly between the support 1 and 2 that is these support between 1 and 2 you will have 5 second that means rij by vs will be replaced by 5 second and x double dot g 2 x g 3 will be again by 5 second. So one can construct a matrix easily because the cross power spectral density functions between the supports are now known this cross power spectral density function is given by this equation if we recall the seismic input that we discussed and this is the cross power spectral density function at support i and this is at support j and it will be multiplied by a coherence function cuhi,j and since this is known now one can obtain the different cross power spectral density function very easily in the case of a single up quick ground motion train of ground motion moving with a velocity the power spectral density function at different supports remains the same that is equal to sx double dot g that is which is specified only the change or the cross power spectral density function arises because of the coherence function. So we write down the cross power spectral density a function between the two excitations i and j as coherence i,j into sx double dot g. So in this particular fashion one can write down all the cross power spectral density function terms note that sx double dot g and x double dot g i that is the cross power spectral density function between j and i is equal to cross power spectral density function between i and j. This happens in this case because sx double dot g is a real quantity there is no complex quantity involved over here and since the real quantity the star that we had used in defining the relationship between the cross power spectral density function between 0.1 and 2 and 0.2 and 1 that star is not required and therefore they become the same. Now with the help of that we obtain the cross spectral density function matrix of excitation that is given by this matrix the r matrix we had computed before for this problem the r matrix was 1 by 3 into 111 and this coherence function we have used sorry in the in the previous discussion I have simply written is as vs it will be 2 pi by vs and the values of rho 1 and rho 2 they are 5 omega by 2 pi and 10 omega by 2 pi. So with that values of rho 1 and rho 2 one can obtain the cross power spectral density function matrix of excitation. So this is the cross power spectral density function between the support 1 and 2 and this is between support 1 and 3. And we have taken sx double dot g a single quantity a common out of them because a single cross spectral density function is defined for the ground motion and the excitations at different points vary because of the phase lag. So this becomes a power spectral density function matrix and this is the r matrix. So therefore with the help of these and the h matrix one can obtain the power spectral density function matrix between of u 1 and u 2 these two displacements and the diagonal terms of that matrix gives you the p s d f of u 1 and u 2. So they are plotted here and the plot again shows that the maximum value occurs around the first fundamental frequency of the structure and the values of the or the rms values of the responses becomes equal to 0.0089 and 0.0045 for u 1 and u 2. So this is the p s d f of u 2. The time history analysis that we obtain for the L centre earthquake note that the power spectral density function input that you have given here is the power spectral density function of the L centre earthquake which is tabulated at the end of the book as an appendix one can use those digitized values. So we have used those digitized values for obtaining this the power spectral density function ordinates at different frequencies. So when we obtained we took the same L centre earthquake as a time history and we did a time history analysis for the monthly support excitation system with a time lag of 5 second between supports we obtained the result as 0.0092 so that was the value of the rms value of the response that we obtained from the time history and the spectral analysis give the result at 0.0089. So you can see that the results are very close to each other that verifies or validates the accuracy of the result. We solved another problem which was a problem of the pitch roof portal frame that is we had a pitch roof portal frame if you recall and there is a degree of freedom here and a vertical degree of freedom 5 and this is having a multi support excitation and this multi support excitation here they are two multi support excitation the time lag between these two points is given as 5 second that is Rij by vs that becomes equal to 5 second. So one can have a correlation coherence function with rho 1, rho 1 defined as exponential of minus 5 omega by 2 pi so with the help of that we can construct the 2 by 2 power spectral density function of excitation Sx double dot g again is the power spectral density function of the L centre earthquake that we have taken and with the properties of the pitch roof portal frame that is the mass matrix defined as this and the K matrix defined as this we obtain the 2 natural frequencies as 5.58 and 18.91 the alpha and beta that was calculated with the help of these 2 frequencies were 0.431 and 0.004 and this was the C matrix that we constructed and the R that we determined before that is the influence coefficient matrix. So with these matrices defined we again used the same equation for finding out the power spectral density function of displacement 4 and 5 and using that multiplication of Amr and Rm and H omega star so on we obtained the power spectral density function matrix of the displacement that was a 2 by 2 and the results are compared for 2 cases that is in one case we assume that it is perfectly correlated and in other case we assume that there is a phase lag of 5 second the results which are given over here is not correct the correct results are shown over here in this paper that is the RMS value that was worked out for these displacement and these displacement was 0.0331 when it was fully correlated that is there was no time lag between these 2 points and when they were partially correlated that is there was a time lag between the 2 the RMS value of the response was 0.023. So one can see that when there is a for fully correlated ground motion that is for the single support excitation the response is more compared to when it is partially correlated for the displacement 5 the when it was perfectly correlated the value was 0 that is expected because the these 2 supports would be moving as a single unit that is it is it becomes a single point excitation system and in that case we expect that the there will not be a vertical motion over here the entire thing would be vibrating in this particular fashion and the for when it is partially correlated so there is a little bit of displacement that occurs at the crown. So this shows the importance of the correlation between the excitations at the 2 supports if we assume it to be perfectly correlated that is we assume it to be a single point excitation system then in some cases we can get some erroneous result for certain degrees of freedom it is better to consider the partial correlation between the excitation produced by the single earthquake travelling at a certain speed however this kind of thing needs to be done only for cases where the 2 supports are spatially separated by a large distance for small distances this particular partial correlation effect of partial correlation is almost negligible though the power spectral density function matrices or the the power spectral density function matrices are shown over here and one can see that the frequencies at which it is speaking is nearly equal to the frequency of the the or the first frequency of the system. Now here these the power spectral density functions they differ a little bit for the case of with time lag and without time lag and that is shown over here this is with time lag and for without time lag there was only one peak whereas with time lag there are 2 peaks in the power spectral density function. Next let us come to the absolute displacement that is how to obtain the power spectral density function of absolute displacements of the responses whenever we have a multi support excitation system we have seen that there are 2 displacements that one need to consider that is a relative displacement of the non support degrees of freedom with respect to the support that is called the relative displacement the other one is the absolute displacement in which the total displacements at different non support degrees of freedom are considered the it is necessary because if we wish to find out the member end forces then we require the relative displacement of one end with respect to the other in terms of the total displacement for a multi support excitation system for single support excitation the since the same displacement takes place due to the ground motion at all supports therefore it is only the relative displacement that is good enough to obtain the member end forces from the displacement. Now the absolute displacement can be written with the help of this equation that is i into x plus r into x g so if we look at this expression over here the i is a matrix of identity matrix that is diagonal terms of 1 1 1 of diagonal terms of 0 that multiplied by the 3 displacements u that gives the relative displacements of the system and the ground displacements are x g 1 to x g 4 therefore non displacements and therefore we have a r matrix which will be 4 by 3 r matrix and when it will be multiplied by this x g vector then we will get the influence of these displacements coming on to this non support degrees of freedom. So the influence of this at these 3 degrees of freedom plus the displacements that is caused at these 3 points because of the vibration of the structure with respect to the support that is given by this addition of these 2 becomes the absolute displacement of the total displacement. Now once we have this equation then using that equation one can write down the power spectral density function matrix of absolute displacement with the help of this equation this becomes s x x itself because this is an identity matrix and therefore the i and i t of that if I multiply with s x x it remains s x s itself this becomes r into s x double dot g into r t then s x double dot g that becomes or x g that becomes equal to i s x g r t and s x g x that becomes into r into s x g x i t. Now what is not known to us this is known to us s x s is known to us from the dynamic analysis of the system s x g that is also known because we know the power spectral density function matrix of the ground acceleration and from that power spectral density function of ground acceleration one can find out the power spectral density function of the displacement this relationship we had shown before that is s x double dot g is equal to omega to the power 4 into s x g. So there exist a relationship between the power spectral density function of ground acceleration and the ground displacement so using that relationship one can find out s x g if s x double dot g is specified. Now here this is the term cross power spectral density function between the response that is the displacement and the ground displacement that is to be obtained and once this is known this s x a matrix is completely known. Now one can write down the relationship between the x g and x from the relationship that exist between this that is s x omega is equal to minus h m r x double dot g omega. So this is a basic equation that we write if we recall for the single degree of freedom system we have written x omega is equal to h omega multiplied by p omega. Now here the p omega is equal to m into r and h is the h matrix. So x omega vector is related to x double dot g vector that is the ground acceleration vector with the help of this relationship. And once we have this relationship then x double dot g omega further can be written as omega square multiplied by x g omega that follows from the relationship basic relationship that exist between the ground displacement and ground excitation if they are assumed to be a harmonic excitation. And for harmonic excitation whether we represent it in the complex form or in the real form we know that the displacement and the acceleration are related through omega square. So that is why this x double dot g omega is replaced by omega square into x g omega. And once we have this relationship that means this is equal to this x omega then from this I can easily get the cross pore spectral density function between x g and x and x and x g. So here it will be of course there will be a since it is a matrix there will be a complex conjugate of transpose. So that star is missing over here. So with the help of these relationships that is once we know s x g x and s x g then one can obtain the value of the pore spectral density function matrix of the absolute displacement. So we stop at this we will solve a problem in the next class.