 In the last lecture we discussed about the frequency domain solution of the single and multi-degree of freedom system and in that we discussed about the obtaining the frequency contents of the ground motion and with the help of that frequency content of the ground motion we solve the problem in frequency domain and obtain the frequency contents of the response and then inverse Fourier transform it to obtain the time history analysis. For the multi-degree of freedom system we require the R matrix whereas for the single point excitation system we require I matrix or I vector whatever be the case. Now here is an example problem which illustrates how we can obtain the response of a multi-degree of freedom system using both time domain and frequency domain. In a multi-degree of freedom system for example in this frame we have three ground accelerations and these three ground accelerations are different although the same earthquake wave is passing through all the three supports but because of the time delay that has been assumed the excitations at the three supports would be different. So in order to understand that we take 30 seconds duration of the ground motion and which is travelling when it arrives here at this particular point then the ground motion at this point would be 0 and next here also it will be 0. When 5 seconds of ground motion has passed through this particular point then we get the ground motion over here and at that point of time ground motion still at this point is 0. When 10 seconds of ground motion has passed through this point then we get the ground motion here as well. So as a result of that the excitation caused by the same travelling wave of ground motion is different at these three different supports and we can obtain this by the following considerations from the 30 seconds of record of ground motion that is available to us. The last 10 seconds of record have zero values for the first support that means when all the 30 seconds have passed through this point then we have the next 10 seconds of the time history of excitation would be 0. In this case the first 5 seconds and the last 5 seconds have zero values and in between we have got the 30 seconds of the ground motion and for the last support we will have the first 10 seconds we will have zero values rest of the 30 seconds we will have non zero values. As a result of that effective total duration of ground motion is taken as 40 seconds and that is how the three different records of excitations differ on a scale of measurement of 40 seconds. Now with these three ground motion distinct three ground motions we write down the equation of motion for the system this has two non support degrees of freedom and the support degrees of freedom are three and they are shown over here that is the these are the three ground accelerations and the R matrix we computed before as one third one one one one one one. So for this particular problem we had seen before that the participation of the ground motions in creating responses at the two non support degrees of freedom where equal and was equal to one third one third one third. So this was of course for this problem for other problems it may not be like that that is what you have seen before the R R matrix could be different for different kinds of structures and it is to be obtained from the matrix condensation relationship. Now with this particular equation now we put on the values of x double dot g 1 x double dot g 2 and x double dot g 3 in time from the different records that we have obtained and these records are used for solving the problem in time domain using numerous beta method and when you use the numerous beta method using direct integration technique in the recursive form we require the C matrix and the C matrix is obtained with the help of these two coefficients alpha and beta and this is the Fn matrix of the numerous beta method and this is the Hn matrix for the numerous beta method the initial condition that was used is equal to 0 velocity and 0 displacement with the help of that we went for the recursive equation solution of the recursive equation to obtain the response of the system. In order to remind you what is the form of the recursive equation let us have a look at the recursive equation so that yeah this was the recursive equation that we use for numerous beta method so this is Fn matrix and this is a Hn matrix. So plugging in the different values that we use in the solution of the problem we get the solution and we get relative displacement velocity and acceleration the displacements at u1 and u2 that is the two non support degrees of freedom the time histories of that are shown over here it can be seen that although the 30 second is the duration of the earthquake but the excitation duration was effectively 40 seconds and therefore the response that we get at u1 and u2 are extends beyond 30 second and it almost goes to 40 second of course in this part the responses are very very small. Next we solve the pitch roof portal frame and here if you recall the in the pitch roof portal frame we had the degrees of freedom one in the horizontal direction and other at the crown in the vertical direction these were the two degrees of freedom and for that we assume the 5 second time delay between the two supports by solving the Eigen value problem we get the frequency as omega 1 and omega 2 as these values and from that we computed alpha and beta as that and delta t was taken as 0.02 and this is how we obtain the R matrix if you recall that is we give a unit displacement over here and find out what is the forces developed here and here that gives the stress stiffness matrix and from that one can obtain the value of the displacements caused at these points due to the unit displacement. Similarly if we give a unit displacement over here one can find out what is the vertical displacement here and what is the horizontal displacement over here and they constitute the R matrix that is what we had shown for this example when you are working out the R matrix for different structures and the degrees of freedom let me repeat again is the horizontal degree of freedom here one and another horizontal vertical degree of freedom over here these are the 4 and 5. So, we calculated the mass matrix this calculation was shown how do we calculate this mass matrix for the pitch to portal frame this is the stiffness matrix condense stiffness matrix corresponding to the two translational degrees of freedom and this is the R matrix that we obtain for the two degrees of freedom again. So, now by writing down the equation as mx double dot plus cx dot plus kx is equal to minus m R into x double dot g with the using that we obtain the response of the system, but before that let me tell you that in this case although the duration of ground motion earthquake record is 30 second, but the duration of excitation at the two supports are taken as 35 seconds because there is a 5 second of time delay between the two support that is what is assumed. Therefore, in the first case there will be the first 30 seconds will be non-zero values and the end 5 seconds will be 0 values for the second support the first 5 seconds will have 0 values and the next 30 second will have non-zero values that is how we construct the 35 seconds of excitation for the two supports and we use these two excitations and obtain the values. Now, in this problem it was solved for two cases in one case we consider a time delay and in another case we obtain that there we assume that the there is no time delay between the two supports. So, this is the result shown for the without time delay case that is in both the supports we had the same ground motion of 30 seconds and we continued the integration till 35 second. So, after 30 second the responses that we see is due to the free vibration of the problem and the u 1 rather the degree of freedom 4 that is the horizontal degree of freedom we see that the ground displacement or the displacement of the response is about 0.075 whereas, the vertical response at degree 5 that is of the order of 2 to the power 10 minus 3 it is expected because since the ground motion is in the horizontal direction therefore, the vertical response of the structure would not be very high and therefore, we get a low value. This is the scenario when we have got the time delay effect considered in the analysis and we see that the there is a difference of course, between this case and the previous case that is in the previous case we had 0.075 almost of the maximum value here the maximum value it goes up to 0.048 or so. So, we see that there is a with time delay the response is reduced that is expected whenever there is not a perfect correlation of the ground motion between the 2 supports the responses at least the displacement responses generally is decreased and, but the important thing is that we see that the vertical in the vertical degree of motion the responses has increased that has become nearly equal to 4 into 10 to the power minus 3 where in the previous case the it was almost near 2 into 10 to the power minus 3. So, we see that the horizontal in the horizontal direction the response of the structure is decreased due to the time delay effect whereas, in the vertical direction in which the ground motion is not acting. So, in that vertical direction the response of the structure is increased. Next let us come to the state space direct analysis and we have written down the equation of the state space before and in that we have an A matrix and F matrix and the form is equal to z dot is equal to A z plus F if you recall where A is a matrix z is the state of the system consisting of x and x dot that is a displacement and velocity. Now, here this is for the single point excitation system this is not R this should be I that is influence coefficient vector and this is for the multi point excitation system this becomes R R into x double dot g that becomes the force and the first part that means this vector is of size 2 n if it is n degree of freedom then the vector will be of size 2 n. So, first n values will be 0 and the last n values will have minus R minus I into x double dot g over here and here it will be minus R into x double dot g and how we construct R into x double dot g that we have explained in the previous problem. So, we can solve the problem for the system in time domain as well as in frequency domain and in the time domain one can solve the problem in using the new mass beta method or do homo integration for a single degree of freedom system, but when we solve the same problem in for multi degree of freedom system generally we are not able to solve the problem using do homo integration. So, there is a problem and therefore, we generally do not try to solve the problem using do homo integration for multi degree of freedom system when we try to solve it in the state space form. So, in that case we generally use some other integration scheme than the do homo integration. These problem in the state space can also be solved in frequency domain and in the frequency domain what we do is that we try to find out the H matrix that is the frequency response function of the set system. So, let me first clarify this. Say this is the equation z dot is equal to a z plus f g and we say that we are able to Fourier synthesize this f g which is described in time mind you f g is a vector and the top values are 0 and the last values have got non zero values corresponding to the time history. Then this f g t can be Fourier synthesized and can be written in this form f g i omega into e to the power i omega t f g i omega can be easily obtained from using the f f t algorithm that we discussed before. Now if the excitation is of this form then it is reasonable to assume that the steady state solution of the system will be also of this form z will be equal to z i omega into e to the power i omega t. Now if we substitute this into this equation that is we differentiate this once then the differentiated form will be equal to this capital i cap omega z i omega e to the power i omega t where the i cap is nothing but a matrix with all diagonals at imaginary imaginary quantity i i i and the non diagonal components are all 0. So, this is the definition of your i cap with omega multiplied by z i omega into i e to the power i omega t. So, that becomes the z dot and in the case of z we simply substitute this value. So, we have got a multiplied by z i omega e to the power i omega t and for f g we write down the f f t of f g. Then if I take this onto the left hand side then we get this particular matrix i cap omega minus a into z i omega that becomes equal to f g i omega because e to the power i omega t cancels from both sides and then one can write down for a particular frequency say j th frequency if we can write down z j i omega to be is equal to i cap omega minus a inverse of that into f g omega and this is what is called h j i omega capital h j i omega for the j th frequency and f g i omega can be obtained from the f f t that I told before. And how we get the individual components or the elements of this vector when you have the r matrix that we had discussed before that means the one element typically one element of this vector would be of the form of r 11 x double dot g 1 plus r 12 into x double dot g 2 and so on. That will be the one element and in time one can compute this and can get a particular value in time and that is how we can have all the elements values of the elements at every instant of time t that we feed in the f f t. So, we have to now carry out f f t for the n number of time histories because here we will corresponding to all n degrees of freedom we will have one excitation. So, we will have n time histories which are inputted into the f f t algorithm to find out the vector of f g i omega. Now, with this background so for the j th response quantity in frequency domain we can write down this relationship and h j i omega is defined like this that is what we explained before. And then once we get the value of j j i omega then for different values of omega we obtain this j j till the Nyquist frequency or the cut off frequency and then after that we add on the complex conjugate values that is again what we discussed before and that entire values of the values of the j j omega at a interval of delta omega. So, all the values including the complex conjugate values that are given in i f f t and i f f t gives the value of j t and the j t contains not only the displacement, but also a velocity as well. So, that is how one can solve the problem in state space in frequency domain. So, now we look into another example in which there is a pipeline and in this this pipeline is lying on the ground, the resistance provided by the ground is represented by the springs over here and the damping provided by the soil that is shown here as a dashpot. So, at these three supports we provide these equivalent spring and dashpots for the soil and these are the masses which are lumped at these three points and we have got three degrees of freedom in the vertical direction one two and three and a rotation here. So, what we do first is that we condense this the rotational degree of freedom we condense out and then have only a three by three matrix corresponding to this vertical three degrees of freedom with the these values which are specified over here. Next week obtained this is the k matrix which is obtained this is the c matrix which is obtained using assuming it to be a classically dam system and in which the values of alpha and beta are these. So, with these values the c matrix is generated then we form the a matrix and this becomes the fg matrix and since we assume that the ground motions at these three supports they are same that means there is no time lag between them that is why we have 1 1 1 over here this is minus 1 1 1 that is the last three values are this and the top three values of course will be 0. Then we Fourier synthesize these fg and once we Fourier synthesize these fg then one can obtain the values of the responses of the three displacements in frequency domain that is j j omega and using ifft we obtained the response of the three degrees of freedom in time domain. So, this is the time history of the displacements for degree of freedom one that is shown then we have the time history of displacement at two and time history of displacement this is one and two these two are shown and this result is for the solution that we obtained by solving the equation not in the state space, but by solving the problem in the as a second order differential equation there also we can obtain the solution in the frequency domain. So, the objective of this problem was to solve this problem by two methods that is using the frequency domain ordinary and state space solution. So, ordinary means the solution of second order differential equation and the state space solution is the one in which the h j matrix is different than the h j small h j matrix that we obtained for the second order differential equation. So, and the results you can see that they compare very well that is the displacement of degree of freedom one over here has a maximum value of almost 0.05 there you can see that again the maximum value is almost 0.05. So, the responses obtained by the ordinary solution of ordinary differential equation in the state space equation they are obtained as same. Next if we are wanting to solve the problem for absolute displacement then we use a different equation and the in that different equation on the left hand side of the equation you have all the terms written in terms of the total displacement that is total displacement total velocity and total acceleration. On the right hand side instead of the acceleration we require the knowledge of displacement and the velocity that we discussed before and generally we assume that this C s g that is the coupling term in the damping matrix they are generally set to 0 because that is very small and in most cases we only consider p g to be is equal to minus k s g into x g. So, by knowing not the ground acceleration but the ground displacement one can obtain the value of p g vector over here and once we know the p g vector then we can solve the problem either as a second order differential equation or in the state space and we can use whatever method we want that is we can solve it in time domain or in the frequency domain. Now, we come to what is known as the modal analysis which is very popular in the dynamics in the beginning of the multi degree freedom system. I mentioned that the solution for the multi degree freedom system can be obtained for two kinds of excitation one is for the single point excitation other for the multi point excitation. Again for both types of excitation we have a direct solution and the solution obtained by modal analysis. When we use the direct solution then we require the C matrix to be defined and for defining the C matrix we assume that the damping matrix is a is of classical nature or we say that the it is a classically damped system and when we assume the it to be a classically damped system then C matrix is written and to be is equal to alpha times m plus beta times k where alpha and beta values are obtained for using the two frequencies of the system generally the first two frequencies are considered to obtain the values of alpha and beta. So, once we have the C matrix for the system then one can go ahead with the direct solution either solving it as a second order differential equation or by solving it as a coupled of first order differential equation which is called the state space equation and one can use both time domain solution and the frequency domain solution. In the time domain solution for multi degree of freedom system we generally adopt numerous beta method for solving the problem. Duhamel integral is somewhat what we call complex if we extend it to the multi degree of freedom system. Now, we come to the modal analysis and in this analysis we try to take advantage of the properties of the motreps or undamped motreps of the structures and I am sure all of you know from your the knowledge of your dynamic analysis how we carry out the modal analysis, but still for the capitulation let me try to summarize what you have already learnt in your dynamics for the modal analysis. We write down the displacement that is the x are the dynamic displacement of a multi degree of freedom system that as equal to a phi matrix that is a motreps matrix multiplied by z which are known as the generalized coordinates or sometimes also known as the modal coordinates. So, with this defined we go to the main second order differential equation we substitute in the main equation the phi into z in place of x. So, once we do that then this becomes m into phi multiplied by z double dot the c matrix becomes c into phi into z dot and this one becomes phi k into phi into z. So, in place of x we write phi z here in place of x dot we write phi z dot and here in place of x double dot we write down phi z double dot. Once we do that then we pre multiply by the transpose of the phi matrix and once we pre multiplied by phi t then here we get a term as phi t here we get phi t here we get phi t and on the right hand side also we multiplied by phi t. Now, for a single point excitation system it will be i for a multi point excitation system it will be r then the comes the property of this phi t m phi and phi t c phi and phi t k phi. The mode shapes of the undamped system obtained by solving the Eigen value problem that is k minus m m omega square absolute value of that will be equal to 0 or in standard form it is written as x is equal to lambda x and we give this a matrix into the in any standard program for Eigen values and then the it gives the Eigen values and Eigen vectors of matrix A. So, those Eigen values and Eigen vectors are nothing but the mode shapes and the frequencies of the system. Now, the property of these Eigen vectors or the mode shapes of that it is orthogonal with respect to k matrix and it is orthogonal with respect to mass matrix. Orthogonality condition means phi t k phi will lead to a diagonal matrix that is will this will be having only a diagonal values as non zero all of diagonal values will become zero. So, this turns out to be a diagonal matrix similarly m matrix becomes a diagonal matrix because this is also orthogonal with respect to the mass matrix. Now, if one assumes that the c matrix is proportional to the alpha times m or is proportional to mass and stiffness matrix then one can write as c is equal to alpha into m plus beta into k. Therefore, phi t c phi also will become diagonal because phi t k phi will be diagonal phi t m phi will be diagonal. So, therefore, this phi t c phi also becomes diagonal. Now, if the left hand side of these things becomes diagonal then we can see that the in the left hand side the equation of motion becomes uncoupled in the sense that only you have diagonal terms over here. So, the each one of these equations are independent of each other that is one can write down a equation only in the form of this that is this m bar i into z double dot i c bar i into z dot i k bar i into z i. This is the ith equation and this is a single variable z i is a single variable and one can write down such equation which is consisting of n number of equation that is i varies from 1 to n where n is the degree of freedom for the multi degree of freedom system. So, this is the basis of the modal analysis and if the k bar i and m bar i that are taken and we divide k bar i by m bar i then we get the frequency natural frequency square that is the ith natural frequency we get like that and one can write down c bar i that is to be is equal to twice into twice z i omega i into m bar i and if we divide all through by m bar i then we land up in a equation in a single degree of freedom equation in which it is defined with respect to frequency of the mode ith mode and the damping of the ith mode. Generally we assume the constant damping for all the modes therefore we do not write down here z i but we write simply z i indicating that it does not vary with modes. Then on the right hand side the entire thing this is phi i t m r into x double dot g that can be written as a summation of lambda i k multiplied by x double dot g k. Now the k indicates the supports that means in a multi degree of freedom system if you have got s number of supports then this s number of supports can be considered in the analysis and these at these n supports we have different ground accelerations that is represented by x double dot g k, k stands for the number of the support. So this x double dot g k varies from support to support and lambda i k is a quantity which also varies from support to support. So here we have got lambda i k defined as phi i t multiplied by m into r k where r k is the ith column of the r matrix. So divided by phi i t m phi i t. So that is the definition of lambda i k so it is support dependent that is it has an index k. So this summation is done for all the supports over here and we generate a time history of the ground excitation for the structure that means for the entire structure for the ith generalized coordinate we will have only one time history of excitation which will be called the generalized excitation for the ith mode and they are the effect of the different types of excitations at different supports will be all summed together to find out the time history of the generalized excitation at the ith mode. So now we can solve this equation written in terms of the general coordinate or we call this modal equation we can solve as a single degree of freedom equation either in time domain or in the frequency domain. In time domain again we have option we can use Duhamel integration or we can use Newmark beta method or we can use any other integration scheme. Let me just try to elaborate this lambda i k if the ground acceleration is not varying from support to support that means at all supports if you have got the same excitation on time history or the ground acceleration then in place of r k it becomes i that is the influence coefficient vector 1 1 1 1 in that case it becomes the k it becomes independent of k and we have a unique value of lambda i defined for the ith mode. So that quantity is known as the mode participation factor that has been discussed I am sure in your dynamic analysis class. So for the single point excitation you have this mode participation factor as a single quantity but if it is a multi support excitation system then there is not a single value of the mode participation factor. This mode participation factor then becomes support dependent that means for each support we have one mode participation factor defined. Obviously if s is equal to 1 that is if you have got single support excitation then it this generalized formulation can be made what you call made same as single point excitation system. Now let us look into the example an example solved for these multi support excitation system with the modal analysis. So here we take the example of the cable state bridge if you recall in that cable state bridge we had the many supports or rather four supports were there in the cable state bridge let me show you the diagram for that for your recapitulation yeah this is a cable supported bridge that we have taken up in order to obtain the arm matrix and in that you can see that we have got one support second support third support and fourth supports at these four supports the ground accelerations were different and the degrees of freedom that we considered were three degrees of freedom that is at the two top of the towers or at the top of the two towers we have got two horizontal degrees of freedom and a vertical degrees of freedom is considered at the center of the deck. So these were the three degrees of freedom dynamic degrees of freedom that we considered. So we condensed out all the rotational degrees of freedom then we obtain a equation of or obtain an expression for the stiffness matrix and the mass matrix corresponding to this translational degrees of freedom and then the from the stiffness matrix we again take took out to the three non support degrees of freedom that is one two three and the rest where the support degrees of freedom and from there we constructed the arm matrix for the system and arm matrix for the system was obtained like this this was arm matrix corresponding to all the four support degrees of freedom and arm matrix obviously was three by four because the three are the non support degrees of freedom. So with these arm matrix in position we try to solve these problem and the solution for the problem were obtained with the K matrix that is the for corresponding to non support degrees of freedom that was the K matrix that was obtained and mass matrix was a diagonal mass matrix because we lamped the masses at the three degrees of freedom and then this was the arm matrix that I had shown you before and then the omega 1 omega 2 omega 3 3 frequencies were obtained the phi 1 t phi 2 t phi 3 t that is the three more trips well like this and then we have the first modal equation that we can write in this particular form. Now the you can look at the nature of the mode shape it is interesting to see the nature of the mode shapes so the nature of the mode shapes is the first mode basically is such that you have these are the first two degrees of freedom that is at the top of the pylons and you can see the in the top of the pylons if one is moving in this direction the other moves in the this direction that is you have a situation like this and with the at the centre we have a downward upward displacement. So the phi 2 basically were such that both the pylons are moving in the same direction and there was no what you call displacement at the centre and this basically is again a mode shape in which the both pylons moved in the opposite direction and the centre of the deck moved down rather than up. So these were the three mode shapes that were obtained so using these three mode shapes so we write down the three modal equation and for these three modal equation these are phi i phi 1 t a m and this one will be equal to r 1 not r 1 that multiplied by g that would give the rho g 1 so this is the value of the generalized ground acceleration or generalized excitation of the first generalized equation and with a delta t defined as 0.025 this thing can be integrated but in order to obtain the values of x double dot g 1 x double dot g 2 and x double dot g 3 and g 4 for that we take the ground acceleration of 30 second which is travelling and there is a 5 second of time delay between the supports as a result of that you will have for the first support the 30 seconds of the record first 30 seconds will be non-zero and rest 15 second will be 0 values similarly for the second support you will have first 5 seconds as 0's then you will have got 30 second of the record and the last 10 seconds will be 0 so that way we can construct the different excitations at different supports from the 30 second of the actual earthquake that is travelling and with a 5 second time delay between each support.