 In the previous lecture, we are discussing about the multi degree of freedom system with multi support excitations. We are looking into the equation of motion. The equation of motion can be written in two different forms. One is in terms of the total displacement that is this equation in which all the quantities on the left hand side, they are in terms of the total displacement, total acceleration and total velocity. And on the right hand side, we have minus k s g into x g. k s g is the coupling matrix between the support degrees of freedom and the ground degrees of freedom that is the k s g is the non support coupling between the non support degrees of freedom and the support degrees of freedom that is the degrees of freedom at the ground. And x g is the ground displacement vector. The other form was in terms of relative displacement that is this equation in which all the quantities over here are relative quantities that is relative displacement, relative velocity and relative acceleration. And in this equation, we had the r a matrix called the coefficient matrix. And x double dot g happens to be the ground acceleration vector. So, the difference between these two equations are that one is written in terms of the total displacement other is written in terms of the relative displacement. On the right hand side, for this equation we require ground displacement to be specified at each support whereas, here at each support we must know the ground acceleration. And then we can solve the problem. Next we wanted to explain how the r matrix is generated. And in the previous slide that was given the r matrix is generated with the help of this equation that is minus k s s inverse into k s g multiplied by x g or the r matrix is equal to minus k s s inverse into k s g k s s inverse is the partitioned stiffness matrix corresponding to the non support degrees of freedom. And k s g is the coupling matrix between the non support degrees of freedom and the ground degrees of freedom. So, once you are able to get these two matrices then one can construct the r matrix. So, we had shown an example for this in the previous lecture. This is a frame in which we had three ground excitations at three different supports. And these were the two degrees of freedom which are the non support degrees of freedom. So, the partitioned matrix for that or k s s matrix was this the k s g was this. And then with the help of the equation and for r we calculated this quantity that is k s s inverse k s g and that turn out to be one third into one one one and one one one. So, that the three ground displacements share equal or equally or share equally in producing the non support responses at the non support degrees of freedom. Next we take another example to illustrate the same you know r matrix that means how we can construct the r matrix for a pitch loop portal frame. In this pitch loop portal frame we have the one two three these three degrees of freedom are the rotational degrees of freedom. And the four and five they are the translational degrees of freedom in addition to that we have two support degrees of freedom that is six and seven. So, first what we do we write down the stiffness matrix for the entire thing that is the for all the seven degrees of freedom we write down the stiffness matrix the way we write down the stiffness matrix for a static analysis. And then we partitioned them for example x one x two x three they are the rotations. And these rotations are taken on the top. So, this is the three by three matrix corresponding to the rotational degrees of freedom. And this is the matrix three by four that we are calling as k r u this is the coupling between the rotational degree of freedom and the translational degree of freedom. And this is these four by four is the translational degrees of freedom and the four by four square matrix correspond to that. So, from this matrix we can obtain a condensed stiffness matrix corresponding to these four degrees of freedom that we are calling as k bar u u. And this k bar u u equal to k u u that is this matrix minus k u g u r k r r inverse multiplied by k r u. So, that is a standard condensation procedure that I think all of you know. And once you do that we get the condensed stiffness matrix corresponding to the translational degrees of freedom. So, this matrix now is the k bar u u matrix is now of size four by four. And in that the degrees of freedom involved are x four x five x six and x seven. And again we partitioned them over here. So, that these are the this is a matrix corresponding to the non support degrees of freedom x four and x five. And this is the matrix corresponding to the support degrees of freedom and they are the coupling matrices. Now, with this we get the value of r using again the previous formulation that is r is equal to the minus k s s or r is equal to minus k u s inverse into k bar u s j. And these two matrices are given over here. So, with the help of that we get the value of r. So, these quantities are shown here in this slide the this portion that is your k r r matrix this is k r u matrix. And then this was the k bar u u matrix and from there we isolated k u s matrix and k u s g matrix. And after that we obtained the value of r by simply multiplying k s k u s inverse with k u s g. And this is the two by two r matrix that is generated. Next we take another problem it is a model of a cable state bridge in that we have degrees of freedom as this at the top we have a degree of freedom here we have a degree of freedom to and at the center of the deck we consider another degree of freedom which is in the vertical direction. So, we will try to find out the response of the structure for these three degrees of freedom. And the entire equation of motion is written for these non support degrees of freedom. So, in addition to these degrees of freedom there are rotations at these points. And we have translations at this support point this support point and this support point and this support point. So, we have in all four plus three seven translational degrees of freedom out of them four degrees of freedom are at the supports three degrees of freedom are the non support degrees of freedom that is the translational degrees of freedom as one two and three. And the rotations are eight nine ten they are the rotations. So, what we do is the first we write down the entire stiffness matrix that is the stiffness matrix. And for the entire system which is a ten by ten matrix out of that the first seven or seven by seven stiffness matrix corresponding to the seven translational degrees of freedom they are grouped at the top. And the three rotational degrees of freedom are grouped at the lower part of this vector. Then we do this matrix condensation using this relationship. And condense the entire stiffness matrix to the translational degrees of freedom x one to x seven. Once we get this condense stiffness matrix then this condense stiffness matrix is further partitioned into the non support degrees of freedom that is the three degrees of freedom which we are acting at the two at the top of the pylons. And one at the centre of the deck these are three translational degrees of freedom. And other four degrees of freedom are at the four supports of the cable supported bridge. Once we partitioned them then using this matrix and this matrix we can obtain the value of the r matrix. So, in this way we constructed the r matrix for this problem. And the details of this is given here that means element by element we generated the different quantities. And then assembled them to this ten by ten stiffness matrix. And after that we condensed them to the translational degrees of freedom. And from that we had taken out the three by three sub matrix corresponding to non support degrees of freedom and obtained the r matrix. And the resulting r matrix was this. So, here again for the four different ground motions we have this three by four matrix. And with the help of this three by four matrix we solve the problem that is the problem in which on the right hand side we have got minus m r x double dot g where x double dot g are the four ground acceleration defined at the four support points. Next week see how we can convert this equation of motion that we had written for the multi degree freedom system with the r coefficient or the r coefficient matrix on the right hand side. In one case in which we are writing down the equation of motion in terms of the relative displacement. And in other case we had the equation of motion in terms of the total displacement. And on the right hand side we had the k s g matrix and we require instead of ground acceleration ground displacements. So, those two second order differential equation or multi degree of freedom second order differential equation they can be written in the state phase form as before. Only difference here will be that in the previous case we had for single degree freedom system we had got minus k by m and minus c by m these were the terms which were written. Now in place of that now we have got minus k s s into m s s inverse. So, this term become this similarly the c by m term becomes the c s s into m s s inverse. So, and this one becomes instead of one it becomes i. If we wish to write down the second order differential equation in terms of total displacement as a as a set of first order differential equation or in terms of the state space formulation. Then the a matrix remain the same only thing that changes is the f bar one and here we see that f bar becomes equal to minus m s s inverse k s g this will be not k s s this will be k s g into x g. So, this will be the only change. Therefore, the for multi degree freedom system with multi support excitation we can write down the equation in 4 different forms 2 equations can be written as a second order differential equation one in terms of total displacement and other in terms of the relative displacement and correspondingly we can have 2 state space formulation for these 2 equations. Now we try to write down the state space form of the equation motion for the problem 3.4 that is this problem yes for this problem we wish to write down the entire equation of motion in the state space form. So, here we have seen that if we were wanting to write down the equation of motion in terms of your state space form we have to generate a matrix called a matrix and that a matrix contains c s s or c by m. So, c matrix must be known. So, that is the first thing that we do for the entire structure now we obtain the c matrix. So, for obtaining the c matrix these approach is already known to all of you that one can obtain the c matrix of a particular structure provided we know the first few frequencies of the system. So, here it was a 2 degree freedom problem in the sense that there are 2 non support degrees of freedom. So, the we had 2 natural frequencies. So, once we know these 2 natural frequencies then with the help of that one can compute the values of alpha and beta which are required to obtain the c matrix considering the c matrix to be mass and stiffness proportional that is we write down c s s to be is equal to alpha times mass matrix plus beta times the k matrix and using this we can obtain this damping matrix for the entire system. And once we have these damping matrix written for the system then we can use the same equation that is this formulation that is k s s we know we know m s s inverse and also we know now the c s s matrix and m s s inverse is also known. So, therefore using this we can write down the entire a matrix and this a matrix turns out to be like this here it is it will not be 182 it will be 102 you can make this correction where rho is equal to root over k by m and on the right hand side the force vector if you look at the expression for the force vector this force vector is equal to minus r into x double dot g. So, r for this problem we obtained as the this was the r that is one third 1 1 1 1 1. So, using this value of r one can obtain the value of the f that has minus 0.33 into x double dot g 1 plus x double dot g 2 plus x double dot g 3 that is what we observed before that all the three ground accelerations equally influenced the non support degrees of freedom. So, that is how the the terms of the load vector over here that is generated. Now, if we wish to write down the equation of motion in terms of the total displacement and again in the state space form then the a matrix remain remains unchanged as I said before. So, a matrix remains this and only thing that changes is the right hand side load vector and the right hand side load vector turns out to be x g 1 plus x g 2 plus x g 3 into rho square and this is nothing but minus k s g into the m s s inverse. So, k s g into m s s inverse turns out to be this one and here we can see that the degrees of freedom that we require are not the degrees of freedom the quantities that are that is required for defining the force is the three displacement at the three supports. Now, once we are able to write down the equation of motion in different forms that I described that is in the state space form and as ordinary differential equation in terms of relative displacement and in terms of the total displacements then we come to how we can solve these equations. So, that is what we are calling as the response analysis. Now, the response analysis can be carried out both in time and frequency domain and first we will take up the single degree of freedom system and the methodology that will be adopting for solving the single degree of freedom system that will be extended to multi degree of freedom system later. Now, in time domain analysis there are many methods which are available out of that we will take up only two methods that is a Duhamel integration and a Numer's beta method. I think in your dynamics course all of you have already done these two time history methods which are very popular in earthquake engineering. Therefore, you are acquainted with the different important things associated with these two integration strategies. However, we will recast these two formulation that is the Duhamel integration formulation and a Numer's beta method formulation in the form of a recursive equations rather than the usual way that you have solved in your dynamics course. Let us take the Duhamel integral first. As you know that a Duhamel integration Duhamel integral we consider the load to be a series of impulses that is the load to be consisting of a series of impulses like this and these series of impulses for that if we consider any at any time t we see what is the response that we obtain and then sum up the effect for or some of the responses for all the impulses to get the final response at this particular point. So, what we do that first we consider a impulse at a time tau from the from the origin that is 0 time and a an impulse of f multiplied by d tau is applied to the system or single degree of system at time tau. Then the elapsed time is t minus tau that is when the impulse is produced over here then we see its effect at this time that is after a time of t minus tau that is what we call as the elapsed time elapsed time of t minus tau. Now as you all of you know that the problem of producing an impulse to a single degree system is equivalent to producing a velocity to the system or imparting a velocity to the system. So we can conceive the problem as a damped free oscillation with a initial condition of displacement and the initial condition of velocity. So in the free vibration equation if we provide the these initial conditions that is the displacement at from here we start counting the time. So at 0 time if we consider what is the displacement and what is the velocity then we can substitute this displacement and velocity into the equation that provides you the response at any instant of time t for a damped free oscillating single degree of freedom system. And that is given over here this is a standard equation that is the x t minus tau x t minus tau is the response at time t and this is given as e to the power minus j omega n t and this is the velocity part that is that if m into x double dot g tau this is the force impulse force that is provided at 0 time so that is f tau that divided by m that becomes the impulse force initial velocity and omega d is the damped frequency. So and sin omega d t minus tau d tau comes because f tau d tau that is the total impulse so that is how d tau is coming into picture over here and this equation is valid for 0 displacement at the initial stage and a velocity is equal to f tau d tau divided by m. So for this velocity we get this is a standard equation for damped free oscillation. Now this x t minus tau as we obtain then one can obtain the response x t for all the impulses that we have obtained. That acts from this point to a time t so integrate from 0 to t this entire expression over here and this is known as the Duhamel integration and using this Duhamel integration one can find out the value of the response at any instant of time t. If this f tau is an integrable function then there is no problem one can obtain the response of this x t in a closed form analytically. However if f tau is not a integrable function then one has to perform a numerical integration for this to get the value of x t. So instead of doing that numerical integration one can recast the entire formulation in a slightly different fashion. Let us consider this as the ground acceleration or the support acceleration for a single degree of freedom system. And for that we have a interval of time delta t so that we can define a time step k and next time step k plus 1 difference between them is delta t. At the these two time steps the values of f k plus 1 and f k they are known that is the ground accelerations at these two points are known. So we can multiply them by m to get the forces at these two points. Now if we look at the response at k plus 1 then the response at k plus 1 can be obtained provided we know the response at k that is the k th time station. And in this kind of formulation in which we obtain the response at a particular time step with the help of the responses at the previous time step this formulation is called the time marching formulation. So in a time marching algorithm what we do is start with zero time and they are zero time the displacement velocity and acceleration they are known. And then we obtain the response at the next time step that is at after interval of delta t and then we proceed in this particular fashion. So this is a time marching scheme but this time marching scheme to be used in the case of a Duhamel integration requires some kind of consideration and that is what is shown over here in this slide that what we do is that at k that is at time t k we know the forcing function f k we also know the forcing function at k plus 1 time then what we consider that the as if the force between f k and f k plus 1 is varying linearly that means it is varying in this particular fashion from this point to this point. And once we assume that there that is linearly varying then at any time tau we can obtain the value of this f tau in terms of f k plus 1 and f k. So this is what we do here in this formulation and we write down the f tau in this particular way and that is it is a sum of a constant force f k and then it is a linear term which is or the triangular part of the equation. So using this equation one can define f tau at any time tau taken or counted from t k. Now if we look at the responses at time t k plus 1 then we see that the response at time t k plus 1 is consist is consisting of three responses. That is the initial condition that exist at k or the time step k for that with that initial condition the single degree of freedom system vibrates as a damp free oscillator and because of this vibration there will be some response which will be produced at t k plus 1 that is the next time step. So this is the first part of the response. Second part of the response is due to f k that is constantly acting over the duration of time delta t and the third portion is the triangular variation of the load between k and k plus 1. So the responses for these three would provide the final response at t k plus 1. So t k plus 1 clearly now depend upon these quantities that is x k x dot k. So they are the initial velocity and displacement at k th time step and f k and f k plus 1. So what we can write down is that x k plus 1 we can write down. To be is equal to some constant multiplied by c k x k some constant multiplied by x dot k some constant multiplied by f k and some constant multiplied by f k plus 1. Similarly one can write down x double dot k plus 1 in terms of this equation where d 1 d 2 d 3 d 4 are the constants that is to be obtained and then one can obtain the value of x double dot k 1 that is the acceleration at k plus 1 at time and that is from the parent single degree of freedom equation that is if we know the displacement and velocity of the system one can find out the acceleration. So using these three equation one can get at any time step k plus 1 all the three quantities that is the displacement, velocity and acceleration in terms of the displacement, velocity at the previous time step that is k and the forces which are acting at k and k plus 1 they are eventually known. Therefore the formulation is centered around finding out this constant c 1 c 2 c 3 c 4 and d 1 d 2 d 3 d 4. Now in this we can see that the first part of the solution that is for the initial condition this will be a function of x k x dot k and delta t. Second part of the solution that is for a constant f k that is a rectangular f forcing function what is the response at this point. So this will be a function of f k and delta t and for the triangular part or third part of the solution requires the knowledge of f k f k plus 1 and delta t. Now this comes out the first solution comes out from the damped free oscillation in that simply substitute the initial condition as x k at x dot k and then find out what will be the response at k plus 1 at time step. This one is a problem of a Duhamel integral for a rectangular type of pulse and for that the analytical solution is available and all of you have done that. Similarly for this one the response can be obtained for a triangular kind of pulse and this also can be obtained using the Duhamel integration because it can be obtained analytically. So for all the three the analytical solutions are available then once we get all the three solutions then these three solutions are summed up together and the like terms that is the terms containing x k terms containing x dot k and terms containing f k and f k plus 1 they are all collected together and multipliers that will be associated with these variables they would form the values of c 1, c 2, c 3, c 4 etcetera. And that is what is done over here and once you are able to find out these quantities that is c 1, d 1, c 2, d 2, c 3, d 3 etcetera then you can write down the equation in this recursive form where q k plus 1 is equal to a matrix A into q k plus h into f k plus 1 where q basically represents for all the three quantities in a vector form and A is a matrix consisting of these constants and the h is again a matrix consisting of some of the elements of some of the constants of those equations. So the c 1, c 2, c 3 etcetera can be computed the way I told you and there for your information given over here in these equations and you can see that these c 1, c 2, c 3 etcetera depend upon a delta t that is a time step and all other quantities are known that is omega d, delta omega t, xi, omega n etcetera all of them unknown therefore one can compute easily the values of c 1, c 2, c 3, c 4, d 1, d 2, d 3, d 4 etcetera. So that is the recursive form of the Duhamel integration and it requires the solution of the problem for certain known cases that is the known cases are that dam free oscillation that is a known case. The solution for a rectangular pulse that also can be obtained using Duhamel integral analytically and the response for a triangular pulse that also can be obtained using an analytical solution. So therefore with these three known solutions one can obtain the constant c 1, c 2, c 3, c 4 of the A matrix and one can write down the entire Duhamel integration in this recursive format. The advantage of this solution is that if f tau or the force if it is a constant is not a integrable one then in any way one has to go for a numerical integration. So instead of doing that numerical integration one is using the known solution for a triangular pulse rectangular pulse pulse and a dam free oscillation. So using these three solutions you are able to tackle the problem. Next is the numerous beta method in the numerous beta method we again solve the problem numerically using a time marching scheme that is you try to find out the response at k plus 1 at time step and with the help of the response at time k which will be taken as known. So we start with 0 time t where all the quantities displacement velocity and accelerations are specified and we solve for the next time step using those values and in that particular fashion we march up to the next time step and in that particular fashion we march up to the next time step and we are able to solve the head. Now the two key equations that are used in the numerous beta method are the ones that is shown in this equation 3.52 and 3.53. If we look at these two equations these two equations are already known to you. For example if I ask you to find out the velocity at a particular time given a velocity at some other time t then you can find out that velocity is equal to the previous velocity plus the acceleration multiplied by the time t that gives you the velocity x dot k plus 1. Only thing what has been done over here is that the instead of a constant acceleration that we assume over the time interval here the velocity that is varying or other time acceleration that is varying from k to k plus 1. So that varying velocity is considered with the help of this particular formulation that is here we assume that as if the acceleration is varying linearly. Now if we assume that the acceleration is varying linearly then it is equivalent to assuming that the displacement is varying quadratically because if you integrate twice the acceleration then you get the displacement. So if there is a linear variation of the acceleration between two points then if the acceleration is integrated this linear variation will then become a sorry cubic variation sorry not quadratic cubic variation. Now with this assumption that is the acceleration is varying linearly then velocity is varying quadratically then velocity is varying quadratically and displacement is varying in the cubic form with this assumption we write down this particular two equation that is given a value of x dot k plus for a given for a given x dot k we can find out x dot x dot k plus 1 using this particular linear variation of acceleration then we can write down from that the displacement at k plus 1 a time step given the displacement at time step k and velocity at time step k. So this is again a very familiar equation with all of you say if we wish to find out the displacement at a time at some time then at the previous time whatever is the velocity that is say u plus v into t plus half f t square this is what you have all done in your physics class. So it in fact is a modification of that equation here these the first two terms remain the same here what we have done we have manipulated the acceleration that is we are not assuming a constant acceleration and acting between the two points but a acceleration which is varying linearly and therefore we take into account the acceleration at both the points and assume to vary linearly between the two points. Now if we provide the value of delta as half then this particular term becomes an average acceleration between the two point that is you can see that it will it turns out to be x double dot k plus x double dot k plus 1 divided by 2. Similarly if we consider beta to be is equal to 1 fourth then this turns out to be half of the average acceleration into delta t square. So we see that for a value of delta is equal to half and beta is equal to 1 fourth these two formulations or these two equations provide us the displacement and velocity at k plus 1 h station using as assumption that the acceleration is varying linearly and we assume that the acceleration to be an average acceleration over the time interval. Now with this assumption we can and go ahead in this particular sequence of the that is we write down the equation of motion in the usual form that is m x double dot k plus 1 c x dot k plus 1 into k x k plus 1 is equal to minus m x double dot g k plus 1. So this is the equation of motion at k th time step we can rewrite it in this fashion by dividing the entire equation by m. So this is the equation which is written in terms of the frequency and damping then what we do is substitute the two equations 3.52 and 3.53 in place of x k plus 1 and x dot k plus 1. So x k plus 1 and x dot k plus 1 is equal to minus m x double dot k plus 1. They are now written using equation 3.52 and 3.53 that we have described before and once we substitute them into the equation then we can see that the equation that will be there on the left hand side would be in terms of x k x dot k x double dot k and the acceleration x double dot k plus 1. So the x double dot k plus 1 this acceleration is not known to you or known to us therefore from this equation one can find out x double dot k plus 1 and this will be a function of the known quantity of the quantities that is the acceleration displacement and velocity at the previous time step and the acceleration ground acceleration at k plus 1 at time station and once we get the value of x double dot k plus 1 then this can be substituted into the equation 3.52 and 3.53 that I have shown that is a two cardinal equation in that equation if you substitute for x double dot k plus 1 then you get the value of x dot k plus 1 and x k plus 1. So all the three quantities that is x k plus 1 x dot k plus 1 and n x double dot k plus 1 can be now written in terms of the known quantities of x k x dot k and x double dot k and of course the load. So this can be written in a compact form like this that is q k plus 1 is equal to a matrix f n into q k multiple h n into f k plus 1. So f k plus 1 is the load that is acting at the k plus 1th station and q k is completely known that is the displacement velocity and acceleration at k th time station and with the help of that we can obtain the value of the response at k plus 1 at time station. Again this is in the recursive form and this recursive equation provides you the responses all the three responses at a particular time station with the help of the responses at the previous time station and the f n matrix and the h n matrix are given over here. The alpha terms out to be these quantity which can be compensated.