 Welcome back everyone. In today's lecture, we are going to discuss the dynamics of a seismically isolated structure under the action of earthquake excitation. So we are first going to learn how to represent a seismically isolated structure in terms of a multidegree of freedom system and then look at its frequencies and the important parameters that actually define a seismically isolated system and then further discuss how those parameters actually represent the behavior in terms of its dynamic characterization. And what are the important things that need to be considered in the analysis of a seismically isolated building. So let us get started. So let us basically start with a very simple formula that we have been like using and utilizing again and again. What we said and that you already know that the time period of any structure we said that it would be 2 pi m by k. We have been utilizing this to understand the concept of base isolated structure as well. Now what is the inherent assumption here in terms of the degrees of freedom? Like if I am writing this formula, what can I say about the degrees of freedom for the structure that I am writing this formula? Can I say it is a single degree of freedom system, only valid for single degree of freedom system or a simple representation could be this mass stiffness and if you want to put damping that is fine as well or so this is the spring mass damper representation or if you want to write in terms of frame representation then you may be, you may consider this interpretation. This is your view. So the horizontal stiffness is represented by the overall stiffness of this frame and basically you can have ground motion here. Now what happens during the earthquake you would have a ground motion like this. So in both direction. Now we said that this is the formula we are going to use but as you might imagine in reality basically no base isolated structure would be a single degree of freedom system. So in the end this is an approximation and like any approximation or assumption it is valid for certain cases and this works out to be good for base isolated structure because of the nature or the relative flexibility of the isolation layer, would not be valid for any other type of structure unless you have and that is what we are going to look in today that when would it be valid, this single degree of freedom representation and if it is not valid can we do a simple analysis of two degree of freedom system so that we understand how the super structure and the isolator basically behave, how and what kind of basically participation do they have in the overall response. So those are the things that we are going to look into today and before we do that I will just quickly go through some concept and like you know brush up some of your concept of and you might already be aware but we will do that anyway of earthquake excitation of a single degree of freedom system and a multi degree of freedom system. So as you might know if let us say I have this frame here and I have this mass here I am assuming that the whole floor mass is actually concentrated at this level. Now due to earthquake there would be a ground movement so the whole structure will move with the ground and then there would be relative deformation in the structure. So first there would be a rigid body motion of the structure due to ground movement let us say this would be UG which is basically the displacement of the ground and we are representing the ground motion as ground motion acceleration history at UG double dot t. Now then we would have this deformation here and let us say this mass moved here. So this relative deformation we are going to denote as the deformation of the structure this is the deformation of the ground and this is the total deformation U t. So basically U t is UG plus relative deformation of the mass. This mass we are considering to remember the degree of freedom that we are representing we are saying that this whole frame can be represented as a single degree of freedom system if there is no actual basically actually if the member are very rigid so that whatever this node is going through actually the same is would be going through the right node. And the overall if we consider the lateral deformation is what contributes to the or what denotes as the degree of freedom. So the total displacement of the mass is UG plus U and if you want the total acceleration would be can differentiate it twice it would be this. Now if you want to write down that basically the equation of motion for this one we know that you can perhaps cut this structure here ok and write it ok. So if you consider the free body diagram of this ok acceleration is always total acceleration ok but deformation or the force in the frame is due to relative deformation ok because of rigid body motion of the structure it does not basically induce any internal forces ok. So the internal forces would be whatever the relative or the lateral stiffness of the structure is let us say this is K times the relative deformation ok. And there is no other force remember whatever force that is being applied if you are thinking from the ground that is coming into this total acceleration. So the equation of motion that I will get is basically this here. And if you substitute U t is equal to this quantity here you can further write it as Mu double dot in terms of relative K u is equal to minus m U g double dot ok and this is basically your effective earthquake force. So I understand many of you already aware of this but I am just repeating it for the sake of completeness ok. So this is for a single degree of freedom system and if you had a damper ok you could perhaps include that as well. So this is the undamped system but let us say you had a damper like this. So remember damper also depends on the relative velocity in the structure between the points ok between the points of connection of the damper. So then you could perhaps write this as C u K u is equal to minus m U g ok and this quantity here the C u basically relative damping force and the stiffness force these are all depend on the relative deformation and relative velocity ok. But acceleration we included as total acceleration because it is always measured with respect to inertial frame. So this is our equation of motion for a single degree of freedom system ok. If you understand this then we can perhaps write it for a multi degree of freedom system as well ok. So let us say I have a multi-storey shear type building ok. The assumption in a shear type building is that the flexural deformations ok are small they actually members are rigid. So overall we can represent the degrees of freedom by deformation at each floor ok. So this is a shear type building. Now let us say masses are again concentrated here this is m1 and this is m2 ok. And you can have again forces U g basically ground motion ground acceleration U g double dot applied like this ok. So like what we had done for this we can also write down the equation of motion for this by considering equilibrium of the masses one at a time. So if you consider the equilibrium of mass m2 which is at this position ok the relative and let us say story stiffnesses we are considering as k1 and k2. So the force here would be in opposite direction of deformation which would be U2 minus U1 because the story stiffness is k2 so the net force would be k2 times the story drift which would be U2 minus U1 ok. And again this would be the total acceleration so it would be m2 U2 double dot plus U g double dot ok. Similarly for the first story if I consider like this the force from the top would be equal and opposite to the force that you are applying here so this would be Q2 U2 minus U1 in the opposite direction. And below it would be k1 due to a story first story and then U1 minus this which would be basically 0 ok and the inertial force would be m1 U1 plus U g double dot ok. So we can write down separately the equation of motion for each of these ok. It would be m1 U1 plus m1 U g double dot plus k1 U1 and this would be minus of k2 U2 minus U1 that should be equal to 0. So what you will get basically m1 U1 in this side plus k1 plus k2 U1 minus k2 U2. So this is for first story ok similarly for the second story if we consider the equation of motion for this we can write it is and remember there is another term here that we have brought from this side it is minus m1 U g double dot ok. So for the second story again it would be m2 U2 plus U g double dot plus k2 U2 minus U1 and that is equal to 0. So it can be further written as m2 U2 plus or let us say here it would be minus k2 U1 plus k2 U2 is equal to minus of m2 U g dot. So this is second story. Now you know that this both equation can be combined as a single matrix equation ok. If you combine this basically this equation here and this equation here ok. This can be combined and written in a matrix form as the coefficient of U1 which is m1 0 0 m2 times U1 U2 and then the stiffness matrix which is basically k1 plus k2 this term here minus k2 again minus k2 and this is k2 and U1 U2 and this is equal to minus m1 U2 m1 0 basically this is minus m1 minus m2 and this is U gt. But we can further write this as in terms of mass matrix m1 0 0 m2 and then another vector which is called a influence vector 1 1 and then U gt you will see that it is the same quantity. So, my final equation of motion is m1 0 0 m2 U1 U2 plus k1 plus k2 minus k2. So like we did for a single degree of freedom system we can generalize the same equation for a multi-degree of freedom system as well where this is mass vector m times the displaced acceleration vector blue double dot then the stiffness vector k times the displacement vector U and this is equal to again the mass vector m sorry mass matrix m and then the influence vector l and then U gt. And this is the question of motion for multi-degree of freedom system. Now, for a single degree of freedom system we did not have any concept of contribution factor or anything because there was only one mode. But as you know for a multi-degree of freedom system basically it is represented by different modes and the overall response is basically some of the contribution of response due to each mode. And that becomes important because let us say your structure has multiple modes. Multiple modes also means that multiple time period or frequencies and dynamic behavior when you try to predict you very well understand what happens to a single degree of freedom system when it is applied to different type of load for example harmonic load earthquake load it is much easier to visualize but it is not so easy for a multi-degree of freedom system. Because you have multiple modes with multiple frequencies and if you say that the overall response is governed by a typical mode then you need to know how much is that mode is contributing and what would be the effect of higher modes. So all these things are like you know topic of interest if you start to go much into the dynamics of it but we can rely on that to explain the behavior or the multi-degree basically multi-degree of freedom representation behavior of a base isolated structure as well. And we are going to use the same concept extended to base isolated structure the equations that we have derived. Let us see what happens to a base isolated structure. We said that till now we have considered base isolated structure to be a rigid mass okay on top of any spring or anything or if you want to draw like this perhaps this would be a good representation. This is the superstructure mass this is the isolator stiffness in the horizontal direction although I am drawing it horizontally this is a spring mass system in reality what I am saying you have a base isolated building let us say like this and you have multiple isolator let us say okay and we assume it to be almost rigid this is a single mass and through because it is rigid whatever like you know displacement you take here the same is the displacement here. So we say that let us assume the displacement as U and that to be the degree of freedom for this base isolated structure and the horizontal stiffness is basically the stiffness of the isolation layer okay and then we say that so this is a stuf representation okay and then we say that Tn would be 2 pi under root m by k where m is the mass of the superstructure okay remember that whereas k is stiffness of isolation layer not the superstructure. But as you can imagine in reality superstructure would not be rigid right it would be I mean it would have some finite flexibility and it is important that we take that flexibility into consideration to completely understand the behavior okay. So let us do that let us see how do we do that so what I am going to draw here okay let us see I had a base isolated structure in which superstructure is being represented by this frame and then there are isolators which I am representing through thus these two isolators okay and basically this mass here let us call it mb which is the mass of the base mat okay and ms or let us just call it m okay which is the mass of the superstructure okay. So mass of the base mat and the mass of the superstructure and we are assuming that all the column mass everything has been lumped with the superstructure mass. Now the stiffness of the isolation layer I will denote it as kb that is the stiffness of the isolation layer and the damping coefficient of the isolation layer let us say cb similarly for the superstructure the total stiffness of the superstructure let us say is damping is cs and the stiffness is ks okay and this isolation is fixed at the ground and it is connected to this base mat and there is a ground excitation. So now this is a two degree of freedom representation where we are going to represent okay deformation at this level and deformation at the superstructure level to represent the displacements at the isolation level and the superstructure okay to represent the overall behavior. Now again of course in reality it won't be a two degree of freedom system but a two degree of freedom system is a much better representation in terms of understanding the dynamics or the contribution of superstructure and isolation. So the overall structure I am dividing between the isolation layer and the superstructure layer okay and I am trying to study the dynamic behavior of that okay. So let us see under the action of this ground motion what will happen first initially if the structure was here okay there would be some ground movement so this is the initial position of the structure okay the ground movement let us say it is ug okay that would be the rigid body motion of the overall structure then there would deformation at the isolator level let us represent it through this deform shape and we will say that this deformation is actually vb okay let me write it here this is v of b this deformation here now the superstructure would also have some relative deformation okay so just write it draw it here and this relative deformation let us say vs don't go by the scale because vb would typically be much higher than vs okay but just for the sake of drawing I have drawn it like this so vb is the relative deformation at the isolation layer with respect to ug or with respect to ground and vs is with respect to the isolation layer okay I hope there is no confusion here now we wish to write down the equation of motion again similarly for this one and it is up to us how do we write down the equation of motion so what I am going to do here in this case first I am going to write down the equation of motion of this whole system the whole structure and in the second case I am going to just write down the equation of motion of the upper slab if you wish you can do isolation layer at one time and then the overall or the top layer at other time like we did in the previous cases but for this cases from the mathematical manipulation point of view it is much easier for me if I first write down the equation of motion for the whole structure and then the top slab but in the end because there are two degree of freedoms basically you would only get two independent equations okay not more than two so it doesn't matter how do you choose to write down the equation of motion you would still get only two independent equation of motion okay so let us first consider the whole structure and the consider the free body diagram of the whole structure okay so when I say the whole structure I mean I mean the whole super structure this is the mass m this is the mass mb okay now if I consider free body diagram of this and basically I am cutting the structure here the elastic forces would be due to the isolation layer only right and that would be kb which is the base isolation stiffness base isolator stiffness times the relative deformation of the isolator which is here vb okay now my structure is moving right so I have to apply inertial forces okay and if I apply the pseudo forces it would be opposite to the direction of motion so for the top mass it would be mass times total acceleration now the total acceleration would be ug plus vb plus vs okay due to each of these three quantities so it would be vb plus vs plus ug acceleration and inertial forces on the mass below mb would be mb vb plus ug are there any other external forces that are acting on it no I mean because we are considering damping now we also need to consider the damping force here in the isolation layer and it would be simply cb times the relative velocity which let me denote by vb dot okay so if I write down the equation of motion for this one okay let me just write it down m vb plus vs remember this is acceleration here plus ug plus mb vb plus ug plus cb vb dot plus kb vb is equal to zero now I need you to do one thing I have written down for the whole structure now what do you do you just cut this structure at this point only the top basically the superstructure and then similarly draw the free body diagram and write down the similar equation of motion for the top layer okay once you're done with that like we did here combine the whole thing or write it the whole two simultaneous equation like this in terms of matrices okay remember on the left hand side the acceleration vector or the displacement vector that we are looking for is v equal to vb and vs okay and the acceleration vector basically would be v double dot is equal to vb vs so accordingly you have to arrange equations like we did for this and write it like that okay let us discuss the solution now okay and we'll compare so we have already written this for the whole structure now we can let us say just isolate the or like you know just draw the free body diagram of the superstructure so for that if you consider the superstructure let us say this is my superstructure it would have basically inertial force opposite to the direction of excitation or the pseudo force pseudo inertial force opposite to the direction of excitation which would be vb plus vs because this is the total displacement and we are differentiating it twice okay and then it would have because of the superstructure so we are considering only relative you know deformations it would have ks times the vs because vs is the drift of the superstructure similarly the damping force as well cs times vs dot okay so we can write down our equation of motion for the superstructure as let me just m vb plus vs plus ug plus cs vs dot plus ks vs okay and let us now further write down these two equations so that are like you know they can be written in the matrix forms to do that what I need to do combine all the terms which are multiplied by vvv dot acceleration term vb double dot and vs double dot together and same for the displacement and then see what do we get so here I would get as m plus mb times vb plus m times vs plus cb vb dot plus kb vb and I need to bring this term on the right hand side which is minus m plus mb okay times vg sorry ug double dot so the ground acceleration term similarly for this I can write mvv plus m vs okay plus cs vs dot plus ks vs and the ground motion term I am going to bring it on the right hand side sorry not this ug double dot so these are the two equations that I want to combine in the matrix form okay and if I do that what would I get if I write it in the matrix form this this coefficient and then this and this coefficient so it would be m plus mb and then m and then m and m okay and it would be vb double dot and vs similarly cb 0 0 cs vb dot and then vs dot plus same as kb 0 0 cs vb vs and this is equal to minus of m plus mb and then here it is m and then it is multiplied with ug now like we did previously we can further write down this as minus of m plus mb then m and then m and m which is the same as the mass matrix I am trying to write it same in the mass mass matrix term and this I will write it as 1 okay and 0 because influence vector if you look at it if you multiply this vector with this matrix then I will get the this vector here so this is my equation of motion with the mass matrix as let me write it here mass times acceleration plus damping times the velocity vector plus the stiffness matrix times the displacement vector and this is equal to mass influence vector and ug where the mass vector and everything is again and damping matrix all are shown here okay so this is the multi degree of freedom equation for my base isolated system okay and this is what we need to solve to find out the frequencies of the base isolated structure and the super structure or like different mode sorry okay and the participation factor in everything so this is what we are going to utilize but before we get into that let us define some useful quantities okay so we say that remember whenever we talk about a fixed base building and base isolated building we always compare the frequency of the fixed based okay fixed base super structure or the building to base isolated structure isn't it we say that if the fixed base frequencies let us say 1 hertz and the base isolated frequencies point hertz how do they going to how they basically contribute to the overall response so that is why we we always talk in terms of the fixed base building and the base isolated building okay so what I am going to do here I am going to consider the super structure as a fixed base with mass m is stiffness ks and this is a single degree of freedom representation of the basically the super structure so for this I know that for the fixed base super structure I can find out the frequency as ks divided by m and damping as cs divided by 2m omega s similarly for the base isolated structure I am going to define this is my base isolated structure right if I consider not a single degree of representation now sorry not a 2 degree of representation that I have considered above but a single degree of representation okay just for the argument why I am doing that I will come back to that later okay with the stiffness kb here and the overall mass of the super structure would be now what it would be a capital M which is m plus mb mass of the super structure at this one so when I consider a single degree of representation of the same building that I have been doing a 2 degree representation I can write that the frequency as omega b equal to kb m plus mb right I hope you understand all of the formula that we are doing so we are going to write this as kv divided by m where m is now the total mass of the super super structure remember initially when the structure is not isolated you only have m but when you need to isolate the structure we need to provide an additional base mat here like we discussed yesterday and the day before that in the class that for the base isolated structure you need to have either a rigid arrangement of plinth beam okay or you can provide a very rigid base mat okay so what do you have here is kb m plus mb is equal to kb divided by m okay where I was the total now these are the two frequencies that we are already discussed that what would be the frequency of the super structure and the base isolated structure but when we consider this equation and 2 degree of freedom representation now we know that from this we are going to get two frequencies omega 1 and omega 2 right because it is a 2 degree of freedom system can you tell me would this omega 1 and omega 2 be same as omega s and omega b what do you think remember these are not the modal frequencies okay these are not the modal frequencies these are just the frequencies of the these two structures that we have defined these are not the modal frequency of 2 degree of freedom system it won't be same okay although it might be close depending upon some situation like you know some parameters but this is a 2 degree representation with frequency omega 1 and omega 2 and these are two separate structure a super structure that been considered as a single degree of freedom system so you got this frequency and then the whole base isolated structure that I have again got the frequency as okay when we consider 2 degree of representation of the same building okay we know that we will get frequencies omega 1 and omega 2 I don't know what this frequency would be okay what I have done below here I have considered two single degree of freedom system okay remember this is the same system the second one this one here is the same system as above but now being considered as a single degree of freedom system okay so in general this single degree of representation frequency would it be same as the multi degree of freedom system frequency if you consider a frame and then you consider as a single degree of freedom system or if you consider a multi degree of freedom system would the corresponding frequency of both these two would be same what do you think now we have considered in the 2 degree of freedom system the flexibility of the super structure like in this case when I am considering single degree representation of the same structure I am considering this whole thing to be rigid I am saying that there is no relative deformation so there are no forces generated because of this one and the overall basically response of the system I can represent it through the deformation at the isolation layer but above I am not doing that so in general the frequencies of this and like you know the 2 degree of freedom representation and this 2 single degree of representation system would not be same okay but we will see that for base isolator structure it might be close okay because of some parameters or some like you know because of the inherent property of the isolators that we use we will see that how do we do that okay if omega b is closer to omega 1 then we can say that the very first approximation that we had considered it is a good approximation in this case what is m here m plus mb and k here is basically kb right and why that is a good representation it depends on the frequency ratio we will come back to that okay so in general can I say the stiffness of the isolation layer kb would be much much smaller than ks the superstructure stiffness this these are the assumption for base isolated structure now do we have this or not we say that the base isolation layer is much flexible compared to the superstructure right in the base isolated structure okay so this assumption is okay okay for a base isolated structure the base isolator stiffness is much smaller than the superstructure stiffness so okay that is why if you consider the equation for omega s and omega b I can say that superstructure frequency would be much larger than the overall base isolated structure frequency or vice versa time period of the base isolated structure so time period of the superstructure would be much smaller than time period of the base isolated structure okay this m is not negligible it is comparable to mb or mass of the base mat is not negligible it is come so maybe I should write it like mb mb is of the same order of the superstructure so the total mass which is m plus mb okay total mass m plus mb okay this is actually not negligible or like you know there is not magnitude of order difference between this and any other quantity basically what we define we define a term gamma let us say which is the ratio of the mass of the superstructure by total mass where total mass is m plus mb so we can write this as m plus mb as well okay and this is not negligible however okay however so I can say that m and mb are comparable but typically kb is much much smaller than ks so that's why I am able to make this assertion assuming that mass are almost of the same order but the stiffnesses are much smaller because I mean imagine superstructure unless it's like a very flexible let us say steel building very tall steel building and the time period is above 1 second or something like that if you have any structure the typical time period is less than 0.5 second or let us say 0.6 second okay and the base isolated time period is typically like above 2 second okay so if I can make this claim let me define okay a parameter epsilon which is the ratio of base isolated frequency to the superstructure frequency and the square of that why I am doing that I will come back to that because what happens when I start solving this equation I tend to get these terms epsilon which is the ratio of this frequencies and gamma and then I need to have some idea of the estimate that what are the order of these terms so that if I have to neglect some terms I can make an informed decision based on these now if omega b which is the base isolated frequency okay if it is much smaller or the omega s which is superstructure frequency it is much larger than the base isolated frequency what we typically observe that this is of the order of 10 to the power minus 2 or a smaller so because now see I am comparing the square of the quantity so if the initial it was 1 by 10 or 0.1 when you consider the square that ratio becomes 10 to the power minus 2 I am considering the square because when I solve this ultra degree of freedom equation then I actually get terms like this okay square terms is this clear everybody I am just defining these term because I am going to encounter them later when I solve this equation and then it becomes helpful to know these equations okay all right so I have two term epsilon which is omega b by omega s square and then I also have gamma okay which is m by m okay so how do we solve a multi degree of freedom system how do we get the frequencies eigen value problem so what do we do basically we take the mat basically determinant of this quantity here right k matrix minus omega square mass matrix and the determinant of this we set it equal to 0 correct and then we get equation in terms of omega and the frequencies actually provide omega square you can think it as eigen value whatever the values of frequencies we get provide the modal frequencies so we can do that here what was this this was kb 0 0 ks isn't it our k was kb 0 0 ks and what was m here sorry this would be omega square m or I will write m plus mb m plus mb the first term is m plus mb I will just write it capital M which is m plus mb here okay so m m m and the small m this is equal to 0 okay so so basically the equation of motion oh sorry not equation the characteristic equation that I will get in fact let me do this why don't you write it the characteristic equation and then solve remember if you have a equation like this okay what is your x minus b plus minus b square minus 4ac divided by 2a right so take the determinant of this get the equation the equation that you will get it would be in terms of omega is 4 and omega square so it would be a quadratic equation in terms of omega square okay please get that and try to get by making this assuming this approximation epsilon and gamma try to get omega 1 and omega 2 okay you will be basically getting the values of omega 1 and omega 2 you will see that okay that how these frequency are relating to the frequencies that we have defined here for two single degrees of freedom systems okay so please do that we are going to discuss that anyway but it is important that you do that so go ahead and try to solve get first the characteristic equation and then solve it and then utilize these approximation to see what do you get as omega 1 and omega 2 and how do they relate to omega s and omega b okay it is a very interesting exercise actually for base isometric structure you will see something like you know very nice okay let us let us see what do we get I hope at least you got the characteristic equation remember if I substitute this and then I solve this basically okay what I would get as my determinant as this would be kb minus omega square m okay and then minus omega square m the second element minus omega square m and this would be ks minus omega square m and this is equal to 0 okay and then I can multiply this and rearrange the term okay and knowing that my gamma is actually m by m okay I can write this as 1 minus gamma omega 4 minus omega b square plus omega s square times omega square plus omega b omega s square 0 so remember this is a quadratic equation in terms of omega square okay now further what we can do or okay let me just do this first write down so it is quadratic in terms of omega square okay so my omega square I can write it as minus b remember this is b here so this would be omega b square plus omega square because my b here is now this one minus of this quantity plus minus b square minus 4 ac so omega square omega s square whole square minus 4 1 minus gamma omega square omega s square divided by okay 2a and a is 1 minus gamma here okay now in this case if you take omega b square outside okay so let us say I take omega b square outside I will have 1 plus omega square by omega b whole square plus minus when I take omega b basically outside here okay what I will have to enter here is basically omega 4 so it would be 1 plus omega s okay or maybe what should I do here now do it like that we had assumed omega b by omega s right so maybe I take omega s outside here so let us do that okay let us take omega s outside so this becomes 1 plus omega b by omega s plus minus 1 plus omega b by omega s square whole square minus 4 okay omega b by omega s square and this whole thing divided by 2 1 minus gamma so this is omega s square now what is this term here this is my zeta that we had used right so I will write it here first let me write down all the terms then we will we will start to make approximations okay 1 plus epsilon 1 plus epsilon square minus 4 1 minus gamma times epsilon divided by 2 1 minus gamma so I will keep writing this as let me take this outside 2 1 minus gamma and inside I would be left with 1 plus epsilon and if I expand this this would be 1 plus zeta square plus 2 zeta or maybe I should write it like this 1 plus 2 zeta plus zeta square and this would be 4 sorry I keep calling epsilon as zeta this is epsilon 4 epsilon plus 4 gamma epsilon okay now let us see what approximations can we make can I say in this case omega square 1 2 1 minus gamma I have 1 epsilon and this one as I have 1 plus or in this case if I subtract this term and this term I would have 2 epsilon minus 2 epsilon okay and then plus 4 zeta sorry gamma epsilon plus this now remember we said that this quantity gamma is m by m which is less than 1 but it is not a very small quantity it is still comparable right we told that however zeta is of order of 10 to the power minus 2 so zeta square would be order of 10 to the power minus 4 so very very small compared to this so utilizing that I can perhaps neglect this and then I have the term 1 minus 2 zeta let us take that common and it would be 1 minus 2 gamma okay now what we can do in this case okay let me just we have to now simplify this so that now we know that if I have something like this 1 plus x to the power n a polynomial expansion of this one if x is a very small quantity can be written as 1 plus nx if x is very very small right so we are going to utilize that as well okay and then see what happens okay so let us do that so I have omega s square 2 1 minus gamma 1 plus epsilon plus minus this term is gone I am remaining with this term here and if I take the whole root of this one it would be 1 minus half times this quantity here okay and that will give me plus minus 1 minus epsilon 1 minus 2 gamma okay now remember my omega n square and omega 2 square would correspond to when I consider first the positive sign here and then the negative sign here and I do not get right now I do not know yet which one correspond to what so let us first take the positive sign and then see what do we get okay the lower frequency would be the omega 1 and the higher frequency would be the omega 2 okay so this is omega square 2 1 minus gamma I am considering positive term now so it would be plus 1 minus zeta plus 2 gamma sorry epsilon plus 2 gamma epsilon so this term this term will cancel off what I will get here omega square 2 1 minus gamma 2 okay 2 plus 1 if I take 2 common 1 plus gamma zeta okay now this will cancel off here and I will get as omega square 1 plus gamma zeta and 1 minus gamma okay now for this one okay I said that zeta is order of 10 to the power minus 2 gamma is less than 1 but not very negligible or not negligible not very small okay so can I say gamma zeta term can be neglected with respect to 1 however the denominator term gamma cannot be respected with respect to 1 but gamma times epsilon can be so this term again would be omega square divided by this term would become equal to or not equal to approximately equal to 1 and I have this here so what I am getting here is actually the higher frequency so maybe this is omega 2 square we will see that what do we get as the other term when I take the minus sign here so omega 2 is omega s 1 minus gamma okay now let us see when we take when we take the negative term then what do we get when we take negative here in fact why don't you do this take negative here and see what do you get simplify like what I have done here and see what do you get as omega in the second case when you take the negative term should not take you much time okay just do the similar kind of manipulation that I have done here and see what do you get okay let us see when I take the negative sign I mean this expression would be omega epsilon times omega s square okay so omega s square 2 1 minus gamma times 1 plus epsilon minus 1 and then this would become plus so I should maybe put this whole thing inside the bracket this is plus zeta 1 minus 2 gamma okay when I do this I will get this as omega square 2 1 minus gamma this this will cancel off I will have this epsilon common I will take it and this would become 2 minus 2 gamma okay so if I take 2 outside it would be 1 minus gamma it would be cancel off here okay 2 2 will cancel off and I will get omega square remember this is omega square now this is omega 1 square okay so this is so this would become omega s square times epsilon now I know that epsilon is omega b square times omega s square right this is the value of epsilon so if I substitute that here what do I get omega square times omega b square times omega s square okay so omega 1 square s omega b square okay and omega 1 s you just take the square root would be omega b what is omega b what we had to be define the omega b as omega b was frequency of the base isolated structure when considered as a single degree of freedom representation is it or is it not and even when I solve a 2 degree of freedom representation of the same building I am still getting my omega 1 as omega b and omega 2 as omega s divided by 1 minus gamma okay so for a base isolated structure even when you consider 2 degree of freedom representation as long as these assumptions are valid that your frequency of the superstructure is much higher than the base isolated structure so that these are valid the first mode is basically the base isolated mode the second mode is the deformation in the superstructure but the frequency is not exactly equal to omega s it might be close to that but not exactly equal to omega s okay now is this a important conclusion or not that even when you solve a 2 degree of freedom representation your first mode was coming that is coming out to be equal to the base isolated mode and the second mode is the deformation in the superstructure because omega s is what it is the frequency of the fixed base superstructure okay now you can go ahead once you have the frequency you can find out the mode shapes as well right that is not difficult to do but we are not going to do that I am just going to write down the final answer you can go and verify if you like your phi 1 would be mode shape and this is the base isolated mode shape 1 plus epsilon okay and phi 2 would be you will get as 1 1 minus gamma zeta divided by gamma okay now as you know this is what this is phi 1 1 right and this is phi 2 1 this is phi 1 2 and this is phi 2 2 okay so if I have to represent these modes draw these mode shapes how would that look like phi 1 1 is equal to 1 and phi 2 1 is equal to epsilon where epsilon is a very small quantity remember okay so if I have to draw it I will draw something like this small deformation in the superstructure compared to large deformation in the bearing or the isolation layer okay so remember we are measuring everything with this relative quantity okay so this one is phi 1 1 which we have obtained as 1 this one is phi 2 1 which we have obtained as epsilon okay where epsilon is a very small quantity so the in the first mode what we are seeing here the superstructure contribution is very less okay the deformation in the superstructure corresponding to the first mode is epsilon very small and we can again go ahead and draw the second mode what happens in the second mode so phi 2 let me draw this remember now phi 2 it is in it is negative phi 2 2 so it is an opposite direction to the phi 1 2 so it would look like something like this this here is basically your phi 2 2 and this here with respect to the relative deformation of the base is phi 1 2 this is equal to 1 and phi 2 2 is basically the second quantity here okay so the in the second mode actually okay remember this is your omega 1 and this is omega 2 in the second mode the superstructure deformation or the it is not negligible compared to phi 1 2 so phi 2 1 is not negligible compared to phi 1 2 however if you try and go ahead and find out the modal participation factor which you can do it by writing down the expression for the modal participation factors okay so you can multiply the mode shape and find out you would get this approximately equal to 1 minus gamma times epsilon and this you will get as gamma times epsilon okay and of course if you sum this up it would be equal to 1 because the sum contribution of each mode should be equal to 1 okay but what I am saying in the second mode the contribution of the superstructure is not negligible compared to the contribution of this phi 1 2 which is the base isolation however when you calculate the participation factor of the each mode you will see that first mode is actually approximately equal to 1 minus gamma zeta which you can I not say that this quantity is very small compared to 1 so this is almost equal to 1 the whole participation is actually coming from the base isolated mode okay is that okay so you have to what we have derived here is basically okay the contribution basically the frequency of each mode which we see that it is coming out to be close to omega 1 and omega 2 okay and the participation of each mode so the omega 1 was close to the base isolated mode omega 2 was not close but the superstructure frequency okay the participation of the base isolated mode is almost very almost close to 1 and you have very less participation of the superstructure mode this would only be valid as long as your approximations basically which we had assumed e omega b by omega s square is very small which means omega b by omega s so as you can see that if you want the whole response to be dominated by the isolated mode you need to ensure this the frequency of the base isolated mode is much smaller than the frequency of the superstructure mode and how we can ensure that if we assume that the super basically the masses are comparable we can ensure this okay by ensuring kb is much smaller than ks so that isolation layer is very flexible but remember in the kb it is sorry omega b is actually kb divided by m and omega s is ks divided by superstructure mass so it is not exactly equal to omega b or not directly proportional to kb it also depends on this mass okay but assuming that mass are comparable not of the order magnitude of difference as long as you ensure that you can ensure that effectiveness of base isolation structure by ensuring that the contribution of the response mostly come from the isolated mode okay is this clear anybody has any doubts very important that you understand you might have been discussing all about this that you know I mean contribution of the isolated mode the first mode is mostly the isolated mode itself but having a mathematical appreciation is also important because then you can have an insight into the overall dynamics of a base isolated structure superstructure should be stiff so typically we like you know based on the observation and like you know I mean anybody who has designed or like you know come through lot of structures they would know that for all the building structures typically the period is between 0.2 second to 0.8 second or 1 second let us now for base isolated structure we just want to ensure that it is spaced apart from that period okay so if the structure period is 0.5 second we go for at least four times that let us say 2 second okay so that my omega by omega s becomes 1 by 4 if I go for 0.5 second fixed and so I just reverse that one now this is what omega no this is omega by omega and then then this is 0.25 right so what does my epsilon becomes 0.25 square which would be 0.0225 I think right yeah what will happen if we consider this now this would be 0.22 I think this would be 625 all right so this is of the order of 10 to the power minus 2 isn't it and basically like you know it's a design basically assumption you know as long as you maintain in that period range you usually get a good iso good spacing between the or basically contribution mostly from the isolation mode remember this is m1 and m2 are basically what these are modal masses so maybe instead of writing it like this I will write it like this one l is your influence vector so I won't write it like this okay so now you have a mathematical proof and the understanding of the assumptions that we make when we talk about the dynamics of base isolated structure you should have complete idea know that how does the super structure like you know behave what role does it play in the overall response what do the frequency depend on in order to ensure that the base isolated structure mostly behaves like a single degree of freedom structure what are the assumptions that are required you know all those things now should be known to you because now you have mathematical background of the same