 Hello everyone, in today's class we are going to continue our discussion from the previous classes in which we basically learned how to find out the response of a multi-degree of freedom system subject to external force. And we used method what is called modal superposition analysis. So we are going to continue discussion of on modal superposition analysis and we are also going to discuss how we can apply modal superposition analysis to seismic analysis of structures and also learned about the response spectrum method which gives us directly the peak response of multi-degree of freedom system. So let us get started. We have talked a lot about modal decomposition of modal decomposition, basically diagonalization of mass matrix, diagonalization of stiffness matrix. Now let us look at the forces. So what I mean to say when I apply any force vector to a structure, the force would also distribute itself among different modes and we said that if we have force Pt which is defined as like P1t and then P2t and so on. The way to find out P and t, I need to be clear about this, this is small p here, P and t is basically the vector or not the vector, the excitation force in the nth mode. So effective force in the nth mode. How do we get that? We simply get that by multiplying like others with transpose of the mode shape times the P vector here, this is this. And this would give me P and t. Now when I take P and t, like any other vectors, like my vector ut here, I know that my force would also be distributed in different modes. So I will have effective forces at different degrees of freedom in different modes. And that's where the concept of expansion of this forces comes into picture that what would be the expansion of my vector P to different modes. And let us see how do we do that. So basically what I am saying, let us say I have a force Pt for which, so if I have something like this, let us say at different degrees of freedom, and all of these have same time variation, same time variation. Then I can write down my force vector P as a vector that represents the spatial distribution of that force times the same time variation Pt. I am writing it like that because it will later help me in the analysis and interpretation of the result. And we will see how do we do that, how do we get the distribution. So basically, let us say I have something like that, this vector what I wish to do for different mode, I want to get the spatial distribution as something like this, okay. So instead of saying like this, let us say some distribution is there, okay. And I want to see and so on. So let us say this is my vector S here. I want to find out a vector S1 plus a vector S2 and so on for each mode. So this is let us say mode 1 and this is mode 2 here, okay. I have a force vector which I have written as Pt and if the Pt has the same time variation at all degrees of freedom, I can write it something like this S which represents the spatial distribution basically if this is 0.1, 0.2, okay. This is let us say 0.15, this is 0.6, okay times the Pt which is the time variation. And like you know why we are doing this because if you remember the seismic excitation has a property something like this where all the degrees of freedom are subjected to the same time variation of the force which is the ug double dot t, okay. So we will utilize this knowledge there. So just keep that in mind, okay. So basically I want to write this as let us say summation r equal to 1 to n and this is SR, okay. So sum of multiple modes, okay. Now this can be written as a modal participation vector which we write it as gamma of r, okay. Let us say this is gamma of r times mass matrix times phi r which is the mode shape vector for that particular mode r and then sum it over. So basically what I am saying that my SR or SR here is basically nothing but this quantity here, okay. And a question might ask that how can I directly writing this equal to this much? Well if you remember to get the modal coordinates q1, q2 we had written, okay. We had written again mass or like you know this vector here, phi times qr summation r equal to 1 to n, okay. The idea is that if I write it in terms of multiplication with some mode shape and multiply with the transpose of another mode shape, I can use the condition of orthogonality and I can find out these vectors here or these factors that we have defined, okay. So I am going to do exactly that. I am going to multiply, okay, this as s here and this r equal to 1 to n, okay. And this I will multiply phi r again the same quantity phi n times mass times phi r. Now as you know this quantity would always be 0 except when r equal to n, okay. And if that is the case I am just only term that is going to survive here would be when r is equal to n. So I am going to write this as phi n times s is equal to this phi n t mass and phi n. And this is nothing but the diagonalized mass element mn, okay. So phi gamma n can be written as s times mn. So remember if I have a vector s which is given remember force is given the external force is given and if I can write down that external force let us say pt equal to a spatial distribution vector s times some time variation, okay let us say small pt. So this is also given I know my mass vector I know my stiffness vector I want to write this as sum of vectors spatial for each mode so that I can see how much is the contribution being made by the vector force vector in each mode. So s can be found out as so remember once you find out this gamma n I can get my spatial distribution in nth mode as how would I get that just get write this expression here gamma n times mass times phi n, okay. And then we can get the total distribution or expansion like this. So once we know the gamma n then Sn we can also obtain like this and this would be the spatial distribution, okay. Now let us see what are the properties of this or what are the physical interpretation of this Sn, okay. Now you know that when you will apply this force external force pt, okay a structure can vibrate in any of its modes, right. Now when a structure in vibrating in each of its modes can I say in each mode it would have some inertial forces associated with each degree of freedom. What Sn basically represents it represents the distribution of those inertial forces associated with the application of the applied force, okay. So again I will repeat it if you apply external forces pt on a multi-degree of freedom system, okay. Let us say something like this, okay. In each mode the structure will vibrate, okay. And when the structure is vibrating it would have inertial forces associated with each degree of freedom, okay. What this Sn represents is basically distribution of those inertial forces in each mode, okay. And when you multiply with the time variation of these force it will give you basically the total external force that is being applied on that mode, okay. But Sn just represents the inertial force distribution in different modes. We did expansion of our mode shapes basically different modes using the displacement vector. Now this is the force that is effective force in each of the modes, okay. And we can do this for earthquake forces and doesn't have to be earthquake forces as long as the time variation at each degree of freedom is same and you are able to represent your force vector as some spatial distribution vector times, okay. Same time variation, okay. Let us say this is not QT, let us say this is some small pt which is the time variation then we can interpret results like this, okay. And like you know we can prove that mathematically as well. Let us say inertial force in my nth mode would be what? In my nth mode the inertial force let us say MIN here would be mass matrix times the acceleration of each degree of freedom in that nth mode, right. So let us say this is the nth mode. The inertial force is mass times acceleration which is UNT, okay. Now UNT is what? Remember UNT is phi n times QNT. So all I need to do is that take the double differentiation of the time function here. So mass times this is phi n here times Q and double t. Now if you compare this remember if you compare this to p, the time variation of the force capital P equal to phi times pt, okay not phi times sorry, S times pt. This is the time variation, this is the time variation. What is S here? S is basically this quantity, okay and this quantity is what? This is the spatial distribution associated with the inertial forces that I have, okay. So I hope this is clear to you. Now what happens that if you have a force, if you are able to find out Sn vector which is the spatial distribution of force, Sn times the time variation pt of this force, okay would give you the response in the nth mode, response of all degrees of freedom in nth mode, okay. And there is no other mode that is going to contribute to this response because now this is the effective force that you have found out assuming that this is how the total force is going to be distributed among different modes. If this is clear now let us look back at our equation of motion in terms of this S, okay. If you remember our equation of motion was mn times Q and double dot t plus Sn times Qn dot t plus Kn times Q and t and this was equal to P and t here, okay. Now I can go ahead and I can divide it by mn so that this would be Qn t and this becomes 2 zeta omega n zeta n sorry plus omega n square Q and t. Now remember P and t I can write as zeta n times the p vector, right. So when I multiply with this mn this is nothing but phi t and pt if it is can be if it can be written as a spatial distribution some time variation I can write it like this, okay the time variation and I am dividing it with by mn, okay. Now what is this quantity here is it not gamma n that we have already derived so our equation of motion can be written as zeta n omega n plus omega n square Q and t is equal to gamma n times the time variation pt, okay. So this is my equation of motion for each degree of freedom where this is a factor gamma n is basically a factor which is a modal participation factor so I will write it anyway this is called modal participation factor, okay. So this factor basically says that of the given force how much is the total force that is going to be applied in the particular in particular mode, okay and this is what basically the excitation force becomes what you are used to do you can solve this equation remember we have done nothing new we are just considering a special case, okay so this is just one of the special case for which it can be written like that, okay so let us say this is a special case, okay and if this is special case is still not there you can still write down the same thing pt as some vector and do the same thing and multiply and like you know solve it, okay it is not that it has to be like this, okay. So using this the modal participation factor we have now written down our equation of motion in this form here which is basically the equation of motion of a single degree of freedom system subject to excitation pt multiplied with gamma n to account for the distribution of the force in its nth mode you know I mean let us say many times what happens, okay you have or you would have the solution of this equation available, okay and I will give you some example, okay let us say I have a building, okay let us say the same two-story building and let us say this is being applied to a blast load or a triangular pulse let us say so this is the load and you know the time variation of this load so it is being applied with the same variation pt here and here similarly instead of this I could have pulse load or I could have like you know ground motion which is something like this, okay for all these type of motion we have a standard result available from the previous chapters, right in which we solved the equation of motion this, right you solve the equation of motion not q let me first write down in terms of u, u equal to pt where pt could be one of these forces here. So equation solution to this one is available which is basically response of a single degree of freedom system to any of these pulses. Now if I know the time variation of these forces are to be the same and I know the solution and the same thing is now being applied to a multi-degree of freedom system what do I need to do to consider response of each mode I just need to multiply whatever response I get here by gamma. So think it in like you know two step first step is to get the time variation of the response, okay for our case it is qt, okay these responses are available using standard results. So solution to this one is solution to this SDUF system is available. Problem is now I have a multiple degree of freedom system, okay. So solution to this is available how do I utilize this for a multiple degree of freedom system. Now I know that multiple degree of freedom system can be represented using n number of if n is the degree of freedom single degree of freedom system, okay. So to utilize this result all I need to do is to multiply response of this using this to get for that particular, okay. So you will also see in the book it might be written as d here instead of u to avoid any confusion, okay. So this is d dot and this is d here. So this is available if you want to find out q and t for a multiple degree of freedom system it would be nothing but this factor accounting for the distribution of the total force in the nth mode time the standard results that you available here. Let us say this is d and t you obtain. Now generalize this to df system instead of zeta sorry instead of gamma n we put gamma equivalent, okay. But here it is for that particular mode, okay. So qn we can get it like this, okay. And if I have to get the contribution of the nth mode to the total displacement at each degree of freedom all I need to do is remember this would be sorry this would be phi n times q and t, right. Now q and t I can write it as zeta times dn. So this I can write it as phi n times dn t, okay. Similarly if I have to find out equivalent static force in the nth mode, okay. So if I have to find out let us say equivalent static force in the nth mode I can find this out simply as writing this as k times u and t, k is here, u and t is my I can write this expression here it would becomes zeta n times phi n times dn t, okay. And if I want I can further write down using eigen value equation this as omega n squared times m times phi n. So this I can write down as m phi n and there would be omega n squared term here as well. So let me just write that times dn here, okay. Now what is my let me again recall if you remember sn was zeta n times m times phi n, isn't it? This was the expression. So can I not write this I can combine this and this term here I can write this as omega n squared times sn times dn, okay. So this is vector here or instead of writing it like this let me combine it sn times omega n squared times dn t, okay. What is this quantity here? This is the acceleration right in the nth mode and this is exactly what I told you when I was discussing the physical interpretation that if you apply a force which will have the distribution s times some time variation, okay sn represents basically the forces in its nth mode, okay the inertial forces in its nth mode due to those applied forces and that becomes clear here. I also told you that if you multiply those forces with the acceleration you would get the equivalent static forces in that particular mode. Then let us look at the most common problem that you might encounter, okay would be earthquake analysis right. So for seismic excitation, okay. Now for seismic excitation what is my applied force vector? This is my applied force p effective is what is my p effective? It is mass matrix times the influence vector times the ground acceleration. Right I hope influence vector you already know how to get influence vector. Basically you apply a unit displacement in the direction of the ground motion and then you see along each degree of freedom due to that unit displacement what is the corresponding displacement and then you form the influence vector. So I had given you example of this, okay if you define this one as u1 and u2 and you have ground motion like this then if you apply a unit displacement ugt equal to 1, okay influence vector is constructed like here it is 1, if the ground displacement is 1 u1 is 1 however u2 is 0, okay. So as simple as that this is my influence vector and you apply multiply with the mass matrix and this is the time variation of force. So basically this is of the form this that we have discussed right. The vector s times some time variation pt, okay where my vector s is basically nothing but m times l. I can repeat the same procedure and I can show you how to get that. So let us do this let me write down here basically the equation of motion that we get mn times qn double dot cn times qn dot plus kn times qn is now equal to phi n or let me write down p effective n which is in this case what would be the pn here. It would be minus phi nt times p effective which is basically this force here or let me first write down this, okay. So this is pnt and p effective is no minus sign here sorry minus signs comes after I substitute this l times ugt, okay. This is what we have here, okay. So if I multiply this whole thing or divide this whole thing by mn what would I get qn plus 2 zeta n omega n divided by this whole thing by mn times ugt. Remember I am just doing the same thing that I have done here, okay. Now pt instead I am writing ugt. It is said my s is m times l if you remember, okay. This thing the numerator actually is defined as ln. This comes from again the equation that you had done in the generalized coordinate system and zeta n or sorry gamma n is basically ln divided by mn where ln is this divided by m. So this whole thing is actually gamma n, okay. And you could have you could have written the same thing without even me having to write all this you know. So basically your expression becomes or expression or the equation that you need to solve for seismic excitation it becomes this the same thing that we have written above becomes gamma n times ugt, okay. So if you are able to solve this equation for a for the nth mode then you can do for all the modes and combine all the modes, okay using this expression here and you can get the displacement as contribution of each mode n equal to 1 to n and this I can write it as phi n times qn here. But now for a special case q and t is represented as especially for ground motion. This can be represented as gamma n times phi n times dnt. My u and t is basically so I have just write down this, okay. Now in addition to that we had also derived my equivalent static force in the nth mode is sn t times the acceleration, okay. So you know the displacement in the nth mode and you know the equivalent static forces in the nth mode, okay. Your job is now basically very easy. You know the equivalent static for let us say this is m3 n, m2n and m1n. These are equivalent static forces, okay. Displacement is known. These equivalent static forces are known. Base shear is very easy to calculate. Storage shear is very easy to calculate. Just consider equilibrium and same goes for the moment as well, okay. For example moment at the base here would be what would be the moment here. Can I not say moment would be simply f1n, this is h1. Let us say this is h2 and the total height is h3. Can I say moment at the base would be f1 times h1, f2n times h2 plus f3n times h3. This would be the total base moment in that structure, okay. Now consider this, okay. Ground motion, this factor represents basically the modal contribution, okay. We know what is the solution to this equation, isn't it? We have the zeta n, omega n, qn plus omega n square qn is equal to ug t, okay. And dt is basically solution to this equation, right. Maybe I should not write sorry. I should not write q and t here. Let us write down dt here. So, this is known to us what is also known to us in fact, okay. What is given to us is response spectra. What is response spectra? It is the peak value of any response quantity as a function of different time period for single degree of freedom system, okay. So, we have this, we have given this from a response spectra perspective. So, let us say and most of the cases our goal of any structural analysis is to find out the peak responses like peak displacements and peak forces, okay. So, I have my un t, I have my fnt, okay. And I have my let us say mnt which is the moment. These are for all the nth mode, okay. So, let us do something. Let me say this is my displacement un t, okay. And this is the time variation. So, as we know, it would look like something like this here, okay. Now, let us say this is first mode instead of saying un, okay. This is the first mode, okay. And the second mode would be something like this, okay. Third mode would be again something like this. Now, different degrees of freedom system. What is response spectra? Let us say this is my response spectra and I write something like this. I mean, if it is d, it looks like something like this here, not like this. Something like this, okay. If it is a, the acceleration, it looks like something like this here, okay. So, this is d is basically what u max if the damping is very small, okay. This is a means my q double dot max. Now, so for different mode we can find out what is the peak response from the response spectra, okay. But let us look at this. For each mode depending upon the time period of this mode, for example, for this one, this is my u1 max. It is happening at this time t1. For this case, let us say this is my u2 max. And for the third case, let us say this is my u3 max. Can you appreciate the fact that if I have n single degree of freedom system, their maximum response would occur at different time. Like for first case, it would occur at this time for the second case at this and the third case at this point, okay. The question becomes, we can combine different mode and we can find out the displacement as a function of time, okay. And the time variation, how do or how can I utilize the response spectra, which is the peak response, so that I can also find out the peak response of a multi degree of freedom system, okay. And understand the issue here that we have. For a single degree of freedom system, I know my peak response. But for multi degree of freedom system, I do not know my peak response because it is a combination of different single degree of freedom system and different single degree of freedom system will achieve their maximum response at different time. So, and the idea is why do we do the response spectra analysis? Do you remember? We do response spectra analysis so that I do not have to solve the equation of motion every time and find out the response, okay. For example, if I do the response spectra, okay, or as a matter of fact any other force, I do not actually solve this equation, is it? Because I'm only interested in the peak response. I only need to know the TN value. If I know the TN value, I can come here and I can find out the DMAX. I do not actually numerically solve this. And the response spectra, they have been, they are already given, okay. So, for each ground motion, people have prepared response spectra and those are given. So, now I'm considering not the combining dot doing the time history analysis or anything, I just want to get the peak response. And I'm not interested in like, you know, at each time combine different systems then get the time variation of each degree of freedom system and then find out the maximum at what point the maximum would occur and all those things. I want to resort to a very simple procedure in which the maximum response for each degree of freedom can be found out. My only challenge is that they happen at different point of time. And the maximum response UMAX cannot be simply U1MAX, U2MAX plus U3MAX and so on. In fact, we always, sorry, it would be always less than equal to this. This goes not only for displacement but also for all the things that you find out. Remember, now if you talk in terms of UMAX, all the quantities, whether it is equivalent static force or like, you know, moment at the base, all things can now be found out in terms of UMAX for that particular mode, okay. And then we need to come up with a procedure so that we can combine these three or like in all different modes, the maximum response and get somehow UMAX because this is not simply equal to the algebraic sum. So what do we do? There are people have come up, like, you know, there are different methods that are available through which the maximum response of different systems can be combined so that I can get UMAX, you know, and those methods so basically methods of combining peak responses of different modes, peak responses of different modes, okay. So people have come up with different method how to actually combine different modes and get the total maximum response. Now as you can imagine, these methods are approximate. These are not exact method, but they are quite like, you know, reasonable in terms of they are like, you know, people have done like, you know, a lot of research and they usually come to like, you know, closer to the exact value, okay. And let us look at two of those methods where people have actually utilized that and have come up with the methods, okay. So the first method is called SRSS. SRSS method is basically sum root of square, okay, of all the sums. So square root, sorry, square root of sum of squares. So basically in this method, if you have multiple responses, D1, D2, D3 and so on, maximum response can be obtained as D1 square, D2 square, D3 square and so on. Now note that it is not the algebraic sum and it would always be less than the algebraic sum, okay. So people have come up with this method and these tend to be like, you know, produce good results. Of course, there are like, you know, cases where the difference between actual peak response and the peak response after in this method would be little bit different, okay. And you don't know to get into that, although I'll tell you what, if the modes are very closely spaced, like one mode is like, you know, 0.5 second time period, other mode is like, you know, 0.55 second time period, then this method might not work out that well. And there are other methods to do that. The other method like, you know, people employ complete quadratic combination, okay. And this method is little bit mathematically complicated. So in our course, we are only going to focus on this because, you know, I mean, you can have for hand calculation, SRSS is only amenable. If you have a software, you can implement CQC and get the total response, maximum response using individual peak responses using complete quadratic combination method. For seismic excitation, I can go ahead and find out the time variation of the response, okay, by combining different modes that we have done previously. But good thing about the seismic response is that we have available response spectra, okay. So peak responses are available to us. And this would apply for any other method for which the response of a single degree freedom system are applied in terms of some chart. So even like, you know, you could do it for pulse type motion where the peak response is available in terms of the time period and the time duration of the pulse. So if the peak responses are available, we can combine them, the peak responses, but not using conventional method that we have been doing so far because in the previous method, we were combining responses for any time t. And the final response was as a function of that time t. Now, I don't want to solve it in any time t, I don't want to find out d1 t, because I know my d1 max from the response spectra here. If d1 max is known, I can combine the maximum responses and approximately get the total maximum response using one of these methods, okay. So this is the summary of, okay. So this method is also called, you will see in codes response spectrum method. And the response spectrum method is nothing but a special case of modal superposition analysis, okay. Remember modal superposition analysis is for any time t, but for maximum response, we use the response spectra and it is called response spectra method, okay. So it is a special case of modal superposition analysis for peak responses or responses. Let us see one example how we can actually utilize this method, okay. So what we are going to do, let us take the example of what we have done and extend it here for the same time variation, okay. So basically what I want to do here, we had done this method, right, 2M, M, 2K and K, okay. Now, let us say or what I want to do here basically, I want to find out if earthquake force is applied, let us say here, something like this. I want to find out the expansion of the earthquake forces, okay. Remember earthquake forces can be written as what is S here, this is the influence vector and remember earthquake forces are nothing but S times UGT. If I write this as M times influence vector L, okay. Can you find me out what is, now here there are 2 degrees of freedom. So I will have S1 here and I will have S2 here, okay. So get me S1 and S2, okay. I will again rewrite the equation, S1 is gamma N times mass times phi N, okay. And gamma N is phi N times, not phi N, sorry, phi N transpose time the S vector divided by MN, okay. So once you have that, remember my F and T, the equivalent static forces are nothing but S times A and T, the nth mode it would be S N times A and T, okay. Similarly, so you would have F1 and F2. Once you have that, can you also get me the base here V1, V2 at each degree of freedom flowing this method, also try to get the same answer just directly from the displacement, that is from U1T and U2 to T for each mode by multiplying the story displacement with the S stiffnesses and utilizing this expression here. Remember A1T is nothing but omega N square D1T and A2T is nothing but omega 2 square D2T and see if you get the using the equivalent static method, okay. Let us try, okay. Let us see, let us try to get the answer. So my S vector is nothing but mass times L vector. Now influence vector in this would be, what would be the influence vector here? If you apply a unit ground displacement in each degree of freedom, the displacement is 11. So the influence vector would be 1 here. So effectively what you have here is 2M0M times 11, okay, which basically is 2MM, okay. And that makes sense. I mean, if you get the effective earthquake force, it is nothing but mass times L times UGT, okay, which is basically minus 2M UGT and MUGT is each mass multiplied by the ground acceleration. And what I want to do is to see how these, because of these earthquake forces, how the inertial forces in each mode is distributed, okay. And let us see how do we do that. To do that, first what I need to do, I need to find out the parameter gamma 1 or before that, let us do this. Let us first find out M1. I think it was 3M by 4, 3M by 2. How about M2? How about K1? I think 3K by 4 and 6K, right. And so we have that, let us say gamma 1 would be phi 1 times T times the mass vector times L divided by M1. So, let us see what do we get? Half 1 mass times L is 2M M, okay. Well, there is a negative sign here. Factor does not have any negative sign. And then M1, M is 3M by 2. So, this one we get as 2M divided by 3M by 2. So, this gives a 4 by 3, yes. Similarly, gamma 2 would be phi 2 is 1 by 3, okay. So, what would be my S1 which is associated with the inertial forces in the first mode? S1 is gamma 1 times mass times phi 1. So, 4 by 3 times 2M 0, 0M times half and 1. And this comes out to be, this would be M and M. So, both would be 4M by 3, okay. Similarly, S2 phi 2M times phi 2. So, this would be 1 by 3, 2M 0, 0M and minus 1 and 1. So, this would be my, this would be minus 1 by 3. So, there would be minus 2M and M. So, this would be 2M by 3 and this would be minus M by 3, correct. So, and remember this is my U1, this is U2, this is U1, this is U2. So, it would be better interpreted in terms of graphical representation. Total mass is 2M and 2M here and M here. And this is my S. Remember S was 2M and M. And if you look at each mode, so this is S1, okay. At mode, at degree of freedom 1, this is 4M by 3 and it is in positive direction, the inertial force, okay. The same mass 4M by 3 in the positive direction. This plus I have for the second case 2M by 3 and in the opposite direction M by 3, okay. And this is S2. And if you look at this, this is just, you know, if you sum this up, 4M by 3 minus M by 3 becomes M. 4M by 3, 2M by 3 becomes 2M, which is this, okay. And if you want to, let us say, you have a earthquake ground motion UGT for which you have found out the acceleration or you could do displacement as well. I will do the second method using displacement. So, let us say, due to this, for each mode you have A1T and A2T, which is basically the solution of this equation except there is no gamma term here. To convert it from that to for each mode, you need to multiply with the gamma. So, basically for each mode, if you remember, what is the equivalent static force? Fn. Fn is nothing but Sn times A and T, right, in short. So, all I need to do, multiply this with A1T for the first mode, A1T for the first mode, A2T for the second mode times A2T for the second mode. And what I have got is basically the equivalent static forces, okay. And to find out the storage here, what do I need to do? Let us consider storage here. If you consider the equilibrium, what would you get? The total storage here. So, this would be your second degree of freedom, first mode. And this would be first degree of freedom, first mode. This would be second degree of freedom, second mode, first degree of freedom, second mode. Okay. So, we can get U21 as 4M pi 3 A1T V11 as, now it will have, if you take a, it will have some of both. So, you will have 4M by 3 A1T plus 4M by 3 A1T, which is basically 8M by 3 A1T. Similarly, so this is corresponding to this one. For the second mode, this is first mode. For the second mode again, V22 is minus M by 3 A2T. Remember, it is just this force. And V12 is sum of both two. So, I will have 2M by 3 A2T plus or let us say minus here, minus M by 3 A2T, which is basically is M by 3 A2T. Now remember, these things we are doing for the A2T, but even if the quantity is A2 max, for that particular mode, at least I would, I can get using the same procedure, right? It's a different story that when I will combine the first and second mode, I will take the SRSS. Okay. But basically the same procedure is what I will follow. I told you that same, okay, the same expressions I could have gotten using the expressions for displacement or, in short, U1T and U2T. What is my U1T? It's gamma 1 times, well, we had written it as phi n times q and t. Remember, once you get used to all these notations and derivations, even if you forget something, you can easily write it. And that's what happens with me as well. You know, I just write down the original equation and then you can just write down everything using that. So phi n times gamma n times d and t. Okay, so this is for n. U1 will become, okay, if you want to write down for U1, it will become gamma 1 times phi n, which is half 1 times d1t. Okay, so I'll just write down here as d1t. Okay. And gamma 1 was what? It was 4 by 3, right? So this basically becomes 2 by 3 d1t or I will keep d1 outside and 4 by 3 d1t. Now, what is d1 here? d1 is nothing but a1 divided by omega n square. What is a1? a1 is k by 2 m. Okay, so this would be 2 m by k times a1. And like you know, I will substitute that later. Similarly, my U2 vector become minus 1 by 3 times the mode shape minus 1 1 and then d2t. So I will get here as 1 by 3 minus 1 by 3 d2t. So if you want to look at in terms of displacement, what basically it is saying? In the first mode, you have this displacement, remember this is nothing but U11 U21. This is U12 U22. So for the first mode, this is U12, this is U, sorry, not U12, U11, this is U21 plus U2, sorry, again 1 2, U12 and this is U22. Okay, so this is first mode. This is second mode. So let us, so I am discussing the first procedure, not the equivalent static procedure. So in this procedure, I know that the storage stiffness here is 2k, storage stiffness here is k. Okay, 2k, k here. So what would be my V21 and V11? Similarly, V22 and V12 would be, remember V21 that is here would be storage stiffness which is k times the storage drift which is U21 minus U11. So U21 minus U11 would be 4 by 3 minus 2 by 3. So it is k times 2 by 3 d1t. Now I can substitute my d1t equal to 2m by k1. So that would give me 2m by k, it would become 4m by 3 a1t. Okay, and V11 become storage stiffness which is 2k times the storage deformation U21, okay, minus 0. So that would be U2k, 2k times U21 which is, where is U21, 4 by 3 times d1t. Again I will write as d1t as this quantity here to get this one as 8 a1 by 3. Okay, we can compare the results actually, we can look at the results that we have got and see what is the, so you can do for the second mode. I am not going to repeat it for second mode, you can just do it and we will see how to now obtain the responses using peak response quantities of a SDF system or the response spectra. Okay, all right. Thank you.