 Welcome back everyone. So we are going to continue our discussion on seismic response of multi degree of freedom system and we are going to look into few examples today and see how we can apply the response spectrum method to a simple shear type building and then using that how can I find out the peak responses for the displacement at different stories as well as the story drifts and the peak values of the base moment using a response spectrum ok. What I am going to demonstrate the response spectrum procedure on a three storey building ok but remember the same procedure can be extended for any other type of multiple multi degree of freedom system doesn't have to be a shear type building or anything like that ok but here because it's easy to draw and everything I'm just going to demonstrate that example using a multi storey building ok. So what we are going to do is basically modal analysis procedure. So modal analysis procedure is basically when we consider this like you know decomposition into several single degrees of freedom system and the response spectrum procedure is when we consider peak responses ok. So modal analysis procedure can be it's a general procedure to basically find out the response of at any point of time ok. Response spectrum procedure is a special case when we are talking about only the peak responses ok so just keep that in mind. So what I basically have here is a let us consider a three storey building ok and this three storey building has masses at each level M1, M2 and M3 ok and the story stiffness is as K1, K2 and K3. Now there is a ground motion now ok let us say ground motion ground excitation UGT because of which ok I have these displacements here which are U1 sorry U1, U2 and U3 ok. So we can consider basically equilibrium at each floor ok I am not going to repeat that but if you like you can just draw the free body diagram of each floor or let me just do that anyway. So I have here M3 ok U3 plus UG total acceleration of mass ok and then because of the floor below I have K3 U3 minus U2 ok. For the second floor I have equal and opposite force on this so K3 U3 minus U2 pseudo force M2 U2 minus not U2 U2 plus UG opposite to the direction of motion and then K2 U2 minus U1. Similarly for the first floor ok I have M1 U1 plus UG and then equal and opposite force here K U2 minus U1 and then K1 U1 here. You can write down the equation of motion basically we will have the three equation of motion ok M1 U1 plus UG plus K1 U1 minus K2 U2 minus U1 equal to 0. Basically M2 U2 plus UG plus K2 U2 minus U1 minus K3 U3 minus U2 is equal to 0 and then the third floor which is basically M3 U3 acceleration U3 plus ground acceleration then K3 U3 minus U2 is equal to 0. So I can also write it in terms of matrix form M1 0 0 0 M2 0 0 0 0 M3 U1 U2 U3 so by now you should be very proficient at least with setting up this equation of motion ok. Building should be very easy for you especially the shear type building it would always be of this form if the degrees of freedom are defined at each level ok. So this is my stiffness matrix U1 U2 U3 and if I bring all these M1 U U1 M2 U2 to the right hand side what do I get basically M1 M2 M3 times UGT. Now I would like you to recall from the last class the things that we did ok. So what did we say we said that this can be further written as mass times acceleration vector plus stiffness times displacement vector and this can be written as mass times the influence vector times UGT ok and like you might have question that why we are using influence vector at all well it is easier when we do multi-storey building that it is simply in terms of this like in a vector 1 the influence vector is 1 1 1 here but you will see in many other type of multiple degree of freedom system your influence vector is not 1 1 1 it would be something different and in those cases writing down the influence vector basically provides you a generalized equation of motion you do not have to worry about whether it is a shear type building or whether it is other type of multiple degree of freedom system ok. So for the general case ok the equation of motion is M U acceleration vector plus K U and then minus mass matrix times influence vector times UGT ok. Now if we discuss like you know how to get the influence vector for different type of systems ok. Let us now see if we have the damping in the system ok all I need to do ok for the damping cases ok just write down this equation of this form additional damping matrix here times the velocity vector ok plus K times U vector but you need to be careful usually damping matrices are not given and many times it need to be constructed from what do you remember it need to be constructed using damping ratios right zeta 1 zeta 2 like that ok and in those cases my C 1 or C n becomes ok what is my C n it is 2 or if I write it like this zeta n is C n by 2 m n omega n so C n becomes 2 zeta n m n omega n ok. So this would be given to you zeta n ok and you need to formulate that from the and we saw that what were the different methods there were mass proportional matrix damping proportional matrix and there were also Rayleigh damping ok in which case we formulated as linear combination of mass and stiffness matrices ok. So in those cases basically if the damping for one or two modes are given we can find out damping for the other modes as well ok. If that is clear let us write down the single degree of freedom equation for the corresponding this equation ok. So if you remember we can write it down as dissociate single degree of freedom system or for the nth mode we can write down this equation of motion as m n times q n plus C n times q n times k n q n is equal to let us say this quantity would here becomes or p effective or p n let us call it p n effective t ok where p n effective t is nothing but for the nth mode ok or I will just delete the subscript it is just called directly p n p n is nothing but phi n t times the vector. So this is in this case would be phi n t times mass times l times u g t ok. So this is what we get for the seismic excitation ok and this is our n uncoupled single degree of freedom system ok these are all uncoupled and solution to these type of equations we already know ok. You can write this as 2 zeta n omega n q n so I am dividing it with m n this would become omega n square ok and this would be basically p n t by m n ok where p n t by m n is nothing but minus no not we will have to write down the phi t here this is phi t times m times the influence vector n u g t and this divided by m n and if you remember ok this is what this is gamma n ok this expression here is gamma n which is also called modal participation vector modal participation vector ok. So the equation that we need to solve is basically let me just zoom it out the equation that we need to solve basically becomes this ok q n t plus 2 zeta n q n not q n omega n q n t so omega n square q n and this is equal to zeta n u g t now except the term here I have the solution available for the rest of the equation ok and let us say the solution that we have available for is this equation here ok because this comes from the this is basically somewhat you can think it like you know relationship between the single degree of freedom system and the multiple degree of freedom system right apart from that if you just consider q to be any variable ok this is a single degree of freedom system ok only this zeta n the modal participation factor is what represents how the my single degree of freedom system is related to the overall multi degree of freedom system ok. So we already have solution for this available from the standard equations or equal like you know charts or whatever if it is not a ground motion ok so we have and this is the equation that we use when we try to find out response of a single degree of freedom system there is of a single degree of freedom system subject to ground motion ok u g t all right so if you want and if you are only considering linear system then we can say that if we know the solution for this the solution for this equation would be simply excitation whatever the solution we get for this multiplied with this factor because it is a linear system so basically if we know the solution for this equation solution for this equation would be simply gamma n times d and t ok all right and so this was the solution in terms of time if you wanted the maximum peak solution as well at least for this single degree of freedom system it would again be related by the same expression instead of d n it would become d m max because remember only d n is varying in time and q n is varying in time ok gamma n is just a function of the mode shapes and all those things ok this expression here there is no time here ok so whenever d n reaches the maximum value q n is basically that multiplication multiplied by this factor here so it would reach the maximum so at least for that single degree of freedom system if I know the d n max q n max is also known ok all right now let us see we also talked about how to do the modal x expansion of the excitation vector remember and we said that that if my applied force modal let us say expansion of excitation vector ok so we said that if somehow for some special cases my p vector the applied force vector can be written as a spatial distribution s times the same time distribution and this is specifically true for seismic excitation then I can perhaps find out what is the contribution or what is the basically expansion of this vector s in each mode and we said that the modal expansion of vector s represented how much is the inertial force in each mode due to applied excitation ok so we said that we can write down the expansion of this as let us say n number of vectors I will call it n here and I have already derived the expression for these in the last class phi n where this gamma n is nothing but phi n to the power t times s times the m n ok and if you consider seismic force what is the p effective p effective is mass times l times u g double dot t so if you compare this becomes your s ok the spatial variation this is your time variation so basically we can expand it and we also we also saw that if that is the case ok then how can we actually expand the same vector basically ml ok so if I substitute s equal to ml it can be expanded using the same expression n equal to 1 n so all I need to do is to find out zeta n here and substitute s equal to m times the influence vector to get the modal participation vector and once we get that it is very easy to actually expand the excitation vector in terms of inertial forces in each ok and we also said that if you want to find out the internal forces or the element forces ok there are two procedures that you can utilize the first procedure is finding out u and t which is the contribution of nth mode to the total displacement response as k times sorry not u and t here may write here as ok directly utilizing this expression here ok so if you have u and t available after you have found out your q and t then and if you have let us say multi-story building you can find out let us say in the nth mode the storage here in the jth mode would be storage here kj times ujn and uj minus 1n ok as the drift ok this is the first procedure find out the u and t and then find out the storage here individually and all the forces the second procedure was equivalent static procedure in which we find out equivalent static forces as stiffness times the displacement remember this is for the nth mode so k times u n is equal to phi n ok times q and t here ok and k phi n is nothing but omega n square mass times phi n times q and t either you can leave it at this point or you can further simplify it ok by utilizing this expression that we had derived here so what you can do in this case let us write down remember your phi n is nothing but if you look at this expression here ok I can write down the total uh response as uh where is sn yeah sn here as phi n m times this expression ok now q and t is basically what here if you look at q and t is nothing but always the modal participation factor gamma n times the tt that you obtain where dt is the response of the single degree of freedom system without any factor gamma n and dt let us call dnt it is also related to acceleration in the same mode these are pseudo acceleration ok utilizing this relationship ok so what we can do here we can write down gamma n and omega n square and dnt is basically a n so this would become a n t now what is this expression here if you look at this expression here this is nothing but sn a n t so if you have found out the dissociation of mass times l vector which is basically s ok all you need to do in each mode just multiply with the modal acceleration a and t and that will give you the equivalent static forces ok and if you want to find out u and t that is not difficult as well ok u and t you can directly find out as gamma n divided by omega n square times phi n and a n t ok so what we are going to do actually we are going to do an example today ok and then see how we actually solve this system and find out the basically the displacement response as well as the internal forces ok and we will also see the response spectrum analysis so response spectrum analysis as we already discussed let me just again summarize that as well so in the response spectrum analysis so till now whatever we have done combining the response at each time instance ok but that is not easy many times i don't want to find out you know the time variation of the response and let us say the peak response is given to me something like this a and t as a function of or not a and t let us just call acceleration here it might be given in s a by g or a by g whatever notations you are using and this is the time period so response vector is always defined for a single degree of freedom system now my different modes will have different time periods ok and they will have different peak responses depending upon the time periods of each mode ok i hope that is clear so i have now a multiple multi degree of freedom system ok in multi degree of freedom system i have decomposed into n single degree of freedom system with different time periods ok and if a response spectra is given i can say that each mode will have different peak response different depending upon the time period let us say t 1 t 2 t n so on ok but they will also occur at different time ok so and if a n is given or like you know displacement is given doesn't matter always remember this expression a n equal to omega n square times t n ok this is a good approximation if damping is smaller than 20 percent ok then this is a good approximation i mean this is not the approximation approximation is this that this is the ut max maximum ok this is a good approximation so you can directly utilize these expressions no need to worry about whether this is exact or not ok now let us look at what is the problem statement here ok so basically we what do we know we know maximum value of u n t ok as u n max for n equal to 1 2 3 ok this i know how because response spectra for that particular ground motion is given to me ok now we what we need actually we need maximum of ut which is the total response ok so basically i need this maximum of summation of all such u n t here ok this is what i need and as i said in the last class this might not be equal to summation of maximum or u n max of individual mode because maximum response might occur at different point of time ok so then what do we do well remember we showed this graph ok if i consider different modes here these are single degree of freedom system ok let us say d 1 this is d 1 remember how is your u and d is related if you remember sorry q and d your q is nothing but this times d n ok so just remember that similarly your a n is omega n square d n ok and d n it becomes if you want to write down q n in terms of a n then you can further write it as omega n square times a n here if the n values directly given ok so i know that my d 1 max occurs as this time this is my d 1 max ok d 2 max occur at this time let us call this d 2 max and let us say d 3 max all the looks like you know the same point occurs at this point so the idea is i know the maximum response of each three mode and like you know all such n modes how do we combine them so that we get approximately the d max or and subsequently the u max ok so as we said there are different methods the method that we would be utilizing is called sr ss method ok square root of sum of squares ok which basically say that d the maximum would be d 1 max square plus d 2 max square and so on ok and there are other type of combination as well but we are not bothered about that ok so let us do an example so basically the same three-story building that we have i am now giving you the value of m 1 m 2 m 3 and k 1 k 2 k 3 ok and t 5 10 to the power 3 kg this is 250 into 10 to the power 3 kg and this is 350 into 10 to the power 3 kg so these are the story masses ok story stiffnesses are 3000 kilo Newton per meter sorry not 3000 30,000 30,000 kilo Newton per meter then this is 20000 kilo Newton per meter and then this is 10000 kilo Newton per meter ok all right now the first step is for you to do here 1 omega 2 omega 3 phi 1 phi 2 and phi 3 ok remember the mass matrix is 35000 what I am going to do because we have to like you know go to the next step as well I am to I am going to give you the frequencies and the mode shapes now just assume that you have come up to this step by doing it and then we will further proceed ok so let me just write down frequencies that you will get is 4.5 to 9.63 14.38 ok and similarly the mode shape I am going to write it here I am normalizing with respect to third story ok so this is 0.3 here 0.64 and 1.0 and this is phi 2 as minus 0.7 minus 0.62 and again 1.0 and this one phi 3 is 2.34 minus 2.62 then 1.0 ok so once you get the mode frequencies and the mode shape remember you have got your single degree of freedom system the next step is to how to relate the response of a single degree of freedom system to multi degree of freedom system and how do we do that through this modal participation vector right this is the vector that relates your single degree of freedom system or establish the relationship between your single degree of freedom system to the multi degree of freedom system for which you actually want to find out the response ok so you already know the expression for this right this is what phi n for seismic excitation this s here would becomes mass times the influence vector ok divided by m n which is basically this and again utilize your calculator to do matrix multiplication as well ok so now what I want you to do give me the value of this factor because I need these so that once I have obtained my single degrees of freedom system I have those factors so that I can relate the response of single degree of freedom system to multi degree of freedom system now many times it would be asked to you or it would be required like you know for you to find out the excitation vector what is your excitation vector here do you remember the excitation vector p effective is what it is nothing but minus m l ok times ug t here ok so what do you need to do find out the expansion of this vector ok so that is the next step ok that vector is what what is your m times l m times l would be simply 350 250 and 175 times 10 to the power 3 ok so this you need to find out what is this vector so let us get the expansion what I need you to do find out s 1 s 2 and s 3 so that this can be written as let us say this is s here as s 1 s 2 s 3 ok how do we get s 1 s 2 s 3 simple you have already found out gamma 1 gamma 2 gamma 3 remember s n is nothing but factor gamma n times 5 times this vector what it is here let us say I write here 10 to the power 5 10 to the power 5 and 10 to the power 5 I just want these elements that's all ok so this is this is what you would get as s 1 s 2 s 3 and what it basically means that we had initially and let me write it like this here this was 1.75 this was 2.5 and this was 3.5 ok the total vector due to applied force now because of this applied force I had inertial force in each mode ok that I can write this is 2.49 2.29 and 1.49 ok the second mode is basically minus 9.0 0.80 and 1.28 and the third mode is basically 0.73 minus 0.58 and then 0.73 ok all of these are multiplied with 10 to the power 5 ok I'm just writing down the internal miracles so this is your s 1 this is your s 2 and this is your s 3 now if you want to follow the procedure of equivalent static force what is the equivalent static force in each mode f n is basically nothing but s n times a and t right so all you need to do find out a and t from q and t ok once you get that all you need to do multiply this with a and t here a and t here n a and t which would be let us a 1 sorry a 1 t a 1 t n a 1 t this a 2 t here and basically this is your equivalent static forces and it becomes so easy to find out the base here in the first mode as or storage here whatever you like some of all these forces ok because you already know similarly in the second mode and the third mode and same goes for the storage here so all you need to do now is find out the a 1 2 a 2 2 and a 3 t ok now remember you know what is your d and t right either for the ground motion or for any other thing this was the equation of motion remember what was the equation of motion d and t plus 2 zeta omega d and t velocity plus d and t and that is equal to for ground excitation it was u g t but it had not been a ground excitation then let us say a blast loading or some triangular pulse or like you know some other type of loading then it would be basically representing that history but as long as you can find out d and t you can find out your q and t as this times d and t ok or this is basically equal to an by omega n square t ok and a and t you can again easily find out if this is your d and t a and t is nothing but whatever d and t you get from there just multiply with the omega square n sorry times d and t just substitute like you know here and you will get the displacement response sorry in this case the force response displacement response and all other responses as a function of the total solve now displacement response you can directly get u as summation of all u and t so phi and t q and t for each mode now q and t for each mode is what q and t for each mode is whatever gamma n times d and t s ok so you can substitute that and get the total response now for the ground motion d and t is not those that easy to find because for ground motion what do we do we do response spectrum analysis so now let me give you the response spectrum ok your response spectrum our design spectrum is a very simple spectrum which is this ok so I am drawing it here this is the pseudo acceleration as unit of g so everything is given in terms of g this is 0.4 this is actually one here ok this coordinate and this is for 5 percent damping ok this t n here 0.125 and this is 0.6 and this is t here ok so now what you need to do here in this case ok or yeah let us just do that for each mode you have omega 1 omega 2 omega 3 you need to find out what is your t 1 t 2 t 3 once you find that out ok it will give you the maximum response that is it will give you d n max ok so it will give you d n max or in this case a 1 max because it is acceleration is given a 1 max can be calculated remember this sorry I forgot to give you this this is basically 0.5 times t divided by t ok so you find out what is your a 1 max a 2 max a 3 max ok 2 max using SRSS method how would you get u max if a 1 max a 2 max a 3 max is known you can find out d 1 max as a 1 max divided by omega 1 square similarly d 2 max as like this or you can directly find out u n max remember u and t is what y n times q n t which I can write it as q n I can write it as gamma n times d n t ok if I write d n t as a n t by omega n square I can write it like this so this is in terms of time now if you want to find out u n max you don't even need to calculate actually d 1 max just do this find out you know gamma n you know omega n square you know phi n just find out a n max for each mode once you know u 1 max u 2 max u 3 max use SRSS as u 1 max square plus u 2 max square plus u 3 max square to get an approximation of the total maximum displacement here I'm telling you here I'm telling you that all three modes have the same damping at 5% ok if it is told otherwise then you can do something else but here I'm telling you that all modes have the same damping value of 5% ok remember when you are finding this all of these are actually vectors right so just keep that in mind in fact this is also a vector this would give you three values here here and here and when you take the SRSS it would be for the first degree of freedom first degree of freedom first degree of freedom then second degree of freedom second degree of freedom second degree of freedom and so on t 1 here comes out to be 1.39 second right now if t 1 is known let us look at where the t 1 is t 1 is somewhere around here right so a max or a 1 max would be basically 0.5 times 1.39 correct and times multiplied by g ok if you want to write it in terms of units so this would be 4.23 meter per second square now if a 1 max is known I can find out d 1 max is required but what I'm going to do I'm going to directly utilize this expression ok what was my gamma 1 if you remember my gamma 1 where is it let me see gamma 1 was wait did I write my gamma 1 here 4266 minus 0.5118 gamma 3 is 0.0892 so it is 1.4266 gamma 1 so just utilize this expression 1.4266 omega 1 square is whatever the value of this one or you can write 2 pi by t or whatever 1.39 I have only in front of me this I'm just utilizing this then 5 1 again just utilize this whatever we had got here 0.3 0.64 and 1.0 okay this is 4.52 here so 0.3 okay 0.64 and 1 and this is 4.52 okay this time a 1 max which is basically 4.23 now it will give me the u 1 1 max u 2 1 max u 3 1 max okay which actually for this case I get that as let me just write it down here okay 88.4 189.4 and 294.6 these are all in mm okay just keep that in mind similarly this is my u 1 max similarly u 2 max I can find out okay that would be u 1 2 max u 2 2 max and u 3 2 max okay some values would be there now to find out similarly u 3 max also to find out the peak response using SRSS the quantities that you need to combine would be this square plus this square plus this square under root so what I'm saying my u max would be under root because remember I have got the maximum at each degree of freedom and this is what the response that I'm going to combine the peak responses it would be u 1 1 max square plus u 2 1 max square plus u 3 1 max square then u 2 1 max square then u 2 2 max square and then u 3 2 max square similarly for third one okay and this will give you at each degree of freedom u 1 max u 2 max u 3 max see it looks complicated but once you like you know go through and like you know the steps and you are aware of it you can quickly do all these calculations