 Let us start our today's lecture on this video course of Geotechnical Earthquake Engineering. Let us look at the slide here. On this video course of Geotechnical Earthquake Engineering for NPTEL, we are going through this module 4 on strong ground motion. Let us do a quick recap what we have learnt in our previous lecture, like we have learnt about what are the amplitude parameters of an earthquake, like it can be acceleration, velocity and displacement record with respect to time, then how to define the peak value of the acceleration we have seen that PHA peak horizontal acceleration and it can have two orthogonal components in north-south direction and east-west direction and the maximum resultant value of the PHA we can take by taking the vector sum of those two orthogonal PHA values. And Trifunik and Bredi in 1975, they have proposed for the design what is the value of PHA with respect to the relationship with the modified Markhali's intensity scale MMI scale and generally the PVA is not that important earlier people used to think about as proposed by Newmark and Hall in 1982, but in recent days people say it is better to estimate PVA also that is peak vertical acceleration and in absence of the data of PVA as per suggestion by Newmark and Hall 1982, PVA can be considered as two-third of the value of that PHA for the design purpose. And peak acceleration data with frequency content and duration is more important, because we have seen that it is not necessary that only that PHA value will be more important, it is also necessary how much is the frequency content and how much is the duration of an earthquake. Then we have seen that relationship between PHA versus MMI and PVA versus MMI. So, this is for the vertical, this is for the horizontal as proposed by Trifunik and Bredi in 1975 corresponding to various modified Markhali intensity scale. So, if you know the MMI value of a particular earthquake at a site using this relationship you can get your value of PHA or PVA which can be further used for the design. Then we talked about the value of PHV that is peak horizontal velocity and we have mentioned that PHV is a better estimation that PHA because of capturing the intermediate frequency also very well, but whereas the peak displacement or peak horizontal displacement and peak vertical displacement though they are associated with the low frequency content, but because of the filtering error that is after filtering out the noises etcetera it will be very difficult to get a reliable results in terms of your peak displacement values. So, that is why in most of our earthquake engineering design either we use PHA that is peak horizontal acceleration or PHV peak vertical acceleration. Then we talked about various other amplitude parameters like sustained maximum acceleration we have seen it can be a three cycle sustained or five cycle sustained depending on your requirement that is how much that highest value of acceleration is continuing through how many numbers of cycles. Then we have seen for the design what is known as effective design acceleration and as per the proposal by Kennedy in 1980 that EDA can be considered as 1.25 times of that three cycle sustained maximum acceleration of PHA and Benjamin and associates proposed that if you filter out all the higher frequencies that is above 9 hertz frequency if you filter out all accelerations then whatever acceleration spectra or acceleration response you get with respect to time that acceleration time history whatever PHA value is gives directly you can use that as your design or effective design acceleration. Then to provide the frequency content parameter we have seen either using Fourier spectra or by power spectra or by response spectra we can find out how much frequency content exist in a particular earthquake motion Fourier spectra is nothing but it is an assumed Fourier series of a randomly behaved earthquake motion that we express in the form of a Fourier series through this periodic function and this value is given as the amplitude and this value is given as the phase angle of whether if you consider it is an acceleration or displacement accordingly this value will correspond to that value of amplitude and phase. We have seen what is known as Fourier amplitude spectra which is nothing but a plot of that amplitude versus frequency and also we have seen Fourier phase spectra which is nothing but a plot of phase versus frequency this is a typical plot of Fourier amplitude spectra Fourier amplitude in log scale in y axis and frequency in the log scale in the x axis there we have found what is the definition of the corner frequency and the cutoff frequency that is the minimum value of the frequency at which that maximum amplitude or maximum value of that Fourier displacement or acceleration whatever you take as a response that will start and cutoff frequency is up to that maximum frequency with up to which it will continue. Then we have seen another mathematical expression of this earthquake motion through power spectra the advantage of this we have discussed that it can consider the random process through using this power spectrum density function like this. Then we have discussed the most widely used frequency content parameter through the response spectra as we have mentioned response spectra can be a acceleration response spectra or displacement response spectra or velocity response spectra that we have seen for various rocky site and soil site acceleration velocity and displacement response spectra how this response spectra is obtained through the single degree of freedom mass spring dashpot model by applying an particular input motion corresponding to a particular exciting frequency and the damping ratio. So, we will derive it today in this lecture. Also we have discussed about the duration in the duration we have seen it is nothing but the time required to release the entire strain energy during an earthquake process whereas the relative duration is nothing but it is the time between that 5 percent to 95 percent release of the energy. And for engineering design we have seen another important duration parameter is nothing but bracketed duration which considers the time duration where earthquake acceleration is more than that minimum value or threshold value of acceleration which causes damage to structure which is 0.05 g. So, that is the time or duration which is known as the bracketed duration with this we had completed our previous lecture. Now, we will continue our today's lecture with the derivation of the response spectra. So, let us go back this is the typical picture of a response spectra how to arrive at the response spectra this is because this is the maximum used frequency content parameter in the earthquake engineering. Let us take a simple mass spring dashpot single degree of freedom model which we have already discussed in our previous video course on soil dynamics. But for clarity let us complete it once again over here. Let us say this is a single degree of freedom system as we know is DOF system with mass m and spring constant k. And U t is nothing but that single degree of freedom which defines it position at any instant of time and P t is nothing but the externally applied dynamic load on the system. Suppose if we have the variation of this P t with respect to time. So, how this P t varies with respect to time let us say it varies like this. Let us take a very simple example first then we can idealize it to a generalized case. Say this is the dynamic load which is acting on the system that is from time 0 to time t d. It is a constant force which is acting on the system is P naught after t d there is no force acting on the system. What does it mean from time 0 to t d it is a forced vibration and beyond time t d it is a free vibration case. So, that is the simple problem which already we have discussed in soil dynamics. So, let us see the solution of it for that we should have the initial conditions be known because that is the basic information should be known to us. Let us say U that is displacement at t equals to 0 say it is 0 and U dot at t equals to 0 that is the velocity at time t equals to 0. Let us say that is also 0 that is I am considering a system which is starting from absolute stationary condition with 0 displacement and 0 velocity at the initial starting of the dynamic time t. Now, as we have already identified it is having two phases of vibration one is forced vibration case up to time t d and then the free vibration case beyond time t d. Let us see how the variation is occurring. So, let us consider for different portion that is for forced vibration phase that means t any time should be less than or equals to that t d because force is acting till the time t d. So, any time if we consider between this range that is nothing but a forced vibration phase. So, what is the solution of this one if we refer to our former course on this soil dynamics we will see that U of t with U s t U s t is nothing but static value of the displacement can be expressed as 1 minus cosine omega n t for a constant force like p naught is acting on the system. So, that was the solution already we have derived in our soil dynamics video course. If you have gone through that you can easily understand I am not again repeating that portion I am trying to derive beyond that what is the response spectra. So, if you are having any doubt you can refer to that course and clarify by yourself. In this case omega n is the natural frequency of the system which is nothing but root over k by m which is nothing but once again 2 pi by t n t n is the natural time period. So, if I express it in terms of t n it will look like 1 minus cosine of 2 pi t by t n and this is valid as I said for a time which is between 0 to t d. So, this is one set of solution as we already know about it. So, now let us see what will be the response in the free vibration phase. So, in the second phase that is for free vibration phase that means the time exceeds that value of that t d. That is nothing but a free vibration because you have taken out the load. What is the equation of motion? You have basic equation of motion m u double dot plus k u equals to 0 from our chosen model. Now, u of t the solution of that t equals to can be written as u at t equals to t d times cosine of omega n t minus t d that is the initial displacement for this case of free vibration. What is the initial displacement for this free vibration? That is the last point of the previous phase or forward vibration phase at t equals to t d is the nothing but the starting point of your free vibration plus the velocity u dot at t equals to t d by omega n times sin of omega n t minus t d. That is the solution as we know for any free vibration phase that initial displacement times cosine of omega n t plus initial velocity by omega n sin of omega n t. In this case t is nothing but beyond that time t d. So, that is why t has been replaced by t minus t d and the initial time has been taken as where that t equals to t d condition was existing. So, at t equals to t d it is the starting time for the free vibration that is the reason as I have already mentioned starting time for free vibration phase. Now, with this what we can rewrite this as a solution let us see over here. So, next we can write that u at t equals to t d will be equals to u of s t 1 minus cosine 2 pi t d by t n times sin of omega n minus sin this is the solution we can write from the forced vibration phase because that was the solution we have seen u of t by u of s t equals to 1 minus cosine 2 pi t by t n. So, I have just placed t equals to t d over there. So, that is the initial condition and initial velocity u dot should be for this second phase of free vibration can be u of s t if I differentiate this omega n sin of omega n t d fine. Therefore, what we can write u of t by u of s t static can be written as the initial value is 1 minus cosine 2 pi t d by t n times cosine of omega n t minus t d. I am writing now the complete solution that u of t for the second phase that is the free vibration phase. I have taken this u s t out in the denominator that is why we can write it like this class sin of omega n t d by t d times sin of omega n t minus t d because omega n was in the denominator. So, it gets cancelled. So, remaining portion is this one I have already taken this out. So, this is the simplified form am I right therefore, if I further simplify it what we I can write it u of t by this u of s t can be written as this is the initial value. This is the trigonometric function as you can see the variation we can simplify it and rewrite is as 2 of sin pi t d by t n times sin of 2 pi times t by t n minus half of t d by t n. So, this complete solution for the second phase is valid for which time. So, this is valid for t greater than equals to t d am I right. So, what we can see in this case that the if you look at the value or the function of that u t by u s t it is a function of this t d by t n ratio that is why I have expressed it in this form t by t n ratio and t d by t n ratio is it clear what I am trying to say. Let me express it in a more simplistic way. Therefore, my final output says that u of t by u of s t that is the dynamic displacement with respect to the static displacement is a function of what parameters that t d by t n ratio and that t by t n ratio. So, that means any dynamic response or dynamic displacement is dependent on not the individual values of this duration up to which the earth is acting or the dynamic load is acting nor it is dependent on the individual value of the total duration or the individual value of the natural time period of the system. But it is dependent on the ratio of this that is how long the dynamic load was acting over the natural period and how long the total time you are considering what does it mean. Suppose earthquake duration was for say 20 seconds and I am interested that after 1 hour of earthquake what is the damage on my structure then this function tells us clearly that in that case in the solution you have to put t d as your 20 seconds but t has to be 1 hour that is 3600 second clear. So, it is clear that the damage we are considering for any structure whether substructure or superstructure anything it is not only a function of your earthquake duration it is a function of your at time where you are interested to but not that individual value it is a related to a ratio of that natural time period of the system. Clear? That is the reason if you look if you have understood this solution clearly this automatically tells us that it is not necessarily that all the earthquake damages are going to occur during the earthquake or soon after the earthquake it may happen that your structural damage or any damage as such due to the earthquake in the soil also may occur after a large time has spent after the stopping of the earthquake also. So that is what I wanted to show through this solution what it says it says that dynamic displacement or the response is not only a function of that ratio of duration of earthquake to the natural period but it is also a dependent parameter on at what time you are interested to know the behavior of your structure. So, we have to be careful about both these two parameters this as well as this ratio. So, once knowing that let us find out what is the maximum response then we can probably be able to appreciate it maximum response of the structure subjected to this case. So, in the forced vibration phase that is the first phase forced vibration phase if we look at at t equals to t n by 2 that means if our time is equals to t n by 2 then what will be the value of that u t by u s t maximum value maximum amplitude of that will be equals to 2 am I right that is if you put it in this solution I will show you once again that is what was my solution for this case if I bring it here yes this was the solution if you look at here within the forced vibration phase so I am talking about within forced vibration phase. So, obvious reason is my t has to be within t d now what I have chosen this t value I have chosen as t n by 2 that is the ratio I have taken as t by t n ratio I have taken as half 0.5 right. So, if it is so if you put that value what it is becoming it is becoming minus of cos pi so plus 1 so 1 plus 1 2 that is the possible maximum value of this solution right. So, that is why I have said it gives us the maximum value of this u t by u s t max is 2 at this point. Now, if your value of this t d is greater than this t n by 2 then what happens then we have to see two solutions what are those we will look at here. So, let us see mention this as a response let us identify this as a response u t by u s t that is with respect to static so if I want to show the response r d let us say response which is nothing but u t by u of s t I am defining it can be expressed by two solutions 1 minus cosine of 2 pi t d by t n and 2 these are the maximum values this is applicable when your t d by t n ratio is less than equals to half and this is applicable when t d by t n ratio is greater than equals to half am I right. So, that is what we said just now that maximum response is very much dependent on this ratio as you can see if this ratio is less than equals to half then your maximum response will be computed like this based on whatever value of cos you are getting that will be coming. But if this ratio this ratio is more than or equals to 0.5 in that case always your maximum response will be two times that is the dynamic response will be two times of your static clear. Now, let us see in the another phase that is the free vibration phase. So, for second phase that is for free vibration phase what is the maximum value of this u t. So, if I write down that u t of max that will be nothing but your amplitude whatever you are getting at t equals to t d that square plus that u dot t equals to t d by omega n that one square am I right are you satisfied with this expression let us look at here. So, as I said the solution of this will be in the form of for free vibration phase I brought it this here u of t is expressed as initial displacement times cosine of omega n t plus initial velocity by omega n times sin of this. So, what can be the maximum value of this u t obviously it will be root over this amplitude square plus this amplitude square right. So, that is why u initial square plus this velocity by omega n this square that will give us the maximum value of this entire solution is not it. So, u of t maximum that is why I said can be written as this. Now, if I simplified. So, u of t this max u of t max can be written as on simplification by putting the known expression 2 of u s t times mod of sin pi t d by t n. Where from I got it let us look at the solution of this then we will able to understand. Yes here was the solution if you remember this one if you see this was the solution for free vibration phase at the end right. So, what is the maximum value of this solution this portion this amplitude remaining is sin. So, if sin is 1 then it will be maximum am I right. So, if it is maximum. So, 2 times u s t sin of pi t d by t n that is what I have written over here and it can be amplitude either minus or plus. So, with this absolute response will be let me write it down once again. Therefore, the absolute maximum response that is r d what we are expressing that can be written as 2 times of that sin pi t d by t n this is for the free vibration phase. Therefore, if I write it in a combined form I can write it as 2 times mod sin pi t d by t n 2 solutions. So, this is the solutions actually this is one case another case is 2 when this case is valid when this t d by t n ratio is less than equals to half and this condition when t d by t n ratio is greater than equals to half is it clear see this is the complete solution what is the maximum value of this again when this sin parameter is 1 am I right that will give us that value of 2 maximum value when it will occur that will occur when this ratio t d by t n is more than equals to half. Otherwise, if it is less than equals to half it has to be 2 times of whatever the sin value gives us. So, this is the response in free vibration zone and in the forced vibration zone if you look at this was the maximum response. So, everywhere it is a function of that t d by t n and that t by n concept will come to you only when you are interested about a particular time when you want to study the effect on your structure as I said after 1 hour after 30 minutes after 10 minutes like that. So, that is the reason if you see over here here I have not considered the damping portion, but if you consider the damping portion if you remember my soil dynamics lectures the typical value of this plot the response spectra which we show that t n and say s a by g it can be acceleration response spectra it can be displacement response spectra it can be velocity response spectra this looks like this for a particular value of your damping ratio say 5 percent damping or something like that for a particular damping ratio this behavior we can obtain and if I want to plot this u t max by s for our case how it should look like let us see. So, I am now plotting this value of u r d versus t d by t n. So, this is the variation of t d by t n with respect to r d r d as I have defined. So, u t by u s t maximum of that I am plotting. So, as I said over there it is 0. So, this is 1. So, this is 2 this is 0 let us say this is half this is 1 then this is 2 then this is 3 like that. So, at half if I draw over here the variation will be something like. So, maximum response it occurs at the t d by t n ratio of this one when it is subjected to a that constant force type vibration for a finite time period, but if it is a random similar process has to be used to get the solution and get this response like this. So, that similar procedure we adopt for the earthquake random motions also which probably you have to solve using the Duhamel's integral process which I have discussed thoroughly in the soil dynamics course for any kind of motion, but everything finally will be a function of this how long it is occurring that earthquake motion with respect to the natural time period and at what point of time you are interested to find out the value or response. So, another parameter I want to highlight over here. Let us look at here like it is called the pseudo value of acceleration response spectra and pseudo velocity response spectra and pseudo displacement how it is defined. Let us say this is a typical structure of height h which is subjected to this force F s t over here. I have a base shear here B naught t let us say and a moment resisting moment let us say M naught t. For this structure and let us denote that U naught as the maximum value of that U of t the function with whatever it is subjected to any kind of dynamic load that F s t is nothing but a spring force acting on the structure with k as the spring constant or the stiffness times U of t which if we simplify it further it can be expressed as M omega n square times U of t why because as we know omega n is nothing but root over k by M. So, automatically k is nothing but M omega n square is not it. So, in this case we call it as M times A of t what is then A of t if you look at here A of t is nothing but as per our written expression A of t is omega n square times U of t. So, this A of t is known as pseudo displacement. So, this is pseudo displacement this A of t. So, A of t can be written as omega n square U of t this A of t is called pseudo acceleration pseudo acceleration response spectra. So, this is a if we plot this value of U naught what is U naught as I said it is a U t max. If you plot this U naught versus this t n that is known as displacement response spectra and if you differentiate it further this U naught with respect to time what you get over here that U naught dot which is nothing but maximum of this U dot t versus that t n that plot will give us velocity response spectra. If you further differentiate it with respect to time U naught double dot is nothing but maximum of this U double dot t versus the plot of T n that gives us the acceleration response spectra. So, for the chosen solution of A of t what we have seen. So, pseudo velocity response spectra can be written as A of t. Can be obtained as let us say v in the unit will be velocity unit semitra per second will be omega n times d or omega n times displacement 2 pi by T n times d and pseudo acceleration response spectra will be A in terms of meter per second square unit that will be omega n square. So, what I have written over here earlier if I correlate you can see over here this is known as pseudo acceleration which is nothing but omega n square natural frequency square times your velocity your displacement profile with respect to time omega n square times U of t. So, that will give you the pseudo acceleration response spectra. So, like that we can draw any response spectra whether it is a pseudo acceleration response spectra or pseudo velocity response spectra or pseudo displacement response spectra. Now, let us come back to our topic now, let us start some more parameters like other spectral parameters what are the others various spectral parameters which generally are used one is called r m s acceleration nothing but root of r m s. Now, let us come back to our topic. Now, let us start some more parameters like other spectral parameters what are the others various spectral parameters which generally are used one is called r m s acceleration nothing but root mean square acceleration. So, this is the parameter that includes the effects of amplitude as well as frequency. So, this acceleration parameter root mean square acceleration it consider the both effect of the amplitude as well as the frequency it is expressed in this form A of r m s equals to root over 1 by t d integrate 0 to t d A of t whole square times d t A of t is nothing but your actual variation of acceleration with respect to time. If you integrate over the time period that t d and then find out the root mean square this is nothing but the you are taking the mean by dividing it and root you are taking. So, root mean square acceleration you will get by this over the time domain and t d is the duration of the strong motion. And A i is another important spectral parameter which are commonly used for design A i is nothing but full form is areas intensity. This areas intensity is a measure of the total energy at the recording station and is proportional to the sum of the squared acceleration. So, it is defined as areas intensity is pi by 2 g integrate 0 to infinity A of t whole square d t. So, that acceleration response with respect to time whatever you have during an earthquake if you integrate it over 0 to infinity by taking a square of that and then multiplied it by pi by 2 g that estimate the total energy which is measured during a recording station which is known as areas intensity. Another spectral parameter is spectrum intensity spectrum intensity or A i. A i is defined as the integral of the pseudo spectral velocity curve just now I have discussed what is pseudo spectral velocity curve which is known as the velocity response spectrum. So, if you integrate that between that time period that is if you have that plot of the pseudo velocity spectrum versus t n between 0.1 to 2.5 seconds between that whatever you get if you integrate that curve these quantities are motivated by the need to examine the response of structure to ground motion as many structures have fundamental periods between 0.1 to 2.5 seconds that is the reason why the spectrum intensity it is calculated for any structural damping ratio to determine their response or determine their damages is it clear. So, this is the typical range of the fundamental period of any structure of our civil engineering structure it will be within 0.1 second to 2.5 seconds clear. And in the reverse way we can see considering the structural period or fundamental period of a structure if you take the in hards how much it is coming about 1 by 0.1 to 10 hards and this is 1 by 2.5. So, it will be inverse of 2.5 that much hards will be the typical range of the natural or fundamental frequency of the system. So, if your earthquake frequency also comes within that then there you have to take care of the structural damages etcetera that is the reason you have to integrate that velocity pseudo velocity response spectra curve between this time period of 0.1 to 2.5. Another spectral parameter is dominant frequency of ground motion f d what is f d it is defined as the frequency corresponding to the peak value in the amplitude spectrum that is whatever amplitude spectrum you have in that whatever is the frequency which corresponding to its highest value or peak value that will be your dominant frequency. Thus f d indicates the frequency for which the ground motion has the most energy because it is giving you the maximum amplitude. The amplitude spectrum has to be smoothened before determining the f d which is quite obvious as we have already understand we have to smoothen it filtering it etcetera. Other parameters like predominant period what is predominant period or t p period of vibration corresponding to the maximum value of the Fourier amplitude spectrum that is you have to express the amplitude response in terms of Fourier series. So, as we have already seen Fourier amplitude versus the period if you plot it like this the something it will like look like this is for two different ground motion this is for ground motion 1 this is for ground motion 2. Can you see for both of them the predominant period is same t p is same that is where it is occurring the maximum value of the Fourier amplitude. So, this parameter represents the frequency content of the motion the predominant period for two different ground motions with different frequency content can be same making the estimation of frequency content very crude. That means suppose for this one if you plot it in the logarithmic scale whatever is your frequency content and if you plot this in logarithmic scale whatever is the frequency content they are quite different. But, their predominant period where the maximum value occurs is same that is what you have to use the combinations of this various spectral parameters to identify the nature of any earthquake motion not only in terms of maximum amplitude, but also in terms of frequency content. What is bandwidth bandwidth or b w is of the dominant frequency measured where the amplitude falls to 0.707 which is nothing but 1 by root 2 of the amplitude of the dominant frequency again this is based on the smooth amplitude spectrum. Central frequency what is central frequency it is a power spectral density function that p s d what we have already discussed within that can be used to estimate the central frequency of the ground motion. The n th central moment and the central frequency can be estimated like this this lambda n is the n th frequency that can be estimated as 0 to omega n omega to the power n g of omega that is the power spectral density function times d w. So, and that central frequency is defined as this second one spectral moment divided by the 0 h 1 that is lambda 2 by lambda naught root of that will give you the central frequency this is another just estimate of a spectral parameter. So, central frequency is used to calculate the theoretical median peak acceleration using this expression this value of this omega capital omega you can use for that to calculate the u max value in this fashion. And the shape factor it is indicating the dispersion of the power spectral density function that p s d about that central frequency how it varies with respect to that central frequency that delta can be obtained using this expression that is the lambda 1 the first spectral moment square divided by 0 h 1 and the second one. Then it lies between the value of 0 and 1 higher value indicates the larger amount of bandwidth the or higher value of the spectral. Another very important spectral parameter is nothing but v max by m x ratio this is important what is v max v max is the maximum velocity. If you consider the simple harmonic motion is applied to a single degree of freedom system like structure what I have already discussed about the derivation etcetera. And a max is the corresponding maximum amplitude that is from your u t max if you find out the velocity max for a single harmonic motion it will be how much omega times u t right is not it that is if I want to express suppose my u of t is having some expression say u sin of say omega times t. Then u t maximum of that is nothing but u then what is the v max v max is nothing but omega times u mod of that am I right because velocity you can find out by differentiating it with respect to time this displacement and maximum value of that will be this will become cosine omega will come over here. So, omega times that u and what will be your a max omega square times u mod of that am I right. So, that says in this slide whatever I have discussed just now look at here. So, v max by a max ratio is related to the frequency content of the motion for simple harmonic motion with a period t this v max by a max can be represented as t by 2 pi is not it. If I look at here let us see here. So, v max by a max for this simple harmonic motion will be nothing but how much v max by a max will be 1 by omega which is nothing but t by 2 pi right that is what it is shown over here you can see here in the slide v max by a max is nothing but t by 2 pi. Now seed and idriss in 1982 proposed average values of that v max by a max for different sites within 50 kilometer of the source that is they have analyze the values of v max by a max between the earthquake epicenter to a point of site of response for different types of site that is rocky site soil site etcetera. And what are the typical values they have mentioned the average values for rock they have mentioned this value of v max by a max ratio. And remember what will be the unit of this it will be the unit of this t that is in second is not it. So, that is why for rocky site they have mentioned the typical value is 0.056 second for stiff soil stiff soil within the 200 feet that is a stiff soil available at relatively shallower depth the value is 0.112 second and deep stiff soil that is the stiff soil is available beyond 200 feet in that case the value of this t by 2 pi or v max by a max ratio is 0.138 second. So, what we can see this value of v max by a max ratio keep on increasing if we go from a stiffer to a softer soil that is what it means. So, this gives us a very good estimation, but remember these are the typical average value the actual value can differ and how this value can be used to estimate for estimation of the magnitude of another location or computation of epicentral distance that we will see through an example problem in the next class. So, with this we have come to the end of today's lecture we will continue further in the next lecture.