 Hello everyone, welcome to the lecture today. In previous chapters in this course, we have seen that how to obtain analytical response of a single degree of freedom system subject to either for the case of free vibration or when the loading is harmonic or pulse type loading or step forces, but we will encounter many cases in which the loading cannot be represented as a closed form solution. And even if the loading can be represented as a closed form representation, the solution cannot be obtained as a closed form analytical solution. So in that case, we have to rely on numerical methods to find out the response. And we are going to learn few of those numerical methods and see how to utilize those numerical methods to find out the response of a single degree of freedom system. Now, because this course only focuses on single degree only on structural dynamics and not numerical methods, our discussion would be brief. And we are going to discuss one explicit method, one implicit method and feel free to explore more because there is a vast literature available on this topic. But I am just going to show you illustration or the examples of those methods and it will give you an idea or a platform that you can use to further explore this topic. So, let us get started. In today's class, we are going to see how to get the response of a system, a single degree of freedom system using numerical methods. So, if we consider our equation of motion that we have been dealing so far is actually can be written like this. And in the previous chapters, we have seen that p t which is also the excitation function, how to get the response u t for a given excitation p t and basically we got this response for different kind of situation. So, we got this where only the initial conditions were provided and no force was acting. So, that was pre-vibration. So, the motion was initiated by initial condition and then we saw cases in which we had a force acting on the system. So, in forced vibration basically we saw different kind of forces. First we considered harmonic excitations, harmonic excitations in which the force can be represented as function of sin or cos. And then after that we also saw how to get response subject to different kind of arbitrary excitation, but for which we could define the excitation as a function of some time variable. So, for example, step excitation or ramp or different kind of pulse loadings we saw how to get the response. So, half sin pulse or the triangular loading. Now for all these cases, we were able to get u analytically as a closed form solution of some function here. Now it might happen that we can encounter cases in which PT cannot be represented as some function or even if it can be represented as some closed form function. It is not easy to derive the solution for UT analytically. And for those situation it is imperative that we study methods through which we can get the response even the analytical solutions are not available ok. So, that is where the numerical response evaluation comes into picture ok. So, numerical response evaluation would be utilized to get the response of the single degree of freedom system or we are dealing here only with single degree of freedom system, but remember this can also be extended to multi degree of freedom system ok. So, we would get the numerical response using different methods and we are going to talk about those methods ok. So, let us consider PT ok which is some arbitrary function cannot be right as a closed form solution or even if we consider these cases here many type many times these loadings that are represented as a nice you know a smooth function of f are actually simplification of the actual loading. For example, a triangular loading is a simplification of a blast loading for which the loading actually like you know varies like this. Because it is not actually a triangular loading for, but for many purposes assuming or approximating the blast loading as a triangular loading gives us comparable results for you know depending upon the response quantities of interest ok. So, the question becomes how to get the response if PT ok. So, let me just draw PT here ok. So, PT let us say is given ok some random arbitrary variation ok something like that. Now subject to that PT we will have some response right ok. So, let us say the response looks like this ok. Let us say this is ut. So, the idea is how to get from this point to this point ok. So, our problem statement here is that if we have been given the equation of motion here ok if equation of motion is something like this ok. Now what I am going to write down my loading or the spring force as some function of u and u dot ok. Just to emphasize that when we use the numerical methods the utility for these kind of methods for the applications are not limited to linear systems ok. It can be extended to non-linear system as well ok. So, in that case my FS what we have been assuming as k times u ok it need not to be like that ok. So, we can consider non-linear system like bi-linear or hysteretic systems as well ok. So, just giving you a like a what are the potential application. However, for the purpose of this course we would limit our discussion to only linear systems ok, but numerical methods can be utilized to get the response for any other type of system as well ok. So, let us say this is a very question of motion ok. So, p t could be either an external force or it could also be ground excitation or support excitation in which case it can be written as s times the acceleration of the ground or the support ok. And these initial conditions are given ok. These initial conditions are given basically u equal to let us say u 0 and u dot 0 is equal to u dot of 0 ok. So, this is our problem statement. So, let us first discuss the fundamental principle of numerical response evaluation how do we do it. So, for numerical response evaluation what do we do? The given function if it is given in a discretized form ok then it is fine. Otherwise even if it is given in some continuous form ok remember as I said this is like you know more useful for complicated functions ok, but it can be utilized for other in any type of function it is not restricted to a particular set of function. So, what do we do? We represent our force in discrete values ok separated by simple time steps ok. So, let us call this p 0 here p 1 p 2 and so on at time t equal to 0 ok t 1 t 2 and so on. And at any time let us say it is step i ok. So, the time is t i and the value is p i let us say after delta t it is t i plus 1 and the force is p i plus 1. So, p t actually we represent it as a set of discrete values ok p 1 p 2 p i p i plus 1 and then p n ok. And our displacement need to be determined at the same instances of time. So, basically u 1 u 2 u i u i plus 1 let us say u n remember all of these are our time instances t 1 t 2 t i t i plus 1 t n ok. So, we discretized our excitation ok and then we get the response as discrete values of the response at those points. So, we do not get it as a function or a closed form solution, but for each time instance say t 1 t 2 we get a value of u 1 u 2 ok at those instances of time. So, let us say this is t 1 0 t 1 t 2 this would be u 1 u 2 and so on let us say at t i and t i plus 1 I get as t i t i plus 1 I get my responses is u i and u i plus 1 ok. So, this is our goal here that how from a discretized points that represents the excitation function how from that we can get our response like this and the way we do it we employ numerical methods to calculate the response. So, using these numerical methods we can get the response like this ok. And basically our idea is that if we knew the state of a system at time t equal to t i. So, let us write down our equation of motion at time t equal to t i ok it would be c u i and then k u i is equal to p i ok. If I have somehow found out the values of this parameter ok. So, the response parameter acceleration velocity and displacement and force p i is already given ok. We want to go from this step to the next step here ok. So, you want to go to the next step here alright and this is basically called time stepping ok. So, numerical methods are also many times referred to time stepping methods ok and many places they are also called as integrators ok. So, keep that in mind ok. So, we want to go from this step to this step. So, if somehow we can devise some numerical method which could assist us in getting the state of system from t equal to t i to t equal to t i plus 1 then I can get the response values at all instances of time ok. And if I can get my numerical response then I can go ahead get the peak response and get the other response quantities as well. So, basically I would get set of data points representing my response here ok and this is whatever goal is ok. So, through these chapters we are going to be discussing methods ok through which we can go from state t equal to t i state t equal to t i plus 1 and through that we are going to obtain the response. Now, the question comes why there are methods ok and not a single method ok. If a single method is there that can do the task then you know why do I need multiple numerical methods? Well, it depends on many criteria ok. Some of those criterias are basically when we get a numerical solution unlike an analytical solution ok which is a closed form solution of a dependent variable a numerical solution might or might not be accurate it might not be stable or it might not be convergent and it depends on many parameters that leads to those criteria being satisfied. So, the reason for having many methods some of those reasons I am listing down here. So, it depends on these criteria stability, convergence ok and accuracy of the numerical response ok. So, let us see what does it mean basically stability means that remember in all these methods we basically go from one step to next step ok and then we use a time step delta t i which is t i plus 1 minus t i ok. Now, a numerical solution is stable when decreasing the time step actually gives me a value that is going to a certain value it is it might or might not be accurate ok. So, the numerical solution should be stable ok in the presence of any kind of round off errors or anything like that ok the solution should not be fluctuating ok. So, if my actual solution let us say analytically have it obtained the actual solution is like this ok. Stability means that there might be inaccuracy like this however the solution is still stable ok. However, if I get a solution like this ok that means the solution is not stable ok. So, stability means that it is basically giving me a realistic value not fluctuating ok. Now, second thing is convergence convergence means that when I decrease the time step delta t i then my solution should approach to a particular value at each and every time instant ok. So, that would convergence means for example, if I decrease my time step then it should move towards a fixed value. So, if I decrease my time step my solution u should move towards a fixed value ok which might or might not be an accurate value, but it should at least like you know converging towards that value ok. And that is why we next we have accuracy if the value towards it is converging ok is equal to the analytical solution ok. It is an exact value then we say that my solution is actually accurate ok. So, in terms of basically numerical solution these three conditions might not always be satisfied ok. So, that is why we require ok different numerical methods which can provide appropriate stability convergence and accuracy for different kind of or different scenarios ok. And you will see that some methods numerical methods work better in certain situation and certain other numerical methods work better in some other situations ok. So, this is from the point of view of stability convergence and accuracy, but there is one also other point which plays a big role and that is computational cost ok. So, remember each of these numerical solutions when they are determined ok computer has to perform certain number of operations to obtain the solution ok. And how efficient a computational solution is it depends on the algorithm for that particular method ok. So, some method might be computationally very efficient ok and it might be utilized in certain scenarios ok, but we might have to give up on certain other features like accuracy ok. So, that is why different numerical methods are utilized so that computational time is manageable ok with some expectation on the accuracy of the solution ok. So, these four criteria determines which method or which numerical method to evaluate the response are better ok alright. If that is clear let us move to our first numerical method which is interpolation excitation method ok interpolation excitation method. Now in an interpolation excitation method adds as the name suggests if I have the system like this here ok what I do I take the function f t or p t ok and then let us say this is the time variation and this is time t equal to t i ok. So, let us say this is state i and this is state i plus 1 ok I draw a straight line between two two states let us say this is p i and this is p i plus 1 ok. So, we discretized our excitation in the interpolation excitation method ok. So, if our system is linear ok this only works for linear system and this is popular because it is computationally very efficient. So, if the system is linear I do not have to go and use like you know complicated numerical methods ok. If my only issue is that my p t cannot be represented in terms of closed form of function or subject to p t my u value cannot be determined ok as an analytical solution then I can utilize this interpolation excitation method for linear systems ok and then it is a very computationally efficient method. So, let us say what do we do in this method given the state i ok I interpolate my function excitation function between p i and p i plus 1. So, let us say I define a time variable which starts at states i ok and goes up to i plus 1. So, that tau here represents ok the time variable between this and this ok. Now, this total time step is delta t i that is already known to be ok. So, p tau which is at any time tau after the step after the time step i can be written as ok p i plus delta p i divided by delta t i times tau this is the interpolation of the excitation function ok. So, my equation of motion that I need to solve becomes ok remember delta p i here is nothing but p i plus 1 minus p i and my delta t i is t i plus 1 minus t i. So, my equation of motion becomes as mu double dot plus k u that should be equal to p i plus delta p i divided by delta t i tau ok and I am first considering undamped system, but same procedure can be extended to damped system as well ok. So, this is my equation of motion subject to initial condition remember my variable tau is starting at time i ok. So, the initial conditions are basically conditions given at u of i ok and u 0 is u of dot velocity is at i again ok. So, the initial conditions are provided at the start of this variable tau ok. So, the total response can I say remember I am trying to find out response subject to this is my excitation function. So, this can be represented as response due to initial condition plus response due to this plus response due to let me just rewrite it again ok. So, it is the sum of response due to this plus. So, a step function then a ramp function and plus the initial condition ok that was acquired up to the step i. This is very important you have to remember my system is not starting from rest there would have been some initial condition up to step i. So, I need to include that ok. So, the total would be sum of response due to all there is this solution and it can be written as u tau ok as equal to first the initial ok free vibration due to initial condition I can write it as u i cos omega n tau ok. There is velocity term here ok sin omega n tau and then the response due to the step function here which we have already derived is given as the value of this one which is ok delta p i divided by k tau delta t i minus sin omega n tau divided by omega n t i ok. This is due to ramp function ok this is due to the loading due to this linearly increasing phase due to a step function let me write it it would be p i by k 1 minus cos omega n tau ok. So, the total displacement can be represented as sum of displacement for each of this combination and I could do that because it is a linear system. So, the principle of superposition is valid. Now, I can go ahead and I can differentiate this ok and I can get certain expression ok. The next thing that I need to do remember what I want given the state at i I want to go to i plus 1. So, what do I do? I substitute my time variable tau equal to del t i ok when tau becomes del t i remember if you look at here when tau becomes del t i I basically reach to step i plus 1. So, u of tau becomes u i plus 1 ok u dot of tau becomes u dot i plus 1. So, that is I am going to do in the expressions that I have obtained here ok that I will get my u i as some function here and u i plus 1 as some function here which would be in the ok in terms of known parameters ok of i known parameters in terms of ok previous steps ok. And then so remember this would be u i plus 1 this would be u dot i displacement and velocity at step i plus 1. Now, these can be arranged ok these can be arranged and the solution can be written in this form here some constant times u i plus another constant times ok and you can actually derive that and see what are those constants ok it is not very difficult to do ok. I am not deriving it here so that you know not to get walked down by with all the equation and mathematics here the important thing is to understand the concept here ok. So, it would come like something like this ok u i at the previous step velocity at the previous step force at the previous step and force at the current step, but remember I already know the force at the I already know what is the force at the current step. In fact, the excitation is given to me which according to this problem ok. Then my force values is it is like they are known at all the time steps ok. Similarly, my velocity would be another constant which is different from this ok in terms of u i plus u i dot and then p i and then again another constant times p i plus 1 ok. So, although we have derived this for the undamped system the general form of this equation is going to remain same even for damped system ok. So, both this is the equation only thing that is going to differ that these constants are going to be different and you can refer to any textbooks to have a look at what these constants are. These constants basically depend on the structural property of the system. So, it would depend on the time period frequencies are damping and also the time step of the system ok. It would not depend on any other parameter ok. So, these parameter these constants would need to be calculated only once ok at the beginning ok at the beginning of when the motion starts ok with the initial condition of course that has given to us and once that is known I can utilize I can utilize these expressions to go from step i to i plus 1 and then I will keep on repeating that. Now, know that we have obtained displacement and velocity here acceleration we can add a differentiate and obtain it or if you consider the equation of motion as this ok you can get the acceleration at step i plus 1 once the displacement and velocities are known to us ok. So, if these are known to us which in fact they are I can get remember these quantities are known that we can also get the acceleration ok. So, generally what do we do we program it in some sort of mathematical programming software. So, you can use you can use like you know matlab you could utilize even excel or you can utilize like you know python or any other language of your choice ok and you can put this inside a four loop a loop that actually goes you know from over a certain values certain iteration and then you can get the response. So, basically given it will start with some initial condition and then basically this is the response you will get you will get the discretized points ok and how closely these points are separated ok those depends on the time step that we choose ok. So, we have considered a constant time step but remember it does not have to be even that if the time step is changing you can still find out utilizing the same method. Only thing is that in that case all these constants would need to be calculated at each and every step. So, this would be changing because if your delta t is changing then these function these constants would be changing ok. So, this is called interpolation excitation method ok, interpolation excitation ok or other way around sorry, excitation interpolation method ok which is useful for linear system ok because it is computationally efficient and works quite well for the linear system ok. But you know I mean that might not be sufficient because in reality we would encounter many systems that are non-linear and I would want to have a method that could be used for linear as well as non-linear system. So, from that part perspective we are going to study two additional numerical methods that can be universally utilized whether the system is linear or non-linear ok. The first of these methods that we are going to discuss is called central difference method ok it is called central difference method ok. Now central difference method ok it is basically based on the finite difference approximation of the time derivatives of the displacement velocity and acceleration ok. Now you might already be aware that given displacement value let us say u i ok and u i plus 1 if I have to find out and let us say these are separated ok these are at time ti and ti plus 1. So, separated by delta ti which is equal to ti plus 1 minus ti. So, if these points are given this velocity is basically the slope of the line joining these two points ok. So, if you want to approximate the velocity ok approximately calculate the velocity it will be simply slope of the line joining these two points. So, I can write down my u i dot as u i plus 1 minus u i divided by delta ti ok and acceleration would again be ok acceleration can written can be written as ok let us call this y plus 1 it can be written as velocity ok divided by delta ti all right. So, this is the concept that we extend in central difference method ok. So, what do we do in central difference method ok we first consider velocity ok joining the displacements which are apart by two delta ti. So, instead of just consecutive points I can also say that I would write my velocity u i as u i plus 1 then u i minus 1 and now remember these points are now separated by two delta ti ok because there are two steps in between. So, the slope joining these two would be two delta ti and the reason why I am considering like you know two steps apart it would clear when I write the acceleration remember by default if I try to express my acceleration here ok as u i dot and then let us say or let me write this as u i as ok. So, this is the acceleration I am trying to find out which would be derivative of velocity ok let us say it is this same expressions that I have just used above ok. So, I can further substitute my u i dot from the expression here ok and I can write this as ok I can go ahead and I would write that expression as u i plus 1 minus u i divided by delta ti and minus u i minus u i minus 1 divided by delta ti and this whole thing would be multiplied with 1 by delta ti which is the denominator here ok. So, we can arrange it and we can write down our acceleration as u i plus 1 minus 2 times u i plus 2 u i minus 1 divided by delta t square ok. So, what I have done here ok I have written down my velocity and acceleration which are time derivatives of the displacement ok I have written down as simply as an expression of displacement ok. Why is it necessary? Well if I have that then the differential equation that I have which is this ok the differential equation that I have ok. Although derivative terms can be removed if I substitute this and we written in terms of only displacements ok and that becomes very useful to us why we will see now ok. So, I am going to write this here as u i plus 1 minus 2 u i plus 2 u i minus 1 divided by delta t square this is my acceleration that is yes and u i minus 1 times 2 delta t then plus k u i and this is p i ok. So, the way I want to arrange this equation so that I can write it as some constant times u i plus 1 is equal to some functions which are known on the right hand side. So, that if this coefficient is known and the functions on the right hand sides are known then I can find out this value ok and which basically means given the state of the system at i I can find out the state of system at i plus 1 because once the u i plus 1 is known then I can find out u i dot plus 1 and then u dot i acceleration using the expressions that we have obtained here ok. So, basically that is our goal ok from this expression how to get that. So, if you rearrange this what you will see this can be written something like m divided by delta t square plus c divided by 2 times delta t ok and this u i plus 1 should be equal to p i minus delta t square minus c by 2 delta t and then it is multiplied with u i minus 1 and then I have the second term here or say third term here which is this. So, if you rearrange this equation ok that we have we will get this. Now, what I am going to do I am going to write this as another expression ok which I will denote as ok k hat it basically some represents some sort of stiffness or the equivalent stiffness and this is equal to p i hat ok. Now, if you look at these carefully k hat depends on the mass and the damping on the system and whatever the time step that you choose ok and if the time step is constant this k needs to be calculated only once not at each and every step. So, if it is a linear system ok and the time step is constant this k needs to be calculated only once. Similarly, on the right hand side p i is known to me from the previous step and u i minus 1 and u i they are also known to me from the previous steps and the other things are basically system properties that can be determined using the provided values. So, I get my equation for the differential equation in this form ok. So, that I can find out my u i plus 1 as p i divided by k hat ok. So, in this one as well given the state of the system at previous steps I have been able to find out the response at the next step ok. Now, the initial condition would again be the same ok it would be u 0 and u dot 0. Now, there is one issue here though ok if I consider this equation if you look at it here ok at i equal to 1 ok let us say if you consider in this expression ok at i equal to 1 or let us say i equal to 0 ok. So, that I am calculating this k hat times u 1 I am calculating the first step ok. So, with i equal to 0 this would become k hat times u 1 equal to p of 0 and what would be the p hat sorry p hat 0 what would be the p hat 0 it would be p 0 minus this coefficient this is minus u minus 1 here and this is u 0. Now, u 0 is known to me as an initial condition p 0 is known to me ok that force at time t equal to 0 ok. The question becomes how do I get this because this is not known to me ok. So, to get q minus 1 consider it to be another constant that need to be determined ok. And the way we find out using the initial condition when we where we write down u 0 is equal to u 1 minus u minus 1 I am using the same expressions that we have used for velocity and acceleration ok. So, I am going to use the same expression here alright to delta t and then u 0 double dot is equal to u 1 minus 2 u 0 u minus 1 divided by delta t square. Now, all the terms are known in this one except ok u 1 and u minus 1. So, we can solve this and find out u minus 1 as ok because there are two equations simultaneous equation I can solve it and I can this get u minus 1 as u 0 minus delta t times u dot 0 plus delta t square by 2 u dot 0 ok. And then my solution becomes complete in the sense that I can now utilize this to start my iterations and get the response. Now, remember how accurate the response would. So, let us say I have found out analytically that response looks like something like this ok. Now, whether my numerical response how close it would be to the actual response ok. It depends on many parameters and one of those parameters are time steps that we use ok. So, it can go something like this ok it can follow like this it is always discretized you know is it not always discretized ok. So, how accurate it is or how close it is to actual solution one of those parameters that determine this is actually the time step ok. And central difference method is stable if it is satisfy this criteria. So, if you have a single degree of freedom system ok my delta t by t n should be smaller than 1 by pi ok. If this condition is satisfied then my solution would be stable ok if I determine using the central difference method ok. And you know this condition basically comes from you know some stress velocity criteria ok which is beyond the scope of this course. But basically some condition need to be satisfied for this method to be stable ok. Which is typically not a problem for the like you know some general type of structures that are there in structural engineering because we typically use time steps which are of order 0.01 or 0.005 like that ok which would always satisfy this condition because time period of a typical structure can be between 0.2 second to let us say 2 second or 3 second like that ok. All the time periods in this zone this condition is satisfied very easily because it would require delta t to be ok between 0.06 ok to the value 2 by ok. So, it would be around 0.66 let us say ok and that is not a problem ok. Because typically the value that we use are satisfied but remember the time period is for a particular mode and here we are only considering single mode because it is a single degree of freedom system. But for multi degree of freedom system ok many modes could be of interest let us say 10th mode and as the mode increase and you will see in the later chapter the time period actually decreases. So, depending upon your mode of interest this condition might not be satisfied. So, it might be problematic for multi-degree freedom system ok. So, that need to be kept in mind all right. So, this was the discussion on central difference method ok which is perhaps one of the easiest method that could be universally utilized for a linear and non-linear ok systems altogether. The next type of method ok that we will discuss ok is called a Newmark method ok. So, Newmark method is another set of numerical procedure used to find out the response of a differential or response of a basic dynamic system ok. Now, Newmark method as we discussed in the central difference method as well the idea is to find out variation of U i or the response parameter at i plus 1 ok ok in some sort of known functions ok. So, that is the goal once we can find that out then I can go from step i to i plus 1. Now, what Newmark method does it assumes ok Newmark methods assumes certain variation certain variations of acceleration ok between step i and i plus 1 step i and i plus 1 and depending upon those variations I get the variation of velocity and displacement ok. So, let us see what those variations are. So, in the first type of Newmark method basically if I go from let us say step i where the acceleration was U i double dot I go to the next step ok i plus 1 ok. So, the first set of numerical in the first set of numerical procedure ok it is assumed that acceleration is constant between the step i and i plus 1. So, it is a constant value and the value of that acceleration actually is ok the value of that acceleration is let us say I represent in terms of variable tau remember this one total is delta t i ok and if I have to represent my acceleration for any point in between i and i plus 1 it would be simply a constant value average of these two values ok. So, I will write it as U double tau is average of U i of tau and then U i of tau double dot ok. So, the acceleration is assumed to be constant ok. In the second set of ok in the second set of numerical procedure what is assumed that again let me draw the step i and i plus 1 here basically a linear variation of acceleration is assumed. So, it is assumed that between the step i and i plus 1 my acceleration is not constant anymore it is varying linearly between the value U i double dot to U i double dot plus 1 ok. So, in that case again let me say this is delta t i and I am representing the time between ok i and i plus 1 is tau. So, I will write my acceleration as ok U i ok times tau divided by delta t U i plus 1 minus U i ok which is basically a linear interpolation between this these two values. So, the acceleration is varying linearly ok. So, this is called linear ok this is constant acceleration this is linear acceleration ok. Once the variation is known I can go ahead and I can integrate both these expressions here to get the velocity as a function of tau is not it. So, I can if I integrate it I will get it as initial velocity plus the constant average acceleration ok times tau ok. And here if I integrate it I will get a velocity as velocity at i plus this times tau plus tau square by 2 delta t ok U i plus 1 minus U i double dot ok. Now, I know that when tau is equal to delta t i then my U dot of tau or U of tau becomes the next step it goes to the next step. So, it becomes U i plus 1. So, if I substitute it here I can write that as U of t i or U i plus 1 the velocity is equal to U i plus delta t i I am substituting tau as delta t i is equal to this ok. Similarly, it can be done for the linear acceleration as well I can write down my U i plus 1 as equal to U i plus delta t by 2 here U i plus 1 and then this value here ok which is sort of same as what we have obtained here in the left and center velocity I follow the similar distribution ok. I can again go and then integrate these expressions for the velocity as a function of tau to get the displacement as a function of tau ok and I can go ahead and write this as U i plus U dot of tau plus tau square by 4 U i plus 1 and then here it becomes U i ok. Similarly, for this I would write my displacement as U i plus U dot of i tau plus U dot of i tau square by 2 and then tau cube ok tau cube by 6 times delta t times U i plus 1 minus U i. So, again I am going to do the same thing substitute tau equal to delta t to get the next step response. So, my U of i plus 1 becomes U i plus U dot of i delta t ok this is delta t square by 4 U i plus 1 and then U i. Similarly, here I can get the expression as U i U i delta t plus delta t square 1 by 6 U i plus 1 plus 1 by 3 U i double dot ok. So, what we have seen here that given or assuming a distribution of acceleration between the step i and i plus 1 I am able to derive ok the expressions for velocity and displacement at step i plus 1 as a function of velocity and displacement and acceleration at previous steps ok. Now, in this method ok the first method is called constant average acceleration method. It is called constant average acceleration method. The second is called linear acceleration method ok. So, this is called new marks constant average acceleration method and again new mark linear acceleration method. Now, the expressions that we have obtained for velocities here and displacements here can be written as a combined expression in terms of some constants and depending upon values of those constants I can either get this constant average method, acceleration method or linear acceleration method. So, if I write down my velocity as ok U i plus 1 dot is equal to U i dot plus 1 minus gamma delta t and then delta t U i plus 1. Similarly, I would combine both these expressions write down my displacement as ok as a this expression here this expression right here ok. So, I have now written in terms of constant gamma and beta ok and what happens when gamma is equal to half ok and beta is equal to 1 by 4 these expressions transform to the expressions that we have obtained for constant average acceleration method ok average acceleration method and if I utilize gamma equal to 1 but beta equal to 1 by 6 then I get the same expressions for velocity and displacement as linear acceleration method ok here acceleration method ok. So, in general new marks method numerical procedures are written like this in which ok I we have assume or we assume certain variations of acceleration and then we get these expressions here ok. Now, if you look at carefully here ok the velocity and velocity and displacement at step i plus 1 depends on the response parameter at the previous step but I also have these parameter here which are basically acceleration at the current step ok. So, remember unless I can write down the expression such that my current step can be calculated wholly based on the previous step ok I would need to use iterative method to solve new mark equation ok. However, if the system is linear for that special case ok there is non iterative new mark method then I do not have to iterate it because I can solve the system for u i plus 1 and then substitute it in the equation of motion ok. So, only for linear system I can use ok non iterative new mark method ok. So, for linear system all I need to do is basically write down my equation of motion at the step i plus 1 and then substitute all the variations that we have obtained ok the expression for u i plus 1 u i plus 1 dot and then u i plus 1 ok displacement ok whatever the expressions that we have obtained here and when we substitute it and eliminate u i plus 1 double dot acceleration at the current step then I can write down my expression again in the similar form that I have obtained for ok that we had obtained for central difference method. So, for new mark method I also get again expression like this ok where again k is some function of system parameters and p i plus 1 is again system of or like you know expressions of known parameters from the previous steps ok. So, through this we can calculate u i plus 1 ok if u i plus 1 is known then u i plus 1 dot can also be found out and then the acceleration can also be found out either using these expressions here or we can just utilize the equation of motion and get and write this as p i plus 1 times c u i plus 1 times k u i plus 1 divided by m ok and then we keep iterating ok going from step i to i plus 1 where i varies between 0 to n or the steps that we are often interested ok. Now, if you look at here there is an important difference between the central difference method ok and new mark method ok and the new mark method if you remember in the central difference method even to get the response parameters u i plus 1 u i plus 1 dot and u i plus 1 double dot ok. We never utilized the equation of motion at i plus 1 the equation of motion that we utilized was actually this ok this is the equation of motion we utilized and we arranged this substituted the term and then wrote as k hat times u i plus 1 is equal to p i hat ok. So, if you look at let us go back where we discussed central difference method and let us see what is the equation of motion we utilized is this step i or i plus 1. So, even to get the value of the response parameter at the next step we never satisfy the equilibrium equation at the next step ok. We got everything from the current step ok and the next step was just found out using the previous step. These kind of methods where the state of the system is actually obtained wholly from the previous steps it is called explicit methods because it is explicitly determined from the previous steps and the equilibrium at the current step is not utilized. However, in Newmark method if you look at it the equilibrium at the current step was utilized and that must be satisfied to get the response at i plus 1. These kind of methods are called implicit method where response at step i plus 1 is determined using the equilibrium at i plus 1. In this case response at i plus 1 is determined using the equilibrium at i ok. So, these are also popularly referred as explicit methods and implicit methods of numerical procedure to get the response. Now, we saw that for a central difference method the criteria for stability was delta T by T n should be less than 1 by pi. Now, for Newmark method again similar procedure can be followed that was followed to determine this which we did not discuss ok. So, that it can be discussed considering the stress propagation theory ok, but finally we get this condition for stability ok. Now, Newmark method is stable if delta T by T n is actually less than equal to 1 by pi root 2 times gamma minus 2 beta ok. If this condition is satisfied then my Newmark method is stable. Now, if you consider constant acceleration method ok. What was the value of gamma equal to 1 by 2 and beta 1 by 4 and if you substitute this we get here as this term becomes 0. So, this becomes infinity. So, delta needs to be smaller than infinity which would always be the case. So, that is why the constant acceleration average acceleration method is unconditionally stable ok. So, it is always stable ok. It does not matter what is the value of delta T ok. However, it might only be accurate remember there is difference between stability and accuracy. So, it might still be stable, but it would only be accurate if delta T is very small ok. Now, for average for the linear acceleration method as opposed to constant average acceleration method. So, for linear acceleration method as well we can substitute the value of gamma equal to half and beta is equal to 1 by 6 ok. So, delta T by Tn we get this criteria as 0.551 ok. So, this must be satisfied for the solution or the response to be stable and you know for most of the systems actually this is not an issue ok. So, this is satisfied for the civil engineer structure ok for most of the cases anyway ok considering typical values of properties of the structures ok. The issue here is accuracy ok. So, while they might still be stable, but for to be accurate the delta T has to be sufficiently small ok. And usually what happens whenever we utilize these integrators we perform some analysis to see we do some sensitivity analysis to find out if the delta T that I am using for the analysis is small enough that it is stable and it is converging towards an accurate solution. So, remember I used all three properties that we discussed that it should be stable and it is it should be converging towards a stable an accurate solution or the accurate solution ok alright. So, these are the three typical you know numerical procedures that are used and there are so much of additional things to learn for these type of methods you know stability, convergence you know all those things you should recognize that there are lot of material that can be studied for this topic. However, we would limit our discussion only to these three just to get an overall idea. What you can do you can go ahead and you can try out some of the excitations that we have already considered for example, harmonic excitation like this ok. Consider some typical values of the properties and the excitation for this we already know analytically what is the response. So, we know the exact response ok from the solution of differential equation, but you can employ the interpolation method, excitation interpolation method or central difference method or Newmark method and then compare the response ok, compare the response for different values of the time steps and see how accurate they are ok alright. So, with this we conclude our chapter here today. Thank you very much.