 Welcome back everyone. Today we are going to discuss the idea of impulse and how the response of a system subject to pulse forces can be approximated using the same expression which only depends on the area of the loading and not the shape of the loading for a specific value of pd by Tn ratio. So, let us see how we can get solution for the response of a single degree of freedom system subject to impulsive forces. So, today we are going to see how to obtain the response of a single degree of freedom system subject to different type of pulse excitation. Till now what we have done last class we found out what is the response subject to step force or step excitation in which a force is applied over a duration of td ok and then it is taken off and then we found out how to what is the response of a single degree of freedom system. So, basically the procedure that we describe that we have to consider the forced duration or forced phase of the response and then free vibration phase ok. So, for any pulse type motion ok we can divide our response into forced response and free vibration response ok basically to represent when the force is applied on the single degree of freedom system and when the force is removed. Now today we are going to discuss different type of pulse excitations ok examples of which could be observed in real life ok. So, one of those pulse excitation is triangular pulse ok. So, basically you have a force of amplitude p naught ok and then it decreases with time and then goes to 0 over the time duration td. So, as I said this is a force that is applied suddenly and then it goes to 0 over time duration td ok. Basically this type of like one of the examples where this type of pulse excitation could be observed or where the loading could be represented as a triangular pulse is basically blast loading ok. So, if you have a surface blast what happens the shock wave actually travels like this ok and when it hits a structure it applies a pressure loading on that structure and that pressure loading the time variation of that pressure loading can actually be approximated using a triangular pulse ok alright. So, let us see how do we find out response to triangular pulse excitation. So, we are going to divide our force pt or the excitation pt and two phases first phase is the forced phase ok which we are going to write as p naught 1 minus t by td. So, this is 1 here ok. This is for time duration smaller than td and for greater than td this is basically 0. Now, to get the response ok we are going to first find out the response during the forced vibration phase and then during the free vibration phase using the initial conditions at the end of forced vibration phase here ok. So, that would be ut and u dot of td ok. So, the procedure remains same ok you could either use to Hamilton integral to find out the response ok or you could just solve like you know utilize the classical method of solving differential equations ok. So, what I am going to do I am just going to write down the final solution ok. So, let us first consider the forced vibration case ok. So, I can write down my ut if I consider as dohamel integral I can write this down as uts p naught by m omega m 0 to t this load would be 1 minus tau by td times sin omega n t minus tau d tau remember I am not doing anything new I am just finding out what we have discussed that ut due to an arbitrary force can be found out using this expression here ok this dohamel integral here in which we write down p tau and then h of t minus tau which is the unit impulse response function and which for this case ok would become something like sin omega t minus tau ok d tau. So, this we had derived in the first lecture of this chapter. So, we are just basically utilizing that. So, once we substitute the values ok we can go ahead and find out the response as ut equal to us t naught which is basically p naught by k ok times 1 minus t by td minus cos omega n t plus sin omega n t by omega n td ok. So, this is for the forced vibration phase now what will happen at the end of the forced vibration phase it would have acquired certain displacement and velocity. So, when the load becomes 0 it still has that velocity and displacement ok. So, it is going to go into free vibration without any application of external load ok and it will keep on vibrating with that because there is no damping in the system ok. So, basically what I am saying for the free vibration phase let us let us get the response. So, for the free vibration case I can just write down my responses u of t is equal to u of td cos omega n t minus td plus u dot of td divided by omega n sin omega n t minus td. This is the free vibration response with initial conditions are now provided at t equal to td not t equal to 0. So, remember the original equation used to be u 0 cos omega n t and then u dot 0 divided by omega n sin omega n t and this was the expression for ut ok when the initial conditions were provided at t equal to 0 ok. Now, the initial conditions are at t equal to td. So, basically I am going to find out my response using this expression here ok. So, we can substitute the value of u of td and u dot of td from this expression here substituting it here and then we can simplify it and get the final response of the system which I can write it as u of t is equal to q s t naught times sin omega n t by omega n td minus sin omega n t minus td divided by omega n td and then we have a cos term here ok alright. So, now, we have obtained the response for forced vibration phase and the free vibration phase ok. So, one of the things that we need to find out is actually subject to a triangular pulse let us say p naught n td ok. What is the peak response of the single degree of freedom system ok. So, this is an important parameter that we need to find out and to find out first we first need to consider whether the peak response occurs during the forced vibration phase or whether it occurs during the free vibration phase because depending upon that I will have to find out the maximum value of u during the forced vibration phase or maximum value of u during the free vibration phase ok. So, that we need to do ok. So, let us see how do we do that ok. So, for the free for the forced vibration phase basically what we need to do we have we can define our r t s ok let us call it r 1 ok we will call this r 1 r 1 of t is basically u of t where r 1 is basically the response modification factor u of t divided by u s t naught ok. So, we will get this expression here which I have here. So, I am just going to write it again ok alright. So, this expression I get and to find out at what time ok this is maximum basically I need to differentiate it with respect to time and find out and basically equate it to 0 and then find out t equal to t max ok from that expression when I substitute it equate it to 0 ok. And then I am going to substitute this time back to this expression here to find out the r 1 max ok and basically what do I get r 1 max as in this situation ok 2 minus t m by t d where t m is the value of the t that I get when I substitute d r by dt equal to 0 and the value of t m that I actually get I am going to write that as well ok. So, t m basically I get it as t n by pi and then tan inverse 2 pi t d by t n. So, this expression here becomes 2 minus t n by t d ok and then tan inverse 2 pi t d by t n. Now, as you can see from this expression here my r 1 is only a function of t d by t n which is the same conclusion that we had obtained for other type of pulse motion which was the step force. So, even for the sine force again t d by t n is the important parameter based on which my peak response depends ok. So, this is for the case when I am assuming that my maximum response occur due to during the forced vibration phase ok. Let us say that is not the case ok. So, what I am saying ok what I am saying here basically let me say this is I am representing the displacement on y axis and time on x axis ok and I am also trying the static displacement curve which is basically starts from p naught by k the peak value. So, this is basically u s t naught equal to p t divided by k. So, this is the time variation ok. So, it might happen that the maximum response can occur during the forced vibration phase or it can occur during the free vibration phase ok. So, in that case let us say it occurs during the free vibration phase the response would still be increasing something like this ok and somewhere during the free vibration phase I get the maximum response. So, at at the end of the forced response whatever response that I will get could not be the maximum value ok that would be the value ok which would further be increasing due to the during the free vibration phase ok. So, in this case basically T d is smaller than the T m and T m is basically obtained during the free vibration phase ok. This is opposed to the case when my maximum response occur during the forced vibration phase in this basically what happens it goes something like this ok. So, during the forced vibration itself the maximum response occurs ok and it keep on going like this. So, again this is my u s t of t equal to p t by k this is p naught by k and this is the u t curve ok and this is my T m and this is T d ok. So, in this case which is the case that we have just discussed the maximum response occurs during the forced vibration phase if that is not the case then let us see what happens if the maximum response occurs during the free vibration phase ok. So, to find that out what we need to do ok for the free vibration r a t I need to write is as u t divided by u s t naught ok this is for the free vibration phase ok. So, let me write it here and I can substitute the expression here ok to find out the expression which I get it as nothing but omega T d and this expression here ok. So, let me just write down the expression for your reference alright. So, this is the expression that you get. Now, again even for this we are going to follow the same procedure ok d r by d t for the maximum response during the free vibration ok I am going to equate it to 0 then find out the value of t equal to t m at which the maximum response occurs and then substitute it back to this expression here to find out the r max ok. So, the t m I will just give you the answer the t m that we get here basically t n by 2 pi and then tan inverse 1 minus cos omega n t d ok times sin omega n t d minus omega n t d ok. So, if I substitute this t equal to t m back in this expression r t ok I would get the sorry this should be here r 2 max I would get the r 2 max here ok. So, and you can go ahead and then again plot basically for each phase ok. So, for forced vibration phase and then the free vibration phase ok both condition you can find out what is the r 1 max and r 2 max and then to find out the overall response you can take the maximum both value. So, let me first draw it the r value ok for both of these expressions you know you can do it numerically or you can do it analytically ok. So, this I am going to write down the horizontal axis as let me draw here this is basically t d by t n because that is the parameter on which my r max depends on ok. So, what happens in this case ok for the first case I am first going to plot the forced vibration response if the my maximum occurs during the forced vibration response I will see that the curve looks like something like this ok and for the free vibration response ok if I try to find out it would look like something like this ok something like this. So, this is basically r 1 max and r 2 max and this point at which this crossover basically is you can find that as the crossover point as t d by t n equal to 0.371. So, when t d by t n is less than 0.371 the maximum response is governed by r 2 max ok when it is less than 0.371 when it is greater than 0.371 it is governed by the r 1 max the forced vibration response ok and then you can find out the overall maximum response of the system. Now, note that this is different from the case for the step excitation for which the t d by t n if it was smaller than 0.5 then my free vibration response governed ok. So, in that case my r 2 governed and when it was greater than then the forced vibration governed ok. So, this value is actually different ok. So, that you need to keep in mind ok. So, once you have the r value basically given the plot of r as a function of t d by t n ok and r is nothing, but what r is basically your peak dynamic displacement divided by the peak static displacement which is p naught by k and if we have this r curve for any pulse pulse type excitation ok given the property t d and the property of the system t n t n we can find out what is the r value and then we can find out the what is the maximum dynamic displacement ok. So, there are several curves like this ok. So, some of those curves like you know that you can encounter in real life are let us say sine pulse excitation ok you could also have double triangular excitation something like this or you could have like you know different type of pulse excitation ok. And like it is not possible to do all find out the response to all of those excitation, but it is worth noting that how to develop the solution. So, you understand the procedure. So, we have discussed step force excitation and we have discussed triangular ok. So, we are not going to discuss these, but you can refer to any standard textbook to find out the basically r for each ok. So, response modification factor is a function of t d ok. So, this is p here, this is p naught, this is p naught, this is p t here ok. Again this is well I should not say like this, let us say it is like this ok. So, it is p t and this is t d here ok. So, for all of these cases you can find out what is the value of r versus t d by t n ok. And it could be of any general shape and that is what is important to understand and also how the response would differ for different different type of pulse excitation ok. Response that is known then the procedure is similar to find out the maximum dynamic response in the system alright. Now, we are going to switch over to different a special situation of these pulse excitations ok and then see how does that actually affect the response. Let us we saw that for all these pulse excitation when t d by t n ok was greater than let us say some value ok, let us say it was greater than 0.5 ok. So, in those cases we saw that the maximum or the peak response ok peak response occurs during the forced vibration phase during the forced vibration phase ok. And if it occurs during the forced vibration phase ok. So, if it occurs during this phase here or this phase here or this phase here then it what is the shape of the pulse it matters a lot it plays a big role in finding out the peak dynamic displacement ok. And that we can demonstrate by comparing the curves for ok. So, let us say I have three type of excitation ok. One type of excitation is the step excitation alright in which the forces p naught and duration is t d. Second type of excitation is let us say sine excitation ok in which again the peak is p naught and this is t d and the third type of excitation ok is actually triangular excitation which is t d and peak p naught ok. Now, before getting into the mathematics of it can you imagine if you have been given these three type of pulse excitation just by looking at the nature of the curve which is going to provide you the maximum response ok. Think about it for a second and then see how to find out well if my t d is greater than 0.5 or let us say the peak response occurs during the forced vibration phase. Then the peak response depends on the shape of the curve ok and which would be higher among three depends on the how fast the load is applied for each of these cases. So, if you look at it for the step type excitation the load is applied here suddenly compared to the sine half sine pulse excitation in which it is little bit more gradual compared to this one. And then if you consider this triangular pulse then the peak excitation or the forces actually the rate of increase of force is again smaller than these two. So, if the peak is say then analytically I can say that my step excitation the step force would provide the maximum response in the system ok. And you can go ahead and like you know compare the actually the curves of r versus t d by t n and that would be pretty much evident. So, if you try to plot this let us say this is the r value and this is t d by t n. So, we know that for a step excitation the r or the maximum response can reach up to 2 or the dynamic displacement could be up to 2 times the static displacement ok. So, I am going to plot it here ok and this looks like something like this alright. And this value here is actually 0.5 ok. Let us say this is 1, this is 2, this is 3 like that ok. And this is here 2 and let me show this one here is 1 ok. So, this curve we had already derived and you can go ahead and see that if that is the case or not ok. And if you consider the sine pulse excitation or triangular excitation again it would increase like this with smaller initial slope ok. And it would give you a dynamic response which is higher than static something like this. And for triangular it will start again like this with again little bit smaller slope and then it would give like something like this ok. So, both of them are smaller ok. And this I am just like you know drawing it from the expressions that you would get and you can refer that to any textbook ok. But this is typically how it looks like ok. So, you can see that depending upon the shape of the curve it matters a lot ok. If the forced vibration response occurs during the forced vibration phase or for case where T d is greater than 0.5. Now, let us consider a case in which T d by T n is actually very very small ok. In that case basically what I am saying that either it is so small that the. So, in that case when it is very small what will happen? My peak response would occur during the free vibration phase. So, the peak response would occur during free vibration phase ok alright. And if the duration T d is very small compared to T n then what we can write the total response u t as whatever the response due to unit pulse times the whatever impulse due to the load or the pulse or the force that you are applying the pulse force you are applying. So, let us say I have some random distribution ok. So, let me say something like this it does not look very good let us say something like this ok of T d here and peak is P naught this is sub general shape ok. If T d is very less what will happen? The response would actually occur much after the time T d during the free vibration phase ok. So, in that case let us first calculate what is the area of this P naught versus T d curve ok. So, the and that we define as an impulse. So, I can say this would be P d P t from 0 to T d that would be the impulse ok. Now, if the duration is very small we know that it can be treated as an impulse ok. And the response can be obtained as I times the response due to unit impulse function which is nothing but m omega n times sin omega n T or you can use the like you know same expression if you assume that due to impulse you get initial velocity ok. So, but no initial displacement remember impulse only provides initial velocity. So, this gives you u dot 0 is equal to i by m and if you consider the free vibration response with 0 initial displacement it would again give you the same expression ok. So, my ut is nothing but i divided by m omega n sin omega n T ok alright. So, for cases where the pulse duration is very small compared to the time period of the system the response can be represented as this ok. And in this case if you look at this does it matter whether it is a let us say a sign pulse or a triangular pulse or a stiff force as long as my i is same for all 3 cases or the area under the force time curve is same or the impulse total impulse is same for all 3 cases. It does not matter whether it increasing at a slower rate or whether it is increasing at faster rate or basically what is the variation of the or the shape of the pulse ok. So, that is a very important conclusion ok. So, you can go ahead and you can find out the response using this and this is basically the upper bound of the response considering the assumption that all the force is actually concentrated at T equal to 0 which is ok if T d is very small ok. Then we do not have to consider time variation of this one I can just consider this whole impulse to be situated at T equal to 0 and this is the response that we get. So, again we can go ahead and we can compare the responses ok to different type of pulse. Now, let us again consider 3 type of pulses which is basically let us say this is a rectangular pulse P naught by 2 and this is T d and this one is a sign pulse which is here is basically pi P naught pi 4 and this is here is T d ok and then I have a triangular pulse which is P naught T d. Now, if you look at carefully all three curves the total area or the impulse for all of these are same as half P naught times T d ok. This is same for all three. So, the impulse is same for all three ok and if you try to plot let us say U naught divided by P naught by k ok. Let us see what do you get ok. So, if I try to plot this as a function of T by T d what you will see ok the curve actually looks like something like this here ok this is 2 here. So, the initial line is actually same for all of them ok this curve is actually also touches this one. So, it goes something like this here ok this might not be like an exact representation, but this is let us say for comparison purposes. So, what do we see when T d by T n is up to like it is a very small this is actually 0.5 ok when it is half of this let us say 0.25 ok in that situation the U naught or P naught by k actually it is same for all the pulse type motion. So, when T d by T n is smaller than let us say 0.25 ok then the shape of the curve does not matter ok as long as the area under the force variation is same for all 3 pulses ok and the total response only depends on the total impulse ok. So, as long as you can calculate impulse as a time integral of the force ok and if this situation is satisfied here then your response only depends on the impulse ok the area and not the shape of the pulse ok. So, remember that alright. So, this is an important conclusion. So, and this is like an extended. So, I need every time you are given a problem in which you have been asked to find out what is the response to a any arbitrary impulse ok let us say this is here T d you first find out what is T d by T n ok if it is a smaller than let us say 0.5 then you can just go ahead and use this expression you do not have to utilize the expression for R d from like you know different curve ok to find out what is the response this would be a reasonable approximation ok sorry this curve here for this condition alright ok. So, these basically we have discussed response to single degree of freedom system to different type of pulses and the methodology to get the response for force vibration and free vibration and then the peak response alright. Now, let us consider a case which is all too common in reality ok cases in which you have a ground excitation ok. So, when you have ground excitation alright being applied as a pulse force. So, let us see what happens in those cases and one of the very common example is actually when you have a sign type curve here ok. So, and this could represent like you know simple bumper on a road or it could represent like you know any type of non-continuous or like you know non-periodic curvature in the road ok. So, let us say this is basically represented as u g t. So, the acceleration due to ground excitation if it is represented as u g of t equal to peak ground acceleration times sin ok. We are going to represent this pi t d by t m because this is how we represent a half sin pulse excitation. This is for t smaller than t d and this is equal to 0 for t greater than t d where t d is basically time taken ok to cross this ground curvature. So, let us say if I have a vehicle which is moving over with velocity v ok and this whole length is given to me as l then t d simply becomes l by v ok. The length of the this curvature here divided by the velocity of the vehicle v alright and I can represent in terms of excitation something like this oh pardon me this is wrong actually this should be pi t by t d ok not t d by t n is the way ok. So, this is how we represent the this curve here ok alright. So, if the ground excitation is given and remember that here this is in terms of displacement ok this is in terms of displacement and we can differentiate and find out what is the ground excitation ok in terms of time, but let us say this for this case this is how it is ok and we could have this available the ground excitation ok and then let us say this is some function of ok u g naught times sin of some function we do not care ok. Now, we know that for this type of excitation alright what is the peak response ok. So, u naught which is the peak dynamic displacement times the peak static displacement we know that for pulse type excitation something like this is represented as r a d ok. Now, u s t naught is nothing, but peak force due to ground excitation or ground excitation divided by k. Now, we have derived if there is a ground excitation then how can we write down the effective force as minus m u g naught this is the peak value of the ground excitation ok divided by k and this can be further written as if I drop the well let us just keep it like this and denominator I would have k by m which can be further write down written down as omega n square ok and we can drop the negative term it is inconsequential here. So, utilizing this I can write my r d as u t divided by u s t naught as equal to omega n square times u naught divided by u g naught I have just substituted the value of u s t naught here alright. So, if r d is given for any type of pulse excitation that is being applied through the ground ok with peak ground acceleration of u g naught double dot ok I can go ahead and find out what is the peak displacement in the any vehicle that is going over that ground excitation or like in any system to which that ground excitation is being applied ok utilizing the r d for that particular pulse type ground excitation ok. So, it could be like you know something like this or it could be also like this we do not know it yet ok, but this expression is for all the general any type of ground excitation which is which can be approximated as a pulse type excitation ok. One more thing we can see here if we have an undamped system ok for an undamped system we know that ok I can write down this expression here equal to 0 ok remember this is for undamped system for the ground excitation ok and this is the same expression that we utilize when we write this as u t as relative acceleration times the ground excitation and which comes on the right hand side to give the minus m times u g double dot ok, but I can write this for any type of single degree of freedom system subject to ground excitation and if you utilize this I can say that minus my total acceleration as minus k by m times u of t ok which is nothing but minus omega n square u of t. So, the peak value would also be related of both these quantities the total acceleration and the relative displacement ok. So, I can further write this as peak value of the total acceleration equal to omega n square times u 0 and I have dropped basically the negative term here ok. So, this is the peak dynamic displacement this is the peak total acceleration ok and I can go ahead and substitute this expression here. So, that my r d also becomes let me first write down the last expression and numerator I will substitute this as u total divided by u g not something like this alright. So, if the ground excitation is given in terms of any of the pulses ok this like you know the sign pulse or the triangular pulse and if we know the r for any of those ok and it could be f n a general shape I do not have to go ahead and solve the differential equation I can just utilize the r value to get the these quantity peak dynamic displacement ok and then peak value of total acceleration in the system alright and this is something similar to what we did for earthquake excitation only thing in this one is that my r d has changed now to this type ok. One more thing to remember you might get confused that we used to write down transmissibility as u t not ok sorry this should be here u t not divided by u g not ok how come this is now r d well remember when zeta is 0. So, we have talked about undamped system my t r is basically equal to r d ok and that you can get from the expression remember t r is nothing, but 1 plus 2 psi omega by remember what was the expression for t r in the numerator I had this term ok I had that term here and then divided was r d basically if damping is 0 then this term is 0 and then basically this becomes equal to sorry r d is not in the numerator this is r d here this becomes r d here ok. So, that is why we get the same expression ok. So, with this the theoretical discussion on the non-periodic excitations are now concluded what we are going to do now we are going to discuss two small problems ok and then we are going to discuss the solution of those problems ok. So, in the first case ok so let us say this is a example one here what do I have basically I have a building alright which is a one story building ok and it has been idealized as a 4 meter high frame ok. So, this is a building here which I am idealizing as a single story frame this is total as 4 meter height is 4 meter ok and it is given that the beam is almost rigid and these columns are actually pinned at the base ok. So, this is like a fixed connection and pin connection at the bottom ok and the properties of the columns are provided ok. So, the properties are columns ok are given as this is as 2772 ok the section modulus which is basically the moment of inertia divided by the distance of neutral axis ok is given as 252 centimeter cube the elastic modulus of steel material for these columns which are these columns are made up of is 200,000 mpa ok and the system basically has natural time period of 0.5 second that is also given to you. So, what has been asked that a pulse type of square pulse load of 20 peak is let us say this is PT here 20 kilo Newton is applied ok the time duration of this is 0.2 second and what do you need to find out the response quantities which are basically the peak displacement of this due to this loading ok and also the peak stress in one of these columns ok. So, the peak stress in those columns alright. So, these are all the data that have been given to you and you need to find out the response of the single story frame subject to this square type pulse motion rectangular pulse motion ok. So, pause here for a second ok and then try to solve this problem alright alright. Let us now discuss the solution for this problem remember as we discussed whenever a pulse type excitation is given ok and the property of the structure is given the first step is to find out what is the value of T d by T n because based on T d by T n we might decide not to utilize the R d at all ok because if T d by T n is very very small let us say smaller than 0.25 then we can just assume it to be a an impulse ok and then calculate the area and find out the response using the expression I by m omega n times sin omega n t ok and if it is greater than that then we will have to resort to using the R d versus the response the response spectra ok the R d versus T d by T n curve. So, T d by T n here is 0.2 times 0.5 which is 0.4 and this is greater than 0.25 second sorry 0.25 not second this is the ratio ok. So, 0.25. So, we cannot assume that the applied forces as a behaves like an impulse and we need to find out what is the R d value. Now, you can go to the chart or the response spectra that we have or for rectangular pulse we know that when T d by T n is smaller than half then the R d is basically given by this expression ok R d is given by this expression 2 sin pi T d by T n and if you substitute all values you will get this one as 1.902 ok. Now, to find out the value of static displacement so that we can find out the dynamic displacement first I need to find out the equivalent lateral stiffness of the system ok lateral stiffness of the system and that I can find out as k equal to remember I have two columns and the columns have boundary condition as fixed here and pinned here ok. So, for fixed pin condition we know that it is 3 EI by L cube and you can substitute all values of EI and L cube and you can find out the this one this comes out to be around 260 kilo Newton per meter ok and we have to well this would be 2 multiplied with this and if you multiply this this comes out to be 520 kilo Newton per meter ok. So, you can now go ahead and find out the peak static displacement which is P naught pi k I know that 20 kilo Newton load was applied. So, this divided by 520 times 10 to the power 3 which basically gives me a value of 3.85 centimeter ok. Now, we know that if I have a single degree of freedom system represented through this frame and if I have a displacement let us say something like this, then the force equivalent to static force is nothing, but the lateral stiffness times the. So, the peak value of the lateral forces lateral stiffness times the peak dynamic displacement. So, the dynamic displacement here is R d times UST naught which is 1.902 times 3.85 and this is equal to 7.32 centimeter ok and we can use that to find out what is the maximum ok what is the maximum lateral force in the system ok that is not difficult to do. Let us see how do we do that ok. So, either we can do this or what we can also do remember if we have something like situation something like this where this one is getting displaced by U naught. So, basically I am saying this is getting displaced by U naught ok. The moment at this point can be written as whatever the lateral stiffness times basically U naught ok. So, this is the sorry this is the rotational stiffness that I need to write down as U naught. So, I can go ahead and substitute the value and then I can find out what is the maximum response of the system. Now, this theta k theta is nothing, but the lateral stiffness times L ok. So, you can go either of these approaches you can also you can either go ahead and find out what is the lateral stiffness times k U naught or you can based on moment you can also find out remember that you will get exactly the same value because k theta ok it is nothing, but k times L ok alright. So, when you do that ok when you do that the value that you would get let me write it down here m you will get as 76.1 kilo Newton per meter and the force F s naught you will get as basically whatever the P naught times R d ok which is 20 times 1.902 ok and this comes out to be approximately as 38.0 kilo Newton ok. Now, remember that this is the force in the total. So, this is the store or this is the force in both columns. So, if you want to consider force in one column you will have to divide it by 2 ok. So, this you need to divide it by 2 ok to find out basically whatever the force you get and that would be 19 kilo Newton ok and same for the moment, moment you can also get as 19. So, the moment now at the top of one of the column ok would be 19 times whatever the length of that column is which is basically again 76.1 kilo Newton per meter which will basically same as this one ok alright. So, this is 76 here ok. So, once you know the moment in any section ok how do you find out the stress ok the stress is nothing, but moment times c divided by i where c is the distance from the neutral axis and if you want to find out the maximum basically stress then this c becomes the half of the section depth ok. So, in this case that would be the section modulus. So, I can directly write this as this value and this gives me a stress of 309.301.9 mega Pascal ok and this is the procedure that we basically utilize. So, this is a very simplistic version of what is actually done during the analysis and design practice, but remember if you have given something like this and you have been told that the yield strength of the steel is 450 MPa just for example ok and these steel columns this steel building is actually subjected to that pulse load. Basically you are going to follow this procedure you are going to find out what is the maximum stress due to the applied load and then you are going to compare with respect to yield stress and then you are going to make the conclusion whether the structure is safe or not whether it is going to yield or not ok. So, this is the procedure that we follow alright ok. Now, let us consider a second example ok. The second example that we have is basically let me just write down here ok. We have an overhead tank alright. So, this is an overhead tank on which a force is being applied at the top and the properties are given this is 20 meters ok and the mass is also being given as ok. So, the mass is given as 50132 kg lateral stiffness is given as 1600 kilo Newton per meter ok and you can calculate the time period as 2 pi m by k equal to 1.11 second ok and the damping is also given 1.24 percent ok. Now, the Pt is basically an arbitrary force for which the variation is given to me. So, it is something like this ok a linear variation up to a peak value of 200. So, this is Pt and the units are in kilo Newton ok. So, it goes to 200 over a time duration of 0.02 second and then it is given like this at 0.05 this is 80 and then 0.06 this is 20 and this Td here at which it becomes 0 is actually 0.08 second ok. So, basically what we need to find out is the maximum force in this alright in the system and the basically maximum base here and the maximum moment for this overhead tank alright. Now, for this type of arbitrary force ok as we discussed the first step is basically find out what is your Td by Tn ok. So, Td is 0.08 Tn is 1.1. So, definitely this is much smaller than 0.25 ok. So, this can be treated as an impulse load ok and we can find out the area under this curve ok and there are multiple ways to do that you can consider to be made up of this triangle and then some of the area like this or you can like you know consider to be like you know made up of multiple using trapezoidal loop rule you can find out the total area. So, I am going to write it as ok remember this is nothing, but 0s to 8 second 0.8 second Pt dt ok and this you can write down first ok all these separations are equal 0.02. So, I can take that out and then I can write it as 0 plus 2 times 20 plus 2 times 80 ok and then 2 times remember this is 220 plus 0 and this gives me as 6 kilo Newton second all right. So, now remember what is the what was the expression if it is a pure impulse ut is equal to i by m omega n times sin omega n t. So, the peak dynamic response is nothing, but i by m omega n which I can write it as i by k and omega n I can write it as 2 2 pi by t n here all right. So, we are going to substitute the values ok and then see what do we get ok. So, 6 into 10 to the power 3 times 2 pi times 1600 times 10 to the power 3 into 1.11 and this gives me a value of 2.12 centimeter. So, once we have the dynamic displacement remember this is the water attack let us say it is deforming like this by u naught. So, the total force at the base would be the lateral stiffness times the dynamic displacement ok and we can substitute those values here ok times 0.2 1 2 ok and this we will get as 33.9 kilo Newton. And once we have the basically the base shear ok we can also find out our moment as ok moment can could be easily found out remember how would it look like if you have this force being applied here your shear force is actually varies like this all right ok. And the moment would actually vary start from 0 and with a constant slope due to constant value of shear force increases up to value m here which is nothing, but whatever the shear force is times the height of this ok which is 20. So, moment is 33.9 times 20 which gives me a value of 678 kilo Newton per meter ok. So, for this arbitrary excitation using the principle of dynamics we have found out what is the total base shear and total base moment ok and then we can go ahead and like you know design the system subject to these forces and moments ok all right. So, with these two examples we are going to conclude this chapter ok all right. Thank you very much.