 Welcome back everyone. In previous lectures in this course we have seen how to obtain the response of a single degree of freedom system subject to either free vibration or harmonic excitation and we did that for Damton and Dam system. But in reality there would be several there are several other type of loading which may not be described using harmonic loading and the example could be step loading, pulse loading and other type of forces. So what we are going to start today is basically how to obtain the response of a single degree of freedom system subject to arbitrary excitation. So let us get started. Today we are going to start a new chapter which is basically the non-periodic excitation to single degree of freedom system. So right now we would only be focusing on single degree of freedom system. Now if you recall from previous chapters till now let us see what we have done. We set up the equation of motion for a single degree of freedom system which we said the equation of motion for a linear system it was something like this and this is equal to P of t and depending upon P of t which is like in a referred to as the excitation or the forcing function we divided our study in free vibration and forced vibration and in free vibration we first study undamped free vibration and then damped free vibration okay. Now in the forced vibration we studied the harmonic excitation or the periodic excitation okay. So till now we have finished up to this part here. Now harmonic and periodic excitations are fine but you will encounter many loads in a real life scenario which might not be either harmonic or periodic okay. So for those type of system it becomes imperative that we study these type of system okay. So response to subject to non-periodic excitation okay and then later of course we will study this but today's chapter is focused on non-periodic excitation okay which is another common set of loading that are encountered in real life and for which the approach to find out the analytical solution is little bit different than what we have studied so far for harmonic excitation okay. So again for this case our problem statement is okay problem statement is basically Mu dot plus Cu dot plus KU is equal to P t okay and the initial conditions okay are given to us. Let us first take this as 0 initial conditions okay. So this is our problem statement and here the P t the forcing function is neither harmonic neither pure periodic okay. It is some arbitrary variation okay. So if you consider let me just say any arbitrary variation of P t with respect to time let me just draw it like this okay and then I need to find out the solution or the response of this single degree of freedom system subject to this okay. So let us see how do we do that okay. Now for these type of functions P t what actually we do in order to find out the response we divide this function in very small intervals okay. So throughout the loading we divided in very small interval okay and then at any time t let us say this is at any time t we try to find out the response U t subject to all these small durations loading okay up to the time t okay. So what do we do? Let us say I want to find out the response at any time U t okay. So I have the excitation function and I divide it in small intervals like this okay. The response at time t U t okay would have contribution from each of these okay. So let us say this this and this each of this small duration loading up to this point okay. All right. So this is the strategy that we are going to employ okay and let us see how that works out but before we get into that we are going to introduce a new concept okay which is called impulse okay and we are going to talk about unit impulse okay response to unit impulse okay and I will show you why do we do that okay. Now an impulse basically it is defined as a force okay. So impulse is a very large force okay that acts for a very small duration okay that acts for a very small duration okay. Okay such that the area under okay the area under the force and the duration is still some finite value okay. So basically the force that acts for a very small time and very large magnitude is characterized as impulse such that it still has finite area under the force time diagram okay. So let us say let me draw an impulse like that. So what I am going to do here I am going to write p t here okay and this is my time axis t. So I am going to consider a force which is at time tau okay at any time tau okay and basically this duration here okay let me call it epsilon okay and this magnitude basically this force is actually I will call it 1 by epsilon. So what happens as epsilon goes to 0 okay the p tau is actually goes to infinity okay. However even in that case my p tau times d tau over the duration okay over the duration tau to tau plus epsilon would still be equal to 1 okay and that I can see it from here if I multiply basically 1 by epsilon with epsilon I still get a unit area okay. So this is one way to define a unit impulse okay. So we said that the time integral is still a finite value even though the force goes to a very large value for a very small duration of time okay. So this is the definition of impulse okay. Now mathematically this kind of functions can be represented using something called delta okay Dirac delta function okay. So let me just write it this is called Dirac delta function okay and if you have come across this function previously this basically says that delta x is equal to 1 for when its argument of this function is actually equal to 0 okay and it is 0 everywhere else okay it is 0 at all other places for x not equal to 0 okay. Now if I have to do this for this function what I do I represent this Dirac delta as this t minus tau okay so that this function is actually equal to 1 when t equal to tau okay. So when t equal to tau this argument again becomes 0 okay same thing that we did here okay and this is 0 for t not equal to tau okay. So this is mathematically how we represent the a unit impulse okay. Now according to the Newton's second law of motion okay if we have a force p right if we have a force p that acts on the body okay the rate of change of momentum of that body is basically equal to the applied force okay. So basically let me utilize the Newton's second law okay the Newton's second law says that t by dt of momentum and as you know momentum is defined as mass times velocity that should be equal to the applied force okay and if mass is constant I can write this as m dt that should be equal to p okay. Now we can go ahead and integrate both side of equation so that we can get it as p okay we can write this as p of dt integrated over time t1 to t2 okay let us say it goes from state 1 to state 2 then I can write this as m du times okay state 1 to state 2 let us say that it went from velocity v1 to okay v2 or let us say it went from u1 dot to u2 dot okay so that I can write this as m okay u2 dot minus u1 dot okay which is nothing but m delta u dot okay. So basically this term here that you see this is the time integral of force right what we defined as impulse okay and as you can see from this expression impulse is basically the change in the momentum okay. So if you apply impulse of any body that is basically it is equal to change in the momentum of that body okay. So let us say if you consider a single degree of freedom system that we have been dealing till now again the same representation okay we have a single degree of freedom system here okay and I apply an impulse on it okay. So what is going to happen okay if somehow I can calculate the magnitude of impulse okay that would lead to the change in momentum okay and let us say this is a unit impulse so that okay this is a unit impulse so that I have let me just go back to that the impulse is basically applied at time t equal to tau okay so I will draw that figure again here okay. This is at any time tau okay the force pt so the impulse is at that time tau this is a unit impulse which basically means that my p tau which we have derived p tau times d tau is equal to 1 over the time duration okay. So when we have this system here the spring mass tamper system and we apply a unit impulse at time t equal to tau what will happen it would lead to change in momentum okay so let us say this unit impulse okay p tau times d tau which is equal to 1 and if initially I am assuming that the system was at rest okay after the time t the velocity at time tau would be equal to okay m times u of tau okay I am assuming that there was no initial velocity okay so that tells me that u of tau is actually equal to 1 divided by m okay. So if we apply an impulse to a system it gives rise to it leads to change in momentum which further gives some initial velocity to the system. However it does not lead to any initial deformation okay because the spring as we discussed impulse is applied for a very short duration so the spring does not get time to actually react to that high magnitude small duration force so what happens okay let me this is velocity actually don't mistake that so initial displacement is still equal to 0 okay so with these conditions okay what we want to do we want to find out the response of this system okay the response of this spring mass tamper system with these initial condition and that would be the response to a unit impulse at any time equal to tau okay so let us see what do we get if you remember okay your expression for response to an undamped system remember it is now like a situation in which the initial conditions have been given to you and you have to find out what is the further motion so it was like when you apply an impulse it provides initial condition and then it is like a free vibration so after this these initial conditions are applied it would undergo basically free vibration okay and if you remember the equation of motion for free vibration for an undamped system let us first do undamped system we will again do the damped system it was u of 0 okay cos omega n t plus u dot of 0 okay times sin omega n t and there was this omega n term here okay there is one small difference though here it was due to initial condition at time t equal to 0 however for us the initial condition is at time t equal to tau okay so for our case basically the expression will become u of tau okay cos remember our motion is starting at t equal to tau okay and there is no impulse or the near there is no any force before t equal to tau okay so I am going to shift my axis so that becomes omega t minus tau plus u dot 0 omega sin omega n t minus tau because our impulse is applied for at t equal to tau and this solution is only valid for time that are greater than tau because if time is smaller than tau the response is actually equal to 0 now we already know that this is equal to 0 so we are only left with this term here okay one correction this should be here u dot of tau okay so we will substitute the value of u dot of tau which is 1 by m omega n and this is sin omega n t minus tau okay and this function can also be written as a function of h of t minus tau okay so this is for undamped system similarly you can write the equation for free vibration of a damped system and similarly create the expression as u of tau is equal to m omega n e to the power minus zeta omega n t minus tau okay sin omega d sorry it should be here omega d here okay t minus tau again this we can write it is another function h t minus tau this is for damped system okay this is for damped system so basically these h t minus tau are unit impulse response function okay these are basically a response of a single degree of freedom system due to unit impulse so these are called unit impulse response functions all right so we have obtained basically the response due to unit impulse and remember this is a response at any time t due to impulse at time t equal to tau okay so let me write that again okay response time t due to unit impulse at time t equal to tau okay now once we have that figured out okay we know that response due to unit impulse okay now we can delve into finding out the total response okay subject to the arbitrary excitation p of t which is varying arbitrary with time okay so once you know the h of t minus tau which is the unit impulse response function remember this is the response due to unit impulse so if you want to find out okay if you let us say if you want to find out response due to any impulse that is non-unit okay so let us say response due to impulse i equal to let us say some other function p tau times d tau okay so how do we do that or let me just don't use the integration sign here let me just say that it is a very small value okay so that i can approximately say that the area is like p tau times t tau just like a rectangular area okay so response due to this impulse okay we can find that as response due to unit impulse times the magnitude of this impulse okay and this works for a linear system so if the system is linear we can employ this technique okay because for a linear y can directly multiply the response okay with with the impulse magnitude to get the proportional response okay so if i have a response because remember when we said okay when we said that this is the variation all right when we said this is the variation of pt with respect to time and we said that the variation looks like something like any random or like an arbitrary function okay we have the function for unit impulse but for these cases these impulse small if i divide it in small time duration these impulses won't be unity all right so let us say this is at any time tau okay and this is the time d tau okay and then like you know this is the one two three so on okay so we want to find out for each of these strip impulse like you know what is the response so that i can directly find out by multiplying with this function okay so multiplying h of t minus tau response due to unit impulse okay times the magnitude of the impulse okay so this is the response at any time t so let me instead of just seeing that i did d of ut due to a small impulse but this is just one strip here this is just due to this at any time t let us say here i want to find out okay time t so what do we do then well as we have previously discussed response at any time t would be the total response due to all the impulses okay up to the point or up to the time t okay so if we integrate this function from this time 0 to t okay it would give me the total response at time t due to all the impulses okay and let us see how does that look like graphically so what i am saying let me say i am trying this graph here okay and then i am trying another graph here okay so let us say in the first case due to first impulse one it will undergo some free vibration okay okay for the second one it will start little bit at time okay after d tau okay and again it will give me some response okay which depends of course on the magnitude of that impulse these two impulses are not same okay it will again give me some response okay and it will keep on doing that okay let us say i draw at time d tau okay so at this point also i will have some response okay and if i keep adding them okay all the point till i get to this point it will give me total response so this is let us say du1 du2 and so on this is basically du at any time t okay or tau let us say let us call this du of okay we will call this du of t okay so the total response as you will you can imagine okay initially it could be only due to this function then it will keep on adding and the response will keep on adding up or subtracting depending upon whether they are in phase or out of phase okay so overall you will get some response you know random response okay which might look like something like you know i am not trying accurately to reflect that response function but it would look like something like this and this would be my total response okay so as i said i need to sum up response due to all the impulses okay so the function that i have basically was this okay it was p tau okay and then i had h of t minus tau d tau okay and i want to integrate this if i want to find out okay the u at any time t i want to integrate this du t up to time t equal to zero to time t so p of tau t t minus tau d tau and this goes from zero to t now remember here my variable is d tau okay so the variable here is tau not the time t t is basically up to the point till which i want to integrate okay so basically this impulse here that i had considered at any time tau that is my integration variable so if i vary this tau from zero to t and sum up all the response due to all the impulses i will get the final impulse so this expression here okay the expression that i have written here it is called okay let me again rewrite it this expression p of tau h t of t minus tau d tau this is called convolution integral it's called convolution integral and like you know it finds a lot of application in like you know multidisciplinary field okay you will see at many places right you know this convolution integral now for our case okay for single degree of freedom system okay we already have the expressions h of t minus tau for a damped and undamped system so we can substitute it here and we can get the expression for the ut due to any arbitrary varying force pt okay so that expression can be used to obtain the response and that expression let me just write it here ut let us first write for a damped system okay i can write this as m omega t zero to t remember integration variable is tau okay i have this exponential term which is t minus tau okay and there is p tau term here as well okay so let me just do this let me first write here the p tau p tau e to the power minus zeta omega n t minus tau and this is sine omega d t minus tau d tau and i integrated from 0 to 2 to get the expression for all the analytical expression for this ut okay remember i am able to do that i am able to simply sum up all the function because i am assuming that all these functions are linear so my structure is linear what basically linear means let us say if my structure is linear elastic okay so the fs versus u is basically like this okay so the response i can directly sum up from the individual responses okay so this convolution integral are strictly for linear systems okay because we are using method of superposition okay we are using method of superposition okay so we have obtained this expression for un-damped system okay and if you put the value of zeta equal to 0 okay you can get the expression for un-damped system as well okay which is not very difficult again we will write it as p tau e to the power minus zeta omega or there won't be any zeta term here what i will just simply get psi omega n okay t minus tau d tau so this is for un-damped system okay now one thing to note here would be in all these scenarios we had considered u 0 or initially that system was at rest so at rest initial condition okay so when we said that my force is actually starting i just considered the effect of force assuming that the system was at rest but how about my system was already had some initial condition like it had some initial displacement from the position of equilibrium and it had some initial velocity okay if those values are non-zero okay so we had obtained the solution for at rest initial condition for non-zero okay for non-zero initial condition okay you need to find out the response due to initial condition like it's a free vibration okay you need to add the response that you get due to free vibration with initial condition of u 0 and u dot 0 okay which is not very difficult we already have derived the expression for this from for un-damped free vibration and free vibration okay so that needed to be if it's like you know had any kind of initial condition that needed to add up to this expression that we have derived here okay this is specialized form of convolution integral that we use for our case is called Duhamel integral okay this is called just giving you some terminologies here so that when you see that you remember this is called Duhamel's integral okay and this is just an special like you know case of convolution integral okay now as you can imagine basically what we are trying to do here okay for any arbitrary excitation which are not periodic or harmonic okay so we had obtained for periodic and harmonic loading okay the analytical expression for u of t but it is not so simple for any arbitrary varying function okay so p tau if it's a very simple function then I can integrate this expression okay and obtain the solution for ut okay but if p tau is very complicated okay if it's very complicated then perhaps I won't be able to evaluate the integration okay analytically and then I'll have to go into numerical integration we will which we will see in a like you know future chapter okay but instead of doing that there are better methods to calculate the response for the numerical response instead of just integrating the Duhamel integral okay so this was just to give you an idea that if there's any arbitrary non-periodic or non-harmonic function loading x function pt then how to get the response okay it might not always be the like you know best method to go about finding the solution of a response of a single degree of freedom system but it's like you know it's good to have a knowledge of this function okay all right once you understand the Duhamel integral let us now go into some special cases of non-periodic loading okay and then we are going to see the calculate the response okay so what I would like to start with is step force okay a step force is typically defined as a force okay that you apply suddenly okay so it is like a step and then you maintain over time okay so let us say a load of amplitude p naught okay is applied suddenly and then it is maintained over time nine very first chapter we had already find the solution to this using the conventional by solving the differential equation okay so basically if you consider an undamped system we can go ahead and we can find out the solution to this using homogeneous sort of complementary solution plus the particular solution and we had seen that for an undamped system we had obtained the ut was coming out to be p naught pi k times 1 minus cos omega nt okay this we had obtained using solving the differential equation okay the same solution can also be obtained okay just to demonstrate you okay the application of Duhamel integral let us find the same solution using the Duhamel integral okay so remember for an undamped system the Duhamel integral say the response is basically 0 to t okay p tau sin omega n t minus d tau this is the expression now the force is actually constant okay so it does not matter what time you consider p tau would always be equal to p naught all right so if you substitute it here and integrate this expression okay let us say this is p naught sin omega n t minus t tau remember we are integrating with respect to integration variable tau okay so this we can write it as let us take p naught outside this would be minus cos omega n and then minus minus would get plus so this would be basically t minus tau omega n 0 to t and after you substitute the value and remember if I take omega n outside m omega n square would be k so this would be equal to p naught by k you substitute the very these limits of the integration I will get this as again 1 minus cos omega n t okay so I mean in this case you just happen to find out that this might be easier to do like that however for as the p tau or the loading function gets complicated Duhamel integral tend to not be a good method to calculate the response okay so in this case what do you see the response to a step function okay p naught by k okay we say that if this is the response like this okay p naught by k is nothing but the peak value of the static displacement okay so my dynamic displacement history is represented like this so this is nothing but basically equilibrium actually shifts from 0 okay if I try to plot the response from 0 it oscillates about us t naught which is basically p naught by k okay so let us say initially it was here okay as you apply this sudden load start oscillating about this load or this static displacement okay this is how the response would look like okay so basically when you apply a step force okay what do you see the system start oscillating about its natural at its natural basically frequency okay about a new equilibrium position which is the static displacement of the system okay due to the load p naught all right once you know that let us see what is the maximum value of this ut or basically the peak dynamic displacement now in this case as you can see this function is actually varying between cos omega n t between plus 1 and minus 1 okay so the maximum value would be when cos omega n is minus 1 or this I can say the maximum would be 2 times us t naught okay that would occur when omega cos omega n t equal to minus 1 so as we have seen if you had a statically applied load p naught okay what would be the deformation p naught by k and that is what you have been studying till now before this dynamic course okay however if you apply this as a step force okay you get a dynamic displacement which is twice the static displacement so it depends how the load is actually being applied okay so if the load is applied suddenly like a step force then you get a displacement which is almost two times the static displacement okay and so this is for a undamped system we can follow the same procedure for a damped system as well okay so for a damped system as well the response to the step force can be calculated using the same expression you can use that to hamel if you like but you will see that to hamel integral becomes very complicated in this case okay and you can go ahead and perform that integration and have a look at it the integrand that you will have here is basically okay 0 to t remember p of tau would be p naught here okay so I am just writing p naught times e to the power minus zeta and t minus tau and then I have the second term here okay and if you go about integrating this function it gets a little bit tricky you know values can still find it okay not okay you will see that it might not be the best possible way to go about it so let me just write down the final solution for this okay okay this has expression for the response of a damped system okay to step force all right so this is what do you get okay and if you see again this is oscillating about this ust however the amplitude of the second term okay so this is now oscillates with that frequency omega d or time period td okay but with time because of this exponential term it starts to decrease so the amplitude starts to decrease so let us say this was the undamped system let us say this was the undamped system okay damp system what will happen depending upon the value of damping the response will start to decrease and go like this and if the damping is very high okay it will go to this one very quickly okay this is u of t here and this is time t okay so this is what happens now in this case you saw that utility of Duhamel integral it won't be that effective I mean in this case if you had this expression let us say to solve for a damped system okay if this is equal to p0 you might be just like it might just be easier to like you know find it using the common method of basically that solve solution of a differential equation okay so if you remember particular solution we write it as p0 by k here and complementary solution we write it as e to the power minus zeta omega nt a cos omega dt plus v sin omega dt okay so the total solution you can write it as p0 by k times e to the power minus zeta omega nt times a cos omega dt plus b sin omega dt and you can substitute the initial conditions which we have assumed at rest initial conditions if you substitute this you can get the values of this constant a and b and you will get the same expression okay and as I said the response looks like so this is undamped damping is equal to 0 and these are the curves for let us say some intermediate value of damping zeta 1 or zeta 2 okay so this is what the response to a step function looks like okay and let us say if your goal okay is to apply a step force okay so that you minimize the vibration okay one of the example would be like you know you take a let us say you take a weight okay and you put it basically you basically put it on a any scale okay to measure this weight now you don't want too much of vibration because the reading would be fluctuating keep on fluctuating and if the damping is very small it would keep on fluctuating okay and it would give you correct reading so let us say if you want to weight in weighing machine what happens okay the weight is designed it is in a spring okay and this is like a step force right what is this step for this is like mg acting suddenly here okay so in this case you assign very high damping okay so that as soon as you drop this weight on the top of this digital weighing machine okay it comes to the rest very suddenly without any vibration to whatever the value mg that is being applied okay so it will converge to the value of mg okay alright so this was the response to a step force now we will delve into a different kind of force which is called basically ramp or linear basically ramp force okay so let us now consider ramp loading okay now when we said that and that in this case remember we said that if you apply this load statically okay and if you'll apply this load as a step function the step function suddenly applied a step function give you displacement which is twice the statically applied displacement okay now what do you mean when you say that you would apply it statically okay okay one way to define is basically apply the load so slowly that it does not produce a lot of dynamic effect okay and to do that we basically apply ramp loading so basically a ramp loading looks like something like this okay so if this is the variation of PT versus T okay it linearly increases up to the value of load that you want to apply let us say this is p0 if you want to apply p0 now you are not applying it suddenly now remember in the previous case you apply it suddenly now you are not applying suddenly you are applying okay slowly or I won't call it slowly it depends on the rise time let us say this is defined as the rise time okay rise time is basically time taken to reach the amplitude of the force in a ramp loading that is we call it rise time okay so in this case basically we are applying something like this and depending upon the value of tr we will see later that our solution differs okay if tr is very small then it is almost like a step force okay if tr is very large then there is like you know constantly linearly increasing function okay now for this let us first find out the response to lean sorry response to linearly increasing force so what I want to find out when the loading is still in this zone right here okay how does the response look like okay so if I apply so we will call this when it is still in the within still the linearly increasing force linearly increasing force remember for this case the PT I can write it as p0 t by tr okay for t less than tr and it is equal to p0 for t greater than tr okay let us put a equal sign here okay so we want to find out when it is still in this range what is the response okay and that is not very difficult to do if I consider a undamped system this is the equation that I get for my equation of motion okay this is the differential equation that we get all right and we can go ahead and find the solution to this again you can utilize Duhamel integral it would not be that difficult in this case okay or you can go using the conventional method now in this case let us go with the conventional method we know that particular solution I can use as okay p0 by k t by tr if we take this as a particular solution then this is one of the solution that satisfy this equation and complementary solution you know that it takes the form of for an undamped system okay we can write this as a cos omega nt plus p sin omega nt so my total solution becomes a cos omega nt plus p sin omega nt plus p0 by k t by tr and if you substitute the at rest initial condition u0 equal to 0 and u dot equal to 0 what you will see ut okay you can get the value of the expression for ut as p0 by k t by tr minus sin omega nt by omega n tr okay now if you look at let us just consider the linear part of the force here and then try to plot it okay this is my t the first part is nothing okay but the particular solution from here okay that is the particular solution from here and this is basically a linearly increasing function okay so the first part is basically p0 t by tr multiplied with a factor of 1 by k okay and this is basically us t okay remember what did we call when we say time variation of us t is when you consider a zero effect of mass in the system okay so you substitute mass equal to 0 and whatever the force basically you get pt divided by k okay that is basically your us t and that is what is it is here okay so remember pt was p0 t by tr okay so this is p0 t by tr okay so this is the force pt that is being applied okay and instead of force let me just write here the response u of t okay this is this line is my us t t and because if I am still considering in the linearly increasing zone it has still not reached the peak okay so there is no nothing like you know it is still increasing there is no peak value of us t yet okay and this is basically an oscillating function sin omega t okay when in when we sum this up these two function okay remember this function the second part is some looks like something like this okay and depending upon value of omega n and tr okay is amplitude would differ so the total response when you sum this kind of function and this kind of function okay it would look like something like this okay so basically the system start to oscillate again at its natural frequency omega n okay however about a about a about its static solution okay if there was no mass in the system no dynamic effect okay this is my u static and this is my total solution okay the difference is basically your this solution here so this is my u of t all right okay so we have seen that for the linearly increasing part this is how the response looks like and depending upon you know the value of omega n and tr it might look like something like this or it might also look like something like this and in many cases you know this is not actually desirable because I want to take the system statically without creating much vibration and if I get something like you know this curve here then it is not desirable so for those kind of system okay we have to apply or we have to increase the value of tr okay increase the value of tr so that this actually reduces so when you increase the value of tr it becomes closer and closer to the static solution okay all right we are going to conclude here in the next class we will see okay to this ramp loading after we consider this phase as well how do we get the total response and we will do that for damped system as well as undamped system okay all right thank you