 Welcome back everyone. In today's lecture we are going to discuss what happens when a dynamic system undergoes pre-vipration. So in the previous lectures we have seen that when a structure subject to dynamic load then we learned how to set up the equation of motion. Now in subsequent classes including today we are going to looking up or we are going to be discussing some special cases that can be analytically solved okay and we are going to start with the simplest case of dynamic loading which would be pre-vipration in which case a structure is basically assigned some initial condition like displacement and velocity and it is allowed to let vibrate without any external force acting on it okay all right. So let us get started. Till the lectures that we have done so far we saw that how to set up basically the equation of motion of a system under action of an external force okay and we derived that the equation of motion can be written as f i plus f t plus f s and that should be equal to the applied force and for linear system we saw that we could write it as m u double dot then c u dot and then k u should be equal to p of t okay. So up to the last chapter our focus was to derive or to finally come up with an equation of motion which is of this form for a different type of system okay so we had different system under the action of external load and we saw that how to derive this equation of motion now what we are going to do we are going to study special cases of this equation of motion for different type of vibration okay now depending upon what is the external load and what my system comprise of we are going to deal each of we are going to discuss about each of these systems one by one and then see what are the application of those type of systems and then how do we actually or like you know how do we physically interpret those systems okay so if you look at these equation of this equation of motion we basically have three components okay now depending on whether we have a value of damping okay zero or non-zero and what is our force of excitation okay based on these two important parameters we are going to discuss different properties of a dynamic system okay so for example let me characterize what happens for each of these cases so in general we saw that we had equation of motion okay now we draw something like this then when damping is zero okay so when damping is zero we had three or let us say let us start with that when the applied force is equal to zero so Pt is equal to zero in this case then we will have what is something called as free vibration okay so this is free vibration and under this free vibration depending upon whether my system is as a zero or non-zero value of damping okay we can either have undamped free vibration or damped free vibration okay so again we are going to characterize between undamped free vibration or damped free vibration so remember that this was for Pt equal to applied force equal to zero this is c equal to zero okay and this is c not equal to zero okay so this is the one categorization one tree branch in the other part we have what we are going to discuss not a zero value of applied force but a non-zero value in that case we are going to discuss some specific cases for example what we are going to start with if we have non-zero value of the applied force we will have forced vibration okay and depending upon the nature of forcing function okay we are going to study where few specific cases and then few generalized cases so we can have like you know harmonic motion so when I say harmonic I mean something of you know I mean something similar something combination of sine and cos okay and then we could have a non-harmonic motion okay so we are going to say let us say yeah non-harmonic motions so we could have let us say so this one would be like you know something like sine cos etc okay non-harmonic could be like you know something like step functions or like you know pulse excitations or things like that and then we will have random excitation okay and random excitation we have let us say earthquake excitation or seismic excitation okay so depending upon what is the nature of the forcing function we are going to study all these topics and then of course like you know in each of these topics we are again going to study you know damped and undamped okay so again we'll have damped and undamped and we are also going to discuss that how we are actually going to obtain the response of those system for example if you have a harmonic excitation like it is feasible to obtain an analytical solution or a mathematical solution which is a closed form solution for non-harmonic cases if for very few specific cases we would be able to obtain analytical solutions okay for random excitation it is not possible the like you know because the excitation cannot be represented as a simple like you know I mean a combination of functions that are amenable for mathematical solution so in that case we are going to study how do we obtain solution using numerical methods okay so we are going to study that as well okay now today what we are going to do okay we are going to focus on this re-vibration okay what we are going to focus on this case today as re-vibration okay now as I said what do you mean by free vibration okay free vibration happens when you basically imparts an initial velocity or displacement to a system and let it vibrate without any action of external load okay so basically in short okay basically in short what we will have as pt equal to 0 all right so this is the condition for free vibration without any action of external load we let it vibrate freely okay and we will see that for this type of system we could have either undamped free vibration okay or we could have damped vibration and we are going to study both of these today okay now the question becomes that you know how do we impart or initiate free vibration okay so one of the ways to do that because we just said that there is no external force the question like in how does the motion start so we to start a free vibration we impart either initial displacement to the system okay so it would be either initial displacement of the system or initial velocity okay so initial displacement means u0 or initial velocity would mean u dot zero and like you know I mean you can see there are a lot of example for this type of system for example if you take a pendulum and then you lift it or if you provide initial angular displacement of let us say hit a knot all right and let it vibrate okay then it would go under the action of it would basically undergo free vibration okay so that is one example there are other examples as well for example initial velocity would be something you know I mean you take a hammer and then you hit something let us say an elastic system with that hammer okay and then that would be basically imparting initial velocity because in that case the hammer imparts lot of force in a very small amount of time it does not have time to react so it would not have initial displacement as such but because of the like you know large force in a short duration of time it would acquire initial velocity and we are going to discuss that like you know that is called impulse and how do we impart that okay so these are the two conditions through which we can initiate free vibration in a system all right now these type of system or these type of motion free vibration as I said because it is without action of any external load okay these type of system can be utilized or that this kind of motion can be utilized to obtain basically important parameters of a oscillatory motion okay for example so let me just write it here to obtain important parameters okay and two important parameters okay and we can we can do that experimentally so two important parameters are basically what is the and we are going to define that what is the natural frequency okay and what is the damping ratio okay so these two parameters can be obtained using the free vibration of a system okay so it is like you know it becomes very imperative that we obtain analytical solution for the free vibration of a system okay so the first thing that we are going to do in this chapter would be un-damped free vibration and then we are going to go to and discuss the damped free vibration okay so first thing we are going to discuss is undamped free vibration so remember this is a free vibration and this is undamped so free vibration means Pt is equal to 0 and damped means C is equal to 0 okay so basically our equation of motion becomes what if you substitute these two equation of motion I basically get this okay with of course we need as I said we need to have an initial condition otherwise if you solve this differential equation without an initial condition you would get a trivial solution okay so the initial condition a non-zero it need to be a non-zero initial condition either what should know sorry or what should be the values of u of 0 and u dot of 0 and one of them must be non-zero to initiate the free vibration okay so these two condition need to be utilized all right so what we are going to do first we are going to solve this system and obtain the analytical solution for u of t okay and then we are going to get into the physical interpretation of the response for the free vibration so solution to a second order linear homogeneous differential equation remember that we now do not have any term on the right hand side of this equality so we discussed that you know the how to get the solution of a homogeneous differential equation basically let me write my general solution u of t as e to the power lambda t and then we are going to substitute it back to the original equation so that I get after substitution plus a and that should be equal to 0 okay now e to the power lambda t term would never be 0 it won't give me any feasible solution so the only thing that can be 0 is what I have with the brackets here and that gives me the value of the lambda so there will be two roots of this equation this is the quadratic equation it would be plus minus under root minus k by m okay which I can write as under root k by n where I represent under root of minus 1 as the complex number i okay and this is nothing but let me just fix the view yeah so that I can write this as i omega n and we are going to discuss the significance of this omega n and this parameter so we are going to do that but let us first substitute our roots to the original equation which is u of t because it has now two roots I am going to write my solution as a linear combination of e to the power lambda 1 2 which represents the solution with respect to the first root and then e to the power lambda 2 which is corresponding to the second root okay so I will get here as and then b e to the power i omega n t if you remember from your complex numbers theory what did you have if you have e to the power i x you could write it as cos x plus i sin x remember and then e to the power i x minus i x can be written as cos x and minus sin x so if you make that substitution in this equation here and then again rearrange the term okay you will get something like a plus b cos omega n t plus let me just fix it here I will get it as cos omega n t plus i a minus b sin n t and again these two parameters can be represented as another constant the values of which need to be determined to obtain a specific solution this equation so I am going to represent as another constant omega n and then d sin omega n t all right so I have obtained the solution to this second order homogeneous equation linear homogeneous equation here now I have two unknown constant and that we are going to determine using the values of or the using the provided initial conditions okay so okay so let us substitute those values before that I need to differentiate it once to obtain the expression for u dot t so if I do that I will have as minus c omega n sorry this should be a sin omega n t here plus e omega n cos omega n t all right so let us substitute these values so u0 is equal to c and remember t equal to 0 cos omega n t is basically it would be 1 plus 0 because sin of omega n t at t equal to 0 would be 0 so I have obtained my value of c as u of 0 okay now u dot 0 would be equal to this term again would be equal to 0 plus I will have d omega n and then the value of cos omega t would be 1 okay so this gives me the value of t as u dot 0 okay if I substitute this to the equation that I had the expression that I had for ut I would get it as u of t is equal to u of 0 cos omega n t plus u dot of 0 sin omega n t all right so this is the solution of free vibration for the initial condition u of 0 and u dot of 0 and you know of course depending upon whatever the values of u0 or u dot 0 you have you can obtain the specific or like the numerical value of u of t okay so this means just let me write it this is as free undamped vibration okay so as you can see I have the expression for u of t and let us just try to plot this function okay and see how does it look like so I have a cos function and then I have a sin function okay so if you plot it if you plot these expression okay so what I have here is all right and then this is u of t and okay so I can write it as so if you plot it here it would look like something like this okay so what do I basically see here my resulted function is actually or the expression for ut is actually a harmonic function and also a periodic function which repeats itself okay it repeats after a specific time okay if you look at this graph here I know that whatever the initial displacement at time t equal to 0 was u of 0 okay and the slope initial slope of this ut is basically the velocity which I am going to write it as a u dot of 0 okay now in this case if you look at this or let me just write it like this take it from here so it has some kind of periodicity this function here okay and this periodicity is actually if you saw this function as you see the function here actually repeats itself after a or after 2 pi by omega n seconds okay and that you can you know prove by substituting or finding out the value of ut at time equal to t as well as t plus 2 pi by omega n t okay you can find that out from this and you can you can prove that the values are still the same okay now this quantity here 2 pi by omega n and then all the time after which it repeats itself is called the natural time period of the system okay so this is called natural time period of the system okay and remember that the value of omega n that we have obtained is under root k by n so now the value of t n would become using this relationship here 2 pi under root m by k okay so as I said t n is called the natural time period of the system okay and omega n is actually called the natural circular frequency of vibration okay so it is called natural circular frequency of vibration okay and as you can see from this relationship omega n is actually related to t n using this relationship here okay and the unit of omega n is actually radian per second okay and then there is another parameter which is called the natural cyclic frequency of vibration okay and we represent it using f and it is called natural cyclic frequency of vibration okay and the unit of this f is actually first let me just write down the relationship between f and omega n so they are related as you can see f is basically first let me in terms of the time period is 1 by t okay so this is the cyclic frequency of vibration okay and it is also related to omega n using this relationship here and the unit of f is actually cycles many places you will see it is written as cycles per second or in short cps but more commonly you will see that hertz as being used as the unit of the natural cyclic frequency of vibration okay now we are using the word natural quite a lot so you just have to keep in mind the significance of the word natural is basically that this is the natural means that it is without any action of external load and it is for an undamped system so these natural properties are so maybe I will just write it here natural properties are defined for undamped system okay without action of external load okay now it is very important to understand you know the physical significance of these parameters and what are the typical values that you would see for these parameters okay now time period as I said it is basically the time required to complete a single cycle of motion okay now if I have this graph here and if I had to represent it let us say for a system the spring mass system that we have discussed okay so let me say this is a this is b and I can take that for like you know for any section of this vibration cycle that I see here okay it would be c e at e okay so the motion is always considered about the equilibrium position okay and for many cases like an equilibrium position is also the initial position or the undefined position which is fine for example we considered when we considered the motion of a hanging the spring mass system hanging from in the vertical direction for that we said that let us consider the motion from this position so the initial position was equilibrium okay and then what we do either we impart an initial displacement or initial velocity okay or a combination of both okay so this starts from let us say a value of if I am measuring you from the equilibrium okay starts from a value of q equal to whatever the initial initial rate is provided so that would be let us say case a here okay then what happens as u is increasing it attains a or goes to a maximum say in this case somewhere around here okay and this is the maximum displacement of the system due to initial velocity and displacement okay and this is the position b that is shown here so this is position a this is position b okay so it will go to this maximum position then it again it will start coming back to the initial equilibrium position okay all right so this would be again c here okay and then it will go with the so when it goes to a maximum value of the displacement right here okay u would be maximum but the velocity would be zero okay and that you can see from this cycle here as it goes from a to b you can see the displacement is maximum however this curve is actually flat here which means that a slope of the this expression is zero the slope is actually the velocity of this spring mass system so the velocity is zero all right and then it starts coming back to its initial position okay and then it passes to its equilibrium position at which its displacement again becomes zero however it passes at the maximum velocity so at this point is what when is your velocity is maximum okay and that velocity carries this system in the opposite direction so the negative u direction as you can see here okay so I am going to represent that let me say here so this is the state d which corresponds to the negative maximum okay and then again after it attains this negative maximum it starts coming back to its uh equilibrium position at which it completes a cycle of motion and this uh you know this actually keeps on repeating okay so you this is the basic like you know the breakdown of a free vibration of an undamped system okay now let us come back to these three parameters that we talked about the time period the circular frequency and then the cyclic frequency okay and as I said that units would be second here it would be radian per second and here it would be hertz okay so uh now what do you think if I have a two if I have two systems a system a and a system b okay now system a if I am saying to you that a system a has very high frequency okay very high frequency okay of course uh you know this would be in comparison with system b here what are the things you can deduce from this information that has been provided to you okay if it has high if it has a system is a high frequency compared to uh system b there could be multiple possibilities one could be or like you know let us just list down those possibilities if the frequency is high remember that how is the frequency how is the time period connected to the frequencies is time period is either 2 pi by omega n or it is also 1 by f so it is inversely proportional to the frequency so that means the time period would be small small here okay large okay what other things you can deduce remember that my omega n is nothing but a by m so the frequency is higher that means it could be a stiff system okay compared to a flexible system and I mean this makes sense isn't it if you consider a flexible system can you physically imagine that a flexible system would take more time to come back to its or like in the finisher cycle of motion compared to a stiff system okay a stiff system if you give initial displacement it vibrates at a very high frequency okay high frequency means that there are a lot of or there are like you know higher number of uh cycles in a single second of motion which is basically consistent with this definition here 1 by f okay other possibility for the given scenario would be instead of saying stiff system or flexible system for a given same stiffness this could also be possible if the mass is small here okay so let us write down a small mass here it is large mass okay so these are the few of the possibilities that might result in high frequency system a system a with a high frequency compared to a system with b with low frequency okay so this you have to keep in mind if a system is stiff that means that it would have high frequency okay if a system has very large mass that means that it would have low frequency okay or I can also say that if a system a very large mass that means or flexible system then it would have large time period of oscillation compared to a system with a small mass and a stiff system all right okay once that is there uh let me come back to our expression for u of t again which we had obtained as u of 0 cos omega nt plus 2 dot 0 omega sin omega of nt and remember that we said that omega nt I can write this as 2 by by t so I can also write this expression as and then I can write this as similarly 2 by by t by tn okay now remember that we know from this graph okay what is u 0 what is u dot 0 uh we have also find out what is the time period of uh vibration or the oscillation okay there is one more thing that is remaining to find out remaining to be found out and that is the amplitude of the oscillation here okay let us represent this using u 0 okay so amplitude would be basically whatever the maximum values value of this expression that I have here okay so I need to find out maximum value of ut okay let us see how do we uh get that uh if you consider any function of type a cos let us say a cos theta plus b sin theta if you remember your trigonometric identities what I can do here let me uh multiply and in this case here I will multiply and divide by this under square root of a square plus d square okay and if you look at this carefully if I consider tan alpha equal to b by a then this expression is nothing but sin of alpha and this expression is nothing but cos of alpha okay but tan alpha as I said is b by a so I can write this further as under root a square plus b square and this would become cos theta into cos alpha plus sin theta into sin alpha and from your trigonometric identities I can write this as cos of theta minus alpha okay and that we can like you know substitute it back here so in this case remember my a is u of 0 b is u dot of 0 so I can write this whole expression as u of 0 cos of theta minus let us say I represent this instead of I will say this as phi okay now I know that maximum value of cos function is actually plus 1 so that the maximum value of ut would be let me write it as u naught with this okay so the free vibration with initial displacement and velocity happens around this static flip game position with an amplitude which is represented by this here okay all right once that is clear okay let us see how do we find out the important dynamic properties of the system okay till now we were deriving our equation of motion as mu double dot plus cu dot plus ku equal to phi of t we said that for a free vibration this equal to 0 for an undamped free vibration this equal to 0 okay so a simple problem on this would be a system an undamped system would be given to you okay and you could be asked to find out what is the value of omega n which you have to find as k by r m remember that there are multiple ways to obtain the equation of motion in terms of different degrees of freedom as well okay however whatever you do the omega n should be same because remember omega n depends on the stiffness and the mass of the system it does not depend on the applied load so it is a property of a system okay it does not depend on the applied load so that you have to keep in mind okay so once I write this expression omega n equal to under root k by n you can obtain the equation of motion differently with different coefficients okay point is what I am trying to make here is whatever the way you obtain your equation of motion your omega n should be in the simplest form coefficient of the stiffness term or stiffness force okay so basically here okay divided by m which would be the again coefficient of instead of saying the stiffness force let me just write it as coefficient of the displacement okay the second term I can write it as coefficients of the acceleration all right so using this I should be able to obtain my omega n which is of again as I said it is in radian per second okay and then I can find out the time period of the system as well as the frequency of the system okay all right I hope up to this point it is clear to you okay now what we are going to do we are going to do an example okay and this is not a problem for a book what we are going to do we are going to take typical property of the values of a system of a building let us say with a column and like you know beam and then see what is the value of the time period that we get okay so let us say a typical single story building is what do you think is the typical size of a room would be okay it might be let us say 10 feet by 10 feet of course if you live in Mumbai then it might even be smaller isn't it okay anyway so let us say I have this and I want to find out the time period of the lateral motion okay there was a problem remember in which we had considered torsional motion but right now we want to consider only the lateral time period of the system okay so let us say you know I mean there are four columns a typical column could be 300 mm by 300 mm and there are four of those remember e of concrete let us say it is 25 mpa concrete okay so let us say it is 25 mpa concrete column remember that e is nothing but 5000 times under root fck okay so I would get as e as 2500 okay 2500 mpa okay and if I consider these not to be actually let me just again write it or let us just say this is the fixed fixed connection although in reality it is not like like you know always fixed fixed like you know many times it would be pain because to like you know provide fixed it we need to do additional detailing and like you know provide rigid support to the ground okay but anyway just for the sake of this calculation let us assume it is a fixed fixed connection okay so remember that what was our stiffness in that case was 12 ei by l cube okay and a typical story height you can assume as let us say 3.5 meter okay so if I substitute it here I will get as 12 times e which is 2500 mpa okay and this I can write in 1 mpa remember 1 mpa is nothing but 10 to the power 6 Newton per meter square and what would be i of a section okay so if I have a section like this bd remember if it is bending about this axis then it would be i would be bd cube by 12 and in this case both are same the square cross sections so I can it would be just 0.3 to the power 4 divided by 12 that is my ei and the length is 3.5 into the power cube okay and this need to be found out so you can calculate this value okay let us see what do we get when we calculate that okay so okay so I get it as 47230 p okay and that would be let us say Newton per meter now in terms of mass what I could do I could lump all the masses here okay so half of the mass of all this and remember this is for one column sorry so what I need to do is use four times which I let us say I will do in the final calculation it is first talk about the mass mass you can lump it at the roof level by assuming that half of the mass of this these columns are lumped to the roof slab here okay and then the dead weight of the roof slab you can add it to that okay and then you can also add the live load okay and all those would come contribute to the total mass of this system okay so for the mass of the system let us say I have four column and I would assume that half of this is actually lumped okay now remember that cross section area is 0.3 by 0.3 and then I would have to multiply this with 2400 which I am assuming as the density of the concrete okay this is for the column for the slab let me assume like you know typical slab of 250 mm okay so in that case the total would be remember that the slab dimensions I can assume this has to be 4 meter by 4 meter okay so it would be 0.25 thickness of slab times 4 times 4 and then again 2400 so this is basically the dead weight okay but in reality there are other type of things as well for example floor finishing is there there might be like you know other things that are installed on top of it because I do not have information regarding that I am going to neglect that however what I am going to do I am going to assume a live load to this slab here okay which I would assume as a live load acting as 3 Newton per meter square okay which is very typical so I will write it as 3 Newton per meter square times the area which I have here is okay and remember that I need to convert this to into mass unit system here okay so this is Newton so to convert it to kg I will multiply this with 9.8 okay so let us see how much I get 4 so I am getting as kgs okay so let us say if I use this what do I get as time period b this is the mass and then I have four of these four of this column here so I get it approximately as 0.5 second and as you like you know do more and more problems and calculate the time period you would see that 0.5 second is very typical of like you know low to medium rise building as the height of the building increases the time period would actually increase but using these typical values we have obtained the time period to be approximately 0.5 second okay so depending upon what kind of structure it is I will give you some typical values of the time period of different type of structure okay so you know buildings depending upon whether it is a low and like you know low rise building or high rise building it could be between let us say 0.4 to 1.5 second actually for high rise building if it's a very stiff structure like nuclear power plant or something like that okay then it's a very stiff structure so the time period would be very small it is around 0.2 seconds okay you have typical values between 0.2 to 0.3 second or you could have very flexible structure like a suspension bridge okay and which could have like you know depending upon which direction you are considering time period even like you know seven to eight second in one direction or like you know three to four seconds so depending upon the direction you could have like you know different time periods which are like of typical values like you know varying around six seven second or three four seconds so you know it's a good practice to keep the value these values in mind to reflect like you know what are the typical values of different type of systems for their time period and frequencies just to get a feel of the these numbers and these parameters all right so with this I would like to conclude the lecture today thank you