 Welcome back everyone. In previous classes, we basically discussed how to formulate equation of motion for undamped system and as well as damped system. Now as we discussed damping that we used in our differential equation of motion, it is of type linearly viscous damping, ok. So the question comes, if we have a structure or a system, how do we actually determine the value of damping in a structure, ok. So in today's class, basically we are going to look into how to experimentally determine the value of damping, what are the different methods that are available, ok. We are going to learn about two methods specifically, logarithmic decrement method and then there is another method which is called half power bandwidth method. And we are also going to see the nature of damping in a structure is not always of viscous type, but we can equate the energy dissipation mechanism in a structure to a viscously damped system and find out the properties or the equivalent viscous properties of the system. Because it allows us the simplified differential equation that can be easily solved, ok. So let us get started with the today's lecture. So till now what we have covered in free vibration is actually two types of free vibration. The first one that we discussed is undamped free vibration, ok. So we discussed undamped free vibration and the second type of vibration, free vibration that we discussed was damped free vibration, ok. And we set up the equation of motion for these two type of free vibrations. So we know that for undamped free vibration we would not have any damping term, ok. So let me write down the equation of motion here. So there is no damping term here and of course applied force is 0 on the right hand side. For damped free vibration we had and then now we also had damping term here. So let me just relate it, ok. So we had this term here and this was equal to 0. And for a given initial condition let us say u0 and u dot of 0, similarly for this type of vibration we derive the solution of this differential equation for the displacement at any time t, ok. And we saw that the ut for this was u of 0 time cos of omega nt plus we had u dot of 0 times sin of omega nt, ok. And for this we got it as u of t equal to u of 0 cos omega dt plus u dot 0 and then theta omega u of 0 divided by omega t sin omega dt, ok. So few important things to note for these two displacement histories. If you look at the displacement history for undamped part, ok. What we have here is a constant amplitude that can be written as basically something as u0 times cos omega nt minus phi where u0 is the amplitude of this motion, ok. And it can be written as times u dot of 0 omega n whole square, ok. As opposed to if we try to write down the equation of motion of this displacement history for damped free vibration we can write it as, ok. And then the inside term I can write as some constant u times cos omega d minus phi where u is nothing but u of 0 times, ok. So this is the expression that we get for u. So basically let me write this one as again u0 can be written as e to the power minus theta omega nt times u this quantity. So you can see if you compare the amplitude of undamped free vibration and damped free vibration you could easily see that for undamped free vibration this does not depend on time it is actually a constant value, ok. Only depends on the initial conditions and omega n, ok and it remains constant with time. As opposed to the second amplitude that we have here for damped free vibration in which I have this exponentially decreasing term that is multiplied with this u to give me the amplitude at any time t. So amplitude is actually decreasing over time, ok. If you wanted to draw both of these displacement histories, ok, on the plot here. So let us say this is my u of t and this is t, alright. So in this case let us say with some initial condition my free vibration starts like this with constant amplitude over time, ok. And then my damped free vibration will start at the same initial condition, however it will decrease over the time, ok. So it is going to decrease over the time, alright. And this is the envelope curve which represents the decreasing amplitude of the damped free vibration, alright. Second thing to notice is that in un-damped free vibration my system is actually vibrating with the frequency omega n, ok, however in damped free vibration my system is vibrating with frequency omega t and of course we saw that relation between omega d and omega n is nothing but this, ok, which for a small values of zeta omega d is approximately equal to omega n, alright. So once we have this clear, ok, before going any further this zeta is nothing but remember this damping ratio, we define this damping ratio zeta, ok, as c by c critical, alright. And where c is the damping coefficient of the system that is being considered and c critical is the critical value of the same damping coefficient for which it would inhibit or it would transition from an oscillatory motion to a non-oscillatory motion, ok. And you know, I mean we derive the value of that c critical, it was nothing but 2m omega n which can be further written as if you write omega n is under root k by m, this you can write as as 2 under root k m, alright, ok. Once we have the expression for zeta we saw that depending upon the value of zeta for example, I could have a system with damping ratio less than 1, I could have a system with damping ratio 1 or damping ratio greater than 1, ok. Then we categorize the system based on these damping ratios either under a damped system, ok, or critically damped system here or over damped here, alright. And most of the, you know, civil engineering structures are actually fall in this category of damped system, ok. Most of the civil engineering structure would have damping coefficients which are smaller than 0.1 or 10 percent, ok. But there might be some systems which should be critically damped and over damped especially if you think about like shock absorber or things like a retracting door mechanism that you have, they utilize critically damped or over damped systems, ok. So these would find the second, the next two damping systems would be more characteristic, more like, you know, it would be more common for mechanical system or aerospace systems, ok. So our focus of study would be limited to under damped system from here on, alright, ok. Because that is clear, let us see how do we utilize the damped free vibration and un-damped free, this type of free vibration to get the system properties, ok. So basically how to determine experimental determination, experimental determination of system properties or structure properties for our case. And the structure properties that I am specifically talking here could be time period, it could be, if it is a un-damped system, let us say un-damped time period, if it is a damped system, it will be a damped time period. And if it is a damped system, another parameter to find would be the damping ratio, ok. Remember it is very critical to find the damping ratio in many cases because we said that there are several sources of energy dissipation in a system. And just because we cannot explicitly, mathematically model those mechanisms, for example, friction at the joints, you know, internal, a little elastic straining of materials. So we said that all of those energy dissipation will represent using viscous damping. And for that viscous damping, what we derived the expressions for zeta and everything. So it is always like, you know, one would be the critical parameter that we can determine using the experimental tests. Now what do we need to determine those would be the displacement history, ok. So let us say displacement history is given to us, all right. So if the displacement history is given to us, ok, we need to determine what would be these properties, ok. And let us see how do we do that, ok. Now for damped system, we said that the displacement at any time t can be represented as u of t equal to u times capital U times zeta t times cos, ok, not a bracket here. So it would be cos omega d minus phi, all right, where the expression for u is already known to us which depends on the initial condition, initial conditions and the frequency of the system, ok. Now let us consider, ok, let us consider the displacement here at the next displacement here, ok. So basically what I am saying, consider two peaks of the displacement history, ok, at time t. So of course the next peak would be at time t plus td, ok. Let us say u1 is at u of t so that my next peak is at u of t plus td, ok. Now let us see what do we get as expressions for these, ok. So if I write u1 by ut, the ratio of these two peaks, ok, we substitute in this expression that you have here, we will get as zeta omega t times cos of omega d minus phi, ok. And the second, the denominator would be basically again same u, however I will have to replace t with the t plus td and then again cos omega d t plus td minus phi, ok. Now remember that cos is a periodic function, right, it is a periodic function with a period of 2 pi, ok. And omega d times td is basically 2 pi, ok. So this denominator costume would be equal to the same as whatever the costume you have the numerator, all right. And then you are left with so u will also, the capital U will also get cancelled off. So if you simplify this ratio, what you will get as this e to the power zeta omega n td, ok. And if you look at this value carefully, see that there is no time term in here. So what I am saying, u1 by u2 is actually constant, ok. And the same would be true for any time t and t plus td, ok. So any successive to successive peaks would have the same ratio, all right. So I can say that this is also same because remember the time t that I had considered here was any random time t, it does not have to be the first peak or second peak, two successive peak. So u1 by u2 would also be equal to u2 by u3, ok, equal to u3 by u4 and so on, all right, ok. So that actually remains constant, ok. Now if you look at this value here, my zeta omega n td can be written as zeta omega n and td is nothing but tn divided by 1 minus zeta square, ok. And omega times tn is 2 pi so I will write this as 1 minus zeta square, all right. Now we represent basically this as a value that is called delta, ok. And this delta plays an important role in further because this ratio is very important. So what do I have here is actually u1 by u2 is equal to e to the power delta here. If I take the logarithmic, the natural log of both sides I will get delta as nothing but equal to ln u1 by u2, ok or it can also be said ln u2 by u3 and so on, ok. Any ratio of any two successive peak and a log of that ratio is the value ln. Now this is called the logarithmic decrement. This ratio is called logarithmic decrement because this is the ratio through which the successive peaks actually reduces due to damping, all right. Now this lambda here that we have, sorry this delta here that we have 2 pi zeta. For a small value of zeta, this actually is denominator that we have here is approximately equal to 1 and we can write this as 2 pi zeta and this is like you know an accurate enough representation for all zeta smaller than 20 percent which covers almost most range of structures for our purposes, ok. If I have that then I can write 2 pi zeta equal to ln u1 by u2. So the damping can be obtained as or the damping ratio can be obtained as ln u1 by u2. Now remember u1 and u2 is they are these quantities are available to you from the experimental time or experimental displacement history that you obtained from your tests, ok. So, you know you can utilize either use of string potentiometer or some other lab testing instrumentation through which you can obtain that but the idea is that your damping is nothing but 1 by 2 pi ln u1 by u2, ok, all right. Once you have that then you this method will help you in determination of the damping, ok. Now let us further discuss one thing, let us say you have a system which has a very small value of damping, ok. Now if the damping is very small then what happens that this decrement between successive peaks are also very small, ok. So those are not as prominent as you see here not by this much but even further small, ok. This decrement is actually much smaller, ok. So let me try to draw this and try to be accurate and so on, ok. Now in this case what happens if you take two successive peaks for example here and here then there could be lot of error due to actually reading, it might be reading error, ok or it might be some other error, ok but it would lead to some error in the estimation of the damping value. So what we try to do for lightly damped systems we do not consider one peak but we consider over multiple peaks so that the error is actually distributed for example in this case let us consider this peak which I would call as peak I and after J number of cycles I would consider another peak which is peak J. So this is I plus Jth peak after J number of cycles, ok. So this is J number of cycles, alright. So I will have I plus J and it will have J number of cycles I get this, alright. Now let us see what happens if I consider any peak Ui, ok and the ratio of this with respect to any peak which is after J number of cycles so I have Ui plus J, ok. So in this case what I can do I can write this as Ui, Ui plus 1 then Ui plus 1, Ui plus 2 and so on then Ui plus J minus 1 then Ui plus J, ok. Now I know that the ratio of successive peaks are nothing but the logarithmic decrement delta, right. So what I am going to do here I am going to write this e to the power delta from this expression here again e to the power delta and so on to the power delta. This would be J number of times, ok because there are J number of cycles, alright. So this would add up in the to the power and it would give me e e to the power J delta. So if I take the log of both side I get delta as 1 divided by J times ln Ui plus Ui plus J, ok and if you write delta as 2 pi zeta ignoring the denominator term then the damping zeta can be calculated as 1 by 2 pi ln Ui Ui plus J, ok. Now this is over J number of cycles. So let us say if you had considered only successive peaks, right to calculate the damping it would lead to the reading error because now you are considering only two peaks so there would be reading error in U1 and U2, ok and then you are just using that ratio to get the damping. However, let us say you are using 10 number of peaks, ok. So the ratio over 10 number of peaks would decrease more and the error, the reading error is actually gets divided over 10 number of cycles so you are likely to get more accurate reading if you consider multiple number of cycles, alright, ok. So basically once we have this method which is called logarithmic decrement method to determine the damping or the viscous damping in a system, ok. We will utilize this to get the value so once you will have zeta, ok. If this is displacement history is given to you from the experiment, this you can also use to get the value of Td, ok and then if the Td is known then the natural vibration frequency can also be obtained using minus zeta square, ok, alright, ok. Once you have this, let us see how we can utilize this. So basically using this, these displacement history you can obtain these two values, ok and you know you can look at some examples where they have provided the displacement history to obtain the damping ratio of the system, ok, alright. Next topic that we are going to study, let me just fix this. So next topic that we are going to study would be the energy dissipation or the energy at any time T in a free vibrating system, ok and let us see how do we do that, ok. So what we are trying to study here is I think it has energy and free vibration, ok and let us first consider undamped free vibration, ok. So we are going to first consider undamped free vibration, ok. If I draw the spring mass representation I basically have this stiffness spring k, this mass m which has going through the displacement ut, ok and the initial conditions are given to you, ok. So if these for these initial condition the initial like the input energy to the system can be written as remember there would be two type of energy due to velocity of this mass there would be kinetic energy in the system, right and due to deformation of in this spring the potential energy is basically obtained as the stiffness energy in the system, ok. So there are two type of energy in the system and we are going to find out the expression for energy at any time T that is our goal here, ok try to find out the energy at any time T in the system, ok. So let us see the energy first the input energy to the system, ok due to the velocity or let me first write down the strain energy. So the potential energy is due to the strain energy in the spring which can be written as half k u0 of square and then the kinetic energy I can write it as, ok, alright. Now let us see at any time T the displacement is basically given as for undamped free vibration as u0 and cos omega nT plus u dot 0 sin omega nT and the velocity is given as you need to differentiate it once so that you get as minus omega u0 sin omega nT plus u dot 0 cos omega nT, ok. So at any time T again the total energy would be sum of the energy in the spring, ok which is the strain energy so I am going to write as E s of T and then the kinetic energy of the mass I am going to write it as E k of T, ok and if you write it as this E k of this square of this quantity here, ok plus half of, again you will you can write this as mass times the square of this u dot T here, alright. And just keep in mind this relationship that we had that omega n is under root k by m so you can write k as also as m omega n square, ok and if you utilize that and you take omega n outside from this here let us see what do we get, ok. So I am writing as m omega n square, right this square is here plus half m omega n square and then I will have a term here which I can write as minus u0 sin omega nT plus u dot of 0, ok cos omega nT, ok and there is this term here, ok. So I can basically take this common, ok and well I will simplify that and add that, ok and use the trigonometric identity that cos square omega nT plus sin square omega nT is actually 1, ok. Basically this will simplify to, in this case u0 square plus u dot 0 square plus the 2 ab term will get cancelled off for both terms, ok. So you can further write it as, so I am going to multiply this with this and then the same quantity with this except in this case what I am going to do here is actually write again m omega n square is k, so the first term that I multiply with this I am again going to write it as k times u0 square, alright and the second term that I am going to write it as would be, now remember that I have this term here, ok. So this is actually not only this much but this whole quantity square, ok. So when I write the second term this omega n square by omega n square will cancel off and I will get this, ok. So this is the total energy at any time T which we see is independent of time and this is what you should expect because this is a free vibration where there is no energy dissipation because we have neglected damping. So theoretically you know the system should have the same energy, ok energy should be conserved at any time T, ok. So this is the energy at any time T which is independent in time and this is also equal to the initial input energy that we had obtained, ok. Now you could do the same thing for the damped free vibration, ok and by using the expression for ut of damped free vibration and exactly following the same procedure, ok. Nothing is going to change here in that case as well, ok. So you can try that out but remember in damped free vibration when you try that you will find that the energy is actually a function of time and it is decreasing with time, ok. So if you consider let us say damped free vibration, ok. You will see that it could actually be decreasing the energy should be decreasing with time, ok and that dissipation in the energy is due to the viscous damping that we have assumed in the system. Now if you have any viscous damping force which is FD and due to this force if there is a dissipation energy and let us say the system is undergoing displacement U, right. The energy dissipation due to viscous damper can be calculated as force times the displacement, do you? Ok, over let us say time t1, ok. Now we know that FD can be written as C of the C time U dot, ok and du is basically nothing but again velocity times dt, ok. Basically du by dt is equal to U dot. So this is the expression I can further write this as Cu dot square times dt and depending upon what the expression is for U dot you can find out how much the energy that is being dissipated in the viscous damping term here, alright. So we have considered now the energy in a free vibration, in the free vibration for a damped system and then we discussed that how you can follow the same procedure to get it for the damped system as well. The same procedure is undamped and damped system, ok and the energy dissipated in the system can be calculated using this expression here, alright, ok. So with this all the topics on free vibrations are finished. What we are going to do now? We are going to do two examples to demonstrate the principles of this, ok and these are practical examples that you might observe and you would have come across this at some point of time, ok. So for the first example let me consider this, ok. Now many of you would have seen that when something is shipped to you for example if it is an expensive package, ok and then you buy it from online retailer or somewhere else or like you know something is getting shipped then it is usually shipped inside a box and if an item is of high value, ok. So if something is shipped to you, it is actually shipped inside a box, this is our example one, ok and they put an item, they put the item inside the box and they try to fill it up either with some kind of material that provides the protection to this. Now if the item is of high value and somebody let us say comes to you and ask you like you know that it is a very expensive item, I want to ship it. So I want to design, I want you would ask you to design the container and the material, this is stuffing material so that it would be protected against any kind of damage due to fall accidental fall or things like that, ok. So how would you actually go about modeling this problem and then solving this problem, ok. That is what we are going to discuss in this example, alright, ok. Now let us take the example of this expensive item inside the box, ok. First as an engineer you have to come up with some design parameters, ok like you cannot or you should not ship it like you know in a very large box otherwise what will happen the shipping cost would increase and you might think that this is just for one item but just imagine if everybody starts shipping in big boxes how much of cost escalation that would lead to in the shipping and the logistics, ok. So you have to come up with an efficient design. So let us talk about some design parameter, ok. First thing here would be what is the weight of the item, ok. Let us say it is a mobile phone, ok and a typical mobile phone let us say it is a mass of this mobile phone is around 0.1 or let us say it is 100 grams. So that first let me write it as 100 grams. So I can write it as 0.1 kgs, ok. Now for this mass, right I have to fill this box with some kind of material, ok. It might be foam, it might be like you know just a stuffed paper or it might be some other kind of material, it might also be bubble wrap. So these are the typical stuffing material that you use and for these materials you would typically have the value of the stiffness constant, ok. So example if I represent this box with this item inside and basically this stuffing material providing some kind of flexibility, ok to this item that is inside, ok. Now remember that for this problem you would only focusing on vertical impact because that is more important, ok. So although there could be it could be represented as horizontal springs we are only worried about the vertical stiffness of the spring, ok. So these two springs that are important to me, right. This and this, ok. So let us say just for the sake of argument it is 10 Newton per meters, ok. The third thing would be what could be the typical height of fall because that is going to determine the initial force or initial conditions that would lead to the damage to or the vibration or like you know the movement of this item inside the box. Now if you consider typical height of a person is around 1.75 meters, right. And let us say he is like you know getting up and down in a vehicle. So vehicle let us say it is around 0.5 meters from the ground, ok. So we are talking about let us say 2.25 meters of total height, alright. And just to add some factor of safety I am saying that the total height over which it can fall might go up to let us say 3 meters, ok. So I am considering that at max let us say 3 meters is the typical height of fall that in case of an accident like an accidental fall. Now what will happen? Let us say this box is made up of simple cardboard, ok. So it is not elastic. As soon as it hits the ground the box itself does not bounce but the material inside it starts to vibrate. So let us say it is falling over height of 3 meter on the ground. So what happens after it falls to the ground both and this like you know the mass of this cardboard is like in its although it would have some realistic mass we are going to assume it is leg cheaper compared to the item inside it, ok. So let us say it is falling so that falling through height of 3 meter this has mass m and there is stiffness k. As soon as it falls what happens all the velocity gets transferred to the system that you have inside that box. So this mass here, ok and that velocity would be what? I can simply write this as under root 2 gh, ok. And if you can calculate that, right you can calculate that as 2 times 9.8 times 3, ok. And you can calculate this value and that could be the initial velocity, ok of the further vibrational or the oscillatory motion of this mass, ok. So let us see what do we get that as. So I am just going to make some quick calculation here, ok. And I basically get that as 7.6, ok meter per second and this is the initial velocity. So what happens that as soon as it hits the ground the container comes to rest but it imparts initial velocity to the mass inside it and the mass now it starts vibrating, ok. So the mass inside it, it start vibrating with some initial velocity, ok. Now the design parameter on the off would be what should be the dimension of this box and how can we determine that? Well I can calculate that maximum displacement of mass due to this initial velocity, ok and that would provide me the approximate dimension of the box, ok. So let us say what is the maximum displacement due to initial velocity. So can I say that the U naught remember what we had calculated? It is U0 square plus U dot 0, this is square. Remember that this is 0 because there is no initial displacement only velocity U0 the maximum displacement would be simply velocity divided by omega n, ok. So it would be 7.66 divided by omega n and omega n is nothing but k by m which is 10 divided by 0.1, alright and that gives me a value of 0.766, ok, meters, alright which is basically 76.6 centimeters, alright. Now of course this is a factitious value and the realistic value would differ. Now remember the maximum displacement is now this U naught, ok and it could go in either direction. So the box dimension of the box height should be at least two times this displacement. So at least two times this 76.6 which gives me 153.2, ok. So let us provide a box height of 160 centimeters, ok. So this I have just demonstrated you like you know a typical application of this kind of system. You might think this is very trivial and I am like you know we are talking about a small stuff here but the same principle could be extended to any kind of system and actually it is like you know the shipping of expensive items is a big thing. Let us say if you are shipping something you know of the order of value of like you know million or billion of dollars you would not want to skimp like you know on just providing like in a proper packaging. So it becomes a very important issue, alright, ok. Now let us come to second example, ok. In the second example we are going to discuss the model of a car, ok. So let me draw here. Let us say I have a car. Now typical mass of a car might be 1000 to 2000 kgs. Of course it depends on the size of the car as well. Let us say it is a mid size car, ok. So let me consider this is a car which has a mass of, ok, 1500 kgs, alright, ok. And the suspension and the tires of this I am going to represent it with a spring here and the damping in that suspension tire I am going to represent with a damper here, ok. Now it is given to you then when it is at rest this car this spring is deflected. So initially it is deflected by let us say 10 centimeter which I can write as 0.01 meters, ok. And it is critically damped and it is critically damped, ok. So what do you need to find out what is the damping coefficient of the system the value of C what is the damping in the system, ok. Now and what is the frequency of vibration. In the second case what you need to do now let us say this car is now being occupied by four people, ok. And let us say average weight of each person is like in or the mass of each person is 75 kgs. So that for four people they would add around 300 kgs of mass. So the total mass becomes 1800 kgs. You now need to tell me after these four people occupy the car what happens to the value of damping ratio, ok. Now important thing to understand here is that the damping coefficient is the property of the system or the structure and it does not depend on the mass or the stiffness constant, ok. However the damping ratio zeta is actually which ratio of C by C critical, ok. And it depends on the mass of the system because C is constant and this I can write as 2m omega n. So it depends on the mass of the system. So initially it was critically damped. So the value of zeta would be 1, right. When we increase the mass of the same system remember the damping would still remain constant C value. However when you increase the mass m the damping ratio or zeta would decrease, ok. So it would become under damped and it might start oscillation in the system, ok, ok. Let us go as to one like you know one step at a time. So for the first like you know for the first part a let us say this is part a it is a critically damped system. So zeta is equal to 1, ok. And this should be equal to C by 2m omega n. But remember it is said that I first need to find out what is the stiffness of the system, right. And it is said that under the weight of the vehicle it deflects by 10 centimeter. So can I say that initially k times 10 centimeter which is 0.01 this is equal to what is the weight of the body 1500 times 9.81, ok. So from here I can calculate k as 0.01 and frequency I can also calculate, alright. Or instead of doing that let us just do this, right omega n by k by m and I can write this as 2 under root km, ok. Now C would be zeta is 1, 2 under root km. So a k I know is nothing but 1500 9.81 divided by 0.01, ok. And then I have m which is again 1500. So I can get the value of C from here, ok. Is that clear? Now once that is known, ok I can also find out the value of omega n from here and then T n from here. In the second part what I am going to do, ok. Now the mass is increased, correct, ok. Once the mass is increased and then the frequency is also going to change because that in terms also depends on the value of k, right or the m. So the mass changes and the frequency changes. So what I am going to do? I am going to take but the thing that is not going to change is the damping coefficient. So zeta 2 by zeta 1 if I take the ratio, ok. I will get that as m2 omega n2 divided or not sorry it would be other way round, ok. It would be m1 omega n1, ok divided by m2 omega n2, alright. Now in this case I can further write this as k1 m1 by k2 m2, ok. And remember that the k1 or the stiffness of the system is not being changed only the mass is being changed. So I can further write this as 1500 divided by 1800, ok which is nothing but so this would be 0.8, correct and then 33 like this, ok. So zeta 2 would be 1, remember zeta 1 is 1 so this would be nothing but 0.833 which you can calculate, ok and you can write this as 0.91, ok. So damping from the value of 1 it decreases to 0.91 it becomes an underdamped system and it would you know under the action of force or under the action of like you know initial displacement or velocity it can now vibrate, alright. So I hope these both problems explained to you the principles of free vibration and helped you like you know apply those concepts to real life problems, ok. You can extend this knowledge to different to solve like you know different type of problems in the free vibration, ok, alright. Thank you.