 Welcome back everyone. In the last few lectures, we saw that how to set up the equation of motion and also saw how to find out the free vibration response of a multidegree of freedom system. Now, we discussed about damped system as well as undamped system. Okay, now we learned few methods when we were discussing single degree of freedom system as how to find out or how to estimate damping in a system. Now, we are going to adapt that procedure for a multidegree of freedom system and see how we can utilize different type of damping models to actually represent viscous damping in a multidegree of freedom system. So, in addition to talking about the significance of damping in a multidegree of freedom system, we are also going to look at few ways in which we assign damping to a multidegree of freedom system which includes Rayleigh damping as well. Okay, let us start the lecture. Now, if you remember, typically for single degree of freedom system to find out damping, what we had done experimentally, we had considered first free vibration response by giving initial dispirations and then taking the ratio of the successive peaks amplitude and find out the damping using logarithmic decrement method. The other method to find out damping was we had found out the peak response of a single degree of freedom system as we change the excitation frequency. So, let us say this is the response or let us say this is acceleration peak acceleration here. Okay, so we had obtained our frequency response curve like this. Okay, whatever peak response was, we basically had to draw our horizontal line which was at a value r by 2 and wherever it cuts the frequency response curve, we needed to find out omega a and omega b and utilizing this omega a and omega b, okay, damping could be determined using this omega a and omega b. Okay, and this was called as half power bandwidth method. Now, the question is when we have response of a multi degree of freedom system, the response is not due to a single mode, either the first mode or second mode. The response is always due to contribution of all the modes. So, how do we utilize this method or can we utilize these methods to actually find out the damping of individual modes? Okay, so we need to modify these methods a little bit or we need to adapt these methods to find out the damping of a multi degree of freedom system for the particular mode. And the limitations of these methods are that typically, we cannot find out damping of all the modes, but maybe only for first mode. Okay, and why is that when we consider response of a multi degree of freedom system, okay, either we can consider the acceleration response at some place. Typically, what happens? As I have mentioned, the fundamental mode is the mode for which the frequency is lowest or the time period is highest. Okay, so let me write it like this, omega 1 is less than omega 2, like, you know, omega n. So, t1 is greater than t2 and so on. So, through these methods, if I measure response, let us say acceleration response at some point, what happens? Initially, the response history is due to contribution of all the modes. Okay, so it would be something like, you know, like this. But as the time progresses, because of the damping, what happens? The higher frequency response is damped first. And that goes back to the chapter when where we have discussed that the damping actually damps out the higher frequencies first, okay, and the lower frequencies greater. So, after some time, the response that is remaining is primarily because of the first mode or the fundamental mode for which the frequency is the smallest. So, utilizing that, maybe let us say after this point, we can find out the damping using logarithmic decrement method here. Okay, or we can also find out the response frequency response curves. But in this case, again, considering the steady state response where the contribution of all the higher modes have actually damped out. So, using these methods, these two methods, we can only find out damping for possibly the fundamental mode, okay, or the primary mode or the first mode. Okay, so we can find out this. But the question is, if we need to find out the total response, then we need these damping value for each and every mode. And how do we get that? I mean, we cannot simply assume it to be zero. Okay. And the only information is that we have the damping maybe from the experiment for maybe the first mode. Or maybe what do we do? We can also assume that the second mode might be like, you know, somewhere close to the first mode. Okay, and the damping might be similar. But the thing is that we only know the damping for maybe at max one or two nodes. So, are there any methods through which knowing the damping for one or two modes can allow us or would allow us to find out damping for all other modes? Okay. So, that is the problem statement here. Okay. So, to overcome this limitation, what do we do? We consider proportional damping. Okay. Now, let us see what is proportional damping. It is a mathematical way to find out the damping for several modes given the damping for one or two mode. Okay. Now, we need this damping matrix or we need the damping ratio. Okay. The requirement, the basic requirement is that the damping matrix should be classical or like, you know, we should be able to diagonalize the damping matrix. Okay. Now, we know that mass and stiffness matrix are symmetric and they can be diagonalized. Okay. So, if we can write down our damping matrix as a linear multiplication of let us say mass matrix. Okay. The resulting damping matrix would be diagonal matrix. Okay. If we can find out this factor alpha here. Similarly, I can write down this as a some multiplication of the stiffness matrix. Again, my damping matrix would be diagonalized, can be diagonalized because my key matrix is a symmetric matrix. Okay. This is the first approach is called mass proportional damping. The second approach is called stiffness proportional damping. All right. But the question might come, what is the physical significance? I mean, how can we simply assume it to be like, you know, proportional to mass or stiffness matrix? Well, in reality, okay, maybe this stiffness proportional matrix can be, you know, justified saying that my damping matrix in the end is basically represents the velocity, the relative velocity between two points. And the stiffness matrix I know is represents the basically the relative stiffness between two points like a story stiffness. Okay. However, how do we justify the mass proportional matrix? Because in reality, like, you know, it is basically means that whatever your masses are, this damping is basically proportional to the mass matrix, or it represents if a heavier mass is there, it would provide you some resistance, so it kind of represents the aerodynamic damping. Okay. So directly, we cannot say either of these are like, you know, can be directly have some physical significance as such, but mathematically, we can say that we can write down our damping matrix as a linear or like, you know, as a linear multiplication of or some multiplication of mass or damp stiffness matrix, and we can find out those factor to see if that indeed would give us a diagonal matrix that can somehow represent the damping. Okay. And we can equate the damping in a particular mode using this approach to the damping for the mode that is available and the rest of the damping can be found out. So let us look at one at a time. Let us first look at mass proportional damping. So for mass proportional damping C is alpha times m. So if I diagonalize this, I can write down the diagonal C matrix as alpha times diagonal m matrix. Okay. Or if I write it in terms of element, I can write down Cn as alpha times mn. All you need to do is to multiply, pre-multiply with the transpose of the modal matrix and then first multiply with the modal matrix. So this is the relationship you get. Cn is equal to alpha mn. Okay. Now if that is the case, then the damping ratio for the nth mode can be found out as Cn divided by 2 mn omega n. So this is alpha mn divided by 2 mn omega n. So my damping ratio actually becomes alpha by 2 divided by 1 by omega n. Okay. The only unknown here is alpha. And how do we get alpha? Well, let us say from experiment, we know the damping for nth mode. Okay. Which would be in most cases the primary mode. Okay. But let us say for the numerical aspect, we specify for any other mode. Okay. So we say that, let us say we specify or assume, we know or we specify the damping for the ith mode. And that is equal to alpha by 2 1 by omega i. Omega is known to us from modal analysis. So my alpha constant can be found out as 2 zeta i omega i. Okay. And once alpha is known from the calculation for one of the modes, rest of the, for the rest of the modes, the damping can be found out as alpha by 2 times 1 by omega n. And like if you look at the variation of the damping here. Okay. Let us say this is zeta versus omega n. Okay. So this is like a hyperbolic distribution. Okay. Where basically zeta n is equal to alpha by 2 some constant times 1 by omega n. So if we specify for let us say first mode here, the damping in the rest of the modes would be smaller than the first mode. Okay. So this is the mass proportional damping. Now let us look at the stiffness proportional damping in which basically the damping matrix is beta, a constant times the stiffness matrix. So that constant is beta here. Cn, I can write the nth element of the diagonalized damping matrix now as beta times Kn. Kn is nothing but mn omega n square. Okay. Omega n is basically under root k by mn, Kn by mn. So I can write that. So the damping in the nth mode can be written as Cn by 2m omega n. And if I substitute that here, Bn mn omega n square 2n. So this would be beta by 2 omega n. Okay. So my damping in the nth mode can be written as beta by 2 omega n. Okay. So again like we did for the mass proportional damping, if we assume, let us say damping for the jth mode or if we know the damping for the jth mode, then zeta i can be written as beta by 2 times omega j. So beta would be 2 zeta j by omega j. Okay. And once beta is known from the damping specification for one of the modes, for the rest of the modes, we can find out zeta n is beta by 2 omega n. Okay. And if we look at the variation for this, if this is zeta n here and this is omega n here, this is actually a linear variation between zeta n versus omega n where the slope is actually beta by 2. Okay. So this is toughness proportional damping. Now as I said, we cannot directly justify in some sense that these actually represent some damping mechanism. But in the end, we have to specify the damping to the system using some mechanism through which we can fix the damping value for the modes that we know the values from experiments and for the other modes, we have to specify using some mathematical formulation. Okay. So in reality, the damping is neither mass proportional, it is neither stiffness proportional. Okay. But if we write it, the overall damping matrix of the linear combination of both mass plus stiffness proportional, maybe then we can sum up to some like you know, to some extent, we can say that the damping is being contributed to the mass and the stiffness always. So this would be closer to the reality than the previous two damping mechanisms that we have considered. Okay. So in this case, my damping matrix is written as a linear combination of mass and the stiffness matrix. And again, because mass is a symmetric matrix and the stiffness is a symmetric matrix. Okay. For typical structures, my C can also be diagonalized here and okay, it can be written as Cn as alpha mn plus beta kn. Okay. This I am writing, you know, if I just write multiply with the pre-multiply with the phi t and the post-multiply with phi, so that this becomes phi t m not capital N here. This is small n the and again beta times phi t times the stiffness matrix times the modal matrix. Okay. Which is basically the diagonal matrix C times alpha times diagonal mass matrix and then diagonal stiffness matrix. Okay. So that is where I get this from just considering the nth element, nth diagonal element. So again, the zeta can be written as the damping ratio in the nth mode, zeta n as Cn divided by 2 mn omega n and that is equal to 2 mn omega n. And this I can write as alpha by 2. Okay. Similarly, the following the calculation that we have done previously, 1 by omega n, this is alpha times 1 by omega n plus beta by 2 times omega n here. Okay. So let me rewrite this. Okay. So this is the expression. Now we know that we have two unknown constants alpha and beta. So we need two equations. So now we need a specification of damping or assumption of damping for two of the modes. So either we can find like, you know, assume it based on some experimental evidence or we can like, you know, just assign it some values which represent some representative value of the actual damping in the system. So what we are going to do is we are going to consider specified damping in two modes. Let us see. Let us say those are ith and jth mode so that my zeta i becomes alpha by 2, 1 by omega n and beta by 2, 1 by, sorry, this would be 1 by omega i now. Okay. Similarly, the jth mode becomes 2 omega j and beta by 2 omega j. So now what you have is basically two simultaneous equation in terms of alpha and beta and it can be solved and the values of alpha and beta can be found out once the alpha and beta are known. Then utilizing this expression here I can find out damping for any other mode. Okay. And in this case, for a special case, let us say when zeta i is equal to zeta j, the solution for alpha and beta can be written as 2 omega i times omega j divided by omega i times omega j times, let us say this value we assume as the same value as zeta. Okay. And beta can be written as 2 by omega i plus omega j times zeta. Okay. So for a special case if zeta and for the two modes if the same damping values are specified, then we can utilize this expression. And then once alpha and beta are known, then we can just use this expression here to find out the damping for any other mode. Okay. Now if you look at the variation for the Rayleigh damping, remember for mass proportional damping it was varying like this. For the stiffness proportional damping it was varying like this. Okay. For Rayleigh damping it is basically some of both damping mechanism it actually varies like this. Okay. So this is mass proportional and this is stiffness proportional and this is Rayleigh damping. Okay. So the Rayleigh damping overcomes few limitations of the mass proportional or the stiffness proportional damping which are if we assign certain value to stiffness proportional damping. Okay. It would mean that if multiple degree of freedom system has large number of modes. Okay. The damping would increase with the mode frequency and it would become unbounded after certain point. Okay. Similarly for the mass proportional damping if we specify based on like you know some modes which are now which is not the primary mode, then what will happen greater than intended damping would be assigned for smaller modes. However through Rayleigh damping we can consider the modes to which the damping used to be assigned and all the frequencies that would lie between the damping would be bounded for those. Okay. So there are like you know extensive research on the different damping mechanisms but for the purpose of this course we are only going to study up to this. Okay. And what we are going to do? We are going to do one example to see how we actually implement these damping mechanisms. So if we have the damping values for few of the modes or if we assume it can we find this out. Okay. For other modes. So let us do an example. In this example a three-story building is given to us. Okay. Like you know there are some masses and stiffnesses. This is m1, m2 and m3. Okay. So in general we need to find out the mode shapes and the frequency but for this case those are given to us. Okay. That the frequencies are 11.57 the first mode then 13, 31.62 second mode and then 43.2 the third mode. Okay. The mode shapes are also given to us. This is 0.289, 0.500 then 0.577 and 52 as minus 0.5770 and 0.57752 as 53 as 0.289 minus 0.5 and then 0.577. So the mode shapes are also given and it will say that for the first two modes the damping values are given as 5%. We need to find out what is the damping ratio. What will be the damping ratio for the third mode. Okay. So take 10 minutes and solve this problem then we can discuss. All right. Let us discuss the solution to this problem. Okay. We can find out our alpha remember this the damping values are same. Okay. So we can just use these expressions here to find out alpha and beta. Okay. When we substitute two times this is the omega 1 and omega 2. So 11.57 times 31.62 divided by 1.57 plus 31.62 and this comes out to be approximately 0.847. Similarly beta equal to 2 by 11.57 plus 31.62 and it comes out to be around 0.0023. So for any other mode okay any other mode my damping can be found out as alpha by 2 1 by omega 3 plus beta by 2 omega 3. So once we substitute the value this actually comes out around 6% or 0.06. Okay. So we saw that the frequencies utilizing the frequencies and the mode shape if we assume the damping values for the two modes shape the damping for the third mode can be determined using the relay damping. Okay. And one more thing to notice here in this case remember relay damping is something like this. We had assumed for the first mode let us say the second mode is around here and let us say third mode is around here. Okay. So the third mode which is outside this range okay it would be the frequency would be greater than omega and omega 2 the damping is around 6% right compared to the first two mode which is around 5%. If we had considered for the first mode and some third mode or the fourth mode and wanted to find out the damping in between of the modes that would be always smaller than the assumed damping. Okay. So this is just to like keep that in mind. Alright. So with this discussion we are going to conclude our discussion on damped free vibration. Okay. In next class we are going to start a new chapter on forced vibration. Okay. Alright. Thank you.