 Welcome back everyone. Today we are going to be taking up from the last lecture and further discussing how to set up the equation of motion for a dynamic problem. So, in today's class the first thing that we will going to discuss is basically how to simplify a typical structure or a system to an idealized system that can be utilized for dynamic analysis and that idealization would help us in obtaining the response of such system and first we will do a single degree of freedom system and we will discuss what a degree of freedom actually means and then see what is the differential equation for obtaining the dynamic response of a single degree of freedom system. So, let us get started. We are going to be starting the first chapter which basically introduces the basic equation of dynamics. Okay. So, we are going to be dividing this lecture in three parts. The first part would be how do we idealize systems for the dynamic analysis? Okay. Then the second part would be the components of a dynamic system and then in the third part we would be setting up the equation of motion. Okay. So, coming to the first part which is basically idealization to a simple system or to a system which can be used for dynamic analysis. Okay. As we discussed in the lecture before this one, if we want to obtain the response of any structure or any system subject to a non-static load then one of the steps that are required is to simplify a structure which can be used in our numerical analysis. Okay. So, let me just start by taking example of a building. Okay. So, in its very detailed form a building might have multiple floors. Okay. And it might have walls, it might have columns. Okay. Now, if we want to simplify this system for analysis, what we need to do? Depending upon what is the goal of our analysis, we might simplify this to a multiple degree of freedom system or what is called multi-degree of freedom system MDOF. Okay. So, what we are going to do? We are going to consider the tributary allocation of the mass. For example, in this case we are going to assume that all this slab is actually situated at this level like this. Okay. So, M1, M2, M3 and then we are going to consider is frame type columns and buildings, columns and beams form the component of this MDOF representation. Okay. So, this is one representation. Now, in this case, depending upon whether I am considering my columns and beam to be actually in extensible or not, I can further simplify it. Okay. So, I can further simplify this system assuming that these columns and beams cannot have axial compression or extension. These are rigid actually. So, in that case, what we will have, if you remember from your structural analysis, in general a 2D frame structure would have how many degrees of freedom? 3 degrees of freedom at each joint. Right. So, I would have 3 degrees of freedom. So, all of these would be at 3 degrees of freedom. However, if I want to represent it with columns and beam that are basically actually rigid, I can further simplify this. Okay. And I can represent it using a single degree of freedom at each level of the building. So, if I have to represent it or let me just go here again, I can represent it something like this. Okay. So, in this case, these would be K1, K2, K3 that represents the story stiffness and then I have all the mass at different levels which are represented by M1, M2, M3. So, what I have done, let us say this much of wall or I should have drawn three story. So, this much would go into the second floor, this much of would go into the first floor and this would go into the top floor. And I have simplified this model to a system that can be used for analysis. Now, this was for multi-degree of freedom system. Now, depending upon what is the goal of our analysis, this can even be simplified to a single degree of freedom system in which the total mass of the building for example, like this can be simplified through a single degree of freedom system. Okay. So, the total mass is here and then K. Now, you could argue whether this one presentation, SDF representation is an appropriate idealization of that system or not. Well, it depends what is the goal of your analysis. For example, if it is a rigid building and you need to find out just the base shear, then perhaps this might work out to be reasonably okay. But in many cases, this might not be an appropriate representation. However, what we are going to do, we are going to start our mathematical formulation with single degree of freedom system. And once we have understood the behavior of single degree of freedom system to different kind of loading, okay, then we are going to see how we are going to idealize a multi-degree of freedom system. Now, I have talked or I have mentioned many times degrees of freedom, right. So, I have said like you know degrees of freedom, okay. So, the question comes what exactly is degrees of freedom? Now, for our course or in dynamics in general, degrees of freedom represent the number of independent displacements of the mass that are required to represent the displaced position of a body with respect to its original position. For example, like in this case, if I have something like this and how many degrees of freedom would I need to represent the displaced position? So, let us say there is a horizontal earthquake and I want to represent the position of this one. So, in general, I would only need one degree of freedom to represent the mass with respect to or the position of the mass in the defined position with respect to its original position, okay. So, that thing you have to keep in mind, okay. So, this is degrees of freedom. Similarly, if I have like you in multiple degree of freedom, remember there could be how many degrees of freedom here, one would be for this, one would be for this and one would be for this, okay. So, I could have single degree of freedom system or I can have in this case three degree of freedom system or which is like you know a multi-degree of freedom system, okay. So, you have to keep in mind that how do we represent degrees of freedom, okay. So, now what we are going to do, I am going to show you some examples, okay, in which you have to find out how many number of independent degrees of freedom you would require to actually represent the displaced position of masses in the body, okay, with respect to its original position, okay. So, let us take our first example. In our first example, I have something like a beam which is spin connected at the left end and then there is a spring at the right end, okay. In the second one, so I am just going to present all four examples and then we are going to go through all of these one by one, okay. In the second example, I have a beam which is supported on two springs, okay. Now, in the third case, I have a fully mass system, okay. Just going to draw that here, okay. So, I have mass m here and then it goes over fully which is like this fixed here, okay. Just assume this to be on the top of it, okay. And then I have another spring coming out to me. So, let me just redraw it again. So, I have this spring in between and I have then another mass here, okay. Now, similar to this, there is also fourth case in which I have similar kind of settle but there is one small difference or like, you know, I mean, in fact, there is a big difference. Just draw it again and then you would know initials, this part of the setup is same, okay. So, this part is same, okay. So, there is another spring here which goes over this, okay. And then there is another spring which comes like this. Then there is a rigid bar here, okay. And it is connected, been connected at this point, okay. Then there is another spring and then there is another bar which is connected here and which is pinned at the support, okay. So, there are four examples in front of you and you have to find out how many number of degrees of freedom or independent degrees of freedom required for each case to represent the displaced position or body or in simple it would be just finding out what are the degrees of freedom of this system for dynamic analysis, okay. So, at this point of time, I would like you to take five minutes and just go through each setup one by one and find that out, okay. All right. Now, let me discuss each problem one by one, okay. So, what I have here, there is a, remember that all of these are rigid bars, right. So, in the first case, if this is constrained to rotate about this point, okay. So, it can only rotate about this point and there cannot be any other movement because bar is rigid, okay. So, the degree of freedom for this is one degree of freedom, okay. Now, let us consider the second example in which I have this bar which is supported on these two springs, okay. Now, in general, this bar can translate and this can also rotate. For example, if you think, let us say it goes, it rotates about its center of mass and then it also translates by this much. So, let us say this is the one degree of freedom and then it rotates about this point. So, this is another degree of freedom, okay. So, for this to completely represent the displaced position, we need two degrees of freedom, okay. Now, for the third example, okay, I have this mass here, okay. Let us call it M1 and this is M2, okay. Now, this mass, let us first start by saying the degree of freedom of this mass is U1. Now, remember whatever this pulley can rotate about this point. However, if you consider the rotation of this pulley to be independent, it is actually not because there is no spring in between. So, it is not flexible or you can see it is inextensible between these two points. So, whatever is U1 by the same amount, okay, the pulley actually rotates related by U1 equal to whatever the radius of pulley times the rotation of pulley. So, there is not an independent coordinate here, okay. However, there is again a spring between this point and this point. So, M2 need to be represented by another degree of freedom, okay. So, in this case, again, I have two degree of freedom system, okay. All right. Now, let us come to the fourth example. In this example, let us again start by saying this is U1, my first degree of freedom system. Now, in this case, if you look at it, this is not inextensible between the mass and the pulley, okay. So, in that case, the rotation of the pulley would be another independent degree of freedom needed to describe the displaced position. So, let us say this pulley is rotated, let us call it theta 1, okay. Now, whatever theta 1 is here, okay, again, whatever is rotated by this amount, there is a spring in between, okay. And this, the first bar is constrained to rotate about its support, okay. So, I would need another theta 1 here or let us call it theta 2. And then there is again a spring in between, okay. So, for this case, again, I need an independent degree of freedom theta 3. So, in general, I have 1 U1 and then 3 degrees of freedom theta 1, theta 2, theta 3. This is 4 degree of freedom system, okay. You could modify this problem a little bit and say if there is no spring in between, for example, right here, then the degree of freedom would reduce by 1 because whatever is theta 1, okay. This displacement here would be theta 1 times the radius of this pulley, okay. So, it is actually pretty much determined here and the rotation at this point would be, okay. So, this one would be theta 1 times the radius of pulley and this rotation would be theta 1 time divided by the whatever the length here is which is let us say L, okay. So, it is again not dependent. So, it could reduce a 3 degree of freedom system, okay. So, I hope now it is clear how do we represent degrees of freedom in a system, okay, all right. So, now let us move on to different components of a dynamic system, okay. And let us see how do we simplify a system to a mathematical model which is amenable for mathematical analysis or setting up the equation of motion. And in this case, I will simply take example of a single story building, a frame type building, okay. So, what I want to draw here, let me say I have a single story frame here, okay, something like this, okay. So, as you could imagine, it would be in three dimension, but I have just considered in frame saying that all the mass are concentrated along this line. So, basically what I am saying, let us say it would be in three dimension, but I am assuming that all of this is actually concentrated on this, okay, all right. So, let me just go ahead and delete this so that, now as you can imagine, this would have certain mass, even the column would have certain mass. But can you imagine if you consider a single story building or for a matter of fact like, you know, multiple story building, most of the masses are situated where in a building, can I say it would be at the floor level, okay. And of course, you can say that there are also masses concentrated at the wall, but what do we do? We say that half of the mass above and below the floor, I can just lump it at that floor level, okay. So, a representation for this single story frame building would be something like I take all the masses and lump it and here, okay. This is just one way to represent it, okay. Now, this system has mass which is, I am assuming it to be here. Now, if I apply a lateral load to this system, what will happen? The system would of course deform, okay. So, let us say the system deforms like this and the displaced position becomes something like this and this mass moves from here to here, okay. And let us say this is the displacement with respect to horizontal position, okay. And the same would be the displacement of the mass, okay. Because this representation is inherently assumes that these columns and the beams are inextensible actually, of course, okay. So, this is a simplified representation or single degree of freedom representation, okay. Now, remember, in general, if you remember from your undergraduate, okay, if I had a frame like this, how many degrees of freedom you needed to represent this system? Well, in this case, let us say 3 degrees of freedom here and 8 degrees of freedom here. Now, I could only reduce this system to a single degree of freedom system. Why? Because I assume if this is inextensible, so this degree of freedom, this degree of freedom would be same. If columns are inextensible, then there are no vertical degrees of freedom here, okay. And if I assume that this beam here or the slab here cannot be bent, okay. Then I can say that I can represent this with a single degree of freedom without any rotation here, okay. So, I can say that, right. So, in this case, let me just redraw it. Basically, when I redraw it, I will get the same figure what you see above, okay. So, what did we talk about? We have a mass here, okay. If I am applying a force, it is deforming. So, I have a stiffness and there is also a third component. So, what did we talk? What did we discuss? We have mass, we have a stiffness and there is a third component which is called damper. Now, can you imagine if you pull an elastic system or a flexible system, okay, and let it vibrate, would it keep on vibrating or infinity? Okay. In reality, it is not possible, okay. So, while for some system that idealization is assumed, in reality, what happens if you pull any system, there is always some dissipation of energy. And if you let a system vibrate, what happens after some time, any system would come to rest, okay. So, it would not keep on vibrating infinitely, it would ultimately come to rest, okay. So, damping basically represents the energy dissipation in the system due to which the amplitude of vibration of a system reduces to 0 over a time, okay. So, we have three components of this system, okay. And we are going to see how we can arrive the mathematical formulation for each of these three components, okay. So, let us see how do we do that, okay. So, the first representation we are going to talk about is actually for the stiffness, okay. Now, if you consider any system that can be deformed, okay. For example, I mean the simplest one is what, if I take a spring like this, there is no mass here, okay. So, just keep that in mind. And if I apply a force, which is the force in the spring, you know that it will deform, right. And how is the, let us say displacement in this is related to Fs. Can I write Fs is equal to k of spring times u, okay. So, this relationship you know, okay. Nothing complicated about this one. Similarly, for any other type of a structure as well, for example, let me take an example of a cantilever beam, okay. So, what I have here is actually a cantilever beam. And when I apply a load, let us say Fs here, you know that it will deform, okay. And what is this deformation here? If you remember from your structure analysis class, can I say u is Fs lq by Tei, okay. You remember this relationship, right, okay. Or I can write it in other way saying my Fs equal to Tei by lq times q, okay, which is similar to this relationship except now here my stiffness is represented by this quantity, Tei by lq, okay. And this I can do for like, you know, different type of structure. For example, I can do it for let us say simply supported beam, okay. I can do it for, you know, I mean I can do it for fixed supported beam, okay. I can do it for different type of system and then equivalently I can derive this relationship, okay. So, this is clear that when we apply a load on a deformable body and it deforms, then the system can be idealized through a spring, okay, with a stiffness and for which the applied force, applied force Fs and the displacement u is related through this relationship, okay, okay. If that is clear, let us come back to our example that we were discussing the single frame building here, okay. So, this is what we are talking about. In this situation, I am not considering any mass, okay. So, mass is not considered here. Just forget about mass for time being. It is not important. I mean if you look at it, mass is an, is it coming in picture in any of the steps here? No. So, let us just forget about the mass now. What is happening? If I apply a force on this, it will deform, okay, or if we draw it again, it will deform. Let us say this is the force applied is Fs, okay. It, due to this applied force, there is a deformation of u and if you take the free body diagram here, how does it look like if you have the system here, okay. The resisting force is whatever force you apply here, okay. Now, we have to find out how is this force related to this displacement, okay. So, we have to find out this Ks factor here. Now, if the system is elastic, we saw that we can directly find out using the principle of structural analysis, okay. So, Ks could be Ti by L cube depending upon what kind of setup you are considering, okay. So, for a small deformation, usually this relationship is linear, okay. So, if I let us say draw Fs versus u, it use, it would be usually linear. However, you can imagine if you keep applying the force or keep increasing the force, there would come a point at which the system would start to break or yield, okay, whatever you want to say and then it won't remain linear. There would be energy dissipation and then you have to consider the system, something like a non-linear system, okay. Pardon this surface spin here. It is not giving good result in the diagonal lines, which is okay. All right. So, we do see that the structures are actually like an integer non-linear. However, for this structural dynamics course, we would only be consider considering linear relationship of the force versus deformation, okay. So, that is Ks times u, okay. Only linear relationship would be considered, okay. All right. Now, let us consider two extreme cases for this frame here, okay. Two extreme cases, by what I mean, let me say in this case, in one case, okay, this is very flexible, okay. So, that the modulus of rigidity is zero, okay. And then again, you apply the force and in the second case, it is very, very rigid. So, that modulus of this one is equal to 180, okay. Now, if you apply the force and if you are asked to find out that tell me what would be the equivalent stiffness of this system, how would you do that? As you can imagine, if there is no flexural rigidity here, so if the column wants to bend here, there is nothing to prevent that rotation, okay. So, it would be just the displacement you applying like this and the columns or let me try it again here, very good. So, column looking like simply like this because there is no restraint at rotation here, okay. So, it would look like something like this, okay. In this case, because it is so rigid, these columns cannot rotate at these points of rotation. So, the deformed shape would look like this, okay. So, it would simply become case of a cantilever here. So, two cantilevers here. In this case, it is also cantilever but it is restrained again rotation, okay. So, basically this look like cantilever which is not restrained against rotation. This is a cantilever which is restrained again rotation, okay. So, if you remember from your structural analysis, okay. What is the displacement or u here? u is nothing but Fs and as we have discussed about Fs l cube by Tei and here this u is Fs l cube by 12 Tei, okay. Now, there are two columns in each of this. So, ks for this system would be 2 times 3EI by l cube and for it would be 2 times 12EI by l cube, okay. Now, in reality, you can imagine that the situation would be somewhere in between and for that there is a way to derive the relationship like in terms of let us say ratio of flexurigidity of the beam and column, what the values are but we are not going to get into that, okay. I am just demonstrating you two very simple cases, okay. All right. So, this is how the force and displacement relationship would be in this case, okay. Now, let us come after discussing the stiffness, okay. So, in this case, we have discussed stiffness. Now, let us discuss damping in the system, the second component, okay. So, we are going to talk about damping, okay. And as I mentioned before, damping is the process, okay. So, damping is the process through which amplitude of a vibrating system diminishes. So, through which amplitude of a vibrating system diminishes, okay. Now, you can imagine if you have a building and if you apply a load, there is various space in which the energy could be dissipated if you apply, you know, displacement to a system, okay. And damping represents here in this case especially what we are discussing, you have to think what are the different type of mechanisms. So, let us say one mechanism could be repeated elastic straining, okay. So, if you have repeated elastic straining leads to energy dissipation due to friction at the molecular level. This is one way. Other could be if you let us say have a building, let us say it is a steel building or other type of, let us say it is a wood building or any type of building. At joints, you could have like in a bolted connection or you could have like in a welded connection if it is let us say wooden building, again you could have different type of connections. So, the second could be friction at joints and connection of a building, okay. So, joints and connections, okay. What other mechanism you think could be there, okay. There could be friction between structural and non-structural component. So, friction between structural and non-structural components. If it is a concrete building, there could also be if you apply a deformation like in a back and forth, there could be closing and opening, okay. So, let us say opening and closing of micro cracks. So, many of these mechanisms involve friction and generation of heat due to which the energy is actually lost, okay. And that is why you cannot have in reality a perfectly, a perfect elastic system, okay. You would have non-conservative forces, okay. Now, let us say we have to model this, we have to model damping. So, you can imagine for the simple mathematical formulation, modeling each of these technique, okay, is very, very difficult and it is not required, you know. I mean, if I want to get into that much of detail then, you know, I mean, I would not do it for just a simple problem like this, okay. So, how do we do, what do we do? We combine all these damping mechanism, okay, into a simple mechanism and we are saying that we are going to represent all these energy dissipation mechanism through a damping called viscous damping, okay. So, viscous damping is there. Viscous damping, as you can imagine, it is also like, you know, called like, you know, viscous damper. What happens in viscous damping? Let us say I have a cylinder in which I have a piston like this and there are like, you know, fluid in each of these orifices, okay. So, fluid in each side of this piston. Now, what happens if you can imagine, if you push this piston, the fluid will flow around these orifices, okay. So, there would be resisting force, okay. Now, can you imagine if you try to push it very quickly, the force would be more okay. And if you try to push it very slowly, then the force would be almost zero. It is like, in real life, let us take an example, you have a fluid and then you try to push a flat object onto the fluid. If you do it very slowly, it won't offer any resistance, okay. But if you try to do it very quickly, it offers lot of resistance. So, a damping force, can I say it is proportional to not the displacement because there are displacements in both cases, whether you apply slowly or whether you apply suddenly. However, it is better correlated to velocity, okay. So, the damping force is proportional to velocity and the coefficient of proportionality is called damping coefficient and represented by c times u0. So, this is called linear viscous damping mechanism. So, although in a building or in a structural system, in reality, the nature of energy dissipation or damping is not actually viscous damping, it is just mathematically easy to combine all these energy dissipation in a single mechanism and represent it through viscous damping. And then we will see in a later chapter, how do we get this viscous damping coefficient or how do we get equivalent viscous damping to represent all those systems, okay. But right now, this is the formulation or this is the expression for viscous damper. So, you will see that later what we are going to do, we are going to represent a damper or viscous damper mechanism through this notation here, which basically like a simplified notation of this damper here, okay. So, I hope that is clear, okay. So, now again, let us come down to this one which we are considering. So, if I have to just represent the energy dissipation mechanism in this frame building, what I am going to do, I am going to connect a damper between two points of this structure. So, let us say I do not have anything, I do not have mass, I do not have stiffness, okay. Just to represent damper, I have something like this here, okay. And the force in this damper can be represented with F, sorry, I have been missing, there should be Fd here, okay, should be Fd forced from damping. So, if you consider, if we apply a force Fd here, the resisting force in this frame would be what? You consider something like this, and I cut the structure at some point, okay. Here, it would be Fd, same force would be applied here, okay. So, this is the damping mechanism that we consider, okay. So, once that is clear, now let us get into the third part which is the equation of motion, okay. So, the equation of motion that we are going to set up, so let us see how do we set up the equation of motion. So, as you know, our system was something like a frame here, okay, on which there was this mass right here, and then to represent all damping mechanism, I have this damper. So, this is how I am representing mathematically or like, you know, including all the components of our dynamic system, okay. Now, if I apply an external force here, okay, what my goal here is to find out the equation of motion for this mass or for this single degree of freedom system subject to this external force Pt, okay. Now, there are two ways in which we could do it and this comes from like, you know, if you have already done a course in engineering mechanics, okay. What do we do? If I have to set up equation of motion, either I can consider Newton's second law which is basically what? Resultant force resultant of all the force on the body is basically mass times acceleration, okay. So, I can use Newton's second law, all right. So, as per Newton's second law, okay, I have to first draw the free body diagram, okay. So, let us consider equilibrium of this mass and I am going to cut my system here, okay. So, what are the forces acting on this system? If you consider here, I have force Pt which is acting here, okay. And if I cut my system, as we saw previously, whatever the stiffness of this system is, I can write fs times s into u and remember that there is this damper here as well, okay. So, this damper here, the force in this damper would be fd, okay, is represented by c times u0 and then there is this mass here, okay. Now, what is the resultant on this mass m here? It is Pt which is and remember that this is actually deforming something like this here, okay. So, I am assuming that it is deformed by amount u. So, Pt, okay, minus fs minus fd the stiffness force and the damping force. This is the net resultant force. This mass should be equal to mass times acceleration which is m times u double dot and if I bring it on the same side, I can write it something like m u double dot plus fd plus fs is equal to Pt, okay. And if I consider this to be a linear made up of linear viscous damper and a linear system, the stiffness is linear, then I can write it this as cu dot and then fs as ku they should be equal to Pt. So, this is my equation of motion, okay. I will keep coming back to this equation of motion again and again so many times in the subsequent chapters. You will see that how to solve it, how to use this for like you know undamped system and damped system, okay. So, this is the equation of motion, okay for a single degree of freedom system for a SDOF system. So, Newton's second law was one way to do it. There is other way you can do it and that is basically using dynamic equilibrium of the mass, okay. So, I need to use the dynamic equilibrium of the mass. So, remember till now you have only considered in your engineering mechanics as statics, okay. You did not consider dynamic equilibrium and here we use what is called the Alambert's principle, okay. Which you might also know through the terminology called pseudo force which says that if a mass is accelerating, okay. I can apply an inertial force on the mass which is opposite to the direction of acceleration, okay. And then solve the equilibrium of the system as a simple static problem. So, what it says in this case, I will again have the same forces Fs, I will have here Fd, okay, due to this damper, okay, this Pt is here and this mass I will say I am going to apply an inertial force Ffi, okay which is opposite to the direction of motion and then I am going to write it as Fi plus Fs plus Fd and that should be equal to Pt, okay. So, I did not use that resultant force should be equal to mass term acceleration, I just apply on the mass and opposite equal, basically force that is opposite to the direction of motion and this is equal to Fi is basically equal to mass times acceleration and then I can have Fs and Fd, let me just bring it here and we have Cu naught Au should be equal to P of t, okay. So, again remember it is up to you which approach do you want to follow, okay. I have found out that for a multi sometimes complicated problem in multi degree of freedom system, this second approach might work out to be better, but you know I mean you just need to be consistent both of them would give you similar result, okay. So, remember setting up the equation of motion is the first step in a dynamic structural dynamics problem, okay and you need to be very accustomed how to do that, okay. So, we have done it through an example of a frame structure, you could have like you know different type of structure like no you could have one of the examples that we just saw here, let me see again, okay that four of the examples that I showed you and you could set up the equation of motion for those examples, all right, okay. Now, after we are done with this, remember the same thing mass damping and this stiffness, okay can be obtained considering a representation which is a simpler representation, okay a mass spring and damper representation, okay and this the mass spring damper representation have been typically used in courses like mechanical vibration or physics, okay. So, the formulation that I have described till now considering in mind that this is intended for structural engineers, but the spring mass damper is a more commonly adapted representation in mechanical vibrations and like you know other physical problems, okay. So, we will go in to discuss that representation as well, okay.