 Hello everyone, so in the last class we basically discussed how to idealize a real structure into a more simplified model that can be used to analyze our structures subject to dynamic loading and we set up a single degree of representation of those systems. So we are going to extend that discussion today and we are going to see different examples and how to represent each and every system through a spring, mass and damper representation and we are going to learn about the inertial forces, the damping forces and the stiffness forces. And first we are going to look into the frame type of structure and then we are also going to consider different type of structures that we might encounter and solve some problems. So let us get started. So as we discussed we first considered the spring mass representation, sorry we first considered the frame representation. Now let us see how we can simplify the same thing to a spring mass damper system. So next thing that we are going to do is the spring mass damper representation, spring mass damper representation which basically is nothing but I mean just another way of looking at a dynamic system. So for example if I had this frame in which I had three component in which I said I had a damper here, I had a frame here which provided me the stiffness and then I also had a mass not here, initially it was here and that let us say displacement and then under the action of load it displaced by you, the same thing can also be represented to a spring mass damper system which is basically nothing but a spring here, a damper here and then I have the mass here on a frictionless surface. I am applying a force Pt, okay this is let us say K, this is C, okay so under the action of this force it goes through a displacement you, okay. So this is just another way to representation of the same system, so just keep that in mind. The single degree of freedom system can be represented with either of these two, okay and for further analysis and derivation we would be just using this representation while keeping in mind that this is actually representation of some sort of system, it might not be a frame type of structure, it might be some other type of structure but in the end this basically is a single degree of freedom representation of some sort of system, okay. Now this is or like you know let us again write down our equation of motion so in this case let me write it down, okay. So this is our equation of motion, okay for both system, okay. Now the same thing you know I mean many places you would also see instead of this you are going to see a vertical, so you are going to see a vertical spring mass tamper, okay. So both are like you know represent the same type of system with just one thing that you need to keep in mind I am just going to discuss that the effect of gravity. Now in this case if it is horizontal you have gravity acting mg and there is a normal reaction, okay. However if you are considering the motion in horizontal direction then these two quantities do not come into play, so you do not need to worry about those, okay. However if you consider now let us just consider only spring here, okay damper we can similarly add just a spring mass system here. So I have mass m, now this is a spring k. Now as you can imagine if you suspend mass m from a spring, okay then the spring would not be undeformed, okay. As soon as you hang a mass m from a spring it will let us say this was initial position, okay initial undeformed position as soon as you hang a mass m it will come to some initial equilibrium position, okay. And let us say let us call this delta s t, okay and this is your equilibrium position, okay. So the body is in equilibrium now and if you just apply again principle of statics I can just write k delta s t which is the force in the spring that should be equal to mg. So initial deformation is nothing but delta s t is mg by k, okay not very difficult. Now what will happen if you apply a displacement further displacement u, okay. The system will start to oscillate, okay because there is a restoring force or there is a force in the spring which will pull this mass back towards its equilibrium position and there is an mg which is acting down. Let us say force p t is again acting here, okay. So now let us say I am applying a displacement I am applying a force p t due to which I have displacement u. So if you consider a free body diagram of this mass in this new position here let us call this is a new position, okay which are the forces that are acting its p t downwards, okay. And because the spring is deformed what is the total deformation in the spring is a delta s t plus this u. So the total deformation would be k times u plus delta s t, alright. So just keep that in mind while writing the equation. So let me write it again here, okay. This force here is fs equal to k times u plus delta s t. And if there would have been a damping force there would have been like a damper here. You could have written similarly the damping force upward, okay. But now for this case let us just write down the equation of motion, okay. So you have fs then you have p t downward. Now there is also gravity that is acting. So I can write mg that is acting downwards. Is there any other force? No, I do not think so, okay. So now write down the equation of just motion. So what do you have here is and sorry yeah there is another force I completely missed that the most important force. If it is moving down either you can put a pseudo force through upward. So let us say m u double dot or like you know you could have used the Newton's second law and then the resultant of all these forces would have been equal to m u double dot, okay. So let us write m u double dot plus a then u plus delta s t is there. This should be equal to p t plus mg, okay. So p t and mg are there, alright. So now if you see I write it like this my k times delta s t is equal to mg. So this cancels off and I can just simply write it as p t, okay. So this gives me an important conclusion that if displacement of the body of course for a single degree of freedom of representation if displacement is considered from equilibrium position, okay. Then gravity forces can be omitted. Then gravity forces can be omitted, okay. So you do not need to consider in this case you do not need to consider gravitational force if you are considering the deformation from the position of equilibrium, okay. Because initially that gravity force was actually being balanced by the initial stored energy or the initial internal energy which is k times delta s t in this system, okay. So the gravity force can be omitted, okay. But there are two exceptions to this rule, okay, okay except. So let us say except when gravity force is the restoring force. We will see an example of that, okay. Or gravity force is the destabilizing force, destabilizing force, okay. So considering the first example where the gravity force is the restoring force, the most and one of the simplest example is a pendulum mass system in which you have a pendulum like this which is been connected. So constraint to rotate about this point, okay. So this rod is here, okay. A rigid rod is here and then it is rotating about this point. So let us say this is theta, okay. So let us derive the equation of motion of this body, okay. So what do we have here? Let us say this is the trajectory of the motion. I have force which is acting mg here, okay. And as it moves in the anticlockwise direction by the rotation theta, what will happen? I have a force acting mg, this force will have a component which is tangent to the direction of motion, okay. Or towards this direction, okay. So and that will act as a restoring force. So as it goes like, you know, oscillates about the position of equilibrium. Remember this is the position of equilibrium, okay. This gravitational force acts as a restoring force, okay. So if I have to write down the equation of motion, okay. Let us draw the free body diagram of this, okay. This is my theta here, okay. So there would be certain force here in this direction, okay. And either you can write down the equation considering the rotational equilibrium or you can consider the translational equilibrium of this mass here, okay. Now if you consider rotational equilibrium about the point of rotation, okay. You have this mg here, okay. And this distance here is if this is theta and the length of the pendulum is l, this would be l sin theta and it is creating a anticlockwise movement. Now I have assumed the direction of theta to be anticlockwise, okay. So if I have to apply pseudo force, it would be opposite to the direction of theta which would be again clockwise, okay. So this would be i equal to whatever the moment of inertia of this about this point which is nothing but m l square, okay times theta dot i alpha, okay. So instead of remember f i, okay. Here I have i times alpha, okay. So let me just write it as m l square theta double dot and then the moment of mg about this point which is nothing but mg times l sin theta that should be equal to 0. And for a small oscillation, okay, I can approximate sin theta is equal to theta and I can write it as theta plus g by l. This is my equation of motion, okay. I could have obtained the same thing if I had considered the translational equilibrium of the mass, okay. So in that case you could have considered, let us say this is, okay, translating about this point, okay. So you can write it as basically what would be the tangential velocity at this point? It would be l times a tangential acceleration l times theta double dot. So the mass times u double dot is actually m times l theta double dot, okay. And if you consider the moment about this point, it would be nothing again l theta double dot which basically is the same thing here, okay. Now this is the case where gravity is acting as a restoring force and it is coming in the equation of motion to give me this. So I could not neglect gravity here, right, because initially gravity was not producing any internal energy in the system, okay, at this position. In the second case, let me take an example again where the gravity is the destabilizing force and I am just going to invert this pendulum and then what I have here is a rotational spring with rotation k, rotational stiffness constant k, I have m here, okay. And if on position it would look like something like this, okay, where the weight of this one would always be acting downwards. So let me just write it as mg here and this is the direction of motion. So again, if I consider the rotational equilibrium, I can have i times theta i that is the moment, second moment of inertia of this body about this point, okay, not, sorry, not second moment of inertia, moment of inertia about, moment of inertia of this body about this point and so okay, let me just consider just, okay. So there would be ml, again I am assuming l to be the length of this, so ml square theta double dot and there would also be restoring force in this rotational spring here which would be k theta but then I have mg acting here, okay and this distance here is, if this is theta, this is l, this is l sin theta, okay. So if I write down the equation of motion, it would be ml square theta double dot plus k theta, both of them are producing anticlockwise and then clockwise moment due to mg, so mg l sin theta and for a small angle of course sin theta becomes theta and so I can write it as this ml square theta double dot plus k minus mg l, yeah, k minus mg l times theta equal to 0, okay. So this is the equation of motion now, okay. Remember the critical value of the weight or the critical value of stiffness at which the oscillation would not happen or this would destabilize would be what? Then k minus mg l becomes 0 or k becomes basically mg l, okay. So in this case the gravity was actually reducing the stiffness of the system, its stiffness would have been k times theta, however gravity is acting against it and reducing the stiffness of the system, so it is acting as a destabilizing force. So in these type of typical cases you could not possibly neglect gravity, however if gravity is balanced by internal forces, balanced by spring forces, if gravity is balanced by spring forces in the equilibrium position then it can be neglected in setting up the equation of motion if, if what? Setting up the equation of motion if the displacement, if the displacement is considered from the equilibrium position, okay, from equilibrium position, yeah, so keep that in mind, okay. Now let us come down to another example in which I will show you how to actually find out for a combined system in which you have multiple springs, not a single spring, a multiple spring but a single degree of freedom system, okay, how to find out or set up the equation of motion. So let me take a very simple example, okay, so I have a cantilever beam here, alright, in this cantilever beam what do you see? I have another spring here, okay, then there is a mass that is attached here, alright, so I have a mass which is attached here, okay. Now as you know in the previous lecture we said that if you have a cantilever beam, I mean there is a point load at the end, it also behaves like a spring, right, what did we say? If I have a cantilever beam and then I have a force acting, okay, and if this is length L and EI then I said that it could be represented by a spring like how? How do I represent it? K, this force Fs, Fs equal to K times U where K is 3EI by LQ, okay, so I mean this is not like a direct showing you directly with the two spring but just like you know you have to actually represent it like that, okay. So in this case I have to set up the equation of motion for this body, okay. Now as I discussed the last problem, we said that in the equilibrium position there would be some initial deformation, okay. Let us say that deformation is delta st, okay, and delta st would have some contribution of this beam here as well as some contribution of this spring here, okay, but I do not care, okay. I am going to represent it as a single quantity delta st, okay. So in equilibrium, let us say it is something like here, this is the equilibrium position and this is my delta st here. Now if you push it further then again it would vibrate, let us say this displacement is U, alright. So this displacement is U here, okay. And now I need to find out the equation of motion. So best thing for any type of problem is first to draw the free body diagram, okay. It makes your job very simple, okay. So if I consider this and if I draw the free body diagram of this mass, how would it look like or the whole system, let us say whole system. It is coming down okay. So there would be inertial force upward, okay. And because you are pulling the spring down there would also be spring force. So let us say this is the spring force Fs here, okay, and this is the spring force Fs, okay. So the same spring force is through the beam and then the same spring force is acting on this mass here, okay. Let me just redraw it again so that they are at the same level because if I draw the free body diagram this, okay. I have Fs that is acting here and then this is spring is the spring force, okay. What is the value of Fs? We are just going to write it down, okay. So I have a Fs here, Fs here and then I have mass m here, again the same Fs is acting on here, okay. And it is coming down so there would be inertial force Mu double dot here, okay. The mg would be acting here, all right. And are there any other forces? I do not think so, okay. So if I write down the equation of motion what would I get, okay. I would get Mu double dot plus Fs is equal to, yeah. And of course if there is any external force Pt then I can write Pt like this. So it could be T of t plus mg, okay. Now if I am considering my Fs to be, Fs would include the component from the initial deformation and the large one as well. But remember can I write here Fs is equal to let us say in this case whatever the deformation it initially had in the displaced position if I am considering U, U would have some component of the spring plus beam, all right. So keep that in mind. So Fs would be Ks times Us, all right. And it would also be equal to if you consider here Kb times Ub, okay. So the same Fs is being applied here which is applying the force Kb times Ub. And the same is there through the spring here, the same force is being transferred here. I could just explain you through the series and the parallel connection but I am not going to do that right now. I will do that as a consequence of this example instead of using that to where I just simply represent it like that, okay. So Fs is equal to Ks times Us and Kb times Ub. Through the Fs I can also represent this whole system as all the deformable part of this system as one system and through a single equivalent stiffness, okay. And through the total displacement which is the sum of because now I am considering for the whole system. So I am considering the U which is the total of Us plus Ub, all right. Okay. So here I can write it Mu plus K equivalent times U. Now remember in this case I also have to consider delta St, okay. So let us say this is the from the position of the deform but here if you considering Mg I have to consider delta St here as well, okay. So this would be Pt plus Mg. But initially I know that my Mg was equal to K effective times delta St. So that I can cancel off. I can get as Mu double dot plus Ke equal to Pt. So my only job is to find out Ke which I can easily do. How? Let us see. I have said my U is Us plus Ub. Can I say my Us is Fs divided by, what is it? Ke equivalent. Us is Fs divided by Ks and then Fs divided by Kb, okay. So my Ke equivalent, if I write it, it would be Kb times Ks divided by Kb plus Ks and then I can substitute it here, the only unknown here to get the final equation of motion, okay. So this system here where I have to deform body it behaves like a two springs in some certain sort of combination. We are going to see what typical type of combination I could have, okay. So let us consider two situation in which I have a spring like this in which this is K1, this is K2 and then of course there is a force that is being applied. So that is one situation and in the second situation I have two springs K1 and K2. Now what can you tell me about this connection and this connection? Can I say in this connection in the first one the force in all the springs are same, okay. So if this is Fs the same would be the force here Fs to all of them, okay. However the displacement is actually distributed between these two springs, okay. This is called a series connection of springs in which force is same in all springs but displacement is the sum of displacement in all the springs, okay in all the springs, okay. So basically what I mean to say u is equal to u1 plus u2, okay. And if you write that you just write if u is basically represented, if I am representing this whole system through a single system on which the force Fs is applied then there is a displacement of u with a spring constant K, can I say I can write it. So this system would have u as what? Fs divided by K is equal to Fs divided by K1, remember force in the spring 1 is also Fs and the force in the spring 2 is also Fs. So that display or deformation would be simply force divided by the spring constant and through this series relationship I can get it as, okay. So K is as K1, K2 divided by K1 plus K2, okay. All right. Now this is the series connection. Now in the second type of connection which is called parallel connection, okay. Displacement is same. So the displacements are same for all the spring. However, force is distributed between the spring. So in this case u is equal to u1 is equal to u2 and if I need to represent it through a single spring which has K equivalent as the spring constant with the same deformation u, okay. Let us see how do I do that. Okay. So as I know my force here, basically if I am applying Fs, Fs is equal to K1 u by plus K2 u. However, the equivalent system Fs is Ke times u. So simply K is equal to K1 plus K2. So the mathematical formulation itself is not very complicated. What I need you to remember when do we say a connection is in series or in parallel is what is important. For a series connection force is same in all the springs but the displacement is distributed among the spring and for the parallel connection displacements are same but the force is distributed. So using these two you can determine if the connection is in series in parallel if you ever get confused. You know and if you have a combination of series in parallel you can always like you know start taking two springs at a time and reduce it to a very simple system. Okay. All right. So I hope it is clear up to this point. All right. So the next thing that we are going to do till now we have said our equation of motion is what Cu Ku is equal to so this is our equation of motion. Okay. Now one of the very common situation especially for structural and earthquake engineers are is a scenario in which actually the base of the structure is excited by a ground acceleration. Okay. So what I am saying I have this frame here or I could do the same thing with the spring mass representation. I will show you how do I do that and this is undergoing a ground acceleration. Okay. So this represents an earthquake ground excitation okay. So this is earthquake ground excitation. All right. So in this case how do we set up our equation of motion? Okay. And similarly in terms of spring mass damper representation it will look like something like this have damper and then this okay. And the support that I have here this one is not fixed support anymore it is actually going through acceleration Ug. Of course because of this there would be some displacement as well. Okay. So let us say it is U and then there is a displacement of the ground due to earthquake. Similarly in this case if I write it like this is happening due to the ground movement before there is some shift due to ground movement and then of course due to internal deformation of the structure. Okay. So this much of component U is relative deformation of the structure. Ug is the ground deformation and then this total deformation is called ut which is the ground deformation Ug plus relative deformation of the structure. Okay. So this thing you need to know. Okay. And then of course in terms of velocity and displacement it would be also similar to the acceleration ut would be Ug of t plus U double dot t. Okay. So again if I have to set up my equation of motion let us consider the equilibrium of this body here. All right. So let us say a force well there is no external force let us say being applied here. Okay. So in this case if it is deforming then I have because of the stiffness component I have FS because of damping component which was there I have FD and then because of inertial component which is basically this not here is I have Fi component. Okay. And my equation of motion would be simply Fi plus FD plus FS is equal to 0. All right. Now let us see how do we write each of these terms here. Okay. So the inertial term remember my ground itself is moving. Okay. So the inertial force Fi I need to have some frame of reference. Correct. So I need to have some inertial frame of reference. Now in this case the ground is also moving. So the total acceleration of the body is what is actually ut here. Okay. So the inertial force on the body is ut which is nothing but mass times ug of t plus the relative deformation or relative acceleration. Okay. Now let us see what happens to the damping term FS and FD. Can you imagine in this case whatever the deformation happens along the two ends of the structure is what is responsible for the stiffness force. Okay. Try to imagine like this because of the rigid motion of the structure there is no deformation that is produced in the structure. Okay. So you have something like this here and then it moves here again there is no deformation produced. So deformation or the stiffness force is because of the relative deformation in the structure. Okay. So because of that reason FS is actually k times the relative deformation ut. Okay. Not the total deformation. Okay. And you know just to explain it I give an example of consider a spring in which or consider a train moving right and you are standing on the ground. Okay. And you are applying a force on the spring. Correct. Now if you apply the force whether you are on the ground or inside the train put the force in the spring be different. No. However consider yourself now you are on the ground and running. Okay. And the second situation you are inside the train and the running. Would you think the total inertial force when you would be different? Yes. Because on the ground if you are running the total inertial force when you would be just your mass times your acceleration. However in the train if you are running the total inertial acceleration on you would be or inertial force would be your mass times your acceleration plus the acceleration of the train. Okay. So there is a difference in terms of what quantities we consider whether it is total or whether it is relative. Acceleration is always total and displacement is always relative. Similarly the damping term is also relative because we define damping something like between the two ends of the structure. So the damping as we previously also discussed is due to the deformation in the structure leading to some energy dissipation. Okay. If you move something from one place to other like you know or like you know in a simple language rigid motion of a body does not lead to any dissipation in the body unless there is a deformation. Okay. So damping force is also related to deformation. Okay. So in this case for example initially my damper was here. Okay. And then it moved here. Okay. So the damping term is also relative. So it would be c times the velocity. So this one is very crucial to understand. Now if you have substituted all these terms in the equation here what do I get? I get m u double dot c u double dot plus as I give c u dot then a u and then I bring this m u u dot term to the right. Okay. So although it did not have any external force in here something like p t being applied at the top of the structure we see I am getting a term on the right hand side. Okay. And if I consider this to the equation of motion here consider these two here this is my effective force that is being applied because of the earthquake which is mass of the structure times the acceleration of the ground and that is why we say if I have a heavy structure okay then it would attract more earthquake forces if I have a lighter structure it would attract less earthquake forces. And one of the very good example is what happened during the Bhuj earthquake in Gujarat. See many of the houses residential houses had this water tank at the top. Okay. Now these water tank actually are very heavy masses okay supported on the top of the structure where the acceleration is highest. So you have ground acceleration then you have structural acceleration and then you have this right. Because of these are heavy concentrated masses the acceleration on these masses or the force due to earthquake on these masses are huge. Okay. And this led to failure of many of the overhead tanks and ultimately the failure of the overhead tanks also led to the failure of the roof and ultimately the failure of the building in many cases. Okay. So a very heavy mass attracts a very large earthquake force. All right. And if that heavy mass is distributed over a larger span or like you know foot print then the earthquake force individual on individual components are relatively lesser. Okay. So remember that we have considered external force as PT and we have also considered for earthquake and we saw that for earthquake I am getting the effective force as minus of m is opposite to the direction of motion. And this u here is actually relative deformation. Okay. With respect to the base of the structure or whatever reference point you are using. All right. All right. So we saw that there are different ways of coming to the equation of motion. Okay. And in the next subsequent chapter we would be dealing with how to solve this equation of motion and some of the physical reinterpretation of the equation of motion and there are different ways to solve this equation of motion. Okay. So with this I would like to conclude this lecture. Okay. Thank you.