 Welcome back everyone. In today's lecture we are going to extend the concept of frequency response factors to another factor that is called transmissibility and then see how transmissibility may be used to characterize the dynamic effect of load on the system. It could very well be the ratio of force transfer to a support or amplification due to in the response of a single degree of freedom system due to ground motion or as such. So, let us get started and see how can we obtain transmissibility for different type of systems. So, till now what we have studied in damped and undamped harmonic vibration of a single degree of freedom system is to find out the response for transient steady state and then we focused on the steady state. Now, we have till now studied response of the single degree of freedom system subject to sinusoidal harmonic load. So, the equation of motion was actually this. So, it was a sinusoidal excitation. But a harmonical load can also take the form basically form of a cosine excitation. So, whether it is P naught sin omega t or whether it is P naught cos omega t, in terms of excitation if you plot the excitation P t here. So, let us first plot sin omega t and then if you plot cos omega t there is no like a major difference except that there is just a phase difference of 90 degree that is the phase difference between sin omega t and cos omega t. So, it makes sense that when you analyze the system for peak responses there will not be much different except that your response would just be shifted by a phase difference of phi. So, you know and if you do not like you know if you do not cannot imagine this analytically or like you know just qualitatively what you can do you can actually solve this equation using the same procedure that we have previously followed. In this case my particular solution would be of the same form because it has cos and sin both terms. So, again it would be C sin omega t plus d cos omega t ok and the value of C and would be actually different here. However, the C square and let me just write it what would you get the value of C. If you solve this substitute this equation here and this and then try to solve it. You will get as 2 zeta omega by omega n and in the denominator you will have this term here ok. Similarly, d would be p naught by k and then it would be 2 zeta sorry. Now, I will have this term here 1 minus omega by omega n ok and the denominator I will again have the same term ok. So, while C and d are interchanged here and like you know. So, if you remember when you had the sinusoidal excitation C was actually the same term p naught by k. However, you had 1 minus omega by omega n square here and in d you had minus 2 zeta omega by omega n divided by this denominator ok. So, while these terms are interchanged the magnitude or the C square plus d square would still be the same ok. And so, if you again try to plot and like you know the homogeneous or the complementary solution would be same because it is just the solution to this equation here which is not changing ok alright. But we are not concerned about that we are mostly concerned about the particular solution because particular solution is the one that is the steady state solution of this excitation ok. So, if you write the steady state solution or for the cosine excitation it would be of the similar form except instead of sin now you will have cos here. So, u 2 would be u naught cos of omega t minus phi alright. And here in this case u naught I can further write as u s t naught times the displacement modification factor times cos omega t minus phi ok. So, already would again be same here and then phi would also be the same here ok. You can again write the same expression ok here alright. So, in terms of the peak responses there is not much of a difference only thing is that now the solution is has a phase difference with respect to the sin term ok alright. So, let us so what we have done till now we have obtained the response of this system dynamic response specifically the steady state response and we try to find out the different response factors ok. So, what are the response modification factor we have found out is Rd, Rv and Ra which are basically for displacement velocity and acceleration ok. And we did that because for some cases depending upon what kind of application that system has you might want to find out what is the change in the displacement due to the dynamic excitation what is the change in the velocity and what is the change in the acceleration ok. Now, what we are going to study ok with these factors now know ok we are going to study something called force transmission because as you can imagine it is all well and good if you try to find out the displacement velocity and acceleration, but many times what you have let us say I have a single degree of freedom system like this ok. And it is supported to some let us say a wall or a roof or let us say it is just a ground does not matter ok. The point is now in this case ok let us say a force is being applied here Pt equal to P0 sin omega t and we are not like you know in this case we are going to focus mostly on what is the force that is transmitted to this support ok. And it is like you know it is a very common case you if you have let us say any kind of rotating system attached to a wall or if you have like you know any kind of fan attached to let us say roof ok. So, it is like you know it is becomes imperative that you find out what is the total force that is generated due to the dynamic motion. So, that you can actually go ahead and design the support for that kind of forces. Now for this case you can understand if this force is applied statically or if I do not consider any mass here then can I say the total force the total support force ok would be in this case here ok. What would be the total support force? It would be P0 sin omega t ok. So, for the static case where there is no dynamic effect of the mass ok and if there is no dynamic effect there is no velocity across the damper. So, I can say that whatever the force that is being applied is the same force that is basically being resisted by the support here ok. Now what I want to find out if my system had the mass and it is applying the dynamic load at certain frequency ok then what happens to the total force? Now if you consider the free body diagram of this system here the total force that would be experienced by the support would be whatever at any time instant of course it would be the force in the stiffness K ut plus whatever the force in the damper is ok. So, the sum of these two forces I am just cutting this and I am considering the free body diagram ok. So, and now we know that given like an excitation PT what is the value of ut and u dot t. So, I am just going to substitute that here ok. So, I am going to write this as u0 sin omega t minus phi and then it would be C omega u0 ok I am differentiating it once to get the velocity ok alright. So, what I can do here I can write down u0 as ust the static amplitude times the RAD value ok. So, let me just do that write this as ust0 times. So, this I am u0 I am just taking outside in writing it as ust0 times RAD. So, I am left with K sin omega t minus phi plus C omega cos omega t minus phi. And as you would know I again have this inside term which is A sin omega t plus B cos omega t. So, the maximum value would be under root the coefficient square plus this square ok and the basically under root of the sum of these squares. So, if I take the maximum f t0 it would be ust0 times RAD and the maximum of this remember what I am saying that if I have A cos theta plus B sin theta the maximum of you know it would be under root A square plus B square ok because I can write it as a cos or sin function ok and the maximum value of that would always be plus 1 ok. So, here I am doing the same thing. So, I have now got this term K plus C square plus omega square ok. So, I am just considering the P like you know the maximum value of force transferred I am not writing the whole expression f t of t equal to that much ok. So, once I have that I need to further simplify let me write this as f t0 and this ust I can write P0 by K RAD and inside what I have here is actually K square plus C square omega square. So, let me take K outside. So, inside I have 1 plus C omega square K square. Now, if you remember my damping ratio is nothing but C divided by C critical ok and C critical can be written as either as 2 m omega n or it can also be written if you write m as K by omega n square I can write this as 2 K ok and this I would write it as omega n ok. So, I can go ahead and substitute C by K as 2 theta by omega n here alright. So, I am going to go ahead and substitute this value here. So, what do I get I will first do this this is P0 RAD and inside root I would get 2 theta omega n and then I have another term this omega square to the power whole square ok. So, if you consider my f t0 the peak force that is transferred to the support due to excitation the harmonic excitation is P0 times RAD times 1 plus 2 theta omega by omega n ok. This quantity here is called the transmissibility ratio or TR ok. So, it is called transmissibility ratio ok which basically represents that how much of force that is force that is being transferred to the support here ok. So, if you write it the peak dynamic force by peak static force is transmissibility ratio here alright and that I can write it as let us say if I have to combine both I can write out write this one as 1 plus 2 theta omega by omega n square divided by the RAD which I can again write as here as and then I can have this term ok. So, this is my transmissibility ratio which is kind of another basically a factor to represent the effect of dynamic load ok like we had the displacement modification factor velocity modification factor and acceleration modification factor we also have TR which basically represents how much of the force that is being transferred here to the support if you apply harmonic load to the mass ok. So, like what we did for this we can again go ahead and plot this function ok. So, let me just copy this figure here ok. So, if you plot it for different values of damping we get a plot like this here ok. So, I plot it here for different values damping. So, as you see here the transmissibility ratio it starts with a value of 1 for very small value of frequency ratio ok and then it increases at and becomes maximum at certain value of omega by omega n and then it again decreases. Now, the transmissibility ratio actually if you look at here. So, this is my line corresponding to 1 which basically says that the support or the force experienced by the support is actually P naught ok. Now, what happens if you keep increasing the value of frequency ratio there would come a point that force transferred to the support would actually be less than P naught or the static or the peak static force ok and that is a beneficial thing to have in many situations ok. For example, if you have a vibrating machine you would ideally like to have the total force that is transferred due to vibration of that machine to the float is smaller or you want to reduce that to whatever the frequency that is being all the force that is being applied by the rotating machine ok and for that this condition need to be satisfied ok that omega this is the value of root 2 ok that if omega by omega n is greater than root 2 ok then transmissibility is less than 1 or I can say that the force f t is actually smaller than the static force alright. Now, there is an important distinction between the graph that we have obtained here and the graph that we had obtained for the response modification factor r d, r v and r a. If you look at carefully what you actually see here ok that when you increase the damping ok these three quantity displacement velocity and acceleration these actually reduce at all values of frequencies. So, what I mean to say ok when we plotted it here we saw that it was reducing at all frequencies ok. So, if you increase the damping it was reducing the response was getting reduced at all values of frequency. However, just look at this here what we have for transmissibility ok. When omega by omega n is greater than root 2 as I increase the damping my transmissibility actually increases that tells me that the damping is not beneficial for the situation when omega is greater than omega by omega n is greater than root 2. So, in this region I cannot say that damping would always be beneficial with respect to the force transferred to the system and why that is happening here well that is happening because I have a damper here ok. So, the force in damper actually increased with the increase in frequency ok. So, when you take the sum of both displacement and velocity term what you see here if omega by omega n exceed this value here in this case damping actually increases the value of transmissibility ratio. So, before this this for different value of damping it was actually getting decreased ok, but if you look at here now the higher damping term is actually higher and the smaller damping term is actually lower here ok. So, this thing you have to keep in mind. So, there is always a trade-off when you are designing a system for example, let us say you start a machine ok, you start a machine which is rotating and it is kept on a floor and let us not say it is a wheel let us say it is a some support system I do not know what those support systems are ok. Let us say it is just supported to a wall or a floor or whatever ok. Now remember that you cannot suddenly go 0 to like you know frequency omega by or whatever the operating frequency is ok, you will start the machine and it will start rotating ok and then it will achieve its operating frequency. Now there is always a trade-off if you design a system which is very very flexible ok. So, in this zone right then what you would ideally like to do have some damping in the system so that as you achieve the operating frequency as you increase the frequency from 0 to its operating frequency which let us say is around here ok. So, I know that at operating frequency that damping is not beneficial here but I need some frequency so that it when it passes through the resonance frequency of this system the response does not become unbounded the response is not too large. So, you need to have some damping in the system ok. So, there is always a trade-off when you have to design systems like this ok. So, this transmissibility in this aspect is little bit different from the other modification factor response modification factors that we have studied till now ok alright. So, let us see after we have done this till now whatever we have considered in equation of motion ok is actually applied load in the form of this ok which was basically the PT ok. But it might just so happens one of the like you know very common situation is that where you have the support excitation ok. So, let us say I have a machine kept on a floor ok and the floor is due to there is a vibration due to someone walking here or there is another machine operating and this machine might get excited due to this vibration here or other example could be let us say there is a vehicle ok. There is a vehicle that is moving and due to ground it is getting excited here or there is a hump because of it it is getting excited or other thing could be you have a structure which is supported on the ground and there could be an earthquake which is applied through the base of the structure. So, in this situation actually the load or the force PT is actually not applied directly to the mass but to the ground ok. So, we are going to study this case next ok. So, basically response to support excitation ok. So, let us see what happens now we studied that if I had a system right. So, let us say I am going to consider a simplified single degree of representation of all these systems. So, I can consider something like this here in which ok this had a support excitation of let us say Ug of t and this had relative acceleration of U of t ok or another representation that we have previously considered ok. This is frame representation. So, in this case we said that ground is moving by certain distance initially it was here and then what happens ok. There is this one is basically the ground displacement ok and then due to deformation of this I would have the relative deformation which would be U of t ok and the total displacement would be basically the sum of ground displacement plus the relative deformation and same goes for the acceleration ok. So, the total acceleration actually is sum of ground acceleration plus the relative acceleration ok and same goes for the velocity and displacement. But we also know that the force in this frame or this here the spring force of the damper force it only depends on the relative deformation however the acceleration is always absolute ok. So, we derive this equation of motion if you remember we derive this equation of motion that in terms of relative deformation ok equal to mass of the structure times the ground excitation alright. So, in this case the p effective is actually this much ok and we can follow the same procedure to find out the response to this system. Now remember let us say if I am able to represent my ground excitation as sinusoidal excitation which I am going to write it like this ok. Then my p effective would be nothing but minus m Ug which is the peak ground acceleration times sin omega t. So, if you consider this as p0 I can write this as p0 sin omega t ok which is nothing but what we have already studied ok or let me say take the negative sign inside. So, that I will say that p0 is minus m times Ug double down ok. So, it is the exactly the same form that for which we have derived the response ok. So, what is the ut for this we know that the displacement ok. So, the steady state displacement steady state ut is nothing but dynamic peak amplitude displacement peak displacement times sin omega t minus phi and this is nothing but rd times Ust0 times sin omega t minus phi all right. And you can substitute the value of rd or like not the rd, but you can substitute the value of p0 by k. So, remember p0 here is now minus m times Ug double dot ok. So, let me write it that here m minus m G k times sin omega t minus phi ok. So, I can using this expression I can find out the relative deformation in the structure subject to the support excitation as well ok. Now, one of the important aspect of this is if I know the relative displacement I can get the relative acceleration as well right. How do I get that? Just double differentiate this expression here to get the relative acceleration which would be rd ok. And if you take the double differentiation of sin it would be minus cos or minus omega square cos omega t minus phi ok. And if I bring this m in the denominator k by m would be omega n square. So, I can write this term as Ug0 times omega by omega n square times sin omega t minus phi ok. So, the relative acceleration is also obtained now ok. Now, the important parameter in terms of acceleration is the total acceleration on the this mass that is attached to the support ok. So, what we are going to find out is actually the total acceleration which is nothing but relative acceleration plus the ground acceleration ok. So, you can substitute that here ok. And then I had the expression for the ground acceleration which was Ug0 sin omega t no phi here ok. And you would see that ok you can take this term outside ok. And then what you need to do just expand this term sin omega t minus phi knowing that my tan phi is 2 zeta omega n omega by omega n by this expression here ok. And then I again have sin omega t. So, similarly you can do that ok. You can expand that term ok. And then you have another term sin omega t here ok. And then if you try to maximize this what you will see this you get as U t double t is Ug0 alright times some factor which is equal to the transmissibility ratio. And then of course there would be some sin term here ok. I can write it as a sin term alright. So, this I can write. So, this is nothing but the maximum value of the total acceleration ok. So, I can write this as maximum value of total acceleration as ground acceleration times t r ok. So, this is equal to t r. So, this is an important conclusion that we have found out here that apart from the force that is being transferred. So, in the first case what we saw I had this system in which a harmonic load was applied which was of the form sin omega t. And we saw that the F t0 by P0 was transmissibility ratio. In the second case what I did I did not apply this force, but actually my support was getting excited due to a force ok which I represented as Ug of t double dot the acceleration as Ug or double dot ok times sin omega t. In this case I saw the acceleration or the total acceleration that is being experienced by the system ok. So, the amplitude of the total acceleration divided by the peak ground acceleration ok is also t r which is a very very important conclusion here. And this is a typical scenario for any kind of support excitation ok. Now, let us see if that is the case and I am again going to copy this figure here. So, let me just copy here ok. So, now firstly we dealt in terms of force transferred. Now, let us consider how much the acceleration is being transferred. So, my t r now let us consider the total acceleration divided by the ground acceleration alright. Now, as we can see if omega by omega n is much smaller than 1 ok. What happens actually ok for this case my t r is actually 1. So, my total acceleration is actually approximately equal to ok the ground acceleration ok. And basically this is the case when the excitation frequency is much smaller than the natural frequency of the system ok. If that is the case or you can say that the structure is rigid ok. So, in this case I can say that if this is a rigid so that omega n is like you know very very high ok. And that would lead to omega by omega n is much smaller than 1. And if you apply ground acceleration with a peak value of u g double dot we see that this would move rigidly with the ground. And that is why the acceleration experienced by this would also be u g naught ok the total acceleration. So, that my transmissibility is 1 ok. And this makes sense like you know if you try to imagine this system in reality as well. If you have a rigid system like this so that omega n is very high and omega by omega n is much smaller than 1. Then for this rigid system if you move the ground your structure would also move with the ground ok. And the amplification is actually 1 all right the transmissibility ratio is 1. Now consider the other extreme case when omega by omega n is much greater than 1. What happens transmissibility ratio is actually approaches to 0. And in this case the total acceleration is approaching to 0. And that you can again imagine let us say I have a system and I have a very flexible column ok. And in this case sorry the direction should be in this direction here ok. So, in this case what happened this mass is very very heavy ok. And when you apply ok so the heavy mass my frequency here is very very small omega n is very very small so that omega by omega n is much greater than 1. And now you apply ground acceleration with the peak value peak ground acceleration as u g naught double dot what will happen this mass there is no acceleration. So, this we have figured out this is actually equal to 0 what that means this will move the ground will move beneath this or the ground will move beneath this and the mass will stay at this point itself ok. So, actually there is no acceleration transfer the total acceleration to the mass is equal to 0 ok. So, this situation when omega by omega n is much greater than 1 what happens whatever the ground acceleration you apply the structure is actually isolated from that acceleration. And this is the principle behind vibration isolation ok this is the principle behind vibration isolation ok. And you would later see it could be simple vibration isolation of machines ok or it could be seismic isolation of structures as well ok. And the basic principle remains the same that the frequency of the structure ok is much much smaller ok than 1 or it is a very small value. So, that omega by omega n is greater than 1 and you can imagine this happens frequency becomes very small when your stiffness is very small or it is a very flexible system ok. So, what do you do you typically know that concrete is not very flexible right or a machine let us say it is a steel machine or something like that it is not very flexible. So, then the question comes how do I get that flexibility so that my frequency is very small compared to excitation frequency well we can use some flexible material. And rubber is one of those materials that can provide me very small value of omega n. So, if you take a structure or a machine and if you keep it on the rubber pad and if you try to move it what will happen this rubber has very small stiffness. If you take a pad of rubber and you try to deform it it is very small however it is still very good in compression. So, you still have the axial load capacity to support the structure, but in the horizontal direction the shear modulus is very less ok. So, it effectively leads to very small value of the stiffness and hence very small value of the natural frequency ok alright. So, we saw that whether it is the force transferred to a system for this case where the load was being applied ok. So, what did we see? We saw that it could be either a system in which the load is being applied to the mass and if you try to find out the maximum force divided by the peak static force this ratio was transmissibility or the other case was I have the same mass, but now if the support itself is moving or shaking or applying a harmonic motion ok. Then what happens in this case transmissibility is let me write this expression here transmissibility is whatever the total acceleration peak total acceleration divided by the peak ground acceleration ok. Now, a typical scenario of support excitation ok that might be encountered let us discuss one example. In some cases what you see ground is actually not plain, but it might be something like this. So, there are undulations on the ground or like you know and in some cases that might be approximated with a let us say a sine or a harmonic function ok. For example, I will give you an example this might be a elevated roadway or a bridge ok and what happens in the long term due to creep effects of the concrete you always get some sag ok. So, at the centre you would have some deflection alright. Now, what happens due to this deflection ok let us say this deflection is U naught I mean if the total deflection is let us say some value ok delta U naught is actually delta by 2. So, if I draw a line which cuts this at the middle point then I can approximate the curvature or the profile of this load road as a harmonic function. Let us see how do I do that. Let us say the span of this elevated roadway or bridge ok is L ok and then there is a vehicle ok let us say there is a vehicle ok not a very good representation let me say it is like this representing the suspension system and it is moving in the horizontal direction with the velocity V ok. So, what my goal here is to represent this profile of the elevated road bridge as a ground excitation. Let us see how do I do that I know that the general form of this function ok would be ok U g of t ok the displacement would be some peak ground displacement times sin of omega t where omega is ok the frequency of excitation ok. And what would be the omega here let us say the total time taken to cover this total span here. Now, if you look at here one span corresponds to basically one cycle here is not it. So, I can say the time taken which would be the total length of the span divided by the velocity would be actually the time period for this profile here alright and so the omega or the frequency which we would consider here is an excitation frequency would be 2 pi by t ok. So, here it would be 2 pi V by L. So, I can substitute here to represent the ground excitation as U g naught times sin 2 pi V by L times t ok. And remember U g naught is nothing, but this quantity. So, instead of saying it as U naught which I use for the structure let me write it as U g naught ok. So, this thing here is actually U g naught half of the total deflection ok alright. So, now, I have represented my ground displacement as an excitation in this form ok. Now, this is not exactly an acceleration remember that expressions that we have derived here these are in terms of acceleration, but it is not very difficult to find out in terms of acceleration right if sorry. If I differentiate it twice ok I will get this as minus ok this let us say whatever the frequency here I am saying omega here it would be omega square times U g naught times again sin omega t ok. And now this would become basically the peak ground acceleration, but you need to be careful here in this term amplitude now depends on the amplitude of the excitation force depends on the frequency of the excitation or the excitation frequency. So, this is not actually constant ok as with the other cases this is actually not constant. So, you need to keep that in mind. Now, for those cases where ground displacement can be represented as if the ground displacement is represented as U g naught sin omega t you will see that I can again follow the same procedure differentiate it with respect to t and then get the U g double dot t substituted then again get the like you know total basically relative acceleration ok. In this case what happens if the ground displacement is given in this form you can substitute and you can find out that the total displacement divided by the ground displacement. So, the peak of total displacement divided by the ground displacement also comes out to be transmissibility ratio t r ok. How do I get that? Well I can write my total displacement at any time t as equal to the relative displacement at any time t plus the ground displacement at any time t and U t I have already figured out it is basically we got that as r d times U s t naught times sin omega t minus phi ok and U g I have got as U g naught ok sin omega t ok and if I write U s t naught as p naught by k which is effectively minus m ok and then acceleration which is omega square times U g naught remember I have already got that ok here this expression. So, this is I can write here as minus r d times ok p naught is remember minus of that effective acceleration. So, I will write this as m omega square U g naught ok and that divided by k there would be m term here as well and then sin omega t minus phi plus U g sin omega t ok. Remember that there is a minus sin here as well. So, this term this minus will go off ok and you can get this one as U g naught ok r d times omega square and if you bring this m to the denominator again I can write this as omega by omega n square sin omega t minus phi plus sin omega t and remember if you remember from the previous what we have done previously here this is the same thing that we had obtained here is not it look at this look at this expression instead of U g double dot ok acceleration term I have just the displacement term here ok. So, the maximum value of this one would be if you just consider the maximum just remove the t term here it would be this times t r ok. So, this is equal to the transmissibility ratio again all right ok. So, we saw that transmissibility ratio can be used to represent basically three type of ratio one is in terms of the force transmitted for the case where the harmonic force is applied to the mass ok. Other in terms of the acceleration transmitted to the mass if you shake the ground and the third is the total displacement of the mass with respect to the displacement of the ground ok. So, these are basically important parameters and you have to make a distinction you have to remember when to use these transmissibility in what scenarios ok. So, what we are going to do now based on what we have studied ok we are going to look at some examples of this ok to see how does it work all right how we can apply this principle. So, in the first case we consider the concept of vibration isolation ok. So, let us say I have we consider example one here in the first case let us say I have some like you know sensitive machines here and I mean this machine is kept on the floor and supported by let us say rubber pad this whole rubber pad here ok. We typically use this in like a laboratory where we do not want any vibration coming into our readings in measurement system ok because of some like you know others the things shaking or vibrating. So, we typically keep this sensitive instruments on the top of these rubber pads ok. Now, it is given that the total mass of this machine is 50 kg all right. The stiffness or the vertical stiffness of this rubber pad is actually 1400 Newton per meter ok or sorry it is 1400 Newton per meter ok. And the vibration due to surrounding can be represented as ok a harmonic excitation with an amplitude of ok. So, with an amplitude of 0.1 g and a frequency of 10 hertz ok. So, this vibration can be represented as an excitation function which is equal to ug double dot t is equal to 0.1 g and frequency is this much. So, k would be 2 pi times 10 which would be 20 pi. So, I can write as cos 20 pi t. This is my harmonic excitation to this machine here ok. What is being asked is that what would be the acceleration transmitted to the machine here due to this vibration given that the damping ratio for this system right now is 0.1 or 10 percent ok. So, how do I do this problem? In this case ground excitation is given to me in terms of acceleration remember acceleration not the displacement and it is asked to find out how much of the acceleration could be transmitted to the structure. So, this is a typical transmissibility problem that ground acceleration is given and you need to find out how much of that acceleration is transmitted to the machine here. Now to do that first I need to find out what is the natural frequency of vibration of this machine which is supported on this rubber pad ok. That I can find out by considering the vertical stiffness of the rubber pad ok which would be 14000 k divided by m. So, 50 kg and this gives me a value of omega n as 16.7 ok. So, the ratio the frequency ratio omega by omega n that I will get would be 20 pi which I have already found out here divided by 16.7 and if you calculate this you will approximately get this ratio as 3.75 all right. So, let us use this ratio to find out transmissibility which is nothing but 1 plus 2 zeta omega by omega n ok this square divided by this. So, in this expression not a square here in this expression I can substitute the value of omega by omega n zeta is given to me as 10 percent ok to get the transmissibility ratio as equal to 0.1 ok. So, this I can write is as the acceleration transferred to the machine ok the peak acceleration transferred to the machine divided by the peak ground acceleration ok which is nothing but this divided by 0.1 g ok remember 0.1 g is the peak ground acceleration or peak support acceleration to be more precise here. So, the acceleration transferred to the system would be 0.1 g times 0.1. Now, this I get as 0.01 g. Now, what do we see here add a rubber pad ok and then the machine was supported on that one my transmissibility ratio is actually 0.1. So, although the ambient vibration on the support was around 0.1 g due to this rubber pad and because it is flexible ok by acceleration that is transferred to the system is actually reduced by 10 fold and I get this as 0.01 g ok whether this is acceptable or not for the operation of that machine is another issue and that brings us brings us to the second part of this question which says just imagine this instrument can only work accurately if the acceleration is up to 0.05 g or not 0.05 g but 0.005 g ok and as a designer you are asked that what you would do ok that if you are using the same. So, rubber pad need to be same. So, are not changing the rubber pad ok. So, what would you do as a designer right now the acceleration transmitted is 0.01 g but it can only work accurately the machine if the maximum acceleration is within the limit of 0.005 g. So, half of this value. So, how would you reduce this given that you need to use the same rubber pad ok. So, I know that if the rubber pad is same then my k is same correct this stiffness is determined by the rubber pad and my damping coefficient is same I am not talking about the damping ratio my damping coefficient is same ok. So, what I can do in this thing remember that I know that my transmissibility actually decreases with for a particular value of damping I have drawn this curve ok. I know that as I increase omega by omega n right ok my transmissibility actually reduces. So, if I want to reduce this further I need to increase the ratio omega by omega n ok. So, this need to be increased ok which also means remember the ground vibration is same ok you still have the same noise from the surrounding only thing that you can do here to get the increased value of this is actually decrease the value ok. So, if you decrease the value of the natural frequency ok you might be able to reduce the acceleration in the system and to reduce the omega n what would you do omega n is nothing but under root k by m remember rubber is fixed rubber pad is same so k is fixed ok. So, reduce the basically you are going to increase the value of mass. So, you are going to add some mass to the system so that its frequency decreases ok and that is so that the frequency ratio increases and transmissibility decreases ok and that would lead to the smaller acceleration in the system. So, let us say the modified mass of the system is m ok which gives me this much of acceleration 0.005 g ok which means that transmissibility ratio is 0.05 g divided by the peak ground acceleration which in this case is 0.1 g ok. So, this is 0.05 so my transmissibility is 0.05 ok. Now, if I increase the mass although the damping coefficient is same because rubber pad is same my damping ratio is going to change because it is c by c critical and c critical is actually 2 m omega n. So, if you are changing the mass if you are changing the frequency of the system consequently you are changing the damping ratio ok. So, let us see what do we get we have the transmissibility ratio ok and I am going to utilize this and see how much I get the value of this one. So, I can write this expression as ok c divided by 2 k times omega n right which I can further write it as substitute the value of c ok to get the value of omega n. Now, question is how do I get the value of c right this is already given to me is not it this c damping was initially given to me as 0.1 ok and initially the mass and frequency was also known to me ok. So, I need to find out c and k which are not changing. So, I can write this expression as c divided by 2 k ok times omega n ok. So, c by 2 k would be 0.1 divided by omega n and omega n we had calculated how much let us say 16.7 all right. Now, after the mass is added to the system let us see what is happen transmissibility ratio which becomes 0.05 ok this is equal to 1 plus 2 zeta omega by omega n all right divided by 1 minus omega by omega n square plus 2 zeta omega by omega n know that in the modified system the zeta and omega n both are different ok. I can also write 2 zeta omega by omega n is equal to omega n times sorry omega times 2 zeta by omega n. Now, we know that 2 zeta by omega n is what if you consider zeta was c by c critical which was 2 k by omega n. So, this I can write is as ok 2 zeta by omega n ok is equal to c by k or zeta by omega n is equal to c by 2 k ok for the modified system all right. So, this I would write it as omega times ok 2 zeta by omega n ok. So, let me keep it 2 here. So, this is become c by k and I know that the value of c by k previously from here ok c by k is 2 times 0.1 divided by 16.7. So, I will write it here 2 times 0.1 divided by 16.7 and this frequency applied frequency is also not changing right. So, this is 20 pi divided by this one here and if you calculate this you will get this one as ok 0.752 ok. So, you can substitute it here 1.0.752 square divided by 1 minus omega by omega n square plus 0.752 square and this is equal to the transmissibility ratio. Now, this is a quadratic equation in omega by omega n square and if you solve this what you would find that the value of omega n this leads to is actually 12.32 which is equal to k by m remember k is not changing here. So, it would be 14000 by m. So, the modified mass is actually 92 kg that leads to this much of transmissibility ratio or the acceleration within the permissible limit and remember that original mass was 4050 kg. So, I had to increase the mass by 42 kg ok. So, that my acceleration transmitted to the system ok. This transmitted to the system is actually reduced to 0.05 g which was specified as the permissible limit alright ok. I hope this problem is clear to you ok. Now, let us move to our second example. The second example is basically let me copy the example itself ok. Let me copy the example. So, I have what I have here is actually a vehicle ok which is moving on the top of a bridge ok and the dimensions of the span a dimension basically the span length is given to you and it is said that there is a because of the creep there is some deflection in the span ok with the total deflection equal to 150 alright mm. So, the half of the deflection would be 75 mm ok and it is being asked that determine the maximum possible force in the spring ok. So, I am representing the vehicle ok using this Sduev system it is moving in horizontal direction with some velocity and because of movement in the horizontal direction it would have vibration in the vertical direction ok. We saw that it is because of the ground profile or the sorry in this case it would be the deck profile ok. So, you need to find out because of this movement what is the maximum possible force in the spring ok and the vehicle velocity at which that would occur ok. Now, I know we have previously discussed that if I have a ground profile like this or deck profile like this I can represent the excitation as this displacement profile ut equal to ug not times sin 2 pi times the velocity of the vehicle by length of span time t ok and ug here is 75 mm. So, let me write 0.05 075 times sin and let us for the time being say that this quantity is omega here. So, I am writing it like this alright ok. This is my excitation function now I know that this excitation function I need to first represent it as ground excitation ok. So, if I want to represent it in terms of acceleration I will have to differentiate it twice. So, that I will get this as minus omega square ug not times sin omega t ok. So, my p effective the force is actually mass times the acceleration sin omega t ok alright. Now, this is p effective what is the displacement now remember displacement would be nothing, but and I want to find out relative displacement not the total displacement why because the force the first thing that is asked the maximum possible force in the spring and force in the spring well as well as damper also force in the spring and damper it depends on the relative deformation and relative velocity ok not the absolute deformation and absolute velocity ok. So, ut I want to find out the relative which I can represent here or write it here as u not times sin omega t minus phi ok and u not is nothing, but rd times us t not ok static displacement which I would again write as p not by k times sin omega t minus phi ok and p not here is this term here that you see. So, it would be minus m omega square ug not divided by k times sin omega t minus phi this is ut. So, the force at any time t in the spring would be k times ut ok. So, if I multiplied with k I can again write this as rd times k ug not I write it multiplied with k and take ug on the outside and what I am going to do here take m ok in the denominator and then write this as omega by omega n square let us delete this from here I am going to write it here sin omega t minus phi ok and of course, there is a negative term here all right does not matter the maximum value is actually we want to find out the magnitude ok. Now, if you look at this term carefully here I have this term k stiffness and ug not. So, both are constant this sinusoidal function is vary ok. Now, one mistake that you might do you might just look at rd and think that the deformation force of the spring force would be maximum when rd is maximum. However, remember that rd is now being multiplied with this omega by omega n square and remember we maximize when we maximize some function omega by omega 1 was actually the variable. So, and omega n is what 2 pi v by l. So, if I want to find out the velocity at which this spring force is maximum thus omega is actually variable. So, I need to find out the maximum value of this other things are constant ok all right. So, the let me write this expression here again peak value with respect to time ok the spring force as k ug not times omega by omega n square times rd ok because maximum value of sin function is 1 ok. So, to get the maximum value of f f s o not I need to find the maximum value of this quantity here which is nothing but remember this is similar to r a which is the acceleration modification factor. And if you remember correctly r a is maximum when omega n is or when excitation frequency is omega n divided by 1 minus 2 zeta square and the maximum value of r a max is actually 1 by 2 zeta 1 minus zeta square ok. So, if you substitute r a max for this whole expression here the f s o max ok would be k ug not times this value here. So, for the given value of the stiffness ok which is how much is here it is 40 this much 140 kilo Newton per meter ok. So, 140 kilo Newton per meter times ug not which is 0.075 times this here damping ratio is 0.4 ok and you can find out this expression here. And remember this maximum value is attained as omega at omega equal to omega n 1 minus 2 zeta square ok and omega n is what 2 pi v by l. Now l is already given to me I need to find out the velocity at which it becomes maximum ok this is 30 meters if you look at here yes this is 30 meters. So, you can find out the velocity at which it becomes maximum ok. So, this value comes out to be approximately equal to 14.3 kilo Newton ok and similarly you can find out the velocity term as well ok. So, these two problems are sorry there is another part to this problem yes what would be the velocity v that would produce the resonant condition for the total displacement. So, in the second part they have asked about the total displacement right. Now remember we discussed that if the ground excitation is given in terms of displacement and not the acceleration then I can write down the total acceleration to the system divided by the ground sorry the total displacement of the system divided by the ground p ground displacement is actually equal to transmissibility ratio ok and let us say what is the problem the velocity v that would produce the resonant condition for the total displacement. So, remember all I need to do here I got this tr ok which would be a function of omega by omega n here. So, what I need to do here again take differentiation of d of t by tr divided by ok d of omega by omega n equated to 0 this give would give me the value of omega by omega n that for that the tr is maximum and once that is known I know that omega n is 2 p 2 pi v by l ok this would be equal to this we already know from the mass of the vehicle and the stiffness of the suspension system and we can again be obtained all right ok. So, these two problems have demonstrated the concept of transmissibility and basically vibration isolation ok with this we would like to conclude the lecture today. Thank you.