 Welcome to the lecture today. We have been discussing so far a spring mass damper representation of different type of system but as you would realize in reality most of the structure are not discrete mass spring and damper right. So you have a structure like you have beam you have column so in order to get the response of this system we need to have a method through which we can simplify a continuous system to a single degree of freedom system. Okay so if you have a beam column then how can we simplify that to a spring mass damper representation and because we have already now in basically in discussions we have already learned how to get the response of a single degree of freedom system is just mass spring damper to different type of loading we can then find out the response of a continuous system once it is reduced to a single degree of freedom system to different type of loads and the way we simplify a continuous system to a single degree of freedom system is through a concept called shape function okay and we are going to learn about that today and we are going to see how we can employ the idea or the concept of shape function to convert single degree of freedom sorry continuous system to a spring mass damper representation of a single degree of freedom system. So let us get started. So till now what we have seen we have considered a single degree of freedom representation of general structures okay so it could be a spring mass damper representation okay. Now most of the system in reality are not single degree of freedom system and for many system single degree of freedom system might not be an appropriate approximation to get the response or desired response quantities of interest. So what we typically would like to do move from analysis of single degree of freedom system to multi degree of freedom system okay now in order to do that to make that transition we first going to learn what how do we analyze if we have a continuous system for example let me take the example of this cantilever beam here okay. Now for this cantilever beam depending upon how many points we consider on this cantilever beam in general a continuous system okay would have infinite degree of freedom okay. Now it might be appropriate for some cases to consider or analyze this using a single degree of freedom system or for more accurate representation we might have to switch over to multi degree of freedom system. Now although multi degree of freedom system supposed to be more accurate however it is also computationally expensive as well as the standard solution might not always be available okay but we have seen that for single degree of freedom system okay if we have this equation of motion we have derived response for different type of excitation and we have learned how to get the response using different type of numerical method. So we have standard results available for single degree of freedom system. Now the question comes is there a way in which or using which I can analyze a continuous system without having to get into the complication of analysis associated with multi degree of freedom system and that is why that is where the generalized SDUF system comes into play okay. So as I have said if you look at any system any continuous system in general it would have infinite degree of freedom. So our goal is that somehow we can reduce the system to a single degree of freedom system and then find out k equivalent and m equivalent. So although I am analyzing a continuous system I would utilize some methods through which I can do this simplification so that I do not have to do the calculation for multi degree of freedom system. Second thing if you remember till now what we have considered movement of a single mass so either mass was simply attached to a spring okay or it could have been a rigid bar that might be hanging okay. So in both cases I need only one degree of freedom system or one degree of freedom to represent a displaced position from its original equilibrium position okay. Now in these cases and let me take the example of this rigid bar here if you look at this rigid bar the rigid bar is constrained to move around this point okay move around the point of rotation. Now on this rigid bar if you look at it let us say we consider a displaced position in this fashion okay. Now in this displaced position depending upon where is the location of a point on this rigid bar I can have different displacement however I would have the same angular velocity or angular acceleration okay. So two things come into play for a continuous system one is displacement okay. So now we are talking about or let us not call it displacement let us say the first thing that comes into picture is the location of a point okay location of a point. And the second thing that is common to single degree of freedom system or a continuous system is the time variation of the response okay. So we are going to consider the time variation of a response. So while for a single degree of freedom system we had considered only location of a single point. Now for a continuous system there could be several points on the continuity of the body and we will have to basically find out the displacement history of each and every point on that body okay. So that is why the generalized stereo system comes into picture. So what do we do for a continuous system we represent the deformation okay. So let us say I have a deformation which is function of now x okay and t this represents the deformation of any point on a continuous system. So let me take the example of this cantilever beam here okay. And let us say under action of some force okay it is deforming so that in the displaced position okay this is how it looks like. Now this deformation is what we are talking about this is our u of x okay this quantity here where x is the distance from the fixed end okay or any reference coordinate system that you have assumed for that cantilever beam. So this is what I need to find out okay and subject to let us say there is a force Pt here okay. So subject to this force I would like to find that out. So the first objective of this chapter is to find out this u of x u of x and t okay subject to this excitation and the way we do that is we decompose the deformation in terms of two function first is phi x which is only function of the location or x and the second is zt which is the only function of time okay and phi x basically is called shape function and zt is called generalized coordinate and let us see what are the significance of these two quantities. Now if you look at the deformation a different instant for a continuous body for example you have this cantilever beam here okay so at different instance of time it would be something like let us say this is t1 then this is t2 here and then so on. Now two things are here considering that the body is vibrating okay there are two things that are happening first the movement of each and every point on the body with respect to time okay. However if you look at it let us say location of this point which is at the far most end okay compared to another point which is here okay. So these two point and then after some another time let us say rotation location or the ratio of the vertical coordinate these two are going to remain same okay and this is the first thing that we need to or we are going to talk about is the shape function. So to look at into the shape function first you need to think about what is shape okay. Shape basically represents the relative aspect ratio you can think it in terms of aspect ratio of any body. So for example if you look at these two these two might have a different size but they have the same shape and whether they have the same shape it depends on the relative aspect ratio okay so the aspect ratio of the sides of these two. So what happens similarly in terms of deformation I am going from t equal to t1 here to t equal to t2. So while my shape is changing or sorry while my size is changing my shape is still maintained. So if I can find out some function that can represent the shape of a structure and multiply this with some other function that represents the time variation then that is going to give me the overall deformation. For example in this case if I represent the shape as a let us call this as c and d okay. So let us say a by b is equal to c by d equal to some quantity let us say some aspect ratio is k okay. So if we can represent the total size as k times some time variation all I need to do is find out this time variation and if I multiply this with the shape of the body it is going to give me the overall size. Now take that analogy and let us revert back to the vibration of this cantilever beam that we are considering. So let us say I am able to represent and since this is one-dimensional body in terms of coordinate x let us say I am somehow able to represent location of each and every point on this deformed shape of the cantilever beam through a function phi x and then there is another coordinate z t which is basically changing with time. So if I multiply phi x with z t it is going to give me the u of x t which is basically the overall deformation at each and every point on this beam okay. Now shape does not change with time as you have observed here or here what actually is changing is the generalized coordinates. So the time variation actually comes from the generalized coordinate z t okay and our goal is somehow is to derive an equation which let us say may be of this form here okay. So we have equation of motion. So our goal is to somehow using this shape function is to find out an equation something like this okay. If I am able to find that out I would be able to get z t as a function of t subject to the excitation p equal n t okay and once I get that all I need to so that would represent how the deformation actually is changing with respect to time and if I multiply that with the shape of the structure which represents the location of each and every point the relative location then I might be able to find out the overall deformation of each and every point of the body at any time instance t and that is what we will try to do here and let us try to understand this through two example. First I will give you an example of a rigid body okay. So if you consider a pendulum okay let us come back to the pendulum. Now say at any time instant t equal to t zero the pendulum was in equilibrium position and then at any time instant t equal to t one it has come here. Now if I consider okay if I consider x to be the coordinate represents the say in this direction okay and this to be the theta t which is changing with respect to time I can find out location of each and every point using this equation here let us say this is u of xt the overall displacement which would be equal to whatever the x is times theta t okay. So whatever theta t we are considering we are going to multiply with that with x now for this case and this we have already been doing remember r theta basically represents the location on the or the radial location or on the circumference of any particle. Now for this case u xt represents the displaced position of any point okay. Now if you consider here basically the x is the phi or the shape function now this is a rigid pendulum okay and the mass is hanging at the end of it. So here x represents the shape function which is a very simple function and if I multiply this with theta t which represents here the z t the time variation. The multiplication of these two gives me the displaced position of each and every point on this body with respect to its original position. So for a rigid okay for a rigid setup or rigid system let us say rigid sdof system okay I can easily find out the shape function and that would be unique okay because there would be only one way in which the shape of the structure can be represented. However if I compare that to let us say a beam which is say simply supported beam let us take example of a simply supported beam. Now as I said if the system is considered to be a continuous system okay there would be a continuous system has infinite degrees of freedom. So by definition it would have infinite modes of vibration and that you might recall from your knowledge of sound theory okay wave theory where you consider different harmonics of vibration it is somewhat similar to that one. So and we will look into that more detail when we talk about the multiple degree of freedom system okay. So a system like this is simply supported beam a continuous system it can vibrate like this or another mode of vibration could be like this or another mode of vibration could be something like this and so on okay and it can have like you know infinite degrees of or infinite modes of vibration. Now the question becomes if I have to find out phi x which of these shape functions do I need to take and how can I get those shape functions or the mode shapes okay. Now we are going to get into multiple degree of freedom system where we are going to do the exact determination of the mode shape however for this chapter we do not want to do that we do not want to do all the analysis to get the exact mode of a system what we are going to do though here is assume that the first mode of vibration is actually represents the deformed shape and then assume any shape function or any function phi x that somewhat represents the deflected shape of the structure okay. For example if I consider a cantilever beam I know that in deformed position it looks like something like this. Now this can be represented by various type of expression so if I consider this horizontal axis to be as x okay and this deformation to be uxt okay to represent the deflected shape let me first write down uxt as phi x times zt. Now the deflected shape can be represented in various ways the deflected shape I can either look at it and see well it represents a parabola. So the deflected shape can be represented using the expression x square by l square okay I can also look at it and see well it does look like a function okay a cosine function which is this much okay so that at x equal to 0 it is 0 and at x equal to l it is actually 1 or this deflected shape can be approximated with the static deflected shape of a cantilever beam subject to the point load at its end so that it can be represented using 3x square by l square minus x cube by l cube okay. So depending upon what shape function we choose we can have multiple shape function to represent the deflected shape of this structure. Now as I have told you I am not trying to determine the exact mode shape which I could if I did the multi-degree freedom analysis using large number of modes but here my goal is to do it without having to get into all the mode determination so I am going to assume the deflected shape to be one of these functions okay. Now these functions as you can imagine would provide you approximate results okay and whether that approximation is good enough or not depend on the shape function that you have selected okay. Now all three shape function can be used to represent the shape of this deflected shape of this cantilever beam. However the only condition that all the shape function must satisfy is that all selected shape function okay must satisfy the displacement boundary condition and what is the displacement boundary condition so for example this cantilever beam has zero displacement at its left end as well as zero rotation okay so whatever shape function I select that must satisfy that. Now what we usually do like in this case when I write u x t equal to phi x times z t okay in these type of scenarios we typically write phi x such that it becomes or we normalize phi x with respect to a point of reference okay. For example the actual deflected shape okay the actual deflected shape of a cantilever beam subject to a point load at its end okay let us say this point load is one so the deflected shape actually looks like something like this u x or this let us say here u x equal to 3L x square minus x cube divided by 6EI okay this is the actual deflected shape however when I will represent the deflection okay I am going to write it as I am going to write this deflection with respect to some convenient reference point for example I know that the deflection at the end is PL cube by 3EI which for unit force becomes L cube by 3EI. So if I write my u x as L cube by 3EI okay times okay so if I take that common this becomes 3 by 2 x square divided by L square minus 1 by 2 x cube by L cube okay so I can normalize such that this represents the shape function and this represents the z at any time instant okay and zt would be this factor multiplied with something so that my phi x is now normalized with respect to the deformation at the end of this cantilever beam okay so usually whatever the shape function that I get the deformation that I get let us say delta x and you divide it with respect to some reference point okay so that let us say here it is L cube by 3EI such that when you substitute phi equal to L it becomes 1 here okay if we do that then basically u of L becomes phi of L times z which is equal to 1 times z okay so if I normalize it then my deformation becomes the generalized coordinate at that reference point so I have done it with respect to the end deformation here okay and you can choose it any convenient point for example if you have a simply supported beam something like this here you can normalize with respect to deformation at the center okay and you can select a shape function so that it becomes okay at L by 2 it becomes 1 okay and it is up to you how to select that okay so once that is known okay I need to first find out those shape function for different type of structure all right now let us get into how do I simplify so I know that to represent the deformation of any point or at any location in a continuous system this would be the deformation would be represented as this now once I have assumed the phi x the question is how do I get the z of t and to do that we have mentioned that I need to simplify the single degree or the continuous system to a single degree of freedom system okay to a SDF system okay and let us see how do we do that so what we are going to do we are going to consider an example of ground excitation or let us say ground that is called a seismic excitation okay so basically what I am saying here I have the same cantilever beam okay and it is undergoing vibration due to ground now due to this it would have some vibrations that is displacement due to ground which is let us say is u gt plus it would have its own deformation the relative deformation and it would be different at different location let us say this is coordinate x and let us say at any time instant the deformation is actually u here okay so this represents the deformation of a structure now to get the equation of motion okay so we need to get the equation of motion here we fix this so to get the equation of motion I know that the total acceleration at any location in this structure can be represented as ut of xt would be equal to ground acceleration plus the relative acceleration of that point okay now knowing that let us first find out what is the inertial force due to this acceleration we know that inertial force at any point in this can be so inertial force would look like something like this okay applying it opposite to the direction of motion so at any height x inertial force I can write as f m f i of x and t and this would be mass times acceleration let us say the mass is a function of x okay so it does not have to be a constant mass so the mass density is actually m of x times the acceleration which I have just written as m of x this ut of x and t and this can be written as m x times u g of t plus u x of t now this is effectively the external force that is acting on this cantilever beam okay to get the equation of motion what we are going to do we are going to use a method called principle of virtual displacement and you might have already come across this in different courses okay so we are going to utilize principle of virtual displacement okay now principle of virtual displacement basically says if a system is in equilibrium and if you apply a small virtual displacement let us say del u of x then work done due to this virtual displacement okay due to this virtual displacement work done by the external forces are equal to work done by internal forces okay so let me repeat it again if a system is under equilibrium given a small virtual displacement then work done on the system due to this virtual displacement and due to external forces is equal to work done by internal forces okay due to this virtual displacement so the external forces as I have mentioned is this inertial force here okay now work done due to this let us write down w e it would be so what I am going to do here I am going to write t w e and internal forces now the external force as you know can be computed as force times displacement okay so let us say this is the defined position and it is given a small virtual displacement which is again a function of x so this is small virtual displacement which is del u of x t the function of x and t and at the top it was let us say it was z okay because I am normalizing it with respect to the displacement at the top and let us say this is del z okay alright so let us see due to that how much is the work done so work done due to the inertial forces would be f i x t okay this is force at any height x and if you consider the force over the height dx it would be this times dx so this is the total force over the height dx now this is undergoing a virtual displacement of del u x t okay and I need to integrate it over the whole height to get the total virtual work done by the external forces now for the internal forces if it is a flexure system and I am assuming that there is no significant shear deformation okay the work done due to flexure you might already know this is done by the internal moment m x t okay due to curvature del k x and I know if u is my deformation then u dash x represents the angle and u double dash x represents the curvature okay so the external internal work done would be work done by this moment m x t times theta or let us say it is del theta okay and this need to be integrated over the whole height now del theta is nothing but curvature times the height that is being considered so I can write this as 0 to l m x t times del k x times dx here okay and you might remember from your undergraduate classes from beam theory m x t is basically written as e i x times u double dash x okay and then there is del k x times dx now before going further okay let me point out that if I consider my u x to be phi x times z t then u dash double dash x would be phi double dash x times z t and u double dot t would be phi x times z double dot t further because u x is this much my del u x should follow the same shape function and it will become del u x by phi x times t z and also del u double dash x would be phi double dash x times del z so we are going to utilize this and substitute in these two expressions here and here and let us see what do we get so this expression is nothing but okay I will have a negative sign m x times u g t plus u of x t double dot times dx okay this is the external force the work done by the external force here okay and of course there is this term here del u of x okay now if I simplify it further I can take this u g double dot out and then substitute these expressions here so that I can get this simplified expression as the external force basically it would become minus u g double dot t m x and then del u x dx okay integrated over of course 0 to l and then minus m x and then substitute the expression for this which would be phi x times z double dot t times del u x which would again be del z here okay so I have to include the del z term or let me just first write it as del u of x times dx okay and then what I can do I can take out or del u x I can write it as this okay and then take out del z outside okay so that I get this as del z times minus u g double dot t 0 to l m x phi of x okay times dx and then here minus z dot double t and then 0 l m x phi x square dx okay similarly I am going to further expand this term here okay this here substituted so that I can write it as e of i x times phi double dash x times z t times the curvature is nothing but del of okay so it would become phi double dash x times del z time dx so this is e i x times phi double dot double dash x okay and this whole times del z so if I quit these two equations expression and then bring it to right side okay the expression that I get is del z here times m x times phi x square dx and this whole times z double dot t okay plus I have this term here this term that I am going to write next is 0 to l e i x okay phi double dot dash x square dx times z of t okay plus this term here which is 0 to l m x phi x dx and this whole term times u g double dot t and this is equal to 0 now I know that my virtual displacement del z cannot be equal to 0 by because I have assumed it to be a finite virtual nonzero displacement so what I have left with the expression inside the bracket to be equal to 0 okay so let me do that okay let me now have a look at carefully these terms that are inside this bracket these terms only depend on the parameter x or the location and it is not a function of time okay so if I consider this expression okay let me write this as m equivalent times z double dot t plus another constant parameter which is k equivalent times z t plus l equivalent times u g of t and this is somewhat familiar to you from the expression of a single degree of freedom system okay where I have a second order differential equation in z okay the generalized coordinate z and that is what I need to solve okay where m equivalent for this system is this expression here okay and the k equivalent is e i x times phi double dot dash x okay and then I also have l equivalent which is 0 to l m x phi x dx okay so I have been utilizing the shape function phi x okay I have been able to reduce a continuous system to a single degree of freedom system and then my job becomes very easy because I can solve this system and I can find out what is z of t once I find out z of t which is the time variation of the response I can multiply with phi of x okay as I have told you this time before to get my u of x of t so displacement of each and every point on that continuous system as a function of time okay and that is how we solve any type of continuous system we assume a shape function utilizing that shape function we find out what is the m equivalent k equivalent and l equivalent and then once that is found out we solve this second order differential equation to get the value of z of t okay so this is the procedure that we would be following to solve a continuous system. Now you note one thing here we have not considered damping okay and as I have previously mentioned and damping the c equivalent there is no term such as like you know c equivalent that I can find out just from the properties the geometrical and mechanical properties of the structure okay so what do we do because damping is a very complicated mechanism so typically if we find out damping ratio from experiment for different type of structure and then use that to solve the system or include the damping in the system okay so first let us write down this expression here that we have okay in this form so let me rearrange that so that I can write it as first for the undamped system okay k equivalent divided by m equivalent z of t this is l equivalent m equivalent okay times u g of t okay so we can write this as z t plus omega n square of z t and thus t is equal to I am going to define it as another parameter u g of t okay now if I have a damping in the system okay you might know the familiar expression I can further write include the damping as zeta omega n times z dot t plus omega n square times z of t minus state equivalent u g of t so this is for undamped system and if I know the value of damping experimentally or let us say assuming some like some ratio I can formulate this equation like this okay and this gamma equivalent is nothing but l equivalent divided by m equivalent which if you like to write in terms of expression in phi x and m x it would be 0 l okay and here it would be m x phi x square dx okay so we have been able to reduce our continuous system to a single degree of freedom system okay and find out and utilizing that we are going to find out the response of the system now once we find out m equivalent and k equivalent it is very easy to find out the frequency of the system as before okay so the frequency square would be k equivalent divided by m equivalent which is 0 to l e i x times phi x square dx divided by 0 to l m x phi x square x times dx okay and then that would give me the frequency and subsequently the time period of the system okay so this is how we reduce to a reduce a continuous system to a single degree of freedom system now the next step is given this how do we get the response of the system okay or before going into that let us first see some example for example typically we said that if we have let us say a cantilever beam something like this okay and a load is being applied let us say at one end we used to simplify the system to a spring mass system saying that I am going to assume my k equivalent to be 3 e i by l cube and then mass to be represented by tributary but many times it would not be as simple as that and I don't know what is my m equivalent here what is my k equivalent here okay and this method that we have just described is going to assist us in finding out this k equivalent and m equivalent and how accurate it would be it depends as I said previously on the shape function okay so here what we are going to do now we are going to consider this cantilever beam and then consider three shape functions okay the first shape function would be basically phi 1 x which represents the static deflected shape of the cantilever beam subject to a point load at one of it ends so let us represent the let us write down the first shape function as 3 by 2 x square by l square minus half x cube by l cube. The second is another shape function which is 1 minus cos phi x by 2 l and then there is a third shape function okay which I am going to write as the parabolic function x square by l square now if you look at all the shape function all of these shape functions satisfy the displacement boundary condition for this cantilever beam which is phi 1 equal to phi of 0 equal to 0 and phi dash of 0 equal to 0 okay so all I need to do substitute the shape function in the equations that we have derived for mass and stiffness here to get the expression or to get the value of m equivalent and then k equivalent and then omega n and compare to see using different shape function how do they actually affect the equivalent properties and the modal properties of this cantilever beam okay so the job is simple basically substituting this phi 1 x phi 2 x phi 3 x in that expression for m equivalent and k equivalent and finding out the equivalent mass and equivalent stiffness so for the first case when we do this value comes out to be approximately as 0.23 times m l where m is the mass per unit length and l is the total length such that the total mass is m times l okay the k equivalent you would find out that it comes out to be equal as 3 a by l cube okay and the omega n that comes out to be is actually equal to okay you can simply substitute the values and then you can find out what is the value of omega n for this case okay for this case here it would simply be 3.57 divided by l square okay and this would be E i by mass per unit length okay now for the second case when you substitute that you will see this value comes out to be around 0.227 m l and this value comes out to be around 3.04 E i by l cube and omega m comes out to be 3.66 E i by m and this is l square okay for the third case if I do that okay find out this comes out to be 0.2 times m l this comes out to be 4 E i by l cube and this comes out to be around 4.47 divided by l square E i by m now if you look at that we see that the mass that is participating in the dynamic response is actually not m by 2 remember m l is equal to the total mass so it is not actually m by 2 but it is actually 0.23 times or approximately one fourth of the mass okay which makes sense because the mass which are further away from the fixed support is going to participate more compared to the mass which are closer okay and that is why the effective mass that actually participate as 0.23 times the total mass okay and the k colon that we get is actually 3 E i by l cube which is expected because we had assumed the static deflected shape to be the shape function in this case and for the static deflected shape the equivalent stiffness okay is actually 3 i by l cube okay for the second case this is again okay any arbitrary function to represent the deflected shape okay so when I use this I look at here again the mass is around 0.23 the stiffness is 3.04 which is again close to the first shape function and the frequency is again 3.66 times that same common factor so first and the second shape function produce result which are quite comparable okay however if you look at the third one there are not significant but like you know I mean you can look there are like you know differences which are much more than the first and the second one so you have 20 percent of the total mass however the stiffness now increases to 4 times E i by l cube and even your frequency is 4.47 so the question comes how do I know which shape function is appropriate so typically a static deflected shape of any structure can be assumed to represents the dynamic deflected shape although it would not be exactly equal however I know that a static deflected shape would always because it it constrained by the same boundary condition it would represent the or it will satisfy the displacement boundary condition okay and then we can check now in this case what I see here all three satisfy the displacement boundary condition however only the first to satisfy the force boundary condition okay and I will show you how now the force boundary condition here is that I know a cantilever beam at the free end the moment should be equal to 0 so the moment at length l should be equal to 0 which means this expression E i x times phi double dash x okay times z at x equal to l should be equal to 0. Now if I look at it phi 1 not here at l is actually equal to 0 phi 2 dash actually equal to 0 however phi 3 dash double dash is actually equal to 2 by l square so it is not equal to 0 so while all three satisfy the displacement boundary condition the first two also satisfy the force boundary condition so it represents the deflected shape okay that also satisfy the additional force boundary condition it is more likely to represent the dynamic deflected shape compared to the third one and that is why third one produces large deviation compared to the first and second one okay so this is how we select the shape function okay and shape function is very critical to the accuracy as long as satisfy the displacement boundary condition usually the results are reasonable okay but if it also satisfy the force boundary condition your results would be very close to the actual solution okay so with that we would like to conclude our lecture today and in the next lecture we are going to further study how to apply this to get the response of a continuous system in terms of internal forces and moment okay thank you.