 Welcome back everyone. We are going to continue our discussion from the previous class in which we are going to extend the concept of shape function to shape vector and employ it for discrete structures ok and then see how to get the response ok. So in today's class we are going to continue our discussion on generalized SDUF system ok and just to recap basically what we said a generalized SDUF system is a continuous system ok. So let us take an example of a simply supported beam here which can be transferred to a single degree of freedom system ok. So let us consider the spring mass representation ok with k equivalent and m equivalent and if there is a force that is being applied either a point load or it could be any general force ok. Let us say it is represented as kpxt. So again I can write this as p equivalent ok. So basically what we said a continuous system can be transferred to a single degree of freedom system using something that we said shape function and shape function is nothing but an assumed deflected shape of the continuous system ok and typically in this procedure we consider any approximate function that can represent its deflected shape. So when I say deflected shape it can be ok any shape that represents the deflection under the applied load ok and then basically we derived the equation of motion and also the expressions for the k equivalent and equivalent and p equivalent utilizing the mass and the stiffness distribution as well as the assumed shape function ok and we said that this method is approximate ok and the accuracy of this method depends on the accuracy of the shape function that we are considering ok. So the expression that we derived for m equivalent was let me just write down ok with m of x times phi of x square times dx and integrated over the whole length of the element the structural element. Similarly k equivalent expression we derived as eix which is the fracture rigidity times double differentiation of the shape function this is square integrated over again the whole length ok and then we also define the expression for l equivalent which is nothing but 0 to l mx phi x times dx ok and basically using these equivalent parameters ok the equation of motion that I had derived was m equivalent times z double dot t the acceleration in terms of generalized coordinate z if you remember the total deformation we had written as phi x times z t where phi x represents the deflected shape or the shape function and z t is basically the time variation ok which we called as generalized coordinate. So we got this ok so if I wanted I could include the damping term as well but remember as I said damping term typically we do not have an expression for the damping term because damping we get from experiments. So we get the damping ratio and we convert it to get actually the c equivalent ok. So let us first let us write down without damping here ok the expressions that we had was k equivalent times z t and that was equal to l equivalent times u g double dot t. Now this was the expression that we had derived for ground excitation ok or earthquake excitation in which basically the u g double dot was the acceleration of the ground. Now I can write I can divide the whole thing by m equivalent and I can write this expression as z t times omega n square z t equal to minus l equivalent times m equivalent u g double dot t and this term I can write down another factor which is gamma equivalent ok. So let my expression or the equation of motion I can write it as this. Now this was one this one was for undamped system ok for a damped system we get the zeta value from experiment and then I can utilize the same expression that we had previously used for single degree of freedom system ok and write down this expression here ok and this is for damped system ok. So up to this point we had derived our equation of motion ok and basically our frequency omega n square is nothing but k equivalent by m equivalent which I can substitute the expressions and I can find out the values ok. So I can write this as e i x times pi x integrated over 0 to l and then I have the expression for m equivalent which is m x times pi x square dx ok. So this makes our job lot more easier this allow us basically this allows us to analyze a continuous system without having to approximate is as a multi degree of freedom system ok. So continuous system can be reduced to a single degree of freedom system which we call generalized SDF system with the help of a shape function ok alright. Now the natural sequence of things once we get the equation of motion is to find out z of t ok after we have obtained z of t the question comes for this remember for a single degree of freedom system that we had considered ok discrete system ok only one degree of freedom so there was no x term here. So wherever the degree of freedom was defined there we had defined our total deformations what I am saying if we had considered this system ok. In this case ok whatever the u value for this was actually ok at this location only now for a continuous system I know that u would be different at different location. So this also ok poses the question what happens to the forces at different location. So I know that my u is now a function of x and t ok. So I would also have a different point what are the internal forces and moments ok. So internal force and moment that I call internal forces and moments and to do that again we are going to utilize the same method that we have done. So we are going to utilize something called equivalent static method in which ok so to get the internal forces ok so to get internal forces alright. Now this method basically allows us to find out to obtain the internal forces. So in this method what do we do actually let us consider ok that we have a beam and then there is a distributed load I do not know what is the variation of that. Now if you might remember from your solid mechanics course ok that if you have a distributed system like that with only flexure rigidity eix then I can write down my omega x which is the applied load is equal to eix times u double dash x times ok so are the differentiation of the whole expression. So this is the applied force. Now equivalent static method basically says that if I have analyzed a system and I found out what is the value uxt ok or let us say at any time instance I have found out what is the ux the internal forces in the system can be found out by applying an external force ok which is equal to omega x here ok and the same displacement ux through the stiffness component of the structure ok so what it is saying and I can say I am only considering the stiffness component it means I am doing a static analysis so there is no mass here. So equivalent static analysis there you do the dynamic analysis you find out the deformation at each and every point or it sorry each and every time instance once you find out then at that at any particular time instance you apply an external force and do the static analysis of the system ok to get the internal forces and this external force is nothing but it could be as a function of the stiffness component times the dynamic deformation ok which in this case basically this external force is this here ok. Now what I am saying this omega x so you might have used the notation omega x I am just going to say this is my fx. So I am going to apply this force external distributed force to my beam and then I am going to find out subject to this force what are the internal forces using static analysis no now no need to do any dynamic analysis ok. So I can do that I can substitute the value of ux equal to phi x time x t ok so that I can write my f force let us say it is at any time instant so that I would write it as e ix times phi double dash down again ok and I can take z t outside ok. So this is the expression that I would be using to get the external force so this is the external force that need to be applied ok. Now we can do some simple manipulation and we can see that the same expression can be converted to ok same expression can be converted to omega n square m of x times phi x times z t ok and it is more than more often this is like you know this is the expression that we would be using to get the external static or the equivalent static force to which the internal forces can be obtained at any time instance ok and the derivation from how to get from here to here can be found out ok anywhere remember to get this we are just going to utilize this expression that we have here and then also use the principle of virtual work such that the work done by the external force x t times del ok. So I am going to write down this times del ux times the x over the whole length it should be equal to work done by the internal forces which is due to the flexural or let us say the moment here ok this is moment times curvature times dx ok and we can substitute and utilize this expression and we will get finally this expression here for equivalent static force ok. Once the equivalent static force is known other part are simple I can just apply this equivalent static force and get the internal forces. Now more often than not as we have previously discussed in earthquake analysis ok so remember this is the response that we are considering for ground vibration ok or earthquake analysis ok. So for earthquake analysis usually peak response is of most important one of the most important parameter ok. So we consider mostly peak response and we know that peak response of any single degree of freedom system can be obtained using response spectra ok or design spectra depending upon what you have been provided or given. So let us say this is the response spectra so that let us call this as pseudo acceleration and this as G similarly I would have displacement ok which would let us say look like something like this ok not G sorry this is A as units of G this is versus time period and this is D versus time period ok. So remember these response spectra are basically response of any system ok of this type. So if I write it ok remember these were drawn for let us say Z double dot 2 zeta omega n Z dot plus omega n square Z and this is equal to minus UG of D. So this is the response spectra is basically correspond to this equation here ok and same for the deformation but remember the equation that we have for generalize the studio system left hand side is same. However in the right hand side I have a factor which is gamma equivalent ok. Now since we are considering linear system if I know the peak response of this system to get the peak response of this system I just need to multiply this with the factor gamma equivalent ok because my ground motion is now multiplied with this. So the response would again be multiplied with this factor. So the peak response would be multiplied with gamma equivalent. So for this case the peak generalized displacement can be written as Z naught equal to gamma equivalent times D where D is the peak deformation ok which can be further written as if you write A if you write the D as pseudo acceleration divided by omega n square. So this can be written as A divided by omega n square ok. Now remember once we have obtained the peak response then I can get the peak value of the displacement as u x t ok. So you want to get the maximum of this one over time of course ok. So this would be u x ok again the peak displacement would be function of x because the peak displacement would be different at different location in the structural element ok. So this can be written as u naught x equal to gamma equivalent times D times phi x ok. Basically all I am doing is writing down this one as phi x or instead of doing that let me just do this. This would be equal to phi x times Z naught and what is Z naught here? Gamma equivalent times D. So that I can write this as phi x ok times D ok. So peak equivalent static force can be written as ok remember this is my unit. So peak equivalent static force if you look at the expression here all I need to do is substitute Z t equal to Z naught and that would give me the peak value of the equivalent static force. So let us write down that as m x phi x times Z naught which can be written as gamma equivalent times m x times phi x times A ok because if you write Z naught equal to this expression here then omega n square omega n square cancels off and you get basically this expression that you have right here ok. So subject to this we can find out the internal forces and the moment and let us see how do we do that. So we are going to take the example of the cantilever ok the cantilever column that we had considered and in this case basically we had said that ok like well the inertial forces would look something like this F i as a function of x and t ok. We had also said that ok for this case let us apply we want to find out the response of the system subject to this. This is the equivalent static force this is 0 at x equal to 0 plus phi x equal to 0 ok. Now if I want to say find out what is my shear at any height y ok then let us see how do we do that or instead of saying at y let us say I want to find out shear at any height x. So to do that what will happen the shear at this point would be due to contribution of all the forces which are above this one. So all the sum of all equivalent static forces which is above this one and this just comes from the free body diagram. So shear at any point which is at height x let us say this is v x that I want to find out would be due to all the forces which is basically F naught x above that location. So to do that let us do one thing let us consider any small element which is at height of y ok and of differential height dy. So this is not F x now let us say this is F naught y ok. So v x basically is integration from ok x ok first let us write down the expression sum of all the forces. So it would be F naught y alright and dy integrated over x to the total height. Let us say it is l or h or whatever you are considering. So this gives you the shear at any height location or whatever you want to say at distance x ok. Similarly the moment at this point ok so let us say this is the direction we are considering correct here. So the moment let us say it is something like this ok or it would be in opposite direction here ok. So m x again become similar ok. So I am going to utilize the same expression here x to l this would be moment of this force F dy and the lever arm would become y minus x. So let me rearrange this expression this becomes y minus x times F naught y dy integrated from x to l ok. So these two expressions we utilize to get the basically forces and or the shear forces and moment at any height x ok. Now the parameters that are of more important importance in this case are the base shear and the base moment which are typically used to find out what is the base shear of a structure in equivalent to static method ok. So in that case what I can do ok I can write down at the base v b all I need to do the lower limit is basically now becomes 0 and this is F naught ok y dy integrated over the whole height ok and you would see that I can write this expression ok in terms of if you substitute the expression for let us say F naught y which is right here ok. I can take or let my first write down like this omega n square or let me use the second expression here. So it is this times m x pi x or the integration variable here is y you know it does not matter really this case because this need to be integrated now this is constant not a function of actually. So I can write this as m y pi y dy ok. Now if you remember this expression is nothing but l equivalent. So I would write this expression as l equivalent times gamma equivalent times a ok. This is the expression for the base shear ok. Similarly base moment I can also write down m d o is equal to 0 to l ok in this case y times F naught y dy. Now again when I substitute this I would get this as ok some l equivalent dash a where l equivalent dash is actually 0 to l y m y pi y dy or you can write in terms of x and does not actually matter. So these expressions can be utilized to find out the base shear and base moment for any continuous system ok. Now remember that till now we have considered basically the ground shaking and we have derived the equation of motion for ground shaking. However instead of the ground shaking if you have the external applied force which may be distributed or concentrated but let us say in general we write the external applied forces ok as distributed force P of x of t ok instead of the ground motion u g double dot t. So in that case the same expression I can derive only thing is that I would have a p equivalent t terms which is nothing but in this case p x t times phi x times dx integrated over the whole length ok and that you can you know just find out. So in this case my p equivalent becomes whatever the force distribution that is given to you ok multiply with the shape function phi x and integrate over the whole length alright. So this is instead of ground excitation if you have a force acting throughout the length of the structure element then we can utilize this expression to find out and then further solve the system ok alright. If this is clear then let us do an example because that would make this discussion like I put it in a perspective ok. So what do we actually have here ok let us consider this ok we are going to consider the same problem of a cantilever basically column here. So let us do this example here ok. So what do we have again a cantilever column which basically representing a chimney which is fixed at the bottom the total height is 200 meters ok and this is of hollow cylindrical shape of total or the external diameter equal to the external diameter equal to 16 meter the internal diameter or instead of internal diameter the thickness of the wall is given as 1 meter ok let me draw it in a larger view. So this whole diameter ok this whole diameter is actually 16 meter however this thickness is 1 meters 1 meter. So in that case the internal diameter would become actually 14 meters alright. Now additional information is given that for this case it is of concrete material of density that you can assume as 2400 kg per meter cube the modulus plastic modulus can be taken as 25000 mpa ok. The damping can be assumed as 5% ok and it is given that the expression for SA by G or the spectral acceleration can be taken as 1.8 divided by the natural time period ok. So I mean this is for any design spectra in which case ok this expression can be given or can be taken to find out the spectral acceleration for this problem ok. So let us now solve the problem see in this type of problem the first step is always to determine what is the basically the section parameter and what is the modulus of rigidity the fracture rigidity and all those things so that we can find out the solution and then assume the shape function. Now for this case you can assume the shape function to be 1 minus cos pi x by 2 L and if you remember from the last class if you assume this to be the shape function ok the expression for the frequency was coming out to be 3.66 divided by the total length ok times EI divided by the mass per unit length ok. So at this point of time pause it and then take 10, 15 minutes to solve this problem ok alright. Let us discuss this problem ok. So let us go step by step the total length is given as 200 meters ok. Now remember we need to find out what is the mass per unit length so we need to find out what is the mass per unit length we also need to find out the inertia. So these two parameters need to be found out to determine the value of omega n so that I can find out Tn because once I get the Tn then I can find out what is the spectral acceleration ok. So let us do that let us find out mass per unit length would be nothing but whatever the density of the concrete times the area not the volume I consider area multiply it with the density of the concrete it would give me mass per unit length. Now in this case density is 2400 kg per meter cube and the area would become 5 by 4 times 16 square minus 14 square ok and this gives me basically the mass per unit length as 113 ok times 100 kg per meter alright. Now moment of inertia of a hollow cylinder ok is basically pi by 64 d 2 to the power 4 and d 1 to the power 4 because it would be pi by 64 times 16 to the power 4 minus 14 to the power 4 and this you can find out comes out to be 1 3 p 1 meter 4 ok. So the flexure rigidity which is EI can be found out as 25000 mPa and I can convert mPa to Newton per meter square as 10 to the power 6 Newton per meter square times 1 3 3 1 meter 4 ok or let us first omit this and this comes out to be approximately as 3.33 times 10 to the power 11 Newton per meter square sorry Newton meter square ok. So let us now find out what is omega n which is 3.66 times L square EI by ok EI by m. So I can substitute all these values and I can get the value of omega n is 1.57 radian per second. So the time period can be easily determined as 2 pi by omega n which comes out to be as 4 seconds ok. Now I know that for 4 second my pseudo acceleration is 1.8 divided by 4 ok and what is also given in this problem that we need to find out the response of the system of a response spectra for which this is the spectral acceleration but is scaled to a pga of 0.05. So in that case I need to multiply this with 0.05 and this times g would be my acceleration ok and if I substitute that then I get this as 0.112 times g ok. Now I know that my displacement is nothing but pseudo acceleration divided by omega n square ok and if I substitute that I get this as 44.6 centimeter ok. So that z naught would be equal to gamma equivalent times d. So I also need to find out gamma equivalent here which is L equivalent times m equivalent and if you remember from the previous classes ok I can get either directly by integrating let us say L equivalent as whatever the mass times pi x times dx and m equivalent ok and you can substitute all those values and we can see that you will get this one as approximately equal to 1.6. So that your z naught becomes 1.6 times d which is 44.6 centimeter and this one comes out to be approximately at 71.5 centimeters ok and the peak displacement now. So this is the general peak value of the generalized coordinate ok. So my actual displacement which represents the deformation as a function of location or x can be found out as pi x times z naught ok. Now pi x I have assumed as 1 minus cos pi x by 12 times 71.5 centimeters ok. So basically or I mean you can write it like you know 71.5 first and this as well centimeters. Now once we have that I can find out what is the equivalent to static forces as f naught x equal to in this case I can directly write this as gamma equivalent ok times mx times pi x times a and I can substitute remember mx is nothing but a constant value of mass per unit length pi x is 1 minus pi x by 12 a is 0.112 times g and gamma equivalent is basically 1.6. So let me write 1.6 times 13 100 times 1 minus cos pi x by 12 times 0.112 times g and if I write it in terms of kilo Newton per meter this would be 200 1 minus cos pi x by 12 times kilo or this expression kilo Newton per meter. Now the base shear could easily be calculated using the expression that we have previously derived as gamma equivalent times l equivalent times a. Now gamma equivalent I have only already found out as 1.6. L equivalent I can find out from the previous classes 0.363 times the mass per unit length times l and the acceleration is 0.112 times g which comes out to be approximately as 0.065 mlg which I when I substitute gives me 14500 kilo Newtons ok. So using this example basically this example demonstrate how to employ the generalize STUF procedure of analysis to find out the response of a continuous system using the similar methods that we have used previously for a single degree of freedom system ok. All right. So this gives the idea of how to analyze continuous system. Now we are going to discuss another type of a specialized system which is basically a lumped mass system. So this is not a continuous system however this is a lumped mass system ok and especially we are going to consider shear type building ok. So first I need to define what is a shear type building ok. Now if I consider any building representation so let me consider a three-story building here ok. I have this building here. Now this building has beam in columns ok and we typically assume that all the masses are actually concentrated at the floor levels. Now for this if I consider that the axial deformation can be neglected ok and at the floor level the beam is actually in combination with the slab. So I can if I assume that the flexure deformation in the beam can be neglected then so the second assumption is that flexure deformation of beams are neglected. So first you try to understand what those assumptions mean. If you consider any column or any beam ok like this if you apply axial force the displacement is actually very small because the axial stiffness is very high. However if you apply a lateral force then you are relatively it takes you smaller force to actually generate some finite displacement ok. So in this case we are doing the same thing. We are saying that for the shear type building I am not going to consider any axial deformation in the beam or column ok and I am not going to consider any flexure deformation in the beam. So if you remember in general ok in general a 2D system at any node would have 3 degree of freedom right. This you might remember from your structural mechanics class. Now if I say that the axial deformation of the beam and columns are neglected this system basically reduces to this. Remember there are no axial deformation so this and this can be removed because there is no vertical deformation due to column in the column ok and there is no axial deformation in the beam then deformation at this point and at this point would be same. So I can write down a single degree of freedom system to represent the horizontal deformation. I still have these 2 deformation ok which is basically the flexure deformation or the deformation along the flexure degree of freedom at these 2 nodes. But if I assume that the flexure deformation of the beams can be neglected which is a reasonable assumption if you consider a fixed fixed column of a structure in which you have slab at each floor level. So that the beams are connected to the slab and it provides us ok very high rigidity to the beam then these flexure deformations can be neglected. So I have at or this is a single story because we had considered a single story for each story I have only 1 degree of freedom. So in this case ok a shear type building can be represented using the horizontal deformation in the shear direction at each level ok. So for example this shear type building ok I would have only 3 degree of freedom to represent the horizontal deformation at 3 stories that we have considered here ok. Now the question is ok we are going to learn about multi degree of freedom system later using exact finding out exact deflector shape and doing all sort of analysis. However my question is is there a way that which I can analyze this multi degree of freedom system ok using methods of single degree of freedom system ok. So I am going to employ the same method that I did for continuous system if somehow I can predict how does this deflector shape look like ok. Then I am going to or I would be able to reduce this system to a single degree of freedom system with k equivalent and m equivalent and then again I can do my job. So again this is also an approximate method so I am analyzing a multi degree of freedom system or a multi story building using single degree of freedom system ok. And the precursor for this one here is that we need to assume the deflector shape ok. Now in this case it is not called a shape ok it is actually called a vector ok because in this case what do we do actually ok. In this case we represent let us say if I have a building initially at this position and then let us say it is deflected like this and the degrees of freedom are defined at these level ok. So somehow if I am able to represent what is the shape represented by these degrees of freedom then I would be able to find out my shape vector ok. So to do that let us say at any degree of freedom j I can write down my deformation coordinate ok u j t as pi j times z t where j is the degree of freedom that I am considering ok. So for in this case remember when we had a cantilever beam I had represented this deflected shape phi x as different shape functions ok one of them was 1 minus cos phi x by 2 l. In this case what I am going to do I am going to represent a shape vector which basically represents let us say if it is a straight line I would say this is 1 ok this is 2 by 3 this is 1 by 3. So phi I can write it as if this is 1 by 3 2 by 3 and 1 something like that and then this represents the shape which basically says that this different coordinate although they would vary in time ok. However at any time instant they would always be in this proportion represented by this shape vector here ok and the total deformation u at any time instance ok can be written as vector. So again this would be vector phi times z t which is z t is just one value ok. So this is somewhat similar to what we have already done. So I am not going to repeat the whole derivation I am just going to write down the final expression ok I am going to write down the final expression for the mass equivalent and the k equivalent ok. Now remember for shear type building if I consider this ok we represent story stiffness as k j at jth story let us say sum of stiffnesses of all the columns at that particular story the jth story ok. And if it is a shear type building I can simply write this one as summation 12 E i by h cube over all the columns ok and we utilize this. And the story shear in that particular story can be written as v j equal to k j times the relative deformation. So what I am saying let us say at any particular story jth story ok there is the relative or instead of doing this let us just do this ok. Again let us draw a multi degree of freedom system. So at any story let us say the shear force in that particular story would be the story stiffness let us say this is k j times the relative deformation ok. So if this is let us say j minus 1 and this is u j ok the drift story drift is defined as delta j is equal to u j minus u j minus 1. So the deformation of the floor above and below or the floors that constitute that particular story this would be equal to del j which is u j and u j minus 1 ok. And remember this expression here ok this in terms of element this expression can be written as u jt is equal to pi j times zt ok. So like the derivation we had done for continuous system using the principle of virtual work I can again derive the similar expression however in this case what I am going to write down just the final expression. So m equivalent comes out to be now there is a summation term because it is a discretized system not a continuous system ok. This comes out to be z equal to 1 to n ok where j is basically the degree of freedom ok. So mj pi j square ok and then similarly I can write down ok I can write down the expression for kj which or sorry k equivalent which comes out to be j equal to 1 to n kj phi j minus phi j minus 1 whole square and l equivalent as summation j 1 to n ok mj phi j all right. So like continuous system we can utilize this expression for a lumped mass system like a shear building to find out the response and all I need to do is assume the deflected shape I can assume to be a straight line I can assume it to be parabola but in each case I would represent the deformation at particular degree of freedom in terms of a vector. This is the only difference between a lumped mass discretized system with the continuous system ok and this would become more clear when we do one example. So let us let us consider in this case 5 story building ok in which I am assuming that at each level the mass is same. So 1, 2, 3, 4 and this one becomes 5 ok 1, 2, 3, 4, 5 all these masses are m actually and all these story stiffnesses are k ok and let us say we assume that it actually deflects linearly. So that if I assume that the topmost coordinate is 1 that this would be 5 4 by 5 this would be 3 by 5 and then 2 by 5 and then 1 by 5. So that I can write down my shape vector as 1 by 5, 2 by 5, 3 by 5, 4 by 5 and then 1 which is basically 5 by 5 ok. So my mass equivalent would become ok so my mass equivalent would become nothing but mj which is m for all the system and then this square. So m times 1 square plus m times 1 by or sorry 4 by 5 square ok then m times 3 by 5 square and so on up to m times 1 by 5 square and once I substitute these values ok I get final mass equivalent mass is 11 m by 5. Similarly k equivalent ok we can write it as k at each level times first let me just write down that expression because this would be j equal to 1 to 5 ok and then this is kj times 5j and 5j minus 1 square ok. So this would be k 1 minus 4 by 5 square plus k 4 by 5 minus 3 by 5 square plus k 3 by 5 minus 2 by 5 square and so on 2 by 5 minus 1 by 5 square ok. We can simplify that and we see that our the k that we get is actually k by 5 ok. And similarly we can find out the L equivalent as well ok which is basically in this case I would get as mj times 5j j equal to 1 to 5 and you can substitute the value and this would basically come out as 3m ok. So we have seen that by assuming the deflected shape vector I have been able to reduce this system to a single degree of freedom system in which it can be represented as k equivalent and m equivalent where k equivalent and m equivalent have been derived like this and the equation of motion would become ok in this case z double dot plus omega n square which I can easily find out ok omega n is nothing but in this case k equivalent divided by m equivalent under root. So this the z equal to minus l equivalent divided by m equivalent ok times u g t if there is a ground acceleration ok. Remember omega n equivalent is nothing but k by 5 divided by 11 m by 5 which comes out to be 0.3 times k by m ok. So we have seen that we solved continuous system and we saw we solved lumped mass multi degree of freedom system a shear type building as well without having to solve a multi degree of freedom system. So this is the power of a generalized system ok by utilizing or assuming a deflected shape or a shape function or a shape vector for a discretized ok. In subsequent classes we are going to see how to solve a multi degree of freedom system without any approximation. So without assuming that what is the deflected shape we look like we would be actually able to find out what would be the deflected shape actually looks like ok. And we will do that in subsequent classes ok. Alright with that I would like to conclude this class ok. Thank you very much.