 In the last class, we started discussing topics related to dynamic analysis of stability and analysis of time varying systems, so we started by discussing the effect of presence of an axial load on natural frequency of a beam like this, and we showed that when P approaches the critical value, the natural frequency drops to 0 and the solutions become, will start growing in time. The next question we considered was what happens if the axial loads are time dependent, we considered few situations, for example straight forward cases when a beam like this is subjected to axial load PFT, or at stack like this subjected to biaxial earthquake ground motions, so there will be a vertical component, we saw that the presence of vertical component appears as a parameter in the governing differential equation, and we call this type of systems as parametrically excited systems, here if you write the equation for the beam oscillations, P of T will appear as a parameter in the equation of motion, in the sense it multiplies a term involving dependent variable, therefore it is called parametrically excited systems. Another important application in civil engineering problems is problems of vehicle structure interaction, that is encountered typically in bridge engineering problems, so here as a vehicle traverses a supporting structure like this, if one writes the combined equation for dynamics of vehicle and the beam, we saw that the governing partial differential equation had time varying terms. So we will continue our discussion with this, and we posed a few problems, one of the problem that we need to, we have agreed to address is that of problems of follower force, so here suppose if you consider a cantilever beam which is loaded actually like this, and as the structure deforms if the direction of the application of the load remains unaltered, we have already analyzed this problem, suppose if you imagine a situation where the point of, the line of action of the load remains tangential to the deformed middle axis of the beam, then we need to revisit this problem and we need to find out what is the critical value of P, so we will consider this problem in today's class and we will show that if you perform a static analysis for this problem it will not yield a satisfactory solution, but on the other hand if you perform a dynamic analysis it gives a different answer, and that seems to be logically agreeable. So we will consider three problems, one is how to characterize resonances in systems governed by equations of the form, M of T X double dot C of T X dot plus K of T X is equal to 0. Now the resonances occur because there are time varying terms in the mass stiffness and damping matrices, and if these terms are periodic in nature then the question of resonance is appropriate because exaltations have a steady state character, so we could expect steady state responses from the system as well and see whether the responses remain bounded or grow in time. Next we will consider another problem where we start with partial differential equations with time varying coefficients, we saw that in the previous lecture that this type of systems are governed by partial differential equations which have time varying terms, we will ask the question how to make finite element models for such systems. Next we will consider if there are any situations in statically loaded systems wherein one needs to use dynamic analysis to infer stability condition, so these are the some of the questions that are on our agenda we will consider this in this and the following lecture. So we will begin by considering qualitative analysis of parametrically exited systems, for purpose of illustrating the basic concepts we will consider a single degree freedom system that is U of t is a scalar variable, scalar function of time and it is governed by a second order differential equation of the form U double dot P1 of t U dot plus P2 of t U t equal to 0. Now let us assume that there are some non-zero initial conditions and we are assuming that P1 and P2 are periodic with period capital T, now the governing equation is a linear second order ordinary differential equation with time varying coefficients, it admits two fundamental solutions. So let U1 and U2 be the fundamental solutions for this of this equation, then any solution U of t can be written as linear superposition of the fundamental solutions, therefore I get C1 U1 of t plus C2 U2 of t, now we will consider the governing equation at t equal to t plus capital T, so if I write this it will be U double dot t plus t P1 t plus t etcetera equal to 0, now since the coefficients are periodic with period capital T I can write for P1 of t plus t and P2 of t plus t I can write P1 of t and P2 of t, so what this means is if U of t is a solution then U of t plus t also is a solution to the governing equation, because if U of t satisfies this equation U of t plus t also satisfies this equation. So this enables us to write U1 that is the two fundamental solutions at t plus t is another solution in terms of U1 and U2, so I get U1 of t plus t as this and U2 of t plus t of this and in a matrix form I get this equation, so we are interested in nature of this solution as t tends to infinity, so this we briefly touched upon towards the end of the previous lecture so we will formulate this problem in greater detail as we go along, now we are interested in finding limit t tending to infinity what happens to U of t, this is equivalent to asking I will write U of t as t plus n t and allow n to infinity where n is an integer 1, 2, 3, 4, 5, 6, etcetera. Now U of t plus t is A U t, therefore U of t plus 2 t is A U of t plus t which is A square U of t so on and so forth, so if I write U of t plus n t it will be A to the power of n U of t, so the behavior of limit of n tending to infinity of U of t plus n t is controlled by the behavior of this behavior of A to the power of n as n tends to infinity, so intuitively one can see that this in turn depends upon the nature of eigenvalues of A. Now so what we do to analyze this problem we consider this equation U of t plus t is equal to A U t, then we introduce a transformation U of t is Q into V of t with a aim to diagonalize A matrix, so U of t plus t is A U t that means Q of V of t plus t is A Q V of t that means in this equation I am substituting for U according to this transformation. Now I will pre-multiply by Q transpose so I get this equation, now the objective of making this transformation is that we wish to select Q such that A is diagonalized, that is we wish to find Q such that Q transpose Q and Q transpose A Q are diagonal, so we can start by considering the eigenvalue problem associated with matrix A, A phi is lambda phi. Now mind you A is not a symmetric, need not be a symmetric matrix, it is real valued but not necessarily symmetric, now we select Q to be the matrix of eigenvectors of A that is phi 1, phi 2, and we can show that V of t plus t is lambda 1 V of t because the eigenvectors will be satisfying the conditions phi t, phi will be I, I can normalize them such that this is true, and then I can Q phi will be some diagonal matrix lambda which is the matrix of eigenvalues of A as shown here. Now that would mean V I of t plus t is lambda I V of t for I equal to 1 to 2, now since the matrix is diagonalized I can do that, now therefore V I of t plus N t is lambda I to the power of N V I of t this, now we want to now investigate what happens to these solutions as N tends to infinity, the original function U of t can be constructed by superposing V 1 and V 2, so therefore behavior of U is controlled by behavior of V 1 and V 2 as N tends to infinity, so we can see that as T tends to infinity that means N tending to infinity of V I of t plus N t it goes to 0 if modulus of lambda is less than 1, because a scalar equation this we saw in the last class this is a condition, because every time you multiply this number starts shrinking and it goes to 0 as N tends to infinity, on the other hand if modulus of lambda is greater than 1 this solution will blow off, if lambda I is equal to 1 then it is periodic with period capital T, on the other hand if it is minus 1 then after every 2 t it returns to the original state, the first time you operate it become minus of the quantity and then square of that quantity which is again returns to the original state, so the period will be 2 t, so depending on the nature of Eigen values there are 4 different behaviors possible, either the solution will go to 0 as T tends to infinity or it will shoot to infinity as T tends to infinity or it will be periodic with period capital T or it will be periodic with period capital 2 t. Now we will try to reduce these results into a standard form, so what we will do is we will consider this equation V I of t plus t lambda I V I of t, now I introduce a new variable gamma I and I multiply the both side of this equation by this quantity exponential minus gamma I t plus t, as I do this lambda and gamma I are related by this equation that you can see that and therefore gamma I is nothing but 1 by t log E lambda I, now therefore exponential of minus gamma I t plus t is V I t plus t which is exponential gamma I t V I of t, now if I call this quantity exponential minus gamma I t into V I t as psi I of t it is clear from this expression that psi I of t is periodic function with period capital T, so therefore I can write V I of t as exponential gamma I of t into a periodic function psi I of t, so this is a periodic function. So what we have done now, we have expressed the solution as a periodic function multiplied by an exponent and this exponent could be periodic with decay or periodic with explosion depending on nature of gamma I, so we will make some remarks, gamma I is 1 by t log E lambda I, these are called as the characteristic exponents or flow K coefficients, any solution U of t can be expressed as linear superposition of V I and V II, behavior of U of t as t tends to infinity is governed by the nature of the exponent, suppose this is gamma I t is, we can expect it to be complex value, so I write gamma I alpha I plus J beta I, J is now square root minus 1, I am using for index, so the complex number is now expressed in terms of J, so if the real part of this you know controls the growth or decay of the solution and this contributes to the oscillatory behavior, so U of t is periodic if alpha equal to 0, that means if this pure imaginary this will be e raised to a pure imaginary exponent, this is sine and cosine terms will emerge and therefore V I of t will be periodic, so when gamma I is pure imaginary U of t is periodic, if lambda I is equal to 1, U of t is periodic with periodic capital T, that means this exponent you have to find out and relate it to gamma I, if lambda I is minus 1 U of t is periodic with period 2T, now this is a well known classical result in study of parametrically exerted systems, so the flow case coefficients help us to determine boundaries of stable solutions. In numerical work how do we determine flow case constants, so what we can do is we revisit this equation U double dot of t plus P1 of t U dot of t plus P2 of t U of t equal to 0, now let U1 and U2 be two solution of this equation with the following initial condition, U1 of 0 is 1, U1 dot of 0 is 0, and U2 of 0 is 0 and U2 dot of 0 is 1, that means I am selecting two solutions which start from linearly a set of linearly independent initial conditions, so I can write U1 of t plus t therefore I can write in terms of A1 1 U1 of t plus A1 2 U2 of t, and similarly U2 of t is written like this. Now if you find our objective is to find A matrix, you see for a given system how do you find A matrix, moment you find A matrix you perform eigenvalue analysis and look at the nature of the eigenvalues that answers our question on stability, so it's as simple as that, so U1 of 0 if you see it is U1 of t which is A1 1 U1 of 0 plus A1 2 U2 of 0 which is nothing but A1 1, similarly U1 dot of 0 is U1 dot of capital T which is A1 2, similarly I can find out A1 1 A1 2 A2 1 A2 2, so these are therefore elements of A matrix are determined by solving this equation with two linearly independent solution over one period of excitation, that is all the numerical work that we need to do, moment we do that we will be able to form this matrix and after we can perform a eigenvalue analysis of this, and then examine the nature of eigenvalues and we will be able to infer whether the solution is periodic DKS to 0 or gross, and it is periodic with period capital T or periodic with period capital 2T, so we find eigenvalues of A and infer the nature of solutions by using the following criteria, at this we have discussed, so if modulus of lambda I is less than 1 for I equal to 1 to 2 then the solution will DK, and if it is greater than 1 for any of I's it if any one of the eigenvalues satisfy this property solution will go to infinity, and it is periodic with this, if both lambdas are equal to 1, and it is periodic with period 2T if both lambdas are minus 1, so this completes the calculation of a flow case coefficient by simple integral, so you can use, if you formulate a finite element model you could use numeric beta or any other method that we have discussed earlier, and integrate this equation over one period of the parametric exaggeration, if the condition lambda modulus of lambda I greater than 1 occurs we say that the system has got into parametric resonance, so here the motion grows exponentially with time, okay, you recall how, if that is here if you recall how a harmonically driven single degree freedom system gets into resonance for that if you consider X double dot plus omega square X equal to P cos lambda T with say 0 initial conditions you can show that solution is given by this, now as lambda tends to omega we get the solution as PT by 2 lambda sin lambda T, so as lambda goes to omega and T tends to infinity X of T goes to infinity and the growth is linear in time, so this is when external excitation creates resonance in the system, but if the system gets into parametric resonance the growth is exponential, now in resonance in external excitation the resonance, response amplitudes are limited by damping, and non-linearity of course would also be important as response grows, but in parametrically excited systems damping has no role, if a system gets into resonance presence of damping does not limit the amplitude of the response, that is you can see here already I have a U dot term, so presence of U dot term is not changing the nature of the solution, so in this case if a damping treatment won't solve problems of parametric resonances, whereas that will solve the problem of resonance under external excitation, now here of course as amplitudes become large for a linear system amplitude tends to infinity, but once amplitude cross certain limits the system non-linear it is kick in and the behavior will be altered due to presence of non-linear terms, how do we deal with multi degree freedom systems? So we can consider equation of this form U double dot P of T U dot plus Q of T U, again let us assume P and Q are periodic with period capital T, so in the equation that we got say M of T X double dot plus C X dot etc., that is in the model that we got we had M of T X double dot plus C of T X dot plus K of T X is equal to 0, so I can reduce this equation to this form by multiplying by M inverse, so it reduces to this form. Now again by using the logic that we have discussed U of T plus T can be written as A into U of T, now size of A will be equal to the suppose U is N cross 1, A will be N cross N, so how do we find A? We solve this equation over one period of excitation with a set of linearly independent initial condition, we have to select a set of N linearly independent initial conditions and integrate the solutions over one period, and by examining the response at capital T we will be able to construct A matrix, again we can examine the eigenvalues of A and infer whether the solution grows in time or decays in time or becomes periodic etc., so this is how a qualitative analysis of a time varying system can be performed. Now the next problem that I mentioned was that of so called follower force problems, so here the line of action of P remains tangential to the deformed beam axis, so work done by P is dependent on path of the deformation, so such forces are called non-conservative forces, so now what is the critical value of P? That is the question we are asking, so what we can do is we can perform a static analysis, see when we performed stability analysis of a beam where direction of the load was unaltered, what was the conceptual framework within which we did that? We started with small values of P and incremented P in steps, at every step of incrementing P we gave a slight perturbation to the system and the system gets into oscillations, and presence of damping ensures that the system either returns to its original state or assumes an neighboring equilibrium position, so there was an element of dynamics even in this, so but it didn't, if you were to perform a dynamic analysis of that problem and examine the nature of fixed points associated with the governing equations we would reach the same conclusion as we would do by simply performing a static analysis, we have discussed the issue of fixed points, etc. in one of the earlier lectures. Now inspired by that what we will do is we will formulate this problem using purely a static consideration, so we can resolve this load as P sin phi and P cos phi, and when I write the equation for bending moment at any section I will include bending moment due to the horizontal component and due to the vertical component, so I will set up a coordinate system, X is measured from this and I want to write bending moment at X, so it will be P cos phi into this lever arm which will be F minus Y, and P sin phi will be, lever arm will be this, okay, so that is sin of phi L minus X. Now if we assume that phi is small we can assume cos phi as 1 and sin phi as phi, so the governing equation becomes, by simplifying this I get EI Y double prime plus PY is equal to PF minus P phi L minus, F is the displacement here and phi is the rotation, so what are the unknowns in this problem, F and phi are unknowns here, we don't know what they are because they have to be determined by analyzing the problem. Now I will divide by EI and introduce K square as P by EI and I will get this equation, and I can, this is a linear equation and I can write the complementary function and particular integral, now what are the boundary conditions? It X equal to 0, Y of 0 is 0, Y prime of 0 is 0, and what are these F and phi? F is Y of L and phi is Y prime of L, that is how we have introduced that. Now therefore there are now four constants, A, B, F and phi which are unknowns, and there are four conditions, what are these four conditions? These are the four conditions, so I can now impose that, so Y of X is this and Y prime of X I can find out by differentiating this, and by imposing the four boundary condition Y of 0 is 0, Y prime of 0 is 0, Y of L is F, Y prime of L equal to phi, I get a set of four equations, and the unknowns are A, F, A, B, F and phi, so I can cast it in a matrix form and write in this form, so for non-trivial solution the determinant of this equation must be 0, so if I impose that we get the determinant of this matrix will be minus 1, if we expand that, and it is independent of the applied load, so that means determinant of this matrix is not 0, so the only possible solution is a trivial solution, so what it means that only trivial solution is possible for all values of K, that means no matter what is the load P, the equilibrium position is always stable, so that is structure state of rest is always stable for all values of P, this defies expectations, so did we miss anything in doing this problem? So the thing that we have missed is we have done a static analysis, so the idea is if we say that loss of structural stability is accompanied by oscillations whose amplitude grows in time, whose amplitude grow in time, if we notice that then we can include inertial effects in considering the stability of the equilibrium state, suppose if you do that there is nothing wrong in trying that out, so let us consider the case when P was applied in a conservative manner, that means direction of P would remain unaltered due to deformation of the structure, so when dynamic analysis was performed and when P equal to P critical the response grew linearly in time with natural frequency when natural frequency was 0, the result from static and dynamic analysis coincided, so this we can recall in this context, now what we will do is we will reformulate this problem, now let's again assume that at some stage in the deformation the beam occupies this position, neighboring equilibrium position and P remains tangential to the deform axis of the beam, now what are the boundary conditions at X equal to L for the deform configuration, the bending moment and shear force must be equal to 0, this is because P has 0 components along A A, okay, so now therefore the boundary conditions at, whereas when we consider this problem the boundary conditions at X equal to L was EI Y double prime equal to 0 plus EI Y triple prime plus PY prime equal to 0, so P appeared in specification of boundary condition here, but whereas here since it appears tangent, it load remains tangential to the axis of the beam, the shear, the boundary conditions involving shear force P won't appear, now equipped with this now I will write this equation EI Y4 plus PY double prime MY double dot equal to 0, so the boundary conditions are at X equal to 0, Y and Y prime are 0, and X equal to L bending moment and shear force are 0, so I will again assume a solution where all points on the structure oscillate harmonically at the same frequency and I get this eigenvalue problem, so I divide this equation by EI and introduce K square and a parameter A which is M by a square root M by EI, so here this is a linear equation with constant coefficients therefore exponential must satisfy this equation, so I get phi of X as phi naught E raise to S X and this is a characteristic equation, so this is bi-quadratic equation, so I can get the roots by solving this quadratic equation and these are the roots, so based on this I will be able to write the solution phi of X is A cos lambda 1 X plus B sin H lambda 1 X plus C cos lambda 2 X plus D sin lambda 2 X, this is negative therefore square root of lambda 2 will be imaginary therefore we will get sin and cosine terms, whereas this will be positive therefore we will get sin and cosine term, sin H and cos H terms, so now I have four boundary conditions and I can do that I am skipping those steps, the condition for non-trivial solution we can obtain in this form, by you know writing the four equations forming the coefficient matrix and demanding that the determinant of that coefficient matrix is 0 I get this equation, now this actually is the characteristic equation and it relates P and omega, so this leads to the relation between P and omega and we have this Y of XT is exponential I omega T, this omega need not be real in this case, so if I assume omega to be A plus IB then Y of X, T will be exponential IA minus B into T, so for B less than 0 there will be instability, because the real part of this exponent will be positive and as T tends to infinity the solution grows and that helps us to determine the critical load which is 19.739EI by L square, so this result contradicts this analysis which showed that for all values of P the response is stable, now if P were to be greater than P critical this is how the solution would grow, it will be oscillatory and it grows, so this is called flutter and things like that, now there are two books which this book by Bolotain gives a classical book which discusses these problems and in the existing literature there has been criticism of this model and the basic question that has been asked is are there any situations where we can apply static loads which obey this hypothesis that they remain tangential to the deformed axis, so apparently no experiment till today has been done to characterize this that is a claim but there is a review paper in which these issues are discussed, so if you wish to understand the issues related to this discussion in the existing literature you can read this reference. Now we will now move on to the next item on our agenda on how to make finite element analysis for systems which are governed by partial differential equations with time varying coefficients, so we will consider this problem we have considered in the previous class I have explained all the basic terminologies of this problem and we have got the governing differential equation, so the coordinate system is a origin is here, X is measured along this axis, Y is the displacement measured from the neutral axis here, and vehicle is taken to enter the bridge at T equal to 0 and it leaves the bridge at T exit, and the time that it spends on the bridge is governed by the its motion parameters, acceleration and velocity we assume that these two are constant the time the vehicle is on the bridge, the vehicle itself is characterized in terms of an unsprung mass and a sprung mass and stiffness and damping characteristics of the isolation, so there will be now a degree of freedom associated with the vehicle and dependent variable Y of X of T associated with deformation of the beam, so these are the governing differential equations, the equation for U will obviously have the displacement of the beam because the force in the spring and the force in the damper depends on the relative displacement and velocity between this point, between this point and this point, so this point itself is deforming, so we will get these terms, and also I pointed out that as the structure deforms and this mass rolls on the bridge, this wheel rolls on the bridge it will be rolling on a deflected profile, therefore when I compute velocity and acceleration is needed to characterize the spring forces and the inertial forces we need to consider the total derivative, so that is why we are writing capital D by DT of Y of X of T, and this is the equation for the vehicle degree of freedom, and this is the equation for the beam oscillations, F of X, T is a wheel force that consists of weight of the vehicle, the force transferred from the spring and the force transferred from the damper and the inertial force of the unsprung mass, and this is a concentrated force which point of application changes with time as the vehicle moves and that is depicted through this direct delta function, so F of X, T is the wheel force, as the vehicle exits the bridge, the bridge undergoes small oscillations and since our interest is primarily on the bridge we will not write the corresponding equation for the vehicle, so and at T exit to solve this equation if you wish to model how the free vibration decay takes place after the vehicle leaves the bridge, you should, this is valid from T exit and the state of the bridge at T exit should be computed by analyzing this problem, so that means as the vehicle leaves the bridge the displacement and velocity field of the bridge must be captured through the model that is applicable for this time regime, so what remains as the status, the snapshot of the bridge response at T exit will serve as initial conditions to solve this problem, so what we will do is, we will try to develop a finite element model for this pair of equation, ordinary differential equation and partial differential equation by using what is known as an integral and weak formulation. In the development of finite element method so far in the course we have started with the variational principle, we didn't start from the governing differential equation, when we started discussing approximate method we saw that Rayleigh-Ritz method the way we apply Rayleigh-Ritz and Galerkin method were somewhat different, Galerkin method we applied on a governing differential equation, whereas Rayleigh-Ritz was on a variational formulation, so here we will assume that the starting point for discussion is a partial differential equation, okay, and how do we analyze this? So we need to prepare some basics for that, before we get into this we can consider some more aspects of this problem, see there are certain idealizations made in arriving at this model, for example we are assuming that the bridge deck is smooth, but in reality the bridge deck could be, there could be guideway and nearness and the vehicle will actually be moving on a bumpy bridge, one the deformation of the bridge along with the roughness of the bridge surface contributes to the input of, to the input to the vehicle, for example if this bridge was undeforming, but it is, that means if this bridge was, this vehicle was running on a rigid pavement but with undulations it will still feel the oscillations, not here the vehicle is running on a deforming elastic medium but it has certain undulations, so this can also be included in our model and there could be series of loads crossing the bridge, so again principle of superposition won't be valid here, by that I mean if you find the bridge response, bridge vehicle system response due to passage of one vehicle and similar response analysis for passage of another vehicle, that cannot be superposed to find response when both the vehicles pass the bridge, that principle of superposition is not acceptable. We can improve upon the vehicle model, we can include translation and pitching of the vehicle, we can model the vehicle in greater details and this type of models have also been studied in the literature. In practical situation of course the supporting structure itself will be a three-dimensional structure lattice girder bridge like this, and moving vehicle itself will be a fairly complicated engineering system, so this should be borne in mind when we analyze this type of models. So in reality to explore the, to realize the full potential of finite element method this type of problems need to be analyzed in the framework of a supporting structure being modeled as a, using as a finite element model and the moving system also using another finite element model and these two models move relative to each other, it is that problem that we should eventually be able to solve as we start with such an ideal situations such as this. Now as a prelude to developing finite element model starting from a partial differential equation we will revisit some basic notions about how to do this by considering a simpler problem, so what we do is we consider situations in which the system to be analyzed is described in terms of a governing differential equation, this is in contrast to our studies so far wherein we started with Hamilton's principle in formulating the problem. So what we do is we consider the equilibrium equation which is a partial differential equation say the inhomogeneous beam EI Y double prime, double prime plus M of X Y double dot equal to F of XT and let's assume that the beam initial condition there is a time varying term at the boundary for example, it's a fixed at the left end and it is free at the other end but it carries a time varying moment and it says assume that it starts from rest. Now our aim is to find an approximate solution to this equation in the form with certain trial function phi N of X and generalized coordinates A N of T, as we have seen earlier the substitution of the assumed solution into the governing equation leads to a residue, so what we do is we write what is known as a weighted residual statement for the problem, what we do is we take all the terms which are on the right hand side to the left hand side and we get on the right hand side 0 and multiply that by a weight function W of X and integrate over the domain of the problem, so we get this equation, this equation is the statement of the is known as the weighted residual statement, where W of X is a weight function, if Y of X comma T is the exact solution the term inside the bracket would be 0, so this will be automatically satisfied, on the other hand if Y of X comma T is replaced by an approximation this term inside the bracket won't be 0, and this can be used to obtain equation for A N of T, so how we do this we select a set of suppose there are capital N generalized coordinates, we select a capital N set of weighting functions and write this equation for those capital N number of weight functions and that leads to the equation for A N of T, so the above statement implies that the error of representation is 0 in a weighted integral sense, by choosing N independent weight functions we get N independent equation for the unknowns A N of T, N equal to 1 to N, and that is how we formulate the problem. Now here if you see here when I am representing Y of, here we have to choose two things now, one the trial function and the weight function, now if we now look at the demands on continuity of this trial function and the weight function, here you will see that the trial function need to be differentiable up to fourth order, whereas weight function need not be differentiable also, it should be simply integrable, okay, so the demand on the weight function and the trial function is not even, if that is not a restriction we can simply go ahead and select capital N number of weight function and solve the problem, but obviously that is not a fair situation, to construct trial functions which are differentiable up to fourth order requires fairly elaborate representation for phi N of X and that leads to increased computational burden, so we want to reduce the demand on differentiability on phi N of X, so what we do? So the observation is that continuity requirements and W of X and phi N of X are different, the requirements on phi N of X are more stringent, the weighted integral statement is equivalent to the governing field equation and does not take into the boundary conditions, because boundary condition issue has not yet come up, the unknowns N of T can be determined by considering N weight functions as shown here, so you write this equation for set of N weight functions and we get the required N equation, so to proceed further with the solution we need to select the trial functions phi N of X, N equal to 1 to N which possess fourth order derivatives and satisfy the prescribed boundary conditions, but there is no such stringent requirements on the weight functions, so this looks a bit unfair because we have to select both of them, so now what we do is we integrate the term by parts, by integrating by parts first time, the first term we integrate by other terms we can retain as it is, so it becomes now, fourth derivative becomes third derivative and here we get the first derivative on the weight function, the other terms remains the same, so we integrate once again then this, this return is following from this and from this term I get W prime of X EI Y double prime, so the derivative on this derivative has now vanished and in the integrand I get now W double prime of X and Y double prime of X DX, plus this term as it is, now this statement is known as a weak form, now if you look, if you decide that you will work with this form of the equation then the original weighted residual form statement, then we see that the trial function now need to be only differentiable up to second order, but now the weight function need to be differential up to second order, so the demand on the trial function and the weight function as far as differentiability goes is now E1, so by integrating by parts we have achieved a trade off from, on differentiability requirement on trial functions and weight function, so we will make these comments, so the differentiability requirement on Y of X and hence on the trial functions has come down to 2 and the requirement on W of X has gone up to 2, the integration by parts has enabled us to trade the differentiability requirements between trial functions and the weight functions, now consider the now the, we have two more terms because of integrating by parts, now let us consider those two terms, now we can based on this we can identify two types of boundary condition, these terms are associated with what happens at the boundaries X equal to 0 and X equal to L, so these are clearly associated with boundary conditions, so based on this we can identify two types of boundary conditions, one set known as natural and the other one known as essential. Now the rule for this division is as follows, we call coefficients of the weight function and its derivatives in the above terms as secondary variables, for example weight function W of X the coefficient is EI Y double prime prime, so EI Y double prime prime is a secondary variable, similarly here weight function derivative is multiplied by EI Y double prime, so EI Y double prime is called a secondary variable, specification of the secondary variables on the boundaries constitute the natural or force boundary conditions, the dependent variables expressed in the same form as the weight function as appearing the boundary terms are called the primary variables, thus we have W of X therefore Y of X of T and we have W prime of X and thus Y prime of X of T are the primary variables, specification of the primary variables on the boundaries constitute the essential or geometric boundary condition. So we divide, in summary we are dividing the boundary conditions into natural and essential and we have now developed a prescription for classifying these boundary conditions, these variables as being primary and secondary and the boundary condition being natural or geometric. Now in this type of formulation if you are dealing with even order differential equations which we are always doing in many of the structural engineering problems the number of primary and secondary variables will be equal, the secondary variables have direct physical meaning as far as the problem is concerned, for example here EI Y double prime is a bending moment, EI Y double prime is a shear force, so each primary variable is associated with a corresponding secondary variable, so there is a natural pairing of primary and secondary variables, for example if you have secondary variable which is bending moment the primary variable is a slope, if the secondary variable is a shear force the primary variable is a displacement, so essential boundary conditions involve specifying displacement and slope at the boundaries and natural boundary conditions involves specifying bending moment and shear force at the boundaries for the beam problem. Now on the boundary either a primary variable can be specified or the corresponding secondary variable can be specified, a given pair of SV and PV cannot be specified simultaneously the same boundary, thus for example at a free end of a beam we can specify bending moment to be 0 but the slope remains unspecified, similarly shear force can be specified to be 0 in a simple supported end and displacement remains, no in a free end shear force can be specified to be 0 but the displacement remains unspecified, okay, so now we use some notation EI Y double prime as V which is shear force, EI Y double prime as M we write the weak form in this way, so the weak form statement is this, now we now require the weight functions to satisfy the essential boundary conditions of the problem, the recall the boundary conditions we are considering is Y at X equal to 0 Y and Y prime of 0 and at X equal to L we have M0 applied bending moment and shear force which are 0, now we demand that at X equal to 0 W of 0 and W prime of 0 is 0, so this is similar to the virtual displacement concept where virtual displacement conforms to the prescribed boundary condition, so this weight functions must also conform to the prescribed boundary condition, thus we have W of X, V of X 0 to L will be simply this, which is this and the weak form with this understanding that the weight functions satisfy the geometric boundary conditions we get as this, this can be now used to proceed further with the problem, so this is equivalent to the original differential equation and the natural boundary conditions, now what we did was we started with this assumed solution, so what we will do now is here A N of T needs to be determined, so now I will substitute this into the weak form I get this and I will now use in one way of proceeding further W of X to be equal to the trial function themselves and we obtain a set of N equations for A N of T and we can proceed further with the analysis. This approach leads to symmetric coefficient matrices when we formulate the problem and the natural boundary conditions are included in the weak form and the approximate solutions need to satisfy only the essential boundary condition, so a discussion on this is available in the textbook by J N Reddy you can see this also, so what is weak here is the requirement on differentiability of the trial function has been weakened, it is in that sense the formulation is called a weak formulation, how do we develop finite element solutions for this? Phi N of X were globally valid shape functions in the previous formulation, now if you want to apply finite element method we divide the domain into say N elements, so and if you consider a typically a Kth element with coordinate XK-1 on the left and XK on the right and suppose each node has 2 degrees of freedom, so what we can do is we can consider the Kth element and introduce a local coordinate system XI as X-I equal to N-1 Li where Li is XI plus 1 minus XI which is the length of the element, so as X varies from XK-1 to XK, XI varies from 0 to LK, so for the Kth element using the notation Y prime as dou I by dou psi I can write the equation in this form, so now the boundary conditions are here at XI equal to 0 the displacements are U1 and U2 and XI equal to 1 displacements are U3 and U4, similarly there will be a shear force and bending moment here, a shear force and bending moment, so these are the set of 8 boundary conditions that we need to you know consider, so the weighted residual statement is given by this, the weak form including the required boundary conditions can be, that means again we differentiate, integrate by parts twice and the order of differentiability on Y and W becomes equal and we get this weak form. Now we have these functions F1, F2, F3, F4 which appear in the boundary I write in this form, so I have this as a weak form for the Kth element. Now this is a weak form and now I assume the solution Y of X, T is I equal to 1, 2, 4 UI of T, phi of XI and we can now select phi of XI to be the cubic polynomials, I don't want to discuss the issues on how to select the polynomials, here we have seen that, so we can select them and we select W of psi to be the trial functions themselves and I can get, I can now perform these integrations and the terms involving EI will lead to the stiffness matrix KIJ as shown here and terms involving mass will lead to the inertial, the mass matrix as shown here and the remaining terms contribute to the equivalent nodal forces as shown here, this is for Kth element, so these are the cubic polynomials I select and the mass matrix and stiffness matrix we have derived earlier in our studies and this is same, the same consistent mass matrix and the elastic stiffness matrix. Now we need to assemble, assembling this is a formulation up to Kth element, now the requirements for assembling is the inter-element continuity of primary variables that is deflection and slope in this case and inter-element equilibrium of secondary variable that is bending moment and shear force, so that is how, I mean this we need not discuss in greater detail because we already seen how to do these things, then imposition of boundary conditions, primary variables are not constrained, the corresponding secondary variables are 0, if primary variables are not constrained, the secondary variables will be 0, for example at a free end the translation and rotation are not constrained, therefore the shear force and bending moment would be 0, unless you have externally applied actions at the free end, that's what I am saying, then primary variables are prescribed, for example they are prescribed to be 0 or they can be prescribed to be specified functions of time as in earthquake ground motion or any other support motion problem, so in that case the corresponding secondary variable will determine the reactions, so this leads to the governing equation of motion, so this is a framework where we start with the governing partial differential equation and develop the finite element model for the problem, so in the next class what we will do is we will consider this framework and consider the problem of vehicle structure interaction, this is a partial differential equation and the ordinary differential equation that we have derived and we will develop the finite element model for this case starting from the integral and weak formulation, so we will close the lecture at this, this lecture at this stage and we will pick up on this in the next lecture.