 In today's lecture we will continue our discussion with plate bending elements, so for what we have done is we have developed various types of elements for both theory based on thin plate hypothesis and thick plate hypothesis, and the element that we have developed were of you know some of the elements were conforming somewhere non-conforming. Now before I start discussing a topic for today that is Tiffin plates, I would like to make few observations on some of the issues related to using of these elements. Now two questions I would like to raise, first question is how do thick elements behave when used for thin elements? Then how to examine convergence characteristics when non-conforming elements are used? Both these questions are widely discussed in the literature, so I will just give some broad indications on what are the issues and how some of the difficulties can be resolved. So to enable this discussion we will revisit the problem of Timoshenko beam, so ask the question what happens if beam depth reduces? So if you recall in the Timoshenko beam this is undeformed geometry, and in the deformed geometry the plane section here remains plane but it rotates by an angle psi, it doesn't remain normal to the neutral axis there will be a shearing strain that we wish to take into account. The expression for strain energy we have already derived will be of this form. Now if you examine the integrand in this expression for the energy we see that the field variables are psi and W and the order of the highest derivative is 1, therefore we can use linear interpolation functions for both representing psi and W, so if we consider two-noded element with degrees of freedom W and psi at each node we can write psi of X as this and W of X as this. Now if I differentiate W of X I will get W1 minus W2 by L, this is a constant for the element. Now as beam depth becomes small the psi of X you know the psi of X should go to dow W by dow X, this is what we assume in thin plate theory, so in the thick plate theory this as beam depth becomes small psi of X we can expect that it will go to this, this according to this representation becomes constant, but this is not admissible since the bending energy is given by this term and this becomes 0 if psi of X is constant, so this problem is known as shear locking, so this is undesirable so how do we, how can we remedy this situation? So there are two possible remedies, one is we can use interpolation polynomials such that psi and DW by DX are interpolated by using the same order of polynomials, for example W I will use a cubic polynomial and for psi quadratic so that dow W by dow X and psi will have the same order of representation, or there is an alternative fix known as reduced integration I will try to explain what these things mean in the following slides. So let's consider a beam element, now I will use the non-dimensional coordinate psi is equal to X by L and I will locate the origin here, so the beam lies between XI equal to minus 1 and XI equal to 1 and its length is actually 2A X varying from minus A to plus A, so this axis is X, psi and this rotation is psi and this is W, now the field equations are given by this and these imply that D4W DX4 is 0, D cube psi by DX cube is 0, now accordingly using psi equal to X by A, W can be represented as a cubic polynomial and psi can be represented as a quadratic polynomial, now in this representation the A1, A2, A3, A4, B1, B2, B3 are not independent, they are connected by this equation and if we impose that constraint we get the constraint equations as B1 given by this, B2 given by this and so on and so forth, and we introduce a parameter beta which is EI divided by KAG A square, now as beta goes to 0, beta goes to 0 as depth of the beam reduces, so this is one handle that we will have to examine the behavior of the element as depth becomes small, now we have only 4 independent constants because of this constraint equation that is fine, so now consequently we can represent W in terms of this shape function N1, N2, N3, N4 and WE is a nodal coordinate W1, psi 1, W2, psi itself is represented in terms of this quantities interpolation function N5, N6, N7, N8, but they are all given by this, there are only enough independent constants as mentioned here. As beta, see one thing we should notice is if we now consider the representation for displacement field W, as beta goes to 0 these functions N1, N2, N3, N4 reduces to the cubic polynomials that we used in analyzing Euler-Bernoulli beam. Similarly if I now take psi as dow W by dow psi I get this and it can be verified that the functions that are listed here N1 to N8 satisfy these requirements, that means in the limit of the analysis being applied to a slender beam we recover what we have analyzed, what we considered for Euler-Bernoulli beam, so that is embedded in the way we are interpolating here. Then the procedure for deriving structural matrices is straightforward, we have the energy expressions and field variables are represented in terms of the nodal values and interpolation functions, so we can do the evaluation of structural matrices and we get the structural matrix, mass matrix as given here, it has two parts and they are listed separately and it is given here. Similarly the stiffness matrix is given by, it can be shown that it will be given by this, now in this case as beta becomes small it can be verified that the mass matrix and stiffness matrix converge to the mass and stiffness matrices that we have already derived for Euler-Bernoulli beam, so here there is no problem, here the thick beam element remains applicable as a beam becomes thinner and no special precautions need to be taken to use this element. But on the other hand suppose if we start with the argument that psi and W are the field variables and the highest order of the derivative that is present in the integrand of the variational formulation is 1, therefore I am at liberty to use linear interpolation for both W and psi that is fine, so we can start with that, so W is interpolated like this and psi is interpolated like this and where N1, N2 are linear functions as shown here. Now the element nodal degrees of freedom will be W1, psi 1, W2, psi 2 and they can be assembled in terms of this N matrix and the nodal values as shown here, now we can compute dou psi by dou X and dou W by dou X which are needed in the evaluation of the strain energy and we see that the strain energy due to bending and shear can be evaluated, we get for the bending this is the expression that is EI dou psi by dou X whole square will be this and this is a constant and the shearing function, shearing energy due to shearing will be given by this. So this integral can be evaluated with one term Gauss quadrature and this to evaluate exactly I need 2 point Gauss quadrature because this will be fourth order in XI, so but if we do this then as I already been pointed out we will face the problem of shear locking, okay, because in this representation as the beam depth becomes small the representation doesn't automatically converge to the Euler-Bernoulli beam limit, so there will be a problem if we do this. What is proposed is we replace the linear shear strain variation by a constant in the sense of minimizing the mean square error across the depth, so following this the integrand equation for US, integrand in equation for US becomes constant and hence can be evaluated using a 1 point Gauss quadrature, moment we do this then the locking problem will be eliminated and this method of overcoming the problem of locking is called method of reduced integration. So this amounts to by doing this integration we are virtually ensuring that dou psi by DW by dou X and dou psi are of the same order, okay, so then by using a reduced quadrature as mentioned here we are achieving basically that, that is why the shear locking problem gets eliminated. Now this is as far as T machinico beam is concerned, now let us return to the four-noded thick rectangular plate element, we saw that energy expression is given by this and we found out the element stiffness matrix in terms of flexure and shear and by using this formula. Now the problem of shear locking is possible here too, this can again this can be overcome by using reduced integration, see if we examine the details of these integrands here these integrals can be evaluated exactly using 2 by 2 Gauss quadrature, this provides acceptable result for thick plates, there is no problem here, but however for thin plates the locking problem occurs if we do this, so what is suggested is we evaluate the flexure component by using 2 by 2 Gauss quadrature, but for evaluating the shear part we use a 1 by 1 Gauss quadrature, so this indirectly ensures that the locking problem is eliminated, so this is one question that we posed about behavior of deep beam and plate elements as the thickness becomes small. Now I wish to make few comments on issues related to convergence, in using any finite element formulation we can ask several questions after we formulate the element, at the stage of completing the formulation of the element and when we are ready to use it we can ask several questions, has the element been formulated correctly and has a computer codes have been developed correctly, how do we know that an element available in a readily available software is correctly formulated and coded, suppose even if we are not the people who have developed the element, we may be using production version finite element softwares and we may like to use a particular element in a given model, how do we know that either we are using that element correctly as intended or whether the element itself has been formulated correctly and coded correctly. Now other question we can ask is when non-confirming elements are used how do we verify if the convergers should be achieved by refining the mesh, so these questions can be answered by considering what is known as a patch test, the story goes something like this, what we do is we assemble a patch of elements under study, the assembling of this patch of element should be such that at least one node must be within the patch and the node should be shared by two or more elements and one or more inter-elemental boundaries must exist. Now we load this patch at the boundary nodes so as to create a state of constant stress, so this state of constant stress can be with respect to one of the stress components, for example in a two dimensional problem it can be state of constant stress in sigma xx or sigma yy or sigma xy, now we support the patch just adequate enough to prevent the rigid body motions. Now we analyze the problem using the element developed and we compute the stresses, if the computed stresses within the entire element agrees with the exactly known value of constant stress and perhaps zero stress for other, I mean some of the components then the patch test is passed by the element, an example for that is suppose you are using plane stress quadrilateral element, so we make a patch, this is a node here and we create four quadrilateral plane stress elements and we load this edge by a constant surface traction, so this surface traction is a constant and for this constant surface traction we evaluate the equivalent nodal forces using the correct formulation and these are the forces and this is also the another force that we have applied and this is the support conditions, you can verify that under this state of loading and the way it is supported this patch would be having constant value of stress sigma xx equal to F, where F is the magnitude of this constant value and sigma xy and sigma yy must be equal to 0, so we can analyze, suppose you have developed these elements you can analyze and find out whether these conditions are met or not, so the example that I mentioned considered the case of sigma xx being constant, for a plane stress element we need to repeat the test for other loading configurations corresponding to constant states of sigma yy and sigma xy, before we can pass the test, they pass the element for further use. If the element passes the pass test it is ensured that a finite element model which uses this element converges to the correct solution as the mesh is refined repeated, if an element fails to pass the pass test I mean we should be very careful in using those elements, it assumes that the element being tested is stable, so what is the meaning of a stable element, a stable element does not exhibit zero energy modes when it is adequately supported so as to avoid rigid body motions, one can study stability of the element by considering the Eigen values of the stiffness matrix, there is another numerical way to do that, so what we do is we assemble the patch, a patch as in the pass test and load in, apply the load as in the pass test, now assuming that the element has passed the pass test what we do is we perturb one of the nodal loads by a small amount, if the computer stresses change by a large amount because of this small perturbation then the patch has failed the stability test, if the test is applied at the single element level then the element has failed the stability test, that means if at the element level we see that a small perturbation of the, to the loads as depicted here produces large changes in the stress values, again we say that that particular element has failed the stability test, there is what is known as weak patch test where the element approaches towards passing the pass test as a mesh of the patch is refined successively, that means for a given mesh configuration like this the element may not pass the pass test, but if we go on refining this as we repeatedly refine the mesh the pass test tends towards success, so I mean the test is passed and then we say that the element passes the weak pass test, there are other issues like higher order pass test, for example we have considered in the pass test states of constant stress there can be other issues, for example modeling of pure bending in plane elements, an element with quadratic expansion for U and V must be able to represent exactly a field of pure bending, so pure bending is not a state of constant stress but it has a specific distribution and this can be thought of as higher order pass test. Similarly there are other issues like robustness, for example a state of pure bending in plane mesh must not be affected by the Poisson ratio of the element, so some of this can be used to verify whether the element has been formulated correctly or are we using the element as intended etc., so these issues are discussed at a tutorial level in these textbooks there are many research papers on this, so this is a subject of research, if you are interested there are many research papers on this. Now one more remark that can be made in the same context, these numerically integrated elements lead to the correct convergent results as the mesh is refined, if the order of integration chosen leads to the correct evaluation of the volume of the element, why is that so? The question we are asking is suppose you have evaluated stiffness matrix using Gauss quadrature and we have seen that the Gauss quadrature would not integrate the elements of KE exactly because there will be ratios of polynomials in the integrand, so the question is if we are using such elements how do we know that we are going to get the correct convergent results as mesh is refined, now the solution that is made here is if the order of integration chosen leads to the correct evaluation of the volume of the element then we are okay, I want you to think about this and answer but I will give a hint, as the mesh is refined and constant strain is reached, as mesh becomes finer within an element strain tends to become constant we get the stiffness matrix to be of this form, so this term becomes constant and what remains is the volume of the element, so the statement made here is that if this volume is corrected correctly then this required condition on convergence is met. Now all these checks have been on static behavior of the system, so how about problems of dynamics, so there are few checks on mass matrix I will return to this sometime later in the course but I will just indicate a few points here, this is something to do with Eigenvalue analysis, the question we can ask is are rigid body modes predicted correctly, if I know that a structure is supported in a way that suppose it admits to rigid body modes the Eigenvalue analysis should reveal that, next the question of ability to characterize closely spaced modes and repeated Eigenvalues, we have seen that in structural displaying symmetry in stiffness and mass characteristics, spatial symmetry the Eigenvalues can repeat, for example in a circular plate we saw in the previous lecture which is fixed all around there are two mode shapes for a given Eigenvalue, so how these things are handled and if the natural frequencies are closely spaced will my Eigenvalue solver be able to handle that correctly. Next are all the Eigenvalues in a given range extracted, that means within a frequency range you should not miss any Eigenvalues, so that is also another question, and then again if you have used model reduction or substructuring there will be further questions on choice of masters and slaves and so on and so forth, so the questions on dynamics require further consideration maybe we will return to some of these issues later provided the time permits. Now in the previous lecture we considered four nodded rectangular quadratic thick plate element, I leave it as an exercise for you to repeat that exercise for the quadrilateral element, the strain energy expressions are given here, and this is the geometry of the quadrilateral element, and we use the isoparametric representation as shown here, and this map is, this element is mapped to this master element, and we use this interpolation functions, NI xi, eta is this, and X and Y are represented using the same interpolation functions, and the displacement field theta X, theta Y are represented in this form, so the exercise is to develop the mass and stiffness matrices for this element. Another exercise, you take a eight-noded curved quadrilateral thick plate bending element, and this is an expression for strain energy, and using these representations there are eight trial functions, and the trial functions have been defined here, for 1, 2, 3, 4 this is the trial function, and for 5, 7, and 6, 8 these are given here, and we again use isoparametric formulation where the coordinates X and Y and the field variable W, theta X and theta Y are interpolated using nodal values using the same set of interpolation functions. Now the problem is to develop the mass and stiffness matrices for this element, so this is suggested as an exercise, this will, if you complete these two exercise you would have learned a great deal about formulation of the structural matrices for plate elements. Now we will now consider the next class of problems associated with plate bending, here we are considering plate structures which are stiffened by beam elements like this, so these are typically observed in bridge decks, building floors, ship hulls, aircraft structures on and so forth. Now the beam central axis that is a line passing through this point C here is accentrically placed with respect to the middle surface of the plate which it aims to stiffen. Now obviously the membrane and bending action of the plate and the flexural action of the beam all of them get coupled and correct formulation should be able to handle these details with care, so how do we do this? So the coordinate systems here we assume that the coordinate origin passes through the mid-surface of the plate element and the centroidal axis of the beam is at a eccentricity E and I consider this to be the X axis, this Y axis and this Z axis and the degrees of freedom that we are considering is U and theta X, V and theta Y and W, okay. So a first cut model for this problem we can begin by ignoring membrane displacements of the plate, so we will consider only the flexural action and of the plate and the beam. So now the strain energy is stored in the beam is given by this expression, we are assuming deep beam element, so we are including rotary moment of inertia as well, so this is the equation. The strain energy is due to axial deformation, bending, twisting, bending about Y, bending in the two directions, what we are going to do is we are going to ignore this bending contribution for the time being, this can be, you know this bending in this direction can be taken to be ignored up. Now based on the geometry of this configuration we can see that Uc is given by E into theta Y, which is E dou W by dou X and Vc is minus E theta X and Wc is W. The beam displacements actually need to be compatible with the plate displacements along the attachment line. The expression for kinetic energy I will use these representations, so this is the expression for kinetic energy and Uc, Vc, Wc are given in terms of W theta X through these equations and if I substitute, make these substitutions I get for Uc dot for example I will write E dou W dot dou X, so that is what is here and similarly other terms are represented and I get the expression for kinetic energy in this form. Similarly strain energy is given by these three terms, again for Uc, Wc and Wc I will use these terms and using that I will get the expression for strain energy in terms of W and theta X to be this. So kinetic energy is also obtained in terms of W and theta X, so these are the expressions for kinetic energy and strain energy. As I already said along the attachment line the beam and plate must deform in a compatible manner. Now if you recall along the edges many of the plate bending elements have cubic variation of, a cubic variation for W, so if you recall if you consider this four-noded beam element, sorry thin plate element we had this expression for strain energy and we saw that the field variable in this expression for strain energy is W and the highest derivative present in the Lagrangian is 2 and therefore degrees of freedom are W dou W by dou X and dou W by dou Y and there are four nodes and number of generalized coordinates were 12, because we needed three degrees of freedom at every node and we use this representation. Now suppose if you consider this representation along one of the edges, you see here that if I put X equal to A I get this cubic polynomial in Y, so that is what I am saying here along the edges many of the plate bending elements have cubic variation for W. So now I will use this representation for W, NW1 psi, theta Y1, NW2 psi, etc., etc., as shown here, where NW1 is N1, NW2 is minus A N2 and other details are as here, and N1, N2, N3, N4 are the Hermite polynomials that we have used. So using this I will write W as NW into WE, so where NW is this and WE is given by this. For twisting theta X the second term I use linear interpolation functions that we have seen NX1 psi, theta X1 plus NX2 psi, theta X2 and these are linear interpolation functions, for theta X I get this. So I have this representation for W and this representation for theta and I can combine the two and write the expression for kinetic energy, W is this and theta X is this and substitute and we can carry out the necessary integrations and we can show that the mass matrix will have first integral can be shown to be, will lead to this form and the second integral will lead to this form. So we have these expressions for the three terms in the expression for kinetic energy and we now define UE as element nodal degrees of freedom, W1, theta X1, theta Y1, W2, theta X2, theta Y2. Now we can write the expression for kinetic energy and assemble these matrices, we can show that this ME can be given by these four matrices and these four matrices are as shown here, this M11 is here, M12 is here and M22 will be this. Now how about strain energy? We have this expression for strain energy and again we will use the representation for W and theta X and we evaluate these two terms separately, first we consider the first integral and this leads to a mass matrix of this kind and the second one leads, sorry stiffness matrix of this kind and second one leads to stiffness matrix of this kind. Again we can assemble in terms of nodal degrees of freedom, the element stiffness matrix in terms of these four matrices K11, K12, K12 transpose and K22 and the details of this are given here. In this formulation we have not included the effect of membrane action of the shell element on possible behavior of this beam. Now if you want to include that, first we need to develop an element for the plate itself where we combine the flexural and membrane actions, so that type of elements are known as facet elements, this combines the action of the membrane action and bending action, so the element configuration will be as shown here for a four-noded element, so the nodes are 1, 2, 3, 4 and at every node I will have, for membrane action I will have two degrees of freedom and for flexural action I will have three degrees of freedom, so therefore W will be this and S is subscript for shell and this will be this, so this will be a 20 degrees of freedom model, there are four nodes, at each node there are five degrees of freedom, the five degrees of freedom are U1, V1, W1, theta X1 and theta Y1 at node 1. So we need to now assemble these matrices, write the correct expression for energies and derive the matrices, this is straight forward because we have already done the problem it is now matter of just bookkeeping and assembling the matrices correctly. So we will write the kinetic energy in terms of membrane action and bending action, the membrane action is given by this and this is U1, U1, U2, U2, U3, U3, U4, U4 and I get this matrix for mass matrix for membrane action where each of these sub matrices is a 2 by 2 matrix, this we have derived earlier so we need not have to get into the details once again. For the bending action again the nodal degrees of freedom are 12 which are W1, theta X1, theta Y1 and similarly the other, so the mass matrix for bending action again is given by this and each of these matrices, sub matrices is 3 cross 3, so now if I write the expression for T I need to appropriately assemble these matrices. So now if I define the nodal coordinates for the shell element subscript S is for shell I will have 20 nodal degrees of freedom and I need a 20 by 20 mass matrix and in terms of the element mass matrix, shell element mass matrix this is given in terms of this nodal, you know kinetic energy is in terms of US dot transpose MS, US as shown here, and this MS itself can be assembled in this form where each of these matrix is a 5 by 5 matrix consisting of 2 sub matrices, a 2 by 2 membrane matrix and a 3 by 3 bending matrix. So this assembled mass matrix now is the mass matrix for the so called facet shell element, similar arguments can be used for stiffness, we have strain energy contribution from membrane action and bending action, again the nodal degrees of freedom for US is same as what we use, what was mentioned for kinetic energy and KS is the shell element stiffness matrix and this again is written in this form, all these matrices are symmetric. Now, so we have now derived the mass and stiffness matrix for the facet shell element, now we will return to the problem of now the stiffened plate, so we will now try to construct a refined model where we include the membrane displacements of the plate in the analysis, the expression for energies will remain the same but the details of displacement will be different, see earlier we use UC as E theta Y, now we are including the membrane displacement, the two quantities shown in the red are the new terms, so consequently now the form of the Lagrangian will change now, so we need to now include these new terms, so this is the expression for kinetic energy, this is the expression for strain energy and these are the displacement fields in terms of W dou W by dou X, theta X and theta Y, so now if we consider the expression for kinetic energy you have to substitute this into this expression and rearrange the terms, I will get the expression for kinetic energy to be given by this, there are 4, 5 integrals now because of the cross terms and other things these new terms will be present, similarly strain energy will have 5 contributing terms, 4 contributing terms as shown here, and now the field variables are UVW, W theta X and so we have UVW, theta X, these are the field variables, so now we represent them in terms of interpolation functions and nodal values, first the membrane action U1, U2, these are the interpolation functions, so for bending action I have these degrees of freedom and I get the mass matrix for bending to be given by this, this is for the beam element, the B is for beam element and this is the elements for M11, M12 and M22 which appear in this matrix, so basically we are considering this Lagrangian made up of this and using these interpolation functions and constructing the solution. Chiffness matrix again given in this form U transpose KB, UB and this is assembled in a form K11, K12 and K22, each one is 5 by 5 matrix as shown here, K11, K12, K22, so all these details need to be worked out and I leave that as an exercise, this is K22. Now so far in the course we have focused on vibration problems and we started by discussing simple actually vibrating bar and then flexuring a given plane and 3 dimensional beam element and with that we constructed planar frames, grids, 3D frames and we have considered several aspects of vibration analysis including time integration methods and substructuring and model reduction and then we moved on to study of 2 dimensional elements, plane stress and plane strain elements and plate bending elements and we have seen today the facet shell element and the stiffen plate element. Now in the next part of module of this course we will take up a new topic that is related to stability of structures, later on after completing the discussion on stability we will return to problems of vibration analysis again and we will consider problems of finite element model updating and some issues about non-linear vibrations and some questions on combining numerical and experimental models in problems of you know finite element model updating as well as structural testing using hybrid simulation, so we will return to those topics after we address few issues related to stability of the structure. Now in the subject of stability of structures we basically consider certain states of the system which could be state of rest or state of periodic motion or it could be as well state of random motion, the idea here is we ask the question whether this what happens to these states if we perturb these states by a small perturbation. Now such perturbations can occur in engineering practice because of various reasons, so suppose these perturbations occur, because of this perturbation the response of the structure would change, now we say that these states are stable if the response to perturbation dies and original state is restored, if not if original state is not restored then if the motion grows without limits then we say that the state is unstable, the issue remains unresolved if motion neither grows nor decays we need to consider further you know approaches to analyze this type of problems. So in the next few lectures we will consider questions related to this and we will begin by considering some familiar problems like problems of beam column, suppose if you consider a simply supported beam carrying transverse load Q of X and also axial loads P, in absence of axial loads P we have studied how this structure vibrates how it displaces on and so forth, now the question we wish to consider is what would be the influence of simultaneous presence of transverse loads and axial loads like a P like this, we will be showing that there exists certain critical values of P for which a slightest transverse load will produce huge responses in the system, so the equilibrium state in the neighborhood of P being close to the value of those critical values is unstable in the sense a small perturbation will produce large responses. In there are two types of issues here, the new equilibrium position that the structure takes could be in the neighborhood of the original equilibrium position or we can have certain types of problems as schematically shown here, this is a problem where there are two rigid links hinged at these places and loaded as shown here, as the load P increases this point starts moving downwards and for a certain critical value of P after the structure displaces and reaches this point it will snap. So from this the equilibrium position in the neighborhood of somewhere here is unstable because a small perturbation pushes the response to a far away equilibrium position, so this phenomenon is known as snap through. Now in a structure that is loaded by external loads it is always a question that we should consider what would happen if there is a slight perturbations in the external loads, if the structure is in the neighborhood of being you know on the verge of losing its stability any slight perturbations will create dramatic increase in the response and that virtually means in engineering system the structure would fail, that is highly undesirable. So we would like to know how close we are to a critical state and in a good design we wish to be sufficiently far away from those states that is what we do in design of metal structures, so the question that we will be addressing is not so much on the phenomenological aspects of stability analysis but more specifically how to analyze problems of stability using finite element method, specifically we will be considering what is the role of imperfections in the structure, what is the role of interactions between imperfections and nonlinearities and if there are dynamic excitations, for example if this P and Q that I have shown here are depicted to be static loads here but they can as well be functions of time, suppose Q is a static load but P is a dynamic load, now we can show that for certain types of loads for example if P is something like P naught plus epsilon cos omega t, if it is a harmonic load then the structure can get into resonance even when this driving frequency capital omega need not coincide with any of the natural frequencies of the system, so these are not the traditional resonances but these are special types of resonances, parametric resonances, there are many systems in engineering practice where we get problems of this kind and problems of this kind are characterized by structural matrices which vary in time, so for these time varying structural matrices the questions on parametric excitations and stability of the structure also needs to be considered, so in the lectures to follow what we will do is we will first look at certain conceptual issues related to study of beam columns and we will be showing that the problem of determining critical loads can be tackled using again in Eigenvalue problem we will derive the elastic stiffness matrix and we will also derive a new matrix associated with the structure known as geometric stiffness matrix, so that an Eigenvalue analysis associated with those matrices will be able to help us to determine the critical axial loads, then we will consider built up structures like continuous beams and frames and so on and so forth which carry axial loads or lateral loads and so on and so forth. The question we will ask is for a given loading configuration, if the entire loading configuration is increased in its magnitude by keeping the relative values of loads at different places the same at what value of the increasing the parameter that increases the load alpha will the structure lose stability, so that type of questions we will consider and we will begin by addressing that type of problems for simple structures like single span beams and then generalize it to built up structures, so with this we will close the present lecture.