 we have been discussing issues related to model reduction and substructuring, so we will continue with this topic in this lecture, so we can quickly recall what we have been discussing, we are considering a large finite element model governed by equation of this kind we are restricting our attention to linear time invariant systems, the idea is that we want to reduce this model to a lower order model, R is for reduced here, through a transformation X of T is some psi into XM of T, that means we will partition this degree of freedom X of T into a set of master degrees of freedom and slave degrees of freedom, and the state vector X of T is related to the master degrees of freedom through this psi matrix, so once we find that we can substitute this into the governing equation and we get this reduced mass damping and stiffness matrices and the reduced forces governing the master degrees of freedom, there are different model reduction schemes they differ in their definition of this matrix psi, in static condensation technique the slave degrees of freedom are taken to be related to the master degrees of freedom through relations which are strictly valid only for static under static conditions, so we derive the relation between X and XM using static equilibrium equations and this is a psi matrix. In dynamic condensation we again partition the degrees of freedom into master and slave, in addition we specify a frequency at which the condensation is done, so the psi matrix is given as here, now if one computes the natural frequencies and mode shapes of the reduced system the set of you know natural frequencies from this model need not agree with any of the natural frequencies of the larger model, that is certainly true in case of static condensation method, in dynamic condensation method if this omega is chosen such that it coincides with one of the natural frequencies of the larger system then one of the frequencies in the reduced model will match with that frequency. Now in a third method that is system equivalent reduction expansion process the psi matrix is constructed from the modal matrix of the larger system, so the modal matrix it can be I mean not all modes need to be evaluated, first P modes suppose if we include the modal matrix will be a rectangular matrix, now the transformation matrix there is synthesized from the partitioning of the available modal information and this is the psi matrix that is obtained here. Here again we need to partition the degrees of freedom to master and slave, in addition we need to also specify the modes of the larger model which needs to be replicated in the reduced model, SCRIP has the feature that it exactly reproduces Eigen solutions of the larger model, a set of Eigen solutions of a larger model are retained in the smaller model, so these are the three techniques that we discussed. Now we move on to the next topic in this that is so known as coupling techniques or the substructuring techniques, the basic idea here is again we deal with large FE models and we want to devise some way of reducing the size of the problem while computing the response, the idea here is that often a large finite element model may be needed to take into account geometric complexities of the structure and a very detailed modeling may perhaps be not needed if one is interested in capturing the behavior of the system, a simpler model can capture the spatial variation of the system behavior, but a finer model is needed to capture all the geometric details, so in which case there is no point in working with a very large model, a smaller model would suffice. The other situation where coupling techniques would be needed is when a built-up structure is made up of different components and different engineers are involved in development of different components, so every engineer therefore if possible would develop a computational model as well as an experimental model for each of the substructures, so we need to synthesize the models developed by individuals to predict the behavior of the global structure, the testing on a global structure may or may not be possible, but finite element model indeed can be constructed, so how to work in such situation, so there are basically the coupling techniques can be classified into two categories, spatial coupling method and modal coupling method, in modal coupling method there are two further classification fixed interface and free interface, the main distinguishing feature of these two methods is that in modal coupling method the coupling matrices are derived in terms of eigen solutions of the substructures, whereas in spatial coupling method the coupling matrices are derived in terms of the substructure structural matrices, mass stiffness damping matrices, so there is no eigen solutions that one need to compute if one is working with spatial coupling method, so we will see some of these details and a good reference for a discussion on this is volume edited by Maya and Silva the details are given here, so we will start with the discussion on spatial coupling method, we will restrict our discussion by and large to free vibration analysis, suppose this is a built-up structure and I assume that this built-up structure is made up of two substructures A and B, so this built-up structure is made up of A plus, so called I mean so this plus operation is combining these two systems, now if we consider substructure A the degrees of freedom in this model can further be classified as interior degrees of freedom and coupling degrees of freedom, coupling degrees of freedom occurs here where this node and this node are to be coupled to produce a built-up structure, so the degrees of freedom at this node constitute UC of A and other degrees of freedom in the interior are the interior degrees of freedom, similarly for system B I have UI of B and UC of B, clearly you can see that UC of A and UC of B need to be equal for compatibility relations to be obeyed, now let's consider the equation for subsystem A, so this subsystem A the degrees of freedom as I already said are partitioned into interior degrees of freedom and coupling degrees of freedom, so this partitioning of degrees of freedom induces a partition on structural matrices and this is depicted here K I I A, I stands for interior, C stands for coupling, it's not the I ith element of mass matrix that is not what is meant here, so the equation of motion is obtained in this form, we assume as I already said there are no external forces but nevertheless I should write FCA because these are the coupling forces that we need to consider later on. Now if NC is the size of UC that is coupling degrees of freedom then UIA will have NEA minus NC degrees of freedom, the size of this vector will be that, similar equation for subsystem B with NB degrees of freedom can also be written which is as shown here. Now the interior degrees of freedom include all the unknown degrees of freedom or could be the master DOFs obtained after a model reduction, so that also could be, that means while arriving at this one may use a condensation technique inbuilt into that if need be. Now how do we get the equation for the coupled system, so the coupled system equation is obtained by considering the compatibility of displacements at the interface which demanded UCA must be equal to UCB, and similarly equilibrium of forces at the interface requires that some this FCA plus FCB must be equal to 0, so if we use these conditions I get this matrix, so this is the built up structural matrix for the, I mean for the built up structure this is a mass matrix and the stiffness matrix. Now the built up structure can now be analyzed using these assembled matrices, the size of assembled structural matrices will be NEA plus NB minus NC, that is across this, I mean this is the size of a vector displacement vector, now this method is well suited if subsystems are studied using a finite element analysis if you have access to all these matrices you can easily construct these matrices for the built up structure. Now I have illustrated this with respect to two subsystems but it can be extended to a built up structure having more than two subsystems then the extension is fairly straight forward. Now I come to the other class of methods that is the so called modal coupling method, we'll start with fixed interface method that is also known as component mode synthesis, so one of the well known methods, so we'll again start with the equation for subsystem A and we'll partition the degrees of freedom as interior and coupling degrees of freedom and this equation is obtained. Now I will denote this NEA minus NC as NIA, NI superscript A, now we begin by assuming that at the interfaces the degrees of freedom are made 0, that means the structure is fixed at the coupling point, okay, then the governing equation therefore UC will be 0, so the governing equation will be UIA double dot plus this, KIA UIA equal to 0, so this is 0 and this is 0 and this force is 0. Now we can do a eigenvalue analysis associated with this equation and determine the system natural frequencies and mode shapes, okay. Now at the interface structure of course is not fixed, so we need to correct for that, so in any case using the modal matrix of the fixed interface system I obtain this representation in terms of this modal coordinates denoted as PK A of T, the subscript K here denotes that I am not retaining all the modes, I am retaining only K modes, so thereby I am already achieving a modal reduction, okay. Now we release the degrees of freedom at the interface, this analysis has been done with degrees of freedom fixed, now once I release that UC of A of T is not equal to 0, now we make an assumption now that UC of A of T is related to interior degrees of freedom through relations that are strictly valid under static condensation, that is we use static condensation to illuminate interior degrees of freedom in terms of coupling degrees of, so by that I get, this is the equation I get interior degrees of freedom in terms of coupling degrees of freedom using this relation. Now what I suggest, what we do is the solution is taken to be the sum of, this is a solution that we assume, the unknowns that we take are the modes corresponding to the fixed interface degrees of freedom, a fixed interface system and the coupling degrees of freedom obtained through condensation, and this is a transformation matrix, is that okay? Now this is how we represent, we assume displacement, now we return to the governing equation and make this substitution, so once I substitute this I will pre-multiply by Psi K A transpose and I get this equation, and this leads to the definition of reduced mass and reduced structural matrices, so if we now consider this matrix transpose M into this and analyze this a bit, we get terms involving Phi I K A transpose into M I I A into Phi I K A which I know is a diagonal matrix, because this is a modal matrix of fixed interface system whose mass matrix is M I I of A, so making some of these simplifications we get the mass matrix for the substructure to be this, similarly this is a mass matrix, if we can work through this I get, I mean if we multiply all this and work through this I get the elements of the mass matrix for subsystem A in this form, now similarly the stiffness matrix we can work through, and here if we see here the eigenvectors here this Phi is the eigenvector and it is orthogonal and it is mass normalized, therefore Phi transpose K Phi is a square of the natural frequency, so at the matrix level we get Phi I K A transpose K I I A into Phi I K A to be a this lambda matrix which is the diagonal matrix of natural frequencies of subsystem A in the fixed interface form, so these are the matrices that I get for substructure A, so this is a equation for substructure A at this stage in terms of let me recall once again the coupling degrees of freedom and PK A of T are the modes emerging from fixed interface model for the substructure, so these together constitute the state vector, similarly for system B, subsystem B I can derive a similar equation, okay, now I have to generate the equation for the couple system, these are equations for subsystem A and B derived independent of each other, now I will impose this compatibility relations and this equilibrium equation and I get the combined equation to be of this form where for the combined system the state vector consists of normal modes of subsystem A, normal modes of subsystem B in the fixed interface format and the coupling degrees of freedom, the coupling degrees of freedom are equal by this compatibility requirement, so this is the state vector and this is the mass matrix and this is the stiffness matrix, the stiffness matrix itself is in terms of the subsystem natural frequencies in the fixed interface form and the matrix at these matrices which we have derived, now this is the equation that we need to analyze to produce the Eigen solutions for the built up structure, now we can make some observations, this method requires a knowledge of structural matrices of substructures, it is not suited for experimental studies since in experimental studies such matrices would not be easily available, what would be available will be a set of natural frequencies and mode shapes if you do an experiment, now also creating fixed interfaces may not be feasible in an experimental substructure, okay, so you can't weld a member and so on and so forth, so this method is suited for studies where substructures are studied using finite element method, but as I said one of the motivations for studying substructuring methods at least in the current scenario where computational power is fairly generously available, the reduction of the model in terms of degrees of freedom may or may not be that crucial, but it becomes indeed crucial if some of the substructures are studied experimentally and they need to be combined with other substructures which may be model mathematic, okay, and also this kind of reduction in computational efforts become relevant if you are doing uncertainty analysis where this running of these programs forms a part of a Monte Carlo simulation run, where you need to repeatedly run these programs for you know nominally identical values of model parameters, so these methods remain relevant in such situations. Now we will move on to the next method known as free interface model coupling method, here the objective is to produce a solution strategy where we bank on the Eigen solutions of substructures to construct the coupling matrices and the model for the coupled system, so how does that work? So here again let us consider the equation for subsystem A, I write it in this form and as before we will partition the degrees of freedom into interior and coupling degrees of freedom and that induces partitioning of structural matrices and the forcing vector as shown here, now the sizes are as shown here, we can perform now the Eigen value analysis corresponding to this system, I am not putting UCA to be 0 now, I am including it in the model, now I will perform the Eigen value analysis and determine NA cross NM model matrix and diagonal matrix of Eigen values, okay, and I will make this model matrix to be mass normalized so that phi transpose M phi is I and phi transpose K phi becomes lambda for the subsystem A, now if I were to consider KA term expansion for the response, this is a full NA by NA matrix but I may not include all the modes in a given calculation, so out of capital N modes that are available I may retain K modes, so I will say that I am going to write UA of T as phi KA PK of T, where PK of T are the generalized coordinates, the subscript K here denotes the fact that I am retaining only K terms in the expansion and KA means I am talking about subsystem A, now I will now partition just the partition, U has been partitioned as interior and coupling degrees of freedom and just as it has induced a partitioning of stiffness and mass matrices it also induces a partitioning of modal matrix that is written here. Now we substitute this in this equation and use the orthogonality relation I get the equation for subsystem A in this form, similarly for subsystem B using the same logic I will get this equation, so just to recall I mean just to emphasize this capital lambda is the natural frequencies of subsystem B and the interface is kept free and this is the corresponding modal matrix, now the combined equation for the two systems is written in this form, now we need to now impose the conditions that UCA of T is UCB of T, so how do I get that, that is not implicit here, so this PK A and PK B contain information on coupling degrees of freedom in some sense this equation need to be now constrained by this additional requirement, how do we do that? So UCA of T is given by this in terms of degrees of freedom, generalized degrees of freedom PK A of T and the modal matrix, so these should be equal, so this produces a constrained equation as shown here SP equal to 0 where S is the modal matrix corresponding to the coupling degrees of freedom for subsystem A and similar entity for subsystem B, so now what I will do is this combined vector P I will now partition into PD and PI, this partitioning is done from a mathematical perspective and I will select D degrees of freedom so that S, this matrix S gets partitioned into SD and SI and SD remains as a matrix that can be inverted, this is a non-singular matrix that can be inverted, consequently I can retain, I mean write PD as in terms of PI as minus SD inverse SI PI, so this is a crucial step here, now once I do this I can write the vector P which is PD PI in this form retaining only PI and this matrix I denote as SI that is SI is minus SD inverse SI by I, now noting that this equation is satisfied I get now the coupled equation in this form, this is purely in terms of PI which degrees of freedom which I am retaining, so I will pre-multiply by SI transpose and carry out this operation and I get the equation for the built up system M is SI transpose SI, K is SI transpose this into SI, Q is PI, FQ is this, now since no forces are taken to act at the interface I take the equation to be MQ double dot plus KQ equal to 0, so to summarize I have got the mass matrix as SI transpose SI, K as SI transpose this and SI itself as this, and this S etc. is as shown here, now what needs to be noted here is that these matrices M and K are now derived purely in terms of the Eigen solutions of substructure A and B in the free interface form, you don't need the structural matrices for A and B while implementing this method. The Eigen solutions for one or more of the substructure can be experimentally established, suppose A can be a finite element model, B can be an experimental model or both A and B can be experimental models, of course B and B can also be finite element models, so the knowledge of substructural matrices for the substructures is not directly needed, if you have it so well and good but it is not needed, so this is the method known as component mode synthesis. Now suppose we are considering a situation where two substructures are coupled elastically, that means there is one more coupling element, the equation for coupled system needs to be modified now to take into account this feature, so if you do free interface method the coupling element will not be easily represented, now the compatibility condition in this case is no longer applicable, by this what I mean, suppose I have a system this is one substructure and this is another substructure and this is a coupling element, this is A and this is B, so I am talking about how to deal with this elastic coupling, this UCA of T will not be equal to UB of T because there is an elastic coupling between the two, okay, so it is not rigidly connected, so this displacement can be different from this, now on the other hand this equation remains valid, so what I do now is I will define a coupling matrix K CPL, let that represent the stiffness matrix such that at the coupling the displacements and their forces are related through this coupling matrix, with K CPL being this equal to this. Now the equation for coupled system needs to be modified to take into account this coupling, the compatibility condition is no longer applicable, now you consider this and we get this. Now UA of T as before I will express in terms of this and UB of T in terms of this, so this is a representation I have, now this forcing is related through to these degrees of freedom through this relation, now consequently if you substitute these relations into the governing equation I get equation of this form and the final equation of motion this is for system A and B together and the final equation for the built-up system is obtained like this where the stiffness matrix for the coupled system is modified through these additional matrices, so this helps in dealing with situations where there is elastic coupling between subsystems. Now there are few other strategies where, I mean strategies become possible when in certain class of problems for example if the structure has certain planes of symmetry, for example if you have a one span beam like this can I construct the solution of this structure by analyzing a smaller problem, okay, so instead of solving the larger problem A we will solve two problems B and C and synthesize a solution of this problem, so what I do is I cut the structure here and I introduce this boundary condition this is a roller and a hinge, so now if you look at mode shapes of this simply supported beam all the modes 1, 3, 5, etc are symmetric about X equal to L by 2, so these modes can be simulated by considering this situation, the displacement here you can see this is 0 is not 0, okay, you follow what I mean, right, this is first mode, this is third mode, this is for fifth mode and so on and so forth, so this can be simulated, symmetric modes can be simulated like this, the even modes that is second, four, six, etc. are anti-symmetric, this is L by 2, you can see here this mode shape is this and this is anti-symmetric, these anti-symmetric modes can be synthesized by considering this problem, so what I can do is I can analyze two small problems, okay, and find out mode shapes for this and this, and by suitably using planes of symmetry I can construct the mode shape for this built up system all the symmetric modes can be obtained from this, anti-symmetric modes can be obtained from this and complete system modes can be thus constructed without solving a larger problem. Now this is again illustrated here, suppose you have a multistory building frame, this frame may require say 500 degrees of freedom to analyze the problem, so what I will do is I will utilize the plane of symmetry which is this red line and analyze two problems, okay, so this will have 250 degrees of freedom, this is 250 degrees of freedom, so two such smaller problems will be solved and I will synthesize the solution for the built up system, okay, that means instead of solving a larger problem A we solve two smaller problems B and C and synthesize a solution for the larger problem, so this is another way of utilizing certain problem features to achieve model reduction, these are of course this is clearly possible only when such symmetries exist in the given structure. Now suppose the symmetric structure is loaded asymmetrically, what happens, okay, so that is not a problem, for example if you consider a simply supported beam loaded asymmetrically by a force P of T, I will now consider two problems this plus, so if you add these two you get solution for this problem, now each of these problems can be handled by solving these two problem, okay, so this is you know advantage of using symmetry, so it can in principle can be used even when if structure is of course symmetric about X equal to L by 2 in terms of boundary condition geometry etc., but it is asymmetrically loaded, okay, so still we can utilize symmetry by following these arguments. We will now return to one of the themes that we have been pursuing that is development of structural matrices, so we considered a few lectures before what we achieved was we developed the mass and stiffness matrices for this 3D generalized, so 3 dimensional beam with two nodes and 6 degrees of freedom per node, okay, so how did we do that, we considered this structure to you know the displacement fields that we assumed was axial deformation, deformation in plane, deformation out of plane, and the rotation about the longitudinal axis, so we accounted for all the energies, so this is energy due to axial deformation due to twisting, this is bending about Z, bending about Y, and similarly we computed the kinetic energy due to axial deformation, twisting, bending about Z and bending about Y, and using this we constructed the Lagrangian and we interpolated these field variables in terms of their nodal values, for U and theta we use linear interpolation functions, and for V and W we use cubic interpolation functions, and we derived the 12 by 12 structural matrices. Now we wish to now carry this exercise further, beyond 3D element now we can think of continuous problems, so in continuous problems we can think of plane to 2 dimensional problems, first of all, so in 2 dimensional problems we have plane stress, plane strain and axisymmetric problems, so then we can consider 3 dimensional solids and then plate bending elements and some discussion on shell elements. So the general template for achieving this is basically to write the appropriate equations for strain energy and kinetic energy, identify the field variables, interpolate and you know take an element, identify the nodes and degrees of freedom, and develop the, represent the field variables using polynomials in terms of the nodal degrees of freedom, substitute into the Lagrange equation and derive the equation for the nodal degrees of freedom, and then question of coordinate transformation, assembling, imposition of boundary conditions, calculating external nodal forces, and thus deriving the final equilibrium equation will have to be done. Many of these steps we have already covered, the step beyond formulation of element level structural matrices has already been covered, okay, the problem of coordinate transformation, assembling of matrices, imposition of boundary conditions, computation of nodal forces and assembling all that has been earlier done, so to be able to extend that framework to do more complicated problems, we need to now expand our repertory of elements, so with that in mind in the next few lectures we will start discussing about some of these elements, and we will begin by talking about 2 dimensional element namely plane stress and plane strain elements. Now in all this I am assuming the structure behaves linearly, and material is isotropic and elastic, we are also going to assume homogeneity within an element, so the governing equations we will quickly recall from 3 dimensional linear elasticity, the state of stress at any point X1, X2, X3 is given by this tensor, and the state of strain is given by this tensor, and the displacement fields U1, U2, U3 all these quantities are functions of X1, X2, X3 and time, so number of independent variables is 4, that is X1, X2, X3 and T, and number of dependent variables is 15, there are 6 stress components, 6 strain components and 3 displacements, so how do we tackle these problems? So upon coordinate transformation the stresses obey this rule of transformation, therefore second order tensors, the displacement is a vector, eta follows this rule of transformation, where T is the coordinate transformation matrix. Now we have 3 equilibrium equations which are given here, and this symmetry rho ij equal to rho ji is assumed, that means we assume that there are no body moments acting on the system, so these are the 3 equilibrium equations. Similarly we have the 6 strain displacement relations as shown here, and as I said we are using basically linear models, so the strain displacement relation would be linear, and we assume material is hookiem and also isotropic, therefore the relationship between stress and strain, there will be 6 constitutive laws relating stress and strains and they are given here, there are 2 elastic constants E and nu which is Young's modulus and Poisson's ratio or Lambert's constant and shear modulus, either we can express stress in terms of strains or strain in terms of stresses. We can also use index notations, this I am not going to do in this discussion, so for completeness I have mentioned here all these equations can be written in short form in this way. There is an alternative notation where the stress instead of being written as a matrix we write it as a vector, 6 cross 1 vector, this is the representation for stress, this is the representation for strain, and this is the matrix D that I will be needing in the formulation, and this 6 cross 1 stress and strain matrices are related, for example stress is related to strain through this 6 by 6 matrix, or strain is related to stress through this C tilde which is this, okay, so this is as I said material that obeys Hooke's law and which is isotropic. Now in terms of this matrix D the equilibrium equation can be written in this form, and this is the strain displacement relations this is the constitutive law, either these two represent constitutive law, and by eliminating using these relations I can write for sigma C into epsilon I get this equation, and for epsilon I can write in terms of DU so I will get this equation, and this is a governing equation for displacement in terms of operator D and matrix C, so this is the equation that we will be solving. We also need to discuss about the energies in a 3 dimensional situation what is the expression for energy, now this can be explained by considering an infinitesimal element as shown here, and suppose if it is loaded you know if you consider this and only the axial stresses the force on this phase is sigma XX into DY into DZ, and this force will deform the object and therefore this force would have done some work on that deformation, and the question is what happens to that work done, it is stored as strain energy in the system, so that is this is given by this integral sigma XX into epsilon XX DX, this is a force and the elongation will be epsilon XX into DX, therefore this into this integrated over the entire volume will give us a total strain energy, so this is total strain energy, kinetic energy is given in terms of displacement U, velocity U dot V dot and W dot as shown here. Now we would like to simplify the problem of 3 dimensional elasticity, so as I said here we have 15 dependent variables and we have 15 equations that is 3 equilibrium equations, 6 strain displacement relations and 6 constitutive laws, so there are 15 equations and 15 unknowns. Now not all these 15 equations need to be solved for all situations, in some situations we can simplify, there are two aspects to this simplification, there are four independent variables XYZ and T, and there are 15 dependent variables, we can cut down on number of dependent variables and also we can cut down on the dependence of these variables on XYZ and T by you know suitable simplification, if there is no dynamics T becomes irrelevant so it will be independent variables will be 3 instead of 4, but can we use certain features of the problem and reduce these quantities further, so one such approach is the so called plane stress model. So to consider a plane stress model in XY plane, we consider a continuum as shown here and the basic problem is this red line shows the surface traction there is a body force acting on this, and we want to analyze this continuum would be supported in some manner and all that will be specified, given the geometry of this continuum the way it is supported and the surface tractions and the body forces and the constitutive law of the material which makes this continuum, what are the stresses, strains and displacements, that is the problem we need to solve. Now the specification of geometry for plane stress model we restrict our attention to prismatic body, okay, so this is what is meant prismatic body along the Z axis the cross sectional area cross sectional property do not change, and lateral dimensions are much greater than thickness, so this thickness is smaller compared to the lateral dimension, this is the restriction on geometry, how about loads, the loads are only in the XY plane and there is no body force in the Z direction, the loads can have, can be surface tractions and body forces, but they are restricted only to lie in the XY plane, so the surface Z equal to this thickness I have taken it as 2H, so Z equal to H and Z equal to minus H is free from surface tractions, so this top surface is free from surface tractions. So for these class of problems we can make a simplification to the three dimensional velocity problem and that simplification is known as plane stress approximation, how does it work? Now let's consider the surface Z equal to plus H and minus H, and as we said there are no surface tractions on that, therefore the stresses acting on those planes namely sigma ZZ, sigma ZX and sigma ZY must be 0, at Z equal to plus minus H, for and for all X and Y, and on omega that means on this surface sigma XX, sigma YY and sigma XY are independent of Z, because loading doesn't change with respect to Z, the loading is uniform across thickness, so that is independent of Z. In developing plane stress model what we do is we assume that these conditions prevail not just at the surface but in the interior also, that means we interpolate these features into the interior, by that what I mean strictly speaking sigma ZZ is 0 only on Z equal to plus minus H, but I assume that sigma ZZ is 0 through out, and sigma ZX similarly is 0 only on these two outer surfaces, but I assume that it is 0 for all XYZ, similarly sigma XX and sigma YY and sigma XY these forces do not change with respect to Z on omega, but what I assume is they don't change anywhere in the interior as well, so this is the approximation. Now therefore at the end of this I have now certain quantities have become 0 and certain quantities have become independent of Z, now let's see whether equilibrium equation is satisfied, so you go to the equilibrium equation, write the three equilibrium equations you can see here the quantities that are written in the red are zeros in our as per our model this is 0, so this drops off, this drops off, sigma XZ is 0, this drops off, YZ is 0, this drops off, ZZ is 0, this drops off, and Z is 0, there is no body force that drops off, so that is 0, so the remaining equations are the equilibrium equations which I need to consider further. In the constitutive law again I have assumed stresses now I have to figure out what the these strains are, so this is a relation between stress and strain, but these quantities shown in red are 0, so once I expand I get epsilon XX and epsilon ZZ to be this and epsilon XY is given this, but epsilon YZ and XZ become 0, so two of the strain components become 0 because certain stresses are 0, so at the end of it the stress matrix with all nonzero elements and dependencies is shown here, the original stress matrix is function of XYZ it has six independent components, now this is 0 and these two are 0, so these are 0 as per our model and these quantities are independent of Z, the nonzero strain components are shown here, so epsilon XZ is 0, YZ is 0, but epsilon ZZ will be 0, so now how many I have come from our stress to strain, now how do I get displacement? I have to use strain displacement relation, so the strain displacement relation epsilon XS is this, epsilon YY is this, epsilon ZZ is this, and epsilon XY is related to this, now epsilon YZ I have got it to be 0, and epsilon XZ I have got it to be 0, therefore these quantities need to be equal to 0, but these we will not be able to honor and we simply ignore these relations in further development, that is where the approximations coming. So now there are 10 unknowns that we are considering, sigma XX, sigma YY, sigma XY, and epsilon XX, epsilon YY, epsilon ZZ, and UVW are the unknown displacement, so what are the equations? I have 2 equilibrium equations, 4 stress strain relations, and 4 strain displacement relations, there are 10 equations, so this is the simplification that we have achieved. What has happened? We have achieved a reduction in number of unknowns from 15 to 10, reduction in number of independent spatial coordinates, so this XYZ becomes only XY, and this because of the approximations in treatment of strain displacement relations is not an exact model. Suppose if you substitute this into 3 dimensional equations of elasticity, by that I mean suppose if you analyze the plane stress problem, and the solution that you obtain you substitute into the 3 dimensional equations of elasticity, you will be able to satisfy equilibrium equations, constitutive laws, but you will have difficulty in satisfying certain compatibility relation, not all 6 compatibility equations will be satisfied, so that is one of the limitations of plane stress model. Now in the next lecture what we will do is we will consider another form of simplification that is known as plane strain model, here the objective is again prismatic, one of the example that we can give is that of a gravity dam subjected to hydrostatic force, so if this dam is in a valley like this we can assume that in this portion the dam cross section is prismatic, and the load doesn't change, suppose this is a Z direction, the load doesn't change in the Z direction, now if at Z equal to 0 and Z equal to L, if we assume that the dam is the displacement W is 0, that means it is held fixed, then in plane strain model we deal with this type of situations where we again consider prismatic objects where the thickness this is much larger than the lateral dimension, it is the opposite of plane stress model where in the plane stress model thickness was small in relation to the lateral dimension whereas here it will be the opposite. Now the way we proceed is as follows, now W is 0 at Z equal to 0 and W is 0 at Z equal to L, now if you consider a mid plane the object is prismatic therefore it is symmetric about this mid plane, and the way the boundary conditions are applied on displacement that is also symmetric, therefore if W is 0 here and W is 0 here, W must be 0 here, because load hasn't changed, geometry has not changed, boundary conditions have not changed, so this is 0, now next what I do is I consider a subsection which is half of this dam, so and consider the quarter point. Now again the same logic at Z equal to 0 and Z equal to L by 2, W is 0, now the object is again symmetric about X equal to L by 4, Z equal to L by 4, boundary conditions are the same, loading is the same etc., etc., I get W equal to 0, so what happens is in this model by repeatedly using this argument we postulate that W of X, Y, Z is 0, and next we assume that U of XYZ is function of U of XY, and similarly V of XYZ is V of XY, so this is a displacement field we will assume, W is 0, U is function of X and Y, V is a function of X and Y, we develop this model by considering from this we go to now, from displacement we go to strain by using strain displacement relations, and from this I use constitutive relations and come to stresses, so this model is known as plane strain model, so in the next lecture we will consider the mathematical theory behind plane strain model, and after having covered the two background to the theory of these two models we will start developing finite element models for plane stress and plane strain continue, so we will take up that problem in the next class.