 Okay, we have been discussing analysis of plate bending elements, in the last class we started discussing about the basic formulation, so we considered a thin structural member as shown here where this thickness is small compared to the least lateral dimension and that is the essential feature of a plate and it can carry load either in its own plane or transfers to its plane. So structures carrying, plates carrying forces in its own plane are said to behave like a membrane, whereas if the structure carries loads transfers to its middle surface as shown here the plate would bend and this is known as bending action. So right now we are focusing on analysis of the bending action, if the membrane action has to be studied the plane stress elements that we have developed earlier can be employed. Now in the last class we formulated a few details, so we considered a four-noded rectangular element and at each node the degrees of freedom were translation W or slope along X and Y axis and we introduced non-dimensional coordinate system XI equal to X by A and eta equal to Y by B. Now to derive the strain energy stored in the plate we invoke the Kirchhoff-Lauw assumptions and based on that we derived these stresses in terms of the strains, so there are three stress components that enter our formulation and the strains themselves are given in terms of curvatures as shown here. So the strain energy stored in the plate is obtained in terms of the curvature and the matrix that relates stress and strain and similarly the kinetic energy is given by this expression and in the kinetic energy only the displacement component W enters our formulation in this model. So the field variable was W in our variational formulation and order of the highest derivative present in the Lagrangian was 2 and the degrees of freedom therefore at any node must be W dou W by dou X and dou W by dou Y and the interpolation function must consist of complete polynomials of at least order 2, so for any choice of elements that one can think of the first few terms in the trial function should be this, a constant term, linear term in X and Y and quadratic terms and the higher order terms, so this completes 1, 2, 3, 4, 5, 6, the remaining 6 terms we have to select depending on the theory that we employ for developing the element. So we started discussing this thin rectangular element with 4 nodes, 3 degrees of freedom per node so that the element has 12 degrees of freedom, the field variable was W and again the number of generalized coordination needed based on these considerations is 12, so we use this polynomial, so after quadratic terms we use a complete cubic terms and 2 fourth order terms X, Q, Y and X, Y, Q, we analyze this problem and in the non-dimensional coordinate system we expanded to determine the shape functions we represented the trial function in this form and since the variational formulation has gradients with respect to X and Y we evaluated this and at every node there are 4 nodes, at every node these 3 quantities are specified and using that fact we were able to determine the shape functions and we were able to represent the vector of nodal displacements W bar E in terms of these shape functions that is that field variable is represented in terms of N1, N2, N3, N4 and this vector of nodal coordinates. This N1, N2, N3, N4, NJ term has this form, now we carried out some analysis of this representation and we considered the behavior along edge 2, 3 the field variable theta X and theta Y and we were able to observe that along the edge 2, 3 W and theta X depend on nodal values of nodal values W2, W3, theta X2 and theta X3, that means along this edge 2, 3 the variable W and theta X depends on nodal values of the field variable and its slope with respect to X at nodes 2 and 3. Now if we are to add one more element on this side here again a point that lies on the edge 2, 3 will be shared by these 2 elements and these nodes also would be shared by the 2 elements, so if the field variable is independent of these nodal values at nodes 1 and 4 then automatically continuity will be satisfied, continuity of the field variable and the appropriate derivative will be satisfied. But we observed that while this condition was satisfied for W and theta X we observed that theta Y which is another field variable along psi equal to 1 depends upon values of W and theta X at nodes 1, 2, 3 and 4 as well as theta Y at nodes 2 and 3, so this means that theta Y will not be continuous across element edges and this element is not a conforming element and questions on convergence of solutions obtained using such elements need to be carefully considered, but still this element is known to give acceptable results, so in view of that we will complete the formulation, so the kinetic energy is given by W dot E transpose ME W dot E and ME is elements of ME are elements of this integral, so in the non-dimensional coordinate system we obtain this and since N consists of polynomials, product of N transpose N will lead to polynomials, so this elements of mass matrix can be evaluated exactly, so that has indeed been done and we get the mass matrix partitioned as shown here and is M11, M21, M21, M22 are themselves matrices, although the subscript may mean that it is 11 element of M, it is not so, M11 is a matrix, it is 6 cross 6 symmetric matrix with elements as shown here, and similarly M21 and M22 are also available, so if we put now M11, M21 transpose M21 and M22 we construct a complete mass matrix, so this mass matrix is consistent mass matrix, it is symmetric, similarly the strain energy can also be evaluated, this is the formula for the strain energy and this is the element stiffness matrix, so here the strain displacement matrix B and D matrix which relates stress and strain appear here, so we need to evaluate the elements of this and that can be done by considering B is actually given by the action of these operators on N and we can show that this is actually this in terms of N of psi, eta, and consequently KE can be evaluated in this form where this K11, 1, 2, 1, 3, 1, 4, etc., all these are again matrices, so they are 3 by 3 matrices, so the entire matrix is 12 by 12, so these matrices have been evaluated, here we see two parameters alpha and beta, these alpha and beta are ratios A by B and B by A, so we get this stiffness matrix as shown here, this alpha is A by B and beta is B by A in these matrices, so first we have evaluated the elements of this first column 11, 12, 13, 14, so 11, 12, 13, 14, and for evaluating the remaining elements we manipulate these 11, 2, 1, 3, 1, 4, 1 matrices through these matrices I1, I2, I3, and these details can be worked out, I leave it as an exercise for you to carry out these evaluations. Once stiffness matrix is determined, it is determined exactly in this case, the stresses in terms of nodal displacements can be used evaluated by multiplying the strain matrix with the D matrix, strain vector with a D matrix and we get this. Now we need to evaluate the equivalent nodal forces, so again we use this virtual work statement and based on that we get the equivalent nodal forces to be given by integral and transpose FZDA, where N is again we have already seen this matrix, and if for example if this forcing is a constant, a constant load then the equivalent nodal forces using this formulation is obtained as shown. Next if you want to compute bending moments, this is how bending moments are related to the stresses, and we can evaluate this, we have this expression for sigma XX, sigma YY and sigma XY, and if we carry out this integration we get this stress resultants that are of interest in terms of nodal coordinates, and strain matrix B and constitutive, matrix of constitutive law D, I is a matrix that has been defined here in one of these places. Now the element that we have developed as a pointed out is not conforming, that is the normal slope is not continuous across the edge, now how to overcome this limitation, so how can we do it, there are various approaches in the literature, so we can introduce additional the nodal degree of freedom, or introduce additional nodes, or ensure that normal slope varies linearly along the edges, if normal slope varies linearly along the edges then automatically it depends on two nodes along the edge and therefore the continuity will be guaranteed, so we will see how that happens. So first we will consider a conforming rectangular element, the idea is basically to use products of beam shaped functions in the two directions, so in the N matrix that we referred to earlier the NJ psi, eta is taken to be in this form, there are two functions of scalar variable psi and eta, and these are the functions that has been used in, the cubic polynomials used in formulating the beam element, now it is in the non-dimensional coordinate system, so what we do is we consider this same degrees of freedom as we used earlier, that means we will assume 3 degrees of freedom per node, translation along Z, rotation slopes along X and Y at the 4 nodes, suppose if you proceed with this and use these shape functions, so these shape functions I have shown what are these F and G functions, you can see here these are the four functions, this is the cubic polynomials that were used in formulating beam element. Now if we now substitute this the field variable is now represented in terms of nodal values as shown here, okay, now let's put this in the details of FJ can be inserted here and we can evaluate now the second order derivative dou square W dou XI dou eta, this provides the twisting, twist that we observe in the plate, and if we evaluate this which is straight forward since these are polynomials we can quickly do that, and if we evaluate the value of this gradient at the nodes it turns out to be 0, so the nodal value of dou square W by dou C eta dou eta is 0 for all the nodes, so the problem would be as the element becomes smaller the plate will tend towards a zero twist condition and we will not have desired behavior incorporated into the element development, so this is a limitation, so how do we overcome that? What we do is we will treat this quantity dou square W by dou X dou Y as an additional nodal degree of freedom, so in addition to W theta X theta Y I will use this call it as WXY as another nodal degree of freedom, so then this element will now have 16 degrees of freedom 4 nodes, 4 degree of freedom per element, and then what we will do is if you carefully see here it has to be seen bit carefully there is F term and G term, if you look at values of F at the nodes, the gradient of F at the nodes it is 0, but this is not true for G function, G is not 0 at nodes, so we introduce for WXY an interpolation in terms of GJ XI and GJ ETA function, so W now consists of as before the same format, but this now has 16 elements, and this is 16 cross 1, and the 16 elements are made up of this is transpose, so 4 of this evaluated at J equal to 1, 2, 3, 4, now if you write in a summation form this is what it's meant by the representation as shown here. Now if you compute the second order gradient dou square W by dou CI dou ETA, dou XI dou ETA we get this function, and if we now evaluate this at the nodes, it turns out that this will be given by this function, and this function is not 0 at the nodal coordinates, so therefore this is not 0, so the limitation that we observed here that this is 0 at all the nodes is overcome here. Now we can ask the question can the element perform rigid body motion without deformation, these are some of the requirements that we enunciated in one of the previous lectures when we talked about convergence and those issues, so now if we consider W1 and W3 to be 1, and W2 and W4 to be minus 1, and theta X2 is theta X3, theta X1 is theta X4 as shown here which is minus of that, and this similarly theta Y has this relation we are giving a rigid body rotation to the object, these deformations ensures that we are giving a rigid body rotation. Now if you put these values into the assumed representation here it turns out that WXYJ will be given by this and this WXY, eta will be psi eta, which is what we expect for such type of motion, so this representation is fine. Now we can go ahead and evaluate the mass matrix, stiffness matrix, equivalent nodal forces, you know these are polynomials so it can be evaluated exactly or you can use an appropriate quadrature rule, so I am not going to provide the details of this, this element is known as CR element, this advantage is that in the analysis of built up structure using this element is difficult because of that additional node dou square W dou X dou Y presented at nodes, most of the other elements would not have that element and there will be problem in dealing with that. So one approximation that has been done in the literature is to represent this dou square W dou X dou Y in terms of theta X and theta Y by kind of a numerical differencing scheme and that overcomes this difficulty, that means we can eliminate the WXY degree of freedom but such an element again turns out to be non-conform, so there will be another penalty that we have to pay. Now we will now consider another element, it's a thick rectangular plate bending element, we have been considering behavior of thin elements, now it turns out that for this element we can develop a conforming rectangular element as we will see shortly, so here what is the consequence of plate being thick, so we have assumed in the earlier theory that plane sections initially normal to the middle plane remain plane and also normal to the middle plane, but now what we will do is we will assume that the plane sections initially normal to the middle plane remain plane, consequently epsilon XZ will be independent of Z and epsilon YZ will be independent of Z, but not necessarily normal to the middle plane, if you insist that this should be also satisfied and these two quantities identically become equal to 0, so now we are permitting a certain rotation, a constant rotation across the thickness, it will have still the same length, so therefore epsilon ZZ continues to be 0, so WA is again function of X and Y, now what we will do is we will consider theta X and theta Y to be the rotations about X and Y axis of the lines which are normal to the middle surface before deformation, and that based on that we can evaluate U and V as Z into theta Y minus Z into theta X, so now U and V also will now come into picture, I have U as Z theta Y and V as minus Z theta X, so strains epsilon XX dou U by dou X it is given by Z dou theta Y by dou X, epsilon YY dou V by dou Y which is minus Z dou theta X by dou Y, shear strain dou U by dou Y dou V by dou X this is given by this, so now if you consider the, we will partition the strains into two components epsilon XX, epsilon YY and epsilon XY is one part and the shearing strains as another part, now this as before is given by minus Z into chi, but now this chi is given in terms of theta Y theta X, this is not curvature immediately it cannot be interpreted as curvature, so this is dou theta X, dou theta Y dou Y, now the shearing strains are given by this, dou U by dou Z is theta Y and dou U by dou Z is minus theta X, we call this vector of shearing strain as gamma and now we have sigma which will include sigma XX, YY and XY and tau which include shearing stresses, sigma is related to epsilon XX, epsilon YY and shearing strain with D being this. Now shearing strains are related, shearing stresses are related to shearing strains with this being the D matrix, so we will now call it as DS and this as D, now this factor K accounts for the variation of shear stresses and strains through the thickness, this we have already seen when we talked about deep beams in one of the earlier lectures, so equipped with this now the expression for strain energy has contributions from sigma XX, sigma YY and sigma XY and contributions from shearing stresses, so I write that separately, this is epsilon transpose D epsilon, this is tau transpose gamma and carrying out the integration across the thickness I get the first term as this and the second term as this. Now similarly kinetic energy has now contribution from UA and W, U dot, V dot and W dot and this is expressed in terms of W theta X and theta Y dot as shown here, so chi is the vector of these quantities and gamma is the vector of these quantities. Now let's examine the Lagrangian which is T minus V, what are the field variables? We will have W and theta X and theta Y, highest derivative of the field variable is 1, see we are having dow W by dow X and dow theta Y by dow theta X and dow theta X by dow X, right, so the degrees of freedom are W theta X theta Y at the nodes and since the highest derivative of the field variable is 1, these are the degrees of freedom and we can represent now the field variables in terms of W has 4 quantities we will write like this, W1, W2, W3, W4, similarly theta X1, theta X2, theta X3, theta X4 so on and so forth, so we have now representation for W theta X, theta Y and we are now going to use the interpolation function that we have encountered while dealing with rectangular plane stress elements, there also we had 4 nodal degrees of freedom and the field variable was required to be interpolated in terms of 4 nodal values and we are encountering similar situation therefore we can use the same trial functions. So we have already seen that element maintains inter-element continuity of a field variable and the required derivative, so this W theta X and theta Y maintain inter-element continuity, therefore this is going to be a conforming element, this is named as STK element named after Hughes Taylor and Kanoko-Nukulacha, so we can now represent therefore the vector of field variable W theta X and theta Y in terms of the nodal degrees of freedom which is 12 cross 1 and this is 3 cross 12 matrix of shape functions, these are the nodal degrees of freedom W1, theta X1, theta Y1 at node 1 and so on and so forth. So expression for kinetic energy now can be evaluated as shown here, this is this mass matrix we can evaluate in terms of 2 different components and this can be evaluated exactly, so we can evaluate this, the 2 components of this matrix exactly and that's what we do and we get the mass matrix. Now the strain energy is half WE transpose KEW, now KE itself we decompose into a flexural component and a shear component, this is the component due to flexure and this is component due to shearing, so we partition now the B matrix as shown here where BFJ here is given by this matrix, and similarly the BS for the shear, B matrix for the shear is given by this, now there are 2 components for strain so this will be having 2 rows whereas this has 3 rows. Now we know the interpolation functions, therefore we can evaluate all these gradients that appear in the B matrix and we will be able to evaluate BFJ and BSJ matrices, so dou NJ by dou X correspond to dou NJ by dou XI and if we carry out the required differentiation we will be able to evaluate this without any problem, similarly for the shear, B matrix for the shear component is given by this and we get this matrix, so you can use 2 by 2 Gauss quadrature it leads to exact solution or you can evaluate this exactly as well. So if we do that again by introducing 2 parameters alpha is A by B and beta is B by A we can get the K matrix in terms of various components K11, K21, K34, K41, F means flexure, superscript F stands for flexure and by carrying out the required integration these elements of these matrices can be reduced and the details of K11F, K21F and K31F and K41F are given here, then the remaining matrices K22, K32, K42, etc. are obtained in terms of this first column by doing these operations and we will complete the K matrix. So similarly we can carry out the exercise for the shear part, we get again KS as this matrix and this again has these components as named here and this is symmetric matrix, so the components of these first matrices in the first column are given here and the remaining elements are computed using these operations where I1, I2, I3 have been defined earlier. Now how about equivalent nodal forces, so again we use this formulation FE is N transpose into the applied loads, we are applying loads only in the transverse direction, so this is the formula for that and if FZ is constant the equivalent nodal forces are obtained as shown here. Now again bending moment and shear forces can be computed as stress resultants we have evaluated all the required stresses, so by integrating across the thickness suitably we get the bending moment, twisting moment and the shear forces, so this completes the formulation of a thick rectangular element which is conforming and it allows for a shear deformation, okay, so this is known as midline plate theory. Now how about triangular elements, now we will consider this triangular element 1, 2, 3 define X axis along one of the edges and Y axis along orthogonal to this in the plane of the plate and Z axis is outside the plane of this plate, now this has 3 degrees of freedom per node therefore it has 9 degrees of freedom, so the field variable must now be represented in terms of these 9 degrees of freedom element. So now a complete cubic polynomial will have 10 terms, so if we now take 9 terms there will be a problem so I can take 1 X and Y that is 3, X square, XY, Y square that will be 6, now from this column we have to select 3 more terms, so the idea here is we will retain X cube, Y cube and use a common generalized coordinates associated with some of these two, okay, so we are retaining the complete cubic term but these are not associated with independent generalized coordinates, okay, so we can develop this element and we can show that this will not be a conforming element, again there will be problem in satisfying the continuity along of slopes normal to the surfaces on the edges, so this will again be a nonconforming element. Now how do we proceed here, there is one approach to develop a conforming triangular element, so what we do is suppose if we consider a triangular element 1, 2, 3, what we do is we identify an internal node, an internal point and form 3 triangles, 1, 2 and 3. Now we will consider now these 3 triangles separately that means one element will be like this, another element will be like this, and yet another element will be like this, so this will be 1, 2, O, 2, 3, 1, 3, O, now each one we will formulate separately, so there will be 9 degrees of freedom here, 9 here and 9 here, so this totally there will be 27 degrees of freedom. Now we have to now, whereas for the element 1, 2, 3 there will be 9 degrees of freedom, so we have to get now 18 relations which will eliminate the, this 27 degree of freedom model reduces to 9 degree of freedom. So what we do is, obviously there will be requirements on compatibility of deformation at these 3 point nodes, so that will give rise to some equations. The additional equations needed, what we will do is we will identify an intermediate point, see as shown here, so this and this, this and this, and this and this. So what we will do is we will compute the normal slope at these points for each of the elements and demand that this is equal to this, this is equal to this, and this is equal to this. If we do that we will get a conforming triangular element, now the way we select the trial functions in analyzing this we will ensure that the variation of the normal slope along the edges is linear, so we will be able to implement the required relations. So these 2 elements we will consider in the next class, and following that we will consider problems of stiffened plates, these are typically for example observed in say bridge decks where shell will be, plate will be stiffened by girders, similarly these are commonly encountered in aircraft structures, automotive structures and so on and so forth, so there is a combination of beam and plate element, so there will be a plate element and a beam element, so we will have to see how we can develop a model for a stiffened plate element. So what we will do in the following classes is we will first complete this formulation of these triangular elements, and then we will consider few numerical examples, and then come to the cases of stiffened plates. We will consider few numerical examples to illustrate the ideas that we have developed so far, there is a monograph by Arthur W. Lisa, it's a NASA special publication, it is on vibration of plates, and this monograph has several examples of free vibration, results of free vibration analysis for various configurations of plates, circular, rectangular, triangular, quadrilateral, polygonal, etc., etc., so it's a catalog of solutions, many of the solutions are exact, and some others are based on weighted residual approximations, so what I will do is I will pick few of these examples from this monograph and apply the finite element modeling tools on them, and see how we are able to produce the answers reported in this monograph. So let's quickly recall, this is the strain energy as per the classical plate theory, we were writing it in a different form, but when expanded it will have this form, this is the kinetic energy, so the governing equation is of the form D del 4W plus M double dot W dot is equal to 0, so to find out free vibration characteristics we assume that all points on the structure vibrate harmonically at the same frequency, and we substitute this equation into the governing equation, and we get this equation D del 4W minus omega square MW equal to 0, and if we introduce the parameter K to the power of 4 as omega square M by D, we get this equation, and this equation itself can be rewritten in this form. Now there are other few other details here, this is a governing equation and these are the stress resultants in terms of the displacement field, this is the bending MX, MY, MXY twisting moment, transfer shearing forces and edge reactions, these are results from classical plate theory, I will not be getting into the details of these equations but I am stating them for sake of completeness. Now let us return to the problem, that is free vibration problem, so we have assumed this solution, harmonic solution and we have got this equation, and this equation in fact constitutes an eigenvalue problem where now the operator here is a partial differential operator, and for a rectangular domain with all edges simply supported the boundary conditions will be W and MX will be 0 at X equal to 0 and X equal to A, and W and MY will be 0 at Y equal to 0 and Y equal to B. Now this dotted line is a convention used to represent simply supported edge conditions. Now as I said this is an eigenvalue problem, we need to find W equal to 0 is a trivial solution we satisfy this equation but we are interested in non-trivial values of W and we ask the question for which value of K such solutions exist, and K as you have seen is related to the frequency of harmonic excitation, so what we do is this problem is amenable for an exact solution and the mode shapes are given in terms of sinusoidal functions and this is the exact expression for natural frequencies. Now the natural frequency will carry 2 indices when we are considering 2 dimensional problems, this doesn't come up in finite element solutions but in analytical solutions this feature would be present. Now for sake of numerical illustration we will consider a 50 mm thick steel plate which is 4 meter by 2.5 meter, and the exact natural frequencies computed as per this formula is given here and these frequencies are in hertz, and these are the depiction of mode shapes, the dotted line indicates lines along which the mode shapes will be 0, so here there will be no 0 in the 1-1 mode, in 2-1 mode there will be 0 in this way, 1-2 mode like this so on and so forth, these are obtained from this exact expression for the eigenfunction. Now suppose if we use now a 4 by 4 mesh rectangular 4-noded plate element and analyze the problem we get the frequencies to be this, first frequency is obtained as 29.09 hertz whereas its exact value is 27.43 hertz. In this analysis we have used thick plates whereas these results are for thin plates that also need to be borne in mind. These are the contours of mode shapes, you can see here that there are no zeros here whereas this is the nodal line, this is a nodal line and there are these nodal lines and these match with the patterns that the exact solutions depict. Now if we refine the mesh now instead of 4 by 4 mesh if we take 16 by 16 mesh then the first frequency becomes 27.4392 and this is approaching the exact natural frequency, and the second, third are these are the other frequencies and the mode shapes these are the exact natural frequencies 27.4392 and what we have got through finite element analysis is shown in the caption here, so 27.5132 needs to be compared with 27.4392 and so on and so on, 51 with 50, 87 with 86 and so on and so on. So as we see as we refine the mesh we are approaching the exact solutions. Now for other, this is all around simply supported boundary condition, for example if you have again rectangular plates with two opposite edges which are simply supported then it is possible to develop solutions for this, so what we do is we consider the partial differential equation and we expand this solution, in X direction we use the exact eigenfunctions and in Y direction we use this and using a Galerkin type of projections we get equations for YN of Y, so we get the governing equation for YN is this, and this is, and again it's an eigenvalue problem, it's a fourth order ordinary differential operator and there will be four boundary conditions specified on Y at plus minus B Y2, so if conditions at Y equal to plus minus B Y2 are identical for example if these two edges both are free or both are fixed or both are simply supported etc., then the solution can be simplified by taking advantage of symmetry. Now I will not get into the analytical solution, so this is results for one such plate I have shown the cross section of the mode shape here and this is the nodal lines and this is the 0, this is again the nodal line here and there are two nodal lines as shown here. Now there is one example considered by Lisa, he has considered a rectangular plate with two edges opposing edges simply supported and the other edges, one edge is simply supported and the other edge is free, so for this case according to the data given in this book the first natural frequency is 14.2585 hertz. Now we have analyzed this problem, okay, we have analyzed this problem and the number we get for 14.2585 for the given you know meshing configuration is 13.47 hertz, so again we expect that I mean this result has to be examined by refining the mesh and this also is based on an approximate solution, so one can only derive an order of magnitude type of comparisons here. Now if the other edges, suppose none of the edges are simply supported, see the case that I mentioned just now is that two opposing edges are simply supported, but if that condition is not there, so then we can use Rayleigh-Ritz or Galerkin technique by using beam eigenfunctions in the two directions and we can derive analytical solution which are again will be approximate, so that can be done, okay. There are details of such results available in the monograph by Lisa. Now circular plates are also other class of problem which are extensively studied in the monograph by Lisa, and for that the strain energy expression in the cylindrical polar coordinates and the governing equations and stress resultants are reproduced in this view graph, this is for sake of completeness. Now again here if we assume for free vibration analysis all points on the structure vibrate harmonically, the eigenvalue problem will be again mathematically will be of this form, but this del square operator will be now quite different. Now if we expand the solution in terms of sine and cosine terms in theta we can get this set of equations for these amplitudes in R which are depicted here, and two similar equation for, this is equation for WN and star can also be obtained. Now solution to this pair of equations can be obtained in terms of Bessel's functions, Bessel's function of first and second kind and the modified Bessel functions of first and second kind, so we can construct for cylindrical polar coordinates solutions in this form, so this can be used to determine the characteristic equation by imposing appropriate boundary conditions. Now let's consider a circular plate with no solid circular plate with no internal holes and center at the center of the circle, and in this case terms involving IN and KN are set to 0 in order to avoid singular behavior at R equal to 0, so if boundary conditions are symmetric with respect to one or more diameters of the circle then sine and theta terms can be also be removed, and we get the representation for the eigenfunction in this form. Now if we consider plate clamped all around the boundary conditions will be this, and we can derive the characteristic equation, and this problem has been solved and exact solutions are available and the monograph by Lisa gives that. So what we have done is we have created two FE models with M.S. details as shown here, in model 1 there are 96 elements, in model 2 there are 384 elements, so this is a more refined model, and the natural frequencies, first few natural frequencies are listed here. Now actually this system because of its symmetry the eigenvalues will be repeating, for example I will show the mode shapes, this is the first mode shape that means the plate deflects all through in this manner, in an axisymmetric manner, in the second mode the mode shape will be like this, and the nodal line is this, and a nodal line which is orthogonal to this will be another mode shape at the same value of the frequency, so that is what this means, the natural frequencies repeat. Similarly the next mode also appears like this, the nodal lines are like this, and these two frequencies also repeat. So in a refined model we get similar features, and we see that the answers are slowly moving towards the exact solutions. Now another problem that is available in Lisa's book is that of a circular annulus fixed at interior and outer edges, and inner radius of 0.8 meters, and outer radius of 2 meter, this is a numerical example that we have considered, the results are available for this case, and we have done an FE analysis of this, and according to Lisa's monograph the three frequencies, first three frequencies are shown here, and according to the model that has been developed we get 200, 204, and 214 respectively as approximation to these three numbers. So these are the mode shapes for the three modes, this is the first mode, second mode, and the fourth mode. Now I leave it as an exercise, this is a parallelogram plate which is all round simply supported, and dimensions, this angle is 30 degree, and dimensions are as shown here, and exact natural frequencies for these are available, and they are given by this expression. The suggested exercise is to make a finite element model for this plate and compute the natural frequencies, and compare it with this exact solutions are given in this monograph. Now in the next lecture we will consider triangular geometries, and develop first a non-conforming element, we will consider three nodes, at each node there will be three degrees of freedom, and this will be a nine-noded element, and we need to include nine terms in the representation for the displacement. So we will develop this element by assuming a shape function of this form, actually if we include a complete cube it will have 10 terms but we need only nine terms, so what we do is we club this X square Y and XY square terms and associate it with only one single generalized coordinate, and we will develop this element in the next class. We will also develop another approach for a thin conforming triangular element where what we will do is for this triangle element 1, 2, 3 we will introduce an internal 0.0 and divide this triangle into three sub-triangles 1, 2, and 3, and we will analyze each one separately, and consequently there will be 27 generalized coordinates, but for this triangular element there will be 9 coordinates, so 18 of them we need to eliminate, we will do so by seeking compatibility at 1, 2, and 3, and also by seeking equivalence of normal slopes at 0.4 for triangle 2 and 3, at 0.5 for triangle 1 and 3, and at 0.6 for 1 and 2, so we will develop these elements in the next lecture and see how this formulation develops, so at this stage we will close this lecture.