 Previous lecture we have considered two-dimensional continuum, problems of two-dimensional continuum we have considered plane stress and plane strain problems, now we will continue with that discussion and extend our discussion to three-dimensional solid elements, so to quickly recall one of the element that we developed was a linear quadrilateral element having geometry as shown here, and we mapped this to a unit square, and this mapping was essentially done to evaluate these integrals, and in doing so we represented the coordinates here X and Y in terms of this new coordinates XI and eta using the same trial function that were used to represent the field variables, so this formulation is known as isoparametric formulation, so this is how we did and this was a representation, so this we showed that the edges of this master element correspond to edges of these quadrilateral elements, and we discussed how to carry out the evaluation of stiffness and mass matrices with these representations. Now in doing so what we observed was for the kind of geometries that we considered the evaluation of this mass and stiffness elements of mass and stiffness matrices required evaluation of quadratures as shown here, and we proposed that these integrals be evaluated using Gauss quadrature, and one of the issue in implementing this is to decide upon the order of Gauss quadrature, so the rule that was mentioned was that a polynomial of order P is integrated exactly by employing N smallest integer greater than 1⁄2 into P plus 1, so I outlined the rational behind making this decision, we showed that the points where we need to evaluate these integrants coincide are actually zeros of Legendre polynomials, and that formulation was discussed in the previous lecture. So the choice of order of integration needs to be made carefully, a wrong choice can immediately lead to poor estimates of elements of structural matrices. Now two references that I have been using in these discussions is books by Petit and S.S. Rao, so the details are mentioned here. Now before we take up the problem of 3 dimensional elasticity continuum elements we can raise few miscellaneous points and have some discussions. Now the first point I would like to discuss is convergence and choice of order of interpolation polynomial. So the question that we can ask is what happens if you reduce the element size successively in a given finite element model for a structure, does the finite element solution converge, so under what condition it happens, so there are certain requirements that needs to be satisfied by the interpolation functions to guarantee that this happens. The first is the displacement field must be continuous within the element domains, this is actually automatically satisfied since we are using polynomials as interpolation functions in terms of nodal values of the field variables, so within an element continuity is guaranteed, so this condition A is not a problem. Now condition B we will consider the Lagrangian T-V, all these conditions I am discussing with reference to structural mechanics problems. Now the Lagrangian would be a function of the field variables and its spatial derivatives, and also time derivatives, but we are not discretizing in time at this stage, so we need not discuss that aspect. Now if N is the highest order of partial derivative of field variable that appears in L, let us define N as that, then all the uniform states of the field variable and its derivative up to order N must be correctly represented in the limit of element size going to 0, this is one of the requirements, I will explain what it means. The next condition is the displacement field and its derivatives up to order N-1 must be continuous at the element boundaries. Now let's consider condition B, now what is condition B? If N is the highest order of partial derivative of the field variable that appears in our Lagrangian, all the uniform states of the field variable and its derivatives up to order N must be correctly represented in the limit of element size going to 0, so now if all nodal displacements are identical the field variable must be constant within the element, so that is the element must permit rigid body state, then requirement on derivative actually translates into requirement that the element must permit constant strain states, see for example we can explain with respect to a 2-noded axially deforming rod with 2 degrees of freedom U1, U2, we know the expression for strain energy is given by this, the field variable is U, now what is the highest order of derivative appearing here is 1, the interpolation used is U1 1-X by L plus U2 X by L, now if U of X, T is constant that means U1 equal to U2, I want that U should be constant, so if I put that in the assumed displacement form I get U of X, T is U0, actually U of X, T turns out to be U0 which is what we are checking, now similarly if I find the first derivative it is U2-U1 by L which is a constant, so that second condition is also satisfied, so when we use this type of element we can expect convergence of the FE solution as we increase number of elements, there is a further caveats on that I will come to that shortly. Now this also we have seen which is now this is a condition C, the displacement field and its derivative up to order N-1 must be continuous at the element boundaries, this we have seen suppose there are 2 elements 1 and 2 with degrees of freedom as named here U of X, T for the first element is given by suppose this end is fixed it is U2 into X by L1, so if I now evaluate the field variable at L1 it is U2 of T, now similarly U of X of T here is given by U2 into 1-X by L2 plus U3 of T X by L2, now at X equal to 0 for this element U of 0, T is U2 of T as you can see here, so the field variable is continuous. Now that means we have to look at continuity of field variable and its derivative up to N-1, N here is 1 therefore I should only look at continuity of the field variables, indeed this is satisfied in this element. Now elements which satisfy conditions A and C are called compatible or conforming elements, elements which satisfy B are called complete elements, the field variable is said to possess CR continuity if its ARTH derivative is continuous, the completeness requirement implies that field variable has CN continuity within the element and the compatibility requirement implies that the field variable has CN-1 continuity across the element interface, so these are the requirements that we need to satisfy. If the requirements A, B and C are satisfied the FE approximation converges to correct solution if the FE mesh is refined, that is if we use increasing number of elements with smaller dimensions following certain requirements, so what is this? When we are refining the mesh the form of the interpolation function must remain unchanged, you cannot change the form of the interpolation function and the mesh refinement must be such that the mesh with larger number of elements contains the mesh with smaller number of elements, so also the mesh refinement must ensure that all points in the structure are within an element, so what all this means? Suppose you use for a rectangular domain a mesh with four elements as shown here, so when you make a mesh with eight elements this mesh should contain these four nodes, okay, so these nodes must be there, that means I have to partition this element, okay. So in the next refinement I should partition each of these elements, okay, suppose this is now 2 by 2 mesh, if I use 3 by 3 then this mesh won't be a subset of this mesh, so when we talk about convergence we cannot talk of convergence from that point of view, similarly all points on the structure you know what it means by saying that they have to be all of the points on the structure must be within an element means, suppose you have a circular domain and you are using straight edged elements to discretize that, clearly this portion is outside the, this say if you consider this element with this discretization this point is within the structure but not within the any of the element, so this type of discretization we cannot expect convergence, so as I refine this still there will be parts of the structure which don't get into our finite element or the domain that the finite elements cover, so the remedy to this would be to use a curved element. Now what are the factors which contribute to the development of an accurate finite element model, so we can say that accuracy with which the structure geometry is represented is one of the issues, for example in that circle example that I showed this type of issue is quite evident, then choice of polynomials used for interpolation, then distribution of elements and nodes for a same degree of freedom there can be two alternative placement of nodes and different shapes of finite elements, so one mesh may be superior to the other, then details of integration used in time marching, and also one could also include how you evaluate stiffness and mass matrices using Gauss quadrature. Now given that the accuracy of the FE model depends on these features if we now ask the question how to refine the FE model to improve accuracy, here you can reduce the element size that means the same domain will be covered with more elements, so therefore the model will have higher degrees of freedom or retain the degrees of elements but increase the order of polynomial, instead of first order interpolation use a higher order interpolation, so this first refinement is known as H refinement, where H refers to the size of the element, this second one is known as P refinement where P refers to the order of the polynomial used. Next possibility locate node point differently in a fixed element topology, so this is known as R refinement, the other thing is alter the mesh having differing element distributions, next improvements to the time integration schemes if you are doing dynamics problem that also you know contributes to the accuracy that you can achieve, so alternatives involving a combination of all these strategies also can be thought of, so what are the issues in selection of the interpolation polynomial, the polynomial should satisfy to the extent possible conditions A, B and C that is one of the requirement, the representation of the field variable must be invariant with respect to change in the local coordinate system of the element, suppose in a local coordinate system you have XYZ coordinate by renaming X as Y and others similarly renaming other axis the behavior of the element should not change, okay, so then this is known as geometric invariance or spatial isotropy or geometric isotropy, next the number of generalized coordinates must match the number of nodal degrees of freedom of the element, this is you know essential, now how to achieve geometric invariance, the geometric invariance can be achieved if the polynomial contains terms which do not violate symmetry in the Pascal triangle, which I will show now in two dimension or a Pascal pyramid in three dimension, what I mean is if you expand X plus Y to the power of N in a binomial expansion you can arrange these terms as shown here, the first term is a constant it is here, the second one will have X and Y, X square, XY, Y square, X cube, X square Y, for example X plus Y to the power of 0 is 1, X plus Y to the power of 1 is X plus Y, X plus Y to the power of 2 is X square plus 2 X Y plus Y square, so this is X square, XY, Y square and so on and so forth, so the last term will be having this distribution. Now whenever you are choosing the interpolation polynomial we have to keep in mind how to do that, for example if you are for a triangle element we had 3 degrees of freedom, I mean 3 nodes and each node there were 2 degrees of freedom and U was represented alpha 1 plus alpha 2 X plus alpha 3 Y, so we needed there were 3 degrees of 3 nodal displacement values therefore I should use a 3 term expansion for U, so that would be 1, X and Y, okay that's what it means, a rectangular element how do you select? I need 4 terms 1, X, Y and when I come here I can select any one of this in principle, if I select X square it won't be alright because if I rename X and Y axis this will be Y square, so that's not right, so what we do is we take the term closest to the axis of symmetry which is XY, so this is alpha 1 plus alpha 2 X alpha 3 Y plus alpha 4 XY, so just to give this emphasize that if you are to take this or this that is either you retain this term or retain this term as a 4th you know for the 4th term that you need use this as a candidate or use this as a candidate you will get these 2, this is not appropriate because interchanging of X and Y would change the representation. Now just to again to give an example suppose you have 4 nodded element with 3 degrees of freedom per node, now how do we, you need 12 terms to represent that field variable, so how do you select 12 terms? 1, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, up to this there is no problem, but when it comes to this axis which one will you select? So you should maintain symmetry therefore you should take X Q Y and X Y Q, you could as well for geometric invariance you could have selected X square Y square, but the next term you will not be able to select symmetrically, and also you could have selected X to the power of 4 and Y to the power of 4, but as a rule we select terms which are closest to the axis of symmetry and therefore we take this term, so we'll come to this, this we'll be using when we discuss plate bending we'll see that, how about in 3 dimensions? So again we consider X plus Y plus Z equal to the power of 0 is 1, so that is 1, so on this I have, on this line I have X, on this I have Z, on this axis I have Y, X plus Y plus Z to the power of 1 is X plus Y plus Z, so that is X, Y, Z, square, X square plus 2X Y plus 2 YZ, etc., that is this, X square, Y square, Z square, 2XY, 2YZ, 2XZ, similarly you have cubic terms and so on and so forth. Now just another point which I am sure we have come across several times I just want to articulate that in clear terms, how do you decide on degrees of freedom? Okay, so the rule is inspect the functional in the variational formulation, identify the field variables and order of their highest derivatives, call it as N, so the degrees of freedom will be the field variables and their derivatives up to order N-1, see for example in actually deforming element the functional is in the variational formulation will be AE dou U by dou X whole square, so which is a field variable U, and what is the highest order N? So we need field variables and their derivatives up to order N-1, so N is 1 here it is 0, so we need only U, so U is the degree of freedom, whereas if you consider beam element, V is the field variable, highest derivative is 2, so I need the field variable and its derivative dou V by dou X, V and dou V by dou X, so they are the degrees of freedom. So as we consider more complicated problems it won't be immediately clear how to do this that's why I thought I must emphasize this fact at this stage. So similarly just to complete the discussion in plane stress element, this is a stress field, this is a strain field and this is how the stress and strain were related, so when we considered the functional we had epsilon transport D epsilon as the, in the variational formulation this was the functional. Now here therefore if you examine these relations carefully we have the field variables are U and V, what is the highest order derivative? You will see here it will be dou U by dou X and dou U by dou Y and things like that, so highest order derivative is 1, so the nodal degrees of freedom therefore will be U and V, okay. Now equipped with this now we will start discussing 3D solid elements, so quickly we will recall the various variables, this is the state of stress at a point it can be expressed either as 3 by 3 matrix or a 6 cross 1 vector, this is state of strain at a point there are 6 variables again this can be represented as a 3 by 3 tensor or a 6 cross 1 vector and we prefer the 6 cross 1 representation in finite element development, so the stress and strain are related through the constitutive law I am assuming material is elastic, linearly elastic and isotropic, so this is the constitutive law for that material behavior model. Strain energy we have seen it is integral over volume element sigma transpose epsilon DV naught multiplied by half, so for sigma transpose I will write epsilon transpose D transpose, now D is a symmetric matrix so D transpose is D I get epsilon transpose D epsilon, kinetic energy rho into DV naught is the mass of an infinitesimal element with dimension DX, DY, DZ and the kinetic energy of that in X direction is half mass into U dot square, in Y direction half mass into V dot square, Z direction half mass into W dot square and you add up this is the kinetic energy. Now we will consider the expression for strain energy, this epsilon we want to express this now in terms of displacements, so we use the strain displacement relations, so strain is given in terms of displacement, when this matrix operates on this displacement we get the strain, using that I represent UVW now in terms of the shape functions and nodal displacements which I write it as NUE and where this epsilon given this the epsilon will now be this matrix of derivatives into N into UV, because this is N into U this is UV, so this operator acting on N is known as B matrix and therefore the strain energy becomes in terms of displacement field it is given by this expression and kinetic energy is given by this, this we have seen but just I am reiterating so that we recall quickly what we need to use now immediately. Now in considering problems of three dimensional continuum various shapes become possible, so whereas in line elements there was no such dilemma it was everything was simple, in plain problems we had quadrilaterals you know rectangles, triangles, quadrilaterals, isoparametric you know quadrilaterals with curved edges and so on and so forth, similar issues will come up in solid elements now, so we can discretize a three dimensional domain using tetrahedrons or using rectangular hexahedrons or pentahedrons or in a more general situation a isoparametric hexahedron, so what we will do is we will try to develop the logic for developing the structural matrices for at least for some of these elements. Now let us start with considering a tetrahedron element, so tetrahedron element has four nodes this is a Cartesian coordinate system 1, 2, 3, 4 are the four nodes and each node has three degrees of freedom, U, V, and W, so this is four noted element with three degrees of freedom per node and hence we have 12 degrees of freedom for the element. Now the field variables are U, V, W which vary with respect to XYZ and T, so what we do is we represent the field variables in terms of we need four terms so we have to go to now the Pascal's pyramid, so the four terms are 1, X, Y, and Z, so that is what we are doing, I have alpha 1 of T, alpha 2 of T into X, alpha 3 into Y, alpha 4 into Z, so similarly I have for V and W similar representations these alpha 1 to alpha 12 are now the generalized coordinates, now these have to be selected by knowing the value of these field variables at the nodes, so at node 1 I will have U1, V1, W1, here U2, V2, W2 and so on and so forth, so at X1, Y1, Z1 I have U is U1, V is V1, W is W1 and so on and so forth. So now I will use these conditions and I will be able to evaluate those 12 generalized coordinates in terms of nodal displacements, this we have done for a triangle element, so I get the interpolation, the formula for interpolating the field variable within an element in terms of nodal displacements and the trial functions as shown here. Now in terms of the nodal coordinates we can derive the shape functions, the formulary is given here, V0 is a volume of the element and the N1 for example is a constant A1 plus B1X plus C1Y plus D1Z, so it's a linear first order polynomial in three variables. So similarly this V0 itself is the volume of the element that will turn out to be in terms of the coordinates of the nodes and there are some constant A1, B1, C1 they are all explained here, actually there is A1, B1, C1, D1 they have been described here and there is a principle for evaluating the other elements, some rules are given here that can be easily followed. So we have now got a representation like this where these N1, N2, N3, N4 are linear functions of XYZ and there is a constant term as well. So now this is a representation UVW N into UE, UE consist of 12 variables now, U1V1W1 up to U4V4W4. So N is this 3 into 12 matrix and ME is our N transpose N matrix we can evaluate this exactly, so you have to decide upon the order of the terms, N is linear N transpose N will be quadratic and you can use the appropriate order of integration and this can be evaluated exactly or you can actually carry out the integration when closed form, there is no problem. So this is a mass matrix, this is symmetric 12 by 12, now strain energy leading to the evaluation of stiffness matrix this is epsilon into U, this operator matrix into U and this is B and now B is this operator acting on this matrix. So we can evaluate B here in this case it becomes a constant you recall what happened for a triangular element in plane problems a similar thing happens here this is a constant, so then this KE can thus be evaluated exactly, there is no need for quadrature because this is simply a constant term. So once we find all this we know the strain which is constant over the volume and stress is DB into UE which is also constant over the volume that we have considered. So a tetrahedron element, formulation of tetrahedron element is straight forward and it follows the same steps that we use for formulating a triangle elements. Now how about a rectangular hexahedron element? So this element has 8 nodes 1, 2, 3, 4, 5, 6, 7, 8, so it is 8-noded element with 3 degrees of freedom per node and therefore it has 24 degrees of freedom, the field variables are UVW which are functions of XYZ and T, so at each node there will be 3 field variables so there are 3 degrees of freedom, therefore it is a 8-noded element with 3 degrees of freedom per node therefore a 24 degrees of freedom element, so all the structural matrices will be 24 by 24. Now we introduce the coordinate transformation XI is X by A, E ties Y by B and G ties Z by C so that this gets mapped to a cube of dimensions 2. Now each of these field variables now need to be, there are 8 nodes therefore each field variable needs to be represented in terms of 8 terms, okay, so each of these variables need to be represented by a polynomial with 8 terms, how do we select that? So we go back to the Pascal's pyramid we have to take 8 terms 1, XYZ, 4 or over, X square, so the term that we are taking is 1, XYZ, XY, XZ, YZ and XYZ, this is what we will select, we are again taking terms closest to the central axis and symmetry is a requirement, okay. So X square, Y square, Z square we are not taking, we are taking XY, XZ, YZ, subsequently we go to the third order term and we don't take X cube or Y cube or Z cube instead we take XYZ, so this ensures geometric invariance, so I get this representation, now these alphas are the generalized coordinates which need to be selected so that the values of U matches with its respective nodal values at the 8 nodes, so assuming that we have done that I get the representation for the field variables, now we have transformed now to XI eta, Zeta coordinate system I get this representation, and we can show that in XI, Zeta, eta, Zeta coordinate these trial functions are given by this, this is again similar to what we did for linear rectangular plane stress element. Now how do you check for continuity of field variables across the element boundaries? I claim that it is ensured and I leave it as an exercise for you to verify, so you need to consider two neighboring elements and a point lying on the interface and argue out why the field variables are continuous across the interface, so the representation is therefore now UVW is N into UE, that means I have combined all this where N is this matrix 3 by 24 and this is the 24 cross 1 nodal degrees of freedom vector, so we go back to the expression for kinetic energy, so I get the mass matrix to be given by this where an ijth element is given by this integral. Now since this is a cubic function, you will have the nonlinear term that will be present here will be products of XI eta, Zeta, second and third order products, so you can evaluate this integral easily. So if this is done in this case it is possible again to evaluate it in closed form I get the element mass matrix in this form which is shown here where this M itself is a huge matrix of this kind, so this is 12 by 12 matrix, the mass matrix will be 24 by 24 matrix, so the partitioning is in terms of 12 by 12 square matrices, now how about strain energy, so I have displacement is N into UE and strain is given by this and this is our expression for displacements in terms of nodal, strains in terms of nodal displacement values and this B matrix is given by this into this, this is operated on N matrix, so as before I get K matrix to be given by integral over the volume B transpose DB DB naught, so how do we get these terms now, so this B matrix I can carry out that differentiation operation and take it through the trial functions I get this and if we spend some effort we can evaluate all the gradients that appears here, so we can evaluate KE element as shown here or we can use Gauss quadrature to evaluate this and you can argue out that a 2 by 2 by 2 quadrature would complete the quadrature here, I mean would offer a good solution here. Now the similar logic can be used now, how about an isoparametric hexahedron element, so here I consider this as my element, so it has 8 nodes 1, 2, 3, 4, 5, 6, 7, 8, so the faces are you know not orthogonal to each other and they are not rectangles and so on and so forth, this is, this I will map through a transformation to a unit cube of dimensions 2, at lateral dimensions 2, so then we will carry out the integrations needed to implement evaluation of KE and ME in this coordinate system. So the transformation that we are looking for is X is represented in terms of the same trial function that I eventually will use for representing displacement fields, field variables, this is Y, this is Z, in terms of nodal values, so I can assemble them in a matrix form, I can write this matrix into this vector of nodal coordinates, in XI eta, Geta plane I know that this is the form of the trial functions, so I can, I have now the representation that I need for the nodal, the transformation from X to XI eta, Geta plane. Now how about displacement field, the same representation is used in the strain following this representation I get B into UE where B is given by B1, B2, B8 where each of these matrices have this form, where I is 1, 2, 1, 2, 8, okay, so we have these matrices now. Now here I have this term dou N i by dou X, dou N i by dou Y, dou N i by dou Z, etc., so that I need to evaluate, so to do that we can start by finding derivative of dou N i by dou XI and N i with respect to Geta, these are straightforward application of rules of differentiation, so N i by XI is N i by X into dou X by dou XI plus dou N i by dou Y into dou Y by dou XI, etc., so you do for all the variables I get this, and this itself I can write it in a matrix form as shown here, and this matrix is a Jacobian matrix of the transformation, so I will get J into this vector of gradients, this is J, J itself I can write in terms of X I will now use this representation, therefore I will be able to write elements of J in this form, so J is this, now since I know the form of this interpolation function I can carry out these integrations, so the differentiation with respect to XI will give me this, with respect to Geta will give me this, Geta will give me this, so that means I can evaluate elements of J matrix, see why I am doing all this is I need to find out which order terms will be present, right, so we discover that elements of J are triquodratic functions. Now this function that I am basically looking for which one, this is now given by J inverse into this, now KE is thus given by this integral over volume element now become B transpose DB determinant of J into this, now therefore if you can now see this the elements of this matrix are ratios of triquodratic functions of XI, eta and Geta, so these Gauss quadrature will not evaluate this integral exactly, so you will have to have some judgment on this and what is recommended is use a 2 by 2 by 2 Gauss quadrature, and accordingly we get this, this is not an exact evaluation, the mass matrix may get evaluated exactly but not the stiffness matrix, and also we will be needing the volume in this calculation and the volume itself can be evaluated again using Gauss quadrature, it can be done exactly but since we are using Gauss quadrature this can be done, and since the integrand here will be a polynomial this can be evaluated exactly by using 2 cross 2 cross 2 Gauss quadrature. So now with some effort we have formulated the elements, so now we can illustrate with some simple examples, suppose I consider a cantilever block which is fixed at the bottom, at these nodes it is fixed, so in the first model that we have used I have 16 elements and 108 degrees of freedom and we have used 8-noded hexahedron elements, so these are the parameters of the structure, simple structure, and we will let us try to find first few natural frequencies and more shapes, so you can easily see that this will, first mode will be a bending mode in this direction, second mode is likely to be this, and depending on the geometry the third mode could be a torsional, so indeed that happens, the first mode is bending along this direction, and the second one is bending along this direction, and third mode is torsional mode, and the frequencies we get are shown here 84 hertz, 122 hertz, 379 hertz. Now what I will do is I will refine the mesh now, again I am using the same, the refinement is such that within an element I am creating more elements, okay, so that leads to a model with 600 degrees of freedom and 128 elements, and again I am using 8-noded hexahedron element, that means I am keeping the interpolation function the same, and the mesh in the previous edition of the model is a subset of this. So here these are the modes, again first mode is bending here, this and twisting, the frequencies of course are now different numbers, so I am getting now 83.33, this is 84.8, 120, 281, 379, so if we follow all these rules that I mentioned this convergence will be from the above, okay. Now in the next part of the discussions we will consider another 2D approximation to problems of solid continuum, if you recall now we had two dimensional approximations already, in the plane stress model the object was so thin that the stresses across the thickness were neglected, and we got a plane stress model, in the next plane strain model the object was so thick that the displacement across the thickness were neglected, and we got a plane strain model. Now in the next part of our discussion we consider objects possessing rotational symmetry about an axis, and loaded and supported in an axisymmetric manner, so here also we will get a 2 dimensional approximation, so how does it work? Now this is an object with rotational symmetry, so you take a curve and an axis and rotate about this, so you take this line and this axis and rotate about Z axis and you get this first term of the cone, okay. So this is an example of an object with rotational symmetry about an axis, such structures are found extensively in many application, for example this is a cross section of a nuclear reactor vessel and the outer shell that you can see here has this type of property, it is rotationally symmetric about the Z axis here, there will be many internals that may not have this type of rotational symmetry, but still the outer dome for example has this property. So if we are analyzing this type of properties it is useful to take advantage of this simplification in formulating the problem, the dimensions of the problem, the numerical, the dimension of the equation that we need to solve reduces. Now what are the characteristics of this type of objects, geometry, there are 3 dimensional axisymmetry, axisymmetric solid, it is not necessarily prismatic and not necessarily thin or thick, by that what I mean, instead of this curve I can rotate this curve about Z axis I get different geometries, okay, so the cross sectional properties will not be the same, at different values of Z, how about the loads? Now we are going to use a cylindrical polar coordinate system, Z axis is this and this is a radial axis and this is the angle, theta axis, so we are going to use actually, suppose this is the origin, this is theta, this is R, and this is Z, that is the axis that we are going to use. Now with reference to that axis the surface tractions are independent of theta, that means they are constant with respect to theta, the body forces in the theta direction are zero, and the surface tractions in the R and Z direction are independent of theta, that means they are uniform for all values of theta, like a cylindrical, I mean a vessel like this internally pressurized for example, suppose it is closed at the bottom and top and it is internally pressurized we will have this type of model for surface tractions. Then using the property of the symmetry we will postulate that there are three displacement fields, U along R, theta is V and W is along Z, because of the symmetry and of loading boundary conditions and surface tractions and body forces it emerges at V is zero, then U is independent of theta and W is independent of theta. Now the material that we are using is linear, homogeneous, elastic and isotropic, so what we will do in the next class is we will develop a finite element model for problems that satisfy these requirements, so that would offer us another two dimensional approximation to a more complicated behavior. Following that we will consider behavior of again thin elements of a different kind, we have now considered in plane stress models prismatic objects which carry loads in their own plane. Now we want to consider what are known as plate bending problems, so suppose you have a folded plate structure and it is loaded, then this plate for example there will be both transverse loads and in plane loads, okay, this could be a wall or a panel in a structure also, so we consider objects like this whose thickness is constant, right, initially they are flat, in addition to the in plane loads we now consider transverse loads like a slab in a building deforming under its own weight, so that is a plate action. Now a problem like this has two components, one is what is known as membrane action where we consider the behavior of the structure only under in-line, in-plane loads, and the bending action which is behavior of this type of structures under transverse loads, the bending action is known as plate action and the response due to in-line loads or in-plane loads is known as membrane action. The analysis of membrane action can be carried out using plane stress models, so we need to next analyze the behavior under transverse loads, this requires generalization of beam, actually it's a generalization of grid, if you recall we analyzed problems grid structures consisting of beam elements and we, it was pointed out that bending in this member is twisting in this member, so a plate can be considered as a continuous analog of a grid where bending in one direction causes twisting in the other direction, so the stress resultants will consist of apart from bending moment and shear forces there will be a twisting moment, so what we will do in the subsequent class is we will consider these two problems, problems of axisymmetry and problems of plate bending, and this we will develop based on two-dimensional theory of velocity, and this we will develop by developing theory based on Kirchhoff-Lauw assumptions and Midlin's assumptions, so in Kirchhoff-Lauw's assumption the thickness of this member is taken to be small and certain assumptions on how a line element like this behaves will be made, whereas in Midlin theory which is generalization of Timoshenko beam theory the beam can be thick and we will include shear deformation and also while computing inertia we will compute, include effects of rotary inertia, so these two problems we will consider in the next class and we will conclude the present lecture at this stage.