CHAPTER 3 CHAPTER OUTLINE 3. Introduction 75 3.2 Polynomial Form of Interpolation Functions 77 3.3 Simple, Comple, and Multiple Elements 78 3.4 Interpolation Polynomial in Terms of Nodal Degrees of Freedom 78 3.5 Selection of the Order of the Interpolation Polynomial 80 3.6 Convergence Requirements 82 3.7 Linear Interpolation Polynomials in Terms of Global Coordinates 85 3.7. One-Dimensional Simple Element 85 3.7.2 Two-Dimensional Simple Element 88 3.7.3 Three-Dimensional Simple Element 9 3.7.4 C -Continuity 95 3.8 Interpolation Polynomials for Vector Quantities 96 3.9 Linear Interpolation Polynomials in Terms of Local Coordinates 99 3.9. One-Dimensional Element 00 3.9.2 Two-Dimensional (Triangular) Element 0 3.9.3 Three-Dimensional (Tetrahedron) Element 04 3.0 Integration of Functions of Natural Coordinates 08 3. Patch Test 09 75 3. INTRODUCTION As stated earlier, the basic idea of the finite element method is piecewise approimation that is, the solution of a complicated problem is obtained by dividing the region of interest into small regions (finite elements) and approimating the solution over each subregion by a simple function. Thus, a necessary and important step is that of choosing a simple function for the solution in each element. The functions used to represent the behavior of the solution within an element are called interpolation functions or approimating functions or interpolation models. Polynomial-type interpolation functions have been most widely used in the literature due to the following reasons:. It is easier to formulate and computerize the finite element equations with polynomialtype interpolation functions. Specifically, it is easier to perform differentiation or integration with polynomials. 2. It is possible to improve the accuracy of the results by increasing the order of the polynomial, as shown in Figure 3.. Theoretically, a polynomial of infinite order corresponds to the eact solution. But in practice we use polynomials of finite order only as an approimation. Although trigonometric functions also possess some of these properties, they are seldom used in the finite element analysis [3.]. We shall consider only polynomial-type interpolation functions in this book. The Finite Element Method in Engineering 20 Elsevier Inc. All rights reserved.
PART 2 Basic Procedure φ() Eact solution φ() = a 0 = constant Subregion or element (a) Approimation by a constant φ() φ() Eact solution φ() = a 0 + a Eact solution φ() = a 0 + a + a 2 2 Subregion or element (b) Linear approimation Subregion or element (c) Quadratic approimation FIGURE 3. Polynomial Approimation in One Dimension. 76 When the interpolation polynomial is of order one, the element is termed a linear element. A linear element is called a simple element if the number of nodes in the element is 2, 3, and 4 in one, two, and three dimensions, respectively. If the interpolation polynomial is of order two or more, the element is known as a higher order element. In higher order elements, some secondary (mid-side and/or interior) nodes are introduced in addition to the primary (corner) nodes in order to match the number of nodal degrees of freedom with the number of constants (generalized coordinates) in the interpolation polynomial. In general, fewer higher order elements are needed to achieve the same degree of accuracy in the final results. Although it does not reduce the computational time, the reduction in the number of elements generally reduces the effort needed in the preparation of data and hence the chances of errors in the input data. The higher order elements are especially useful in cases in which the gradient of the field variable is epected to vary rapidly. In these cases the simple elements, which approimate the gradient by a set of constant values, do not yield good results. The combination of greater accuracy and a reduction in the data preparation effort has resulted in the widespread use of higher order elements in several practical applications. We shall consider mostly linear elements in this chapter. If the order of the interpolation polynomial is fied, the discretization of the region (or domain) can be improved by two methods. In the first method, known as the r-method, the locations of the nodes are altered without changing the total number of elements. In the second method, known as the h-method, the number of elements is increased. On the other hand, if improvement in accuracy is sought by increasing the order of the interpolation of polynomial, the method is known as the p-method. Problems involving curved boundaries cannot be modeled satisfactorily by using straight-sided elements. The family of elements known as isoparametric elements has been developed for this purpose. The basic idea underlying the isoparametric elements is to use the same interpolation functions to define the element shape or geometry as well as the variation of the field variable within the element. To derive the isoparametric element equations, we first introduce a local or natural coordinate system for each element shape. Then the interpolation
CHAPTER 3 or shape functions are epressed in terms of the natural coordinates. The representation of geometry in terms of (nonlinear) shape functions can be considered as a mapping procedure that transforms a regular shape, such as a straight-sided triangle or rectangle in the local coordinate system, into a distorted shape, such as a curved-sided triangle or rectangle in the global Cartesian coordinate system. This concept can be used in representing problems with curved boundaries with the help of curved-sided isoparametric elements. Today, isoparametric elements are etensively used in three-dimensional and shell analysis problems. The formulation of isoparametric elements, along with the aspect of numerical integration that is essential for computations with isoparametric elements, is considered in the net chapter. 3.2 POLYNOMIAL FORM OF INTERPOLATION FUNCTIONS If a polynomial type of variation is assumed for the field variable ϕ() ina one-dimensional element, ϕ() can be epressed as ϕðþ= α + α 2 + α 3 2 + + α m n (3.) Similarly, in two- and three-dimensional finite elements the polynomial form of interpolation functions can be epressed as ϕð, yþ = α + α 2 + α 3 y + α 4 2 + α 5 y 2 + α 6 y + + α m y n (3.2) ϕð, y, zþ = α + α 2 + α 3 y + α 4 z + α 5 2 + α 6 y 2 + α 7 z 2 + α 8 y + α 9 yz + α 0 z + + α m z n (3.3) where α, α 2,, α m are the coefficients of the polynomial, also known as generalized coordinates; n is the degree of the polynomial; and the number of polynomial coefficients m is given by m = n + for one-dimensional elements ðeq: 3:Þ (3.4) m = n + j = j for two-dimensional elements ðeq: 3:2Þ (3.5) 77 m = n + j = jn+ ð 2 jþ for three-dimensional elements ðeq: 3:3Þ (3.6) In most practical applications, the order of the polynomial in the interpolation functions is taken as one, two, or three. Thus, Eqs. (3.) to (3.3) reduce to the following equations for various cases of practical interest. For n = (linear model) One-dimensional case: Two-dimensional case: Three-dimensional case: For n = 2 (quadratic model) One-dimensional case: Two-dimensional case: ϕðþ= α + α 2 (3.7) ϕð, yþ = α + α 2 + α 3 y (3.8) ϕð, y, zþ = α + α 2 + α 3 y + α 4 z (3.9) ϕðþ= α + α 2 + α 3 2 (3.0) ϕð, yþ = α + α 2 + α 3 y + α 4 2 + α 5 y 2 + α 6 y (3.)
PART 2 Basic Procedure Three-dimensional case: For n = 3 (cubic model) ϕð, y, zþ = α + α 2 + α 3 y + α 4 z + α 5 2 + α 6 y 2 + α 7 z 2 + α 8 y + α 9 yz + α 0 z (3.2) One-dimensional case: ϕðþ= α + α 2 + α 3 2 + α 4 3 (3.3) Two-dimensional case: ϕð, yþ = α + α 2 + α 3 y + α 4 2 + α 5 y 2 + α 6 y (3.4) + α 7 3 + α 8 y 3 + α 9 2 y + α 0 y 2 Three-dimensional case: ϕð, y, zþ = α + α 2 + α 3 y + α 4 z + α 5 2 + α 6 y 2 + α 7 z 2 + α 8 y + α 9 yz + α 0 z + α 3 + α 2 y 3 + α 3 z 3 + α 4 2 y + α 5 2 z (3.5) + α 6 y 2 z + α 7 y 2 + α 8 z 2 + α 9 yz 2 + α 20 yz 78 3.3 SIMPLEX, COMPLEX, AND MULTIPLEX ELEMENTS Finite elements can be classified into three categories as simple, comple, and multiple elements depending on the geometry of the element and the order of the polynomial used in the interpolation function [3.2]. The simple elements are those for which the approimating polynomial consists of constant and linear terms. Thus, the polynomials given by Eqs. (3.7) to (3.9) represent the simple functions for one-, two-, and three-dimensional elements. Noting that a simple is defined as a geometric figure obtained by joining n + joints (nodes) in an n-dimensional space, we can consider the corners of the elements as nodes in simple elements. For eample, the simple element in two dimensions is a triangle with three nodes (corners). The three polynomial coefficients α, α 2, and α 3 of Eq. (3.8) can thus be epressed in terms of the nodal values of the field variable ϕ. The comple elements are those for which the approimating polynomial consists of quadratic, cubic, and higher order terms, according to the need, in addition to the constant and linear terms. Thus, the polynomials given by Eqs. (3.0) to (3.5) denote comple functions. FIGURE 3.2 Eample of a Multiple Element. y The comple elements may have the same shapes as the simple elements but will have additional boundary nodes and, sometimes, internal nodes. For eample, the interpolating polynomial for a two-dimensional comple element (including terms up to quadratic terms) is given by Eq. (3.). Since this equation has si unknown coefficients α i, the corresponding comple element must have si nodes. Thus, a triangular element with three corner nodes and three mid-side nodes satisfies this requirement. The multiple elements are those whose boundaries are parallel to the coordinate aes to achieve interelement continuity, and whose approimating polynomials contain higher order terms. The rectangular element shown in Figure 3.2 is an eample of a multiple element in two dimensions. Note that the boundaries of the simple and comple elements need not be parallel to the coordinate aes. 0 3.4 INTERPOLATION POLYNOMIAL IN TERMS OF NODAL DEGREES OF FREEDOM The basic idea of the finite element method is to consider a body as composed of several elements (or subdivisions) that are connected at specified node points. The unknown solution or the field variable (e.g., displacement, pressure, or temperature) inside any finite
CHAPTER 3 element is assumed to be given by a simple function in terms of the nodal values of that element. The nodal values of the solution, also known as nodal degrees of freedom, are treated as unknowns in formulating the system or overall equations. The solution of the system equations (e.g., force equilibrium equations or thermal equilibrium equations or continuity equations) gives the values of the unknown nodal degrees of freedom. Once the nodal degrees of freedom are known, the solution within any finite element (and hence within the complete body) will also be known to us. Thus, we need to epress the approimating polynomial in terms of the nodal degrees of freedom of a typical finite element e. For this, let the finite element have M nodes. We can evaluate the values of the field variable at the nodes by substituting the nodal coordinates into the polynomial equation given by Eqs. (3.) to (3.3). For eample, Eq. (3.) can be epressed as ϕðþ =! T η! α (3.6) where and!t η = 2 n, 8 α >< α! 2 α =. >: α n + 9 >= >; The evaluation of Eq. (3.6) at the various nodes of element e gives 8 9 ϕðat node Þ >< ϕðat node 2Þ >= ðeþ. >: >; ϕðat node MÞ 2 =! ðeþ Φ = 6 4!T η ðat node Þ!T η ðat node 2Þ..!T η ðat node MÞ 3! α ½~ η Š 7! α (3.7) 5 79 where Φ! ðeþ is the vector of nodal values of the field variable corresponding to element e, and the square matri ½ ~ ηš can be identified from Eq. (3.7). By inverting Eq. (3.7), we obtain! α = ½~ ηš! ðeþ Φ Substitution of Eq. (3.8) into Eqs. (3.) to (3.3) gives ϕ = η! T α! = η!t ½~ ηš Φ! ðeþ = ½NŠΦ!ðeÞ (3.8) (3.9) where ½NŠ = η! T ½~ ηš (3.20) Equation (3.9) now epresses the interpolating polynomial inside any finite element in terms of the nodal unknowns of that element,! ðeþ Φ. NOTE A major limitation of polynomial-type interpolation functions is that one has to invert the matri ½ ~ ηš to find ϕ, and ½ ~ ηš may become singular in some cases [3.3]. The latter difficulty can be avoided by using other types of interpolation functions discussed in Chapter 4. The following eample illustrates the use of Eq. (3.20) in finding the shape functions of an element.
PART 2 Basic Procedure EXAMPLE 3. The nodes of a one-dimensional element are located at = 20 in and 2 = 25 in. The values of the field variable at the two nodes (nodal unknowns) of the element are denoted Φ and Φ 2. Assuming a linear interpolation model for the field variable as where ϕðþ = f g α α 2 η! T α! (!T η = f g and! α = α ) α 2 determine the matri of shape functions [N()] using Eq. (3.20). Solution Approach: Use a linear interpolation model in the form of Eq. (E.). The nodal values of the field variable can be epressed, using Eq. (E.), as so that Φ = ϕð = Þ = f 20g α! Φ 2 = ϕð = 2 Þ = f 25g α!!ðeþ Φ Φ = Φ 2 = 20 25! α ½~ ηš! α (E.) 80 The inverse of the matri ½ ηš is given by ~ ½ ηš = 20 = ~ 25 " # 5 4 5 5 Thus, the matri of shape functions of the element is given by Eq. (3.20): " # ½NðÞŠ =! T η ½~ ηš 5 4 n = f g = 5 4 + o 5 5 5 5 fn ðþ N 2 ðþg where the shape functions associated with nodes and 2, N () and N 2 (), can be identified as N ðþ = 5 5, N 2ðÞ = 4 + 5 3.5 SELECTION OF THE ORDER OF THE INTERPOLATION POLYNOMIAL While choosing the order of the polynomial in a polynomial-type interpolation function, the following considerations have to be taken into account:. The interpolation polynomial should satisfy, as far as possible, the convergence requirements stated in Section 3.6. 2. The pattern of variation of the field variable resulting from the polynomial model should be independent of the local coordinate system. 3. The number of generalized coordinates (α i ) should be equal to the number of nodal degrees of freedom of the element (Φ i ). A discussion on the first consideration, namely, the convergence requirements to be satisfied by the interpolation polynomial, is given in the net section. According to the second consideration, as can be felt intuitively also, it is undesirable to have a preferential coordinate direction. That is, the field variable representation within an element, and hence the polynomial, should not change with a change in the local coordinate system (when a linear
CHAPTER 3 transformation is made from one Cartesian coordinate system to another). This property is called geometric isotropy or geometric invariance or spatial isotropy [3.4]. In order to achieve geometric isotropy, the polynomial should contain terms that do not violate symmetry in Figure 3.3, which is known as Pascal triangle in the case of two dimensions, and Pascal tetrahedron or pyramid in the case of three dimensions. Total number of terms involved Constant model 2 y y y 2 Linear model 3 Quadratic model 6 4 3 3 y 2 y 2 y 2 y 2 y 3 y 3 y 4 Cubic model 0 Quartic model 5 5 4 y 3 y 2 2 y 3 y 4 y 5 Quintic model 2 (a) In two dimensions (Pascal triangle) Total number of terms involved Constant model 8 y Linear model 4 z 2 y y 2 z yz z 2 2 y y 2 3 y 3 2 z z y yz 2 z 2 y z 2 z 3 4 3 y 2 y 2 y 3 y 4 3 zy 3 z 2 yz y 2 z z 2 y 2 2 z 2 yz 2 z 3 y z 3 z 4 Quadratic model 0 Cubic model 20 Quartic model 35 (b) In three dimensions (Pascal tetrahedron or pyramid) FIGURE 3.3 Array of Terms in Complete Polynomials of Various Orders.
PART 2 Basic Procedure Thus, in the case of a two-dimensional simple element (triangle), the interpolation polynomial should include terms containing both and y, but not only one of them, in addition to the constant term. In the case of a two-dimensional comple element (triangle), if we neglect the term 3 (or 2 y) for any reason, we should not include y 3 (or y 2 ) also in order to maintain geometric isotropy of the model. Similarly, in the case of a threedimensional simple element (tetrahedron), the approimating polynomial should contain terms involving, y, and z in addition to the constant term. The final consideration in selecting the order of the interpolation polynomial is to make the total number of terms involved in the polynomial equal to the number of nodal degrees of freedom of the element. The satisfaction of this requirement enables us to epress the polynomial coefficients in terms of the nodal unknowns of the element as indicated in Section 3.4. 82 3.6 CONVERGENCE REQUIREMENTS Since the finite element method is a numerical technique, we obtain a sequence of approimate solutions as the element size is reduced successively. This sequence will converge to the eact solution if the interpolation polynomial satisfies the following convergence requirements [3.5 3.8]:. The field variable must be continuous within the elements. This requirement is easily satisfied by choosing continuous functions as interpolation models. Since polynomials are inherently continuous, the polynomial type of interpolation models discussed in Section 3.2 satisfy this requirement. 2. All uniform states of the field variable ϕ and its partial derivatives up to the highest order appearing in the functional I(ϕ) must have representation in the interpolation polynomial when, in the limit, the element size reduces to zero. The necessity of this requirement can be eplained physically. The uniform or constant value of the field variable is the most elementary type of variation. Thus, the interpolation polynomial must be able to give a constant value of the field variable within the element when the nodal values are numerically identical. Similarly, when the body is subdivided into smaller and smaller elements, the partial derivatives of the field variable up to the highest order appearing in the functional I(ϕ) approach a constant value within each element. Thus, we cannot hope to obtain convergence to the eact solution unless the interpolation polynomial permits this constant derivative state. In the case of solid mechanics and structural problems, this requirement states that the assumed displacement model must permit the rigid body (zero strain) and the constant strain states of the element. 3. The field variable ϕ and its partial derivatives up to one order less than the highest order derivative appearing in the functional I(ϕ) must be continuous at element boundaries or interfaces. We know that in the finite element method the discrete model for the continuous function ϕ is taken as a set of piecewise continuous functions, each defined over a single element. As seen in Eamples.2 to.4, we need to evaluate integrals of the form Z d r ϕ d r d Finite element method can be considered as an approimate method of minimizing a functional I(ϕ) in the form of an integral of the type IðϕÞ = I ϕ, dϕ d, d2 ϕ d 2,, dr ϕ d r The functionals for simple one-dimensional problems were given in Eamples.2.4.
CHAPTER 3 to derive the element characteristic matrices and vectors. We know that the integral of a stepwise continuous function, say f(), is defined if f() remains bounded in the interval of integration. Thus, for the integral Z d r ϕ d r d to be defined, ϕ must be continuous to the order (r ) to ensure that only finite jump discontinuities occur in the r-th derivative of ϕ. This is precisely the requirement stated previously. The elements whose interpolation polynomials satisfy the requirements () and (3) are called compatible or conforming elements and those satisfying condition (2) are called complete elements. If r-th derivative of the field variable ϕ is continuous, then ϕ is said to have C r continuity. In terms of this notation, the completeness requirement implies that ϕ must have C r continuity within an element, whereas the compatibility requirement implies that ϕ must have C r continuity at element interfaces. 2 In the case of general solid and structural mechanics problems, this requirement implies that the element must deform without causing openings, overlaps, or discontinuities between adjacent elements. In the case of beam, plate, and shell elements, the first derivative of the displacement (slope) across interelement boundaries also must be continuous. Although it is desirable to satisfy all the convergence requirements, several interpolation polynomials that do not meet all the requirements have been used in the finite element literature. In some cases, acceptable convergence or convergence to an incorrect solution has been obtained. In particular, the interpolation polynomials that are complete but not conforming have been found to give satisfactory results. 83 If the interpolation polynomial satisfies all three requirements, the approimate solution converges to the correct solution when we refine the mesh and use an increasing number of smaller elements. In order to prove the convergence mathematically, the mesh refinement has to be made in a regular fashion so as to satisfy the following conditions:. All previous (coarse) meshes must be contained in the refined meshes. 2. The elements must be made smaller in such a way that every point of the solution region can always be within an element. 3. The form of the interpolation polynomial must remain unchanged during the process of mesh refinement. Conditions () and (2) are illustrated in Figure 3.4, in which a two-dimensional region (in the form of a parallelogram) is discretized with an increasing number of triangular elements. From Figure 3.5, in which the solution region is assumed to have a curved boundary, it can be seen that conditions () and (2) are not satisfied if we use elements with straight boundaries. In structural problems, interpolation polynomials satisfying all the convergence requirements always lead to the convergence of the displacement solution from below while nonconforming elements may converge either from below or from above. 2 This statement assumes that the functional (I) corresponding to the problem contains derivatives of ϕ up to the r-th order.
PART 2 Basic Procedure A A H E D B D B G F C (a) Idealization with 2 elements C (b) Idealization with 8 elements A H E D B G F C (c) Idealization with 32 elements 84 FIGURE 3.4 All Previous Meshes Contained in Refined Meshes. (a) Idealization with 6 elements FIGURE 3.5 Previous Mesh Is Not Contained in the Refined Mesh. (b) Idealization with 2 elements NOTES. For any physical problem, the selection of finite elements and interpolation polynomials to achieve C 0 continuity is not very difficult. However, the difficulty increases rapidly when higher order continuity is required. In general, the construction of finite elements to achieve specified continuity of order C 0, C, C 2,..., requires skill, ingenuity, and eperience. Fortunately, most of the time, we would be able to use the elements already developed in an established area such as stress analysis for solving new problems.
CHAPTER 3 2. The construction of an efficient finite element model involves (a) representing the geometry of the problem accurately, (b) developing a finite element mesh to reduce the bandwidth, and (c) choosing a proper interpolation model to obtain the desired accuracy in the solution. Unfortunately, there is no a priori method of creating a reasonably efficient finite element model that can ensure a specified degree of accuracy. Several numerical tests are available for assessing the convergence of a finite element model [3.9, 3.0]. Some adaptive finite element methods have been developed to employ the results from previous meshes to estimate the magnitude and distribution of solution errors and to adaptively improve the finite element model [3. 3.5]. There are four basic approaches to adaptively improve a finite element model: a. Subdivide selected elements (called h-method). b. Increase the order of the polynomial of selected elements (called p-refinement). c. Move node points in fied element topology (called r-refinement). d. Define a new mesh having a better distribution of elements. Various combinations of these approaches are also possible. Determining which of these approaches is the best for a particular class of problems is a comple problem that must consider the cost of the entire solution process. 3.7 LINEAR INTERPOLATION POLYNOMIALS IN TERMS OF GLOBAL COORDINATES The linear interpolation polynomials correspond to simple elements. In this section, we derive the linear interpolation polynomials for the basic one-, two-, and three-dimensional elements in terms of the global coordinates that are defined for the entire domain or body. 3.7. One-Dimensional Simple Element Consider a one-dimensional element (line segment) of length l with two nodes, one at each end, as shown in Figure 3.6. Let the nodes be denoted as i and j and the nodal values of the 85 φ() φ() = α + α 2 Φ j Φ i φ() 0 i j i = ( j i ) j FIGURE 3.6 One-Dimensional Simple Element.
PART 2 Basic Procedure field variable ϕ as Φ i and Φ j. The variation of ϕ inside the element is assumed to be linear as ϕðþ = α + α 2 (3.2) where α and α 2 are the unknown coefficients. By using the nodal conditions and Eq. (3.2), we obtain ϕðþ = Φ i at = i ϕðþ = Φ j at = j Φ i = α + α 2 i Φ j = α + α 2 j 86 The solution of these equations gives α = Φ i j Φ j i l α 2 = Φ j Φ i l 9 >= >; (3.22) where i and j denote the global coordinates of nodes i and j, respectively. By substituting Eq. (3.22) into Eq. (3.2), we obtain ϕðþ = Φ i j Φ j i + Φ j Φ i (3.23) l l This equation can be written, after rearrangement of terms, as where and!ðeþ Φ = ϕðþ = N i ðþφ i + N j ðþφ j = ½NðÞŠΦ! ðeþ Φ i Φ j (3.24) ½NðÞŠ = ½N i ðþ N j ðþš (3.25) N i ðþ = 9 j >= l N j ðþ = i >; l (3.26) = vector of nodal unknowns of elements e (3.27) Note that the superscript e is not used for Φ i and Φ j for simplicity. The linear functions of defined in Eq. (3.26) are called interpolation or shape functions. 3 Note that each interpolation function has a subscript to denote the node to which it is associated. Furthermore, the value of N i () can be seen to be at node i ( = i ) and 0 at node j ( = j ). Likewise, the value of N j () will be 0 at node i and at node j. These represent the common characteristics of interpolation functions. They will be equal to at one node and 0 at each of the other nodes of the element. 3 The original polynomial type of interpolation model ϕ =! T η! α (which is often called the interpolation polynomial or interpolation model of the element) should not be confused with the interpolation functions N i associated with the nodal degrees of freedom. There is a clear difference between the two. The epression! T η! α denotes an interpolation polynomial that applies to the entire element and epresses the variation of the field variable inside the element in terms of the generalized coordinates α i. The interpolation function N i corresponds to the i-th nodal degree of freedom Φ ðeþ i and only the sum i N i Φi ðeþ represents the variation of the field variable inside the element in terms of the nodal degrees of freedom Φ ðeþ i. In fact, the interpolation function corresponding to the i-th nodal degree of freedom (N i ) assumes a value of at node i, and 0 at all the other nodes of the element.
CHAPTER 3 EXAMPLE 3.2 The nodal temperatures of nodes i and j (same as local nodes and 2) of an element in a onedimensional fin are known to be T i = 20 C and T j = 80 C with the -coordinates i = 30 cm and j = 50 cm. Find the following: a. Shape functions associated with the nodal values T i and T j. b. Interpolation model for the temperature inside the element, T(). c. Temperature in the element at = 45 cm. Solution. The shape functions N i () and N j () are given by Eq. (3.26): N i ðþ = j l N j ðþ = i l = 50 = 2:5 0:05 (E.) 50 30 = 30 = 0:05 :5 (E.2) 50 30 2. The interpolation model for the temperature inside the element can be epressed, using Eq. (3.24), as TðÞ = N i ðþt i + N j ðþt j = ð2:5 0:05 Þ20 + ð0:05 :5Þ 80 C 3. The temperature at = 45 cm can be determined from Eq. (E.3) as (E.3) Tð = 45Þ = ð2:5 0:05ð45ÞÞ20 + ð0:05ð45þ :5Þ 80 = 90 C (E.4) EXAMPLE 3.3 A one-dimensional tapered fin element has the nodal coordinates i = 20 mm and j = 60 mm with the area of cross section changing linearly from a value of A i = 20 mm 2 at i to a value of A j = 0 mm 2 at j as shown in Figure 3.7. () Determine the matri of shape functions and (2) epress the area of cross section of the fin element in terms of the shape functions. 87 Approach: () Use the shape functions corresponding to linear variation of the field variable. (2) Epress linear variation of cross-sectional area in terms of shape functions similar to the variation of the field variable. Solution. The linear variation of the field variable ϕ() can be epressed by Eq. (3.2) or, equivalently, by Eq. (3.24): ϕðþ = α + α 2 = N i ðþφ i + N j ðþφ j = ½NðÞŠΦ! ðeþ (E.) where the matri of shape functions [N()] is given by Eq. (3.25): ½NðÞŠ = ½N i ðþ N j ðþš j i j i j i h = 60 i h 20 = 60 i (E.2) 20 60 20 60 20 40 40 A i = 20 mm 2 A ( ) A j = 0 mm 2 0 FIGURE 3.7 A Tapered Fin Element. i = 20 j = 60 (mm) (Continued )
PART 2 Basic Procedure EXAMPLE 3.3 (Continued ) 2. The linear variation of the cross-sectional area of the element can be epressed as AðÞ = β + β 2 where the values of the constants β and β 2 can be found by using the known areas of cross section at the two nodes: Að = i = 20Þ = A i = 20 mm 2 and Að = j = 60Þ = A j = 0 mm 2 (E.3) This gives β = A i j A j i j i = 20ð60Þ 0ð20Þ 60 20 = 25, β 2 = A j A i = 0 20 = 0:25 (E.4) j i 60 20 Using Eq. (E.4), Eq. (E.3) can be epressed in terms of the shape functions as AðÞ = A i N i ðþ + A j N j ðþ ½NðÞŠ A! ðeþ (E.5) where the matri of shape functions, [N()], is given by Eq. (E.2) and! ðeþ A is the vector of nodal areas of cross section of the element:!ðeþ A i A = = 20 mm 2 (E.6) 0 Note Equation (E.4) gives β = 25 and β 2 = 0.25 so that the variation of A() can also be epressed as A j AðÞ = 25 0:25 (E.7) 88 3.7.2 Two-Dimensional Simple Element The two-dimensional simple element is a straight-sided triangle with three nodes, one at each corner, as indicated in Figure 3.8. Let the nodes be labeled as i, j, and k by proceeding counterclockwise from node i, which is arbitrarily specified. Let the global coordinates of the nodes i, j, and k be given by ( i, y i ), ( j, y j ), and ( k, y k ) and the nodal values of the field variable ϕ (, y) byφ i, Φ j, and Φ k, respectively. The variation of ϕ inside the element is assumed to be linear as ϕð, yþ = α + α 2 + α 3 y (3.28) The nodal conditions lead to the system of equations The solution of Eqs. (3.29) yields ϕð, yþ = Φ i at ð = i, y = y i Þ ϕð, yþ = Φ j at ð = j, y = y j Þ ϕð, yþ = Φ k at ð = k, y = y k Þ Φ i = α + α 2 i + α 3 y i Φ j = α + α 2 j + α 3 y j Φ k = α + α 2 k + α 3 y k (3.29) α = 2A ða iφ i + a j Φ j + a k Φ k Þ α 2 = 2A ðb iφ i + b j Φ j + b k Φ k Þ α 3 = 2A ðc iφ i + c j Φ j + c k Φ k Þ (3.30)
CHAPTER 3 φ(, y) Φ j φ(, y) = α + α 2 + α 3 y Φ i Φ k φ(, y) j ( j, y j ) y (, y) i ( i, y i ) k ( k, y k ) FIGURE 3.8 Two-Dimensional Simple Element. where A is the area of the triangle ijk given by A = i y i j y j 2 k y = 2 ð iy j + j y k + k y i i y k j y i k y j Þ (3.3) k a i = j y k k y j a j = k y i i y k a k = i y j j y i b i = y j y k b j = y k y i (3.32) b k = y i y j c i = k j c j = i k c k = j i Substitution of Eq. (3.30) into Eq. (3.28) and rearrangement yields the equation ϕð, yþ = N i ð, yþφ i + N j ð, yþφ j + N k ð, yþφ k = ½Nð, yþšφ! ðeþ (3.33) where ½Nð, yþš = ½N i ð, yþ N j ð, yþ N k ð, yþš (3.34) 89 and!ðeþ Φ = 8 < : Φ i Φ j Φ k N i ð, yþ = 2A ða i + b i + c i yþ N j ð, yþ = 2A ða j + b j + c j yþ N k ð, yþ = 2A ða k + b k + c k yþ 9 = (3.35) = vector of nodal unknowns of element e (3.36) ;
PART 2 Basic Procedure NOTES. The shape function N i (, y) when evaluated at node i ( i, y i ) gives N i ð i, y i Þ = ð 2A a i + b i i + c i y i Þ = 2A (3.37) jy k k y j + i y j i y k + k y i j y i = It can be shown that N i (, y) = 0 at nodes j and k, and at all points on the line passing through these nodes. Similarly, the shape functions N j and N k have a value of at nodes j and k, respectively, and 0 at other nodes. 2. Since the interpolation functions are linear in and y, the gradient of the field variable in or y direction will be a constant. For eample, ϕð, yþ = ½N, ð yþšφ! ðeþ = ðbi Φ i + b j Φ j + b k Φ k Þ/2A (3.38) Since Φ i, Φ j, and Φ k are the nodal values of ϕ (independent of and y), and b i, b j, and b k are constants whose values are fied once the nodal coordinates are specified, ( ø/ ) will be a constant. A constant value of the gradient of ϕ within an element means that many small elements have to be used in locations where rapid changes are epected in the value of ϕ. 90 EXAMPLE 3.4 The temperatures at the nodes of a triangular element are given by T i = 20 F, T j = 270 F, and T k = 250 F. If the nodal coordinates are ( i, y i ) = (50, 30) in, ( j, y j ) = (70, 50) in, and ( k, y k ) = (55, 60) in, determine (a) the shape functions of the element and (b) temperature at the point (, y) = (60, 40) in inside the element. Solution. From the known nodal coordinates, the area of the triangular element and the constants a i, b i, c i,... involved in the shape functions can be determined as A = 2 iy j + j y k + k y i i y k j y i k y j = 50 50 + 70 60 + 55 30 50 60 70 30 55 50 2 ð Þ = 250 in2 a i = j y k k y j = 70 60 55 50 = 450 a j = k y i i y k = 55 30 50 60 = 350 a k = i y j j y i = 50 50 70 30 = 400 b i = y j y k = 50 60 = 0 b j = y k y i = 60 30 = 30 b k = y i y j = 30 50 = 20 c i = k j = 55 70 = 5 c j = i k = 50 55 = 5 c k = j i = 70 50 = 20 The shape functions can be found as N i ð, yþ = ð 2A a i + b i + c i yþ = ð450 0 5yÞ = 2:9 0:02 0:03y 500 N j ð, yþ = 2A a j + b j + c j y = ð 350 + 30 5yÞ = 2:7 + 0:06 0:0y 500 N k ð, yþ = ð 2A a k + b k + c k yþ = ð400 20 + 20yÞ = 0:8 0:04 + 0:04y 500 2. The temperature distribution in the element can be epressed as T, ð y Þ = N i ð, yþt i + N j ð, yþt j + N k ð, yþt k = 20ð2:9 0:02 0:03yÞ+ 270ð 2:7 + 0:06 0:0yÞ+ 250ð0:8 0:04 + 0:04yÞ The temperature at the point (, y) = (60, 40) in can be found as Tð60, 40Þ = 20ð2:9 :2 :2Þ+ 270ð 2:7 + 3:6 0:4Þ+ 250ð0:8 2:4 + :6Þ = 240 F
CHAPTER 3 3.7.3 Three-Dimensional Simple Element The three-dimensional simple element is a flat-faced tetrahedron with four nodes, one at each corner, as shown in Figure 3.9. Let the nodes be labeled as i, j, k, and l, where i, j, and k are labeled in a counterclockwise sequence on any face as viewed from the verte opposite this face, which is labeled as l. Let the values of the field variable be Φ i, Φ j, Φ k, and Φ l and the global coordinates be ( i, y i, z i ), ( j, y j, z j ), ( k, y k, z k ), and ( l, y l, z l ) at nodes i, j, k, and l, respectively. If the variation of ϕ (, y, z) is assumed to be linear, ϕð, y, zþ = α + α 2 + α 3 y + α 4 z (3.39) the nodal conditions ϕ = Φ i at ( i, y i, z i ), ϕ = Φ j at ( j, y j, z j ), ϕ = Φ k at ( k, y k, z k ), and ϕ = Φ l at ( l, y l, z l ) produce the system of equations Φ i = α + α 2 i + α 3 y i + α 4 z i Φ j = α + α 2 j + α 3 y j + α 4 z j Φ k = α + α 2 k + α 3 y k + α 4 z k Φ l = α + α 2 l + α 3 y l + α 4 z l (3.40) Equation (3.40) can be solved and the coefficients α, α 2, α 3,andα 4 can be epressed as α = 6V a iφ i + a j Φ j + a k Φ k + a l Φ l α 2 = 6V b iφ i + b j Φ j + b k Φ k + b l Φ l α 3 = 6V c (3.4) iφ i + c j Φ j + c k Φ k + c l Φ l α 4 = 6V d iφ i + d j Φ j + d k Φ k + d l Φ l where V is the volume of the tetrahedron ijklgiven by i y i z i V = j y j z j (3.42) 6 k y k z k l y l z l 9 z ( i, y i, z i ) i k ( k, y k, z k ) j ( j, y j, z j ) 0 (, y, z ) y FIGURE 3.9 A Three-Dimensional Simple Element.
PART 2 Basic Procedure and j y j z j a i = k y k z k l y l z l y j z j b i = y k z k y l z l j z j c i = k z k l z l j y j d i = k y k l y l (3.43) (3.44) (3.45) (3.46) with the other constants defined by cyclic interchange of the subscripts in the order l, i, j, and k. The signs in front of determinants in Eqs. (3.43) to (3.46) are to be reversed when generating a j, b j, c j, d j, and a l, b l, c l, d l. By substituting Eq. (3.4) into Eq. (3.39), we obtain ϕð, y, zþ = N i ð, y, zþφ i + N j ð, y, zþφ j + N k ð, y, zþφ k + N l ð, y, zþφ l = ½N, ð y, zþšφ! ðþ e (3.47) 92 where and ½N, ð y, zþš = N i ð, y, zþ N j ð, y, zþ N k ð, y, zþ N l ð, y, zþ N i ð, y, zþ = ð 6V a i + b i + c i y + d i zþ N j ð, y, zþ = 6V a j + b j + c j y + d j z N k ð, y, zþ = ð Þ 6V a k + b k + c k y + d k z N l ð, y, zþ = ð 6V a l + b l + c l y + d l zþ! ðþ e Φ = 8 >< >: Φ i Φ j Φ k Φ l 9 >= >; (3.48) (3.49) PROBLEM 3.5 Consider a tetrahedron element with node numbers i, j, k, and l as shown in Figure 3.9. Noting that any of the nodes can be considered as the first (local) node, and the net three (local) nodes must follow a counterclockwise direction as viewed from the first node, enumerate the 2 different ways in which the node (local) numbers of the element can be assigned. Solution If node i is labeled as the first (local) node, the other nodes j, k, and l can be numbered as j, k, l, ork, l, j or l, j, k to satisfy the counterclockwise requirement (as viewed from node i). Similar considerations when node j (or k or l) is labeled as the first (local) node lead to the permissible numbering schemes as indicated in Table 3..
CHAPTER 3 TABLE 3. Proper Node Numbering Schemes for Tetrahedron Element Local Node Local Node 2 Local Node 3 Local Node 4 i j k l i k l j i l j k j k i l j i l k j l k i k i j l k j l i k l i j l i k j l k j i l j i k EXAMPLE 3.6 A tetrahedron element with global node numbers 7, 8, 2, and 7 is shown in Figure 3.0. Determine which of the following (local) numbering sequences satisfy the node numbering convention. 8,2,7,7;7,7,8,2;2,7,8,7 Solution The numbering scheme 8, 2, 7, 7 (also the scheme 2, 7, 8, 7) does not satisfy the node numbering convention because the nodes 2, 7, and 7 (7, 8, and 7) correspond to a clockwise order as seen from node 8 (2). Only the numbering scheme 7, 7, 8, 2 satisfies the node numbering convention because the nodes 7, 8, and 2 correspond to a counterclockwise order as viewed from node 7. 93 2 8 7 7 FIGURE 3.0 A Tetrahedron Element. EXAMPLE 3.7 The nodal coordinates and nodal temperatures of a tetrahedron simple element are given by Node i: ( i, y i, z i ) = (0,0,0) mm, T i = 00 C Node j: ( j, y j, z j ) = (20,0,0) mm, T j = 80 C Node k: ( k, y k, z k ) = (0,30,0)mm, T k = 20 C Node l: ( l, y l, z l ) = (0,0,40) mm, T l = 50 C (Continued )
PART 2 Basic Procedure EXAMPLE 3.7 (Continued ) Epress the temperature variation in the element T(, y, z), in terms of the shape functions. 94 Approach: Epress the linear variation of temperature in the element using Eq. (3.47). Solution The volume of the tetrahedron element is given by Eq. (3.42): i y i z i 0 0 0 V = j y j z j = 20 0 0 = 20 0 0 0 30 0 6 k y k z k 6 0 30 0 6 = 4000 mm3 l y l z l 0 0 40 0 0 40 The constants defined by Eqs. (3.43) (3.46) can be computed as j y j z j 20 0 0 a i = k y k z k = 0 30 0 = 24,000 l y l z l 0 0 40 k y k z k 0 30 0 a j = l y l z l = 0 0 40 = 0 i y i z i 0 0 0 l y l z l 0 0 40 a k = i y i z i = 0 0 0 = 0 j y j z j 20 0 0 i y i z i 0 0 0 a l = j y j z j = 20 0 0 = 0 k y k z k 0 30 0 y j z j 0 0 b i = y k z k = 30 0 = ð200þ = 200 y l z l 0 40 y k z k 30 0 b j = y l z l = 0 40 = 200 y i z i 0 0 y l z l 0 40 b k = y i z i = 0 0 = 0 y j z j 0 0 y i z i 0 0 b l = y j z j = 0 0 = 0 y k z k 30 0 j z j 20 0 c i = k z k = 0 0 = f20ð40þg = 800 l z l 0 40 k z k 0 0 c j = l z l = 0 40 = 0 i z i 0 0 l z l 0 40 c k = i z i = 0 0 = f40ð 20Þg = 800 j z j 20 0
CHAPTER 3 i z i 0 0 c l = j z j k z = 20 0 k 0 0 = 0 j y j 20 0 d i = k y k l y l = 0 30 = 20 30 0 0 f ð Þg = 600 k y k 0 20 d j = l y l i y i = 0 0 0 0 = 0 l y l 0 0 d k = i y i j y j = 0 0 20 0 = 0 i y i 0 0 d l = j y j k y k = 20 0 0 30 = 600 Thus, the shape functions of the element can be epressed, using Eq. (3.48), as ½N, ð y, zþš = N i ð, y, zþ N j ð, y, zþ N k ð, y, zþ N l ð, y, zþ (E.) where N i ð, y, zþ = ð 6V a i + b i + c i y + d i zþ = = 0:05 0:0333 y 0:025 z = N j ð, y, zþ = 6V a j + b j + c j y + d j z N k ð, y, zþ = ð 6V a k + b k + c k y + d k zþ = N l ð, y, zþ = ð 6V a l + b l + c l y + d l zþ = 24,000 200 800 y 600 z 6ð4000Þ ð Þ 6ð4000 Þ ð200 Þ = 0:05 6ð4000Þ ð800 yþ = 0:0333 y 6ð4000Þ ð600 zþ = 0:025 z 95 Thus, the temperature distribution inside the element is given by Eq. (3.47): T, ð y, zþ = N i T i + N j T j + N k T k + N l T l = ð 0:05 0:0333y 0:025zÞ00 + ð0:05þ80 + ð0:0333yþ20 + ð0:025zþ50 = 00 + 0:6667y :25 z C 3.7.4 C -Continuity The one-, two-, and three-dimensional simple elements considered in this section satisfy the following two properties that imply C -continuity:. The shape function corresponding to any specific node, such as node i, varies linearly from a value of at that node i to a value of 0 at each of the remaining nodes of the element. Thus, the shape function N i will have a value of at node i and a value of 0 at each of the remaining nodes of the element. 2. The sum of all the shape functions at any point within the element, including its boundaries, will be equal to.
PART 2 Basic Procedure EXAMPLE 3.8 Show that the sum of the shape functions of a three-noded triangular element is equal to one at any point in the element. Approach: Find the sum of the epressions of the shape functions. Solution Using Eq. (3.35), the sum of the shape functions of the triangular element ijk can be epressed as N i ð, yþ+ N j ð, yþ+ N k ð, yþ = ð 2A a i + b i + c i yþ+ 2A a j + b j + c j y + ð 2A a k + b k + c k yþ = (E.) a i + a j + a k + b i + b j + b k + ci + c j + c k y 2A Using the epressions given in Eq. (3.32), Eq. (E.) can be rewritten as 8 9 N i ð, yþ+ N j ð, yþ+ N k ð, yþ = >< j y k k y j + k y i i y k + i y j j y i >= + y j y k + y k y i + y i y j 2A >: + k j + i k + j i y >; N i ð, y Þ+ N j ð, yþ+ N k ð, yþ = 2A jy k k y j + k y i i y k + i y j j y i (E.2) (E.3) Noting that the epression in the parenthesis is equal to 2A (see Eq. (3.3)), the sum of the shape functions can be seen to be equal to : N i ð, y Þ+ N j ð, yþ+ N k ð, yþ = 2A f2ag = (E.4) 96 3.8 INTERPOLATION POLYNOMIALS FOR VECTOR QUANTITIES In Eqs. (3.2), (3.28), and (3.39), the field variable ϕ has been assumed to be a scalar quantity. In some problems the field variable may be a vector quantity having both magnitude and direction (e.g., displacement in solid mechanics problems). In such cases, the usual procedure is to resolve the vector into components parallel to the coordinate aes and treat these components as the unknown quantities. Thus, there will be more than one unknown (degree of freedom) at a node in such problems. The number of degrees of freedom at a node will be one, two, or three depending on whether the problem is one-, two-, or three-dimensional. The notation used in this book for the vector components is shown in Figure 3.. All the components are designated by the same symbol, Φ, with a subscript denoting the individual components. The subscripts, at any node, are ordered in the sequence, y, z starting with the component. The, y, andz components of the vector quantity (field variable) ϕ are denoted by u, v, and w, respectively. The interpolation function for a vector quantity in a one-dimensional element will be same as that of a scalar quantity since there is only one unknown at each node. Thus, where u ðþ= N i ðþφ i + N j ðþφ j = ½N ðþšφ! e ½N ðþš = N i ðþn j ðþ ( )! ðþ e Φ i Φ = Φ j ðþ (3.50) and u is the component of ϕ (e.g., displacement) parallel to the ais of the element that is assumed to coincide with the ais. The shape functions N i () and N j () are the same as those given in Eq. (3.26).
CHAPTER 3 Φ 2i = v i i ( i, y i ) Φ 2i = u i i i Φ i = u i j j Φ j = u j Φ 2j = v j Φ 2k = v k 0 (a) One-dimensional problem y j ( j, y j ) Φ 2j = u j k ( k, y k ) Φ 2k = u k 0 (b) Two-dimensional problem Φ 3i = w i ( i, y i, z i ) i Φ 3i = v i Φ 3i 2 = u i Φ 3 = w Φ 3j = w j (, y, z ) Φ 3 = v ( j, y j, z j ) j Φ 3j = v j Φ 3k = w k Φ 3 2 = u z Φ 3j 2 = u j k Φ 3k = v k ( k, y k, z k ) 0 y Φ 3k 2 = u k 97 (c) Three-dimensional problem FIGURE 3. Nodal Degrees of Freedom When the Field Variable Is a Vector. For a two-dimensional triangular (simple) element, the linear interpolation model of Eq. (3.33) will be valid for each of the components of ϕ, namely, u and v. Thus, and u, ð y v, ð y Þ = N i ð, yþφ 2i + N j ð, yþφ 2j + N k ð, yþφ 2k (3.5) Þ = N i ð, yþφ 2i + N j ð, yþφ 2j + N k ð, yþφ 2k (3.52) where N i, N j,andn k are the same as those defined in Eq. (3.35); Φ 2i, Φ 2j, and Φ 2k are the nodal values of u (component of ϕ parallel to the ais); and Φ 2i, Φ 2j, and Φ 2k are the nodal values of v (component of ϕ parallel to the y ais). Equations (3.5) and (3.52) can be writteninmatriformas! u, ð yþ ϕ ð, yþ = v, ð yþ = ½N, ð yþšφ! e ðþ (3.53) where ½N, ð yþš = N ið, yþ 0 N j ð, yþ 0 N k ð, yþ 0 0 N i ð, yþ 0 N j ð, yþ 0 N k ð, yþ (3.54)
PART 2 Basic Procedure and 8 Φ 2i 9! ðþ e Φ = >< >: Φ 2i Φ 2j Φ 2j Φ 2k Φ 2k >= = vector of nodal degrees of freedom (3.55) >; Etending this procedure to three dimensions, we obtain for a tetrahedron (simple) element, 8 9! < u, ð y, zþ= ϕ ð, y, zþ = v, ð y, zþ = : ; N, y, z ðþ e (3.56) w, ð y, zþ where 2 N i ð, y, zþ 0 0 N j ð, y, zþ 6 ½N, ð y, zþš = 4 0 N i ð, y, zþ 0 0 0 0 N i ð, y, zþ 0 98 0 0 N k ð, y, zþ 0 N j ð, y, zþ 0 0 N k ð, y, zþ 0 N j ð, y, zþ 0 0 3 0 N l ð, y, zþ 0 0 0 0 N l ð, y, zþ 0 7 5 (3.57) N k ð, y, zþ 0 0 N l ð, y, zþ 8 9 Φ 3i 2 Φ 3i Φ 3i Φ 3j 2! ðþ e Φ = >< Φ 3j Φ 3j Φ 3k 2 Φ 3k Φ 3k Φ 3l 2 >= (3.58) >: Φ 3l Φ 3l and the shape functions N i, N j, N k,andn l are the same as those defined in Eq. (3.48). >; EXAMPLE 3.9 The nodal coordinates of a triangular plate element subjected to in-plane loads follow: Node i: ( i, y i ) = (0, 0) mm Node j: ( j, y j ) = (20, 50) mm Node k: ( k,y k ) = ( 0, 30) mm The in-plane displacement components of the nodes are given by ðu i, v i Þ = ð, Þmm, u j, v j = ð 2, Þmm, ð uk, v k Þ = ð0:5, 0:5Þmm
CHAPTER 3 Epress the variations of u(, y) and v(, y) in the element in terms of the shape functions. Approach: Epress the linear variations of u(, y) and v(, y) in the element using Eqs. (3.5) and (3.52). Solution The shape functions of the element can be derived as (see Problem 3.56) N i ð, yþ = : + 0:02 0:03y (E.) N j ð, yþ = 0:4 + 0:02 + 0:02y N k ð, yþ = 0:3 0:04 + 0:0y The variation of the u-displacement inside the element can be epressed, using Eq. (3.5), as: (E.2) (E.3) uð, yþ = ð: + 0:02 0:03yÞu i + ð 0:4 + 0:02 + 0:02yÞu j + ð0:3 0:04 + 0:0yÞu k = ð: + 0:02 0:03yÞðÞ + ð 0:4 + 0:02 + 0:2yÞð 2Þ + ð0:3 0:04 + 0:0yÞð0:5Þ = 2:05 0:04 0:065y (E.4) Similarly, the variation of the v-displacement inside the element can be epressed, using Eq. (3.52), as: vð, yþ = ð: + 0:02 0:03yÞv i + ð 0:4 + 0:02 + 0:02yÞv j + ð0:3 0:04 + 0:0yÞv k = ð: + 0:02 0:03yÞð Þ + ð 0:4 + 0:02 + 0:02yÞðÞ + ð0:3 0:04 + 0:0yÞð 0:5Þ = :65 + 0:02 + 0:045y (E.5) EXAMPLE 3.0 The global nodal coordinates of a triangular plate element used in the stress analysis of a plate subjected to in-plane loads are given by ( i, y i ) = (50, 30) in, ( j, y j ) = (70, 50) in, and ( k, y k ) = (55, 60) in. If u(, y) and v(, y) are the field variables with nodal unknowns given by (Φ 2i, Φ 2i ) = (u i, v i ), (Φ 2j, Φ 2j ) = (u j, v j ) and (Φ 2k, Φ 2k ) = (u k, v k ), find the matri of shape functions of the element. Solution The matri of shape functions, [N(, y)], is given by Eq. (3.54) with N i (, y), N j (, y), and N k (, y) defined by Eq. (3.35). Using the given nodal coordinates, the shape functions are given by (see Eample 3.4): N i ð, yþ = 2:9 0:02 0:03y N j ð, yþ = 2:7 + 0:06 0:0y N k ð, yþ = 0:8 0:04 + 0:04y 99 3.9 LINEAR INTERPOLATION POLYNOMIALS IN TERMS OF LOCAL COORDINATES The derivation of element characteristic matrices and vectors involves the integration of the shape functions or their derivatives or both over the element. These integrals can be evaluated easily if the interpolation functions are written in terms of a local coordinate system that is defined separately for each element. In this section, we derive the interpolation functions of simple elements in terms of a particular type of local coordinate systems, known as natural coordinate systems. A natural coordinate system is a local coordinate system that permits the specification of any point inside the element by a set of nondimensional numbers whose magnitude lies between 0 and. Usually, natural coordinate systems are chosen such that some of the natural coordinates will have unit magnitude at primary 4 or corner nodes of the element. 4 The nodes located at places other than at corners (e.g., mid-side nodes and interior nodes) are called secondary nodes.
PART 2 Basic Procedure 3.9. One-Dimensional Element The natural coordinates for a one-dimensional (line) element are shown in Figure 3.2. Any point P inside the element is identified by two natural coordinates, L and L 2, which are defined as L = l l L 2 = l 2 l = 2 2 = (3.59) 2 where l and l 2 are the distances shown in Figure 3.2, and l is the length of the element. Since it is a one-dimensional element, there should be only one independent coordinate to define any point P. This is true even with natural coordinates because the two natural coordinates L and L 2 are not independent but are related as L + L 2 = l l + l 2 l = (3.60) A study of the properties of L and L 2 reveals something quite interesting. The natural coordinates L and L 2 are also the shape functions for the line element (compare Eq. (3.59) with Eq. (3.26)). Thus, N i = L, N j = L 2 (3.6) Any point within the element can be epressed as a linear combination of the nodal coordinates of nodes and 2 as 00 = L + 2 L 2 (3.62) where L and L 2 may be interpreted as weighting functions. Thus, the relationship between the natural and the Cartesian coordinates of any point P can be written in matri form as = L (3.63) 2 L 2 or L L 2 2 = = 2 ð 2 Þ l (3.64) If ƒ is a function of L and L 2, differentiation of ƒ with respect to can be performed, using the chain rule, as df d = f L L + f L 2 (3.65) L 2 where, from Eq. (3.59), L = and L 2 2 = (3.66) 2 2 0 Node P Node 2 2 (, 0) (L, L 2 ) (0, ) FIGURE 3.2 Natural Coordinates for a Line Element.