Hutton - Fundamentals of Finite Element Analysis (523155), страница 34
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Figure 6.10 shows athree-node triangular element divided into three areas defined by the nodes andan arbitrary interior point P (x , y) . Note: P is not a node. The area coordinates ofP are defined asL1 =A1AL2 =A2AL3 =A3A(6.40)179Hutton: Fundamentals ofFinite Element Analysis1806. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulation3A2 A1PA321Figure 6.10 Areas used to define areacoordinates for a triangular element.L1 ⫽ 03L1 ⫽312P⬘L1 ⫽ 1P2121(a)(b)Figure 6.11(a) Area A1 associated with either P or P is constant.(b) Lines of the constant area coordinate L1.where A is the total area of the triangle. Clearly, the area coordinates are notindependent, sinceL1 + L2 + L3 = 1(6.41)The dependency is to be expected, since Equation 6.40 expresses the location ofa point in two-dimensions using three coordinates.The important properties of area coordinates for application to triangularfinite elements are now examined with reference to Figure 6.11.
In Figure 6.11a,a dashed line parallel to the side defined by nodes 2 and 3 is indicated. For anytwo points P and P on this line, the areas of the triangles formed by nodes 2 and3 and either P or P are identical. This is because the base and height of any triangle so formed are constants. Further, as the dashed line is moved closer to node1, area A 1 increases linearly and has value A 1 = A , when evaluated at node 1.Therefore, area coordinate L 1 is constant on any line parallel to the side of the triangle opposite node 1 and varies linearly from a value of unity at node 1 to valueof zero along the side defined by nodes 2 and 3, as depicted in Figure 6.11b.
Similar arguments can be made for the behavior of L 2 and L 3 . These observationsHutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.5 Triangular Elementscan be used to write the following conditions satisfied by the area coordinateswhen evaluated at the nodes:Node 1: L 1 = 1L2 = L3 = 0Node 2: L 2 = 1L1 = L3 = 0Node 3: L 3 = 1L1 = L2 = 0(6.42)The conditions expressed by Equation 6.42 are exactly the conditions thatmust be satisfied by interpolation functions at the nodes of the triangular element.
So, we express the field variable as(x , y) = L 1 1 + L 2 2 + L 3 3(6.43)in terms of area coordinates. Is this different from the field variable representation of Equation 6.36? If the area coordinates are expressed explicitly in termsof the nodal coordinates, the two field variable representations are shown to beidentical. The true advantages of area coordinates are seen more readily indeveloping interpolation functions for higher-order elements and performingintegration of various forms of the interpolation functions.6.5.2 Six-Node Triangular ElementA six-node element is shown in Figure 6.12a. The additional nodes 4, 5, and 6 arelocated at the midpoints of the sides of the element.
As we have six nodes, a complete polynomial representation of the field variable is(x , y) = a0 + a1 x + a2 y + a3 x 2 + a4 x y + a5 y 2L1 ⫽ 03L2 ⫽ 0L3 ⫽ 13L1 ⫽L2 ⫽1256(6.44)5612L3 ⫽12L2 ⫽ 1L1 ⫽ 121421(a)L3 ⫽ 04(b)Figure 6.12 Six-node triangular elements: (a) Node numbering convention.(b) Lines of constant values of the area coordinates.181Hutton: Fundamentals ofFinite Element Analysis1826. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulationwhich is ultimately to be expressed in terms of interpolation functions and nodalvalues as6(x , y) =N i (x , y)i(6.45)i=1As usual, each interpolation function must be such that its value is unity whenevaluated at its associated node and zero when evaluated at any of the other fivenodes.
Further, each interpolation is a quadratic function, since the field variablerepresentation is quadratic.Figure 6.12b shows the six-node element with lines of constant values of thearea coordinates passing through the nodes. Using this figure and a bit of logic,the interpolation functions are easily “constructed” in area coordinates. Forexample, interpolation function N 1 (x , y) = N 1 ( L 1 , L 2 , L 3 ) must have value ofzero at nodes 2, 3, 4, 5, and 6. Noting that L 1 = 1/2 at nodes 4 and 6, inclusionof the term L 1 − 1/2 ensures a zero value at those two nodes.
Similarly, L 1 = 0at nodes 2, 3, 4, so the term L 1 satisfies the conditions at those three nodes.Therefore, we propose1N1 = L 1 L 1 −(6.46)2However, evaluation of Equation 6.46 at node 1, where L 1 = 1 , results inN 1 = 1/2 . As N 1 must be unity at node 1, Equation 6.46 is modified to1N 1 = 2L 1 L 1 −= L 1 (2L 1 − 1)(6.47a)2which satisfies the required conditions at each of the six nodes and is a quadraticfunction, since L 1 is a linear function of x and y.Applying the required nodal conditions to the remaining five interpolationfunctions in turn, we obtainN 2 = L 2 (2L 2 − 1)(6.47b)N 3 = L 3 (2L 3 − 1)(6.47c)N 4 = 4L 1 L 2(6.47d)N 5 = 4L 2 L 3(6.47e)N 6 = 4L 1 L 3(6.47f)Using a similarly straightforward procedure, interpolation functions for additional higher-order triangular elements can be constructed.
The 10-node cubicelement is left as an exercise.6.5.3 Integration in Area CoordinatesAs seen in Chapter 5 and encountered again in later chapters, integration of various forms of the interpolation functions over the domain of an element areHutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.5 Triangular Elements183required in formulating element characteristic matrices and load vectors. Whenexpressed in area coordinates, integrals of the formL a1 L b2 L c3 d A(6.48)A(where A is the total area of a triangle defined by nodes 1, 2, 3) must often beevaluated. The relationa!b!c!L a1 L b2 L c3 d A = (2 A)(6.49)(a + b + c + 2)!Ahas been shown [7] to be valid for all exponents a, b, c that are positive integers(recall 0! = 1 ).
Therefore, integration in area coordinates is quite straightforward.EXAMPLE 6.2As will be shown in Chapter 7, the convection terms of the stiffness matrix for a 2-D heattransfer element are of the formki j =h Ni N j d AAwhere h is the convection coefficient and A is element area. Use the interpolation functions for a six-node triangular element given by Equation 6.47 to compute k 24 .■ SolutionUsing Equation 6.47b and 6.47d, we haveN 2 = L 2 (2L 2 − 1)N 4 = 4L 1 L 2so (assuming h is a constant)k 24 = hL 2 (2L 2 − 1)4L 1 L 2 d A = hA8L 1 L 32 − 4L 1 L 22 d AAApplying Equation 6.49, we haveh8L 1 L 32 d A = 8h(2 A)(1!)(3!)(0!)96h A2h A==(1 + 3 + 0 + 2)!720154L 1 L 22 d A = 4h(2 A)16h A2h A(1!)(2!)(0!)==(1 + 2 + 0 + 2)!12015AhATherefore,k 24 =2h A2h A−=01515Hutton: Fundamentals ofFinite Element Analysis1846.
Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulation6.6 RECTANGULAR ELEMENTSRectangular elements are convenient for use in modeling regular geometries, canbe used in conjunction with triangular elements, and form the basis for development of general quadrilateral elements. The simplest of the rectangular family ofelements is the four-node rectangle shown in Figure 6.13, where it is assumedthat the sides of the rectangular are parallel to the global Cartesian axes. By convention, we number the nodes sequentially in a counterclockwise direction, asshown.
As there are four nodes and 4 degrees of freedom, a four-term polynomialexpression for the field variable is appropriate. Since there is no complete fourterm polynomial in two dimensions, the incomplete, symmetric expression(x , y) = a0 + a1 x + a2 y + a3 x y(6.50)is used to ensure geometric isotropy. Applying the four nodal conditions andwriting in matrix form gives 1 x1 y1 x1 y1 a0 1 2 1 x2 y2 x2 y2 a1=(6.51)1 x3 y3 x3 y3 3 a2 41 x4 y4 x4 y4a3which formally gives the polynomial coefficients as a1 x1 y1 x1 y1 −1 1 0 a12 1 x2 y2 x2 y2 =a1xyxy23333 3 a31 x4 y4 x4 y44(6.52)In terms of the nodal values, the field variable is then described by 1 x1 y1 x1 y1 −1 1 2 1 x2 y2 x2 y2 (x, y) = [1 x y x y]{a} = [1 x y x y] 1 x3 y3 x3 y3 3 1 x4 y4 x4 y44(6.53)from which the interpolation functions can be deduced.4 (x4, y4)3 (x3, y3)1 (x1, y1)2 (x2, y2)yxFigure 6.13 A four-noderectangular element defined inglobal coordinates.Hutton: Fundamentals ofFinite Element Analysis6.
Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.6 Rectangular ElementsThe form of Equation 6.53 suggests that expression of the interpolationfunctions in terms of the nodal coordinates is algebraically complex. Fortunately,the complexity can be reduced by a more judicious choice of coordinates. Forthe rectangular element, we introduce the normalized coordinates (also known asnatural coordinates or serendipity coordinates) r and s asr =x − x̄as=y − ȳb(6.54)where 2a and 2b are the width and height of the rectangle, respectively, and thecoordinates of the centroid arex̄ =x1 + x22ȳ =y1 + y42(6.55)as shown in Figure 6.14a.
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