Hutton - Fundamentals of Finite Element Analysis (523155), страница 38

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Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.8 Isoparametric Formulation■ SolutionThe partial derivatives of x and y with respect to r and s per Equations 6.56 and 6.80 are4∂x∂ Ni1=x i = [−(1 − s)x 1 + (1 − s)x 2 + (1 + s)x 3 − (1 + s)x 4 ]∂r∂r4i =14∂ Ni1∂y=yi = [−(1 − s) y1 + (1 − s) y2 + (1 + s) y3 − (1 + s) y4 ]∂r∂r4i =141∂ Ni∂x=x i = [−(1 − r )x 1 − (1 + r )x 2 + (1 + r )x 3 + (1 − r )x 4 ]∂s∂s4i =14∂ Ni1∂y=yi = [−(1 − r ) y1 − (1 + r ) y2 + (1 + r ) y3 + (1 − r ) y4 ]∂s∂s4i =1The Jacobian matrix is then[J] =1 (1 − s)(x 2 − x 1 ) + (1 + s)(x 3 − x 4 )4 (1 − r )(x 4 − x 1 ) + (1 + r )(x 3 − x 2 )(1 − s)( y2 − y1 ) + (1 + s)( y3 − y4 )(1 − r )( y4 − y1 ) + (1 + r )( y3 − y2 )Note that finding the inverse of this Jacobian matrix in explicit form is not an enviable task.

The task is impossible except in certain special cases. For this reason, isoparametric element formulation is carried out using numerical integration, as discussed inSection 6.10.The isoparametric formulation is by no means limited to linear parent elements. Many higher-order isoparametric elements have been formulated andused successfully [1]. Figure 6.23 depicts the isoparametric elements corresponding to the six-node triangle and the eight-node rectangle. Owing to themapping being described by quadratic functions of the parent elements, theresulting elements have curved boundaries, which are also described by quadratic functions of the global coordinates. Such elements can be used to closelyapproximate irregular boundaries.

However, note that curved elements do not,in general, exactly match a specified boundary curve.(a)(b)Figure 6.23 Isoparametric mapping of quadratic elements into curved elements:(a) Six-node triangle. (b) Eight-node rectangle.201Hutton: Fundamentals ofFinite Element Analysis2026. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulation6.9 AXISYMMETRIC ELEMENTSMany three-dimensional field problems in engineering exhibit symmetry aboutan axis of rotation. Such problems, known as axisymmetric problems, can besolved using two-dimensional finite elements, which are most conveniently described in cylindrical (r, ␪, z) coordinates.

The required conditions for a problemto be axisymmetric are as follows:1. The problem domain must possess an axis of symmetry, which isconventionally taken as the z axis; that is, the domain is geometricallya solid of revolution.2. The boundary conditions are symmetric about the axis of revolution;thus, all boundary conditions are independent of the circumferentialcoordinate ␪ .3. All loading conditions are symmetric about the axis of revolution; thus,they are also independent of the circumferential coordinate.In addition, the material properties must be symmetric about the axis of revolution.

This condition is, of course, automatically satisfied for isotropic materials.If these conditions are met, the field variable ␾ is a function of radial andaxial (r, z) coordinates only and described mathematically by two-dimensionalgoverning equations.Figure 6.24a depicts a cross section of an axisymmetric body assumed to bethe domain of an axisymmetric problem. The cross section could represent thewall of a pressure vessel for stress or heat transfer analysis, an annular regionof fluid flow, or blast furnace for steel production, to name a few examples. In(r3, z3)zz␪(r2, z2)r(a)r(r1, z1)(b)Figure 6.24(a) An axisymmetric body and cylindrical coordinates. (b) A threenode triangle in cylindrical coordinates at an arbitrary value ␪.Hutton: Fundamentals ofFinite Element Analysis6.

Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.9Axisymmetric ElementsFigure 6.24b, a three-node triangular element is shown having nodal coordinates(r i , z i ) . In the axisymmetric case, the field variable is discretized as3␾(r, z) =N i (r, z)␾i(6.94)i=1where the interpolation functions N i (r, z) must satisfy the usual nodal conditions.Noting that the nodal conditions are satisfied by the interpolation functionsdefined by Equation 6.37 if we simply substitute r for x and z for y, the interpolation functions for the axisymmetric triangular element are immediatelyobtained. Similarly, the interpolation functions in terms of area coordinates arealso applicable.Since, by definition of an axisymmetric problem, the problem, therefore itssolution, is independent of the circumferential coordinate ␪, so must be the interpolation functions.

Consequently, any two-dimensional element and associatedinterpolation functions can be used for axisymmetric elements. What is the difference? The axisymmetric element is physically three dimensional. As depictedin Figure 6.25, the triangular axisymmetric element is actually a prism of revolution. The “nodes” are circles about the axis of revolution of the body, and thenodal conditions are satisfied at every point along the circumference defined bythe node of a two-dimensional element.

Although we use a triangular elementfor illustration, we reiterate that any two-dimensional element can be used toformulate an axisymmetric element.As is shown in subsequent chapters in terms of specific axisymmetricproblems, integration of various functions of the interpolation functions over thevolume are required for element formulation. Symbolically, such integrals arerepresented asF (r, ␪, z) =f (r, ␪, z) dV =f (r, ␪, z)r dr d␪ dz(6.95)Vzr␪Figure 6.25 A three-dimensionalrepresentation of an axisymmetricelement based on a three-nodetriangular element.203Hutton: Fundamentals ofFinite Element Analysis2046. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulationd␪dzrdrFigure 6.26 Differential volumein cylindrical coordinates.where V is the volume of an element and dV = r dr d␪ dz is the differential volume depicted in Figure 6.26.

For axial symmetry, the integrand is independent ofthe circumferential coordinate ␪, so the integration indicated in Equation 6.95becomesF (r, ␪, z) = F (r, z) = 2␲f (r, z)r dr dz(6.96)AEquation 6.96 shows that the integration operations required for formulation ofaxisymmetric elements are distinctly different from those of two-dimensionalelements, even though the interpolation functions are essentially identical. Asstated, we show applications of axisymmetric elements in subsequent chapters.Also, any two-dimensional element can be readily converted to an axisymmetricelement, provided the true three-dimensional nature of the element is taken intoaccount when element characteristic matrices are formulated.EXAMPLE 6.5In following chapters, we show that integrals of the formN i N j dVVwhere N i , N j are interpolation functions and V is element volume, must be evaluated informulation of element matrices.

Evaluate the integral with i = 1, j = 2 for an axisymmetric element based on the three-node triangle using the area coordinates as the interpolation functions.■ SolutionFor the axisymmetric element, we use Equation 6.96 to writeN i N j dV = 2␲VN i N j r dr dz = 2␲AL i L j r dr dzAwhere A is the element area. Owing to the presence of the variable r in the integrand, theintegration formula, Equation 6.49, cannot be applied directly.

However, we can expressr in terms of the nodal coordinates r 1 , r 2 , r 3 and the area coordinates asr = L 1 r1 + L 2 r2 + L 3 r3Hutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.9Then, noting that dr dz = d A , we haveA205L i L j r dr dz = 2␲2␲Axisymmetric ElementsL i L j ( L 1 r1 + L 2 r2 + L 3 r3 ) d AAwhich is of the appropriate form for application of the integration formula. For i = 1,j = 2, the integral becomesL 1 L 2 r dr dz = 2␲2␲AL 1 L 2 ( L 1 r1 + L 2 r2 + L 3 r3 ) d AAL 21 L 2 d A + 2␲r 2= 2␲r 1AL 1 L 22 d A + 2␲r 3AL1L2L3 dAAApplying the integration formula to each of the three integrals on the right,2␲L 1 L 2 r dr dzA= 4␲ A r 1(2!)(1!)(0!)(1!)(2!)(0!)(1!)(1!)(1!)+ r2+ r3(2 + 1 + 0 + 2)!(1 + 2 + 0 + 2)!(1 + 1 + 1 + 2)!␲A2r 2r32r 1=++(2r 1 + 2r 2 + r 3 )= 4␲ A12012012030The integration technique used in Example 6.5 is also applicable to higherorder, straight-sided triangular elements, as shown in the next example.EXAMPLE 6.6For an axisymmetric element based on the six-node, quadratic triangular element havinginterpolation functions given by Equation 6.47, evaluate the integralI =N 2 N 4 dVV■ SolutionUsing Equation 6.93,I =N 2 N 4 dV = 2␲VN 2 N 4 r dr dz = 2␲AL 2 (2L 2 − 1)(4L 1 L 2 )r dr dzANow observe that, even though the interpolation functions vary quadratically over theelement area, the area coordinates, by definition, vary linearly.

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