Hutton - Fundamentals of Finite Element Analysis (523155), страница 41
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Stroud, A. H., and D. Secrest. Gaussian Quadrature Formulas. Englewood Cliffs,NJ: Prentice-Hall, 1966.PROBLEMS6.1Verify that Equation 6.6, with the interpolation functions given by Equation 6.16,allows for rigid body translation of a two-dimensional beam element. (Hint: Forthe rigid body translation, v1 = v2 , 1 = 2 = 0 .)Hutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004TextProblems6.26.36.46.5Verify that Equation 6.6, with the interpolation functions given by Equation 6.16,allows for rigid body rotation of a two-dimensional beam element.Consider the case of a two-dimensional beam element subjected to pure bending.Show that Equation 6.6 results in d2 v/dx 2 = constant, as required.Show that shear force in a two-dimensional beam element is constant regardlessof the values of the nodal displacements.Show that interpolations function N 1 (s) in Example 6.1 is the same asN 1 (s) =(s − s2 )(s − s3 )(s − s4 )(s1 − s2 )(s1 − s3 )(s1 − s4 )(Note: N 1 (s) , as just given, is a particular case of the Lagrange polynomial oforder nnL j (s) =i =1,i = js − sis j − si 6.66.7where the symbol indicates the product of all terms.)Use the definition of the Lagrange polynomial given in Problem 6.5 to find thecubic polynomials corresponding to j = 2 and j = 3 .Use the Lagrange polynomial to determine the interpolation functions for thefive-node line element shown in Figure P6.7.12345xx1 ⫽ 2x2 ⫽ 2.25x3 ⫽ 2.5x4 ⫽ 2.75x5 ⫽ 3Figure P6.76.8The quadratic polynomialP (x , y) = a0 + a1 x 2 + a2 x y + a3 y 26.96.106.116.126.13is incomplete but symmetric.
Is this function suitable for representing the fieldvariable in a four-node rectangular element having 1 degree of freedom pernode? Explain.Determine all possible symmetric, incomplete fourth-order polynomials in twodimensions. Which of these might be useful for C 0 problems in finite elementanalysis?Repeat Problem 6.9 for cubic polynomials in three dimensions.Derive Equation 6.38.Show that Equations 6.36 and 6.43 are identical.Using area coordinates, develop interpolation functions for the 10-nodetriangular element shown in Figure P6.13.
Note that the nodes are equally spacedon their edges.215Hutton: Fundamentals ofFinite Element Analysis2166. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulation37861092541Figure P6.136.146.15Show that node 10 in Figure P6.13 is located at the element centroid.Use the integration formula for area coordinates to evaluate the followingintegrals in terms of total area A:L 22 (2L 2 − 1) 2 d Aa.A4L 1 L 2 L 3 (2L 1 − 1) d Ab.AL 31 L 2 L 3 d Ac.AL 1 L 32 L 3 d Ad.A6.16The interpolation functions for the four-node rectangular element as given byEquation 6.56 are such that4N i (r, s) = 1i =16.176.18What is the significance of this observation?Show that Equation 6.56 is such that constant values of the partial derivatives∂ /∂ r, ∂ /∂ s are possible in an element.
Note that this ensures satisfaction ofthe completeness requirement for C 0 problems.Two three-node triangular elements share a common boundary, as shown inFigure P6.18. Show that the field variable is continuous across thisinterelement boundary.Hutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004TextProblems3412Figure P6.186.19Repeat Problem 6.18 for the two four-node rectangular elements shown inFigure P6.19. The interpolation functions are as given by Equation 6.56.436125Figure P6.196.206.21Examine the behavior of the partial derivatives ∂ /∂ r, ∂ /∂ s across and alongthe interelement boundary defined by nodes 2 and 3 in Problem 6.19.Determine the continuity conditions on and the partial derivatives ∂ /∂ r,∂ /∂ s on the boundary between the two eight-node rectangular elements shownin Figure P6.21.
The interpolation functions are given by Equation 6.59.91061112137128345Figure P6.216.22Verify that the interpolation functions given by Equations 6.59 satisfy thecondition8N i (r, s) = 1i =16.236.24Verify that Equation 6.62 correctly defines the volume of a tetrahedron.In Example 6.5, the expression r = L 1 r 1 + L 2 r 2 + L 3 r 3 is used to allowapplication of the integration formula in area coordinates.
Prove that thisexpression is correct.217Hutton: Fundamentals ofFinite Element Analysis2186. Interpolation Functionsfor General ElementFormulationCHAPTER 66.256.26© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element FormulationUse the integration formula of Equation 6.49 to confirm the result of Example 6.6.Use the integration formula for area coordinates to show thatA=dA =dx dy = 1 1− L2dL 1 dL 2A6.2700Consider the isoparametric quadrilateral element in Figure P6.27. Map the pointr = 0.5, s = 0 in the parent element to the corresponding physical point in thequadrilateral element.3 (2.4, 2.6)4 (2, 2.5)y2 (2.5, 2.1)x1 (2, 2)Figure P6.276.286.296.30Again referring to the element in Figure P6.27, map the line r = 0 in the parentelement to the physical element. Plot the mapping on a scaled drawing of thequadrilateral element.Repeat Problem 6.28 for the line s = 0 in the parent element.Consider the two-node line element in Figure P6.30 with interpolation functionsN 1 (r ) = 1 − rN 2 (r ) = rUsing this as the parent element, examine the isoparametric mappingx = N 1 (r )x 1 + N 2 (r )x 2for arbitrary values x 1 and x 2 such that x 1 < x 2 .a.
What has been accomplished by the mapping?b. Determine the Jacobian matrix for the transformation.r1r1 ⫽ 02r2 ⫽ 1Figure P6.306.31Consider the three-node line element in Figure P6.31 with interpolation functionsN 1 (r ) = (2r − 1)(r − 1)N 2 (r ) = 4r (1 − r )N 3 (r ) = r (2r − 1)Hutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004TextProblemsUse the element as the parent element in the isoparametric mappingx = N 1 (r )x 1 + N 2 (r )x 2 + N 3 (r )x 3with x 1 < x 2 < x 3 but otherwise arbitrary nodal coordinates.a. How does the x coordinate vary between nodes of the isoparametric element?b. Has the basic element geometry changed from that of the parent element?c. Determine the Jacobian matrix for the transformation.d. Find the inverse of the Jacobian matrix.r2r2 ⫽ 0.51r1 ⫽ 03r3 ⫽ 1Figure P6.316.32Consider again the three-node line element of Figure P6.31 as the parent elementfor the two-dimensional mapping defined byx = N 1 (r )x 1 + N 2 (r )x 2 + N 2 (r )x 3y = N 1 (r ) y1 + N 2 (r ) y2 + N 2 (r ) y36.336.34where (x i , yi ) are the coordinates of node i.a.
Let (x 1 , y1 ) = (1, 1), (x 2 , y2 ) = (2, 3), (x 3 , y3 ) = (4, 2) and plot thegeometry of the isoparametric element to scale.b. Could the resulting isoparametric element be used in a finite element analysisof heat conduction (refer to Chapter 5) through a curved solid with ideallyinsulated surfaces? Explain your answer.Show by analytical integration that the result given in Example 6.9 is exact.Use Gaussian quadrature to obtain exact values for the following integrals.Verify exactness by analytical integration.3(x 2 − 1) dxa.06( y 3 + 2y) dyb.11(4r 3 + r ) drc.−11(r 4 + 3r 2 ) drd.−11(r 4 + r 3 + r 2 + r + 1) dre.−1219Hutton: Fundamentals ofFinite Element Analysis2206.
Interpolation Functionsfor General ElementFormulationCHAPTER 66.35Text© The McGraw−HillCompanies, 2004Interpolation Functions for General Element FormulationUse Gaussian quadrature to obtain exact values for the following integrals in twodimensions. Verify exactness by analytical integration.1 2a.x y dx dy001 1(r 2 + 2r s + s 2 ) dr dsb.−1 −11 1(r 3 − 1)(s 2 + s) dr dsc.−1 −11 1(r 5 − 2r 3 )(s 3 + s) dr dsd.−1 −11 1(6r 4 − 1)(s 2 + s + 1) dr dse.−1 −16.36Use Gaussian quadrature to obtain exact values for the following threedimensional integrals.
Verify exactness by analytical integration.2 2 2a.x yz dx dy dz0001 1 1t (r − 1)(s 2 − 2) dr ds dtb.−1 −1 −11 1 1t 3 r 3 s 2 dr ds dtc.−1 −1 −11 1 1t 2 (r − 2) 4 (s 2 − 1) dr ds dtd.−1 −1 −11 1 1(r 3 − r )(s + 4)(t 2 − 1) dr ds dte.−1 −1 −16.37Evaluate each of the following integrals using two-point Gaussian quadrature.Compare each result with the corresponding analytical solution.1cos 2 r dra.−1Hutton: Fundamentals ofFinite Element Analysis6.
Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004TextProblems1b.−1r2rdr+11sin r cos r drc.−11 1d.−1 −16.386.39r 2sdr ds(r 3 + s 2 )Repeat Problem 6.37 using three-point Gaussian quadrature. Are the resultsconverging to the exact solutions?The integral1(r 3 + 2r 2 + 1) dr−1can be evaluated exactly by using two-point Gaussian quadrature. Examine theeffect on the result if three-point integration is applied.221Hutton: Fundamentals ofFinite Element Analysis77.
Applications in HeatTransferText© The McGraw−HillCompanies, 2004C H A P T E RApplications inHeat Transfer7.1 INTRODUCTIONIn this chapter, the Galerkin method introduced in Chapter 5 and the interpolation function concepts of Chapter 6 are applied to several heat transfersituations. Conduction with convection is discussed for one-, two-, and threedimensional problems. Boundary conditions and forcing functions includeprescribed heat flux, insulated surfaces, prescribed temperatures, and convection.
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