Hutton - Fundamentals of Finite Element Analysis (523155), страница 40
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In finite element analysis, integrals of the forms1 1I =f (r, s) dr ds−1 −1(6.103)1 1 1I =f (r, s, t ) dr ds dt−1 −1 −1are frequently encountered. Considering the first of Equation 6.103, we integratefirst with respect to r (using the Gaussian technique) to obtain1 1I =f (r, s) dr ds =−1 −11 n−11[W i f (r i , s)] ds =i=1g(s) ds(6.104)−1which, in turn, is integrated via quadrature to obtainI =m(6.105)W j g(s j )j=1combining Equations 6.98 and 6.99, we find1 1I =f (r, s) dr ds =−1 −1m nj=1 i=1W j W i f (r i , s j )(6.106)Hutton: Fundamentals ofFinite Element Analysis6. Interpolation Functionsfor General ElementFormulationText6.10 Numerical Integration: Gaussian Quadrature© The McGraw−HillCompanies, 2004211Equations 6.104–6.106 show that, for integration in two dimensions, we simplyapply the Gaussian procedure sequentially, just as when we integrate formally.At each step, if we desire an exact result, the number of sampling points (hence,the weighting factors) is chosen by the order of the respective polynomial termsin r and s.
The numerical result is exact. In practice, the number of samplingpoints is chosen to be the same for each integration step, with the higher-orderprevailing, as illustrated in the following example.EXAMPLE 6.8Use Gaussian quadrature to obtain an exact value for the integral1 1I =(r 3 − 1)(s − 1) 2 dr ds−1 −1■ SolutionConsidering first the integration with respect to r, we have a cubic order that requires twosampling points, which from Table 6.1 are given as r i = ±0.5773503, and each of thecorresponding weighting factors is unity. Similarly, for the integration with respect to s,the order is quadratic so the factors are the same.
(In the following solution,√we note, forsimplicity of presentation, that the sampling points are numerically equal to 3/3.) Equation 6.106 is then, for this example,I =22 W j Wi f (ri , s j )j =1 i =1 √ √√2 √ 3233333−=−1−1−1 +−13333 √ √ √2 √ 323333−−− 3+− 1− 1−1 +− 1 = 5.333333333333And, as again may be verified by direct integration, the result is exact.An analogous procedure shows that, for the three-dimensional case,1 1 1l m nI =f (r, s, t ) dr ds dt ≈W k W j W i f (r i , s j , tk )(6.107)−1 −1 −1k=1 j=1 i=1As in the case of one- and two-dimensional integration, the approximation described by Equation 6.107 can be made to give an exact value for a polynomialintegrand if the sampling points are selected as previously described. The following example illustrates the procedure.Hutton: Fundamentals ofFinite Element Analysis2126.
Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element FormulationEXAMPLE 6.9Use Gaussian quadrature to obtain an exact value of the integral1 1 1I =r 2 (s 2 − 1)(t 4 − 2) dr ds dt−1 −1 −1■ SolutionIn this case, we have a √quadratic polynomial in r, so two sampling points are required, withri = ±0.5773503 = ± 3/3 and W i = 1 per Table√ 6.1.
The quadratic in s similarly requirestwo sampling points, s j = ±0.5773503 = ± 3/3, with weighting factors W j = 1.0. Forthe quartic function in t, three sampling points are required for exactness and the valuesand weighting factors per Table 6.1 aret1 = 0.0W 1 = 0.8888889t2 = 0.7745967W 2 = 0.5555556t3 = −0.7745967W 3 = 0.5555556For an exact solution, we then haveI =2 23 W k W j W i f (r i , s j , tk )k=1 j =1 i =1so a total of 12 terms is required. The required calculations are summarized in Table 6.2.So we obtain1 1 1I =r 2 (s 2 − 1)(t 4 − 2) dr ds dt = 3.2−1 −1 −1and this result is exact.Table 6.2 Sampling Points, Weighting Factors, and Calculations for Example 6.9PointrisjtkWiWjWkf(ri, sj, tk)CumulativeSum1234567891011120.577350.577350.577350.577350.577350.57735−0.57735−0.57735−0.57735−0.57735−0.57735−0.577350.577350.577350.57735−0.57735−0.57735−0.577350.577350.577350.57735−0.57735−0.57735−0.5773500.774597−0.77459700.774597−0.77459700.774597−0.77459700.774597−0.7745971111111111111111111111110.888888890.555555560.555555560.888888890.555555560.555555560.888888890.555555560.555555560.888888890.555555560.555555560.3950620.2024690.2024690.3950620.2024690.2024690.3950620.2024690.2024690.3950620.2024690.2024690.3950620.5975310.81.1950621.3975311.61.9950622.1975312.42.7950622.9975313.2Hutton: Fundamentals ofFinite Element Analysis6.
Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.10 Numerical Integration: Gaussian Quadrature213Lest the reader be lulled into the false impression that numerical integrationcan always be made exact, we present the following example to illustrate that(1) numerical integration is not always exact but (2) numerical integration converges to exactness as the number of integration points is increased.EXAMPLE 6.10Evaluate the integral1I =−1r2 − 1(r + 3) 2drusing Gaussian integration with one, two, and three integration points.■ SolutionThe integration procedure requires that we evaluate the integrand at discrete points andsum the results as follows (we do not present all the calculation details: the reader is urgedto check our calculations)One Integration PointW1 = 2r1 = 0−1I ≈2= −0.6666673Two Integration PointsW1 = W2 = 1r1 =√33r2 = −r1I ≈ −0.16568Three TermsW 1 = 0.8888888r1 = 0W 2 = 0.5555555 .
. .W 3 = −0.5555555 . . .r 2, 3 = ±0.7745966I ≈ −0.15923Continuing with the four-point integration results inI ≈ −0.15891Hence, we see a convergence to a value. If the exact result is obtained by formal integration, the result is I = −0.15888. As illustrated by this example, the Gaussian integrationprocedure is not always exact but does, indeed, converge to exact solutions as the number of sampling or Gauss points is increased.Hutton: Fundamentals ofFinite Element Analysis2146. Interpolation Functionsfor General ElementFormulationCHAPTER 6Text© The McGraw−HillCompanies, 2004Interpolation Functions for General Element Formulation6.11 CLOSING REMARKSThe developments presented in this chapter show how interpolation functions forone-, two-, and three-dimensional elements can be obtained via a systematic procedure.
Also, the algebraically tedious procedure can often be bypassed usingintuition and logic when natural coordinates are used. The interpolation functions discussed are standard polynomial forms but by no means exhaustive of theinterpolation functions that have been developed for use in finite element analysis. For example, we make no mention of Legendre or Hermite polynomials forapplication to finite elements. These and other forms are well known and discussed in the finite element literature. As the objective of this text is to present thefundamentals of finite element analysis, the material of this chapter is intended tocover the basic concepts of interpolation functions without proposing to be comprehensive. The treatment here is intended to form a basis for formulation offinite element models of various physical problems in following chapters.
In general, every element and the associated interpolation functions discussed here canbe applied to specific problems, as is illustrated in the remainder of the text.REFERENCES1.Huebner, K. H., and E. A. Thornton. The Finite Element Method for Engineers,2nd ed. New York: John Wiley and Sons, 1982.2.
Bathe, K. J. Finite Element Procedures. Englewood Cliffs, NJ: Prentice-Hall,1996.3. Burnett, D. S. Finite Element Analysis. Reading, MA: Addison-Wesley, 1987.4. Zienkiewicz, O. C. The Finite Element Method, 3rd ed. London: McGraw-Hill,1977.5. Stasa, F. L. Applied Finite Element Analysis for Engineers. New York: Holt,Rinehart, and Winston, 1985.6.
ANSYS Element Reference Manual. Houston, PA: Swanson Analysis Systems, 1999.7. Eisenberg, M. A., and L. E. Malvern. “On Finite Element Integration in NaturalCoordinates.” International Journal for Numerical Methods in Engineering 7(1973).8. Logan, D. L. A First Course in the Finite Element Method Using ALGOR.Boston: PWS Publishing Company, 1997.9. Kreyszig, E. Advanced Engineering Mathematics, 8th ed. New York: John Wiley,1999.10.
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