Hutton - Fundamentals of Finite Element Analysis (523155), страница 39
Текст из файла (страница 39)
Since the element sidesare straight, the radial coordinate can still be expressed asr = L 1 r1 + L 2 r2 + L 3 r3Therefore, we haveI = 2L 2 (2L 2 − 1)(4L 1 L 2 )( L 1 r 1 + L 2 r 2 + L 3 r 3 ) d AAHutton: Fundamentals ofFinite Element Analysis2066. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element FormulationorI = 16A− 8L 21 L 32 r 1 + L 1 L 42 r 2 + L 1 L 32 L 3 r 3 d AL 21 L 22 r 1 + L 1 L 32 r 2 + L 1 L 22 L 3 r 3 d AAApplication of the integration formula, Equation 6.49, to each of the six integrals represented here (left as an exercise), we findI =A(6r 2 − 4r 1 − 2r 3 )3156.10 NUMERICAL INTEGRATION:GAUSSIAN QUADRATUREPrevious chapters show that integration of various functions of the field variableare required for formulation of finite element characteristic matrices.
Chapter 5reveals that the Galerkin method requires integration over the element domain(and, as seen, physical volume), once for each interpolation function (trial solution). In fact, an integration is required to obtain the value of every component ofthe stiffness matrix of a finite element. In addition, integrations are required toobtain nodal equivalents of nonnodal loadings.In this chapter, we focus primarily on polynomial representations of the discretized representations of the field variable.
In subsequent formulation of element characteristic matrices, we are faced with integrations of polynomial forms.A simple polynomial is relatively easy to integrate in closed form. In many cases,however, the integrands are rational functions, that is, ratios of polynomials; andthese are quite tedious to integrate directly. In either case, in the finite elementcontext, where large numbers of elements, hence huge numbers of integrations,are required, analytical methods are not efficient. Finite element software packages do not incorporate explicit integration of the element formulation equations.Instead, they use numerical techniques, the most popular of which is Gaussian(or Gauss-Legendre) quadrature [10].The concept of Gaussian quadrature is first illustrated in one dimension inthe context of an integral of the formx2I =h(x ) dx(6.97)x1Via the change of variable r = ax + b , Equation 6.97 can be converted to1I =f (r ) dr(6.98)−1Hutton: Fundamentals ofFinite Element Analysis6.
Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.10 Numerical Integration: Gaussian Quadraturewith dr = a dx . The coefficients a and b are determined so that the integrationlimits become minus and plus unity. This is conventional for the numerical integration procedure and also accords nicely with the range of the natural coordinates of many of the elements discussed in this chapter.Per the Gaussian integration procedure, the integration represented by Equation 6.98 can be approximated byI =m(6.99)W i f (r i )i=1where W i are Gaussian weighting factors and r i are known as sampling pointsor Gauss points. The weighting factors and sampling points are determined [9]to minimize error, particularly in terms of polynomial functions.
Of particularimport in finite element analysis, a polynomial of order n can be exactly integrated. Referring to Equation 6.99, use of m sampling points and weightingfactors results in an exact value of the integral for a polynomial of order 2m − 1 ,if the sampling points and weighting factors are chosen in accordance withTable 6.1. This means, for example, that a cubic polynomial can be exactly integrated by Equation 6.99, using only two sampling points and evaluating theintegrand at those points, multiplying by the weighting factors, and summingthe results.To illustrate how the sampling points and weighting factors are determined,we formally integrate a general polynomial in one dimension as1(a0 + a1r + a2r 2 + a3r 3 + · · · + an r n ) dr−1= 2a0 +222a2 + a4 + · · · +an35n+1(6.100)Table 6.1 Sampling points and weighting factors for Gaussian quadrature numericalintegration of1−1f ( r ) dr ≈mi =1Wi f ( r i ) .
This is an abridged table, giving valuessufficient for exact integration of a polynomial of order seven or lessmriWi120.00.577350269189626. . .−0.577350269189626. . .0.00.774596669241483−0.7745966692414830.339981043583856−0.3399810435838560.861136311590453−0.8611363115904532.01.01.00.8888888888888890.5555555555555560.5555555555555560.6521451548625260.6521451548625260.3478548451374540.34785484513745434207Hutton: Fundamentals ofFinite Element Analysis2086. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulationand we observe that, owing to the symmetry of the integration limits, all oddpowers integrate to zero. Also note that we assume here that n is an even integer.The approximation to the integral per Equation 6.99 ismI ≈W i f (r i ) = W 1 a0 + a1r 1 + a2r 12 + a3r 13 + · · · + an r 1ni=1+ W 2 a0 + a1r 2 + a2r 22 + a3r 23 + · · · + an r 2n+ W 3 a0 + a1r 3 + a2r 32 + a3r 33 + · · · + an r 3n(6.101)...+ W m a0 + a1r m + a2r m2 + a3r m3 + · · · + an r mnComparing Equations 6.100 and 6.101 in terms of the coefficients a j of the polynomial, the approximation of Equation 6.101 becomes exact ifmWi = 2i=1mWi ri = 0i=1mW i r i2 =i=1mi=1mi=1m23W i r i3 = 0W i r i4 =...(6.102)25W i r in−1 = 0i=1mi=1W i r in =2n+1where m is the number of sampling (Gauss) integration points.Equation 6.102 represents n equations in 2m + 1 unknowns.
The unknownsare the weighting factors W i , the sampling point values r i , and most troublesome, the number of sampling points m. While we do not go into the completetheory of Gaussian quadrature, we illustrate by example how the sampling pointsand weights can be determined using both the equations and logic. First, note thatthe equations corresponding to odd powers of the polynomial indicate a zerosummation. Second, note that the first equation is applicable regardless of theorder of the polynomial; that is, the weighting factors must sum to the value of2 if exactness is to be achieved.Hutton: Fundamentals ofFinite Element Analysis6.
Interpolation Functionsfor General ElementFormulation© The McGraw−HillCompanies, 2004Text6.10 Numerical Integration: Gaussian Quadrature209If, for example, we have a linear polynomial n = 1 , the first two of Equation 6.102 are applicable and lead to the conclusions that we need only one sampling point and that the appropriate values of the weighting factor and samplepoint to satisfy the two equations (in this case) are W 1 = 2 and r 1 = 0 .
Next, consider the case of a cubic polynomial, n = 3 . In this case, we havemWi = 2i=1mi=1mi=1Wi ri = 0W i r i2 =23representing three equations in 2m + 1 unknowns. If we let m = 1 , the first twoequations lead to W 1 = 2, r 1 = 0 , but the third equation cannot be satisfied.
Onthe other hand, if m = 2 , we haveW1 + W2 = 2W 1r 1 + W 2r 2 = 0W 1r 12 + W 2r 22 =23a system of three equations in four unknowns. We cannot directly solve theseequations, but if we examine the case W 1 = W 2 = 1 and r 1 = −r 2 , the first twoequations are satisfied and the third equation becomes√213222== 0.57735. . .r1 + r2 = 2r1 = ⇒ r1 =333corresponding exactly to the second entry in Table 6.1. These weighting factorsand Gauss points also integrate a quadratic polynomial exactly. The reader isurged to note that, because of the zero result from integrating the odd powers inthe polynomial, exact results are obtained for two polynomial orders for each setof sampling points and weighting factors.This discussion is by no means intended to be mathematically rigorous interms of the theory underlying numerical integration.
The intent is to give someinsight as to the rationale behind the numerical values presented in Table 6.1.EXAMPLE 6.7Evaluate the integral1f (r ) =(r 2 − 3r + 7) dr−1using Gaussian quadrature so that the result is exact.Hutton: Fundamentals ofFinite Element Analysis2106. Interpolation Functionsfor General ElementFormulationCHAPTER 6© The McGraw−HillCompanies, 2004TextInterpolation Functions for General Element Formulation■ SolutionAs the integrand is a polynomial of order 2, we have, for exact integration, 2m − 1 = 2 ,which results in the required number of sampling points as m = 3/2 . The calculatednumber of sampling points must be rounded up to the nearest integer value, so in thiscase, we must use two sampling points. Per Table 6.1, the sampling points are r i =±0.5773503 and the weighting factors are W i = 1.0, i = 1, 2 .
Therefore,1(r 2 − 3r + 7) dr = (1)[(0.5773503) 2 − 3(0.5773503) + 7]−1+ (1)[(−0.5773503) 2 − 3(−0.5773503) + 7]1(r 2 − 3r + 7) dr = 14.666667−1The result is readily verified as, indeed, being exact by direct integration.The Gaussian quadrature numerical integration procedure is by no meanslimited to one dimension.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
