Numerical Integration

Transcription

1 MCEN 473/573 Chapte 0 Numeical Integation Fall, 2006 Textbook, 0.4 and 0.5

2 Isopaametic Fomula Numeical Integation [] e [ ] T k = h B [ D][ B] e B Jdsdt In pactice, the element stiffness is calculated numeically. Numeical Integation Schemes:. The Newton-Cotes Fomulas 2. The Gauss Fomulas

3 Numeical Integation Numeical integation 0 xdx = x 0 = ( 2 0) = 2 π 2 0 π sin xdx = cos x 2 0 π = cos + 2 cos(0) = π 2 0 ( sin ) 2 x dx??? We can use numeical method to find the answe.

4 Numeical Integation Integation using numeical methods: Example: F ( 2 x + 6)dx = Exact solution: ( 2 ) 3 F = x + 6 dx = x + 6x = y y = x x The integation epesents the aea unde the cuve

5 Numeical Integation Integation using numeical methods: Numeical integation Scheme :. Divide the inteval of integation into N section; 2. Choose a function to appoximate the vaiation of f (x) in each section; the simplest such function is a constant function that equals to the value of f (x) at the mid-point of each section. 3. The poduct of this constant function and the length of the section appoximates the integation of f (x) ove this section. 4. Summing the poducts fo all sections gives an appoximate answe to the integation of f (x) ove (-,) y y = x y y = x N=, F=2, Eo= -5.26% x N=2, F=2.5, Eo= -.32% x

6 Numeical Integation Integation using numeical methods: Numeical integation y y = x y y = x x x N=4, F=2.625, Eo=-0.33% N=8, F=2.656, Eo=-0.08% As the numbe of sections inceases, the eo deceases.

7 Numeical Integation Scheme 2: Integation using numeical methods: Numeical integation Same as Scheme, except that we choose a linea function in each section to appoximate the vaiation of f (x). This linea function takes the same value and slope of f (x) at the mid-point of that section. y y = x y = x y x Diffeent functions can be chosen to appoximate f (x). x

8 Numeical Integation Integation using numeical methods: Two key steps:. Divide the inteval of integation. 2. In each sub-inteval, choose pope simple functions to appoximate the tue function. Two key featues:. The numeical esult is an appoximation to exact solution. 2. The accuacy of numeical esult depends on the numbe of sub-inteval and appoximate function. One key dawbacks:

9 Isopaametic Fomula Numeical Integation Intepolation using a polynomial Assuming a function F() F() can be calculated at (n+) distinctive points, 0,,, n We can ceate a n-th ode polynomial passing these (n+) points. ψ 2 n () = a + a + a a n 0 But we need to solve (n+) linea equations. (We don t like this!!) Lagangian intepolation 2

10 Isopaametic Fomula Numeical Integation Intepolation using a polynomial Lagangian intepolation We want: Lagangian Intepolation:

11 Isopaametic Fomula Numeical Integation Lagangian intepolation Intepolation using a polynomial () ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) n n l = + + L L L L 0 0

12 Isopaametic Fomula Numeical Integation. The Newton-Cotes Fomulas (one dimensional) In the newton-cotes fomulas, the sampling points of F ae spaced at equal distance. b a F ( ) d The Newton-Cotes Fomulas 0 = a n = b h = b a n b n F( ) d = a i= 0 b a l i () d Fi + R n If F() is a n-th ode polynomial function, Newton-Cotes fomulas should give exact solution.

13 Isopaametic Fomula. The Newton-Cotes Fomulas (one dimensional)

14 Isopaametic Fomula 2. The Gauss Fomulas (one dimensional) In the Newton-Cotes fomulas b n F( ) d = a i= ( b a) C n i F i + R n The sampling points ae known, so we only need to detemine C i n If F() is a n-th ode polynomial function, Newton-Cotes fomulas should give exact solution. In the Gauss fomulas, we take the sampling points as unknowns, so we totally have 2n unknowns. Clealy,

15 Isopaametic Fomula 2. The Gauss Fomulas (one dimensional) To detemine the sampling points, we impose This time (in the Gauss Fomulas), the sampling points ae not equally spaced.

16 Isopaametic Fomula 2. The Gauss Fomulas (one dimensional) b n b F( ) d = l a a i i= = 0 n b () d Fi + P() d β b b F d ( ) = l i a a i = = 0 i a n b () d F + P() d β + R a If F() is a (2n-)-th ode polynomial function, the Gauss fomulas should give the exact solution.

17 Isopaametic Fomula 2. The Guass Fomulas (one dimensional)

18 Isopaametic Fomula Numeical Integation Summay:. Both Newton-Cotes fomulas and Gauss fomulas use polynomial functions to appoximate the integand function.

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20 Isopaametic Fomula Numeical Integation 4 t 3 ( s, t ) ( x, y) 4 y 3 - s x b n a i= F(, s) d = αiα F( si, s ) + R Questions:. Which integation schemes we should pick? 2. What ode of integation do we need? O how many integation points do we need?

21 Isopaametic Fomula Numeical Integation Fo n integation points, since Gauss fomulas offe (2n-) polynomial appoximation wheeas Newton-Cotes fomulas offe n- polynomial appoximation, Gauss fomulas ae moe effective, and moe commonly used. In geneal, we use Gauss fomulas. But, since Gauss fomulas do not sample the bounday points, Newton-Cotes sometimes ae used. P Plastic defomation stats fom the top-bottom suface of the beam.

22 Isopaametic Fomula Numeical Integation The ode of integation scheme, o the numbe of integation points. Rule of Thumb e [ ] T k = h [ B] [ D][ B] e B Jdsdt Assuming the highest ode of polynomial in shape functions is p. Fo displacement based finite elements, we take fist ode deivative to get B matix. Theefoe, we have p-. If we take J as constant, then O ([ ] [ ][ ] ) T B D B J = 2( p ) If we take n integal points in one diection, then we need 2n 2( p ) ( 2 ) / 2 n p

23 Isopaametic Fomula Numeical Integation 4 node isopaametic element: N = ( s)( t) p=2 n ( 2 p ) / 2 = 3/ 2 = 2 We theefoe need 2x node isopaametic element: N = Nˆ 2 N 5 2 N 8 ˆ = ( s)( t) N = ( s )( t ) N p=3 n ( 2 p ) / 2 = 5/ 2 = 3 We theefoe need 3x3.

24 Isopaametic Fomula Numeical Integation 4 node isopaametic element: t y 2x2 s x 8 node isopaametic element: 3x3 t y s x

25 Isopaametic Fomula Numeical Integation Discussions on integation scheme. The integation schemes mentioned above ae minimum equiements to achieve the pecise solutions. 2. High ode integation schemes ae possible, (fo example, 4x4 fo 8 node element). In geneal, fo linea poblem, this is not necessay. But fo nonlinea poblem, such as lage defomation, this may (may not) be helpful. 3. Reduced integation schemes sometime ae used. Fo example, 2x2 fo 8 node elements ae vey commonly used. But educed integation schemes sometime can intoduce spuious non-zeo enegy mode. So one should be vey caeful when use educed integation schemes.

26 Isopaametic Fomula Numeical Integation Accuacy of stesses and stains Obsevations: Since the stains ae obtained by taking fist ode deivatives of displacements, stains and stesses have lowe accuacy than displacements. Since the integals in FE ae caied out on Gauss integal points, the stesses and stains have highe accuacy when calculated on Gauss points than when calculated on nodal points.

27 Isopaametic Fomula Numeical Integation Accuacy of stesses and stains Example: Ele Ele 2 A Vaiations of displacements and stains along line AB B Displacements Stains

28 Isopaametic Fomula Numeical Integation Accuacy of stesses and stains Vaiations of displacements and stains along line AB Displacements Stains Displacements ae continuous acoss the elements. The stains ae non-continuous acoss the elements.

29 Isopaametic Fomula Numeical Integation Accuacy of stesses and stains Stains calculated based on element displacement fields: non-smooth.

30 Isopaametic Fomula Numeical Integation Smoothing (o polishing) stesses and stains Vaiations of displacements and stains along line AB Smoothing the stain fields and stess fields by simply aveaging the nodal stains and stesses Thee ae many othe methods to impove the element stains and stesses, and nodal stains and stesses.