MANE 4240 & CIVL 4240 Introduction to Finite Elements

1. MANE 4240 & CIVL 4240Introduction to Finite Elements Prof. Suvranu De Numerical Integration in 1D

2. Reading assignment: Lecture notes, Logan 10.4 • Summary: • Newton-Cotes Integration Schemes • Gaussian quadrature

3. y F x x x=L x=0 2 1 d2x d1x Axially loaded elastic bar A(x) = cross section at x b(x) = body force distribution (force per unit length) E(x) = Young’s modulus x2 x1 For each element Element stiffness matrix where

4. Only for a linear finite element Element nodal load vector Question: How do we compute these integrals using a computer?

5. Any integral from x1 to x2 can be transformed to the following integral on (-1, 1) Use the following change of variables Goal: Obtain a good approximate value of this integral 1. Newton-Cotes Schemes (trapezoidal rule, Simpson’s rule, etc) 2. Gauss Integration Schemes NOTE: Integration schemes in 1D are referred to as “quadrature rules”

6. g(x) f(1) f(-1) f(x) x -1 1 Trapezoidal rule: Approximate the function f(x) by a straight line g(x) that passes through the end points and integrate the straight line

7. Requires the function f(x) to be evaluated at 2 points (-1, 1) • Constants and linear functions are exactly integrated • Not good for quadratic and higher order polynomials • How can I make this better?

8. f(1) g(x) f(x) f(-1) f(0) x -1 1 Simpson’s rule: Approximate the function f(x) by a parabola g(x) that passes through the end points and through f(0) and integrate the parabola

9. Requires the function f(x) to be evaluated at 3 points (-1,0, 1) • Constants, linear functions and parabolas are exactly integrated • Not good for cubic and higher order polynomials How to generalize this formula?

10. Integration point Weight Notice that both the integration formulas had the general form Trapezoidal rule: M=2 Accurate for polynomial of degree at most 1 (=M-1) Simpson’s rule: M=3 Accurate for polynomial of degree at most 2 (=M-1)

11. Generalization of these two integration rules: Newton-Cotes • Divide the interval (-1,1) into M-1 equal intervals using M points • Pass a polynomial of degree M-1 through these M points (the value of this polynomial will be equal to the value of the function at these M-1 points) • Integrate this polynomial to obtain an approximate value of the integral f(1) f(x) f(-1) g(x) x -1 1

12. With ‘M’ points we may integrate a polynomial of degree ‘M-1’ exactly. Is this the best we can do ? With ‘M’ integration points and ‘M’ weights, I should be able to integrate a polynomial of degree 2M-1 exactly!! Gauss integration rule See table 10-1 (p 405) of Logan

13. Integration point Weight Gauss quadrature How can we choose the integration points and weights to exactly integrate a polynomial of degree 2M-1? Remember that now we do not know, a priori, the location of the integration points.

14. Example: M=1 (Midpoint qudrature) How can we choose W1 and x1 so that we may integrate a (2M-1=1) linear polynomial exactly? But we want

15. Hence, we obtain the identity For this to hold for arbitrary a0 and a1 we need to satisfy 2 conditions i.e.,

16. For M=1 f(x) f(0) g(x) x -1 1 • Midpoint quadrature rule: • Only one evaluation of f(x) is required at the midpoint of the interval. • Scheme is accurate for constants and linear polynomials (compare with Trapezoidal rule)

17. Example: M=2 How can we choose W1 ,W2x1 and x2 so that we may integrate a polynomial of degree (2M-1=4-1=3) exactly? But we want

18. Hence, we obtain the 4 conditions to determine the 4 unknowns (W1 ,W2x1 and x2 ) Check that the following is the solution

19. For M=2 f(x) * x * -1 1 • Only two evaluations of f(x) is required. • Scheme is accurate for polynomials of degree at most 3 (compare with Simpson’s rule)

20. Exercise: Derive the 6 conditions required to find the integration points and weights for a 3-point Gauss quadrature rule Newton-Cotes Gauss quadrature 1. ‘M’ integration points are necessary to exactly integrate a polynomial of degree ‘2M-1’ 2. Less expensive 3. Exponential convergence, error proportional to 1. ‘M’ integration points are necessary to exactly integrate a polynomial of degree ‘M-1’ 2. More expensive

21. Example Exact integration Integrate and check! Newton-Cotes To exactly integrate this I need a 4-point Newton-Cotes formula. Why? Gauss To exactly integrate this I need a 2-point Gauss formula. Why?

22. Gauss quadrature: Exact answer!

23. Comparison of Gauss quadrature and Newton-Cotes for the integral Newton-Cotes Gauss quadrature

24. 2 1 In FEM we ALWAYS use Gauss quadrature Linear Element Stiffness matrix Nodal load vector Usually a 2-point Gauss integration is used. Note that if A, E and b are complex functions of x, they will not be accurately integrated

25. 1 3 Quadratic Element 2 Nodal shape functions You should be able to derive these! Stiffness matrix Assuming E and A are constants

26. Need to exactly integrate quadratic terms. Hence we need a 2-point Gauss quadrature scheme..Why?