1. Interval Finite Element as a Basis for Generalized Models of Uncertainty in Engineering Mechanics Rafi L. Muhanna Georgia Institute of Technology NSF workshop on Reliable Engineering Computing NSF workshop on Reliable Engineering Computing September 15-17, 2004, Savannah, Georgia, USA Hao Zhang Georgia Institute of Technology Robert L. Mullen Case Western Reserve University

2. Outline  Interval finite element analysis  Example  Conclusion

3. Center for Reliable Engineering Computing (REC) We handle computations with care

4. Uncertainty and errors in FEA  Mathematical model  Discretization of the mathematical model into a computational framework  Parameter uncertainty (loading, material properties)  Rounding errors

5. Why interval?  Generalized models of uncertainty  Imprecise probability  Fuzzy set and possibility theory  Dempster-Shafer evidence theory and Random Set  Probability bounds  Fuzzy randomness  others  Computation tool: interval analysis

6. Uncertain Data Geometry Materials Loads Interval Stiffness Matrix Interval Load Vector K u = p Interval Finite Element Analysis

7. Interval arithmetic  Linear interval equation Ax = b ( A  A, b  b)  Solution set  (A, b) = {x  R |  A  A  b  b: Ax = b}  Hull of the solution set  (A, b) A H b := ◊  (A, b)

8. Interval arithmetic  Finding the enclosure  Interval Gauss elimination  Interval Gauss-Seidel iteration  Krawczyk’s iteration  Fixed-point iteration  Others

9. Fixed point theory  Find the solution of Ax = b  Transform into fixed point equation g(x) = x g (x) = x – R (Ax – b) = Rb+ (I – RA) x (R nonsingular)  Brouwer’s fixed point theorem If Rb + (I – RA) X  int (X) then  x  X, Ax = b

10. Fixed point theory  Solve AX=b  Brouwer’s fixed point theorem w/ Krawczyk’s operator IfR b + (I – RA) X  int (X) then  (A, b)  X  Iteration  X n+1 = R b + (I – RA) εX n (for n = 0, 1, 2,…)  Stopping criteria: X n+1  int( X n )  Enclosure:  (A, b)  X n+1

11. Dependency problem

12. k 1 = [0.9, 1.1], k 2 = [1.8, 2.2], p = 1.0

13. Dependency problem  Two k 1 : the same physical quantity  Interval arithmetic: treat two k 1 as two independent interval quantities having same bounds

14. Naïve interval finite element  Replace floating point arithmetic by interval arithmetic  Over-pessimistic result due to dependency Naïve solution Exact solution

15. Dependency problem  How to reduce overestimation?  Manipulate the expression to reduce multiple occurrence  Trace the sources of dependency

16. Present formulation  Element-by-Element  K: diagonal matrix, singular

17. Present formulation  Element-by-element method  Element stiffness:  System stiffness:

18. Present formulation  Lagrange Multiplier method  With the constraints: CU – t = 0  Lagrange multipliers: λ

19. Present formulation  System equation: Ax = b rewrite as:

20. Present formulation  Solve residual iteration (Rump 1983) with overestimation control: x *n+1 = R b – R A x 0 + (I – RA) εx *n x *n+1 = R b – x 0 – RSD x 0 – RSDεx *n x *n+1 = R b – x 0 – RS D ( x 0 + εx *n ) x *n+1 = R b – x 0 – RS M n δ D ( X 0 + εX *n ) = M n δ

21. Present formulation  Rewrite D x = Mδ

22. Present formulation  x = [u, λ] T, u is the displacement vector  Calculate element forces  Conventional FEM: F=k u ( overestimation)  Present formulation: Ku = P – C T λ λ= Lx, p = Nb P – C T λ = p – C T L(x *n+1 + x 0 ) P – C T λ = Nb – C T L(Rb – RS M n δ) P – C T λ = (N – C T LR)b + C T LRS M n δ

23. Examples  Two bay truss  A = 0.01 m 2  E 0 = 200 GPa, with 1% uncertainty, E = [199, 201] GPa  Load [19, 21] kN

24. Two bay truss Displacement of selected nodes v 2 (LB)v 2 (UB)u 4 (LB)u 4 (UB) Comb  10  5 – 21.034– 18.8423.70294.2043 Present  10  5 – 21.043– 18.8223.69424.2075 Naïve  10  5 – 22.762– 17.1033.2224.679 Present error0.04%0.10%0.23%0.08% Naïve error8.21%9.23% 12.98%11.3%

25. Two bay truss Element forces of selected elements N 2 (LB)N 2 (UB)N 4 (LB)N 4 (UB) Comb– 8.347– 7.46111.44812.753 Present– 8.351– 7.45211.43912.758 Naïve– 9.691– 6.127– 10.33634.542 Present error0.05%0.12%0.08%0.03% Naïve error16.1%17.88% 190%170%

26. Two-bay two-floor frame  E 0 = 200 GPa, with 1% uncertainty, E = [199, 201] GPa  w 1 = [24, 26] kN/m; w 2 = [24, 26] kN/m;  w 3 = [48, 52] kN/m; w 4 = [48, 52] kN/m;

27. Two-bay two-floor frame Displacement of selected nodes v 4 (LB)v 4 (UB)  9 (LB)  9 (UB) Comb  10  6 – 6.764– 6.1555.6336.269 Present  10  6 – 6.766– 6.1495.6226.277 Naïve  10  6 – 8.704– 4.2103.7318.168 Present error0.03%0.10% 0.20%0.12% Naïve error28.7%31.6% 33.8%30.3%

28. Two-bay two-floor frame Axial and shear forces of column 1 N 1 (LB)N 1 (UB)V 1 (LB)V 1 (UB) Comb– 149.68– 137.355.2615.879 Present– 149.72– 137.275.24185.894 Naïve– 194.39– 93.10– 30.3841.57 Present error0.03%0.06% 0.38%0.26% Naïve error29.9%32.2% 678%607%

29. Twenty-bay truss  Large scale truss  E 0 = 210 GPa, with 1% uncertainty  A 0 = 0.0025 m 2, with 1% uncertainty (648 interval parameters)

30. Twenty-bay truss Displacement of corner D u(LB)u(UB)v(LB)v(UB) Pownuk’s solution*7.5497.845 – 5.824 – 5.657 Present solution7.5067.886 – 5.843 – 5.637 *sensitivity analysis

31. Example # interval parameters Iteration number Iteration time (sec) Computational time (sec) Variation in typical displ* 24650.1721.042.24% 39250.4533.972.47 % 64861.48415.052.67 % 89073.70440.692.92 % 119278.03195.83.23 % 1452814.331723.38 % 1932826.083823.79 % *defined as ratio of radius to midpoint value (corner D)

32. Example

33. Conclusion  Formulation of interval finite element method is introduced  Element-by-element  Lagrange method  Fixed point iteration  Newly developed overestimation control to reduce dependency problem  Displacement and element internal forces obtained  Accurate and efficient