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

Outline  Interval finite element analysis  Example  Conclusion

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

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

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

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

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)

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

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

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

Dependency problem

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

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

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

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

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

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

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

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

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

Present formulation  Rewrite D x = Mδ

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

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

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

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

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;

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

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– – Present– – Naïve– – 93.10– Present error0.03%0.06% 0.38%0.26% Naïve error29.9%32.2% 678%607%

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

Twenty-bay truss Displacement of corner D u(LB)u(UB)v(LB)v(UB) Pownuk’s solution* – – Present solution – – *sensitivity analysis

Example # interval parameters Iteration number Iteration time (sec) Computational time (sec) Variation in typical displ* % % % % % % % *defined as ratio of radius to midpoint value (corner D)

Example

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