Uncertainty is unavoidable in engineering system. Probabilistic approach is the traditional approach

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1 Interval Finite Elements as a Basis for Generalized Models of Uncertainty in Engineering Analysis Rafi L. Muhanna Georgia Institute of Technology USA Collaborative Research Center 528 German National Science Foundation (DFG) TU Dresden, Germany, June 19, 2007

2 Outline Introduction Interval Arithmetic Interval Finite Elements Element-By By-Element Examples Conclusions

3 Outline Introduction Interval Arithmetic Interval Finite Elements Element-By By-Element Examples Conclusions

4 Introduction- Uncertainty Uncertainty is unavoidable in engineering system structural mechanics entails uncertainties in material, geometry and load parameters (aleatory-epistemic) Probabilistic approach is the traditional approach requires sufficient information to validate the probabilistic model criticism of the credibility of probabilistic approach when data is insufficient (Elishakoff, 1995; Ferson and Ginzburg, 1996; Möller and Beer, 2007)

5 Introduction- Interval Approach Nonprobabilistic approach for uncertainty modeling when only range information (tolerance) is available t = t0 ± δ Represents an uncertain quantity by giving a range of possible values t = [ t δ, t + δ ] 0 0 How to define bounds on the possible ranges of uncertainty? experimental data, measurements, statistical analysis, expert knowledge

6 Introduction- Why Interval? Simple and elegant Conforms to practical tolerance concept Describes the uncertainty that can not be appropriately modeled by probabilistic approach Computational basis for other uncertainty approaches (e.g., fuzzy set, random set, imprecise probability) Provides guaranteed enclosures

7 Outline Introduction Interval Arithmetic Interval Finite Elements Element-By By-Element Examples Conclusions

8 Interval arithmetic Background Archimedes ( B.C.) Α circle of radius one has an area equal to π

9 Interval arithmetic Background Archimedes ( B.C.) Α circle of radius one has an area equal to π 2 < π < 4 r = < π < r=1 π = [ , ]

10 Interval arithmetic Background Modern interval arithmetic Physical constants or measurements g [9.8045, ] Representation of numbers 1/ π / 3 [0.3333, ] 2 [1.4142, ] π [3.1415, ] Rounding errors 1 / / [8.1004, ] R. E. Moore, E. Hansen, A. Neumaier, G. Alefed, J. Herzberger

11 Interval arithmetic Interval number represents a range of possible values within a closed set x [ x, x]: = { x R x x x}

12 Interval Operations Let x = [a, b] and y = [c, d] be two interval numbers 1. Addition x + y = [a, b] + [c, d] = [a + c, b + d] 2. Subtraction x y = [a, b] [c, d] = [a d, b c] 3. Multiplication xy = [min(ac,ad,bc,bd), max(ac,ad,bc,bd)] 4. Division 1 /x= [1/b, 1/a]

13 Properties of Interval Arithmetic Let x, y and z be interval numbers 1. Commutative Law x + y = y + x xy = yx 2. Associative Law x + (y + z) = (x + y) + z x(yz) = (xy)z 3. Distributive Law does not always hold, but x(y + z) ) xy + xz

14 Sharp Results Overestimation The DEPENDENCY problem arises when one or several variables occur more than once in an interval expression f (x) = x x, x= [1, 2] f (x) = [1 2, 2 1] = [ 1, 1] 0 f (x, y) = { f (x, y) = x y x x, y y} f (x) = x (1 1) f (x) = 0 f (x) = { f (x) = x x x x} }

15 Sharp Results Overestimation If a, b and c are interval numbers, then: a (b ± c) ab ± ac If we set a = [ 2,[ 2]; b = [1, 2]; c = [ 2,[ 1], we get a (b + c) = [ 2, 2]([1, 2] + [ 2, 1]) = [ 2, 2] [ 1, 3] = [ 6, 6] However, ab + ac = [ 2, 2 2][1, 2]+[ + [ 2, 2 2][ 2, 21] = [ 4,[ 44]+[ 4] + [ 4, 44] 4] = [ 8, 88] 8]

16 Sharp Results Overestimation Interval Vectors and Matrices An interval matrix is such matrix that contains all real matrices whose elements are obtained from all possible values between the lower and upper bounds of its interval components m n A = { A R A A for i = 1,..., m; j = 1,..., n} ij ij

17 Sharp Results Sharp Results Overestimation Overestimation Let a, b, c and d be independent variables, each with interval [1, 3] b ] 2 2 [ 2] 2 [ 1 1 B, d c b a B = = = 2] 2 [ 2] 2 [ 2] 2 [ 2] 2 [, ,,,, A A b b b b b b b b B b b b b B = = = ] [ ] [ ] [ ] [,, phys phys A A = = = , , * * phys B phys A A B b

18 Outline Introduction Interval Arithmetic Interval Finite Elements Element-By By-Element Examples Conclusions

19 Finite Elements Finite Element Method (FEM) is a numerical method that t provides approximate solutions to differential equations (ODE and PDE)

20 Finite Elements- Uncertainty& Errors Mathematical model (validation) Discretization of the mathematical model into a computational framework Parameter uncertainty (loading, material properties) Rounding errors

21 Interval Finite Elements Uncertain Data Geometry Materials Loads Interval Stiffness Matrix T K = B CB dv Interval Load Vector Element Level K U = F

22 Interval Finite Elements K U = F K = B T C B dv = Interval element stiffness matrix B = Interval strain-displacement matrix C = Interval elasticity matrix F = [F 1,... F i,... F n ] = Interval element tload vector (traction) ti F = N t da i i N i = Shape function corresponding to the i-th DOF t = Surface traction

23 Interval Finite Elements (IFEM) Follows conventional FEM Loads, geometry and material property are expressed as interval quantities i System response is a function of the interval variables and therefore varies in an interval Computing the exact response range is proven NP-hard The problem is to estimate the bounds on the unknown exact response range based on the bounds of the parameters

24 IFEM- Inner-Bound Methods Combinatorial i method (Muhanna and Mullen 1995, Rao and Berke 1997) Sensitivity analysis method (Pownuk 2004) Perturbation (Mc William 2000) Monte Carlo sampling method Need for alternative methods that achieve Rigorousness guaranteed enclosure Accuracy sharp enclosure Scalability large scale problem Efficiency

25 IFEM- Enclosure Linear static finite i element Muhanna, Mullen, 1995, 1999, 2001,and Zhang 2004 Popova 2003, and Kramer 2004 Neumaier and Pownuk 2004 Corliss, Foley, and Kearfott 2004 Dynamic Dessombz, 2000 Free vibration-buckling Modares, Mullen 2004, and Billini and Muhanna 2005

26 Interval Finite Elements Interval Linear System of Equations A x = b x 1 2 [ 1,0] 1.2 = [ 1,0] x Then A A iff A : 2 α = with, [0,1] 2 αβ β

27 Interval Finite Elements x 1 H [0.3, 0.6] A b= S ( A, b ) = [ 0.6, 0.3] S (A, b) P 4 P 3 S sym (A, b) [0.4, 0.6] S sym ( Ab, ) = [ 0.6, 0.4] S sym (A, b) P 1 = (0.3, 0.6) 0.5 P 2 = (0.6, 0.6) P 1 P 2 P H 3 = (0.6, 0.3) but A b S (A,, b) ) [ 0.8, 0.2] P 4 = (0.4, 0.4) 1 [0.2, 0.8] A b= A 1 b 1.0 H A b x 2

28 Outline Introduction Interval Arithmetic Interval Finite Elements Element-By By-Element Examples Conclusions

29 Naïve interval FEA E1, A1, L1 2 p 1 E2, A2, L2 3 p 2 EA / L = k = [0.95, 1.05], EA p / L = k = [1.9, 2.1], = 0.5, p = k1+ k2 k2 u1 p1 [2.85, 3.15] [ 2.1, 1.9] u1 0.5 = = k k u p [ 2.1, 1.9] [1.9, 2.1] u exact solution: u 2 = [1.429, 1.579], u 3 = [1.905, 2.105] naïve solution: u 2 = [-0.052, 3.052], u 3 = [0.098, 3.902] interval arithmetic assumes that all coefficients are independent uncertainty in the response is severely overestimated (1900%)

30 Element-By-Element Element-By-Element (EBE) technique elements are detached no element coupling structure stiffness matrix is block-diagonal l (k 1,, k Ne ) the size of the system is increased u =(u 1,, u Ne ) T need to impose necessary constraints for compatibility and equilibrium u 1 u 3 u 4 u 2 E 1, A 1, L 1 E 2, A 2, L 2 Element-By-Element model

31 Element-By-Element Suppose the modulus of elasticity is interval: E = Eˆ(1 + δ ) δ: zero-midpoint interval The element stiffness matrix can be split into two parts, k = kˆ( I + d) = kˆ+ kˆd kˆ : deterministic part, element stiffness matrix evalued using Eˆ, kd ˆ : interval part d: interval diagonal matrix, diag( δ,..., δ).

32 Element-By-Element : Element stiffness matrix: k = ki ˆ( + d) : Structure stiffness matrix: K = Kˆ( I + D) = Kˆ + KˆD or K k 1 ˆk 1 d1 = = I + ˆ Ne k k N dne e

33 Constraints Impose necessary constraints for compatibility and equilibrium Penalty method Lagrange multiplier method u 1 u 2 u 3 u 4 E 1, A 1, L 1 E 2, A 2, L 2 Element-By-Element model

34 Constraints penalty method Constraint t conditions: cu = 0 Using the penalty method: ( K + Q ) u = p T Q: penalty matrix, Q= c ηc η : diagonal matrix of penalty number η i Requires a careful choice of the penatly number A spring of large stiffness is added to force node 2 and node 3 to have the same displacement. E 1, A 1, L 1 E 2, A 2, L 2 p

35 Constraints Lagrange multiplier Constraint conditions: c u = 0 Using the Lagrange multiplier method: T K c u p = c 0 λ 0 λ : Lagrange multiplier vector, introdued as new unknowns

36 Load in EBE Nodal lload p b = 1 p ( p,..., p ) b T where p i = N φ( ( x ) dx Suppose the surface traction φ( x) is described by j an interval function: φ( x) = a x. p b can be rewritten as p b m j= 0 N j = W F W: deterministic matrix F: interval vector containing the interval coefficients of the surface tractiton e T

37 Fixed point iteration For the interval equation Ax = b, preconditioning: RAx = Rb, R is the preconditioning matrix transform it into g (x * ) = x * : R b RA x 0 + (I RA) x * = x *, x = x * +x 0 Theorem (Rump, 1990): for some interval vector x *, if g (x * ) int (x * ) then A H b x * +x 0 Iteration algorithm: *( l+ 1) *( l) iterate: x = z + G( ε x ) where z = Rb RAx, G = I RA, R = Aˆ, Ax ˆ = bˆ No dependency handling 1 0 0

38 Fixed point iteration, Interval FEA calls for a modified d method which h exploits the special form of the structure equations, ( K + Q ) u= p with K= Kˆ + KˆD R = Kˆ + Q 1 Choose ( ), construct iterations: *( l+ 1) u p K 0 K ˆ *( l ) = Rp u0 RKD ( u0+ε u ) if *( l ) = R R( + Q) u + ( I R( + Q) )( ε u ) ( l ) = Rp u0 RKM Δ ˆ *( l+ 1) *( l) *( l+ 1) ( l) int( ), then + u0 = R RK u u u= u p M Δ: interval vector, Δ=( δ,..., δ ) 1 N e T The interval variables δ,..., δ appear only once in each iteration. i 1 N e Δ

39 Convergence of fixed point The algorithm converges if and only if ρ ( G ) < 1 smaller ρ( G ) less iterations required, and less overestimation in results To minimize ρ( G ): choose R= Aˆ so that G = I RA has a small spectral radius 1 : 1 reduce the overestimation in ˆ 1 ˆ ˆ ˆ G= I R A=I ( K + Q ) ( K + Q + K D ) = RK D G

40 Stress calculation Conventional method: σ = CBu e,(severe overestimation) C: elasticity matrix, B: strain-displacement matrix Present method: E = (1 + δ) E ˆ, C = (1 + δ) C ˆ σ = CBLu ˆ ( l ) =CBL( Rp RCM Δ) = + δ ˆ p ˆ ˆM ( l ) (1 )( CBLR CBLRK Δ) L: Boolean matrix, L u= ue

41 Element nodal force calculation Conventional method: f = ( ku p ), (severe overestimation) T e e e Present method: ( Te) 1( k1( ue) 1 ( pe) 1) in the EBE model, T ( Ku pb ) = ( e) N ( ( ) ( ) ) T k e N u e e N p e e Ne from ( K + Q) u= p + p T( Ku p ) = T( p Qu) Calculate T( p for all elements. c c b b c Qu) to obtain the element nodal forces

42 Outline Introduction Interval Arithmetic Interval Finite Elements Element-By By-Element Examples Conclusions

43 Numerical example Examine the rigorousness, accuracy, scalability, and efficiency of the present method Comparison with the alternative methods the combinatorial method, sensitivity analysis method, and Monte Carlo sampling method these alternative methods give inner estimation x i x x i xo x: exact solution, x: inner bound, x : outer bound o

44 Examples Load Uncertainty Four-bay forty-story frame

45 Examples Load Uncertainty Four-bay forty-story frame Loading A Loading B Loading C Loading D

46 Examples Load Uncertainty Four-bay forty-story frame Total number of floor load patterns = If one were able to calculate kn/m (1.2 kip/ft) 10,000 patterns / s there has not been sufficient time since the creation of the universe (4-8 ) billion years? to solve all load patterns for this simple structure Mt Material ila36, Beams W24 x 55, Columns W14 x m (48 ft)

47 Examples Load Uncertainty Four-bay forty-story frame Four bay forty floor frame - Interval solutions for shear force and bending moment of first floor columns Elements Nodes Combination solution Total number of required combinations = Interval Axial force (kn) [ , 185.7] [ , 0.0] [ , 0.0] solution Shear force (kn) [-5.1, 0.9] [-5.8, 5.0] [-5.0, 5.0] Moment (kn m) [-10.3, 4.5] [-15.3, 5.4] [-10.6, 9.3] [-17, 15.2] [-8.9, 8.9] [-16, 16]

48 Examples Load Uncertainty Ten-bay truss 1 A = m 2 E = kpa kN 20 kn 20kN 20kN 20kN 20kN 20 kn 20kN 20kN 4 = 40 m F = [-4.28, 28.3] kn m F min = -( )( ) 20= kn F max = ( ) 20 =28.3 kn

Examples Load Uncertainty Three-Span Beam 1 2 6 kn /m 3 4 4 m 4 m 4 m 12 m 11.2kNm 4.

49 Examples Load Uncertainty Three-Span Beam kn /m m 4 m 4 m 12 m 11.2kNm 4.8 knm 11.2 knm knm knm knm m m 2 m 2 m m m

50 Truss structure P m P 2 P 1 P 3 4.5m 4.5m 4.5m 4.5m A A A,A A A cross-sectional area 2 1, 2, 3 13, 14, 15: [9.95, 10.05] cm (1% uncertainty) 2 of all other elements: [5.97, 6.03] cm (1% uncertainty) modulus of elasticity of all elements: 200,000 MPa p = [190, 210] kn, p = [95,105] kn p 1 2 = [95,105] kn, p = [85.5,94.5] kn (10% uncertainty) 3 4

51 Truss structure - results Table: results of selected responses Method u 5 (LB) u 5 (UB) N 7 (LB) N 7 (UB) Combinatorial Naïve IFEA δ % % 362% 327% Present IFEA δ 0.19% 011% 0.11% 0.19% 0.15% unit: u 5 (m), N 7 (kn). LB: lower bound; UB: upper bound.

52 Truss structure results u 5 (m) Comb. LB Comb. UB IFEA LB IFEA UB % 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% Uncertainty in cross-sectional area for moderate uncertainty ( 5%), very sharp bounds are obtained for relatively large uncertainty, reasonable bounds are obtained in the case of 10% uncertainty: Comb.: u5 = [ , ], IFEM: u5 = [ , ] (relative difference: 2.59%, 1.80% for LB, UB, respectively)

53 Truss with a large number of interval variables p C p p p p D story bay N e N v p p n@l A B A i E i m@l = [0.995,1.005] A, = [0.995,1.005] E for i = 1,..., N 0 0 e

54 Scalability study vertical displacement at right upper corner (node D): v D Table: displacement at node D = a PL EA 0 0 Sensitivity Analysis Present IFEA Story bay LB * UB * LB UB δ LB δ UB wid/d % 0.10% 2.64% % 0.16% 2.84% % 0.22% 3.02% % 0.29% 3.19% % 0.37% 3.37% 37% % 0.45% 3.56% δ = LB- * * = UB- * * = (LB- * * LB LB / LB, δ LB UB / UB, δ LB LB )/ LB

55 Efficiency study Table: CPU time for the analyses with the present method (unit: seconds) Story bay N v Iteratio n t i t r t t i /t t r /t % 78.4% % 80.5% % 89.1% % 89.7% % 90.1% % 90.0% t i : iteration time, t r : CPU time for matrix inversion, t : total comp. CPU time majority of ftime is spent on matrix inversion i

56 Efficiency study CPU time (sec) Computational time: a comparison of the senstitivity analysis method and the present method Sensitivity Analysis method Present interval FEA Number of interval variables Computational time (seconds) N v Sens. Present hr 9 min

57 Plate with quarter-circle cutout thickness: 0.005m Possion ratio: 0.3 load: 100kN/m modulus of elasticity: E = [199, 201]GPa number of element: 352 element type: six-node isoparametric quadratic triangle results presented: u, v, σ and σ at node F A E xx yy

58 Plate with quarter-circle cutout Case 1: the modulus of elasticity for each element varies independently in the interval [199, 201] GPa. Table: results of selected responses Monte Carlo sampling * Present IFEA Response LB UB LB UB u -5 A (10 5 m) v E (10-5 m) σ xx (MPa) σ yy (MPa) * 10 6 samples are made.

59 Outline Introduction Interval Arithmetic Interval Finite Elements Element-By By-Element Examples Conclusions

60 Conclusions Development and dimplementation ti of IFEM uncertain material, geometry and load parameters are described by interval variables interval arithmetic is used to guarantee an enclosure of response Enhanced dependence problem control use Element-By-Element t technique use the penalty method or Lagrange multiplier method to impose constraints modify and enhance fixed point iteration to take into account the dependence problem develop special algorithms to calculate stress and element nodal force

61 Conclusions The method is generally applicable to linear static FEM, regardless of element type Evaluation of the present method Rigorousness: in all the examples, the results obtained by the present method enclose those from the alternative methods Accuracy: sharp results are obtained for moderate parameter uncertainty (no more than 5%); reasonable results are obtained for relatively large parameter uncertainty t (5%~10%)

62 Conclusions Scalability: the accuracy of the method remains at the same level with increase of the problem scale Efficiency: the present method is significantly superior to the conventional methods such as the combinatorial, Monte Carlo sampling, and sensitivity analysis method The present IFEM represents an efficient method to handle uncertainty in engineering applications

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

64 Frame structure 7 4 w 3 w 4 Member Shape B 8 3 B 4 C 4 C 5 C 6 w 1 w 2 B 1 5 B 2 C 1 C 2 C m 14.63m m 6.1m C 1 W12 19 C 2 W C 3 W C 4 W10 12 C 5 W C 6 W B 1 W27 84 B 2 W B 3 W18 40 results listed: nodal forces at the left node of member B B 4 W

65 Frame structure case 1 Case 1: load uncertainty w w = [105.8,113.1] kn/m, w = [105.8,113.1] kn/m, 1 2 = [49.255,52.905] ] kn/m, w = [49.255,52.905] ] kn/m, 3 4 Table: Nodal forces at the left node of member B 2 Combinatorial Present IFEA Nodal force LB UB LB UB Axial (kn) Shear (kn) Moment (kn m) exact solution is obtained in the case of load uncertainty

66 Frame structure case 2 Case 2: stiffness uncertainty and load uncertainty 1% uncertainty introduced to AI,, and Eof each element. Number of finterval variables: 34. Table: Nodal forces at the left node of member B 2 Monte Carlo sampling * Present IFEA Nodal force LB UB LB UB Axial (kn) Shear (kn) Moment (kn.m) * 10 6 samples are made.

Interval Finite Element Methods for Uncertainty Treatment in Structural Engineering Mechanics Rafi L. Muhanna Georgia Institute of Technology USA

Interval Finite Element Methods for Uncertainty Treatment in Structural Engineering Mechanics Rafi L. Muhanna Georgia Institute of Technology USA Second Scandinavian Workshop on INTERVAL METHODS AND THEIR