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NANO-MECHANICS BASED ASSESSMENT OF FAILURE RISK, SIZE EFFECT AND LIFETIME OF QUASIBRITTLE STRUCTURES AT DIFFERENT SCALES ZDENĚK P. BAŽANT COLLABORATOR: JIALIANG LE, SZE-DAI PANG SPONSORS: DoT, NSF, BOEING, CHRYSLER, DoE CapeTown, 3 rd Int. Conf. on Struct. Eng., Mech, & Computation (SEMC), 9/10/2007 Bangalore, IIS, 11/7/07; Milan 11/12/07 NANO-MECHANICS BASED ASSESSMENT OF FAILURE RISK, SIZE EFFECT AND LIFETIME OF QUASIBRITTLE STRUCTURES AT DIFFERENT SCALES ZDENĚK P. BAŽANT COLLABORATOR: JIALIANG LE, SZE-DAI PANG SPONSORS: DoT, NSF, BOEING, CHRYSLER, DoE CapeTown, 3 rd Int. Conf. on Struct. Eng., Mech, & Computation (SEMC), 9/10/2007 Bangalore, IIS, 11/7/07; Milan 11/12/07

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Energetic ( Quasibrittle ) Mean Size Effect Laws and Statistical Generalization 1c, 2c, 3c – based on cohesive crack model, 1s – statistical generalization log (Size D ) Type 2 Type 3 log ( Nom. Strength N ) c 3c LEFM 2 1 Type m n Weibull r 1 Statistical 1s 1c

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Failure at Crack Initiation: Type 1 Energetic-Statistical Size Effect on Strength and Lifetime

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Importance of Tail Distribution of P f Prob. of Failure 0 1 Tolerable failure prob. P f = Gaussian cdf Weibull cdf Same mean, same σ/ Mean 1 TGTG TWTW = function of P f and P f = 1 – exp[-( σ/ m ] Tail Offset Ratio T W / T G Load

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I. Interatomic bond energies have Maxwell-Boltzmann distribution, and activation energy depends on stress. II. Tests of lab specimens < 5 RVEs do not disagree with Gaussian pdf. Hypotheses: A structure is quasibrittle if Weibull cdf applies for sizes > 10 4 equivalent RVEs Definition:

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physical justification of Flaw Size Distribution: Why? Merely relates macro-level to micro-level hypotheses: 1. Noninteracting flaws, one in one volume element 2. Griffith (not cohesive!) theory holds on micro-level. 3. Cauchy distribution of flaw sizes: Both fatigued polycrystalline metal and concrete are brittle, follow Weibull pdf, yet the flaws cannot be identified concrete. Weibull statistics? …NO

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1) pdf of One RVE RVE = smallest material volume whose failure causes the whole structure to fail 1) pdf of One RVE RVE = smallest material volume whose failure causes the whole structure to fail

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Atomistic Basis of cdf of Quasibrittle RVE Maxwell-Boltzmann distribution frequency of exceeding activation energy Q 3) cdf of break surface: 2) Critical fraction of broken bonds breaksrestorations 1) Net rate of breaks reached within stress duration Q E = 0 x 1 PfPf 1 0 F s Tail = -assumed linear Interatomic pot.

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Power Law Tail of cdf of Strength a) Series Coupling If each link has tail m, the chain has the same tail m If each fiber has tail p, the bundle has tail np additive exponents 12 n 1 2 N softening (reality) plastic brittle b) Parallel Coupling The reach of power-law tail is decreased drastically by parallel coupling, increased by series coupling. Parallel coupling produces cdf with Gaussian core. Power-law tail with zero threshold is indestructible ! long chains (c)

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Power Tail Length for Bundles & Chains 1)Brittle bundle with n = 24 fibers (Daniels' model, 1945) having Weibull cdf with p = 1 … Gaussian core down to 0.3 Power tail up to …irrelevant! (D (l.y.) 3 ) 2)Plastic bundle with n = 24 fibers (Central Limit Theorem) having Weibull cdf with p = 1 Gaussian core down to 0.01 Power tail up to …irrelevant! 4)Plastic bundle with n = 2 fibers having Weibull cdf with p = 12 Gaussian core down to 0.3 Power tail up to 3x10 -3 Plastic fibers extend Weibull tail to 3x OK! 3)Brittle bundle with n = 2 fibers having Weibull cdf with p = 12 Gaussian core down to 0.3 Power tail up to 5x10 -5 longer but not enough Hence, a hierarchy of parallel series couplings is required! Chains tend to extend the power tail!

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2) pdf of Structure as a Chain of RVEs, with Size and Shape Effects 1 – exp[-(σ/ m ] (Infinite chain - Weibull) (finite chain) N eq = equivalent N, modified by stress field (geometry effect)

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Nano-Mechanics Based Chain-of-RVEs Model of Prob. Distribution of Structural Strength, Including Tail Chain model (structure of positive geometry) cdf for 1 RVE Gaussian Weibull (power tail) 1 PfPf 0 RVE strength cdf for 10 4 RVEs brittle 1 PfPf 0 Large structure strength cdf for 500 RVE quasibrittle Gaussian Weibull 1 PfPf 0 Structure strength = = % Gaussian Weibull = RVE causes the structure to fail (Type 1 size effect) grafting pt. 1 RVE Structure Note: If power-law tail reaches only up to P f = , a chain of RVEs would be needed to produce Weibull cdf. 1 2 N

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5.3 Quasi-Brittleness or Threshold Strength? Despite using threshold to optimize fit, Weibull theory can only fit tail Optimum fit by Weibull cdf with finite threshold Optimum fit by chain–of– RVEs, zero threshold Age 2 days 1 m =16 1 m =20 Weibull (1939) tests of Portland cement mortar 1 m =24 ln ln( u ) 7 days 28 days Weibull scale ln[ ln(1 P f )] n data (2 days) = 680 n data (7 days) = 1082 n data (14 days) = 1106 P f 0.65 RVE size cm Specimen vol cm 3 Weibull cdf with finite threshold: KINK - classical Weibull theory cant explain P gr

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cdf of Structure Strength in Weibull Scale ln( / S 0 ) N eq = Kink used to determine size of RVE and P gr Increasing size Weibull Gaussian 1 RVE 1 2 N Structure 1 – exp[-(σ/ m ] ( Infinite chain - Weibull) ( finite chain ) N eq = equivalent N, modified by stress field (geometry effect) Now: Fit by chain–of– RVEs, zero threshold m =24 ln P f 0.65 KINK - classical Weibull theory cant explain Previously: Fit by Weibull cdf with finite threshold ln( u ) ln[ ln(1 P f )] Age 2 days Weibull (1939) tests of Portland cement mortar 7 days 28 days

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Consequences of Chain-of-RVEs Model for Structural Strength 1) Threshold of power-law tail must be 0, i.e. cdf of strength have kinks at the grafting points, moving up with size (# of RVEs) 2) Failure Prob. N eq = Strength 3) log (strength) log (size) Mean size effect - deviation from power law sets P gr P gr = log (size) C.o.V 0 = = = 0.1 C.o.V. of strength decreases with structural size (# of RVEs) 4) 5) Calculate safety factor for P f = as a function of equiv. # of RVEs log (size) log (strength) m Weibull Asymptote can increase or decrease

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Consequences of Chain-of-RVEs Model for Structural Strength 1) Threshold of power-law tail must be 0, i.e. cdf of strength have kinks at the grafting points, moving up with size (# of RVEs) 2) Failure Prob. N eq = Strength 3) log (strength) log (size) Mean size effect - deviation from power law sets P gr P gr = log (size) C.o.V 0 = = = 0.1 C.o.V. of strength decreases with structural size (# of RVEs) 4) 5) Calculate safety factor for P f = as a function of equiv. # of RVEs log (size) log (strength) m Weibull Asymptote can increase or decrease

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Reinterpretation of Jacksons (NASA) Tests of Type 1 Size Effect on Flexural Strength of Laminates - Energetic-Statistical Theory f r / f r 0 D/D b r = 0.8, m = 35, = Type I Size Effect Energetic-Statistical Size Effect Law: Nominal Strength: = constants, = char. size of structure, = Weibull modulus, = no. of dimension for scaling

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Best Fits of Jacksons (NASA) Individual Data Sets of Laminates Stat. Theory Alone Weibull theory m = 3 and CoV = 3 % ? Weibull theory m = 30 and CoV = 23 % ? Laminate stacking sequence: 1.angle-ply 2.cross-ply 3.quasi-isotropic 4.unidirectional m = 3m = 30 Energetic

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Optimum Fit of Existing Test Data Numerical Simulations by Nonlocal Weibull Theory log(D/D b ) (Size) DbDb D D/DbD/Db asymptote-small asymptote-large Nonlocal Weibull (III) Statistical formula, m =24 Numerical : n m log(f r /f r, ) After Bazant, Xi, Novak (1991, 2000)

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Classical (local) theory: – weakest-link model if one RVE is a continuum point: Weibull Size effect: Nonlocal generalization (finite RVE): = nonlocal strain over one RVE. local averaged Nonlocal Weibull Theory = way to combine statistical & energetic size effects (Ba ž ant and Xi, 1991) failure probability of structure --- to capture stress redistribution approximately (1991): = spatial density of failure probability of continuum point

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RVE – defined by homogenization? – – averaging ~ central limit theorem …captures only low- order statistical moments - misses the crucial cdf tail captured by homogenization 1 0 RVE strength PfPf matters for softening damage & failure of large structure New RVE definition: Smallest material volume whose failure causes failure of the whole structure (of positive geometry). homogenization theory is useless for tail

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Can RVE be largely or mostly Weibullian? NO! 1 0 RVE strength PfPf Assume RVE to be largely Weibullian But then the RVE must behave as a chain But then damage must localize into one sub-RVE So the sub-RVE must be the true RVE Assumed Weibull RVE? NO! = This must be true RVE! Proof:

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ln{ln[ 1/(1– P f )]} Weibull distribution (finite threshold) ln Chain of Gaussian RVEs – Gumbel Weibull distribution (finite threshold) ln( u ) ln(mean strength) ln( eq ) 3.5x P f = Present Theory (zero threshold) ln{ln[ 1/(1– P f )]} ln Present Theory Weibull distribution (finite threshold) N eq = N eq = Dental Ceramics Alumina-glass Composite Lohbauer et al Comparison of Present Theory (No Threshold) to Weibull Model with Finite Threshold 1 m Present Theory

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ln{ ln [ 1/(1– P f ) ] } Optimum Fit by Weibull Theory with Finite Threshold ln( u ) P f = mm 18.6mm n data = 21n data = design =196 u = u = 586 design =586 n data = 107 n data = 27 u = 577 design =577 u = 588 design =588 S.F = 1.84 = 361 = 733 S.F = 1.25 = 691 S.F = 1.18 = 662 S.F = 1.15 u = 13.4 S.F = 1.22 = 16.4 n data = 102 design =13.5 (incorrect) design =230.1 n data = 27 u = = 398 S.F = pt Bend Test on Porcelain (Weibull 1939) 4-pt Bend Test on Dental Alumina-Glass Composite (Lohbauer et al., 2002) 4-pt Bend Test on Sintered –S i C (Salem et al., 1996) 4-pt Bend Test on Sintered Si 3 N 4 with Y 2 O 3 /Al 2 O 3 Additives (Santos et al., 2003) 4-pt Bend Test on Sintered Si 3 N 4 with CTR 2 O 3 /Al 2 O 3 Additives (Santos et al., 2003) 4-pt Bend Test on Sintered Si 3 N 4 (Gross, 2003)

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ln{ ln[ 1/(1– P f )]} Optimum Fit by Chain–of–RVEs, Zero Threshold ln( ) (stress) n data = 102 P f mm 18.6mm 3-pt Bend Test, Porcelain 24 1 n data = pt Bend Test, Sintered –S i C P f n data = 27 P f pt Bend Test on Sintered Si 3 N 4 with Y 2 O 3 /Al 2 O 3 Additives 4-pt Bend Test on Sintered Si 3 N 4 with CTR 2 O 3 /Al 2 O 3 Additives P f pt Bend Test on Sintered Si 3 N P f 0.25 n data = 21n data = 27 (correct) 4-pt Bend Test on Dental Alumina-Glass Composite n data = 27 P f

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ln{ln[1/(1– P f )]} Optimum Fits by Chain-of-RVEs (zero threshold), Weibull cdf with Finite Threshold, and Gaussian cdf ln (strength in expanded scale) 3-pt Bend Test on Porcelain pt Bend Test on Sintered –S i C pt Bend Test on Sintered Si 3 N 4 with Y 2 O 3 /Al 2 O 3 Additives 4-pt Bend Test on Sintered Si 3 N 4 with CTR 2 O 3 /Al 2 O 3 Additives pt Bend Test on Sintered Si 3 N P f = des,G =115 P f = des,G = S.F G = 1.30 S.F R = 1.70 S.F W = 1.22 Chain-of-RVEs Gaussian cdf Weibull cdf, finite threshold Asymptotic cdfs des,R =170 des,W =196 S.F G = 3.14 S.F R = 2.12 S.F W = 1.84 S.F G = 1.53 S.F R = 1.57 S.F W = 1.18 des,R =440 des,W =587 S.F G = 1.55 S.F R = 1.42 S.F W = 1.15 des,G =426 des,R =465 des,W =577 des,G =323 des,R =419 des,W =588 S.F G = 2.27 S.F R = 1.75 S.F W = 1.25 des,R =9.70 des,G =12.62 des,W =13.51 With Threshold (wrong) No Threshold (correct) Gaussian S.F G = N.A. S.F R = 5.53 S.F W = des,W =68 des,R = pt Bend Test on Dental Alumina-Glass Composite

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Lifetime

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Lifetime cdf via Morse Interatomic Potential Mean time between interatomic scission under a constant stress: Unstressed and stressed energy well Atomic vibration Period Nonl. stress dependence of activation energy barrier Morse interatomic potential for 1 bond: Dissociation energy barrier Energy barrier as function of stress (Phoenix): Failure probability of atomic lattice: log ( P f ) log ( ) (1/s 10 to 50) Unstressed bond r E Q Q0Q0 Morse Potential Stressed bond

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Distribution of Lifetime for 1 RVE 1 2 N a) Series Coupling If each link has tail m, the chain has the same tail m b) Parallel Coupling 12 n If each fiber has tail p, the bundle has tail np additive exponents Extent of power-law tail is shortened by parallel coupling: n = 2 : P tail n = 3 : P tail c) Hierarchical Model of Lifetime Distribution long chains Parallel coupling produces cdf with Gaussian core. Series coupling increases the power-tail reach. Power-law tail with zero threshold is indestructible !

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Implications for Lifetime cdf at Macro-Scale 1) Weibull moduli for strength cdf m s and lifetime cdf m 1 n 2) Dependence of cdf of lifetime on structure size 1 m Weibull asymptote 3) Size effect on mean structural lifetime log D ( Size) m s /m 10~50 Much stronger size effect! PfPf 1 0 Structural lifetime N eq =1 N eq =500 N eq =10 4 Gaussian Weibull proportional to activation energy barrier Increasing size

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Failure probability distribution as a function of applied stress and load duration PfPf log( / ref ) /s 0 = Weibull Grafted Weibull - Gaussian /s 0 PfPf Weibull Grafted / ref = 1 log( / ref ) / s 0 PfPf

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Effect of loading duration on mean structural strength µ log( N eq ) log( / ref ) log ( mean strength, µ ) log( / 0 ) log(N eq ) log( /µ 0 ) log( / ref ) N eq = s = 1/50 / ref = m =24

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Simple Probabilistic Analysis via Asymptotics

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Type 1 Size Effect on Mean and pdf via Asymptotic Matching Small ( D 0 ) log D ( Size ) log N ( Nom. Strength ) Gaussian pdf Intermediate Asymptote m ndnd Weibull pdf Large D D Larger D Small Size Asymptote Large Size Asymptote Higher T Longer Each RVE = one hierarchical model

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Malpasset Dam, failed 1959 size effect must have contributed

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Ruins of Malpasset Dam c Photos by Hubert Chanson and Alain Pasquet Failed 1959, at Frejus French Maritime Alps

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Cause of Failure of Malpasset Dam Sudden Localized Fracture Fracture Tolerable movement of abutment would today be 77% smaller.

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Deterministic Computations by ATENA with Microplane Model for Scaled Dams of Various Sizes - progressive distributed distributed cracking cracking - sudden localized fracture REAL MODEL

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Predicting Energetic-Probabilistic Scaling without Nonlocal Analysis Fit of deterministic computations for at least 3 sizes gives One evaluation of Weibull probability integral gives … asymptotic matching formula fixed: ( l 0 neglected )

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Verification by Energetic-Probabilistic Finite Element Simulations assumed Energetic -probabilistic formula matches perfectly!

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For quasibrittle materials, not only the mean nominal strength, but also the strength distribution, the safety factors and lifetime distribution, depend on structure size and shape. Google Bazant, then download 455.pdf,464.pdf, 465.pdf, 470.pdf. CONCLUSION

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