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Diapositive 1 University of Liège Aerospace & Mechanical Engineering Fracture Mechanics, Damage and Fatigue Non Linear Fracture Mechanics: J-Integral Ludovic Noels Computational & Multiscale Mechanics of Materials – CM3 http://www.ltas-cm3.ulg.ac.be/ Chemin des Chevreuils 1, B4000 Liège [email protected] Fracture Mechanics – NLFM – J-Integral Linear Elastic Fracture Mechanics (LEFM) y • Cracked body: summary s – 3 failure modes ∞ Mode I Mode II Mode III s s (opening) (sliding) (shearing) yy yy 2a x s∞ – Asymptotic solution governed by stress intensity factors syy 퐾 Asymptotic s 흈mode 푖 = 푖 푓mode 푖 휃 + ℴ 푟0 Zone of asymptotic yy 2휋r solution (in terms of 1/r1/2) dominance True syy mode 푖 푟 mode 푖 0 풖 = 퐾푖 푔 휃 + ℴ 푟 2휋 x Plasticity Structural response 2021-2022 Fracture Mechanics – NLFM – J-Integral 2 Linear Elastic Fracture Mechanics (LEFM) • Cracked body: summary – Potential energy PT = Eint - Qu before – Crack closure integral • Energy required to close crack tip 풖+횫풖 ′ ′ ΔΠ푇 = 풕 ⋅ 풖 푑풖 푑퐴 Da Δ퐴 풖 after – Energy release rate ti 0 • Variation of potential energy in case of crack growth t i 퐺 = −휕A 퐸int − 푊ext = −휕퐴ΔΠ푇 • In linear elasticity 0 . 1 1 ퟎ t Du/2 퐺 = −휕퐴ΔΠ푇 = − lim 풕 ⋅ 횫풖 푑퐴 Δ퐴→0 Δ퐴 Δ퐴 2 0 u0 u0 +Du u – In linear elasticity & if crack grows straight ahead i i i i 퐾2 퐾2 퐾2 퐺 = 퐼 + 퐼퐼 + 퐼퐼퐼 퐸′ 퐸′ 2휇 2021-2022 Fracture Mechanics – NLFM – J-Integral 3 Linear Elastic Fracture Mechanics (LEFM) • Cracked body: summary B – J-integral y r D q • Strain energy flow x a G – Exists if an internal potential exists • Is path independent if the contour G embeds a straight crack tip • No assumption on subsequent growth direction • Can be extended to plasticity if no unloading (see later) – If crack grows straight ahead G=J – In linear elasticity (independently of crack growth direction): 2021-2022 Fracture Mechanics – NLFM – J-Integral 4 Linear Elastic Fracture Mechanics (LEFM) • Analytical y y y – SIF from full-field solution s∞ • Limited cases t∞ t∞ s x x s∞ x ∞ 2a 2a 2a t∞ t∞ s∞ – From energetic consideration 2 2 2 퐾퐼 퐾퐼퐼 퐾퐼퐼퐼 • Growing straight ahead crack 퐺 = −휕 퐸 − 푊 퐺 = + + A int ext 퐸′ 퐸′ 2휇 • From J-integral • Numerical (e.g. FEM) – bi depends on geometry & crack length • Tabulated solutions (handbooks) – http://ebooks.asmedigitalcollection.asme.org/book.aspx?bookid=230 2021-2022 Fracture Mechanics – NLFM – J-Integral 5 Linear Elastic Fracture Mechanics (LEFM) • Small Scale Yielding assumption – LEFM: we have assumed the existence of a K-dominance zone syy Does not always syy K-dominance zone Asymptotic syy exist LEFM asymptotic ( asymptotic -1/2 syy in r solution in 1/r1/2) True syy True syy x x Plasticity • This holds of if the process zone (in which irreversible process occurs) – Is a small region compared to the specimen size & – Is localized at the crack tip – Validity of this approach? • We check the dimensions ∝Process 2 퐾퐼 zone size 푎, 푊 − 푎 > 2.5 0 휎푝 • Non-linear fracture mechanics – Derivation of the LEFM validity criterion – Providing solutions when LEFM criterion is not met 2021-2022 Fracture Mechanics – NLFM – J-Integral 6 Elastoplastic behavior • Power law – This law can be rewritten in terms of the total deformations • Yield stress is replaced by 2 equivalent stress 1.5 • Plastic strain is replaced by p 0 0 p s s / 1 equivalent strain / = 2; n = 1 s s = 4; n = 1 = 2; n = 4 • The governing law becomes 0.5 = 4; n = 4 = 2; n = 11 = 4; n = 11 or 0 0 2 4 6 8 10 0 E/s 0 • Parameter n E / sp p – n→∞: perfect plasticity – n→1: “elasticity” – Doing so requires 2 assumptions • There is no unloading • As elastic strains are assimilated to plastic strains, the material is incompressible – Which are satisfied if • We are interested only in crack initiation and not in crack propagation • The stress components remain proportional with the loading • Elastic deformations are negligible compared to plastic ones 2021-2022 Fracture Mechanics – NLFM – J-Integral 7 HRR field • Summary 2.5 s~ ; n = 3 rr – Assumptions Plane ~s ; n = 3 2 qq ~s ; n = 3 • J2-plasticity with power law description rq ~s ; n = 3 • Small deformations 1.5 e ~ s • There is no unloading and loading is 1 proportional in all the directions (ok for crack initiation and not for crack propagation) 0.5 • Elastic strains are assimilated to plastic strain 0 0 50 100 150 (material is incompressible) q [degq [°].] • Semi-infinite crack 7 – HRR results for semi-infinite mode I crack Plane • Asymptotic stress, strain and displacement fields 6 Plane s 5 4 n I 3 2 1 0 0 5 10 15 n n • J is a “plastic strain intensity factor” 2021-2022 Fracture Mechanics – NLFM – J-Integral 8 HRR field • Summary (2) 0.2 n = 3; Plane – HRR results for semi-infinite mode I crack (2) n = 3; Plane s • Process zone 0.1 y ~ 0 with -0.1 – If SSY • CTOD -0.2 – New definition -0.2 -0.1 0 0.1 0.2 ~x y 1 s0/E = 0.001 p ux Plane s0/E = 0.002 u 0.8 p y s0/E = 0.004 p s0/E = 0.008 x 0.6 p r* n d 0.4 – 0.2 – Also function of J 0 0 0.1 0.2 0.3 0.4 0.5 – We did not assume SSY !!! 1/n 2021-2022 Fracture Mechanics – NLFM – J-Integral 9 HRR field • Validity in SSY – We have two asymptotic solutions • HRR field is valid in the process zone • LEFM is still valid in the elastic zone close to the crack tip HRR asymptotic HRR asymptotic yy -1/(n+1) s s in r -1/(n+1) yy syy in r syy n+1 LEFM asymptotic Log 1 s in r-1/2 yy 2 True syy LEFM 1 asymptotic x -1/2 Plasticity syy in r Log r – Conditions • This is the case if all sizes are 25 times larger than the plastic zone t a L 2021-2022 Fracture Mechanics – NLFM – J-Integral 10 HRR field • Validity in elasto-plastic conditions – Deformations are small – We still have one asymptotic solution valid • HRR field is valid in the process zone – LEFM is NOT valid in the elastic zone close to the crack HRR asymptotic yy -1/(n+1) HRR asymptotic s syy in r s in r-1/(n+1) n+1 yy Log syy 1 LEFM asymptotic -1/2 syy in r 2 True syy LEFM 1 asymptotic -1/2 x syy in r Plasticity Log r – Conditions • This is the case if all sizes are 25 times larger than CTOD t a L 2021-2022 Fracture Mechanics – NLFM – J-Integral 11 HRR field • Validity in large yielding – Example: ligament size is too small – Small deformations assumption does not hold – Neither HRR field nor LEFM asymptotic fields are valid HRR asymptotic HRR asymptotic yy -1/(n+1) -1/(n+1) s s in r syy in r yy syy n+1 LEFM asymptotic Log 1 -1/2 syy in r True syy 2 LEFM 1 x asymptotic -1/2 Plasticity syy in r Log r – There is no zone of J-dominance – Plastic strain concentrations depend on the experiment 2021-2022 Fracture Mechanics – NLFM – J-Integral 12 HRR field • Crack initiation criteria – In SSY: • Criteria based on J or dt are valid: J ≥ JC or dt ≥ dC – J & dt depend on a, the geometry, the loading, … • But as the LEFM solution holds, we can still use K(a)≥KC 0 – May be corrected by using the effective length aeff if s∞< 50% of sp – In Elasto-Plastic conditions: • Criteria based on J or dt are valid: J ≥ JC or dt ≥ dC – J & dt depend on a, the geometry, the loading, … • The LEFM solution DOES NOT hold, we CANNOT use K(a)≥KC – In Large Scale Yielding • Plastic strain concentrations depend on the experiment • Zones near free boundaries or other cracks tend to be less stressed – Solution is no longer uniquely governed by J – Relation between J & dt is dependent on the configuration and on the loading – The critical JC measured for an experiment might not be valid for another one – A 2-parameter characterization is needed 2021-2022 Fracture Mechanics – NLFM – J-Integral 13 Relation J-G • For elasto-plastic materials – J is a useful concept as it can be used as crack initiation criterion – Relation J-G • J = G if an internal potential is defined & if the crack grows straight ahead • For an elasto-plastic material – Since no internal potential can be defined J ≠ G – But let us consider two specimens 1: Non-linear 2: Elasto- » One with a non-linear elastic material (1) elasticity plasticity » One with an elasto-plastic material (2) with the same loading response J1=G1 J ? G – If the crack will grow straight ahead 2 2 s » J1 = G1 as there is a potential defined Q Q » Before crack propagation (which True would involve possible unloading) sTS the stress state is the same for the two materials J2 = J1 = G1 = G2 • If the crack will grow straight ahead, before crack propagation: J = G for an elasto-plastic material E p e True 2021-2022 Fracture Mechanics – NLFM – J-Integral 14 Computation of J • Available methods of computing J – Numerical (see previous lectures) • Contour integral • Domain integral – Energetic approach • Numerical • Experimental – Experimental • Multiple specimen testing • Deeply notched SENB • Eta factor approach – Engineering • Use of handbooks • If J expression is known it can be applied to – Fracture toughness test – Fracture criteria and stability prediction 2021-2022 Fracture Mechanics – NLFM – J-Integral 15 Computation of J : Energetic approach • Prescribed displacement – Let us consider a specimen • With a prescribed displacement • A compliance depending on the crack length – Using compliance curves • Energy release rate: Q Q+dQ • Internal energy: • Energy release rate in terms of displacements – We have Q’ Q’ A A Q Q’ du A’ > A Eint E int u' u' u u 2021-2022 Fracture Mechanics – NLFM – J-Integral 16 Computation of J : Energetic approach Q • Prescribed displacement (2) Q* – Interpretation of Loading Crack A=A Q*+dQ 0 growth for an elastic material • Body with crack surface A0 loaded up to Q* Unloading • Crack growth dA at constant grip A=A0+dA the specimen becomes more flexible u u* the load decreases by •
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