Fracture Mechanics, Damage and Fatigue Non Linear Fracture Mechanics: J-Integral

Transcription

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 Chemin des Chevreuils 1, B4000 Liège Fracture Mechanics NLFM J-Integral

2 LEFM Definition of elastic fracture Strictly speaking: During elastic fracture, the only changes to the material are atomic separations As it never happens, the pragmatic definition is Therefore The process zone, which is the region where the inelastic deformations Plastic flow, Micro-fractures, Void growth, happen, is a small region compared to the specimen size, and is at the crack tip Linear elastic stress analysis describes the fracture process with accuracy Mode I Mode II Mode III (opening) (sliding) (shearing) Fracture Mechanics NLFM J-Integral 2

3 LEFM SIFs (K I, K II, K III ) define the asymptotic solution in linear elasticity Crack closure integral Energy required to close the crack by an infinitesimal da If an internal potential exists with AND if linear elasticity AND if straight ahead growth J integral Energy that flows toward crack tip If an internal potential exists Is path independent if the contour G embeds a straight crack tip BUT no assumption on subsequent growth direction If crack grows straight ahead G=J If linear elasticity: Can be extended to plasticity if no unloading (see later) Fracture Mechanics NLFM J-Integral 3

4 Elastoplastic behavior Power law This law can be rewritten in terms of the total deformations Yield stress is replaced by equivalent stress Plastic strain is replaced by equivalent strain The governing law becomes or Parameter n 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 Fracture Mechanics NLFM J-Integral 4 0 / / 0 p p = 2; n = 1 = 4; n = 1 = 2; n = 4 = 4; n = 4 = 2; n = 11 = 4; n = E E/ / p 0 p 0.5

5 HRR field ~ Summary Assumptions J2-plasticity with power law description Small deformations There is no unloading and loading is proportional in all the directions (ok for crack initiation and not for crack propagation) Elastic strains are assimilated to plastic strain (material is incompressible) Semi-infinite crack HRR results for semi-infinite mode I crack Asymptotic stress, strain and displacement fields Plane ~ rr ; n = 3 ~ ; n = 3 ~ r ; n = 3 ~ e ; n = [ ] [deg.] Plane Plane I n n n J is a plastic strain intensity factor Fracture Mechanics NLFM J-Integral 5

6 HRR field ~ Summary (2) HRR results for semi-infinite mode I crack (2) 0.2 n = 3; Plane n = 3; Plane Process zone with y If SSY -0.1 CTOD New definition u x u y r* y x d n ~ x Plane p 0 /E = p 0 /E = p 0 /E = p 0 /E = Also function of J We did not assume SSY!!! /n Fracture Mechanics NLFM J-Integral 6

7 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 in r -1/(n+1) yy in r -1/(n+1) yy n+1 LEFM asymptotic 1 yy in r -1/2 2 True yy LEFM asymptotic x Plasticity yy in r -1/2 Conditions This is the case if all sizes are 25 times larger than the plastic zone Log yy 1 Log r a L t Fracture Mechanics NLFM J-Integral 7

8 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 Plasticity yy HRR asymptotic yy in r -1/(n+1) LEFM asymptotic yy in r -1/2 True yy x Conditions This is the case if all sizes are 25 times larger than CTOD Log yy HRR asymptotic yy in r -1/(n+1) n LEFM asymptotic yy in r -1/2 1 Log r a L t Fracture Mechanics NLFM J-Integral 8

9 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 Plasticity yy HRR asymptotic yy in r -1/(n+1) LEFM asymptotic yy in r -1/2 True yy x There is no zone of J-dominance Plastic strain concentrations depend on the experiment Log yy HRR asymptotic yy in r -1/(n+1) n LEFM asymptotic yy in r -1/2 1 Log r Fracture Mechanics NLFM J-Integral 9

10 HRR field Crack initiation criteria In SSY: Criteria based on J or d t are valid: J J C or d t d C J & d t depend on a, the geometry, the loading, But as the LEFM solution holds, we can still use K(a) K C May be corrected by using the effective length a eff if < 50% of p 0 In Elasto-Plastic conditions: Criteria based on J or d t are valid: J J C or d t d C J & d t depend on a, the geometry, the loading, The LEFM solution DOES NOT hold, we CANNOT use K(a) K C 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 & d t is dependant on the configuration and on the loading The critical J C measured for an experiment might not be valid for another one A 2-parameter characterization is needed Fracture Mechanics NLFM J-Integral 10

11 Relation J-G True 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 As no internal potential can be defined J G But let us consider two specimen» One with a non-linear elastic material (1)» One with an elasto-plastic material (2) with the same loading response If the crack will grow straight ahead» J 1 = G 1 as there is a potential defined» Before crack propagation (which would involve possible unloading) the stress state is the same for the two materials J 2 = J 1 = G 1 = G 2 If the crack will grow straight ahead, before crack propagation: J = G for an elasto-plastic material TS Q 1: Non-linear elasticity p J 1 =G 1 E e Q 2: Elastoplasticity J 2? G 2 True Fracture Mechanics NLFM J-Integral 11

12 Computation of J Available methods of computing J Numerical 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 Fracture Mechanics NLFM J-Integral 12

13 Computation of J : Numerical approaches Finite element model: direct computation of J-integral For linear elasticity and for any contour G embedding a straight crack U exists if no unloading G 3 G 2 G 1 y x Fracture Mechanics NLFM J-Integral 13

14 Computation of J : Numerical approaches Finite element model: J-integral by domain integration For linear elasticity and for any contour G embedding a straight crack Let us define a contour C=G 1 +G - +G + -G G 1 Interior of this contour is region D Define q(x, y) such that q = 1 on G G + y q = 0 on G 1 - G - x G As crack is stress free Fracture Mechanics NLFM J-Integral 14

15 Finite element model: J-integral by domain integration (2) Computation of Computation of J : Numerical approaches C is closed divergence theorem C q=0 But & y q=1 x q is discretized using the same shape functions than the elements This integral is valid for any annular region around the crack tip As long as the crack lips are straight D Fracture Mechanics NLFM J-Integral 15

16 Prescribed loading Let us consider a specimen With a prescribed loading (dead load) A compliance depending on the crack length Using compliance curves Energy release rate: Computation of J : Energetic approach Complementary energy: Q du Q Q* Q u dq A Energy release rate in terms of displacements As E int u* u Fracture Mechanics NLFM J-Integral 16

17 Computation of J : Energetic approach Prescribed loading (2) Interpretation of for an elastic material Body with crack surface A 0 loaded up to Q* Crack growth da at constant load the specimen becomes more flexible displacement increment Unload to zero The area between the 2 curves is then G da For an elasto-plastic material Q Q* Loading A=A Crack growth 0 Unloading A=A 0 +da u u* u*+du As J = G only before crack growth, this method can be used Either experimentally with 2 specimen with a different initial crack length Or by computing the curve Q(a, u)» Analytically» Using FEM Fracture Mechanics NLFM J-Integral 17

18 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: Internal energy: Energy release rate in terms of displacements As Q Q+dQ Q A <A A E int u* u Fracture Mechanics NLFM J-Integral 18

19 Prescribed displacement (2) Interpretation of for an elastic material Computation of J : Energetic approach Q Q* Loading Q*+dQ A=A 0 Crack growth Body with crack surface A 0 loaded up to Q* Unloading Crack growth da at constant grip the A=A 0 +da specimen becomes more flexible u* the load decreases by Unload to zero The area between the 2 curves is then - G da For an elasto-plastic material As J = G only before crack growth, this method can be used Either experimentally with 2 specimen with a different initial crack length Or by computing the curve u(a, Q)» Analytically» Using FEM u Fracture Mechanics NLFM J-Integral 19

20 Experimental determination of J Multiple specimen testing (Begley & Landes, 1972) Consider 4 specimens with 4 different crack lengths a 1 < a 2 < a 3 < a 4 Under displacement control No fracture taking place This leads to compliance curves Integration to obtain the internal energies W u,q Thickness t a L=4W Q a 1 a 2 a 3 a 4 E int /t u 4 u 3 u 2 u 1 u a u 1 u 2 u 3 u 4 a 1 a 2 a 3 a Fracture Mechanics NLFM J-Integral 20

21 Experimental determination of J Multiple specimen testing (2) As Which corresponds to saying that the area between the curves u(a, Q) and u(a+da, Q) is equal to - G t da The slopes of the extrapolated curves E int (u, a) lead to J E int /t J u 4 a 1 u 3 a 2 u 2 a 3 u 1 J 1 a a 4 u a 1 a 2 a 3 a 4 Small displacements: LEFM holds u 1 u 2 u 3 u 4 For large displacements This technique is very time consuming but it avoids unloading at the crack Fracture Mechanics NLFM J-Integral 21

22 Deeply notched specimen testing Consider 1 Single Edge Notch Bend specimen Prescribed loading Q Non-dimensional analysis 9 input variables: Experimental determination of J W Q, u Thickness t a [ p0 ] = kg. s -2. m -1, [W-a] = m, [t] = m, [E] = kg. s -2. m -1, [n] = -, [n] = -, L [L] = m, [W] = m, [Q] = kg. m. s -2 3 dimensions 9-3 = 6 independent non-dimensional inputs n, n, E / p0, QL / Et(W-a) 2, L/(W-a), W/(W-a) 1 output variable: [u] = m 1 relation in terms of the non-dimensional inputs Prescribed loading Before crack propagation: Fracture Mechanics NLFM J-Integral 22

23 Experimental determination of J Deeply notched specimen testing (2) J-integral with But So the J-integral is rewritten Fracture Mechanics NLFM J-Integral 23

24 Experimental determination of J Deeply notched specimen testing (3) So Can I 1 and I 2 be neglected so that? This would be convenient as it involves only the load-displacement curve Q a Jt(W-a)/2 u Fracture Mechanics NLFM J-Integral 24

25 Deeply notched specimen testing (4) If the Single Edge Notch Bend specimen Is deeply notched Has a remaining ligament W-a fully plastic There is a plastic hinge Experimental determination of J The curve Q u represents essentially the plastic deformations Crack is deep enough so the plastic hinge is localized between the applied load and the crack tip Example: perfectly plastic material L u y x When a plastic hinge exists Q/2 u Static equilibrium: QL/4 = M M And we have L/2 The J-integral becomes I 1 & I 2 = 0 OK Fracture Mechanics NLFM J-Integral 25

26 Experimental determination of J Eta factor approach Expression Is convenient as it involves only the load-displacement curve Is valid for SENB with plastic hinge u Q Jt(W-a)/2 a u Elastic materials SENB with L/W = 4 & a/w > 0.5: I 1 and I 2 can be neglected* Elastic-plastic materials SENB at low temperature, for L/W = 4 & 0.6> a/w > 0.4: I 1 and I 2 can be neglected (Comparison with multiple specimen testing**) How can this expression be generalized to Other geometries, loadings Elastic-plastic materials L *Srawley JE (1976), On the relation of J to work done per unit uncracked area: total, or component due to crack, International Journal of Fracture 12, **Castro P, Spurrier P, and Hancock J (1984), Comparison of J testing techniques and correlation J-COD using structural steel specimens, International Journal of Fracture 17, Fracture Mechanics NLFM J-Integral 26

27 Eta factor approach (2) Experimental determination of J How can be generalized? For other specimens and through-cracked structures But Assuming deformations are largely plastic Eta factor: With h J depending on Geometry & loading Crack length (a/w) Material hardening (n) Materials are not rigidly plastic, so there is an elastic-plastic response How to determine h J? Fracture Mechanics NLFM J-Integral 27

28 Experimental determination of J Eta factor approach (3) How can be generalized (2)? For elastic-plastic behavior Split of elastic and plastic parts of The displacement The J-integral The Crack Mouth-Opening Displacement Q, u Q a W v/2 Thickness t v/2 J p t(w-a)/h p L=4W u Factor h p still remains to be evaluated u p u e Fracture Mechanics NLFM J-Integral 28

29 Experimental determination of J Eta factor approach (4) How can be generalized (3)? For elastic-plastic behavior (2) It is more accurate to measure the CMOD v than the displacement u But v and Q are not work conjugated the new h p factor has to be evaluated with FE Q, u Q a CMOD, v W v/2 Thickness t v/2 J p t(w-a)/h p L=4W v v p v e Fracture Mechanics NLFM J-Integral 29

30 Eta factor approach (5) Factor h p in relation Numerical part: Experimental determination of J Build FE model of the specimen Non-linear analysis Characterize non-linear material response from uniaxial tensile test Load the specimen incrementally, and at each increment» Extract J (with contour or domain integral)» Compute Linear analysis Extract K I and then J e (Q) Factor h p is the slope of J p =J-J e vs W p /t(w-a) Q, u Q a J-J e a W v/2 Thickness t v/2 W p v 1 h p L=4W v p v e W p /t(w-a) Fracture Mechanics NLFM J-Integral 30

31 Engineering determination of J Use of handbooks Split of J: Use of handbooks to determine J e (see SIF lecture) and J p : HRR field This solution holds for fully plastic solutions J becomes J p r becomes W-a As» becomes Q» p0 becomes yielding load Q 0 Tabulation of the values for different geometries or depending on the test Similar relations for n p, u p and d t Fracture Mechanics NLFM J-Integral 31

32 Engineering determination of J Determination of J p Example of centered crack plate y 2W h 2a x With Other h i tabulated p 0 /E = p 0 /E = p 0 /E = p 0 /E = Plane p 0 /E = p 0 /E = p 0 /E = p 0 /E = d n 0.4 d n Plane /n /n Fracture Mechanics NLFM J-Integral 32

33 Engineering determination of J Determination of J p (2) Example of centered crack plate (2) Plane n=1 n=2 n=3 n=5 n=7 y a/w = 1/8 h h h W 2a h x a/w h = 1/4 h h Fracture Mechanics NLFM J-Integral 33

34 Engineering determination of J Effective crack length Evaluation of is required in the engineering method Effective crack length a eff =a + h r p (h not to be mistaken for the previous one) Recall in SSY, h =1/2 & For Large Scale Yielding this would lead to lengths larger than the ligament Correction of h: Use of the same plastic zone expression Fracture Mechanics NLFM J-Integral 34

35 Fracture toughness testing for elastic-plastic materials Procedure detailed in the ASTM E1820 norm What is measured? Plane strain value of J prior to significant crack growth: J C Plane strain value of J near the onset of stable crack growth: J IC J vs Da resistance curve for stable crack growth: J R Specimen that can be used Either SENB or Compact Tension specimen (for CT: CMOD measured at the axes of the loadings) Pre-cracked by fatigue loading Correct sizes Q, u W v/2 Q, u Thickness t a v/2 v/2 h W L=4W v/2 a Thickness t Q, u Fracture Mechanics NLFM J-Integral 35

36 Fracture toughness testing for elastic-plastic materials R-curve method 1 st step: Determine crack length a 0 after fatigue loading 3 cycles of loading-unloading between Q fat and Q fat /2 with Q fat the maximal loading during pre-cracking The crack length is obtained using the tabulated compliance curves C(a) during these cycles (see lecture on Energetic Approach) Example: SENB» If» Then W v/2 L Q, u Thickness t a v/2 In elasticity: C(a) = C e (a) = v/q» In terms of Load Line displacement: Fracture Mechanics NLFM J-Integral 36

37 Fracture toughness testing for elastic-plastic materials R-curve method (2) 2 nd step: Proceed with the testing: increase loading At regular intervals i: unload In order to determine a i with the compliance Do not unload too much in order to avoid reverse plastic loading At least 8 unloadings are required Q Q i C(a i ) 1 a 0 After the final loading step Unload to zero and Mark final crack length Break open (cool if required to have brittle fracture) Measure final crack length v Fracture Mechanics NLFM J-Integral 37

38 Fracture toughness testing for elastic-plastic materials R-curve method (3) 3 rd step: Data reduction For each pair (a i, Q i ) Calculate J e,i : (actually using a eff )» With for SENB Calculate plastic displacement Increment in plastic work is the area below the Q-u p curve J As for SENB h-factor is equal to 2: Average ligament size Eventually: This allows drawing The J vs Da=a-a 0 curve This requires accurate determination of a Fracture Mechanics NLFM J-Integral 38 Da

39 Fracture toughness testing for elastic-plastic materials R-curve method (4) 4 th step: Analysis Even before physical crack growth there is a blunting of crack tip Due to plasticity Which results in an apparent Da Is corrected by plotting the blunting line Using data points in between the 0.15 and 1.5-mm exclusion lines, extrapolate J : Intersection between the 0.2mm offset line and this extrapolation gives the fracture toughness J IC, if the sizes are correct J R J IC J blunting u x u y r* 0.2mm y J extrapolated x Intersection between the blunting line and this extrapolation gives J C 0.15mm 1.5mm Da Values beyond J IC gives the resistance curve J R Fracture Mechanics NLFM J-Integral 39

40 J I [kj/m 2 ] Resistance curves J I [kj/m 2 ] Experimental curves Example: steel A533-B at 93 C* Thickness has an important influence Slight influence of the initial crack length Side grooving suppresses the plane stress state J blunting t =25 mm t =100 mm t =100 mm, side grooved a =116 mm a =134 mm a =146 mm Δa [mm] From these curves one can extract Δa [mm] The toughness J IC crack initiation criterion J (a, Q) = J IC The resistance curve J R (Da) crack stability criterion? *Andrews WR and Shih CF (1979), Thickness and side-groove effects on J- and δ-resistance curves for A533-B steel at 93 C,ASTM STP 668, Fracture Mechanics NLFM J-Integral 40

41 Reminder: LEFM crack grow stability Non-perfectly brittle material: Resistance curve For non-perfectly brittle materials G C depends on the crack surface Therefore G C will be renamed the resistance R c (a) Elastoplastic behavior In front of the crack tip there is an active plastic zone Resistance increases when the crack propagates as Blunting the required plastic work increases This effect is more important for thin Plastic specimens as the zone elasto-plastic behavior 0 is more pronounced in plane A steady state can be reached or not Stability of the crack also depends on the variation of G with crack length Stable crack growth if Unstable crack growth if R c Initial crack growth R c ductile t 3 << (Plane ) R c ductile t 2 <t 1 Steady state R c ductile t 1 R c brittle Plastic wake Fracture Mechanics NLFM J-Integral 41 Active plastic zone DA=tDa

42 Reminder: LEFM crack grow stability Example: Delamination of composite with initial crack a 0 R c, G Dead load Q a Q Q 2 3 Q 1 h h Thickness t R c ducile vs Fixed grip u R c, G u 2 u 1 u 3 a h Thickness t h R c ducile u increasing Q increasing G C brittle G C brittle a 0 a* a** a a 0 a Dead load Q 1 Perfectly brittle materials: unstable Ductile materials: stable, but if a is larger than a** it turns unstable Fixed grip u 1,u 2 or u 3 The crack is stable for any material Fracture Mechanics NLFM J-Integral 42

43 J-controlled crack growth & NLFM crack grow stability Under which conditions is the crack growth J-controlled? Recall that in order to use J Unloading is prohibited So the solution is valid before crack growth But it would be convenient to use the J R curve When is the crack growth still J-controlled? HRR solution holds in an annular region As HRR solution is singular, there is a region not controlled by J There is an annular region between r* and r** controlled by J If the crack grows by da There is an elastic unloading in a region that scales with da So for J to remains relevant if the crack grows For a potential to be defined we required no unloading One clear condition is therefore da << r** Other conditions? Elastic unloading da J-controlled region r* r** Non-proportional loading x Fracture Mechanics NLFM J-Integral 43

44 J-controlled crack growth & NLFM crack grow stability Under which conditions is the crack growth J-controlled (2)? HRR strain field (between r* and r**): As depends on J and a: With As the crack is moving to the right: With & Da y x r x So one has with At the end of the day: Fracture Mechanics NLFM J-Integral 44

45 J-controlled crack growth & NLFM crack grow stability Under which conditions is the crack growth J-controlled (3)? We have with Term is in 1/r it changes the proportionality of the solution But & are of the same order We have J-controlled crack growth if or From the J R curves, the length scale can be extracted as The J-controlled growth criteria are & r** remains to be determined Stability is not established yet Fracture Mechanics NLFM J-Integral 45

46 J-controlled crack growth & NLFM crack grow stability Crack growth stability The J-controlled crack growth criteria are & r** is a fraction Of the remaining ligament W-a Or other characteristic length (e.g. up to r p / 4) For SENB: da < 6% (W-a) and D < 10% (W-a) (obtained by FE analysis) If these criteria are satisfied the stability of the crack can be assessed J depends on the crack length and loading Toughness variation with Da is evaluated from J R curves Stable crack if In terms of the non-dimensional tearing: Fracture Mechanics NLFM J-Integral 46

47 J-controlled crack growth & NLFM crack grow stability Crack growth stability (2) Tabulated values Material Specimen T [ C] ( p0 + TS )/2 [MPa] J IC [MPa. m] dj/da [MPa] T [] D [mm] ASTM 470 (Cr-Mo-V) ASTM 470 (Cr-Mo-V) ASTM 470 (Cr-Mo-V) ASTM A453 Stainless steel ASTM A453 Stainless steel ASTM A453 Stainless steel 6061-T651 Aluminum CT CT CT CT CT CT CT Fracture Mechanics NLFM J-Integral 47

48 Steady State crack grow Moving crack and elasto-plastic material A moving crack sees a plastic wake The nonlinear elasticity assumption (power law) is not valid anymore Ex: perfectly plastic material HRR Steady state Singularity is weaker in the steady state, so the apparent J limit is larger J SS is the steady state limit of J = J IC for brittle materials Many times J IC for ductile materials This is characterized by R c Blunting Plastic zone 0 Initial crack growth R c ductile t 1 R c brittle Plastic wake Steady state Active plastic zone DA=tDa Fracture Mechanics NLFM J-Integral 48

49 2-parameter theory A single parameter cannot always fully described the process Example 1: Thickness effect In LEFM, recall T-stress is the 0-order term, which is dominant at radius r c In general, if T < 0, the measured fracture K will be larger than for T > 0 If thick specimen T > 0 HRR solution is also an asymptotic behavior Measure of J limit is also thickness-dependant Example 2: Large Yielding There is no zone of J-dominance Plastic strain concentrations depend on the experiment Log yy K rupture K c T>0 Plane 2a HRR asymptotic yy in r -1/(n+1) n+1 1 Plane t 2 LEFM asymptotic yy in r -1/2 1 Log r Fracture Mechanics NLFM J-Integral 49

50 2-parameter theory This motivates the accounting for second order effects in the HRR theory J-Q theory (Q in order to be distinct from Q the general loading) All the second order effects are accounted for by a scalar Q Full field can be obtained using finite element method: HRR solution is known: Q correction is then deduced: This method applied to different cases shows that Exponent q ~ 0 Correction is mainly hydrostatic ahead of crack tip (for <p/2) Q is then obtained at one location, and using the FEM e.g. for & at Fracture Mechanics NLFM J-Integral 50

51 2-parameter theory Applications of the J-Q theory Using In SSY there is a unique relation between Q and T Biaxial stress on Centered Crack Panel (CPP) Applying an external tension parallel to the crack results in constraining Q to zero In the absence of such tension, as J increases we get closer to Large Scale Yielding and Q decreases Fracture Mechanics NLFM J-Integral 51

52 2-parameter theory Applications of the J-Q theory (2) SENB specimen Q, u W v/2 Thickness t v/2 L=4W Q depends on the size of the remaining ligament in order to measure J IC we want a deeply cracked specimen (Q ~0) As J increases we get closer to Large Scale Yielding and Q reaches -1 As the stress will be lower for Q <0, the measured fracture toughness will be higher Cleavage fracture Fracture Mechanics NLFM J-Integral 52

53 Exercise 1: SENB specimen Consider the SENB specimen Made of steel E = 210 GPa, p0 = 700 MPa Power law: =0.5, n=10 Geometry t = 10 mm, W = 75 mm, L = 300 mm (Plane state) a = 28 mm W Q, u v/2 v/2 L=4W Thickness t Assuming a toughness of J IC = 200 kpa. m, dj/da = 87 MPa Compute the maximal loading before fracture using LEFM Compute the maximal loading before fracture using NLFM Evaluate the crack growth stability Fracture Mechanics NLFM J-Integral 53

54 Exercise 2: ASTM E813 Consider two alloys Both failed in inter-granular mode Voids growth and coalescence at Material 1 grain boundaries Yield p0 = 300 MPa Grain size 10 mm Material 2 Yield p0 = 300 MPa Grain size 220 mm Questions = 85 MPa a = 25 mm W = 100 mm t = 30 mm Extent and validity of J-dominance? ASTM E813 for elasto-plastic fracture requires that initial crack size a > W/2, why? = 85 MPa Fatigue of Materials, S. Suresh, Cambridge press, Fracture Mechanics NLFM J-Integral 54

55 Lecture notes References Lecture Notes on Fracture Mechanics, Alan T. Zehnder, Cornell University, Ithaca, Other references «on-line» Fracture Mechanics, Piet Schreurs, TUe, Book Fracture Mechanics: Fundamentals and applications, D. T. Anderson. CRC press, An Engineering Approach for Elastic-Plastic Fracture Analysis, V. Kumar, M.D. German, C.F. and Shih CF (1981), Tech. Rep. NP-1931, Electric Power Research Institute. Fatigue of Materials, S. Suresh, Cambridge press, Fracture Mechanics NLFM J-Integral 55

56 Exercise 1: Solution LEFM solution SIF (handbook) with Effective crack length with & with (handbook) Example Q=20 kn Starting point Yield load Fracture Mechanics NLFM J-Integral 56

57 Exercise 1: Solution LEFM solution (2) Example Q=20 kn (2) First iteration Second iteration Fracture Mechanics NLFM J-Integral 57

58 K I [Mpa m 1/2 ] Exercise 1: Solution LEFM solution (3) Whole Q-range with Or in terms of J = K I2 / E K I K C 250 J e J IC K I [MPa. m 1/2 ] J [kpa. m] Q [kn] Q [kn] Rupture load in LEFM (with corrected h) is 63.8 kn Fracture Mechanics NLFM J-Integral 58

59 Exercise 1: Solution NLFM solution Handbook J p : Yield load: Coefficient: Fracture Mechanics NLFM J-Integral 59

60 Exercise 1: Solution NLFM solution (2) Example Q = 60 kn: Whole range of Q: Limit load kn Validity: J dominance if a, W-a > 25 J / p0 = 7.1 mm OK K dominance if a, W-a > 2.5 (K / p0 ) 2 = m KO J [kpa. m] J e J p J=J e +J p J IC Q [kn] Fracture Mechanics NLFM J-Integral 60

61 Exercise 1: Solution Crack growth stability Region of J-controlled crack growth Material length scale: For SENB, crack growth is J-controlled if D < 0.1 (W-a) = 4.7 mm OK Then, stability can be studied for Da < 0.06 (W-a) = 2.8 mm 2 studies: a = 28 mm: done Q rupt = kn, J = 200 kpa. m Dead load: Q = kn but with a = 30 mm J = 286 KPa. m T for this geometry and dead load J [kpa. m] J(a=28 mm) J(a=30 mm) Tearing of the material Q [kn] stable crack (for Da < 2.8 mm) Fracture Mechanics NLFM J-Integral 61

62 Exercise 2: Solution Extent of J-dominance SIF handbook (SSY assumption) = 85 MPa a = 25 mm W = 100 mm t = 30 mm = 85 MPa Fatigue of Materials, S. Suresh, Cambridge press, Fracture Mechanics NLFM J-Integral 62

63 Exercise 2: Solution Extent of J-dominance (2) SIF (SSY): Plastic zone yy HRR asymptotic yy in r -1/(n+1) LEFM asymptotic yy in r -1/2 True yy All the size are not > 25 r p Plasticity x SSY almost holds so the size of the process zone J-controlled region is representative and we use it as an approximation We should use J-criterion Elastic unloading da r* r** x Typical J-dominance radius Non-proportional loading Fatigue of Materials, S. Suresh, Cambridge press, Fracture Mechanics NLFM J-Integral 63

64 Exercise 2: Solution Validity of J-dominance Criterion satisfied for standard E But we also want grain size << zone r** As For 220 mm-grain size Only a few crystals will be in the process zone Crystal plasticity will dominate rather than inter-granular void growth For 10 mm-grain size Continuum assumption holds J-criterion holds Elastic unloading J-controlled region da r** r* x Non-proportional loading Fatigue of Materials, S. Suresh, Cambridge press, Fracture Mechanics NLFM J-Integral 64

65 Exercise 2: Solution ASTM E813 for elasto-plastic fracture For a/w = 0.25: = 85 MPa We need a larger plastic zone If a larger: combination of bending/traction will extend this plastic zone a = 25 mm W = 100 mm t = 30 mm = 85 MPa Fatigue of Materials, S. Suresh, Cambridge press, Fracture Mechanics NLFM J-Integral 65