plastic zone

1. plastic zone CH-6 Elastic-Plastic Fracture Mechanics J Laurent HUMBERT Thursday, May 27th 2010

2. B L a D Introduction LEFM: Linear elastic fracture mechanics Applies when nonlinear deformation is confined to a small region surrounding the crack tip Effects of the plastic zone negligible, asymptotic mechanical field controlled by K Elastic-Plastic fracture mechanics (EPFM) : Generalization to materials with a non-negligible plastic zone size: elastic-plastic materials Elastic Fracture Full yielding Contained yielding Diffuse dissipation LEFM, KIC or GIC fracture criterion Catastrophic failure, large deformations EPFM, JC fracture criterion

3. Plastic zones (dark regions) in a steel cracked plate (B  5.9 mm): Halfway between surface/midsection Surface of the specimen Midsection • Plastic zones (light regions) in a steel cracked plate (B  5.0 mm): Slip bands at 45° Plane stress dominant Net section stress  0.9 yield stress in both cases.

4. Outline 6.1Models for small scale yielding : - Estimation of the plastic zonesize using the von Mises yield criterion - Irwin’s approach (plastic correction) - Dugdale’s model or the strip yield model 6.2 CTOD as yield criterion 6.3 Elastic –plastic fracture 6.4 Elastoplastic asymptotic field (HRR theory) 6.5 Applications of the J-integral

5. y r θ O x 6.1 Models for small scale yielding : Estimation of the plastic zone size • Von Mises equation: se is the effective stress and si (i=1,2,3) are the principal normal stresses. • Recall the mode I asymptotic stresses in Cartesian components, i.e. LEFM analysis prediction: KI : mode I stress intensity factor (SIF)

6. y r θ O x and in their polar form: Expressions are given in (4.36). One has the relationships(1), and Thus, for the (mode I) asymptotic stress field:

7. Substituting into the expression of sefor plane stress Similarly, for plane strain conditions

8. Yielding occurs when: is the uniaxial yield strength increasing n= 0.1, 0.2, 0.3, 0.4, 0.5 • Using theprevious expressions (2)ofseand solving for r, plane stress plane strain • Plot of the crack-tip plastic zone shapes (mode I): Plane strain Plane stress (n= 0.0)

9. Remarks: 1)1D approximation (3)Lcorresponding to Thus, in plane strain: in plane stress: Triaxial state of stress near the crack tip: Evolution of the plastic zone shape through the thickness: B 2)Significant difference in the size and shape of mode I plastic zones For a cracked specimen with finite thickness B, effects of the boundaries: - Essentially plane strain in the in the central region - Pure plane stress state only at the free surface

10. Stress equilibrium not respected Alternatively, Irwin plasticity correction using an effective crack length … 3)Similar approach to obtain mode II and III plastic zones: 4) Solutions for rpnot strictly correct, because they are based on a purely elastic:

11. The Irwin approach s yy r2 ? r1: Intersection between the elastic distribution and the horizontal line • Mode I loading of a elastic-perfectly plastic material: Elastic: (1) Plane stress assumed (1) (2) Plastic correction (2) crack x r1 r2 To equilibrate the two stresses distributions (cross-hatched region)

12. fictitious crack X • In plane strain, increasing of sY.: Irwin suggestedin place of sY • Redistribution of stress due to plastic deformation: Plastic zone length (plane stress): real crack x • Irwin’s model = simplified model for the extent of the plastic zone: - Focus only on the extent of the plastic zone along the crack axis, not on its shape - Equilibrium condition along the y-axis not respected • Stress Intensity Factor corresponding to theeffective crack of lengthaeff=a+r1 (effective SIF)

13. Application: Through-crack in an infinite plate (Irwin, plane stress) with ry ry aeff Solving, closed-form solution: Effective crack length 2 (a+ry) 2a

14. More generally, an iterative process is used to obtain the effective SIF: Initial: Do i = 1, imax : No Yes Y: dimensionless function depending on the geometry. Algorithm: Convergence after a few iterations… Application: Through-crack in an infinite plate (plane stress): s∞= 2 MPa, sY = 50 MPa, a = 0.1 m KI = 1.1209982 4 iterations Keff= 1.1214469

15. Dugdale / Barenblatt yield strip model Very thin plates, with elastic- perfect plastic behavior c c 2a application of the principle of superposition (see chap 4) • Assumptions: Long, slender plastic zone from both crack tips Perfect plasticity (non-hardening material), plane stress Crack length: 2(a+c) Plastic zone extent • Elastic-plastic behavior modeled by superimposing two elastic solutions: Remote tension + closure stresses at the crack

16. c c • No stress singularity (i.e. terms in) at the crack tip • Principle: 2a Stresses should be finite in the yield zone: • Length c such that the SIF from the remote tension and closure stress cancel one another • SIF from the remote tension:

17. A B Closure force at a point x in strip-yield zone: x a c Recall first, crack tip A • SIF from the closure stress crack tip B Total SIF at A: By changing the variable x = -u, the first integral becomes,

18. Thus,KIAcan be expressed by (see eq. 6.15) Same expression for KIB(at point B): One denotes KIfor KIAor KIB thereafter

19. Recall that, The SIF of the remote stress must balance with the one due to the closure stress, i.e. By Taylor series expansion of cosines, Irwin and Dugdale approaches predict similar plastic zone sizes Thus, Keeping only the first two terms, solving forc 1/p = 0.318 andp/8 = 0.392

20. -By setting • Estimation of the effective stress intensity factor with the strip yield model: Thus, and tends to overestimate Keffbecause the actual aeffless than a+c - Burdekin and Stone derived a more realistic estimation: (see Anderson, third ed., p65)

21. sharp crack LEFM inaccurate : material too tough !!! 6.2CTOD as yield criterion CTOD : crack tip opening displacement Initial sharp crack has blunted prior to fracture Non-negligible plastic deformation blunted crack Irwin plastic zone

22. for plane stress conditions, Wells proposeddt (CTOD ) as a measure of fracture toughness Estimation of dt using Irwin model : crack length: a + ry whereuyis the crack opening

23. q=p uy and for plane stress, One has Crack opening uy (see eq. 6.20)

24. From Irwin model, radius of plastic zone Replacing, and also, CTOD related uniquely toKIandG. CTOD appropriate crack-tip parameter when LEFM no longer valid → reflects the amount of plastic deformation

25. C B Material behavior is strain history dependent ! A D • Simplification: non-linear elastic behavior C B Consider onlymonotonic(non cyclic) loading A/D 6.3 Elastic-plastic fracture • J contour integral = more general criterion than K (valid for LEFM) • Derive a criterion for elastic-plastic materials, with typical stress-strain behavior: A→B : linear B→C : non-linear curve C→D : non-linear, same slope as A-B non-reversibility: A-B-C ≠ C-D-A Non unique solutions for stresses elastic-plastic law replaced by the non-linear elastic law reversibility: A-B-C = C-D-A

26. Definition of the J-integral Rice defined a path-independent contour integralJ for the analysis of cracked bodies showed that its value =energy release rate in a nonlinearelastic material J generalizes Gto nonlinear materials : → nonlinear elastic energy release rate As G can be used as a fracture criterion Jc reduces to Gc in the case of linear fracture

27. Recall : Load-displacement diagram and potential energy P P0 U* U D D0 D Dead-load conditions Fixed-grips conditions: U : elastic strain energy U* : complementary energy U ≠ U*(for nonlinear materials) 27

28. OA and OB : loading paths with crack lengths aanda+da A B possible relationship between the load Pand the displacement Dfor a moving crack D C O Definition of J (valid for nonlinear elastic materials): A = Ba : for a cracked plate of thickness B • Geometrical interpretation: E One has Jobtained incrementally Thus,

29. D • In particular , dD D D0 At constant displacement: At constant force (dual form): Useful expressions for the experimental determination of J Generalization of eqs3.38a and 3.43a to nonlinear elastic materials

30. (1) (2) • Experimental determination of J: • Ex :multiple-specimen method - Begley and Landes (1972) : Procedure (1) Consider identical specimens with different crack lengthsai (2) For each specimen, record the load-displacement P-ucurve at fixed-grips conditions

31. (3) (4) (3) Calculate the potential energy Pfor given values of displacementu = area under the load-displacement curve here → plots of P – a curves (4) Negative slopes of the P – a curves : Plots of J versus u for different crack lengths ai → critical value JIcof J at the onset of crack extension (material constant)

32. strain energy density with n T M y crack x • J as a path-independent line integral : T : traction vector at a point M on the bounding surface G , i.e. u : displacement vector at the same point M n : unit outward normal The contour G is followed in the counter-clockwise direction

33. → Equivalence of the two definitions T Gt S Gu u Proof : Y • 2D solid of unit thickness and of area S with a linear crack of length aalongOX(fixed) y • Crack faces are traction-free O x X a • Total contour of the solid G0including the crack tip: Imposed tractions on the part of the contour Gt Applied displacements on Gu

34. Recall for the potential energy (per unit thickness), work of the surface tractions Strain energy Tractions and displacements imposed on GtandGuare (assumed to be) independent of a path of integration extended to

35. : total derivative/crack length Consider now the moving coordinate system x , y(attached to the crack tip), Thus, However, since

36. Thus, from equilibrium equation Using the divergence integral theorem, The derivative of J reduces to,

37. Using Green Theorem, i.e. J derives from a potential End of the proof

38. Properties of the J-integral y x 1)J is zero for any closed contour containing no crack tip. Closed contour around A A G Consider Using the Green Theorem, i.e. thus, Again from the divergence theorem,

39. The integral becomes, However, since Invoking the equilibrium equation, 0 Replacing in the integral, Measure of the energy dissipation due to the crack

40. y x The crack faces are traction free : on and 2) J is path-independent G3 G4 Consider the closed contour One has and dy = 0 along these contours

41. y x G3 G4 Note that, G2followed in the counter-clockwise direction and Any arbitrary (counterclockwise) path around a crack gives the same value of J J is path-independent Choose a convenient contour for calculation

42. Evaluation of Jfor a circular path of radius raround the crack tip G r crack Gis followed from q = -pto q = p One has, J integral becomes, when r → 0only the singular term remain and the integrand can be calculated For LEFM , one obtains : (if mode I loading)

43. = yield strength 6.4 HRR theory → Near crack tip stress field for elastic-plastic fracture derived first by Hutchinson, Rice and Rosengren assuming the constitutive law : Ramberg-Osgood equation a : dimensionless constant material properties n : strain-hardening exponent Power law relationship between plastic strain and stress. For a linear elastic material n = 1.

44. Asymptotic field given by Hutchinson Rice and Rosengren: Aiare regular functions that depend on qand the power law parameters HRR stress singularity and strain singularity singularity recovered for stress and strain in the linear case Jdefines the amplitude of the HRRfield as Kdoes in the linear case.

45. J and K dominated regions near crack tip : K J Large strain region where crack blunting occurs : where HRR based upon small displacements is not applicable

46. 6.5 Applications the J-integral B´ C´ y y ds O A´ D x A x B C J-integral evaluated explicitly along specific contours Loads and geometry symmetric / Ox ? for a plane stress, linear elastic problem

47. From stress-strain relation, Expanded form for (2D problem) Simplification : nx = -1, ny = 0 and ds=-dy ≠ 0 Along AB or B´ A´ nx = 1, ny = 0 and ds=dy ≠ 0 Along CD or DC´

48. Along BC or C´B´ BC : nx = 0, ny = -1 and ds=dx ≠ 0 C’B’ : nx = 0, ny = 1 and ds=-dx ≠ 0 Along OA and A´O J is zero since dy = 0 and Ti = 0 Finally,

49. uy uy Example 6.5.2 Example 6.5.3

50. G Relationship between J and CTOD Consider again the strip-yield problem, y dt x a c (slender zone) the first term in the J integral vanishes becausedy= 0 but