6. Elastic-Plastic Fracture Mechanics
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6. Elastic-Plastic Fracture MechanicsIntroductionApplies when non-linear deformation is confined to a small region surrounding the crack tipLEFM: Linear elastic fracture mechanicsElastic-Plastic fracture mechanics (EPFM) :Effects of the plastic zone negligible, linear asymptotic mechanical field (see eqs 4.36, 4.40).Generalization to materials with a non-negligible plastic zone size: elastic-plastic materialsElastic FractureContained yieldingFull yieldingDiffuse dissipationLEFM, KIC or GIC fracture criterionEPFM, JC fracture criterionCatastrophic failure, large deformations
Plastic zones (light regions) in a steel cracked plate (B 5.0 mm):Slip bands at 45Plane stress dominant Net section stress 0.9 yield stress in both cases. Plastic zones (dark regions) in a steel cracked plate (B 5.9 mm):
6.2 CTOD as yield criterion. 6.1 Models for small scale yielding : - Estimation of the plastic zone size using the von Mises yield criterion. - Irwins approach (plastic correction). - Dugdales model or the strip yield model.Outline6.5 Applications for some geometries (mode I loading).6.3 The J contour integral as yield criterion.6.4 Elastoplastic asymptotic field (HRR theory).
Von Mises equation:6.1 Models for small scale yielding : 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)Estimation of the plastic zone size
We have the relationships (1) ,Thus, for the (mode I) asymptotic stress field: and in their polar form:Expressions are given in (4.36).and
Substituting into the expression of se for plane stressSimilarly, for plane strain(see expression of sY2 p 6.7 )
Using the previous expressions (2) of se and solving for r, plane stressplane strain Plot of the crack-tip plastic zone shapes (mode I): Plane strainPlane stress (n= 0.0)
Remarks: 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.
4) Solutions for rp not strictly correct, because they are based on a purely elastic: 3) Similar approach to obtain mode II and III plastic zones:
The Irwin approach Mode I loading of a elastic-perfectly plastic material:Plane stress assumed(1)crackxr1r2(1)(2)(2)To equilibrate the two stresses distributions (cross-hatched region) Elastic:Plastic correction
Redistribution of stress due to plastic deformation:Plastic zone length (plane stress): Irwins 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.real crackx Stress Intensity Factor corresponding to the effective crack of length aeff =a+r1 (effective SIF)
Application: Through-crack in an infinite plateEffective crack length 2 (a+ry) 2a
More generally, an iterative process is used to obtain the effective SIF: Convergence after a few iterationsAlgorithm:Application:Through-crack in an infinite plate (plane stress):s = 2 MPa, sY = 50 MPa, a = 0.1 mKI = 1.1209982Keff= 1.12144694 iterationsY: dimensionless function depending on the geometry.
Dugdale / Barenblatt yield strip modelLong, slender plastic zone from both crack tips.Perfect plasticity (non-hardening material), plane stress Assumptions:Plastic zone extent Elastic-plastic behavior modeled by superimposing two elastic solutions:Crack length: 2(a+c)Remote tension + closure stresses at the crack
2a Principle: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:
SIF from the closure stressTotal SIF at A:crack tip B(see eqs 5.3)(see equation 5.5)By changing the variable x = -u, the first integral becomes,
Thus, KIA can writtenThe same expression is obtained for KIB (at point B):We denote KI for KIA or KIB thereafter.(see eq. 6.15)
Thus,Keeping only the first two terms, solving for c1/p = 0.318 and p/8 = 0.392
Estimation of the effective stress intensity factor with the strip yield model: andtends to overestimate Keff because the actual aeff less than a+c- Burdekin and Stone derived a more realistic estimation:Thus,(see Anderson, third ed., p65)