6. Elastic-Plastic Fracture Mechanics. Plastic zone. crack. B. L. a. D. Introduction. LEFM : Linear elastic fracture mechanics. Applies when non-linear deformation is confined to a small region surrounding the crack tip.
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6. Elastic-Plastic Fracture Mechanics
LEFM: Linear elastic fracture mechanics
Applies when non-linear deformation is confined to a small region surrounding the crack tip
Effects of the plastic zone negligible, linear asymptotic mechanical field (see eqs 4.36, 4.40).
Elastic-Plastic fracture mechanics (EPFM) :
Generalization to materials with a non-negligible plastic zone size: elastic-plastic materials
LEFM, KIC or GIC fracture criterion
Catastrophic failure, large deformations
EPFM, JC fracture criterion
Halfway between surface/midsection
Surface of the specimen
Slip bands at 45°
Plane stress dominant
Net section stress 0.9 yield stress in both cases.
6.1 Models 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 The J contour integral as yield criterion.
6.4 Elastoplastic asymptotic field (HRR theory).
6.5 Applications for some geometries (mode I loading).
6.1 Models for small scale yielding :
Estimation of the plastic zone size
se is the effective stress and si (i=1,2,3) are the principal normal stresses.
LEFM analysis prediction:
KI : mode I stress intensity factor (SIF)
and in their polar form:
Expressions are given in (4.36).
We have the relationships (1),
Thus, for the (mode I) asymptotic stress field:
Substituting into the expression of se for plane stress
Similarly, for plane strain
(see expression of sY2p 6.7 )
is the uniaxial yield strength
increasing n= 0.1, 0.2, 0.3, 0.4, 0.5
Plane stress (n= 0.0)
1) 1D approximation (3)L corresponding 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:
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.
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 rp not strictly correct, because they are based on a purely elastic:
The Irwin approach
Intersection between the elastic distribution and the horizontal line
Plane stress assumed
To equilibrate the two stresses distributions (cross-hatched region)
Plastic zone length (plane stress):
- 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.
(Irwin, plane stress)
Solving, closed-form solution:
Effective crack length 2 (a+ry)
Do i = 1, imax :
Y: dimensionless function depending on the geometry.
Convergence after a few iterations…
Through-crack in an infinite plate (plane stress):
s∞= 2 MPa, sY = 50 MPa, a = 0.1 m
KI = 1.1209982
Dugdale / Barenblatt yield strip model
Very thin plates, with elastic- perfect plastic behavior
application of the principle of superposition (see chap 5)
Long, slender plastic zone from both crack tips.
Perfect plasticity (non-hardening material), plane stress
Crack length: 2(a+c)
Plastic zone extent
Remote tension + closure stresses at the crack
Stresses should be finite in the yield zone:
Closure force at a point x in strip-yield zone:
crack tip A
(see eqs 5.3)
crack tip B
Total SIF at A:
(see equation 5.5)
By changing the variable x = -u, the first integral becomes,
Thus, KIA can written
(see eq. 6.15)
The same expression is obtained for KIB (at point B):
We denote KIfor KIAor KIB thereafter.
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.
Keeping only the first two terms, solving for c
1/p = 0.318 and p/8 = 0.392
tends to overestimate Keff because the actual aeff less than a+c
- Burdekin and Stone derived a more realistic estimation:
(see Anderson, third ed., p65)