1. 6. Elastic-Plastic Fracture Mechanics
crack
Plastic zone
Introduction
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
Elastic Fracture
Contained yielding
Full yielding
Diffuse dissipation D L a B
LEFM, KIC or GIC fracture criterion
EPFM, JC fracture criterion
Catastrophic failure, large deformations

2. Plastic zones (dark regions) in a steel cracked plate (B  5.9 mm):
Surface of the specimen
Midsection
Halfway between surface/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.

3. Outline 6.1 Models for small scale yielding :
- Estimation of the plastic zone size 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).

4. 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. x y O θ r
LEFM analysis prediction:
KI : mode I stress intensity factor (SIF)

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

6. Substituting into the expression of se for plane stress
Similarly, for plane strain
(see expression of sY2 p 6.7 )

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

8. Remarks:
1) 1D approximation (3) L corresponding to
Thus, in plane strain:
in plane stress:
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:
Triaxial state of stress near the crack tip:
- Essentially plane strain in the in the central region.
- Pure plane stress state only at the free surface.
Evolution of the plastic zone shape through the thickness: B

9. 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:
Stress equilibrium not respected.
Alternatively, Irwin plasticity correction using an effective crack length …

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

11. Redistribution of stress due to plastic deformation:
Plastic zone length (plane stress): X
fictitious crack
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.
In plane strain, increasing of sY. : Irwin suggested in place of sY
Stress Intensity Factor corresponding to the effective crack of length aeff =a+r1
(effective SIF)

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

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

14. Dugdale / Barenblatt yield strip model
Assumptions:
Long, slender plastic zone from both crack tips.
Perfect plasticity (non-hardening material), plane stress
Crack length: 2(a+c)
Very thin plates, with elastic- perfect plastic behavior 2a c
Plastic zone extent
Elastic-plastic behavior modeled by superimposing two elastic solutions:
Remote tension + closure stresses at the crack
application of the principle of superposition (see chap 5)

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

16. a c x A B
SIF from the closure stress
Closure force at a point x in strip-yield zone:
Recall first,
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,

17. Thus, KIA can written
(see eq. 6.15)
The same expression is obtained for KIB (at point B):
We denote KI for KIA or KIB thereafter.

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

19. Estimation of the effective stress intensity factor with the strip yield model:
-By setting
Thus,
and
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)