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

2. 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 D L a B
LEFM, KIC or GIC fracture criterion
Contained yielding
EPFM, JC fracture criterion
Full yielding
Diffuse dissipation
Catastrophic failure, large deformations

3. 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.

4. 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 Elastic –plastic fracture
6.4 Elastoplastic asymptotic field (HRR theory)
6.5 Applications of the J-integral

5. 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)

6. x y O θ r
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 se for plane stress
Similarly, for plane strain conditions

8. 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)

9. 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

10. 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 …

11. 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)

12. 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)

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

14. 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

15. 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 4)

16. 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:

17. 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
crack tip B
Total SIF at A:
By changing the variable x = -u, the first integral becomes,

18. Thus, KIA can be expressed by
(see eq. 6.15)
Same expression for KIB (at point B):
One denotes KI for KIA or KIB thereafter

19. 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

20. 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)

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

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

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

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CTOD appropriate crack-tip parameter when LEFM no longer valid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

25. 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 C A B
B→C : non-linear curve
C→D : non-linear, same slope as A-B
non-reversibility: A-B-C ≠ C-D-A
Material behavior is strain history dependent ! D
Non unique solutions for stresses
Simplification: non-linear elastic behavior
elastic-plastic law replaced by the non-linear elastic law
A/D B C
reversibility: A-B-C = C-D-A
Consider only monotonic (non cyclic) loading

26. Definition of the J-integral
Rice defined a path-independent contour integral J for the analysis of cracked bodies
showed that its value = energy release rate in a nonlinear elastic material
J generalizes G to 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 PU* U D D0 D
Fixed-grips conditions:
Dead-load conditions
U : elastic strain energy
U* : complementary energy
U ≠ U* (for nonlinear materials) 27

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

29. At constant displacement: At constant force (dual form):
In particular , D dD D D0
At constant displacement:
At constant force (dual form):
Useful expressions for the experimental determination of J
Generalization of eqs 3.38a and 3.43a to nonlinear elastic materials

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

31. (3) Calculate the potential energy P for given values of displacement u
= 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
(3)
(4)
→ critical value JIc of J at the onset of crack extension (material constant)

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

33. → Equivalence of the two definitions
Proof : X x Y y O S a Gu u Gt T
2D solid of unit thickness and of area S
with a linear crack of length a along OX (fixed)
Crack faces are traction-free
Total contour of the solid G0 including 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),
Strain energy
work of the surface tractions
Tractions and displacements imposed on Gt and Gu are (assumed to be) independent of a
path of integration extended to

35. Consider now the moving coordinate system x , y (attached to the crack tip),
: total derivative/crack length
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
1) J is zero for any closed contour containing no crack tip.
Closed contour around A x y 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,
Replacing in the integral,
Measure of the energy dissipation due to the crack

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

41. x y G3 G4
Note that,
G2 followed 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 J for a circular path of radius r around the crack tipG
crack r
G is followed from q = -p to q = p
One has,
J integral becomes,
when r → 0 only the singular term remain and the integrand can be calculated
For LEFM , one obtains :
(if mode I loading)

43. 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
= yield strength
a : dimensionless constant
n : strain-hardening exponent
material properties
Power law relationship between plastic strain and stress.
For a linear elastic material n = 1.

44. Asymptotic field given by Hutchinson Rice and Rosengren:
Ai are regular functions that depend on q and the power law parameters
HRR stress singularity and strain singularity
singularity recovered for stress and strain in the linear case
J defines the amplitude of the HRR field as K does in the linear case.

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

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

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

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. Example 6.5.2 uy
Example 6.5.3

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

51. Unique relationship between J and CTOD:
m : dimensionless parameter depending on the stress state and materials properties
The strip-yield model predicts that m=1 (non-hardening material, plane stress condition)
This relation is more generally derived for hardening materials (n >1) using the HRR displacements near the crack tip, i.e.
blunted crack
90° dt
Shih proposed this definition for dt :
m becomes a (complicated) function of n
The proposed definition of dt agrees with the one of the Irwin model
Moreover,
in this case

52. J integral along a rectilinear bi-material interface:
SEN specimen geometry (see Appendix III.1): s
a = 40 mm
W = 100 mm
h = 100 mm
a/W = 0.4
Material 1
and h/W = 1
s = 1 MPa.
Remote loading: 2h a W
Materials properties (Young’s modulus, Poisson’s ratio):
Material 1:
E1 = 3 GPa
n1 = 0.35
Material 2 x y
Material 2:
E2 = 70 GPa
n2 = 0.2 s
Plane strain conditions.

53. Typical mesh using Abaqus :
Material 1= Material 2
Material 1
Material 2
Refined mesh around the crack tip
Number of elements used: 1376
Type: CPE8 (plane strain quadratic elements)

54. Results: Remarks: Material 1 Material 2 Bi-material N/mm Abaqus
0.1641
0.0077
0.0837
(*)
(*) same values on the contours 1-8
N/mm=J/mm2
Isotropic
Bi-material KI
KII
Annex III
0.746 0. /
Abaqus
0.748
0.747
0.183
SIF given in
Remarks:
for the isotropic case (i =1,2).
One checks that:

55. Relationship between J and the SIF’s for the bi-material configuration:
- For an interfacial crack between two dissimilar isotropic materials (plane strain),
where
and
plane strain, i = 1,2
KI and KII are now defined as the real and imaginary parts of a complex intensity factor, such that
with

56. Notes for CH6

57. (1) Principal stresses and invariants in plane state problems
The 3-axis (z-axis) is considered as a principal axis:
s : stress tensor
s3 : associated principal stress
The matrix of s in any orthogonal basis (n, t, e3) is
with the invariants,
In principal coordinate system, the stress tensor is diagonal,
and the two principal stresses in the plane 1-2 are given by,

58. (2) Other expression for rp :
plane strain
plane stress
plane stress

59. (3) 1D approximation of plastic zone length
Along the x- axis (mode I),
Plane stress
Plane strain