1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J
Feb 23, 2016
1
plastic zone
CH-6 Elastic-Plastic Fracture Mechanics
Laurent HUMBERT
Thursday, May 27th 2010
J
Non linear elastic phenomenon
Introduction
Applies when nonlinear deformation is confined to a small region surrounding the crack tip
LEFM: Linear elastic fracture mechanics
Elastic-Plastic fracture mechanics (EPFM) :
Effects of the plastic zone negligible, asymptotic mechanical field controlled by K
Generalization to materials with a non-negligible plastic zone size: elastic-plastic materials
Full yielding Diffuse dissipation
L D a
Elastic Fracture
D
L
a
, ,L a D B
B
LEFM, KIC or GIC fracture criterion
Contained yielding
L D a
EPFM, JC fracture criterion
Catastrophic failure, large deformations
3
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.
Plastic zones (dark regions) in a steel cracked plate (B 5.9 mm):
4
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 - Irwin’s approach (plastic correction) - Dugdale’s model or the strip yield model
Outline
6.5 Applications of the J-integral
6.3 Elastic –plastic fracture
6.4 Elastoplastic asymptotic field (HRR theory)
5
Von Mises equation:
6.1 Models for small scale yielding :
1 22 22
1 2 1 3 2 312e
e is the effective stress and i (i=1,2,3) are the principal normal stresses.
Recall the mode I asymptotic stresses in Cartesian components, i.e.
3cos 1 sin sin ...
2 2 223
cos 1 sin sin ...2 2 22
3cos sin cos ...
2 2 22
Ixx
Iyy
Ixy
Kr
Kr
Kr
yy
xy
xx
x
y
Oθ
r
LEFM analysis prediction:
KI : mode I stress intensity factor (SIF)
Estimation of the plastic zone size
6
One has the relationships (1) ,
1 222
1,2 2 2xx yy xx yy
xy
Thus, for the (mode I) asymptotic stress field:
1,2 cos 1 sin2 22
IKr
rrr
x
y
Oθ
r
and in their polar form:Expressions are given in (4.36).
1 222
2 2rr rr
r
31 2
0 plane stressplane strain
and
, ,rr r
3
0 plane stress2 cos plane strain
22IKr
7
Substituting into the expression of e for plane stress
1 22 2 212cos sin cos (1 sin ) cos (1 sin )
2 2 2 2 2 22 2I
eK
r
Similarly, for plane strain conditions
1 2
2 21 31 2 1 cos sin22 2
Ie
Kr
1 22 2 21
6cos sin 2cos2 2 22 2
IKr
1 221 3
1 cos sin22 2
IKr
1 2
2 2cos 4 1 3cos2 22
IKr
1 22 2 2 2 21
4cos sin 2cos 2cos sin2 2 2 2 22 2
IKr
1 22cos 4 3cos
2 22IKr
8
Yielding occurs when: e Y Y is the uniaxial yield strength
Using the previous expressions (2) of e and solving for r,
22 2
22 2 2
1cos 4 3cos
2 2 2
1cos 4 1 3cos
2 2 2
I
Yp
I
Y
K
rK
plane stress
plane strain
Plot of the crack-tip plastic zone shapes (mode I):
2
14
p
I
Y
r
K
Plane strain
Plane stress (= 0.0)
increasing 0.1, 0.2, 0.3, 0.4, 0.5
9
1) 1D approximation (3) L corresponding to 0pr Remarks:
221 22
I
Y
KL
Thus, in plane strain:
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:
in plane stress:2
12
I
Y
KL
- Essentially plane strain in the in the central regionTriaxial state of stress near the crack tip:
- Pure plane stress state only at the free surface
B
Evolution of the plastic zone shape through the thickness:
10
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:
Alternatively, Irwin plasticity correction using an effective crack length …
Stress equilibrium not respected
11
The Irwin approach Mode I loading of a elastic-perfectly plastic material:
02
Iyy
Kr
Plane stress assumed (1)
crackx
r1
Y
r2
(1)
(2)
(2)
To equilibrate the two stresses distributions (cross-hatched region)
r2 ?
Elastic:
Plastic correction 2,yy Y r r
r1 : Intersection between the elastic distribution and the horizontal line yy Y
12I
YK
r
1
20 2
rI
Y YK
r dxx
yy
21 2
11222
I IY
Y
K Kr r
2 1r r
2
11
2I
Y
Kr
12
Redistribution of stress due to plastic deformation:
Plastic zone length (plane stress):
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
2
112 I
Y
Kr
- Equilibrium condition along the y-axis not respected
yy
real crack x
Y
fictitious crack
Stress Intensity Factor corresponding to the effective crack of length aeff =a+r1
1 1,I effK a r K a r
2
112
3I
Y
Kr
2eff
yyK
X
X
In plane strain, increasing of Y. : Irwin suggested in place of Y 3 Y
(effective SIF)
13
Application: Through-crack in an infinite plate
Effective crack length 2 (a+ry)
eff yK a r
21
2Y
effeff
KK a
Solving, closed-form solution:2
112
Y
effaK
(Irwin, plane stress)2
12
Iy
Y
Kr
with
2a
ry ry
aeff
14
More generally, an iterative process is used to obtain the effective SIF:
eff eff effK Y a a
Convergence after a few iterations…
Initial: IK Y a a
2
01
2I
Y
Ka am
1 plane stress3 plane strain
m
YesNo
1i i
Algorithm:
( )Ii
eff i iK K Y a a
Application:
Through-crack in an infinite plate (plane stress):
∞ = 2 MPa, Y = 50 MPa, a = 0.1 m KI = 1.1209982Keff= 1.1214469 4 iterations
Y: dimensionless function depending on the geometry.
Do i = 1, imax :
( )1 1I
ii iK Y a a
2( )1
2Ii
iY
Ka a
m
( ) ( 1)I Ii iK K ?
15
Dugdale / Barenblatt yield strip model
2a cc
Long, 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:
application of the principle of superposition (see chap 4)
Crack length: 2(a+c)
Very thin plates, with elastic- perfect plastic behavior
Remote tension + closure stresses at the crack
16
2ac c
Principle:
Stresses should be finite in the yield zone:
• No stress singularity (i.e. terms in ) at the crack tip1 r
• Length c such that the SIF from the remote tension and closure stress cancel one another
SIF from the remote tension:
,IK a c a c
YY
17
SIF from the closure stress Y
Closure force at a point x in strip-yield zone:
YQ dx a x a c
Total SIF at A:
IB YQ ( a c ) xK ,a c
( a c ) x( a c )
YY
a cx
AB
I A YQ ( a c ) xK ,a c
( a c ) x( a c )
Recall first,
crack tip A
crack tip B
a a c
Y YIA Y
( a c ) a
( a c ) x ( a c ) xK ,a c dx dx( a c ) x ( a c ) x( a c ) ( a c )
By changing the variable x = -u, the first integral becomes,
1a
Y
( a c )
( a c ) u ( ) du( a c ) u( a c )
a cY
a
( a c ) u du( a c ) u( a c )
18
a c
Y YIA Y
a
( a c ) x ( a c ) xK ,a c dx( a c ) x ( a c ) x( a c ) ( a c )
a cY
a
( a c ) x ( a c ) x dx( a c ) x ( a c ) x( a c )
2 22
a cY
a
a c dx
( a c ) x
Thus, KIA can be expressed by
Same expression for KIB (at point B):
One denotes KI for KIA or KIB thereafter
(see eq. 6.15)
a a c
Y YIB Y
( a c ) a
( a c ) x ( a c ) xK ,a c dx dx( a c ) x ( a c ) x( a c ) ( a c )
1a
Y
( a c )
( a c ) u ( ) du( a c ) u( a c )
a cY
a
( a c ) u du( a c ) u( a c )
19
Recall that,1
2 2
a c
a
dx acosa c( a c ) x
12I Y Ya c aK ,a c cos
a c
The SIF of the remote stress must balance with the one due to the closure stress, i.e.
0I I YK ,a c K ,a c
12 cosYa c aa c
a c
Thus, cos2 Y
aa c
By Taylor series expansion of cosines,
2 4 61 1 11 ...2! 2 4! 2 6! 2Y Y Y
aa c
Keeping only the first two terms, solving for c22 2
2 88I
YY
a Kc
1/ 0.318 and /8 = 0.392
Irwin and Dugdale approaches predict similar plastic zone sizes
20
eff effK a
Estimation of the effective stress intensity factor with the strip yield model:
cos2eff Y
aa
-By setting effa a c
andcos
2
eff
Y
aK
tends to overestimate Keff because the actual aeff less than a+c
- Burdekin and Stone derived a more realistic estimation:
Thus,
1/ 2
28 lnsec
2eff YY
K a
sec2 Y
a
(see Anderson, third ed., p65)
21
6.2 CTOD as yield criterion
CTOD : crack tip opening displacement
Irwin plastic zone
Initial sharp crack has blunted prior to fracture
Non-negligible plastic deformation
blunted crack
LEFM inaccurate : material too tough !!!
sharp crack
22
Wells proposed d t (CTOD ) as a measure of fracture toughness
Estimation of d t using Irwin model :
crack length: a + ry
for plane stress conditions,
2 att y yu r rd where uy is the crack opening
23
21 22 2 2 2
Iy
K ru sin - cos
Crack opening uy
12 2
IK r
uy
(see eq. 6.20)
and for plane stress, 31
2 1
E
One has
82
yIt
rKE
d
42
Iy
K ruE
24
From Irwin model, radius of plastic zone2
12
Iy
Y
Kr
24 I
tY
K
Ed
CTOD related uniquely to KI and G.4
tY
Gd
and also,
CTOD appropriate crack-tip parameter when LEFM no longer valid
→ reflects the amount of plastic deformation
Replacing,
25
J contour integral = more general criterion than K (valid for LEFM)
6.3 Elastic-plastic fracture
Derive a criterion for elastic-plastic materials, with typical stress-strain behavior:
A→B : linear
A
BC
B→C : non-linear curveC→D : non-linear, same slope as A-B
D
non-reversibility: A-B-C ≠ C-D-A
Simplification: non-linear elastic behavior
reversibility: A-B-C = C-D-A
Material behavior is strain history dependent !Non unique solutions for stresses
elastic-plastic law replaced by the non-linear elastic law
A/D
BC
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
2727
Fixed-grips conditions: Dead-load conditions
0
0U P( )d
Recall : Load-displacement diagram and potential energy
P 0
0
U
U*
U* : complementary energy
U : elastic strain energy
U ≠ U* (for nonlinear materials)
0
0
P*U ( P )dP
28
• Geometrical interpretation:
Definition of J (valid for nonlinear elastic materials):
dJdA
A = B a : for a cracked plate of thickness B
loading paths with crack lengths a and a+daOA and OB :
, :P a
possible relationship between the load P and the displacement for a moving crack
Thus,
One has J dA dU dF
dU OBC OAD OAE EBCD
dF ABCD
J da
,P a
a daa
d
P
B
C
A
DO
E
J dA OAE EAB OAB J obtained incrementally
29
• In particular ,
At constant force (dual form):
1 *
P
UJB a
At constant displacement:
1 UJB a
0
0
1 P dB a
0
0
1 P
PdP
B a
Generalization of eqs 3.38a and 3.43a to nonlinear elastic materials
Useful expressions for the experimental determination of J
d
0
30
• Experimental determination of J:
Ex : multiple-specimen method - Begley and Landes (1972) :
(1) Consider identical specimens with different crack lengths ai
Procedure
(2) For each specimen, record the load-displacement P-u curve at fixed-grips conditions
(1) (2)
31
(3) Calculate the potential energy for given values of displacement u
(4) Negative slopes of the – a curves :
→ critical value JIc of J at the onset of crack extension (material constant)
= area under the load-displacement curve here
Plots of J versus u for different crack lengths ai
(3) (4)
→ plots of – a curves
32
J as a path-independent line integral :
ii
uJ w dy T dsx
T : traction vector at a point M on the bounding surface , i.e. i ij jT n
The contour is followed in the counter-clockwise direction
n : unit outward normal
0
mn
mn ij ijw d
with strain energy density
w dy dsx
uT
u : displacement vector at the same point MT
n
crack
M
x
y
33
→ Equivalence of the two definitions
• 2D solid of unit thickness and of area Swith a linear crack of length a along OX (fixed)
• Crack faces are traction-free
Imposed tractions on the part of the contour t
Applied displacements on u
• Total contour of the solid 0 including the crack tip:
X x
Y
y
O
Sa
u
u
t
T
0 S
Proof :
dJda
ii
uJ w dy T dsx
34
Recall for the potential energy (per unit thickness),
t
i iS
a wdS T u ds
Tractions and displacements imposed on t and u are (assumed to be) independent of a
0
ii
S
d ud d w dS T dsd a d a d a
0
0
it
iu
dT , ondadu onda
ijij
w
i ij jT n
Strain energy work of the surface tractions
0 u t
u t
,
path of integration extended to 0
35
Consider now the moving coordinate system x , y (attached to the crack tip),
x X a
d xda a a x
a x
x X a
Thus,
0
i ii
S
u ud w w dS T dsda a x a x
However,
ij
ij
w wa a
ijij a
1
2ji
ijj i
uua x x
iij
j
ua x
since ij ji
iij
j
ux a
dda
: total derivative/crack length
36
Using the divergence integral theorem,
iij
S S j
uw dS dSa x a
Thus,
0
i iij ij j
S j
u udS n dsx a a
0
ii
uT dsa
The derivative of J reduces to,
0
i ii
S
u ud w w dS T dsda a x a x
0 0
i i ii i
S
u u uw dS T ds T dsx a a x
0
ii
S
uw dS T dsx x
iij
S j
u dSx a
from equilibrium equation
37
Using Green Theorem, i.e. A
Q PP x,y dx Q x,y dy dx dyx y
0
ii
S
ud w dS T dsda x x
0
ii
uJ wdy T dsx
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.
x
y A
ii
uJ wdy T dsx
ii
A
uwJ dxd y T dsx x
Using the Green Theorem, i.e.
Closed contour around A
A
Q PP x,y dx Q x,y dy dx dyx y
Consider
thus,
Again from the divergence theorem,
iij j
A
uw dxd y n dsx x
iij j
u n dsx
i
ijA j
u dxd yx x
39
However,
ij
ij
w wx x
ijij x
1
2ji
ijj i
uux x x
iij
j
ux x
since
iij
j
ux x
ij ji
The integral becomes,
iij
A j
uwJ dxdyx x x
Replacing in the integral,
Invoking the equilibrium equation, 0ij
jx
0J
iji i iij ij
j j j
u u ux x x x x x
i
ijj
ux x
0
Measure of the energy dissipation due to the crack
40
2) J is path-independent
x
y
3
Consider the closed contour21 3 4*
2Γ*
4
and 0J 1 2 3 4
*J J J J J One has
3
The crack faces are traction free :
0i ij jT n on 4and
dy = 0 along these contours
3 40J J
41
x
y
3
1 3 2 4 2
Γ*
4
and
2 2*J J
Note that,
1 20J J J
1 2J J
Any arbitrary (counterclockwise) path around a crack gives the same value of J
J is path-independent
2 followed in the counter-clockwise direction
Choose a convenient contour for calculation
42
Evaluation of J for a circular path of radius r around the crack tip
crack
r
is followed from to =
One has, ds rd dy r cos d
J integral becomes,
,, cos , i
iu r
J w r T r r dx
when r → 0 only the singular term remain and the integrand can be calculated
For LEFM , one obtains :2I'
KJ G
E (if mode I loading)
43
6.4 HRR theory
0 0 0
n
Ramberg-Osgood equation
Power law relationship between plastic strain and stress.
derived first by Hutchinson, Rice and Rosengren assuming the constitutive law :
→ Near crack tip stress field for elastic-plastic fracture
0 0 E 0
= yield strength
For a linear elastic material n = 1.
: dimensionless constant
n : strain-hardening exponent material properties
44
1 1
1
n
ijJAr
1
2
n n
ijJAr
Asymptotic field given by Hutchinson Rice and Rosengren:
Ai are regular functions that depend on and the power law parameters
J defines the amplitude of the HRR field as K does in the linear case.
1 1 13
n n niu A J r
singularity recovered for stress and strain in the linear case12r
HRR stress singularity and strain singularity1
1nr
1n
nr
45
where HRR based upon small displacements is not applicable
Large strain region where crack blunting occurs :
JK
J and K dominated regions near crack tip :
46
6.5 Applications the J-integral
Tyn
xn
B´
A´
A
B
C´
D
C
ds
x
y
O
x
y n
ij ij ij ij xx xx yy yy xy xy1 1= σ dε = σ ε = σ ε +σ ε +2σ ε2 2
w
Loads and geometry symmetric / Ox
for a plane stress, linear elastic problem
J-integral evaluated explicitly along specific contours
iij j
uJ w dy n dsx
?
47
2 2 2xx yy xx yy xy
1 1+ν= σ +σ -2νσ σ + σ2E E
w
From stress-strain relation,
y yx xxx x xy y yx x yy y
u uu u=σ n ds+σ n ds+σ n ds+σ n dsx x x x
iij j
uσ n dsx
Expanded form for
Along AB or B´ A´ nx = -1, ny = 0 and ds=-dy ≠ 0
yxxx yx x
uu=σ dy + σ n dyx x
Along CD or DC´
yxxx yx
uu=σ dy+σ dxx x
nx = 1, ny = 0 and ds=dy ≠ 0
Simplification :
(2D problem)
48
Along OA and A´O J is zero since dy = 0 and Ti = 0
B C Dy y yx x x
xx xy xy yy xx xyA B C
u u uu u uJ=2 w-σ -σ dy+2 σ +σ dx+2 w-σ -σ dyx x x x x x
Along BC or C´B´
yxxy yy
uu=-σ dx-σ dxx x
BC : nx = 0, ny = -1 and ds=dx ≠ 0
C’B’ : nx = 0, ny = 1 and ds=-dx ≠ 0
Finally,
49
Example 6.5.3
Example 6.5.2
uy
uy
2y(1-ν)Eu
J=2hw=(1+ν)(1-2ν)h
2
2 3
12P aJEB h
2
50
Relationship between J and CTOD
ca
y
iij j
uJ n dsx
x
d t
Consider again the strip-yield problem,
the first term in the J integral vanishes because dy= 0 (slender zone)
Y
Y
iij j
un dsx
but
yyy y
un ds
x
yY
udx
x
yY
uJ dx
x
t
tY ydu
d
d Y t d
51
Unique relationship between J and CTOD:
Y tJ m d
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.
1 1 13
n n niu A J r
m becomes a (complicated) function of n
blunted crack 90° d tShih proposed this definition for d t :
4 Y tG d
The proposed definition of d t agrees with the one of the Irwin model
Moreover,4
m in this case
J integral along a rectilinear bi-material interface:
x
y
Material 1
Material 2
Wa2h
and h/W = 1
a = 40 mm
W = 100 mm
h = 100 mm
• SEN specimen geometry (see Appendix III.1):
Material 1:
a/W = 0.4
1 MPa.
Remote loading:
Materials properties (Young’s modulus, Poisson’s ratio):
Plane strain conditions.
E1 = 3 GPa1 = 0.35
Material 2: E2 = 70 GPa2 = 0.2
• Typical mesh using Abaqus :
Material 1= Material 2
Refined mesh around the crack tip
Number of elements used: 1376Type: CPE8 (plane strain quadratic elements)
Material 1
Material 2
Isotropic Bi-material
KI KII KI KII
Annex III 0.746 0. / /
Abaqus 0.748 0. 0.747 0.183
Results:Material 1 Material 2 Bi-material
N/mm N/mm N/mm
Abaqus 0.1641 0.0077 0.0837
MPa mSIF given in
(*) same values on the contours 1-8
for the isotropic case (i =1,2).2
21I
i i
KJE ( )
• One checks that:
(*)
Remarks:
N/mm=J/mm2
and 3 4i i 2 1
ii
i
EG
21i
ii
EE
• Relationship between J and the SIF’s for the bi-material configuration:
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
- For an interfacial crack between two dissimilar isotropic materials (plane strain),
where
56
Notes for CH6
57 2 2
1,21 42
a b a b c
3 3 3σe e
The 3-axis (z-axis) is considered as a principal axis:
(1) Principal stresses and invariants in plane state problems
The matrix of in any orthogonal basis (n, t, e3) is
: stress tensor 3 : associated principal stress
3, ,
3
00
0 0
a cc b
n t eσ with the invariants,
1 3
2 22 3
23 3
12
det
I tr a b
I tr tr a b ab c
I ab c
2
σ
σ σ
σ
In principal coordinate system, the stress tensor is diagonal,
1 2 3
1
2, ,
3
0 00 00 0
e e eσ
and the two principal stresses in the plane 1-2 are given by,
2 21 3 1 2 3 1 3
1 4 (2 3 )2
I I I I
58
22
22 2
1 31 cos sin
4 2
1 31 2 1 cos sin
4 2
I
Y
I
Y
K
K
22 2
22 2 2
1cos 4 3cos
2 2 2
1cos 4 1 3cos
2 2 2
I
Yp
I
Y
K
rK
(2) Other expression for rp :
2
4 1max( )
3 2I
pY
Kr
plane stress
plane strain
arctan (2 2)
plane stress
59
Along the x- axis (mode I),
1 22 22
1 2 1 3 2 312e
1 2 02
Ixx yy
Kx
Plane stress 3 0
(3) 1D approximation of plastic zone length
12I
e YK
r
Plane strain
3 1 2
1 22 2
1 11
2 1 2 1 22e
1
1 22
Ie Y
Kr
1 221 1
12
2
2
11
2I
Y
Kr
22
11 2
2I
Y
Kr