1 plastic zone ch-6 elastic-plastic fracture mechanics laurent humbert thursday, may 27th 2010 j
TRANSCRIPT
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 31
2e
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.
3cos 1 sin sin ...
2 2 22
3cos 1 sin sin ...
2 2 22
3cos sin cos ...
2 2 22
Ixx
Iyy
Ixy
K
r
K
r
K
r
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
IK
r
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 stress
plane strain
and
, ,rr r
3
0 plane stress
2cos plane strain
22IK
r
7
Substituting into the expression of se 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 sin
22 2I
eK
r
1 22 2 21
6cos sin 2cos2 2 22 2
IK
r
1 221 3
1 cos sin22 2
IK
r
1 2
2 2cos 4 1 3cos2 22
IK
r
1 22 2 2 2 21
4cos sin 2cos 2cos sin2 2 2 2 22 2
IK
r
1 22cos 4 3cos
2 22IK
r
8
Yielding occurs when: e Y Y is the uniaxial yield strength
Using the previous expressions (2) of se 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
1
4
p
I
Y
r
K
Plane strain
Plane stress (n= 0.0)
increasing = n 0.1, 0.2, 0.3, 0.4, 0.5
9
1) 1D approximation (3) L corresponding to 0pr Remarks:
221 2
2I
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
1
2I
Y
KL
- Essentially plane strain in the in the central region
Triaxial 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
K
r
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
yys
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
11
2 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
11
23
I
Y
Kr
2eff
yyK
X
X
In plane strain, increasing of sY. : Irwin suggested in place of sY 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
11
2Y
effa
K
(Irwin, plane stress)2
1
2I
yY
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 a
m
1 plane stress
3 plane strainm
Yes
No
1i i
Algorithm:
( )I
ieff i iK K Y a a
Application:
Through-crack in an infinite plate (plane stress):
s∞ = 2 MPa, sY = 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
2I
i
iY
Ka a
m
( ) ( 1)I I
i 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 ) x
K ,a c( a c ) x( a c )
YY
a c
xAB
I A YQ ( a c ) x
K ,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 ) udu
( 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 ) xdx
( a c ) x ( a c ) x( a c )
2 22
a c
Ya
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 ) udu
( a c ) u( a c )
19
Recall that,1
2 2
a c
a
dx acos
a c( a c ) x
12I Y Ya c a
K ,a c cosa 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 a
a ca c
Thus, cos
2 Y
a
a c
By Taylor series expansion of cosines,
2 4 61 1 1
1 ...2! 2 4! 2 6! 2Y Y Y
a
a 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
a
a
-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
2
8lnsec
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, 3
1
2 1
E
One has
8
2yI
trK
E d
4
2I
yK r
uE
24
From Irwin model, radius of plastic zone2
1
2I
yY
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 curve
C→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):
dJ
dA
A = B a : for a cracked plate of thickness B
loading paths with crack lengths a and a+da
OA 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
D
O
E
J dA OAE EAB OAB J obtained incrementally
29
• In particular ,
At constant force (dual form):
1 *
P
UJ
B a
At constant displacement:
1 UJ
B a
0
0
1 Pd
B 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
dD
D D0D
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 ds
x
T : traction vector at a point M on the bounding surface G , i.e. i ij jT n
The contour G is followed in the counter-clockwise direction
n : unit outward normal
0
mn
mn ij ijw d
with strain energy density
w dy dsx
u
T
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 S
with a linear crack of length a along OX (fixed)
• Crack faces are traction-free
Imposed tractions on the part of the contour G t
Applied displacements on G u
• Total contour of the solid G 0 including the crack tip:
X x
Y
y
O
Sa
G u
u
G t
T
0 SG
Proof :
dJ
da
ii
uJ w dy T ds
x
34
Recall for the potential energy (per unit thickness),
t
i iS
a wdS T u ds
Tractions and displacements imposed on G t and G u are (assumed to be) independent of a
0
ii
S
d ud d wdS T ds
d a d a d a
0
0
it
iu
dT, on
dadu
onda
G
G
ijij
w
i ij jT n
Strain energy work of the surface tractions
0 u t
u t
,G G GG G
path of integration extended to 0G
35
Consider now the moving coordinate system x , y (attached to the crack tip),
x X a
d x
da a a x
a x
x X a
Thus,
0
i ii
S
u ud w wdS T ds
da a x a x
However,
ij
ij
w w
a a
ijij a
1
2ji
ijj i
uu
a x x
iij
j
u
a x
since ij ji
iij
j
u
x a
d
da: total derivative/crack length
36
Using the divergence integral theorem,
iij
S S j
uwdS dS
a x a
Thus,
0
i iij ij j
S j
u udS n ds
x a a
0
ii
uT ds
a
The derivative of J reduces to,
0
i ii
S
u ud w wdS T ds
da a x a x
0 0
i i ii i
S
u u uwdS T ds T ds
x a a x
0
ii
S
uwdS T ds
x x
iij
S j
udS
x a
from equilibrium equation
37
Using Green Theorem, i.e. A
Q PP x,y dx Q x,y dy dx dy
x y
0
ii
S
ud wdS T ds
da x x
0
ii
uJ wdy T ds
x
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
G
ii
uJ wdy T ds
xGG
ii
A
uwJ dxd y T ds
x x
Using the Green Theorem, i.e.
Closed contour around A
A
Q PP x,y dx Q x,y dy dx dy
x y
Consider
thus,
Again from the divergence theorem,
iij j
A
uwdxd y n ds
x x
iij j
un ds
x
i
ijA j
udxd y
x x
39
However,
ij
ij
w w
x x
ijij x
1
2ji
ijj i
uu
x x x
iij
j
u
x x
since
iij
j
u
x x
ij ji
The integral becomes,
iij
A j
uwJ dxdy
x x xG
Replacing in the integral,
Invoking the equilibrium equation, 0ij
jx
0J
iji i iij ij
j j j
u u u
x x x x x x
i
ijj
u
x x
0
Measure of the energy dissipation due to the crack
40
2) J is path-independent
x
y
G 3
Consider the closed contour21 3 4*GG G G G
2Γ*
G 4
and 0J 1 2 3 4
*J J J J JG G G G G One has
3G
The crack faces are traction free :
0i ij jT n on 4Gand
dy = 0 along these contours
3 40J JG G
41
x
y
G 3
1 3 2 4GG G G G2
Γ*
G 4
and
2 2*J JG G
Note that,
1 20J J JG G G
1 2J JG G
Any arbitrary (counterclockwise) path around a crack gives the same value of J
J is path-independent
G 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
Gr
G is followed from = -q p to q = p
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
ijJ
Ar
1
2
n n
ijJ
Ar
Asymptotic field given by Hutchinson Rice and Rosengren:
Ai are regular functions that depend on q 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 case
1
2r
HRR stress singularity and strain singularity1
1nr
1
n
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 xy
1 1= σ dε = σ ε = σ ε +σ ε +2σ ε
2 2w
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 ds
x
?
47
2 2 2xx yy xx yy xy
1 1+ν= σ +σ -2νσ σ + σ
2E Ew
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 ds
x x x x
iij j
uσ n ds
x
Expanded form for
Along AB or B´ A´ nx = -1, ny = 0 and ds=-dy ≠ 0
yxxx yx x
uu=σ dy + σ n dy
x x
Along CD or DC´
yxxx yx
uu=σ dy+σ dx
x 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 xy
A B C
u u uu u uJ=2 w-σ -σ dy+2 σ +σ dx+2 w-σ -σ dy
x x x x x x
Along BC or C´B´
yxxy yy
uu=-σ dx-σ dx
x 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 aJ
EB h
2
50
Relationship between J and CTOD
ca
y
iij j
uJ n ds
xG
x
d t
Consider again the strip-yield problem,
G
the first term in the J integral vanishes because dy= 0 (slender zone)
Y
Y
iij j
un ds
x
but
yyy y
un ds
x
y
Yu
dxx
yY
uJ dx
x
t
t
Y ydu
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 GPa
n1 = 0.35
Material 2: E2 = 70 GPa
n2 = 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
KJ
E ( )
• 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 s in any orthogonal basis (n, t, e3) is
s : stress tensor s3 : associated principal stress
3, ,
3
0
0
0 0
a c
c b
n t eσ with the invariants,
1 3
2 22 3
23 3
1
2
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 0
0 0
0 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
14 (2 3 )
2I 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 31
2e
1 2 02
Ixx yy
K
x
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 22
e
1
1 22
Ie Y
K
r
1 221 1
12
2
2
11
2I
Y
Kr
22
11 2
2I
Y
Kr