Elastic-Plastic Fracture Mechanics Introduction •When does one need to use LEFM and EPFM? •What is the concept of small-scale and large-scale yielding? Contents of this Chapter •The basics of the two criteria used in EPFM: COD (CTOD), and J- Integral (with H-R-R) •Concept of K- and J-dominated regions, plastic zones •Measurement methods of COD and J-integral •Effect of Geometry Background Knowledge •Theory of Plasticity (Yield criteria, Hardening rules) •Concept of K, G and K-dominated regions •Plastic zone size due to Irwin and Dugdal
Transcript
Elastic-Plastic Fracture Mechanics
Introduction•When does one need to use LEFM and EPFM?
•What is the concept of small-scale and large-scale yielding?
Contents of this Chapter•The basics of the two criteria used in EPFM: COD (CTOD), and J-Integral (with H-R-R)
•Concept of K- and J-dominated regions, plastic zones
•Measurement methods of COD and J-integral
•Effect of Geometry
Background Knowledge•Theory of Plasticity (Yield criteria, Hardening rules)
•Concept of K, G and K-dominated regions
•Plastic zone size due to Irwin and Dugdal
LEFM and EPFM
LEFM•In LEFM, the crack tip stress and displacement field can be uniquely characterized by K, the
stress intensity factor. It is neither the magnitude of stress or strain, but a unique parameter that describes the effect of loading at the crack tip region and the resistance of the material. K filed is valid for a small region around the crack tip. It depends on both the values of stress and crack size.
We noted that when a far field stress acts on an edge crack of width “a” then for mode I, plane strain case
xx
yy
xy
IK
r
RS|T|UV|W|
L
N
MMMMMMM
O
Q
PPPPPPP2 2
12
3
2
12
3
2
2
3
2
cos
sin( )sin( )
sin( )sin( )
sin( )sin( )
zz zz xx yy 0 for plane stress; for plane strain( )
u
uK r
k
k
x
y
IRSTUVW
L
N
MMMM
O
Q
PPPP2
21 2
2
21 2
2
2
2
2
cos ( sin ( ))
sin ( cos ( ))
LEFM concepts are valid if the plastic zone is much smaller than the singularity zones.
Irwin estimates
Dugdale strip yield model:
rK
pI
ys
1
22
( )
rK
pI
ys
1
82( )
ASTM: a,B, W-a 2.5 , i.e. of specimen dimension. ( )K I
ys2 rp
1
50
LEFM cont.
Singularity dominated region
xx
yy
xy
IK
r
RS|T|UV|W|
L
NMMMO
QPPP2
1
1
0
For =0
For =2
all ij , 0
EPFM•In EPFM, the crack tip undergoes significant plasticity as seen in the following diagram.
sh a rp tip
Ideal elastic brittle behaviorcleavage fracture
P: Applied loadP : Yield loady Displacement, u
Loa
dra
tio,
P/P y
1.0
Fracture
Blunt tip
Limited plasticity at cracktip, still cleavage fracture
Displacement, u
Loa
dra
tio,
P/P y
1.0Fracture
Blunt tip
Void formation & coalescencefailure due to fibrous tearing
Displacement, u
Loa
dra
tio,
P/P y
1.0Fracture
large scaleblunting
Large scale plasticityfibrous rapture/ductilefailure Displacement, u
Loa
dra
tio,
P/P y
1.0 Fracture
EPFM cont.
•EPFM applies to elastoc-rate-independent materials, generally in the large-scale plastic deformation.
• Two parameters are generally used:
(a)Crack opening displacement (COD) or crack tip opening displacement (CTOD).
(b) J-integral.
•Both these parameters give geometry independent measure of fracture toughness.
Sharp crack
Blunting crack
y
x
ds
EPFM cont.
•Wells discovered that Kic measurements in structural steels required very large thicknesses for LEFM condition.
--- Crack face moved away prior to fracture.
--- Plastic deformation blunted the sharp crack.
Sharp crack
Blunting crack
• Irwin showed that crack tip plasticity makes the crack behave as if it were longer, say from size a to a + rp -----plane stress
From Table 2.2,
Set ,
rK
pI
ys
1
22
( )
uK r
kyI
2 2 21 2
22
sin( )[ cos ( )]
= uk
Kr
y Iy
1
2 2 a ry
24 2
2uK
EyI
ys
Note:
since
k E
3
12 1
and ( )
CTOD4 G
ysG
K
EI2
CTOD and strain-energy release rate
• Equation relates CTOD ( ) to G for small-scale yielding. Wells proved that
Can valid even for large scale yielding, and is later shown to be related to J.
• can also be analyzed using Dugdales strip yield model. If “ ” is the opening at the end of the strip.
CTOD4 G
ys
ys
Consider an infinite plate with a image crack subject to a
Expanding in an infinite series,
28
ua
Eyys
ys
lin sec(2
)
8 1
122 4ys
ys ys
a
E[1
2(
2(
2) ) ...]
If , and can be given as:
In general,
K
EI
ys ys
22[1
1
6(
2) ]
ys
ysysE
G 0 ( then =
K I2
ys
),
G
m ys
, m = 1.0 for plane stress; m = 2.0 for plane strain
Alternative definition of CTOD
Sharp crack
Blunting crack
Blunting crack
Displacement at the original crack tip Displacement at 900 line intersection, suggested by Rice
CTOD measurement using three-point bend specimen
W
P
p
z
Vp
p
''' pl
p p
p
r W a V
r W a a z
( )
( )
displacement
expanding
Elastic-plastic analysis of three-point bend specimen
el plI
ys
p p
p
K
m E
r W a V
r W a a z
2 ( )
( )
Where is rotational factor, which equates 0.44 for SENT specimen. pl
• Specified by ASTM E1290-89 --- can be done by both compact tension, and SENT specimen• Cross section can be rectangular or W=2B; square W=B
KI is given by
el
I
ys
K
E
2 21
2
( )
KP
B Wf
a
WI ( )
plp p
p
r W a V
r W a a z
( )
( )
loa
d
Mouthopening
p e
V,P
CTOD analysis using ASTM standards
Figure (a). Fracture mechanism is purely cleavage, and critical CTOD <0.2mm, stable crack growth,(lower transition).
Figure (b). --- CTOD corresponding to initiation of stable crack growth. --- Stable crack growth prior to fracture.(upper transition of fracture steels).Figure (c) and then ---CTOD at the maximum load plateau (case of raising R-curve).
c
i
i m
u
loa
d
Mouthopening
Pc
fracture
(a) (b) (c)
PiPu Pm
fracture
Pi
More on CTOD
The derivative is based on Dugdale’s strip yield model. ForStrain hardening materials, based on HRR singular field.
By setting =0 and n the strain hardening index based on
*Definition of COD is arbitrary since
A function as the tip is approached*Based on another definition, COD is the distance between upper and lower crack faces between two 45o lines from the tip. With thisDefinition
2
or ICOD
y y
K JE
11
1 ,
nn
ni y i
y y n
Ju r u n
I
1
32
n
y ije
y y y
,0 ,0y yu x u x
1
1nx
COD ny
Jd
Where ranging from 0.3 to 0.8 as n is varied from3 to 13 (Shih, 1981)
*Condition of quasi-static fracture can be stated as the Reaches a critical value . The major advantage is that this provides the missing length scale in relating microscopic failure processes to macroscopic fracture toughness.
*In fatigue loading, continues to vary with load and is a function of:(a) Load variation
(b) Roughness of fracture surface (mechanisms related)(c) Corrosion(d) Failure of nearby zones altering the local stiffness response
, ,n n yd d n
tip
COD
2
2I
y
k
3.2 J-contour Integral
• By idealizing elastic-plastic deformation as non-linear elastic, Rice proposed J-integral, for regions beyond LEFM.• In loading path elastic-plastic can be modeled as non-linear elastic but not in unloading part.• Also J-integral uses deformation plasticity. It states that the stress state can be determined knowing the initial and final configuration. The plastic strain is loading-path independent. True in proportional load, i.e.
• under the above conditions, J-integral characterizes the crack tip stress and crack tip strain and energy release rate uniquely.• J-integral is numerically equivalent to G for linear elastic material. It is a path-independent integral.• When the above conditions are not satisfied, J becomes path dependent and does not relates to any physical quantities
d d d d d dk
1
1
2
2
3
3
4
4
5
5
6
6
3.2 J-contour Integral, cont.
y
x
ds
Consider an arbitrary path ( ) around the crack tip. J-integral is defined as
J wdy Tu
xds w di
i
iij ij
ij
zz( ),
0
It can be shown that J is path independent and represents energy release rate for a material whereis a monotonically increasing with
ij
ij
Proof: Consider a closed contour:
Using divergence theorem:
J wdy Tu
xdsi
i
i
* ( )*
z
Jw
x x
u
xdxdy
iij
i
A
* ( )*
z
where w is strain energy density, Ti is component of traction vector normal to contour.
A*
*
*
* ( )iij
jA
uwJ dxdy
x x x
Evaluate
w
x
w
x xij
ijij
ij
Note is only valid if such a potential function exists
Again,
ij
ij
w
w
x xu
xu
x
u
x x
u
x
ij i j i j
ijj
i
i
j
1
2
1
2
[ ( ) ( )]
[ ( ) ( )]
, ,
Since
ij ji
ijj
i
x
u
x
( )
Recall
ij
j
j
i
jij
i
x
x
u
x x
u
x
0 (equilibrium) leads to
ij ( ) ( )
Evaluation of J Integral ---1
w
(equilibrium) leads to
,j i
Hence, Thus for any closed contour J * .0 J * .0
Now consider
1
2
3
4
1 2 3 4
0J J J J J
Recall J wdyw
xdsi
* ( ) z
On crack face, (no traction and y-displacement), thus , leaving behindThus any counter-clockwise path around the crack tip will yieldJ; J is path independent.
i dy 0 0, J J3 4 0 J J1 2
Evaluation of J Integral ---2
1 2 3 4
t i
t i
'
a
y
x2D body bounded by '
In the absence of body force, potential energy
z zwdA u dsA
i i' ''
Suppose the crack has a vertical extension, then
d
da
dw
dadA
du
dads
A
ii
z z' '
(1)
Note the integration is now over '
Evaluation of J Integral ---3
t i
t i
Noting that d
da a
x
a x a x
x
a
since 1
d
da
w
a
w
xdA
w
a
u
xds
A
ii
z z( ) ( )
' '
(2)
w
a
w
a x
u
aij
iij
j
i
( )
Using principle of virtual work, for equilibrium, then fromeq.(1), we have
d
da
0
ijjA
ii
i
x
u
adA
u
ads
z z( )
' '
Thus, d
da
du
dxds
dw
dxdAi
i
A
z z' '
Using divergence theorem and multiplying by -1
z zd
dawn
du
dxds wdy
w
xdsx i
ii
( )' '
Evaluation of J Integral ---4
j
t it i
t i
t i
t i
Therefore, J is energy release rate , for linear or non-linear elastic material
d
da
In general
Potential energy; U=strain energy stored; F=work done by external force and A is the crack area.
U F JA
and
a
u p
Evaluation of J Integral ---5
-dP
*dU dU d
Displacement
*U P U Complementary strain energy = dP0
p
Loa
d
For Load Control
For Displacement Control
The Difference in the two cases is and hence J for both load Displacement controls are same
*
p
dUJ
da
dUJ
da
0 0
0
. .
.
pD
p p
J dp dpa a
or
pJ pd d
a a
J=G and is more general description of energy release rate
2
'IK
JE
Evaluation of J-Integral
More on J Dominance
J integral provides a unique measure of the strength of the singular fields in nonlinear fracture. However there are a few important Limitations, (Hutchinson, 1993) (1) Deformation theory of plasticity should be valid with small strain behavior with monotonic loading(2) If finite strain effects dominate and microscopic failures occur, then this region should be much smaller compared to J dominated
region Again based on the HRR singularity
11
,n
Iijij y
y y n
Jn
I r
Based on the condition (2), we would like to evaluate the inner radius ro of J dominance. Let R be the radius where the J solutions are satisfied within 10% of complete solution. FEM shows that R
or
3o CODr
•However we need ro should be greater than the forces zone (e.g. grain size in intergranular fracture, mean spacing of voids)•Numerical simulations show that HRR singular solutions hold good for about 20-25% of plastic zone in mode I under SSY • Hence we need a large crack size (a/w >0.5) . Then finite strain region is , minimum ligament size for valis JIC is
• For J Controlled growth elastic unloading/non proportional loading should be well within the region of J dominance
• Note that near tip strain distribution for a growing crack has a logarithmic singularity which is weaker then 1/r singularity for a stationary crack
3 COD25 IC
y
Jb
and a RdJ Jda R
Williams solution to fracture problem
Williams in 1957 proposed Airy’s stress function
As a solution to the biharmonic equation
For the crack problem the boundary conditions are
Note will have singularity at the crack tip but is single valued
Note that both p and q satisfy Laplace equations such that
R r
2 24 2
2 2 2
1 10 where
r r r r
0 for r
2 , ,r p r q r
2 2 0p q
Now, for the present problem.
1
2 21
21 1
22 2
21 1
2
1 12
cos sin
cos 2 sin 2
Then
cos cos 2
sin sin 2
Consider only mode I solution with
cos cos 2
1 2 cos cos 2
z
z
r
p A r A r
q B r A r
r A B
r A B
r A B
r A Br
r
1 1
1
1 sin 2 sin 2
r
r A B
Williams Singularity…3
Applying boundary conditions,
1 1
1 1
A +B cos 0
2 sin 0A B
Case (i) cos 0 2 1, Z=0,1,2...
2
Z
1 1B2
A
or,
sin 0 Z 1 1B ACase (ii)
Since the problem is linear, any linear combination of the above two will also be acceptable.Thus Though all values are mathematically fine, from the physics point of view, since
2 with Z= ... 3, 2, 1,0,1,2,3...Z
and ij ijr r
Williams Singularity…4
0
0
212
2120
2A 2 1
0
=
ij ij
R
ij ijr
R
r
U r
rdrd
r drd
Since U should be provided for any annular rising behavior and R , 0r
0 ˆ as 0, 1 ( 1 makes 0)ijU r
1
1
31 1 Z2 2 2 2
112 3
, needs > 1. Thus
=- ,0, ,1, ,2... with = Z=-1,0,
positive number.
The most dominant singular form
=- and B
ri
A
Also u r
where
32
52
1 12 2
311 2 3 2
2
01
Now cos cos
+ ...
...
and
where indicates the order of
Iij ij ij ij
r A
r
r
A r r r
Williams Singularity…4
0
1
Note the second term in is a non-singular
and non-vanishing term. However, higher order vanish as r 0
with 2
(no sum on x)2
ij ij
I
IIij ij ix jx
r
KA
KT
r
32
52
1 12 2
311 2 3 2
2
01
Now
cos cos
...
...
and
where indicates the order of
Iij ij ij ij
r A
r
r
A r r r
Williams Singularity…5
0
1
Note the second term in is a non-singular
and non-vanishing term. However, higher order vanish as r 0
with 2
(no sum on x)2
ij ij
I
IIij ij ix jx
r
KA
KT
r
Williams Singularity…6
For in-plane stress components,
0
0 02
I Ixx xy xx xyI
I Iyx yy yx yy
TK
r
I
Second-term is generally termed as "T-stress" or
"T-tensor" with
For brittle crack of length 2a in x-z plane
with & applied
K and
xx
yy xx
yy
T
a
T= yy xx
x
y
2a
z
HRR Singularity…1
0 0 0
Hutchinson, Rice and Rosenbren have evaluated the character of crack tip
in power-law hardening materials.
Suppose the material is represented by Ramberg-Osgood model,
