05_epfm
TRANSCRIPT
ELASTIC-PLASTIC FRACTURE MECHANICS
Suggested readings:
Chapter 3 T.L. Anderson
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Concept of Small-Scale Yielding (1)
I II III IV
I: plastic zoneII: elastic region, elastic solution is influenced by the nerighbouring plastic zoneIII: K dominance zone
IV: non-singular terms become important
2
Kf
r
H.O.T2
Kf
r
SSY holds when the plastic zone of the crack tip is sufficiently small compared to the crack length and other relevant geometric length parameters, such that the elastic singular fields still give good approximation of the actual fields in an annular region surrounding the crack tip!
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Concept of Small-Scale Yielding (2)
ry
a W
,yr a W Under SSY conditions, the load-displacement response is still linear
pP
P
i
ii
iii
i ii iii
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Size of the Plastic Zone: Irwin’s Model (1)
ys
x
y
rp Plastic zone
Crack tip
i) Irwin’s modelFor plane-stress conditions, the size of the plastic zone ry, can be estimated from the stress-field equation by setting the linear-elastic stress equal to the yield strength, ys.
ry
Linear-elasticElastic-plastic
0.52 22
1 2 1 3 3 22
2ys
Based on Mises yield criterion,
31 sin sin
2 21 3
cos 1 sin sin2 2 22
3sin cos
2 2
xx
yy I
xy
Kr
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ry: first-order plastic zonerp: second-order plastic zone
Size of the Plastic Zone: Irwin’s Model (2)
ys
x
y
rp Plastic zone
Crack tip
ry
Linear-elasticElastic-plastic
At = 0, f = 1 and xy = 0, therefore, 1 2yy xx
31 sin sin
2 21 3
cos 1 sin sin2 2 22
3sin cos
2 2
xx
yy I
xy
Kr
0.52 22
2ys yy yy yy
3 0zz
Plane-stress
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0.52 22
1 2 1 3 3 22
2ys
Size of the Plastic Zone: Irwin’s Model (2)
ys
x
y
rp Plastic zone
Crack tip
02 2
I Iyy ys
y y
K Kf
r r
21
2I
yys
Kr
ry
Linear-elasticElastic-plastic
31 sin sin
2 21 3
cos 1 sin sin2 2 22
3sin cos
2 2
xx
yy I
xy
Kr
0.52 22
2ys yy yy yy
Therefore,Plane-stress
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Size of the Plastic Zone: Irwin’s Model (3)
x
y
rp
Plastic zone
Crack tip
ry
Linear-elastic
Elastic-plastic
Assumption: the effect of the plastic zone is to translate the stress distribution towards the right by the amount of distance necessary to restore equilibrium.
Equilibrium requires that Area A equals Area B
0
yr
yy ys ys p ydx r r
A
B
ys
0 2
yrI
ys ys p yK
dx r rr
0
22
yrI
ys y ys p yK
r r r r
21
2I
yys
Kr
21
2Ip y
ys
Kr r
Plane-stress
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Size of the Plastic Zone: Irwin’s Model (4)
For plane strain conditions,
ys
x
y
rp Plastic zone
Crack tip
ry
Linear-elasticElastic-plastic
31 sin sin
2 21 3
cos 1 sin sin2 2 22
3sin cos
2 2
xx
yy I
xy
Kr
at = 0, f = 1 and xy = 0, therefore,
1 2yy xx 3 1 2zz v
0.52 221 2 1 3 3 2
0.52
2
2
22 1 2
2
1 2
ys
yy
yy
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Plane-strain
Size of the Plastic Zone: Irwin’s Model (5)
Assume υ= 1/3 for metals,
3yy ys
Plane-strain
ys
x
y
rp Plastic zone
Crack tip
ry
Linear-elasticElastic-plastic
3 02 2
I Iyy ys
y y
K Kf
r r
21
6I
yys
Kr
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Corrected Crack Length (1)
To account for the crack-tip plasticity, Irwin suggested that the stress distribution at a moderately large distance from the crack tip be found by considering a “corrected crack length”,
eff ya a r
2
effI
yy yK
r rr
a ryry effI effK a
21
'2
effeff II
ys
KK a
This is an iterative process!
Plane stress
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Corrected Crack Length (2)2
1'
2
effeff II
ys
KK a
The difference between KIeff’ and KI
eff should be negligibly small
'eff effI IK K
221
2
eff effI I
ys
K Ka
21
12
effI
ys
aK
Plane stress
Note: KIeff > KI, the difference
becomes dramatic as ∞ → ys, whence the entire procedure of crack-tip zone correction becomes susceptible!
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Corrected Crack Length (3)
21
12
effI
ys
aK
For a reasonable upper limit to the applied nominal stress in a structure, say,
3
4 ys
21.18
0.8511 0.75
2
effI
a aK a
22
0.75 101 12.8 mm
2 2ysI
yys ys
Kr
For a crack with a = 10 mm in an infinite plate,
/effI IK K
/ ys
1.0
1.0
1.4
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Dugdale’s Model (aka: Strip-Yield Model)
• Irwin’s correction is limited to small-scale yielding (SSY) conditions, Dugdale’s model is not limited to SSY• Consider a crack in an infinite thin sheet of materials with very low hardening• The crack length is increased from 2a to 2(a+R), where R is the length of the plastic zone, with a closure pressure equal to ys applied at each crack tip
2a RR
R
Plastic zone
ys
p app RI I IK K K
appIK a R
?RIK
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Dugdale’s Model(1)
Dugdale argues that,
0p app RI I IK K K
i.e., when the closure pressure just closes the crack completely over the length R, the crack singularity at a + R should vanish since there is no crack there!
This relationship becomes the basis to find the distance R (plastic zone size) with respect to the ys, 2a and KI
app
Dugdale, D. S. (1960). Yielding of steel sheets containing slits. Journal of the Mechanics and Physics of Solids. Volume 8, Issue 2 , Pages 100-104
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Dugdale’s Model(2)
Using the weight function technique, we can find the SIF values. Recall that KI for a crack under a pair of point forces equals,
p
x 2 / 2
/ 2Il x
K pl l x
y
x
l
2( )l a R
2 / 2 / 2
/ 2 / 2
a R aRI ys
a a R
l x l xK dx dx
l l x l x
over , and ,ysp R a a a R a
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RR2a
Dugdale’s Model(3)
2 / 2 / 2
/ 2 / 2
a R aRI ys
a a R
l x l xK dx dx
l l x l x
For 2nd integral, assume x = -x’
/ 2 '2 / 2
'/ 2 / 2 '
2 / 2 / 2 ''
/ 2 / 2 '
a R aRI ys
a a R
a R a R
ysa a
l xl xK dx d x
l l x l x
l x l xdx dx
l l x l x
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RR2a
ys
2( )l a R
Dugdale’s Model(4)
2 / 2 / 2 ''
/ 2 / 2 '
a R a RRI ys
a a
l x l xK dx dx
l l x l x
Grouping both integrals,
2
1
2
1
1
2 / 2 / 2
/ 2 / 2
2
/ 2
2 sin2 / 2
2 si
co
n2
2 s
a RRI ys
a
a R
ysa
a R
s
s
ya
y
y
s
l x l xK dx
l l x l x
l dx
l x
l x
l
a R
a R a
a
a
R
R
a
p
xy
x
l
2( )l a R
ysp
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Dugdale’s Model(5)
12 cosRI ys
a R aK
a R
cos2 ys
a
a R
0app RI IK K Since,
12 cosysa R a
a Ra R
R
a R1 cos
2 ys
Dugdale’s paper was not accepted despite the good agreement with experiments, probably because he did not provide clear evidence that yielding was occurring in a narrow strip!Hahn and Rosenfield showed that in plane-stress conditions, the plastic zones do conform to Dugdale’s model. (Acta Metal, Vol 13, pp 296-306, 1965)
Comparison with experiments18Qian X/CE5887
Dugdale’s Model(6)
Based on Taylor series,2 4
1 5sec 1 ...
2 2! 2 4! 2ys ys ys
For ∞<< ys, i,e, SSY conditions,
sec2 ys
a R
a
2 222
20.393
8 8I I
ys ysys
a K KR
Irwin’s results,2 2
10.318I I
ys ys
K KR
Close agreement
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Dugdale’s Model(7)
The effective SIF for Dugdale’s model can be estimated using, aeff = a + R, or
seceffI
ys
K a
However, the above equation tends to over-estimate the effective KI. A more realistic estimate (See Appendix 3.1 of Anderson 2005 for the derivation) is,
2
8lnsec
2effI ys
ys
K a
0 0.2 0.4 0.6 0.8 1.00
0.5
1.0
1.5
2.0/eff
I ysK a
/ ys
Linear-elastic
Irwin’s correction
Dugdale’s correction
2a
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Dugdale’s Model: Example
A steel plate with a central through-thickness flaw of length 16 mm is subjected to a stress of 350 MPa normal to the crack plane. If the yield strength of the material is 1400 MPa, what is the plastic-zone size and the effective stress-intensity level at the crack tip?
350 8 1755 MPa mmIK a
Assume the plate is infinitely large, 2a = 16 mm
2
0.62 mm8
Ip
ys
Kr R
2
8 3501400 8 lnsec 1778 MPa mm 56.2 MPa m
2 1400effIK
8 350cos
8 2 1400R
R = 0.66 mm
Alternatively,
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Application of Dugdale’s Model
Dugdale models are often used to describe quantitatively the role of toughening mechanism in otherwise brittle materials.
The background brittle material is toughened by the ductile incldusions. These can be rubber, metal fibers, etc.
The SIF at the macroscopic crack tip thus equals,
Qian X/CE5887 22
tip app tougheningK K K
Toughening mechanism
Ktoughening
(MPa•m0.5)
Firber (glass, carbon, silica-carbon,etc.)
~ 20
Ductile network (Al2O3, AL, etc.)
12 ~ 20
For comparison, structural steels at very cold temperatures have toughness 20 – 40 MPa•m0.5
Shape of the Plastic Zone (1)
2cos 1 sin2 22
Irr
K
r
3cos22
IK
r
2sin cos2 22
Ir
K
r
• So far, we have considered the size of the plastic zone along the x-axis!• The size of plastic zone varies with • We will utilize the Mises yield criterion to investigate the shape of the plastic zone
Mises yield criterion,
0.52 22
1 2 1 3 3 22
2ys
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How to express the principal stresses in terms of σrr σθθ and σrθ?
Shape of the Plastic Zone (2)
2cos 1 sin2 22
Irr
K
r
3cos22
IK
r
2sin cos2 22
Ir
K
r
C
R
Based on Mohr’s circle,
1
2
rr
r
1cos
2 22I
rrK
Cr
22 1cos sin
4 2 22I
r rrK
Rr
1 C R 2 C R
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Shape of the Plastic Zone (3)
1 cos 1 sin2 22
IK
r
C
R
Based on Mohr’s circle,
1
2
rr
r
2 cos 1 sin2 22
IK
r
3
0 for plane stress
2 cos for plane strain22
IK
r
3 1 2
At = 0, the principal stresses 1 and 2 are equal and act in x and ydirections xx and yy are the principal stresses.At other values, the boundary of the plastic zone is obtained by setting mises = ys
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Shape of the Plastic Zone (4)
For plane stress,
0.52 22
1 2 1 3 3 2
0.52 22 2 22 2 2 2
0.52 2 2
0.52
2
2
42cos sin cos 1 sin cos 1 sin
2 2 2 2 2 2 2 2 2 2
22cos 6sin cos
2 2 2 22
2 31 sin cos
2 22
ys
I I I
I
I
K K K
r r r
K
r
K
r
2
21 31 sin cos
4 2I
yys
Kr
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Shape of the Plastic Zone (5)
For plane strain,
0.52 221 2 1 3 3 2
0.52 22 2 2 2
0.522
2
2
24cos sin cos 1 sin 2 cos 1 sin 2
2 2 2 2 2 2 22
2 3sin 1 cos 1 2
2 22
ys
I
I
K
r
K
r
2
221 3sin 1 cos 1 2
4 2I
yys
Kr
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Shape of the Plastic Zone (6)
0 0.4 0.8-0.8 -0.4-0.8
-0.4
0.4
0.8
0
21 I
ys
x
K
21 I
ys
y
K
Plane strain
Plane stress
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Effect of Thickness (1)
zzxx
B
crackr
yy
• Materials far away from the crack front experience a plane-stress condition• The large normal stress yy near the crack tip will try to contract in the x and zdirection, but is prevented from doing so by the surrounding materials• The crack front near the free surface of the plate experiences less constraint by the surrounding material
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Effect of Thickness (2)
B
z
x
y
z B/20
zz
High triaxiality
(plane strain)
Low triaxiality
plane stress
Mid-thickness
Free surface
Increasing thickness of the specimen30Qian X/CE5887
Effect of Thickness (3)
0
7075-T6 Aluminum
10
20
30
40
50
60
70
0 0.2 0.4 0.6 0.8 1.0 1.2
ksi incriticalK
Specimen thickness (in)
t
High triaxiality zone
Low triaxiality zoneB
Test specimen Surface defect in structures
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1 ksi in 1.1 MPa m
Crack-Tip Opening Displacement (CTOD) Developed initially in the U.K. by Wells and colleagues at The
Welding Institute
Very physically appealing approach based on critical elastic-plastic stretching at the crack front (no abstract energy concepts or advanced math!)
Plastic Deformation
X
yy
a X
yy
a CTOD is an indirect measure of severe stretching at crack tip
CTOD ( )
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CTOD
2
21
8 1 22 ( )
2
I
ysy I
K
v r r KE
2 24 1I
ys
K
E
CTOD increases as K2
(this is a nonlinear effect!)
( )v r
yr
plastic zone
24 1( )
2Ir
v r KE
r
( )v r
yr
Suggestion: re-derive for plane-strain conditions, verify physical units are correct
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Plane-stress
CTOD: Definition
A
A’
45o
45o
v
CTOD
u
Tracy, D. M. (1976). Finite element solutions for crack-tip behavior in small-scale yielding conditions. Journal of Engineering Materials and Technology. 98, 146-151.
CTOD is defined at the intercept of the two 45o lines originating from the deformed crack tip and intersecting the crack profile.
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CTOD(2)
The strip-yield model provides another means to determine CTOD.
y
x
R
A
• A: fictitious tip• R << a• No singularity at A 0app R
I IK K
02I
c
p xK dx
x
p(x)
x
y
c
020app
IR
p xK dx
x
B
At A:
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CTOD(3)y
x
R
A
020app
IR
pK dx
x
B
At A:
The crack-tip opening displacement,
01
2 ln2 2
ysy app R
r x xx u K dx
G x x
2
8app
ys
KR
ysp x
1ln
4
l
yc
x xu p x dx
G x x
Linear elastic fracture mechanics From weight function
1
2 2y I
ru K
G
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CTOD(4)
for –R < x < 0
111 ln
21
xx x x Rg
xR R R
R
where
0ln 2
R
xx xdx Rg
Rx x
2
2 2
1 1
2 2 8
1 1 1 1 1
2 8 4 8
appys ysapp app I
I Iys
app appI Iys
appys I ys
x x xKrx K Rg K g
G R G R
K Kx x xrg g
G K R G R R
2
8app
ys
KR
2
8app
ys
KR
8ys
appK R
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CTOD(5)
2
2
2 2
1
1 1 1
4 8
11 1 1 11 ln
4 8 21
11 1 11 ln
8 2 81
appI
ys
appI
ys
app appI I
ys ys
K x xx g
G R R
xK x x x R
xG R R R
R
xK Kx x xR g
xG R R G R
R
111 ln
21
xx x x Rg
xR R R
R
At x = -R,
2
1
8
appI
ys
KR
G
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CTOD(6)
21 1
8 IKB a G
G
2
1
8
appI
ys
KR
G
But we have derived:
ys
x
G
The strip-yield model assumes a non-hardening material. The actual relationship between CTOD and G depends on the strain hardening and is given by a more general form,
ys
xm
G
1.0 ≤ m ≤2.0
Try to plot (x) v.s. x to observe how the crack tip deforms!
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CTOD (7) Use nonlinear finite element analysis to compute CTOD as function
on applied loading
Thermal loading, residual stresses, etc. introduce NO complications
Just extract CTOD from displacements of nodes behind crack tip at each applied load level
Le
More elements than this needed to resolve variation of fields
13 nodes at this front location-collapse to a point
Extract CTOD at each front location using 45o intercept method
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J-Integral: Energy Release Rate
A
ds
For a 2-D body of area A, with tractions Ti
applied over the bounding curve the potential energy of the body is (assume unit thickness),
i iA
U P WdA T u ds
0 0 0 0 0... ...
ij xx xy yy zz
ij ij xx xx xy xy yy yy zz zzW d d d d d
W is the strain energy density, and ui is the displacement vector
If the 2-D body encloses a crack, the change in the potential energy with respect to crack extension becomes,
ii
A
dud dWdA T ds
da da da
Note: under linear-elastic conditions,
G = -
d= J
da41Qian X/CE5887
J-Integral (2)
“Advanced Calculus for Applications” by F.B. Hilderbrand
in our case: x = a, B = x, A = 042Qian X/CE5887
J-Integral (3)
A
ds
ii
A
dud dWdA T ds
da da da
d x
da a a x a x
a
x
yx0
0x x a 1x
a
dW
da
idu
da
d
da
Assumption: a is parallel to x
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J-Integral (4)
ij ijij
ij
W W
a a a
Recall the principle of virtual work, 0ij ij i i
A
dA T u ds
i i
i
A
u ud W WdA T ds
da a x a x
=0
ii
A
ud WJ dA T ds
da x x
ii
udJ Wdy T ds
da x
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Internal energy=external energy
J Integral: Path Independence
,
,
ii
iij j
iijA
j
i iij j ijA
j
uJ Wdy T ds
x
uWdy n ds
x
uWdxdy dA
x x
u uWdA
x x x x
A
ds
0Equilibrium
iijA
j
uWJ dA
x x x
Divergence theory
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J Integral: Path Independence (2)
ij ijij
ij
W W
x x x
1
2ji
ijj i
uu
x x
1
2ij ji i
ij ijij j i j
uu uW W
x x x x x x x
iijA
j
uWJ dA
x x x
ij ji use
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J Integral: Path Independence (3)
0
iijA
j
i iij ijA
j j
uWJ dA
x x x
u udA
x x x x
iij
j
uW
x x x
Path independency can be restored for the various effects by subtracting integrals for those contributions
J = 0 on the closed path when,• the path encloses no cracks or singularity• there is no body force• the material is isotropic and elastic• E, υ have no gradation in x• the stresses are the simple derivative of W (no residual stresses, thermal strains, etc.)
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J Integral: Path Independence (4)
1
2 3
4
1 2 3 4 0J J J J J
For a cracked body,
Any arbitrary (counterclockwise) path around a crack will yield the same value of J, which is path independent!
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J Integral: Path Independence (5)
• The path independence of J requires a one-to-one relationship between the stress and strain values, valid for linear elastic and nonlinear elastic materials. • For elastic-plastic materials, one strain value may correspond to one or more stress values if the material is unloaded and cyclically loaded.• However, elastic-plastic material behavior may assume the same behavior as that of the nonlinear-elastic materials, provided no unloading takes place.
Nonlinear elastic
Elastic-plastic
Rice, J. R. (1968). “A path independent integral and the approximate analysis of strain concentration by notches and cracks.” J. App. Mech., 35, 397-386.
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J Integral: Path Independence (6)
Conditions for path independence in elastic-plastic materials:1. stress must non-decreasing everywhere2. stress components remain in fixed proportion
However, plasticity induced stress redistribution may violate the 2nd condition. However, J still retains an approximate path independence if its values are obtained from a region outside the plastic deformation, i.e., a far-field J value.
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Example (1)
A narrow strip of material with elastic modulus E and height h is rigidly attached to parallel platens, as shown in the figure below. If the upper platen is displaced by an amount , determine the geometric stress intensity factor for an edge crack midway between the platen faces.
xy
1 2
3
45
1 2 3 4 5J J J J J J
ii
uJ Wdy T ds
x
• At 2 and 4, dy = 0 and displacements are constant, i.e., ∂ui/∂x = 0, J3 = J4 = 0• At 1 and 5 near the free surface, tractions are zero, J1 = J5 = 0
h
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Example (2)
xy
1 2
3
45
h
At 3, sufficiently far from the crack tip, xx = 0, xy = 0, yy ≠ 0, tractions are zero. Therefore, the contribution to J only comes from the W term. Along 3
(assume plane stress condition),
22
3 2 2
0 0 0
1 1 1
2 2 21 1
h h h
Iyy yy
KE EJ Wdy dy dy
h h h E
10xx xx yyE
/yy h
xx yy 21 1
yy yy xx yyE E
2 21 1yy yy
E E
h
22 1
IE
Kh
52Qian X/CE5887
HRR Singularity (1)
Ramberg-Osgood constitutive relationship
0 0 0
n
0
0
E
0
E
0
0
n
E E
0 0
0 0
n
E E
0
0 E
0
E
is the yield offset, equal to 0.002
for steels.
For large plastic deformation, the elastic strain is negligible:
0 0
n
53Qian X/CE5887
HRR Singularity (2)
cos
ii
ii
uJ Wdy T ds
x
ur W T d
x
dsd
Evaluate J integral over a circular path with radius r,
sindy r ds rd
cos ii
uJW T d
r x
J is independent of r! In addition, ij ij ij ijW d ii ij ij
uT
x
1
rshould be cancelled out on the left hand side and on the right hand side!
1ij ij r
0 0
n
combined with
/1
1n n
C
r /
21 1n
C
r
54Qian X/CE5887
HRR Singularity (3)
/1
1n n
C
r /
21 1n
C
r
1. Linear-elastic materials, n = 12. Elastic-perfectly-plastic, n→∞3. Stronger singularity for strain, weaker singularity for stress
Hutchinson (1969), Rice and Rosengren (1969) (HRR solution) derive independently the solution for the crack-tip stress field as,
/1 1
00 0
n
ij ijn
Jf
I r
/ 1
00 0
n n
ij ijn
Jg
I r
. . . .2 36 568 0 4744 0 040 0 001262nI n n n
. . . .2 4 34 546 0 2827 0 0175 0 4516 10nI n n n
for plane strain
for plane stress
55Qian X/CE5887
HRR Singularity (4)
/
i
n
in
jj
J
I rf
1 1
00 0
0 /2 -1
-0.5
0.0
0.5
1.0
1.5
0 /2 -1
-0.5
0.0
0.5
1.0
1.5
ijf ijf
rrf
ef
rf f
rrr
n = 13n = 3
f
rrf
rf
ef
Plane stress condition
56Qian X/CE5887
HRR Singularity (5)
/n n
ijijn
J
I rg
1
00 0
0 /2 -0.1
0.1
0.3
0.5
0.7
ijg
0 /2
-0.2
0
0.2
0.4
0.6
0.8 ijg
n = 13n = 3
Plane stress condition
rg
g
rrgg
rg
rrg
57Qian X/CE5887
J and CTOD (1)
45o
45oCTOD
/ 1
00 0
n n
ij ijn
Jg
I r
Integrating the strain expression yields the displacement vector on the traction-free crack face =
v
u
/n n
n
n
hu Jr
v I h
1 111
00 0 2
r
. CTOD r u v 0 5
The point at behind the deformed crack tip is the un-deformed location of the intercept.
r
r u v
From definition of CTOD,
58Qian X/CE5887
rrr
u
r
1r uu
r r
J and CTOD (2)
/n n
n
n
Jr r h h
I
1 1
10 1 2
0 0
n
n
n
n
Jr h h
I
11
0 1 20
n
nCTOD
n
Jv h h h
I
11
0 1 2 20
2 2
CTOD n
Jd
0
dn depends on n and 0 = 0/E
59Qian X/CE5887
J and CTOD (3)
CTOD n
Jd
0
0 0.1 0.2 0.3 0.4 0.50
0.2
0.4
0.6
0.8
1.0
/ n1
nd
0 0.1 0.2 0.3 0.4 0.50
0.2
0.4
0.6
0.8
1.0
/ n1
nd
/ E 0
0.0080.0040.0020.001
/ E 0
0.0080.0040.0020.001
Plane stress Plane strain
• The plane strain condition reduces the CTOD by about 20%• Less deformation under plane-strain conditions than that in the plane stress• More constraint in plane-strain state!
60Qian X/CE5887
About HRR Solutions
• The HRR solution depends on the applicability of Ramberg-Osgood constitutive model to the material. Ramger-Osgood is not a good representation for some of the mild steels which exhibit distinctive upper and lower yield points.
• HRR solution ignores the linear portion of the stress-strain relationship. The strains near the crack tip are large enough that the linear portion is negligible, but meanwhile small enough that the small strain theory is still applicable.
• The HRR stress field is not unique. There are infinite number of ways to satisfy the governing condition, ijij 1/r, corresponding to an infinite number of constitutive models.
61Qian X/CE5887
J Dominance (1)
• J is more than an energy parameter. It also characterizes the crack-tip stress fields.• J is also a fracture parameter: crack extension should take place if J exceeds a critical value, JIc.• Since the crack tip stress field is dependent on J everything happens near the crack tip should also depend on J a similitude concept
Boundary of K dominance
Inelastic region
For linear-elastic materials, when the plastic zone remains small compared to any dimension of the structure, K governs the stress, strain, displacement around the crack tip.
Any two cracks will be in the same state if they have equal K, regardless of the geometry, type of loading, crack size A K similitude!
62Qian X/CE5887
J Dominance (2)Similar to K similitude, the finite-strain zone around the crack tip should be small compared to the dimension of the structure. Hutchinson showed that the valid J dominance zone starts at,
CTODR 3
K dominance zone (outside plastic zone)
Finite-strain zone
R
J dominance zone
Stresses governed by remote loading
Nonlinear elastic
Elastic-plastic
63Qian X/CE5887
J Dominance (2)
For nonlinear-elastic materials, where the stress-strain curve for unloading is the same as that for loading, J remains path independent.
For real materials, crack extension causes unloading near the crack tip, which violates the assumption of deformation plasticity
Material immediately ahead of the unloading zone experiences highly non-proportional loading!
J similitude only applies to stationary cracks, or practically very small crack extensions, with the Jrising sharply!
Elastic unloading
Non-proportional plastic loading
J-dominance
For ductile materials, a small finite-strain zone restricts the crack extension in the J dominance zone.
64Qian X/CE5887
J From Lab Test (1)
Recall from energy principle: for a fracture specimen under tension, the strain energy and complimentary strain energy follows,
, LLDP
aU
*U
P
LLD
0
, , LLDU a P a d
*
0
, ,
P
LLD LLD LLDU a P U P P a dP Or use the complimentary energy:
65Qian X/CE5887
J From Lab Test (2)
P
LLD
dP
d daa
a
a da
Fixed load P during da
0
0
00
Total potential energy for load control:
= ,
= ,
,,
P
LLD LLD LLD
P
LLD
PP
LLDLLD P
P
U P P a dP P
P a dP
P add P a dP da dP
da a
0
,1 1P
LLD
P
P adJ dP
B da B a
J is the area between the load-displacement curves for different crack lengths, divided by the thickness (and the differential crack extension).
66
Load controlled loading
Qian X/CE5887
J From Lab Test (3)
67
Displacement controlled loading
0
= , LLDU P a d
P
LLD
d
PdP da
a
a
a da
Fixed during da
J is the area between the load-displacement curves for different crack lengths, divided by the thickness (and the differential crack extension)
0
,1 1LLD
P adJ d
B da B a
The potential energy of external forces becomes zero, or Ω = 0
Qian X/CE5887
J From SE(B) Test (1)
68
• Two equivalent expressions to compute J• These expressions prompted research to find approximate, analytical
expressions for J for lab test specimens• Jim Rice (at Brown), John Merkle (at ORNL) and Herb Corten (TAM
Dept. UIUC) collaborated to derive the first simple, analytical expression for the J-integral specifically for the SE(B)
Suppose the remaining ligament, b, is fully plastic
The plastic moment is then
2L
W
a
b = W - a
P, Thickness, B 0
0
0 2Pb
F B
0 2Pb
F B
2
02 4P Pb b
M F B Qian X/CE5887
J From SE(B) Test (2)
69
Pl PlUnder large deformation, the beam deforms into a mechanism with two rigid bars.The plastic rotation can be expressed as a dimensionless function,
200
, ,PPl
M Ef n
b B
20
0
, ,PlPE
M Bb h n
Thinking of the reverse form …
... is a non - dimensional functionh
2
2Pl Pl
Pl L L
From rigid-body approximation,
Plastic hinge
20
0
, ,2
4Pl
PPL E
M Bb hL
n
From equilibrium,4P
PLM
Qian X/CE5887
J From SE(B) Test (3)
70
20
0
, ,2
4Pl
PPL E
M Bb hL
n
20
0
24, ,PlBb h E
P nL L
0
,1LLD
P aJ d
B a
00
28 , ,
2
Pl Pl
PL EBb h n
LP
L b
P
a b
P
Assuming fixed Pl.
0Pl
b
2
Pl
PP
ba
Qian X/CE5887
J From SE(B) Test (4)
71
2
Pl
PP
ba
0
,1LLD
P aJ d
B a
0 0
1 2 2LLD LLD
PJ d Pd
B b Bb
P
LLD
0
LLDPd
2 measured area under load-displacement curve
area of remaining ligamentJ
Qian X/CE5887
The Approach (1)
72
A J formula similar to that for the SE(B) can be derived for the C(T) specimen
Finite element analyses show that the formulas are surprisingly accurate – provided deformations are largely plastic
More accurate expressions may be written for other types of test specimens and through-cracked structures in the form
where the non-dimensional “eta” ( ) factor varies with specimen type, a/Wratio, W/L and, in general, material strain hardening (n)
0
UJ Pd
Bb Bb
Qian X/CE5887
The Approach (2)
73
John Sumpter and Cedric Turner in the late 1970s proposed an even more accurate framework to compute J values from measured experimental quantities
They decomposed the total (measured) load-line displacement into elastic and plastic components
Likewise, the total J value is separated into elastic and plastic parts
e plJ J J
LLD e pl
2 21I pl plK U
JE Bb
P
LLD
02pl
PU Pd
Qian X/CE5887
Begley-Landes Experiments (1)
74
Begley and Landes (1972) conceived of some simple tests they could perform to “measure” applied J values directly from lab tests.
Fabricate and test 4 identical SE(B) specimens with different crack lengths
1 2 3 4a a a a a
1. Plot the measured load-load line displacement curves from the tests
2. They used a “tough” material so that no crack growth occurs during the tests
3. Measure the area under the load-displacement curves at each prescribed
P
L L D
1a
2a
3a
4a
1 2 3 4
No fracture events
Qian X/CE5887
Begley-Landes Experiments (2)
75
1 2 3 4
1 2 3 4
1
2
At we have: , , ,
At we have: , , , , .
a a a a
a a a a
U U U U
U U U U etc
Make the following plot:
J
LLD
2IK
JE
l imitJ P U
B
4
a1
…
• Early in the loading, the contribution from plastic deformation is small and J values quadratically with LLD (since LLD varies linearly with load)
• This multi-specimen technique is very time-consuming and costly but it does work !
Begley, J.D. and Landes, J.D., “The J-integral as a Fracture Criterion” ASTM STP 514, 1972, pp. 1-24
Qian X/CE5887
Resistance Curve (R-Curve)
• Many materials with high toughness do not fail catastrophically at a particular value of J or CTOD. They display a rising J or CTOD with crack extension, which is called R curve.
JR
a
JIc
Crack blunting
Crack initiation
ASTM E-1290: Standard test method for crack-tip opening displacement fracture toughness measurementASTM E-1820: Standard test method for fracture toughness
76Qian X/CE5887
R-Curve
JIC characterizes the material fracture toughness at the onset of ductile tearing (end of crack-tip blunting). Its value, however, is a somewhat arbitrary number since the onset of ductile tearing is nearly impossible to distinguish from continued blunting.
However, JIC is a conservative design limit since most J-R curve rises sharply beyond JIC.
The full J-a curve is more descriptive of the material fracture resistance.
At a larger a, the J dominance breaks down, J no longer characterizes the crack-tip stress-strain-displacement fields.
77Qian X/CE5887
R Curve: Tearing Modulus
• For ductile materials, the slope of the J-a curve indicates the relative stability of the crack growth. • The slope of the J-R curve is often given as the dimensionless tearing modulus.
RR
E dJT
da 2
0
, RJ J
Crack size
P1
P2
P3
12
3
4
• Under load control, the criterion for stable crack growth (instability occurs when the driving force curve becomes tangent to the J-R curve):
RJ J
RE dJ E dJ
da da 2 2
0 0
78Qian X/CE5887
Crack-Tip Constraints (1)
1.Is JIC independent of the specimen geometry?2.Is J-R curve unique for a given material???
J-dominance argument: if the J-R curve derives from two specimens that satisfy all the requirements for J-dominance, J for the same amount of (small) crack extension should not exhibit a dependence on the geometry, i.e., a unique J-R curve for a given material!
79Qian X/CE5887
Crack-Tip Constraints (2)
Constraint: or sometimes termed as “plasticity constraint”, defines the
capability of the crack tip in constraining the plastic deformation around the crack tip within a localized volume of materials around the crack tip!
High-constraint: small-scale yielding configuration
Low-constraint: large-scale yielding condition
80Qian X/CE5887
Crack-Tip Constraints (3)
600
500
400
300
200
100Specimen type
C(T)
Deep SE(B)
Shallow SE(B) SE(T)
DE(T)
A533 Grade B steel
JIC (kJ/m2)
• JIC does not indicate a clear dependence on the specimen geometry!• Reason being:
• JIC represents the energy for very small crack extension and for crack blunting• The process zone for JIC remains well within the finite strain zone• For high-constraint and low-constraint, the energies for crack blunting and crack extension within the finite strain zone are approximately the same
High-constraint specimensC(T), deep SE(B)
Low-constraint specimens:shallow SE(B), SE(T), DE(T)
81Qian X/CE5887
Crack-Tip Constraints (4)
350
300
250
200
150
100
Tearing Modulus TR at a = 1 mm
Specimen type
C(T)
Deep SE(B)
Shallow SE(B)
SE(T)DE(T)
A533 Grade B steel
RR
E dJT
da 2
0
• The J-a curve rises more sharply for low-constraint conditions!• Since the low-constraint condition re-distributes the near-tip stresses to the adjacent materials, spreading the plastic zone to a large volume of material, the near-tip stresses in the low-constraint conditions remain relatively lower than those around a high-constraint crack tip subjected to the same remote J. Consequently, the material damage (separation) takes place at a slower rate than the high-constraint condition.
82Qian X/CE5887
Crack-Tip Constraints (5)
SE(B) specimens
83Qian X/CE5887
Crack-Tip Constraints (6)
Two-Parameter Crack-Tip Fields
• Under large deformation, the single-parameter approach using J or K breaks down in characterizing the crack-tip stress-strain-displacement fields• Extension of the fracture mechanics theory beyond the small-scale yielding condition often utilizes the linear-elastic T-stress or the elastic-plastic Q stress
84Qian X/CE5887
Crack-Tip Constraints (7)
The linear-elastic T-stress is the second term in the Williams’ crack-tip stress solutions in isotropic materials, and represents a uniform stress acting parallel to the crack plane.
T-Stress
Iij ij
TK
fr
T
0 0
0 0 02
0 0
• For linear-elastic materials, the first term in the above equation exhibits the singularity.• The 3rd and higher order terms vanish near the crack tip, but the T-stress remains finite.• Research evidence reveals that the T-stress imposes a profound effect on the stresses deep within the plastic zone.
r1 /
85Qian X/CE5887
Crack-Tip Constraints (8)
R
K-T field
10 2 3 4 5 61
2
3
4
5
HRRn = 10
/yy y
/yr J
/ yT +1.0
0-0.2
-0.4-0.6-0.8-1.0
• Negative T-stresses reduce significantly the opening stress within the plastic zone, while positive T-stresses impose very little effects on yy
• Negative T-stresses decreases the stress triaxiality near the crack tip larger plastic zone size!
Modified boundary layer model for mode I loading
86Qian X/CE5887
Crack-Tip Constraints (9)
Biaxiality ratio :I
T a
K
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
-1
-0.5
0
0.5
1.0
a/W
SE(B)
SE(T)
DE(T)
M(T)
• Shallow cracked specimens often experience negative T-stresses• For M(T) specimens, = -1.0
87Qian X/CE5887
Crack-Tip Constraints (10)
J-Q approach (1):
0ij ij ijT diff
Based on small-strain assumption, the crack-tip stress can be expressed as the summation of the crack-tip stress at T = 0 and the deviation from T = 0
Research evidence indicates the magnitude of the stress shift for nonzero T-stresses remains approximately constant with both distance and angular position in the forward sector of the crack-tip region.
10 2 3 4 5 61
2
3
4
5
HRRn = 10
/yy y
/yr J
/ yT
-0.8
01( )yy diff
2( )yy diff
1 2( ) ( )yy diff yy diff
( ) ( ) ( )yy diff xx diff xy diff
88Qian X/CE5887
Crack-Tip Constraints (11)
J-Q approach (2):
O’Dowd and Shih designated the amplitude of this approximate difference field by the Q stress.
0ij ij ijT diff
00ij ij ijTQ
where,
0
0
yy yy TQ
at = 0 and 0 2
r
J
Q stress is a direct measure of the relative stress triaxiality (constraint) at the crack tip.
89Qian X/CE5887
Crack-Tip Constraints (12)
-1 -0.5 0 0.5 1.0-2.0
-1.5
-1.0
-0.5
0.0
0.5
n = 3n = 5n = 10n = 20n = ∞
0/T
Q
Modified boundary layer model
R
K-T field
For well-contained plastic zones near the crack tip, there exists a unique relationship between Q and T-stress.
90Qian X/CE5887