University of Liège
Aerospace & Mechanical Engineering
Fracture Mechanics, Damage and Fatigue
Non Linear Fracture Mechanics – HRR Theory
Ludovic Noels
Computational & Multiscale Mechanics of Materials – CM3
http://www.ltas-cm3.ulg.ac.be/
Chemin des Chevreuils 1, B4000 Liège
Fracture Mechanics – NLFM – HRR Theory
From elastic fracture to elastic-plastic fracture
• Small Scale Yielding assumption– The process zone, which is the
region where the inelastic
deformations• Plastic flow,
• Micro-fractures,
• Void growth, …
happen, is a small region compared to
the specimen size, and is at the crack tip
• How can plasticity be accounted for?– One approach is the cohesive model
• Non-linearities are localized at crack tip
• Dugdale assumed a perfectly plastic material
• Dugdale model’s validity is limited to thin
sheets of low-C steel or glassy polymers
– The plastic zone has actually a different shape in most cases
• When plasticity is not localized, elasto-plastic solution has to be computed
syy
x
Asymptotic syy
True syy
Zone of asymptotic
solution (in terms of
1/r1/2) dominance
Structural
responsePlasticity
a rp
dt
2a rprp
Crack tipCohesive
zone tip
x
y
sy
dT
2016-2017 Fracture Mechanics – NLFM – HRR Theory 2
From elastic fracture to elastic-plastic fracture
• Dugdale’s model findings
– Thin sheets of perfectly plastic materials
– Crack in an infinite plane
• Size of plastic zone
• J is related to CTOD
– SSY approximation
• If s∞< 40% of sp0 & a > 5 rp then first order SSY holds
• If SSY, a > 5 rp corresponds to
• Use classic LEFM
– Effective crack length method
• If s∞< 60% of sp0 then a second order SSY assumption holds
with the effective length
• For a general crack configuration
a rp
dt
2a rprp
Crack tipCohesive
zone tip
x
y
sy
dT
2016-2017 Fracture Mechanics – NLFM – HRR Theory 3
• Material law (small deformations)
– Yield surface
f < 0: elastic region
f = 0: plasticity– Plastic flow
• Assumption: deformations can be added
• Normal plastic flow
• Von Mises surface with isotropic plastic flow (J2-plasticity)
– Deviatoric part of the stress tensor
– Yield surface
– Normality: since
– Then the plastic flow becomes
– Path dependency (incremental equations in d )
Elastoplastic behavior
True e
Tru
es
sTS
sp0
ep ee
E
2016-2017 Fracture Mechanics – NLFM – HRR Theory 4
• Material law: J2-plasticity with non-linear hardening
– Hardening law
• Parameters a & n
• Term sp0/E represents the elastic
deformation before yield
– Plastic strain during plastic flow
• This law can be inverted as
• As
Elastoplastic behavior
2016-2017 Fracture Mechanics – NLFM – HRR Theory 5
0 2 4 6 8 100
0.5
1
1.5
2
e E/sp
0
s/ s
p0
a = 2; n = 1
a = 4; n = 1
a = 2; n = 4
a = 4; n = 4
a = 2; n = 7
a = 4; n = 7
s/
s 0
p
e E / s 0p
• Material law: J2-plasticity with non-linear hardening (2)
– Plastic strain
– Assume no unloading & plastic flow
• The d can be omitted:
• An internal potential can be defined
Elastoplastic behavior
2016-2017 Fracture Mechanics – NLFM – HRR Theory 6
• Power law– This law can be rewritten in terms of the total deformations
• Yield stress is replaced by
equivalent stress
• Plastic strain is replaced by
equivalent strain
• The governing law becomes
or
• Parameter n
– n→∞: perfect plasticity
– n→1: “elasticity”
– Doing so requires 2 assumptions• There is no unloading
• As elastic strains are assimilated to plastic strains, the material is incompressible
– Which are satisfied if• We are interested only in crack initiation and not in crack propagation• The stress components remain proportional with the loading• Elastic deformations are negligible compared to plastic ones
Elastoplastic behavior
2016-2017 Fracture Mechanics – NLFM – HRR Theory 7
0 2 4 6 8 100
0.5
1
1.5
2
e E/sp
0
s/ s
p0
a = 2; n = 1
a = 4; n = 1
a = 2; n = 4
a = 4; n = 4
a = 2; n = 11
a = 4; n = 11
s/
s 0
p
e E / s 0p
• Power law (2)
– Stress increment
• Using the governing law
• The strain increment is
• Or, as there is no unloading
– Internal potential
• No unloading:
Elastoplastic behavior
2016-2017 Fracture Mechanics – NLFM – HRR Theory 8
• The HRR theory
– From Hutchinson, Rice and Rosengren, 1967-1968
– Assumptions
• Semi-infinite crack
• Loading increased monotonically only for crack initiation
• Power law only if plastic deformations are large compared to elastic ones
– J-integral
• Elasto-plastic model & these assumptions = non-linear elasticity
• The J-integral can be used
• It is path independent, so one may choose
HRR theory
x
y B
n
Gr
q
2016-2017 Fracture Mechanics – NLFM – HRR Theory 9
• Near fields
– Evaluation of J-integral
• As J is path independent, it is the same whatever the value of r
– So the integrant should be independent of r
– As the terms involve stress times strain, it yields
– Generally speaking, the stress tensor can be expanded in a power series
where depends only on q and s’
• Let s’ be the dominant exponent near the crack tip
• Therefore as &
the strain tensor becomes
– Using theses considerations the dominant exponent is:
HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 10
HRR theory
• Near fields (2)
– Near fields are obtained using s’ = -1 / (n+1)
• Stress field
– kn is a plastic stress intensity factor which
» Allows defining independently of the loading amplitude
» Depends on n
– “Elasticity” (n→1): LEFM solution in 1/r1/2 recovered
– Perfect plasticity (n→∞): stress remains finite
2016-2017 Fracture Mechanics – NLFM – HRR Theory 11
• Near fields (3)
– Near fields are obtained using s’ = -1 / (n+1) (2)
• Stress field
• Strain field
– As &
– Eventually with
• Energy
HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 12
• Solution in terms of J-integral
– Let us assume that the solution is known
&
•
• The J-integral is rewritten
where is deduced from
• Let us define
• So stress and strain fields can be formulated in terms of J
&
HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 13
• Solution in terms of J-integral (2)
– Stress field reads
• J is governing the stress field near the crack tip
– It is the plastic analog of K in LEFM
– Assuming SSY and if the crack tip fields are governed by K
» Then a criterion on K is still valid
– If SSY does not hold, the criterion for crack initiation may be J ≥ JC
• N.B. if n→∞ only the strains will depend on J and not the stresses
– So J plays the role of an equivalent “plastic strain intensity factor”
HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 14
• Solution in terms of J-integral (3)
– Validity of this solution
• Stationary crack (no propagation) as unloading is prohibited
• In the region of dominance of the HRR field
– Near the crack tip
– As n increases, this region decreases (FE simulations)
• Small deformations
• Incompressible materials as we are using a power law with deviatoric strain field
– Questions
• How do we compute In, , ... ?
• What is the shape of the process zone ?
HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 15
• Resolution for a semi-infinite crack in plane e state
– Airy functions
• Linear momentum
– If b = 0, there exists an Airy function F:
– As we are not in linear elasticity we cannot say
• In polar coordinates
• As we are seeking a solution in r -1/(n+1), we choose
• Therefore, the following stress expressions are satisfying the balance equation
Semi-infinite mode I crack in plane e state
2016-2017 Fracture Mechanics – NLFM – HRR Theory 16
• Resolution for a semi-infinite crack in plane e state (2)
– Stream function
• As we are incompressible &
with plane strain assumption (if not: ezz≠0 )
• So displacements derive from a stream function Y with
• In polar coordinates
– As &
the displacements become
– Strains are obtained from
• As we are seeking a solution in r -n/(n+1), we choose
Semi-infinite mode I crack in plane e state
2016-2017 Fracture Mechanics – NLFM – HRR Theory 17
• Resolution for a semi-infinite crack in plane e state (3)
– Strain tensor
• Stream function
• So the strain components are
– Function f and g are dependant on each other as they have to satisfy
• The power law
• The normality equation
Semi-infinite mode I crack in plane e state
2016-2017 Fracture Mechanics – NLFM – HRR Theory 18
• Resolution for a semi-infinite crack in plane e state (4)
– Power law with normality
• These two equations lead to (see annex 1 for details)
–
– &
– Back to HRR field
• Near the crack tip, the fields were normalized by introducing an intensity factor
&
• Therefore, the functions f and g are rewritten
&
Semi-infinite mode I crack in plane e state
2016-2017 Fracture Mechanics – NLFM – HRR Theory 19
• Resolution for a semi-infinite crack in plane e state (5)– Using the new non-dimensional functions leads to
• &
•
– These equations can be reduced to a differential equation of the 4th order in
• See appendix I
• Range-Kutta resolution
• Initial boundary conditions
– Mode I symmetry
– Stress free lips
iteration on until these 2 conditions are satisfied
– It remains to choose
Semi-infinite mode I crack in plane e state
2016-2017 Fracture Mechanics – NLFM – HRR Theory 20
• Resolution for a semi-infinite crack in plane e state (6)
– Resolution of the plane e problem
• The differential equation is valid for any multiple of
– Initial condition is chosen so that
– The problem fields can now be determined
• &
• J characterizes the loading intensity on the crack
– Problem dependant
• In is a normalizing factor allowing to express the dependency with respect to J
– Still needs to be evaluated
Semi-infinite mode I crack in plane e state
2016-2017 Fracture Mechanics – NLFM – HRR Theory 21
0 50 100 1500
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 1.1
sqq
; n = 1.1
srq
; n = 1.1
se; n = 1.1
~
~
~
~
~
q [deg.]0 50 100 150
0
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 3
sqq
; n = 3
srq
; n = 3
se; n = 3
~
~
~
~
~
q [deg.]0 50 100 150
0
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 13
sqq
; n = 13
srq
; n = 13
se; n = 13
~
~
~~
~
q [deg.]
• Determination of stress and strain fields
– We sought the approximations &
• Similarly
• As J is path independent
– With obtained from
• Let us define , which
is evaluated numerically
– Eventually
Semi-infinite mode I crack in plane e state
2016-2017 Fracture Mechanics – NLFM – HRR Theory 22
0 5 10 150
1
2
3
4
5
6
7
nI n
Plane e
n
Semi-infinite mode I crack in plane e state
• Process zone shape
– The elastic-plastic boundary corresponds to
• As
• In terms of ,
the boundary is obtained for
2016-2017 Fracture Mechanics – NLFM – HRR Theory 23
Semi-infinite mode I crack in plane e state
• Process zone shape (2)
– The elastic-plastic boundary corresponds to
• In terms of ,
the boundary is obtained for
2016-2017 Fracture Mechanics – NLFM – HRR Theory 24
-10 -5 0 5
x 10-3
-2
-1
0
1
2x 10
-3
y
n = 1.1n = 3n = 13
~
~
x x 10-3
-0.2 -0.1 0 0.1 0.2
-0.2
-0.1
0
0.1
0.2
x
y
n = 1.1n = 3n = 13
r
~
~
~
q
x x 10-3
• Process zone shape (3)
– On particular interest is the size of the
process zone
• Ahead of crack tip (q=0)
• Behind crack tip (q=p)
• At its maximum value
– SSY assumption
• For linear elasticity
– n = 1 & a = 1
–
Plastic zone size
– Here we have assumed n=1/2
Semi-infinite mode I crack in plane e state
100
101
102
0
0.05
0.1
0.15
0.2
0.25
max
r
n
~1/(2p)
2016-2017 Fracture Mechanics – NLFM – HRR Theory 25
~
100
101
102
0
0.002
0.004
0.006
0.008
0.01
r
q=0
q=p
n
r
• Stress field at crack tip
– As &
the stress field at crack tip is
• The method does not remove the stress singularity (except for n→∞)
• As the abscissa is , it appears that J is a measure of the intensity
of the singular crack tip field (except for n→∞)
Semi-infinite mode I crack in plane e state
2016-2017 Fracture Mechanics – NLFM – HRR Theory 26
r0 0.02 0.04 0.06 0.08 0.1
0
2
4
6
8
10
r
syy
/ sp0
n = 1.1n = 3n = 13
~
syy
/ s
0p
r0 1 2 3 4
0
2
4
6
8
10
rs
yy/ s
p0
n = 1.1n = 3n = 13
~s
yy/
s 0
p
• Crack opening displacement (COD)
– Displacement field on lips:
• The COD is defined by
• Using
• It can be seen that
– The COD is of course >0
– Displacement along x is also >0
– So is not the opening
measured at r
– For Dugdale’s model, a CTOD has
been defined. Quid for HRR model?
Semi-infinite mode I crack in plane e state
x
y
uy
ux
r
2016-2017 Fracture Mechanics – NLFM – HRR Theory 27
-0.1 -0.08 -0.06 -0.04 -0.02 00
0.1
0.2
0.3
0.4
0.5
x
u s
p0/J
uy; n = 1.1
uy; n = 3
uy; n = 13
ux; n = 1.1
ux; n = 3
ux; n = 13
~x
us
0p
/ J
• Crack tip opening displacement (CTOD)
– The CTOD is defined as such
that a 90°-angle is intercepted
• Let r* be the value of r satisfying this criterion:
• As
• So
Semi-infinite mode I crack in plane e state
x
y
uy
ux
r*
2016-2017 Fracture Mechanics – NLFM – HRR Theory 28
• Crack tip opening displacement (2)
– The CTOD is defined as such
that a 90°-angle is intercepted
• Let
• So the CTOD becomes
– dn depends on
» n
» but also on
• As for a given material J is uniquely
related to the CTOD, this last one can
be used as a crack initiation criterion
Semi-infinite mode I crack in plane e state
x
y
uy
ux
r*
2016-2017 Fracture Mechanics – NLFM – HRR Theory 29
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
1/n
dn
asp
0/E = 0.001
asp
0/E = 0.002
asp
0/E = 0.004
asp
0/E = 0.008
0.78
1/n
• Summary
– Assumptions
• J2-plasticity with power law description
• Small deformations
• There is no unloading and loading is
proportional in all the directions (ok for crack
initiation and not for crack propagation)
• Elastic strains are assimilated to plastic strain
(material is incompressible)
• Semi-infinite crack
– HRR results for semi-infinite mode I crack
in plane e state
• Asymptotic stress, strain and displacement fields
– The J-integral plays the role of an
equivalent “plastic strain intensity factor”
– Stress field evolves in a proportional
way, so this is applicable to incremental
plasticity as long as J increases
Semi-infinite mode I crack in plane e state
2016-2017 Fracture Mechanics – NLFM – HRR Theory 30
0 5 10 150
1
2
3
4
5
6
7
n
I n
Plane e
n
0 50 100 1500
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 3
sqq
; n = 3
srq
; n = 3
se; n = 3
~
~
~
~
~
q [deg.]
• Summary (2)
– HRR results for semi-infinite mode I crack in
plane e state (2)
• Process zone
with
– If SSY
• CTOD
– New definition
–
– Also function of J
• We did not assume SSY !!!
– Questions
• What is happening for other configurations?
• What is the range of validity?
Semi-infinite mode I crack in plane e state
x
y
uy
ux
r*
2016-2017 Fracture Mechanics – NLFM – HRR Theory 31
-0.2 -0.1 0 0.1 0.2
-0.2
-0.1
0
0.1
0.2
x
y
n = 1.1n = 3n = 13
r
~
~
~
q
x
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
1/n
dn
asp
0/E = 0.001
asp
0/E = 0.002
asp
0/E = 0.004
asp
0/E = 0.008
0.78
1/n
• Mode I crack in plane s state
– Analysis is the same with other , & fields
Other HRR solutions
0 50 100 150-0.5
0
0.5
1
1.5
q [°]
s
srr
; n = 3
sqq
; n = 3
srq
; n = 3
se; n = 3
~
~
~~
~
Plane s
q [deg.]0 50 100 150
0
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 3
sqq
; n = 3
srq
; n = 3
se; n = 3
~
~
~
~
~
Plane e
q [deg.]
2016-2017 Fracture Mechanics – NLFM – HRR Theory 32
0 50 100 1500
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 13
sqq
; n = 13
srq
; n = 13
se; n = 13
~
~
~~
~ Plane e
q [deg.]
0 50 100 150-0.5
0
0.5
1
1.5
q [°]
s
srr
; n = 13
sqq
; n = 13
srq
; n = 13
se; n = 13
~
~
~
~
~
Plane s
q [deg.]
• Mode I crack in plane s state (2)
– Stress field is
– Asymptotic syy stress
• So for the same J, a thin specimen is less stressed at crack tip
Other HRR solutions
2016-2017 Fracture Mechanics – NLFM – HRR Theory 33
0 50 100 150-0.5
0
0.5
1
1.5
q [°]
s
srr
; n = 3
sqq
; n = 3
srq
; n = 3
se; n = 3
~
~
~~
~
Plane s
q [deg.]0 50 100 150
0
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 3
sqq
; n = 3
srq
; n = 3
se; n = 3
~
~
~
~
~
Plane e
q [deg.]
0 5 10 150
1
2
3
4
5
6
7
n
I n
Plane e
Plane s
n
• Mode I crack in plane s state (3)
– Crack Tip Opening Displacement
•
Other HRR solutions
x
y
uy
ux
r*
2016-2017 Fracture Mechanics – NLFM – HRR Theory 34
0 0.1 0.2 0.3 0.4 0.50
0.2
0.4
0.6
0.8
1
1/n
dn
asp
0/E = 0.001
asp
0/E = 0.002
asp
0/E = 0.004
asp
0/E = 0.008
Plane e
1/n
0 0.1 0.2 0.3 0.4 0.50
0.2
0.4
0.6
0.8
1
1/n
dn
asp
0/E = 0.001
asp
0/E = 0.002
asp
0/E = 0.004
asp
0/E = 0.008
Plane s
1/n
• Mode I crack in plane s state (4)– Process zone shapes
• The process zones
– Have a different shape
– Are more diffuse in plane s
• Mode III
– In SSY, the process zone has a circular
shape
• Perfectly plastic
• With hardening
Other HRR solutions
2016-2017 Fracture Mechanics – NLFM – HRR Theory 35
-0.2 -0.1 0 0.1 0.2
-0.2
-0.1
0
0.1
0.2
x
y
n = 3; Plane e
n = 3; Plane s
~
~
x
• Mode II and mixed mode crack
– In SSY, the solution depends on the
elastic mixity parameter
– Examples
• Process zones in plane e state
• SSY
Other HRR solutions
2016-2017 Fracture Mechanics – NLFM – HRR Theory 36
𝑥𝜎𝑝02
𝐾𝐼2 + 𝐾𝐼𝐼
2
𝑦𝜎𝑝02
𝐾𝐼2 + 𝐾𝐼𝐼
2n=1
n=3
n=13
𝑀𝑒=0.7
𝑥𝜎𝑝02
𝐾𝐼2 + 𝐾𝐼𝐼
2
𝑦𝜎𝑝02
𝐾𝐼2 + 𝐾𝐼𝐼
2n=1
n=3
n=13
𝑀𝑒=0
• Compressibility
– Elastic part of the deformations in not incompressible
– Considering this effect will diffuse the plastic zone
• Example: Mode I, plane e state & SSY
• The plastic zone size rp is defined as the
length of the plastic zone ahead of the crack
Other HRR solutions
2016-2017 Fracture Mechanics – NLFM – HRR Theory 37
rp
𝑥𝜎𝑝02
𝐾𝐼2
𝑦𝜎𝑝02
𝐾𝐼2
n =0.4
n =0.3
n =0.1
• HRR solution in SSY can explain 3D effect
– Let us assume that the solution is
• Plane-s-like near free surfaces
• Plane-e-like near mid-plane of specimen
– Even if K and s are not constant through
the thickness, it remains that
• There is a transition in the plastic zone
shape and size
• This transition is responsible for shear lips
• Shear lips form a 45-degree-angle since
szz=0 (so max shear at 45° in plane Oyz)
What can be learned from these models
Mid
surface
Free
surface
x
y
z
x
y
z
2016-2017 Fracture Mechanics – NLFM – HRR Theory 38
-0.2 -0.1 0 0.1 0.2
-0.2
-0.1
0
0.1
0.2
x
y
n = 3; Plane e
n = 3; Plane s
~
~
x
• HRR solution in SSY can explain 3D effect (2)
– During a toughness test, the K measured is
an average one
• It is more important for thin specimen
• The process is never really plane e
– There are actually complex 3D effects
• In SSY, even for a thin specimen, near the
mid-plane a plane-e-state is developed
• If the load increases, the plastic zone can
be plane-s-like near the mid-plane
– Effect of T-stress should also be considered
• Recall T-stress is the 0-order term obtained
with the asymptotic solution, which is dominant
at radius rc
• In general, if the test is such that T < 0, the
measured fracture K will be larger than for
T > 0, independently of the thickness *
• For ASTM toughness tests, the thickness
is large so that T > 0
What can be learned from these models
2aT>0
*D.J Smith, M.R Ayatollahi and M.J Pavier,Proc. R. Soc.
A, 2006, vol. 462, pp 415-2437
t
Kc
K rupture
Plane s
Plane e
t1
t2<<t1
t3<t2
t
2016-2017 Fracture Mechanics – NLFM – HRR Theory 39
• Effective crack length for SSY
– If SSY assumption holds
• J can be expressed in terms of K
• Then the plastic size can be written
• However, there are dependencies on
– Parameters n & n
– Whether it is plane e or plane s
– Rice’s model for perfectly plastic material
• Intersection of the linear elastic solution
with the yield stress leads at h rp
• But there is a redistribution of the stresses
so that the traction remains the same
– First order approximation: outside
the plastic zone, the stress
distribution is shifted
What can be learned from these models
x
syy
sp0
h rp
LEFM asymptotic
syy in r-1/2
Redistributed syy
rp
2016-2017 Fracture Mechanics – NLFM – HRR Theory 40
rp
𝑥𝜎𝑝02
𝐾𝐼2
𝑦𝜎𝑝02
𝐾𝐼2
n =0.4
n =0.3
n =0.1
• Effective crack length for SSY (2)
– Rice’s model (2)
• 2 equations with 2 unknowns
• So the plastic zone size is &
• So everything is as if the crack had an effective length a +h rp = a + rp/2
– From HRR models, numerical simulations, etc …
• Considering blunting, compressibility, hardening, …, an estimation is
What can be learned from these models
x
syy
sp0
h rp
LEFM asymptotic
syy in r-1/2
Redistributed syy
rp
2016-2017 Fracture Mechanics – NLFM – HRR Theory 41
• Effective crack length for SSY (3)
– If s∞< 50% of sp0 then a second order SSY assumption holds
• The cohesive zone remains small compared to crack size
• The effective crack size can be stated as a +h rp = a + rp/2
with for cracks in finite plates
• So there is an iterative procedure to follow:
– a) compute K from a
– b) compute effective crack size
– c) compute new K from aeff and back to b) if needed
– This method is a correction for linear fracture mechanics, but does not allow
considering problems with large yielding
What can be learned from these models
2016-2017 Fracture Mechanics – NLFM – HRR Theory 42
• Validity in SSY
– We have two asymptotic solutions
• HRR field is valid in the process zone
• LEFM is still valid in the elastic zone close to the crack tip
– Conditions
• This is the case if all sizes are 25 times larger than the plastic zone
Validity of HRR field
syy
x
LEFM asymptotic
syy in r-1/2
True syy
Plasticity
HRR asymptotic
syy in r-1/(n+1)
Log
syy
LEFM
asymptotic
syy in r-1/2
HRR asymptoticsyy in r-1/(n+1)
1
2
1n+1
Log r
t
a L
2016-2017 Fracture Mechanics – NLFM – HRR Theory 43
• Validity in SSY (2)
– Crack initiation criteria
• Criteria based on J or dt are valid: J ≥ JC or dt ≥ dC
– J & dt depend on a, the geometry, the loading, …
• But as the LEFM solution holds, we can still use K(a)≥KC
– Might be corrected by using the effective length aeff if s∞< 50% of sp0
– Sizes for K–based toughness test
• Examples:
– Titanium alloy 6%Al-4%V
» Yield: 830 MPa
» Toughness: 55 MPa · m½
» a, t, L > 1.1 cm
– Strength steel
» Yield: 350 MPa
» Toughness: 250 MPa · m½
» a, t, L > 1.27 m !!!
Validity of HRR field
t
a L
2016-2017 Fracture Mechanics – NLFM – HRR Theory 44
• Validity in elasto-plastic conditions– Deformations are small
– We still have one asymptotic solution valid
• HRR field is valid in the process zone
– LEFM is NOT valid in the elastic zone close to the crack
– Conditions
• This is the case if all sizes are 25 times larger than CTOD
Validity of HRR field
syy
x
LEFM asymptotic
syy in r-1/2
True syy
Plasticity
HRR asymptotic
syy in r-1/(n+1)
Log
syy
LEFM
asymptotic
syy in r-1/2
HRR asymptoticsyy in r-1/(n+1)
1
2
1n+1
Log r
t
a L
2016-2017 Fracture Mechanics – NLFM – HRR Theory 45
• Validity in elasto-plastic conditions (2)
– Crack initiation criteria
• Criteria based on J or dt are valid: J ≥ JC or dt ≥ dC
– J & dt depend on a, the geometry, the loading, …
• The LEFM solution DOES NOT hold, we CANNOT use K(a)≥KC
– Sizes for J–based toughness test
• If KC is computed from JC:
• Examples:
– Titanium alloy 6%Al-4%V
» Yield: 830 MPa
» Toughness: 55 MPa · m½
» Young: 110 GPa
» a, t, L > 0.75 mm
– Strength steel
» Yield: 350 MPa
» Toughness: 250 MPa · m½
» Young: 210 GPa
» a, t, L > 1.93 cm
Validity of HRR field
t
a L
2016-2017 Fracture Mechanics – NLFM – HRR Theory 46
• Validity in large yielding
– Example: ligament size is too small
– Small deformations assumption does not hold
– Neither HRR field nor LEFM asymptotic fields are valid
– Crack initiation criterion?
• As there is no zone of J-dominance can J still be used?
Validity of HRR field
syy
x
LEFM asymptotic
syy in r-1/2
True syy
Plasticity
HRR asymptotic
syy in r-1/(n+1)
Log
syy
LEFM
asymptotic
syy in r-1/2
HRR asymptoticsyy in r-1/(n+1)
1
2
1n+1
Log r
2016-2017 Fracture Mechanics – NLFM – HRR Theory 47
• Validity in large yielding (2)
– Plastic strain concentrations depend on the experiment
– Zones near free boundaries or other cracks tend to be less stressed
– Crack initiation criterion?
• Solution is no longer uniquely governed by J
• Relation between J & dt is dependant on the configuration and on the loading
• The critical JC measured for an experiment might not be valid for another one
• A 2-parameter characterization is needed
Validity of HRR field
s
s
M
M
s
s
2016-2017 Fracture Mechanics – NLFM – HRR Theory 48
• Thick specimen with centered crack:
– Steel
• Yield: 350 MPa
• Toughness: 250 MPa · m½
• Young: 210 GPa
• Hardening exponent >>
– Loading
• P = 25, 50 & 150 kN
– Compute the stress intensity factor
in terms of crack size a
• Compare the solutions obtained by
– SSY
– Effective crack length
• Check validity
Exercise 1: Specimen with centered crack
2a2W=1.5 in.
=3.8 cm
L = 3 in.= 7.62 cm
t = 0.5 in
=1.27 cm
P P
2016-2017 Fracture Mechanics – NLFM – HRR Theory 49
• A533-B1 steel:
– Exhibit brittle to ductile transition fracture
– Upper-shelf: Ductile fracture
• Observation: voids separated by 50 mm at T > 0°C
– Lower-shelf:
• Assumption: slip-induced cleavage at T< -30°C
• Critical fracture at
– 1830 MPa at strain rate = 0.01/s
– 2000 MPa at strain rate = 5/s
– Un-irradiated material
• Quasi-static, room temperature tensile test
– Yield sp0 = 481 MPa
– Ultimate sU = 642 MPa at e = 0.254
– Grain-size = 25 mm
• Yield (ksi, Rankine) in terms of temperature
and strain rate
– Questions
1) KIC in terms of temperature & strain rate in lower-shelf region
2) Validity?
Exercise 2: Nuclear reactor vessel
2016-2017 Fracture Mechanics – NLFM – HRR Theory 50
References
• Lecture notes
– Lecture Notes on Fracture Mechanics, Alan T. Zehnder, Cornell University,
Ithaca, http://hdl.handle.net/1813/3075
• Other references
– « on-line »
• Fracture Mechanics, Piet Schreurs, TUe, http://www.mate.tue.nl/~piet/edu/frm/sht/bmsht.html
– Book
• Fracture Mechanics: Fundamentals and applications, D. T. Anderson. CRC press, 1991.
• Fatigue of Materials, S. Suresh, Cambridge press, 1998.
2016-2017 Fracture Mechanics – NLFM – HRR Theory 51
• P= 50 kN
– Far stress field
– Solution for a = 7 mm
• SIF
with
Exercise 1: Solution
2a2W=1.5 in.
=3.8 cm
L = 3 in.= 7.62 cm
t = 0.5 in
=1.27 cm
P P
2016-2017 Fracture Mechanics – NLFM – HRR Theory 52
• P= 50 kN (2)
– Solution for a = 7 mm (2)
• Effective crack length
– First iteration
– Second iteration
– It has converged
• The correction is of about 1%
• Validity:
– s∞< 0.5 sp0: OK
– a = 7 mm, t = 12.7 mm, & W-a = 12 mm > 25 rp = 6.2 mm: OK
Exercise 1: Solution
2016-2017 Fracture Mechanics – NLFM – HRR Theory 53
• P= 50 kN (3)
– For other a
• For a = 9.2 mm
– rp = 0.383 mm 25 rp = 9.6 mm > a : solution not valid anymore
Exercise 1: Solution
rp > W-a
2016-2017 Fracture Mechanics – NLFM – HRR Theory 54
0 0.005 0.01 0.015 0.020
20
40
60
a [m]
K [
MP
a . m
1/2
]K(a); P=50kNK(a
eff); P=50kN
a [m]0 0.005 0.01 0.015 0.02
0
0.01
0.02
0.03
0.04
0.05
a [m]
[m]
25 rp; P=50kN
a; P=50kN
W-a; P=50kN
a [m]
• P= 25 kN
– Solution
• For a = 13.7 mm
– rp = 0.235 mm 25 rp = 5.9 mm > W- a = 5.3 mm: solution not valid
anymore
Exercise 1: Solution
rp > W-a
2016-2017 Fracture Mechanics – NLFM – HRR Theory 55
0 0.005 0.01 0.015 0.020
5
10
15
20
25
30
35
a [m]
K [
MP
a . m
1/2
]
K(a); P=25kNK(a
eff); P=25kN
a [m]0 0.005 0.01 0.015 0.02
0
0.005
0.01
0.015
0.02
0.025
a [m]
[m]
25 rp; P=25kN
a; P=25kN
W-a; P=25kN
a [m]
[m]
• P= 150 kN
– Solution
• For this loading
– 25 rp > a always
– As s∞= 0.88 sp0 SSY theory does not hold, the size of the plastic zone
cannot be approximated by
Exercise 1: Solution
rp > W-a
2016-2017 Fracture Mechanics – NLFM – HRR Theory 56
0 0.005 0.01 0.015 0.020
50
100
150
a [m]
K [
MP
a . m
1/2
]K(a); P=150kNK(a
eff); P=150kN
a [m]0 0.005 0.01 0.015 0.02
0
0.01
0.02
0.03
0.04
0.05
a [m]
[m]
25 rp; P=50kN
a; P=50kN
W-a; P=50kN
a [m]
• Hardening law
– Quasi-static, room temperature tensile test
• Yield sp0 = 481 MPa
• Ultimate sU = 642 MPa at e = 0.254
• Power law fit:
– For a = 1, n = 16
e (s = 642 MPa) = 0.244
– To use HRR theory, we take n = 10
Exercise 2: Solution
s
ep
0.002 0.254-ee
642
481
Suresh, Subra, 3.35 Fracture and Fatigue, Fall 2003. (Massachusetts Institute of Technology: MIT OpenCourseWare),
http://ocw.mit.edu (Accessed 14 Jul, 2010). License: Creative Commons BY-NC-SA
2016-2017 Fracture Mechanics – NLFM – HRR Theory 57
• HRR theory
– Small-scale HRR (no ductile failure)
• Stress field
• Plane strain assumption
– in small-scale
– For a = 1, n = 10: In ~ 4.5
– Mode I
~2.5
Exercise 2: Solution
2016-2017 Fracture Mechanics – NLFM – HRR Theory 58
0 50 100 1500
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 13
sqq
; n = 13
srq
; n = 13
se; n = 13
~
~
~~
~
q [deg.]
• Fracture by slip-induced cleavage
– Small-scale HRR:
– We know that
• Critical fracture from experiments at
– 1830 MPa at strain rate = 0.01/s
– 2000 MPa at strain rate = 5/s
– These values correspond to syy at a characteristic micro-scale distance
r = twice the grain size = 50 mm *
– KIC (Mpa m1/2):
• Quasi-static (strain rate = 0.01/s):
• Dynamics (strain rate = 5/s):
Exercise 2: Solution
* R.O. Ritchie, J.F. Knott, J.R. Rice, J. Mech. Phys. Solids 21, 395-410, 1973
2016-2017 Fracture Mechanics – NLFM – HRR Theory 59
• Fracture by slip-induced cleavage (2)
– KIC (Mpa m1/2):
• Quasi-static (strain rate = 0.01/s):
• Dynamics (strain rate = 5/s):
• In these expressions, sp0 should be expressed in Mpa
– In English unit
– Temperature from Rankine to Celsius
– Stress
Exercise 2: Solution
2016-2017 Fracture Mechanics – NLFM – HRR Theory 60
• Fracture by slip-induced cleavage (3)
– KIC (Mpa m1/2) (2):
• Quasi-static (strain rate = 0.01/s):
• Dynamics (strain rate = 5/s):
Exercise 2: Solution
T [°C]
KIC [MPa m1/2]
-200 -30
3.53.1
36
32
5/s
0.01/s
2016-2017 Fracture Mechanics – NLFM – HRR Theory 61
• Validity
– Upper shell: voids growth small-scale HRR not valid
• We need a model including damage **
– Lower shell
• To use SIF: maximal size for max (KIC/sp0)2
– QS test and T =-30°C : (KIC/sp0)2 = 0.0037 m
– a, t, L > 2.5 (KIC/sp0)2 = 9.3 mm
• To use J: maximal size for max (KIC2/Esp
0)
– QS test and T =-30°C : (KIC2/Esp
0) = 1.1 10-5 m
– a, t, L > 25 (KIC2/Esp
0) (1-n2) = 0.25 mm
Exercise 2: Solution
** R.O. Ritchie, A.W. Thompson, Metallurgical transactions A, 16A, 233-248, 1985
2016-2017 Fracture Mechanics – NLFM – HRR Theory 62
• Resolution for a semi-infinite crack in plane e state (A1)
– Power law with
• Deviatoric tensor, plane e & incompressible material:
– As or
– Out of plane stress &
– Non zero components of deviatoric tensor:
• Equivalent von Mises stress
– In terms of r and f
Annex 1: Mode I crack and HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 63
• Resolution for a semi-infinite crack in plane e state (A2)
– Power law (2)
• Equivalent deformation in terms of r and g
• So the power law becomes
• We have a differential equation in terms of functions depending on q & n
• But we are still missing a relation between the functions f & g
We will now study the normality relation
Annex 1: Mode I crack and HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 64
• Resolution for a semi-infinite crack in plane e state (A3)
– Normality
• Strain-stress relations &
imply
• In terms of f & g:
– Back to HRR field
• Near the crack tip, the fields were normalized by introducing an intensity factor
&
• Therefore, the functions f and g are rewritten
&
• Which allows writing the differential equations in terms of &
Annex 1: Mode I crack and HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 65
• The two differential equations are
– &
–
– Let then, these equations become
•
•
Annex 1: Mode I crack and HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 66
• Theses equations can be reduced to a single ODE as
–
– After differentiating and
substituting and its derivatives, it yields the fourth order ODE
Annex 1: Mode I crack and HRR theory
2016-2017 Fracture Mechanics – NLFM – HRR Theory 67
• Semi-infinite mode I crack in plane e state
– Von Mises stress (Plane e)
– Perfectly plastic material
• Von Mises stress:
• The Mohr’s circle is centered on with radius
– In Region II in frame• er, eq, we have pure shearing in the directions 0 and j=p/2
• ex, ey, we have pure shearing in the directions q and j=q+p/2
Annex 2: Slip line solution for perfectly plastic material
s
srq
2j
2016-2017 Fracture Mechanics – NLFM – HRR Theory 68
0 50 100 1500
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 13
sqq
; n = 13
srq
; n = 13
se; n = 13
~
~
~~
~ I IIIII
q [deg.]
• Semi-infinite mode I crack in plane e state (2)
– In Region I for q → 0: in frame
• er, eq, we have pure shearing in the
directions j=p/4 and j=-p/4
• ex, ey, we have pure shearing in the
directions j=p/4 and j=-p/4
– In Region I for q → p: in frame
• er, eq, we have pure shearing in the
directions j=p/4 and j=-p/4
• ex, ey, we have pure shearing in the
directions j=-3p/4 and j=3p/4
– Slip directions are the ones of maximal shearing
Annex 2: Slip line solution for perfectly plastic material
s
srq
-2j
2j
2016-2017 Fracture Mechanics – NLFM – HRR Theory 69
0 50 100 1500
0.5
1
1.5
2
2.5
q [°]
s
srr
; n = 13
sqq
; n = 13
srq
; n = 13
se; n = 13
~
~
~~
~ I IIIII
q [deg.]
• Mode I crack in plane s state
– Slip lines (perfectly plastic material)
Annex 2: Slip line solution for perfectly plastic material
2016-2017 Fracture Mechanics – NLFM – HRR Theory 70