ALERT School, Aussois, 12 October 2007
Mathematical Analysis
of Strain Localization
Milan Jirásek
1 Examples of strain localization
2 Inobjectivity of softening continuum models
3 Localization analysis
1
Examples of Strain Localization
Examples of strain localization on different scales
railway track after an earthquake (Turkey)
Examples of strain localization on different scales
conjugate shear bands in a perlite natural deposal
(Melos island, Greece)
Examples of strain localization on different scales
660 µ m
330 µ m
rock
(sandstone)
El Bied, 2002
Examples of strain localization on different scales
Desrues and Mokni
Colliat-Dangus, 1988
shear bands in soils …
Johansson, 2002
… and in
confined concrete
Examples of strain localization on different scales
Desrues and Mokni
Colliat-Dangus, 1988
Johansson, 2002
Examples of strain localization on different scales
shear bands in sand
under plane strain compression
(Alshibli et al., 2003)
Transition from diffuse to localized failure pattern
courtesy of Prof. Jacques Desrues, 3S Grenoble
Examples of strain localization on different scales
clay specimens under triaxial extension and compression
Examples of strain localization on different scales
Bésuelle et al., 2006
argillite (clay rock)
triaxial compression
digital image
correlation
X-ray microtomography
Examples of strain localization on different scales
sand under triaxial compression
(X-ray tomography)
Examples of strain localization on different scales
shear failure of masonry
(Lourenço, 1996)
Examples of strain localization on different scales
shear failure of masonry
(Lourenço, 1996)
Examples of strain localization on different scales
shear failure of masonry
(Lourenço, 1996)
Examples of strain localization on different scales
Examples of strain localization on different scales
2
Inobjectivity of Standard
Strain-Softening Continuum Models
Fracture process zone
(Murakami, Ohtani, 2000)
Single-edge-notched beam under four-point shear
Specimen tested by Arrea and Ingraffea (1981)
Single-edge-notched beam under four-point shear
Specimen tested by Arrea and Ingraffea (1981)
Four-point shear: simulated crack trajectory
Pathological mesh sensitivity in uniaxial FE simulation
softening
unloading
Pathological mesh sensitivity in uniaxial FE simulation
softening
unloading
ε f = 20 ε 0
P
softening
unloading
Af t
unloading
h = L/N
Lε 0
hε f = Lε f / N
u
Pathological mesh sensitivity in uniaxial FE simulation
number
of elements
N = 1,3,5,...
ε f = 20 ε 0
P
number of elements
Af t
1
3
5
9
u
Pathological mesh sensitivity in uniaxial FE simulation
ε f = 20 ε 0
P
number of elements
Af t
1
3
40 20
9
5
u
Pathological mesh sensitivity
3
Localization Analysis
Weak discontinuities
Weak discontinuities
Weak discontinuities
Weak discontinuities
3.1
Localization:
General Analysis
Incipient weak discontinuity
traction continuity
−
σ nn+ = σ nn−
σ nt+ = σ nt−
σ tt−
+
σ
−
+
tt
t
+
n
Incipient weak discontinuity
−
traction continuity
+
⋅n =
+
−
⋅n
−
t
n
−
+
+
compatibility
in 3D:
∂u + ∂u −
=
∂t
∂t
∂u + ∂u −
=
∂s
∂s
∂u + ∂u −
Incipient weak discontinuity
∂u + ∂u −
=
∂t
∂t
+
∂u
∂u −
=
∂s
∂s
∂u + ∂u −
=
+c
∂n
∂n
−
+
compatibility
∂u + ∂u −
∂n ∂u −
=
+c⊗
=
+c⊗n
∂x
∂x
∂x ∂x
+
∂u +
=
∂x
sym
∂u −
=
+c⊗n
∂x
=
−
+ ( c ⊗ n )sym
sym
Incipient weak discontinuity
traction continuity
−
+
⋅n =
+
+
∂u +
=
∂x
−
⋅n
−
constitutive
equations
sym
+
= Dt+ :
+
−
= Dt− :
−
∂u −
=
+c⊗n
∂x
+
compatibility
=
sym
−
+ ( c ⊗ n )sym
Incipient weak discontinuity
traction continuity
+
−
⋅n =
n ⋅ Dt+ :
−
⋅n
+ n ⋅ Dt+ : ( c ⊗ n )sym = n ⋅ Dt− :
(n ⋅ D
+
t
⋅ n) ⋅ c
m =1
me,
(n ⋅ D
+
t
−
⋅ n ) ⋅ me = n ⋅ ( Dt− − Dt+ ) :
−
Localization analysis
General equation describing localization:
(n ⋅ D
+
t
⋅ n ) ⋅ me = n ⋅ ( Dt− − Dt+ ) :
−
Simplified form for the same tangent stiffness
on both sides of the discontinuity:
( n ⋅ Dt ⋅ n ) ⋅ m e = 0
Nontrivial solution exists iff
det (n ⋅ Dt ⋅ n ) = 0
Qt = n ⋅ D t ⋅ n
localization tensor
(acoustic tensor)
Localization analysis
General equation describing localization:
(n ⋅ D
+
t
⋅ n ) ⋅ me = n ⋅ ( Dt− − Dt+ ) :
−
For plasticity and isotropic damage models, the general
equation has a nontrivial admissible solution iff there exists
a unit vector n such that
det ( n ⋅ Dt ⋅ n ) ≤ 0 for Dt = Dt+ or Dt = Dt−
For det=0 the solution is of the loading/loading type,
for det<0 it is of the loading/unloading type.
Elastic acoustic tensor
for isotropic elasticity:
stiffness tensor
unit second-order tensor
De = λ ⊗ + 2µI s
unit symmetric
fourth-order tensor
Lamé coefficients
acoustic tensor
Qe = λ n ⊗ n + µ ( + n ⊗ n ) = µ + ( λ + µ ) n ⊗ n
positive-definite !
inverse of acoustic tensor
Qe−1 =
1
µ
−
λ+µ
1
n⊗n =
µ ( λ + 2µ )
G
−
n⊗n
2 (1 − ν )
3.2
Localization Analysis:
Isotropic Damage Model
Simple isotropic damage model
stress-strain relation
= (1 − ω )De : = Du : = (1 − ω )
damage evolution
ω = g (κ )
… damage law in explicit (total) form
f ( , κ ) = ε~ ( ) − κ
f ≤ 0, κ ≥ 0,
ε~ (
)
… loading function
fκ = 0
… loading-unloading
conditions
… equivalent strain, e.g.
ε(
)=
1
: De :
E
Elasto-damage stiffness and localization tensor
= (1 − ω )D e : − De : ω ,
Ded = (1 − ω ) De −
dg ∂ε
:
dκ ∂
ω=
dg
∂ε
De : ⊗
dκ
∂
rank-1 correction
of secant (unloading) stiffness
Ded = Du − g ′ ⊗
localization tensor:
Qed = n ⋅ Ded ⋅ n = n ⋅ Du ⋅ n − g ′ n ⋅ ⊗ ⋅ n
Qu
Qed = Qu − g ′
n
⊗
n
n
rank-1 correction
of unloading acoustic tensor
n
Localization condition for simple isotropic damage model
necessary condition for discontinuous bifurcation:
∃n, m : Qed ⋅ m = 0,
( Qu − g '
n
⊗
n
Qu ⋅ m = g '
n
⊗
m = g '(
n
−1
u
Q ⋅
n
n
⋅m
⋅ m) Q ⋅
= g'
n
)⋅m = 0
−1
u
nontrivial solution
n
n = 1, m = 1
Qu−1 ⋅
= −1
Qu ⋅
?
n
n
n
m ≠ 0 exists iff:
−1
n ⋅ Qu ⋅ n
e.g., iff 1 = g '
Qu−1 ⋅ n
n
⋅ Qu−1 ⋅
n
Localization condition for simple isotropic damage model
For a given state of the material, we know
,ω →
, , g ', Ded
and we need to check whether there exists
a direction n for which
g'
n
⋅ Qe−1 ⋅
n
= 1−ω
Consequently, if
g ' max (
n =1
n
⋅ Qe−1 ⋅
n
) < 1−ω
then a discontinuous bifurcation is impossible.
Equivalently:
g ' < g 'crit ≡
1−ω
max ( n ⋅ ⋅ Q e−1 ⋅ ⋅ n )
n =1
Example – one-dimensional damage model
σ = 1 − g ( ε ) Eε
(during monotonic loading)
σ = 1 − g ( ε ) − ε g ' ( ε ) Eε
Eed … tangent modulus
g 'crit =
1−ω
1−ω
1−ω
=
=
ε
max ( n ⋅ ⋅ Qe−1 ⋅ ⋅ n ) 1 × 1 × 1 × Eε × 1
n =1
E
Eed ,crit = (1 − ω − ε g 'crit ) E = 0
discontinuous bifurcation expected
at the peak of the stress-strain curve
Example – isotropic damage, energy-based equiv. strain
1
: De :
E
∂ε
1
1
=
=
2D e : =
∂
E
Eε (
1
2
: De :
E
n⋅
n
=
=
n = n⋅
Eε ( ) Eε ( )
ε(
)=
Qe−1 ⋅
n
=
−1
n ⋅ Qe ⋅
1
G
n
−
=
n⊗n
⋅
2 (1 − ν )
n
Eε (
)
⋅
1
G
n
n −
=
1
G
n
−
)
σ nn n
2 (1 − ν )
1
σ nn n
=
2 (1 − ν )
EGε (
)
2
n
−
σ nn2
2 (1 − ν )
Example – isotropic damage, energy-based equiv. strain
Uniaxial tension:
= Eε1e1 ⊗ e1
n
n
= ⋅ n = Eε1e1n1 = Eε1e1 cos θ
= Eε1 cos θ
σ nn =
εe(
2
angle between n
and the loading direction
n
⋅ n = Eε1n12 = Eε1 cos2 θ
σ 1 = (1 − ω ) σ 1 = (1 − ω ) Eε1
) = ε1
−e
11
n ⋅ Qe ⋅
n
=
1
cos4 θ
2
2
E
ε
cos
θ
−
=
n ( 1)
EGε1
2 (1 − ν )
= 2 (1 + ν ) ε1 cos2 θ −
cos4 θ
2 (1 − ν )
Example – isotropic damage, energy-based equiv. strain
Uniaxial tension:
= Eε1e1 ⊗ e1
n
n
= Eε1 cos θ
σ nn =
ε(
angle between n
and the loading direction
= ⋅ n = Eε1e1n1 = Eε1e1 cos θ
n
⋅ n = Eε1n12 = Eε1 cos2 θ
) = ε1
−1
n ⋅ Qe ⋅
n
=
1
cos4 θ
2
=
( Eε1 ) cos2 θ −
EGε1
2 (1 − ν )
cos4 θ
= 2 (1 + ν ) ε1 cos θ −
2 (1 − ν )
2
Example – isotropic damage, energy-based equiv. strain
Uniaxial tension (continued):
cos4 θ
1 −ν
max cos θ −
=
θ
2 (1 − ν )
2
2
g 'crit
crit
= arccos 1 − ν
1 −ν
= (1 − ν 2 ) ε1
2
1−ω
1−ω
=
=
−1
max ( n ⋅ ⋅ Q e ⋅ ⋅ n ) (1 − ν 2 ) ε1
max (
n =1
( at θ
n
⋅ Q e−1 ⋅
n
) = 2 (1 + ν ) ε
1
n =1
Eed ,crit = (1 − ω − ε g 'crit ) E = −
Euν 2
<0
1 −ν 2
discontinuous bifurcation expected
after the peak of the stress-strain curve
)
Example – isotropic damage, Rankine equiv. strain
1
) = σ max ( )
ε(
∂ε 1 ∂σ max
=
∂
E ∂
1
ν
=
1 + ν 1 − 2ν
=
= n⋅ =
n
∂
1
= ( p1 ⊗ p1 ) : D e =
∂
E
:
+ p1 ⊗ p1
1
ν
n + p1 cos θ
1 + ν 1 − 2ν
−1
e
= n⋅ , Q ⋅
n
direction of maximum
principal stress
E
n
(ν − cos θ ) σ
=
2
−1
e
⋅Q ⋅
n
n
angle between n
and direction of maximum
principal stress
σ nn n
n −
2 (1 − ν )
1
=
G
nn
+ 2 (1 − ν ) σ max cos2 θ
E (1 − ν )
Example – isotropic damage, Rankine equiv. strain
Uniaxial tension:
= Eε1e1 ⊗ e1 , p1 = e1
σ nn = Eε1n12 = Eε1 cos2 θ
σ max = Eε1
n
⋅ Q e−1 ⋅
n
=
ε1
2 − ν − cos2 θ ) cos2 θ
(
(1 − ν )
max ( 2 − ν − cos θ ) cos θ = 1 −
2
2
θ
max (
n =1
ε1
ν
1−
n ⋅Q ⋅ n) =
1 −ν
2
−1
e
2
ν
2
2
at θ crit = arccos 1 −
ν
2
Example – isotropic damage, Rankine equiv. strain
Uniaxial tension (continued):
g 'crit =
1−ω
1 − ω 1 −ν
=
ε1 (1 − ν / 2 )2
max ( n ⋅ ⋅ Qe−1 ⋅ ⋅ n )
n =1
Eed ,crit = (1 − ω − ε g 'crit ) E =
ν
2 −ν
discontinuous bifurcation expected
before the peak of the stress-strain curve
2
Eu > 0