8/16/2019 Notes on Continuum Damage Models
1/23
8/16/2019 Notes on Continuum Damage Models
2/23
NOTES ON CONTINUUM DAMAGE MODELS 2
1.2 The Isotropic Damage Model in a SmallDeformation Regime
Continuum damage models have been widely accepted for simulating the behavior ofmaterials whose mechanical properties are degrading due to the presence of small cracksthat propagate during loading. To fully describe this phenomenon, we will first use a one-dimensional model (1D) which we will then extrapolate to three dimensional ones (3D).
With regard to continuum kinematics, our study in this section will be carry out in a smalldeformation regime, and will be based on the lecture notes of Prof. Javier Oliver,Universitat Politècnica de Catalunya.
1.2.1
Description of the Isotropic Damage Model in UniaxialCases
Let us now suppose that a material point is subjected to the stress state as shown in Figure11.1, whose apparent stress ( σ ) acts on the section s and due to the presence of faults(microcracks), only the undamaged region will be considered, i.e. the effective section ( s )on which the effective stress ( σ ) acts.
Figure 11.1: Continuum with microcracks.
Then, if we consider the force balance in Figure 11.1, we obtain:
σ=σ ss (11.1)
The equation (11.1) can also be rewritten without altering its outcome as follows:
σ
−=σ
−−=
+
−σ=σ=σ 111
s
s
s
ss
s
ss
s
s d (11.2)
where d s is the damaged section.
Note, the expressions
sd represents the amount of the original section which is corrupted,
which in extreme cases, assumes the follows values:
s
P
B
σ - effective stress
σ - apparent stress
s
σ
σ
σ
microcrack
material point
8/16/2019 Notes on Continuum Damage Models
3/23
NOTES ON CONTINUUM DAMAGE MODELS 3
σ=σ⇒=⇒= 00s
ss d d - The section is not damaged;
0 1 =σ⇒=⇒=s
sss d d - The section is completely damaged.
The amount d s depends on the stress state σ or indirectly on ε . The dimensionless ratio
s
sd represents the damage variable and is denoted bys
sd d = . Then, the equation in (11.2)
can be written as:
( ) 10;1 ≤≤σ−=σ d d (11.3)
where σ is the effective stress .
1.2.1.1 The Constitutive Equation
The effective stress σ and strain, in the undamaged area element, are interrelated byHooke’s law as:
ε=σ E (11.4)
where E is Young’s modulus. Then, by substituting (11.3) into (11.4) we can obtain theconstitutive equation for stress in the one-dimensional isotropic damage model:
( ) ε−=σ 1 E d 10 ≤≤ d (11.5)
We can now verify that as the damage variable evolves, the state no longer returns to itsoriginal value. Physically speaking, we can interpret this as once the material has suffered
damage this will be permanent. Hence, we can conclude that 0≥d , which characterizes anirreversible process. Now, the equation in (11.5) can still be written as:
( ) E d E E sec_d sec_d 1−=ε=σ with (11.6)
where sec_d E is the damage secant stiffness modulus with which we can observe that the damage variable can be interpreted as a measure of the loss of stiffness modulus of the material.
In general, materials have a yield stress that separates the elastic (reversible process) fromthe inelastic zone (irreversible process). In the strain space, we can represent the elasticlimit by the variable 0ε , (see Figure 11.2), in which the damage process has not yet begun,i.e.:
0=d if 0ε
8/16/2019 Notes on Continuum Damage Models
4/23
NOTES ON CONTINUUM DAMAGE MODELS 4
( ) )10(;1 ≤≤ε−=σ d E d
0=d if 0ε
8/16/2019 Notes on Continuum Damage Models
5/23
NOTES ON CONTINUUM DAMAGE MODELS 5
of internal variables ( k a ). Let us also consider there is a process independent of
temperature, and the internal variable associated with the problem is characterized by thedamage variable d . Furthermore, as seen in previous chapters, as the Helmholtz freeenergy must satisfy the principle of objectivity (see Chapter 6), we can express Y in termsof the Green-Lagrange strain tensor ( E ), which in turn collapses with the infinitesimal
strain tensor in a small deformation regime, i.e. ε≈ E . Then, if we consider all of theabove, the Helmholtz free energy can be expressed in terms of:
( ) ,d εY Y = (11.10)
or explicitly as follows:
( ) ( ) εε 2
1 11 :: ee d d −=−= Y Y
Helmholtz free energy forisotropic damage model
(11.11)
where )(εeY is the elastic strain energy density, which is a function of strain only, and e
is the elasticity tensor (or elastic stiffness tensor).
1.2.2.2 Internal Energy Dissipation and the Constitutive Equations
The damage model has thermodynamic consistency, and so, entropy inequality is fulfilled.One way to express this entropy inequality is by means of the alternative form of theClausius-Planck inequality, (see Chapter 5), which is expressed by:
0≥+−= Y h T int Dσ :D
3m
J (11.12)
Note that the terms Dσ : , T h , Y have the unit of energy per unit volume (density
energy). In a small deformation regime εD ≈ holds, and by considering the isothermalprocess we have 0=T , so, the equation in (11.12) becomes:
0≥−= Y εσ :int D (11.13)
Then, the rate of change of the free energy ( ) ,d εY Y = can be evaluated as follows:
( ) d d
d ∂
∂+
∂
∂=
Y Y Y ε
ε
ε :, (11.14)
Next, by substituting (11.14) into the internal energy dissipation given in (11.13) we obtain:
0),( ≥
∂
∂−
∂
∂−=
∂
∂−
∂
∂−=−= d
d
d
d
d int
Y Y Y Y Y ε
ε
σ ε
ε
εσ εεσ ::::D (11.15)
Note that the above inequality must hold for any admissible thermodynamic process, so, let
us assume there is one where 0=d . Here, we obtain 0≥
∂
∂−= ε
ε
σ :Y
int D , which in turn
must also be true for any process. Additionally, if we have a process such that εε −→ , the
only way for the entropy inequality to be satisfied is whenε
σ
∂
∂= Y
holds with which we
obtain the constitutive equation for stress . Thus, the entropy inequality becomes:
0≥∂
∂−=
∂
∂−
∂
∂−=
=
d d
d d
int
Y Y Y ε
ε
σ
0
:D (11.16)
8/16/2019 Notes on Continuum Damage Models
6/23
NOTES ON CONTINUUM DAMAGE MODELS 6
Now, if we consider the energy equation ( ) ed Y Y 1−= , we obtain ed
Y Y
−=∂
∂ , thus
0≥= d eint Y D (11.17)
where by definition 0≥eY . Then, to satisfy the inequality (11.17), the rate of change of the
damage parameter must satisfy:
0≥d (11.18)
Then, by means of thermodynamic considerations we can draw the conclusion that:
0; ≥∂
∂= d
ε
σ Y
(11.19)
We can also express the rate of change of the Helmholtz free energy by means of theequation in (11.11), i.e.:
( ) ( ) eeeee d d d d d Y Y Y Y Y 11 −=−−=−−= εσ εε ::: (11.20)
Next, the rate of change of the elastic strain energy, εε 21
:: ee =Y , was obtained as
follows:
( ) 2
1εεεεεε ::::::
eeee ++=Y (11.21)
where 0=e , since e is constant, and as the elasticity tensor features major symmetry
( eijkleklij CC = ), the equation in (11.21) becomes:
( ) ( )
( ) εσ εσ εε
1
1
2
1
2
1
::::
d
ekl
eijklij
kleijklijij
eklijklkl
eijklijkl
eijklij
e
−===εε=
εε+εε=εε+εε=
C
CCCCY
(11.22)
Note that due to the major symmetry of e , εε :: ee = is fulfilled.
Then, starting from the equation in (11.11) we can obtain the stress by taking the derivativeof the strain energy with respect to strain, i.e.:
( )( ) ( ) [ ]
( )
( ) ( )[ ] ( )[ ]
( ) ( ) ( )
( ) ( ) ( ) ( ) { }e pqij pqeijklkle pqjie pqij pqe jikleijklkl
kjliljki pq pjqiqj pikle pqkl
ij
lk kl
pq
ij
qp pq
kle pqkl
ij
kl
pqij
pq
kl
e
pqkl
kl pq
ij
e pqklkl
e pqkl pq
ijij
ij
d d
d
d
d
d d d
CCCCCC
C
C
C
CC
ε+ε−=
+ε++ε−=
+ε++ε−=
ε∂
ε+ε∂ε+
ε∂
ε+ε∂ε−=
ε∂
ε∂ε+
ε∂
ε∂ε−=
εεε∂
∂−=
εε−
ε∂
∂=
ε∂
∂=σ
2
1 1
2
1
2
1
2
1 1
2
1
2
1
2
1 1
2
1 1
2
1
1
2
1 1
2
1 1
,
21
21
d d d d d d d d
Y ε
(11.23)
where we have taken into account the minor symmetry of the elasticity tensor, i.e. e
jikl
e
ijkl CC = ,e
pqji
e
pqij CC = . Note also that the indexes p , q are dummy indexes, so we can
8/16/2019 Notes on Continuum Damage Models
7/23
NOTES ON CONTINUUM DAMAGE MODELS 7
exchange them for k and l without altering the expression. Additionally, by taking into
account the major symmetry of the elasticity tensor, eklijeijkl CC = , we obtain:
( ) kleijklij d ε−=σ C1 (11.24)
which in tensorial notation becomes:
( )( ) ( )σ ε
ε
εσ 11
,d d
d e −=−=∂
∂= :
Y
The constitutive equations for isotropicdamage model
(11.25)
where σ is the effective Cauchy stress tensor and is defined as:
εσ :e = The effective Cauchy stress tensor (11.26)
and e is the elasticity tensor (fourth-order definite positive tensor) which contains the
elastic mechanical properties. Remember that e can be represented in terms of the Laméconstants ( l , m ) as follows:
( ) jk il jlik klijeijkle d d d d md ld ml ;2 ++=+⊗= CI11 (11.27)
where 1 is the second-order unit tensor, and sym ≡I is the symmetric fourth-order unittensor, whose components are expressed in terms of the Kronecker delta ( ijd ) as follows:
≠
===
jiif
jiif ijij0
1)( d 1 ; ( ) jk il jlik ijklijklsymijkl d d d d
2
1)( +==≡ ΙII (11.28)
Then, by analyzing the constitutive equation in (11.25) we can put in evidence thefollowing sentences:
• Since the damage parameter is a scalar, the stiffness degradation is isotropic;
• We can calculate the stress immediately once we know the current values ofε (strain) and d (internal variable);
• We can interpret the equation in (11.25) as the sum of elastic and inelastic parts, i.e.:
( ) ie
inelastic
e
elastic
ee d d σ σ εεεσ −=−=−= 1 ::: (11.29)
The Elastic-Damage Secant Stiffness Tensor
We can then define the elastic-damage secant stiffness tensor for the isotropic damagemodel as:
( ) esec_d d 1−= The elastic-damage secant stiffness tensor (11.30)
Let us now consider a uniaxial case, (see Figure 11.4), where the material is loaded until thestress state reaches the point P represented in Figure 11.4, after which unloading occurs,
with the unloading path being that indicated by the slope E d E sec_d )1( −= defined in
Figure 11.4.
1.2.2.3 “Ingredients” of the Damage Model
The damage constitutive model is completely determined when the damage variable t d is
known at each time step t of the loading/unloading process. Then, we can define thefollowing elements of the constitutive equation:
8/16/2019 Notes on Continuum Damage Models
8/23
NOTES ON CONTINUUM DAMAGE MODELS 8
The energy norm of the stress (or strain) tensor;
The damage surface and damage criterion. The damage surface defines the elasticlimit, and the damage criterion establishes when the material is in a loading or in aelastic process, and;
A set of evolution laws for internal variables.
Figure 11.4: Stress-strain curve.
The Energy Norm in the Stress/Strain Space
The norm is a measure of distance and so is a scalar. Next, we will define a simple norm inthe stress space denoted by
σ t (equivalent stress), and in the strain space denoted by
εt .
The latter is also known as the equivalent strain :
εσ
εσ εεεσ σ σ
t
t t
t )1(
2;1
1
d
eeeee
−=⇓
===== −
−
Y ::::
(11.31)
Note thatσ
t andε
t are surface equations (ellipsoids) that characterize the stress state at
the current point (see Figure 11.5). The proof of (11.31) now follows:
( ) ( ) ( )( )
εσ
ε
σ
εσ εε
εσ εσ εσ σ σ t t
t
t d
d d d
e
e
−=⇒
==
−=−=−== −
1111
21
:::
:::::
(11.32)
In order to better describe material behavior, others norms will be introduced (see
subsection 11.2.4).
The Damage Criterion
Next we will define the damage criterion in the stress and strain space:
space stress
r qq 0)(),( ≤−= σ σ
t t F and
spacestrain
r r 0),( ≤−= εε
t t G (11.33)
where r is an internal variable (current damage threshold), and q is a stress-like
hardening/softening variable which is a function of r . Note that each material in itsundamaged state is characterized by the initial value of r which is denoted by 0r (the
material parameter), which defines the initial yield in the strain space. Then, the material
σ
E
ε
E d E sec_d )1( −=
Y σ
Dissipated energy
P
1 1
8/16/2019 Notes on Continuum Damage Models
9/23
NOTES ON CONTINUUM DAMAGE MODELS 9
starts to fail (initial damage) when the energy norm exceeds the value 0r . Later we will
relate the variables r and q to the damage variable.
Figure 11.5: Strain and stress state in the principal space.
The damage criterion requires that the current stress state must be on or inside the damagesurface. When the stress state lies inside said damage surface, the material shows elasticbehavior, which can be elastic loading or unloading.
Then we can define the admissible strain space as follows:
{ }0),(: ≤= r εε
ε t GE (11.34)
and the admissible stress space as:
{ }0),(: ≤= qσ σ
σ t F E (11.35)
When it holds that 0),( =qσ
t F , in the stress space, the stress state is on the surface as
indicated in Figure 11.5( b).
The stress space ( σ E ), (see Eq. (11.35)), can be decomposed into the inner domain
( )σ
E int (when the stress state is inside the surface), and other by the surface itself,σ
E ∂ .
We can define then the elastic region in strain and stress respectively as:
{ } { }0),(:;0),(: ≤=
8/16/2019 Notes on Continuum Damage Models
10/23
NOTES ON CONTINUUM DAMAGE MODELS 10
Said damage evolves when the normε
t exceeds the maximum value reached by r . Then,
considering (11.33) and (11.31) we can also conclude that:
r d r q )1()( −= (11.40)
In uniaxial cases, damage starts whenε
t exceeds the first damage threshold value 0r .
Then, from the equation in (11.31) and by means of Figure 11.2, we can obtain:
E r r
E E
E E E
Y
Y Y e
σ=⇒=−
σ=
σ=ε=εε= → =
00
000uniaxial
0
ε
εε εε
t
t t ::
(11.41)
where Y σ is the yield stress (obtained in the laboratory). Then, ),(0 E r Y σ can be
interpreted as a material mechanical property also obtained in the laboratory.
Figure 11.6: The evolution ofε
t and r over time t .
The Internal Variable Evolution Law. The Kuhn-Tucker and Consistency Conditions
2( 0)2(
r r ← )
3( ε
t ←)3(r )
4( ε
t ←)4(r
ε
σ
Y σ
0ε
1ε
1
1 2 3 4
5( )4()5(
r r ←
3r
1ε
1 2 3 4,
0r
5
54 r r =
0=r
t
r
0r
0r
t
εt
1 2
3 4 5
1
2
34
5
6
6
0
>r
8/16/2019 Notes on Continuum Damage Models
11/23
NOTES ON CONTINUUM DAMAGE MODELS 11
The constitutive equation described above uses three types of variables, namely: the free variable { }ε ; the internal variable { }r ; the dependent variables { })(),,(),,( r d d r εσ εY .
Now, to establish how the internal variable r evolves, let us take the example described inFigure 11.6. As we can observe, the discretized r between points 2-3 and 3-4 are positiveand between points 1-2 and 4-5 are equal to zero, so we can conclude that r is amonotonically increasing function, i.e.:
0≥r (11.42)
Graphically, we can see in Figure 11.6 how the variables r andε
t evolve. Furthermore,
we can also verify that in the range between the points 4-6 0),( r ε
t G hold, this
implies that 0),( >∆+ t t r ε
t G , which thereby violates the condition }t r t 0),( ∀≤ε
t G , so
00 =⇒> Gz must be satisfied. Another possible situation is when the current state is
inside the damage surface, i.e. 0),( <t
r εt G , and if in the next loading step 0),( <∆+ t t
r εt G
8/16/2019 Notes on Continuum Damage Models
12/23
NOTES ON CONTINUUM DAMAGE MODELS 12
is satisfied, this implies that 0),(0 ==⇒< r r εz G . We can gather these previous
conditions by means of the loading/unloading condition, also called the Kuhn-Tucker conditions :
0),(;0),(;0 =≤≥ r r εε
t t GG z z The Kuhn-Tucker conditions (11.48)
and by the consistency (persistency) condition:0),( =r
εt Gz The consistency condition (11.49)
If we are undergoing loading, this implies that 0>z , then by means of the Kuhn-Tucker
conditions 0),( =r ε
t G must be fulfilled. Here, the value of z can be obtained by means
of the consistency condition:
r r r =⇒== εεε
t t t 0),(),( GG (11.50)
Schematically, we can summarize the above loading/unloading states as follows:
(11.51)
NOTE: If the parameter ),( d ε
t H , given in (11.46), is not a function of d , we can
express it by means ofε
ε
ε
t
t t
∂∂= )()( GH , where we have introduced the scalar function G
which is a monotonically increasing function, which has proven to be a convenient way toexpress the damage criteria:
( ) ( )
( ) ( ) 0 ;0),(
0 ;0),(
≥∀≤−=
≥∀≤−=
t qF F q
t r GGr
σ σ
εε
t t
t t
F
G (11.52)
Here the loading/unloading condition becomes:
r
r d r r
∂
∂==
),(;),( εε
t Gz z (11.53)
0),(;0),(;0 =≤≥ r r εε
t t GG z z The Kuhn-Tucker conditions (11.54)
0),( =r ε
t Gz The consistency condition (11.55)
The Damage Variable
The parameter q is the stress-like hardening/softening parameter, and is defined in terms
of r as follows:
r
r q
r d r d r q
)(
1)()1()( −=⇒−=
(11.56)
Now, by using the equations in (11.56) and (11.25) we can obtain:
=
<
=
<
0
0
0
0
G
G
G
G
⇒
⇒
⇒
>
=
0
0
z
z
⇒
⇒
⇒
0=z
0=z
⇒
0=d
0=d
0=d
0>d
⇒
⇒
⇒
⇒
(elastic)
(unloading)
(neutral loading)
(loading)
8/16/2019 Notes on Continuum Damage Models
13/23
NOTES ON CONTINUUM DAMAGE MODELS 13
σ σ
r
r q )(= (11.57)
in which the following holds:
[ ]∞∈≤≤ ,10 0r r d (11.58)
Note that with the new definition of the damage parameter given in (11.56), we canrestructure the equation in (11.46) as follows:
r
r qr d
)(1)( −= (11.59)
⇒
∂
∂−
=
−
∂
∂=
∂
∂= r
r
r
r qr q
r r
r q
r r
r
r d d
2
)()(
)(1
)(r
r
r r qd
d
−=
2
)()( H (11.60)
where we have defined a new parameter)(
)(
r
r qr d
∂
∂=H , which is the hardening/softening
parameter.
1.2.2.4 The Hardening/Softening Law
The expressionr
r q
∂
∂ )( defines the hardening/softening parameter, thus:
[ ) [ ] ; ,; )1(),0(; )( 0000 E
r qar qd d r r r r q Y d σ
==∈=∞=∈= H (11.61)
where d H is the continuum hardening/softening parameter and which is characterized by:
0)(Softening withDamage
0)(DamagePerfect
0)(Hardening withDamage
⇒
r
r
r
d
d
d
H
H
H
(11.62)
Here, we will consider the relationship between q and r to be linear or exponential.
The Linear Hardening/Softening Law
Now, by assuming that q varies linearly with r , we have:
>>/−−−
≤
=−=
0
0
111
0
1
00 r r
r r
r
qd
r
r d
r
r H
(11.64)
0>d H
)(r q
0r
0
8/16/2019 Notes on Continuum Damage Models
14/23
NOTES ON CONTINUUM DAMAGE MODELS 14
Figure 11.7: The linear hardening/softening law.
The Exponential Hardening/Softening Law
The exponential law is described by Figure 11.8. Then we can express )(r q as follows:
( ) 0)( 01
0 >−−=
−
∞∞ Awithr qqr q r
r A
exp (11.65)
in addition to this we have:
( )
−
∞ −=∂
∂0
1
0
0)( r r
A
r
r q A
r
r qexp (11.66)
Figure 11.8: The exponential hardening/softening law.
Table 11.1: Summary of the Isotropic Damage Model in a small deformation regimedescribed in the strain space.
ISOTROPIC D AMAGE MODEL IN A SMALL DEFORMATION R EGIME
Helmholtz free energy ( ) [ ] ( )εεε :: eeer d r 2
1with)(1, =−= Y Y Y (11.67)
Damage parameter [ ] [ ]1,0;,,;1)( 0 ∈∞≠∈−= d aar qr
qr d (11.68)
The constitutive equations ( ) ( ) εσ ε
σ :ed d 11 −=−=
∂
∂= Y
(11.69)
Evolution law z =r
[ )
σ==
∞∈
= E
r r
r r
Y
t 00
0 ,
(11.70)
Damage criterion ( ) r r r e −=−= εεε ε :: t ,G (11.71) Hardening Law ( )0)(;)( ≤′== r qr r q d d HH (11.72)
)(r q
0r
0r q ∞ ∞q
8/16/2019 Notes on Continuum Damage Models
15/23
NOTES ON CONTINUUM DAMAGE MODELS 15
Loading/unloading condition 0;0;0 =≥< GG z z (11.73)
Consistency condition 0=Gz (11.74)
1.2.3
The Elastic-Damage Tangent Stiffness Tensor
Next, we will obtain the elastic-damage tangent stiffness tensor, which gives us anadvantage, from a computational point of view, when we are dealing with the incremental-iterative solution procedures and as a result of this, convergence is improved considerably.
The relationship between σ and ε give us this tensor denoted by tan_d , i.e. εσ :tan_d = .
Now, by considering the equation in (11.25), ( ) εσ 1 :ed −= , we can obtain the rate ofchange of the stress as follows:
( ) ( ) ( )
( ) d d
d d d d d d
d
e
eee
⊗−−=
−−=−−=∂
∂+
∂
∂=
σ ε
σ εεεσ
ε
ε
σ εσ
1
1 1 ,
:
::::
(11.75)
in which there is the following:
a) A process with elastic loading or unloading
00 =⇒= d z , thus the equation in (11.75) becomes ( ) ( ) εεσ :ed d 1, −= , with which theelastic-damage tangent stiffness tensor coincides with the elastic-damage secant stiffnesstensor when we are dealing with elastic loading:
( ) )1(1 d whered eetan_d sec_d −==−== x x (11.76)
b) A process with damage loading
r r =⇒= εε t t , and the rate of change of the damage parameter )(r d d = becomes:
ε
ε
t
t
∂
∂=
∂
∂=
∂
∂
∂
∂=
d r
r
d
t
r
r
d d (11.77)
where the rate of change ofε
t can be evaluated as follows:
( ) ( )
εσ εεεε
εε
εεεεεεεε
εε
εε
:::::
::
::::::::
t t
t t
1112
1 2
1
===
+= → =−
ee
e
eeee
(11.78)
Now, by substituting (11.78) into the equation in (11.77) we obtain:
εσ
εε
:t t
1
∂
∂=
d d (11.79)
Then, taking into account the equations (11.79) and (11.75), we can find the relationshipbetween the rates of stress and strain change:
( ) ( )
( ) ( ) εσ σ
εσ σ εσ εσ
εε
εε
:
:::
⊗
∂
∂−−=
⊗∂
∂−−=⊗−−=
t t
t t
1 1
1 11
d d
d d d d
e
ee
(11.80)
which thus defines the elastic-damage tangent stiffness tensor :
8/16/2019 Notes on Continuum Damage Models
16/23
NOTES ON CONTINUUM DAMAGE MODELS 16
( ) ( )
⊗
∂
∂−−= σ σ
εε t t
1 1
d d etan_d (11.81)
and by considering that in a loading process r =εt holds we then find:
32
)(1)(11
r
r r q
r r
r r q
r r
d d d d HH −=
−=∂∂=
∂∂
εε t t
(11.82)
where we have taken into account the equation in (11.61), 2
)(
r
r r q
r
d d H−=
∂
∂.
Then, by substituting the equation in (11.82) into that in (11.81) we can obtain tan_d interms of q and r :
⊗
−−−=
σ σ
εε ee
d etan_d
r
r r qd ::
)( )1(
3
H (11.83)
Now, the general equation for the elastic-damage stiffness tensor tan_d (symmetricfourth-order tensor) is given by:
( ) 00
=→=
=
⊗−=
K
K
r
loading unloading
0)(d withelastic
eee
e
tan_d
:: εεx
The elastic-damage stiffnesstensor for isotropic damagemodel
(11.84)
where,3
)(
r
r r q d HK
−= and ( )d −= 1x .
1.2.4
The Energy Norms
Next, we will define some energy norms, which together with the damage criteria, play animportant role in defining the yield damage surface.
In order to adequately represent the materials different norms will need to be defined so asto describe how these materials really behave. For example, in a simple model for concrete,if we only want to simulate the process of failure caused by tension, the tension-only damagemodel is used which means that it cannot capture the other type of failure caused bycompression. Next, we will define some models used in the isotropic damage process.
1.2.4.1
The Symmetrical Damage Model (Tension-Compression) –Model I
This type of model shows when the material behave the same both with tension or andcompression. The energy norm of this model is then represented by:
εσ σ σ σ
::: )1(1
d e I −== − t (11.85)
We can also define the energy norm of the strain tensor (also known as the equivalentstrain), proposed by Simo&Ju(1987), (see equation (11.31)):
ee I Y 2== εεε
:: t (11.86)
8/16/2019 Notes on Continuum Damage Models
17/23
NOTES ON CONTINUUM DAMAGE MODELS 17
To better illustrate this model, let us consider the state of plane stress ( 03 =σ i ). In this
case, the yield surface is represented by an ellipse, (see Figure 11.9), where 0>σY is thestress limit for tension and compression and the damage surface evolves symmetrically.
Figure 11.9: Damage surface in 2D and the uniaxial stress-strain curve for model I.
1.2.4.2 The Tension-Only Damage Model – Model II
The tension-only damage model does not take into account failure by compression, i.e. thematerial can only fail by tension and here we can define the following stress field:
2
σ σ
σ σ
+=〉〈=+ (11.87)
where 2
•+•
=〈•〉
def
is the Macaulay bracket whose graphical representation can beappreciated in Figure 11.10.
Figure 11.10: Ramp function.
Now, by means of spectral representation, we can represent the stress tensor in terms ofeigenvalues (principal stresses) and eigenvectors as follows:
)()(3
1
ˆˆ aa
a
a n n ⊗σ= ∑=
σ (11.88)
thus:
)()(3
1
ˆˆ aa
a
a n n ⊗σ= ∑=
+σ (11.89)
Note, the relationship between the real and effective stress remains valid, i.e.: ++ −= σ σ )1( d (11.90)
a) Norm in the principal stress space-2D.
ε
σ
0r =σ t
Y σ
Y σ−
Y σ
Y σ
Y σ−
1σ
2σ
E
b) Stress-strain curve
1
Elasticregion
〉〈 x
x
≥
<=〉〈
0
0 0
xif x
xif x
8/16/2019 Notes on Continuum Damage Models
18/23
NOTES ON CONTINUUM DAMAGE MODELS 18
Then, the norm for the isotropic damage model defined previously becomes:
εσ εεε
::: === ee Y 2t (11.91)
Next, in the tension-only damage model +← σ σ , it follows that:
σ σ σ σ σ σ εσ ε
::::::: 112
1
)1(1
)1(1 −+−+−++
−=
−=== eee II
d d
t (11.92)
Then, if we consider the equation in (11.31), we can conclude that:
σ σ σ
::
1−+= e II t (11.93)
Finally, in Figure 11.11 we can visualize the damage surface for two-dimensional cases(2D).
1.2.4.3 The Non-Symmetrical Damage Model – Model III
The non-symmetrical damage model is useful to simulate materials, such as concrete, whose tension domain differs with respect to compression. This model uses the followingnorm:
σ σ σ
::
1
1 −
−+= e III
n
qqt (11.94)
where the parameter q is the weight factor dependant on the stress state σ which is givenby:
∑
∑
=
=
σ
〉σ〈
=3
1
3
1
i
i
i
i
q (11.95)
The parameter n is defined by means of the ratio of the compression elastic limit cY σ to
the tension elastic limit t Y σ , i.e.:
t Y
cY n
σ
σ= (11.96)
In the case of concrete n is approximately equal to 10≈n .
ε
σ
Elastic region
Y σ
1σ
2σ
E
a) Norm in the principal stress space-2D. b) Stress-strain curve.
1
0r =σ t Y σ
Y σ
8/16/2019 Notes on Continuum Damage Models
19/23
NOTES ON CONTINUUM DAMAGE MODELS 19
Figure 11.11: Damage surface in 2D and the uniaxial stress-strain curve for model II.
Figure 11.12: Damage surface in 2D and the uniaxial stress-strain curve for model III.
1.3 The Generalized Isotropic Damage Model
Note that the elasticity tensor e can be written in terms of the following sets ofmechanical parameters ),( ml , ),( ν E , ),( Gκ :
part isochoric partvolumetric
3
1 2
)1(
)21)(1( 2
⊗−+⊗κ=
ν+
ν+⊗
ν−ν+
ν=+⊗= 11I11I11I11 mml
E E e
(11.97)
where )( E =Young’s modulus, )(ν =Poisson’s ratio, ),( ml =Lamé constants, )(κ =bulkmodulus, and m=G is the shear modulus.
In the isotropic damage model the elastic-damage secant stiffness tensor can berepresented as follows:
I11I11 )1(
)21)(1(
)1(
)1(
)21)(1(
)1()1(
ν+
ν+⊗
ν−ν+
ν=
ν+
−ν+⊗
ν−ν+
−ν=−=
sec_d sec_d esec_d E E E d E d d
Note that, in this model the damage variable affects only one of the mechanical parameters,namely, the Young’s modulus. We can also verify that the same damage parameter equallyaffects both the spherical and deviatoric part:
⊗−−+⊗κ−=−= 3
1 2)1()1()1( 11I11 md d d esec_d (11.98)
Another model described by Carol et al. (1998) generalizes the isotropic damage model by
considering independent degradation of the spherical and deviatoric parts and because ofthis the model requires two independent damage variables.
ε
σ
Elastic
region
t Y σ
t Y σ
t Y σ
1σ
2σ
E
cY σ−
cY σ−
t Y cY n σ−=σ−
a) Norm in the principal stress space-2D. b) Stress-strain curve.
0r =σ t
8/16/2019 Notes on Continuum Damage Models
20/23
NOTES ON CONTINUUM DAMAGE MODELS 20
Now, the elasticity tensor components can be expressed by means of their spherical anddeviatoric parts as follows:
( )
−++κ= klij jk il jlik klij
eijkl d d d d d d md d
3
1
2
1 2C (11.99)
Then, with klijV ijkl d d
31=P and ( ) V ijkl jk il jlik Dijkl PP −+= d d d d
21 , the above equation becomes:
Dijkl
V ijkl
eijkl PPC 23 m+κ=
DV e 23 m+κ= (11.100)
Let us now consider that the material parameters κ and m can be degraded by means of
the variables V d and Dd , respectively, and according to the following equations:
00 )1(;)1( mm DV d d −=κ−=κ (11.101)
with which the elastic-damage secant stiffness tensor becomes:
De
ijkl DV e
ijklV D
ijkl DV
ijklV sec_d
ijkl d d d d _ _
00 )1()1()1(2)1(3 CCPPC −+−=−+κ−= m (11.102)
where we have introduced:
( )
−+=
=
κ=
κ=
klij jk il jlik
Dijkl
De
ijkl
klij
V ijkl
V e
ijkl
d d d d d d m
m
d d 3
1
2
12
2
;3
0
0 _
0
0 _ PC
PC (11.103)
1.3.1
The Strain Energy Function
Now, if we consider (11.100), the elastic strain energy function can be rewritten as follows:
( ) ( ) ( )devevole
DV DV ee
_ _
22
13
2
1 23
2
1
2
1
Y Y
mmY
+=
+κ=+κ== εεεεεεεε :::::::: P (11.104)
where we have introduced:
( )
( )devevolee
De Dvole
V eV vole
_ _
_ _
_ _
)(
2
1 2
2
1
2
13
2
1
Y Y Y
mY
Y
+=⇒
==
=κ=ε
εεεε
εεεε
::::
::::
(11.105)
after which it becomes:
[ ]
),(),()1()1(
2
1)1(
2
1)1(
)1()1(2
1
2
1),,(
_ _
_ _
_ _
DdevV vol
dev
deve D
vol
voleV
De DV eV
De DV eV sec_d DV
d d d d
d d
d d d d
εε
εεεε
εεεεε
Y Y Y Y
Y
Y Y
+=−+−=
−+−=
−+−==
==
::::
::::
(11.106)
Additionally, the following holds:
8/16/2019 Notes on Continuum Damage Models
21/23
NOTES ON CONTINUUM DAMAGE MODELS 21
),(),()1()1(
2
1)1(
2
1)1(
2
1)1(
2
1)1(),,(
_ _
_ _
_ _
DdevdevV sphvol
dev
deve D
vol
voleV
dev Dedev DsphV esphV
De DV eV DV
d d d d
d d
d d d d
εε
εεεε
εεεεε
Y Y Y Y
Y
Y Y
+=−+−=
−+−=
−+−=
==
::::
::::
(11.107)
1.3.2
Spherical and Deviatoric Effective Stress
Note that the following equations hold:
dev DsphV De DV eV sec_d d d d d σ σ εεεσ )1()1()1()1( _ _ −+−=−+−== ::: (11.108)
where sphσ , devσ are the spherical and deviatoric effective stresses, respectively and wherethe following is valid:
devsph
dev Ddev
sphV sph
d d σ σ σ
σ σ
σ σ +=⇒
−=−=
)1()1( (11.109)
It is noteworthy that the following equations hold:
( ) ( )dev De DsphV eV
devsph De DdevsphV eV
De DV eV
d d
d d
d d
εε
εεεε
εεσ
::
::
::
_ _
_ _
_ _
)1()1(
)1()1(
)1()1(
−+−=
+−++−=
−+−=
(11.110)
Then, the relationship between stress and strain in rate is given by:
( )
==⇒
+=
+=+
=
devd tandev
sphd tansph
devd tansphd tan
devsphd tandevsph
d tan
εσ
εσ
εε
εεσ σ
εσ
:
:
::
:
:
_
_
_ _
_
_
(11.111)
where d tan _ is the elastic-damage tangent stiffness tensor.
1.3.3
Thermodynamic Considerations
In a small deformation regime εD ≈ holds and in isothermal processes 0=T is satisfied,so, it then follows that the expression for internal energy dissipation given in (11.13)becomes:
0≥−= Y εσ :int D (11.112)
Then, by evaluating the rate of change of the strain energy function given in (11.106), De DV eV d d _ _ )1()1( Y Y Y −+−= , we can obtain:
D DeV V e D DeV V e
D De D DeV V eV V e
d d d d
d d d d
_ _ _ _
_ _ _ _
)1()1(
)1()1(
Y Y Y Y
Y Y Y Y Y
−−−+−=
−−+−−= (11.113)
and by using the stress equation given in (11.108) we have:
[ ]εεεε
εεεεσ
::::
::::
De DV eV
De DV eV
d d
d d
_ _
_ _
)1()1(
)1()1(
−+−=
−+−=
(11.114)
8/16/2019 Notes on Continuum Damage Models
22/23
NOTES ON CONTINUUM DAMAGE MODELS 22
Note that εε :: V eV e _ _ =Y and εε :: De De _ _ =Y , thus:
De DV eV
De DV eV
d d
d d _ _
_ _
)1()1(
)1()1(
Y Y
−+−=
−+−= εεεεεσ ::::: (11.115)
Then, together the equations (11.115), (11.113), and the internal energy dissipation given in(11.112), yields:
0
0)1()1()1()1(
0
_ _
_ _ _ _ _ _
≥+=
≥++−−−−−+−=
≥−=
D DeV V e
D DeV V e D DeV V e De DV eV
int
d d
d d d d d d
Y Y
Y Y Y Y Y Y
Y εσ :D
(11.116)
Since (11.116) must be satisfied for any admissible thermodynamic process, it follows that:
0;0 ≥≥ DV d d (11.117)
where we have taken into account that 0
_ ≥V e
Y and 0
_ ≥ De
Y .
1.3.4
The Elastic-Damage Tangent Stiffness Tensor
Initially we adopt the following norms:
sphV esphsphsphsphV eV εεεσ εσ
ε ::::
_ _ 2 ==== Y t (11.118)
dev Dedevdevdevdev De Dεεεσ εσ
ε ::::
_ _ 2 ==== Y t (11.119)
where the following holds:
( ) ( ) ( )εσ εσ εεεε εε
ε ::::::
sphV
sphsphV
sphV esphsphV esph
V
t t
t 111 _ _ === (11.120)
( )εσ ε
ε :dev
D
D
t
t 1
= (11.121)
Next, we obtain the rate of change of the Cauchy stress tensor:
devsph
D
D
devV
V
sph
D
D
V
V
devsph D
D
V
V
DV
d d
d d
d d
d d
d d
d d
d d
σ σ
σ ε
ε
σ σ ε
ε
σ
σ σ εε
ε
σ σ σ ε
ε
σ εσ
+=
∂
∂+
∂
∂+
∂
∂+
∂
∂=
∂
∂+
∂
∂++
∂
∂=
∂
∂+
∂
∂+
∂
∂=
::
:: )(),,(
(11.122)
where the following holds, (see equation (11.109)):
dev
D
sph
V d d σ
σ σ
σ −=
∂
∂−=
∂
∂; (11.123)
and
D
D
D D
D
D D
D
D DV
V
V V
V
V V
V
V V d r
r
d
t
r
r
d d
d r
r
d
t
r
r
d d
ε
ε
ε
ε
t
t
t
t
∂
∂=
∂
∂=
∂
∂
∂
∂=
∂
∂=
∂
∂=
∂
∂
∂
∂= ; (11.124)
Then, we can express the rates of change sphσ and devσ as follows:
8/16/2019 Notes on Continuum Damage Models
23/23
NOTES ON CONTINUUM DAMAGE MODELS 23
( )
sphsphsph
V V
V V eV
sphsph
V V
V V eV
sph
V V
V sphV eV
V
V
V sphV eV V
V
sphsph
d d
d d
d d
d d d
d
εσ σ
εσ σ
εσ σ ε
σ εσ
ε
ε
σ σ
εε
εε
εε
ε
ε
:
:
::
::
⊗
∂
∂−−=
⊗∂∂−−=
∂
∂−−=
∂
∂−−=
∂
∂+
∂
∂=
t t
t t
t t
t
t
1)1(
1)1(
1)1(
)1(
_
_
_
_
(11.125)
and
( )
devdevdev
D D
D De D
devdev
D D
D De D
dev
D D
Ddev De D
D
D
Ddev De D D
D
devdev
d d
d d
d d
d d d
d
εσ σ
εσ σ
εσ σ ε
σ εσ
ε
ε
σ σ
εε
εε
εε
ε
ε
:
:
::
::
⊗
∂
∂−−=
⊗
∂
∂−−=
∂
∂−−=
∂
∂−−=
∂
∂+
∂
∂=
t t
t t
t t
t
t
1)1(
1)1(
1)1(
)1(
_
_
_
_
(11.126)
with which we can define the following equation:
εσ σ σ σ σ
εεεε
:
⊗
∂
∂−⊗
∂
∂−−+−= sphsph
V V
V devdev
D D
DV eV De D d d d d
t t t t
11)1()1( _ _
(11.127)
and by comparing the above with (11.111), we can conclude that:
sphsph
V V
V devdev
D D
DV eV De Dd tan d d d d σ σ σ σ
εεεε
⊗∂
∂−⊗
∂
∂−−+−=
t t t t
11)1()1(
_ _ _
(11.128)
Top Related