modeling plasticity/damage using internal state variables douglas j. bammann center for materials...
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MODELING PLASTICITY/DAMAGE USING INTERNAL STATE VARIABLES
DOUGLAS J. BAMMANN
CENTER FOR MATERIALS AND ENGINEERING SCIENCES
SANDIA NATIONAL LABS/CA
Phenomenological Modeling Using State Variables
• Kinematics• Thermodynamics• Physical basis (micromechanics)• Parameter determination
– Microstructural (physical) constants– fixed state experiments– evolving state experiments
• analytical solutions - simple tests - nonlinear least squares• numerical integration - optimization
• Numerical implementation• Validation
Coleman-Gurtin applied to plasticity (Kratochvil-Dillon, Werne-Kelly, Teodosiu-Sidoroff, Perzyna,Anand ...
• Assume multiplicative decomposition of deformation gradient into elastic and plastic parts (unloading elastically defines stress free or intermediate configuration - this assumption couples kinematics with constitutive model)
• Assume that the free energy, defined with respect to the intermediate configuration, depends upon the elastic strain and deformation type (or defects) state variables
• Determine restrictions of symmetry and invariance• Develop evolution equations for the state variables (generally motivated from the
microstructural effects they represent)• Conjugate thermodynamic driving forces are defined as the derivative of the the free
energy w.r.t. the state variables and are the stress like variables in the flow rule– Chosen as a result of the appearance of the product of these two terms in the
dissipation inequality• Determine material parameters
– Microstructural constants– Fixed state experiments– Evolving state experiments
The plasticity variables (those related to deviatoric plastic flow from dislocations) are motivated as follows
• Evolution of variables cast in hardening minus recovery format– hardening is based upon a dislocation storage mechanism and has temperature
dependence of shear modulus– dynamic recovery is motivated by cross slip and operates on same time scale as
plastic flow and therefore its rate dependence is determined by the kinetics of slip– thermal recovery is motivated by diffusional climb and operates on a completely
different time scale and introduces a strong rate dependence• Scalar variable represents statistically stored dislocations and gives rise to most of the
hardening– Dislocations are stored inversely proportional to mean free path
• Tensor variable describes geometrically necessary dislocations– Tensor variable hardens in direction of plastic flow and recovers in the direction of the
current value of the tensor variable• Under constant temperature, strain rate and loading direction both variables reach steady
state when hardening is balanced by recovery
Configurations
Kinematics of continuum mechanics begins by introducing a map from material space (manifold) to physical space (R3). Thus the map is from a reference configuration in material space to the current configuration in physical space. The reference configuration is not to be confused with the Lagrangian configuration which is simply a previously occupied configuration in physical space.
X
BR
Bt
Bt0
x
x0
To simplify matters we will consider the reference configuration to be the initial or Lagrangian configuration as we develop our 1D continuum
KINEMATICS
F =FeFp
Following Bilby (1956), Kroner (1960) decompose the deformation gradient into elastic and plastic parts
Figure 5.1: The deformation gradient is multiplicatively decomposed intoelastic and plastic parts. The elastic part is then decomposed by the polardecomposition into an elastic stretch and an elastic rotation.
Strain can be defined with respect to any configuration based upon the Lagrange change in square length per unit square length
E =12(F TF −I ) Ep =
12(Fp
TFp −I )
Ee =E −Ep =12 Fp
T(FeT Fe−I )Fp
E e =12(Fe
TFe−I )
Ep =12(I −Fp
−TFp−1)
E = Ee+ Ep =Fp−TEFp
−1
Define a Lagrange total and plastic strain w.r.t. reference configuration
Then the elastic strain in this configuration is defined as
These can be mapped forward to the intermediate configuration
The velocity gradient is defined as usual and is naturally composed of elastic, plastic, damage and thermal parts. These can be mapped to any other configuration.
L = ˙ F F−1 = ˙ F eFe−1 + Fe
F pFp−1Fe
−1 + c˙ θ 1
= Le + Lp +Lθ
L p =˙ FpFp−1
Le =De+WeL p = Dp+ Wp
˙ E e = D − Dp − Ee L p− LpT Ee
Velocity gradient
Notice that the plastic part of the velocity gradient is naturally defined with respect to the natural configuration as
And the velocity gradient w.r.t any configuration can be split into elastic and plastic parts
And then from algebra
• The decomposition of the deformation gradient into elastic and plastic parts
• Results in additional degrees of freedom
• Requires specifying expressions for plastic stretching (strain rate) and plastic spin
•These are easy to specify in crystal plasticity, but more difficult to motivate in phenomenological models
•Most models ignore plastic spin
•This can be shown to be critical in attempting to predict evolving anisotropy (from texture)
INTERNAL STATE VARIABLE THEORY
• What are state variables?– Variable whose current value represents some observable state of the
material– Can be initialized (can be measured without knowing anything about the
past)• Field theory
– How are kinematics related to the state variables?– Define thermodynamics with respect to specifically defined kinematic
configuration– How does kinematics or the geometric structure of the state variables affect
the degrees of motion of the continuum
The free energy of the crystal depends upon the dislocations present while the plastic deformation is governed by the transport of dislocations
0
0
p
+ +
p b/2
p b
Thermodynamics - Coleman and Gurtin (1967) (Formulation will be small strain, 1D to simplify concepts)
ρ e = σ˙ ε + h
h = −∂q∂x
+ ρr
Rate of change of internal energy is the sum of the rate at which work is being done on the body, , balanced by rate at which heat is supplied to the body, h
Kinematics:
ε = εe + εp
˙ ε = ˙ ε e + ˙ ε p
Thermodynamics - 1st Law
2nd Law
−ρ ˙ ψ + η˙ θ ( ) +σ˙ ε −
qθ
∂θ∂x
≥ 0
Assume that the free energy depends upon the elastic strain, the temperature, and two state variables, the elastic strain resulting from the distribution of geometrically necessary dislocations leading to kinematic hardening, gnand the elastic strain associated with the density of statistically stored dislocations giving rise to isotropic hardening, ss
ψ = ˆ ψ εe, θ, εgn, εss( )
σ = ρ
∂ ˆ ψ ∂εe
, η =∂ ˆ ψ ∂θ
α = ρ
∂ˆ ψ ∂εα
, κ = ρ∂ ˆ ψ ∂εκ
Where, and are conjugate thermodynamic stresses
It follows from the 2nd law
˙ θ =
1ρC
σ˙ ε p −α˙ ε gn −κ˙ ε ss( )
Then neglecting elastic heating effects and neglecting conduction for high strain rate applications, the energybalance reduces to
And the dissipation inequality reduces to
σ˙ ε p − α˙ ε gn −κ˙ ε ss +
qθ
∂θ∂x
≥ 0
The kinematics of plasticity introduced degrees of freedom requiring more”constitutive” equations. Similarly, the extra degrees of freedom also requires temporal evolution equations to complete the system.
This is where the physics of the smaller length scales enters!
Analogous to elasticity, assume a free energy of the form
ρˆ ψ = Eεe2 +
12
kαμεα2 +
12
kκμεκ2
σ = Eεe
α = kαμεα
κ = kκμεκ
Then,
Now let’s take a very brief look at dislocation models of plasticity to determine the necessary forms
To complete the system we need
1. An expression for the plastic strain rate
2. Evolution equations for the state variables
Statistically Stored Dislocations
Zero net burgers vector
b
-b
b
b
Geometrically Necessary Dislocations
Net burgers vector is 2 b
void
overlap
Polycrystal deforms, grains rotate, lose continuity. Geometrically necessary dislocations permit reassembly of polycrystal
Geometrically or Kinematically Necessary Dislocations and/or Boundaries
Thermally Activated
Athermal Diffusion Controlled
Drag Mechanisms
200 400 600 800
TEMPERATURE K
SH
EA
R S
TR
ESS
ε
ε
ε
Some Important Deformation Mechanisms - Klahn, Mukherjee, Dorn
Frost-Ashby deformation mechanism map
Example - Thermally Activated Motion (reaction-rate theory, Eyring (1936))
•Assume the number of times per second that a dislocation segment overcomes an energy barrier under the action of an applied stress can be written as a thermally activated process.
ν+=ν0 exp −
Q0−σbΔA
kθ⎡ ⎣ ⎢
⎤ ⎦ ⎥
• • Q = height of energy barrier• bA = work done by applied stress in overcoming energy barrier or effectively lowering barrier• k = Boltzman constant• = temperature• A = area swept in glide plane when dislocation segment moves to top of barrier• b = Burgers vector
Frequency of jumps backward over the barrier
ν−=ν0 exp −
Q0 +σbΔA
kθ⎡ ⎣ ⎢
⎤ ⎦ ⎥
Net forward reaction rate
ν =ν+−ν−=ν0 exp −
Q0kθ
⎡ ⎣ ⎢
⎤ ⎦ ⎥ sinh
σbΔAkθ
⎡ ⎣ ⎢
⎤ ⎦ ⎥
If l is the length of the dislocation freed after each successful jump, and V is the volume of the crystal, the strain after each jump is
εp =lb2
V
And the strain rate is then
ε p = Nlb2ν
Where N is the number of dislocation segments (activation sites) per unit volume
Force - Distance Diagram(constant rate and temperature)
lb lb - force of applied stressLb - force of average value of spatially fluctuating long range internal stress from other dislocationslb - short range obstacle strength
Force
Distance along glide path
lblb
lb
lb
A/l
Ono (1968) and Kocks et al. (1975) showed that a large range of obstacle shapes could be described by the following
ΔG = ΔF 1 −σκ
⎛ ⎝
⎞ ⎠
p⎡
⎣ ⎢
⎤
⎦ ⎥
q
Where F is the activation energy characterizing the strength of a single obstacle. It also determines the rate sensitivity of the internal strength, .
˙ γ 1 = ∞ if σs ≥ αμ
˙ γ 1 = 0 if σs ≤ αμ
˙ γ 2 =αb
σsμ
⎛ ⎝ ⎜
⎞ ⎠ ⎟
2
βν exp −ΔFkθ
1 −σsκ
⎛ ⎝
⎞ ⎠
⎡ ⎣ ⎢
⎤ ⎦ ⎥
˙ γ 3 = ˙ γ pσsμ
⎛ ⎝ ⎜
⎞ ⎠ ⎟
2
exp −ΔFp
kθ1 −
σsκ p
⎛
⎝ ⎜
⎞
⎠ ⎟
3 4⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
4 3⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
˙ γ 4 =
A2Dv 1 +10ac
b2σsμ
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2
DcDv
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ μb
kθσsμ
⎛ ⎝ ⎜
⎞ ⎠ ⎟
n
Elastic collapse - stress exceeds ideal shear strength
Thermally activated motion
Peierls force - lattice resistance
Power Law Creep
Deformation Mechanisms - Frost and Ashby
˙ γ 5 = AHDDvμbkθ
σsμ
⎛ ⎝ ⎜
⎞ ⎠ ⎟
˙ γ 6 = A2
Dv 1 +10ac
b2σsμ
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2
DcDv
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ μb
kθsinh α
σsμ
⎡
⎣ ⎢
⎤
⎦ ⎥
⎧ ⎨ ⎩
⎫ ⎬ ⎭
n
˙ γ 7 =42σsΩ
kθd2 Dv 1 +πδd
DbDv
⎡
⎣ ⎢
⎤
⎦ ⎥
˙ γ plas = Least˙ γ 2, ˙ γ 3{ }
˙ γ net = ˙ γ 1 + Max ˙ γ plas, ˙ γ 4 or ˙ γ 6{ } + Max ˙ γ 5, ˙ γ 7{ }
Harper-Dorn Creep
Power Law Breakdown
Diffusional Flow
Net plastic Flow
˙ γ drag =ρb2μ Bp
Be Bp +θ 300σsμ
⎛ ⎝ ⎜
⎞ ⎠ ⎟
=5 ×106
0. 1 +θ 300σsμ
⎛ ⎝ ⎜
⎞ ⎠ ⎟
Viscous Drag
Phonon and Electron
dρssdεp
= c11λ
− c2ρss
= c1 ρss − c2ρss −c3ρsssinh c4 ρss
˙ ε p
Motivate evolution equations from Kocks-Mecking where dislocation density evolves as a dislocation storage minus recovery event. In an increment of strain dislocations are stored inversely proportional to the mean free path , which in a Taylor lattice is inversely proportional to the square root of dislocation Density. Dislocations are annihilated or “recover” due to cross slip or climb in a manner proportional to the dislocation density
Evolution of dislocation density
A scalar measure of the stored elastic strain in such a lattice is
εss = b ρss
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
FLOW STRESS
HOMOLOGOUS TEMPERATURE
Model Development
Rather than introducing several flow rules, we propose a temperature dependence for the initial value of the internal strength that emulates all of the mechanisms at a very low strain rate
Y θ( ) =C3 1 + tanh C19 C20 − θ( )[ ]{ }
2 C21 + exp −C24θ
⎡ ⎣ ⎢
⎤ ⎦ ⎥
⎛ ⎝
⎞ ⎠
Introduce a flow rule of the form
€
˙ ε p = f θ( ) sinhσ −α
κ + Y θ( )
⎡
⎣ ⎢
⎤
⎦ ⎥
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪
n
Linear Elasticity
˙ σ = E ˙ ε − ˙ ε p( ) − εe
∂E∂θ
˙ θ
From dislocation mechanics, (statistically stored dislocations)
˙ κ = Hκμ θ( )˙ ε p − Rdκ θ( )˙ ε pκ − Rsκ θ( )κ sinh Qsκ[ ]
Heat conduction reduces to
˙ θ =1
ρCσ˙ ε p −κ˙ ε ss −α˙ ε gn( )
=1
ρCσ˙ ε p −
κ˙ κ 2μ
−α˙ α 2μ
⎛ ⎝ ⎜
⎞ ⎠ ⎟
As dislocations are stored, the dissipation is reduced until recovery becomes dominant. Then the heat dissipated approaches the assumption that 90-95% of plastic work is dissipated as heat.
dαdt
= Hαμ θ( )˙ ε pσ −ασ −α
− Rdα θ( )˙ ε p + Rsα θ( )[ ]αα
The tensor variable is motivated by the continuum theory of dislocations (recall the nonlocal workshop where we introduced the elastic curvature as a state variable resulting in a natural gradient and natural bacck stress). Here we simplify to a local form and choose
ddt
=H ˙ p−−
−Rd ˙ p
€
˙ ε p = f θ( )sinhσ −α
Y θ( ) + κ
⎧ ⎨ ⎩
⎫ ⎬ ⎭
nσ − α
σ − α
The tensor variable is motivated by the continuum theory of dislocations (recall the nonlocal workshop where we introduced the elastic curvature as a state variable resulting in a natural gradient and natural back stress). Here we simplify to a local form and choose
For a wide range of temperatures and strain rates, plastic flow is a thermally activated strain rate. By choosing an appropriate form for the activation energy we get a flow rule of the form
Two parameter fit of 304L SS data at 800C for both large and small strain offset definitions of yield. For two parameters, the model reduces to rate independent bilinear hardening.
EHE +H
DataLarge offsetSmall offset
€
Y θ( ) + κ( ) sinh−1 ˙ ε
f θ( )
⎛
⎝ ⎜ ⎞
⎠ ⎟
1n
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
ST
RE
SS
MP
a
STRAIN
Recovery included for the same compression curve. In this case the model accurately captures both the hardening and recovery through the isotropic hardening variable .
data
€
Y θ( ) + κ( ) sinh−1 ˙ ε
f θ( )
⎛
⎝ ⎜ ⎞
⎠ ⎟
1n
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
HRd
STRAIN
ST
RE
SS
MP
a
HRd
HRd
The small strain fit can be improved by including the short transient a which saturates at small strains as a function of its hardening and recovery parameters
€
Y θ( ) +HdκRdκ
⎛ ⎝ ⎜
⎞ ⎠ ⎟ sinh−1 ˙ ε
f θ( )
⎛
⎝ ⎜ ⎞
⎠ ⎟
1n
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟+
HdαRdα
10-1[1/s]
10 [1/s]
model10 [1/s]
model10-1[1/s]
Six parameter fit of 304L SS compression data with only the long transient k but including the effects of rate dependence of yield through the parameters V and f . The strain dependent rate effect is captured by the static recovery parameter R
sk .
Five parameter fit of 304l SS compression curve including the short transient a . This fit will more accurately capture material response during changes in load path direction
Model prediction for 304L stainless steel tension tests or 304L stainless steel is depicted in Figure 1.
0
200
400
600
800
1000
1200
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
STRESS [MPa]
STRAIN
= 1−5.
= 10-3.
ε = 10−3.
ε = 10−1.
ε = 10−5.
Model prediction of compression tests and compression reload demonstrating temperature and history effects. The 800C test was quenched after being strained to 23% and reloaded at room temperature. The temperature history effect is demonstrated by the reload curve being much softer than the 20C curve.
0
200
400
600
800
1000
1200
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
STRESS (MPa)
STRAIN
20C
800C
20C reload
Kinematic vs. Isotropic Hardening
1
2
3
•If all hardening occurs uniformly by statistically stored dislocations, (and the texture is random), the yield surface would grow isotropically “the same in every direction, independent of the direction of loading”. The radius of the yield surface, is given by , the internal strength of the material.
•This type of loading is illustrated in the figures. The material deforms elastically and the stress increases linearly until the initial yield surface is reached and the material hardens and the yield surface grows until unloading begins at point 1. Upon reversal of load the material deforms elastically until point 3 is reached.
•If geometrically necessary dislocations form pileups at grain boundaries (small effect) or at particles (larger effect), the material exhibits an apparent softening upon load reversal
•To model this, the yield surface is allowed to translate to the same stress point 1 (red surface). Now upon load reversal, plastic flow begins at point 2. Real material would begin a combination of these two exaggerated figures. This is a short transient and represents the center of the yield surface.
•In some cases, we used to use as long transient to model texture effects. But now we introduce a structure tensor for this effect.
1
2
3
The proportion of kinematic and isotropic hardening can be determined from reverse loading tests
f
r
Y
•To satisfy yield at both unloading and reverse yielding (assuming state variables don’t change during elastic unloading)
f − α −κ − Y = 0
r − α −κ − Y = 0
1
2
1
2•Then the kinematic and isotropic hardening proportions can be determined as
= f − σ r
2− Y =
f + σ r
2
The offset used to to determine yield has a large effect upon the proportion of kinematic to isotropic hardening
f
Y
r
r
•A small offset definition of yield stress (such as a strain of 0.005 %) results in a greater proportion of kinematic hardening
•A larger offset (the standard 0.2 % strain) will generally result in a prediction of a domination isotropic hardening
•The truth (experimental data) is most closely approximated by the smallest offset and is most important in problems involving unloading at small strains
J. Hodowany’s Ph.D. dissertation (Caltech, 1997) showed quantitative measurements of the conversion of plastic work into heat as a function of plastic strain
Hodowany’s measurements indicate that in the early stages of deformation, the amount of plastic work converted to heat may be quite small
As deformation progresses, the heat conversion more nearly equals the plastic work rate
Results of J. Hodowany
(Ph.D. dissertation, Caltech, 1997)
Kolsky (split Hopkinson) bar arrangement
High speed measurements of
•Temperature (IR detector)
•Plastic work (strain gages on input and output bars)
Primary results involve a parameter which is the fraction of plastic work converted to heat
Fraction of plastic work converted to heat
€
˙ = ρC
σ ˙ ε p
ρ = material density, C = heat capacity
β = % of plastic working converted to heat
Fraction of plastic work converted to heat for 2024 Al
0
1 0 0
2 0 0
3 0 0
4 0 0
5 0 0
6 0 0
0 5 1 0 1 5 2 0
P l a s t i c S t r a i n ( % )
Stress (MPa)
0
0 . 2
0 . 4
0 . 6
0 . 8
1
0 5 1 0 1 5 2 0
P l a s t i c S t r a i n ( % )
Simulation for a “fully annealed” material
0
1 0 0
2 0 0
3 0 0
4 0 0
5 0 0
6 0 0
0 5 1 0 1 5 2 0
P l a s t i c S t r a i n ( % )
Stress (MPa)
Simulation for an initially hardened material
The effect of damage on the elastic moduli can be determined from considering a voided cylinder loaded in tension
•The effect of the porosity can be viewed as a reduction of the load bearing area
•The stress is then
=P
A˜ σ =
P
A 1−φ( )=
σ
1−φ( )
˜ σ =E 1−φ( )εe = ˆ E εe
•This derivation treats voids as an area distribution. The effective elastic moduli for a volumetric distribution has been determined using self consistent techniques by Budiansky and OConnel
μ φ( )=μ 0 1− 53κ 0 + 4μ0
9κ 0 +8μ0
φ ⎡
⎣ ⎢
⎤
⎦ ⎥ φ( ) = 0 1 −
3κ 0 + 4μ0
4μ0
φ ⎡
⎣ ⎢
⎤
⎦ ⎥
In the same manner, the voids tend to concentrate the stress in the matrix material, thereby enhancing the plastic flow
=P
A =
P
A 1 −φ( ) •For metals, the plastic flow will generally have a strong dependence upon the stress, for example
•In the presence of damage, the stress will be concentrated and the resulting flow rule takes the form
˙ p =˙
⎛
⎝ ⎜
⎞
⎠ ⎟n
˙ p =˙
1−φ( )
⎛
⎝ ⎜
⎞
⎠ ⎟
n
The evolution of damage is based upon the analytic solution of Cocks and Ashby
˙ =˙
⎛
⎝ ⎜
⎞
⎠ ⎟n
1
1
2
•Growth of spherical void in a power law creeping material under a three dimensional state of stress
•Cocks and Ashby using a bound theorem calculated the approximate growth rate of the void
•We utilize the functional form in the evolution of our damage state variable
˙ φ =sinh1−n( )1 +n( )
p'
⎡
⎣ ⎢
⎤
⎦ ⎥
1
1−φ( )n− 1−φ( )
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪dp
•Failure occurs when a critical level of damage has accumulated and the material becomes unstable
Assume that the free energy depends upon the elastic strain, the temperature, and two state variables, the elastic strain resulting from the distribution of geometrically necessary dislocations leading to kinematic hardening, gnand the elastic strain associated with the density of statistically stored dislocations giving rise to isotropic hardening, ss
€
ψ = ˆ ψ e, θ, εgn, ε ss, φ( )
σ = ρ
∂ ˆ ψ ∂εe
, η =∂ ˆ ψ ∂θ
€
= ρ∂ ˆ ψ ∂
, = ρ∂ˆ ψ ∂
, χ = ρ∂ ˆ ψ ∂φ
Where, , χ and are conjugate thermodynamic stresses
It follows from the 2nd law
€
˙ =1
ρCσ ˙ ε p − α ˙ ε gn − κ˙ ε ss − χ ˙ φ ( )
Then neglecting elastic heating effects and neglecting conduction for high strain rate applications, the energybalance reduces to
And the dissipation inequality reduces to
€
˙ p − α ˙ ε gn − κ˙ ε ss − χ ˙ φ +qθ
∂θ∂x
≥ 0
The kinematics of plasticity introduced degrees of freedom requiring more”constitutive” equations. Similarly, the extra degrees of freedom also requires temporal evolution equations to complete the system.
This is where the physics of the smaller length scales enters!
The deformation gradient is decomposed multiplicatively into thermal, elastic, deviatoric plastic and damage parts and make simplifying assumptions about the directionality of thermal and damage gradients
F = FeFpFd F
Fd =1
1−φ( )131
F =F 1≅1+( )1
E =(1−φ)2 / 3
F2 Fp
−TEFp−1 = Ee+ Epd
Fp
Fd
Fe
F
F
=0
natural configuration
Define strain with respect to natural configuration
det Fd =1
1−φ( )
The velocity gradient is defined as usual and is naturally composed of elastic, plastic, damage and thermal parts. These can be mapped to any other configuration.
L = ˙ FF−1 =˙ FeFe−1 +Fe
˙ FpFp−1Fe
−1 +˙ φ
3 1−φ( )1 + c˙ 1
=Le+L p+ Ld +L
L p =˙ FpFp−1
Le =De+WeL p = Dp+ Wp
˙ E e = D − Dp − Ee L p− LpT Ee
Velocity gradient
Notice that the plastic part of the velocity gradient is naturally defined with respect to the natural configuration as
And the velocity gradient w.r.t any configuration can be split into elastic and plastic parts
And then from algebra
Since natural configuration is most closely related to lattice, we must relate velocity gradients in current configuration (where we solve momentum balance) to strain rates in the lattice (where we model material response)
Le = ˙ FeFe−1 Lp =Fe
˙ FpFp−1Fe
−1 =Fe LpFe−1 Ld =
˙ φ 3(1−φ)
I L =˙ FF
I
L = ˙ FF −1 =Le +Lp + Ld + L
E e⋅
≈FeT DeFe
Fe =I + H sup || H ||= <<1
˙ E e ≈De
The current configuration velocity gradient is defined as
where
Now
And assuming, that elastic strains are small
Then, the elastic strain rate in the intermediate configuration is approximately equal to the current configuration elastic velocity gradient
The thermodynamics are formulated with respect to the natural configuration following Coleman and Gurtin. The free energy is assumed to depend upon the elastic strain, temperature, temperature gradient, and internal variables related to dislocations in cells and cell walls, and voids.
Assume the free energy with respect to the natural configuration
ψ = ˆ ψ E e ,K ,A ,Φ ,θ,∇θ( )
−˙ ψ − η ˙ θ +1
ρ σ T ⋅L −
q ⋅gradκ θ
ρ θ≥ 0
T − ρ ∂ψ∂Ε e
⎛
⎝ ⎜
⎞
⎠ ⎟ ⋅ ˙ Ε e − ρ η +
∂ψ∂θ
⎛ ⎝ ⎜
⎞ ⎠ ⎟ ⋅ ˙ θ − ρ
∂ψ∂ gradθ
⋅gradθ⋅
+σ T ⋅ Fe L p Fe−1
( ) − q ⋅gradθ
θ− ρ
∂ψ∂A
⋅ ˙ A − ρ ∂ψ∂K
⋅ ˙ K
−ρ ∂ψ
∂Φ⋅ ˙ Φ ≥ 0
Recall the Clausius-Duhem inequality
Substituting, the dissipation becomes
The balance of energy becomes
ρ c ˙ θ = ρ r − divκ q + σ ⋅ D p + E e L p + L pT E e( ) − α ⋅ ˙ A − κ ⋅ ˙ K − Γ ⋅ ˙ Φ
+ ρ θ∂ 2ψ
∂θ∂E e⋅ ˙ E e +
∂ 2ψ∂θ∂A
⋅ ˙ A +∂ 2ψ∂θ∂K
⋅ ˙ K +∂ 2ψ∂θ∂Φ
⋅ ˙ Φ ⎡
⎣ ⎢
⎤
⎦ ⎥
Neglecting thermal-elastic effects and for high rate processes where conduction is negligible
˙ = 1ρ c
⋅ Dp+ Ee Lp + L pT Ee( )−⋅˙ A−⋅ ˙ K−Γ⋅ ˙ Φ { }
As a result of thermodynamic restrictions we define “thermodynamic stresses” as the derivative of free energy with respect to state variables. The stresses appear in our flow rule and the evolution of the state variables is motivated from micromechanics. These are related through the free energy.
=ρ Fe∂ψ
∂A Fe
T
κ = ρ Fe∂ψ
∂K Fe
T
Γ = ρ Fe∂ψ
∂Φ Fe
T
∂ψ∂grad θ
= 0
η = −∂ψ
∂θ
σ = ρ Fe∂ψ
∂E eFe
T
To satisfy thermodynamics:
Define conjugate stresses to the deformation like state variables:
The elastic strain associated with the dislocations is motivated by considering the energy of a Frank network
The energy of a Frank network of dislocations of density ρd is
W =1+ 16
μb2ρd
Since the stress associated with the dislocations is given by
τ ∝μb ρd
We can identify the strain associated with the dislocation network as,
=b ρd ⇒ W =1+ ν
16με κ
2
The dislocation velocity is assumed to depend upon the magnitude of the “net stress” (effective stress) which is the sum of the applied stress and any stresses due to the state variables which were introduced in the thermodynamics
= ′ −− +Γ
=′ −′ − Γ =φ ′ −( ) 1 −
κ
′ σ − α
⎛
⎝ ⎜ ⎞
⎠ ⎟
τ = =⋅= ′ − −( ) 1 + φ( )
v =v τ( )=v | ′ − |−[ ] 1+φ[ ]( )
The net stress is the Cauchy stress minus the two dislocation stresses plus the void stress
To model isotropic hardening assume dislocation cell stress is in the direction of Cauchy minus back stress due to dislocations in cell interiors and that the void stress concentrates this stress
;
Then the effective stress is
The dislocation velocity would then depend upon
And for small porosityv =v τ( )=v
| ′ − |−1−φ
⎛ ⎝ ⎜
⎞ ⎠ ⎟
We now have reduced the problem for determining the free energy as a function of two scalar “strains” related to a scalar dislocation density and a scalar porosity as well as a tensor strain related to dislocation distribution
ρ ψ (E e ,θ ,φ,k , A ) =1
2μ(θ ,φ) ′ E e
2+
1
2
K(θ ,φ)
2tr 2E e +
1
2aφγ snv
1/ 3φ2 / 3 +
12
aκ μ(θ,φ)k 2 +12
aα μ(θ,φ) A 2
+12
g(θ )
Now assume linear elasticity with respect to the natural configuration, mapping forward to the current configuration, and take the material derivative of the Cauchy stress. The Budiansky solution is utilized to describe the dependence of the elastic moduli on the damage.
μ φ( ) =μ0 1 − 53κ 0 + 4μ0
9κ0 + 8μ0φ
⎡
⎣ ⎢ ⎤
⎦ ⎥ φ( ) =0 1 −3κ 0 + 4μ0
4μ 0φ
⎡
⎣ ⎢ ⎤
⎦ ⎥
o =˙ σ − Weσ + σWe ≈ λ φ( ) tr De( )1 + 2μ φ( )De −
˙ φ
1 − φσ
We=W −WpDe=D−Dp−Dd−D
Linear elasticity
Elastic moduli degraded by voids
Velocity gradient decomposition
The thermal and damage parts of the velocity gradient were discussed in the kinematics, hence we need to address the plastic part (dislocation related) of the velocity gradient
Motivation for evolution comes from Kocks-Mecking
Motivate evolution equations from Kocks-Mecking where dislocation density evolves as a dislocation storage minus recovery event. Hardening is proportional to dislocation cell size which is inversely proportional to square root of dislocation density while recovery is proportional to dislocation density
dρddp
=c11−c2ρd =c1 ρd −c2ρ
ddt
=dρddp
d˙ pdt
=H ˙ p−Rd ˙ p
Then,
For small concentrations of damage the flow rule reduces to the more familiar one. The evolution of the state variable comes form dislocation micromechanics and ductile void growth of Cocks and Ashby. The plastic spin is assumed zero.
Dp=f()sinh| ′ − |−−Y (){ }
(1−φ)V() ⎧ ⎨ ⎩
⎫ ⎬ ⎭
′ −| ′ − |
˙ =H()Dp −Rd() |Dp| −Rs()2
˙ −L e +L e=h()Dp− rs() +rd() |Dp|{ } | |
˙ φ =1
1−φ( )n− 1−φ( )
⎧ ⎨ ⎩
⎫ ⎬ ⎭Dp sinh
tr
⎡
⎣ ⎢ ⎤
⎦ ⎥
Flow rule
Evolution of hardening state variables
Evolution of void growth
For constant true strain rate, isothermal, viscoplastic uniaxial stress states (and in the absence of damage effects - compression), an analytical solution is obtained
=h ˙ ε
rd˙ ε +rs
tanhh rd
˙ ε +rs( )
˙ ε ε
⎧
⎨ ⎪
⎩ ⎪
⎫
⎬ ⎪
⎭ ⎪
= + + Y (θ ) + V(θ )sinh−1 | ˙ ε |
f (θ )
⎧ ⎨ ⎩
⎫ ⎬ ⎭
sat =h ˙ ε
r d˙ ε +r s
+H ˙ ε
Rd˙ ε +Rs
+ Y(θ ) + V(θ)sinh−1
| ˙ ε |
f (θ)
⎧ ⎨ ⎩
⎫ ⎬ ⎭
sat ∝˙ ε
H Rd
˙ ε +Rs( )
=H ˙ ε
Rd˙ ε +Rs
tanhH Rd
˙ ε +Rs( )
˙ ε ε
⎧
⎨ ⎪
⎩ ⎪
⎫
⎬ ⎪
⎭ ⎪
Solution for hardening variables
Resulting flow stress
Saturation stress and strain
Analogous to elasticity, assume a free energy of the form
€
ρ ˆ ψ =12
E 1 − φ( )εe2 +
12
kα μεα2 +
12
kκ μεκ2
σ = E 1 − φ( )εe
α = kα μεα
κ = kκ μεκ
χ =12
Eεe2 =
12
σεe
Then,
Then the dissipation inequality takes the form
€
˙ p − α ˙ ε gn − κ˙ ε ss − χ ˙ φ +qθ
∂θ∂x
≥ 0
E 1 − φ( )εe˙ ε p − kα μεα ˙ ε gn − kκ μεκ ˙ ε ss −12
Eεe2˙ φ +
qθ
∂θ∂x
≥ 0
SUMMARY - CURRENT CONFIGURATION
Linear Elasticity
€
˙ − Weσ + σWe = λ θ( ) 1 − φ{ }trDe1 + 2μ θ( ) 1 − φ{ }De +2∂μ θ( )
∂θ˜ E e˙ θ
+2∂λ θ( )
∂θtr˜ E e˙ θ − σ
˙ φ
1 − φ− p
˙ φ
1 − φ
State Variables
€
˙ = Hκ μ θ( )D p + Hκ∂μ θ( )
∂θ˙ θ Dp − Rdκ θ( )D p κ − Rsκ θ( )κ sinh Qsκ[ ]
€
˙ − Leα + αLe = Hα L p′ σ − α
′ σ − α− Rdα L p α α
Flow Rule -(Cocks, 1989) (Marin, McDowell-1996)
€
F 3h1J2 + h2I12 − h3κ 1 + Sinh−1
f
⎧ ⎨ ⎩
⎫ ⎬ ⎭
1n ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
Dp = f θ( )Sinh3h1J2 + h2I1
2
h3κ
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪
′ σ − α′ σ − α
+˙ φ
1 − φ1
€
h1 = 1 +2
3φ
h2 =12
11 + m
φ1 + φ
€
h3 = 1 − φ( )1
1 + m( )
where
0
50
100
150
200
250
300
0 0.1 0.2 0.3 0.4 0.5
Data 1
R=0.390 PredictionR=0.390 Test DataR=0.178 PredictionR=0.178 Test Data
LOAD [lbs]
DISPLACEMENT [in.]
Radius Test Analysis
0.390 0.043 0.044
0.156 0.021 0.023
0.078 0.014 0.015
0.039 0.011 0.013
Elongation at Failure
Aluminum disks were impacted by hardened tool steel rods. At 3600 in/s this loading resulted in partial failure of the back surface.
Experiment Simulation
Aluminum disks were impacted by hardened tool steel rods. At 4200 in/s this loading resulted in complete penetration of the disk..
SimulationExperiment
Application of the BCJ State Variable Model to Submarine Vulnerability Experiments (Steel Plates Loaded by Focused Blasts)
Problem:Predict response of steel (HY100) plate with initial 1 inch diameter hole loaded by blast wave produced by 38 grams of explosive.
Technical Approach:• Model blast pressure time history with Eulerian code.• Apply the BCJ constitutive model, utilizing the strain rate and temperature dependence and ductile failure capabilities
Analysis Experiment
The model has been used to predict response of steel (HY100) plate with initial 1 inch diameter hole loaded by blast wave produced by 38 grams of explosive. The blast pressure time history is modeledwith Eulerian code
.Analysis Experiment