Adiabatic Shear Bands Theocharis Baxevanis Thodoros Katsaounis University of Crete, Greece Athanasios Tzavaras University of Maryland and University of Crete
Adiabatic Shear Bands
Theocharis BaxevanisThodoros Katsaounis
University of Crete, Greece
Athanasios TzavarasUniversity of Maryland
andUniversity of Crete
Outline
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What is a shear band?
Model in thermoviscoplasticity context
Nature of the instability - Linearized analysis
Nonlinear stability - shear localization
Numerical Results - adiabatic case
Numerical Results - effect of thermal diffusion
Onset of instability - Paradigm Arrhenius model
Effective equation at localization
A Shear Band
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Shear Band in a Aluminum alloy
In high strain-rate loading, for certain metals (aluminum, steel ...), theshear strain localizes in narrow regions called shear bands
Characteristics of bands
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A narrow layer of intense shearing that develops when materials (metals,polymers, powders) are deforming at high rates.
■ Typical shear band widths : 10− 100µm
■ High local values of shear strain : 5− 100
■ Ultra high local shear strain rates : 104/s− 106/s
■ Local temperature rises several hundred degrees
■ High propagation speeds : ∼ 1000m/s
■ Not a crack : material preserves its integrability
■ Shear Bands are precursors to rupture
Experiments performed in experimental device Kolsky bar(groups of Duffy, Clifton - Brown 80’s )
What is a Shear Band?
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In high strain-rate loading of metals the shear strain localizes in narrowregions called shear bands
x = 0
x = dV
shear band
Uniform shearing vs. shear band
Localization of plastic strain in narrow band
Elevated temperature inside the band
Modeling of shear bands
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Adiabatic plastic shearing of an infinite plate occupying the regionbetween the planes x = 0 and x = d.
0
x
d V
y(x, t) displacement in shearing direction
v = ∂y
∂tvelocity in shearing direction
θ temperature
γ = ∂y
∂xshear strain
σ shear stress
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Balance of momentum and energy
ρ vt = σx
c ρ θt = β σ vx + k θxx
γt = vx
ρ density, c specific heat, β portion of plastic work converted to heatκ thermal diffusivity, G shear modulus
yield surface or plastic flow rule
σ = G θ−α γm γnt
α thermal softening, m strain hardening, n strain rate sensitivity
Typical values of the parameterscold-rolled steel (AISI 1018) α = 0.38, m = 0.015, n = 0.019
Constitutive law - yield rule
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The constitutive lawσ = θ−α γm γn
t
is rewritten asγt = θ
α
n γ−m
n σ1
n
Then
■ αn
> 0 thermal softening
■ mn
> 0 strain hardening
■ n > 0 strain rate sensitivity
Constitutive law - yield rule
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The constitutive lawσ = θ−α γm γn
t
is rewritten asγt = θ
α
n γ−m
n σ1
n
Then
■ αn
> 0 thermal softening
■ mn
> 0 strain hardening
■ n > 0 strain rate sensitivity
Elastic effects
γ = γe + γp γe =1
Ge
σ
γt = γet + γpt=
1
Ge
σt + θα
n γ−m
n σ1
n
Mechanism to localization
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■ Under isothermal conditions metals strain harden
■ in large deformation speeds(i) conditions change from isothermal to nearly adiabatic,(ii) strain rate has an effect per se
■ Destabilizing mechanism is induced by thermal softening:(i) Nonuniform strains induce nonuniform heating.(ii) Material is softer at hotter spots, harder at colder spots,further amplifying the nonuformities in strain.
■ Opposed to that are two effects:(i) Momentum diffusion induced by strain rate dependence(ii) Heat diffusion
If heat diffusion is too weak to equalize temperatures in the time-scale ofloading then shear bands may occur.
Zener and Hollomon 44, Clifton 78
Mathematical issues
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vt =1
rσx
θt = κθxx + σγt
γt = vx
σ = θ−αγmγtn
non-dimensional numbers
{
r ratio of inertial versus viscous stresses
κ thermal diffusivity
Adiabatic assumption κ = 0
Special cases
m = 0 α = 0 m < 0
vt = ∂x
(
θ−αvxn)
θt = θ−αvxn+1
vt = ∂x
(
γmvxn)
γt = vx
Uniform shearing solutions
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The system with boundary conditions
v(0, t) = 0, v(1, t) = 1
admits the following Uniform Shearing Solutions
vs = x
γs(t) = t + γ0
θs(t) =
{
θα+10 +
α + 1
m + 1
[
(γ0 + t)m+1 − γm+10
]
}1
α+1
σs(t) = θs(t)−α(t + γ0)
m
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t+γ0
σ
Hardening Softening
for −α + m < 0, σs(t) initially increases (hardening) but eventuallydecreases (softening) with t and can produce net softening
thermal softening + strain hardening → net softening
Nature of the problem
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Regularized ill-posed problem
γ
φ(γ)
vt =(
ϕ(γ)vnx
)
x
γt = vx
Uniform shearing
vs(x, t) = x, γs(x, t) = t + γ0
is a universal solution for any value of n.
■ For n = 0 it is a hyperbolic-elliptic initial value problem. Instabilityinitiates at the max of curve
■ When n > 0 when does instability occur ?
Linearized problem
v = x + V
γ = t + γ0 + Γ
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linearized equations
Vt = nϕ(t + γ0)Vxx + ϕ′(t + γ0)Γx
Γt = Vx
Question: Are the uniform shearing solutions stable?
vs(x, t) = x, γs(t) = t + γ0
It is natural to require that the uniform shear is
asymptotically stable if γ(x,t)γs(t)
= o(1) and vx − 1 = o(1)
unstable if γ(x,t)γs(t)
and vx − 1 grow in time
Linearized analysis - relative perturbation
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Vt = n(t + γ0)−mVxx −m(t + γ0)
−m−1Γx
Γt = Vx
■ ϕ(γ) = γ−m m > 0 n << 1,
(i) −m + n > 0 thenuniform shear is stable in a relative perturbation sense
Γ(x, t)
t + γ0= 1 + O
(
(t + γ0)−α
)
Vx(x, t) = x + 1 + O(
(t + γ0)−α
)
(ii) −m + n < 0 thenuniform shear unstable in relative perturbation sense
maxx|Γ(x, t)
t + γ0
| ≥ O(
(t + γ0)δ)
for some δ > 0
Nonlinear stability
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■ ϕ(γ) = γ−m m > 0
−m + n > 0uniform shear is nonlinearly asymptotically stable
Γ(x, t)
t + γ0= 1 + O
(
(t + γ0)−α
)
Vx(x, t) = x + 1 + O(
(t + γ0)−α
)
Nonlinear stability
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■ ϕ(γ) = γ−m m > 0
−m + n > 0uniform shear is nonlinearly asymptotically stable
Γ(x, t)
t + γ0= 1 + O
(
(t + γ0)−α
)
Vx(x, t) = x + 1 + O(
(t + γ0)−α
)
■ On the complementary region −m + n > 0 we know that theuniform shear is linearly unstable in the relative perturbation sense
Nonlinear stability
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■ ϕ(γ) = γ−m m > 0
−m + n > 0uniform shear is nonlinearly asymptotically stable
Γ(x, t)
t + γ0= 1 + O
(
(t + γ0)−α
)
Vx(x, t) = x + 1 + O(
(t + γ0)−α
)
■ On the complementary region −m + n > 0 we know that theuniform shear is linearly unstable in the relative perturbation sense
■ But is instability the same as localization ?
Collapse of diffusion across a band
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Suppose the initial data are localized in strain.Will it persist ? What happens in the band ?
−m + n < 0 v0(x) = x, γ0(x) is localized
γ
v0
(x)
V=1
V=0
?
either limt→T ∗
sup0≤x≤1
γ(x, t) =∞, T ∗ <∞
or vx(x, t) = O(t−1), t→∞ outside the band
and γmax <∞ inside the band (unloading)
Collapse of diffusion across the band
Numerical Simulations - adiabatic
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Adiabatic case κ = 0 - system of three equations. Initial data
0 0.2 0.4 0.6 0.8 1x
1
1.0005
1.001
1.0015
1.002
Vx
(a) v0,x
0 0.2 0.4 0.6 0.8 1
1
1.05
1.1
1.15
1.2
1.25
1.3
(b) θ0
0 0.2 0.4 0.6 0.8 10.008
0.012
(c) γ0
−α + m + n < 0
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vx and θ at t = 0.32, κ = 0
0 0.2 0.4 0.6 0.8 11e-09
1e-06
0.001
1
1000
1e+06
(d) vx(γt)
0 0.2 0.4 0.6 0.8 1
1
10
100
1000
10000
(e) θ
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v at t = 0.32, κ = 0
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
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stress at t = 0.32, κ = 0
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
(f) σ
0.49999 0.5 0.50001
0.1223
0.12235
0.1224
(g) σ at x = 0.5
Numerical simulations - effect of thermal
diffusion
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Numerical run: velocity versus time – κ = 10−8
log(TIME)
0
2
4
6
8 X0
0.20.4
0.60.8
1
v
0
0.2
0.4
0.6
0.8
1
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Numerical run: stress versus time – κ = 10−8
log(TIME)
0
2
4
6
8
X
0
0.2
0.4
0.6
0.8
1
σ
0
0.2
0.4
0.6
Suggestions of numerical runs
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■ Instability of uniform shearing at the adiabatic case
■ Instability leads to localization which is clearly a nonlinearphenomenon.
■ Localization associated with collapse of momentum diffusion acrossthe band.
■ Heat conduction can diffuse and supress shear bands
■ Numerical runs suggest metastable response for the heat conductingcase
Analytical study - Arrhenius model
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System in non-dimensional form
vt =1
rσx,
σ = e−αθvnx
θt = kθxx + σvx
Uniform shearing solution
vs = x
θs =1
αln (αt + κ0)
σs = e−αθs =1
αt + κ0
by solvingd
dtθs = e−αθs
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Perform a rescaling of dependent variables based on uniform shearingsolution. This corresponds to doing a non-dimensionalization of themodel based on the uniform shearing solution.
v = V (x, t)
θ = θs(t) + Θ(x, t)
σ = σs(t) Σ(x, t)
and a rescaling of time using
τ̇ = σs(t) ⇐⇒ τ(t) = 1α
ln(
ακ0
t + 1)
get
ut =1
rΣxx,
Σ = e−αθun
Θτ = keατΘxx + (Σu− 1)
where u = vx
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Note that
■ Uniform shearing solution is mapped to equilibrium point
u0 = 1 Θ0 = 0 Σ0 = 1
■ If we consider the system consisiting of only the last two equationsthis forms a relaxation system whose solution relaxes to theequilibrium manifold
Θ =n + 1
αln u
Linearized stability analysis
u = 1 + δu1 + O(δ2),
Σ = 1 + δΣ1 + O(δ2)
Θ = 0 + δΘ1 + O(δ2)
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obtain linearized system
∂u1
∂τ=
n
ru1,xx −
α
rΘ1,xx,
∂Θ1
∂τ= keατΘ1,xx − αΘ1 + (n + 1)u1
Σ1 = −αΘ1 + nu1
with boundary conditions
u1,x(0, t) = u1,x(π, t) = 0 Θ1,x(0, t) = Θ1,x(π, t) = 0
System admits solutions in form of Fourier modes
u1(x, t) = v̂j(t) cos(jx),
Θ1(x, t) = ξ̂j(t) cos(jx)
which satisfy a non-autonomous system when k 6= 0.
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We do an analysis of the latter in the case that we freeze the timecoefficient, which shows
■ The eigenvalues are always real
■ There is one positive eigenvalue if and only if
kj − α < 0
where j = 1, 2, ... is the Fourier mode.
■ As a result, if k = 0 there is always a positive eigenvalue for anyFourier mode. That is the equilibrium is linearly unstable.
■ For k past a certain threshold all Fourier modes decay. This offers aheuristic explanation that eventually the uniform shearing solutionbecomes stable. Recall that k → keατ .
Quantitative criterion to instability - power
law
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Adiabatic system (κ = 0) in non-dimensional form
vt =1
rσx,
θt = σγt,
γt = vx,
σ = θ−αγmγnt .
Stress reformulation, form of a reaction-diffusion system,
σt =n
rθ−
α
n γm
n σn−1
n σxx +
(
−ασ
θ+
m
γ
)
θα
n γ−m
n σn+1
n ,
γt = θα
n γ−m
n σ1
n ,
θt = θα
n γ−m
n σn+1
n .
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Time rescaling. Motivated by the form of the uniform shearingsolutions we introduce a rescaling :
θ(x, t) = (t + 1)m+1
α+1 Θ(x, τ(t)), γ(x, t) = (t + 1)Γ(x, τ(t)),
σ(x, t) = (t + 1)m−α
α+1 Σ(x, τ(t)), v(x, t) = V (x, τ(t)), τ = ln(1 + t).
In the new variables (V,Θ,Γ,Σ) system becomes:
Vτ =1
re
m+1
1+ατ Σx,
Γτ = Vx − Γ,
Θτ = ΣVx −m+11+α
Θ,
Σ = Θ−αΓmV nx .
Theory of relaxation systems. First equation viewed as a momentequation the others describe dynamics towards an equilibrium manifold
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Flow of the o.d.e. system
ΓΘ
Σ
(h) Γ, Σ-plane flow
Γ
Θ
Σ
(i) Θ, Σ-plane flow
Closure equation for U = Vx is
Uτ =1
re
m+1
1+ατ∂xx(cU
p)
p = −α+m+n1+α
, p > 0 forward parabolic, stablep < 0 backward parabolic, unstable
Effective equation
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Time rescaling and Change of time scale
θ(x, t) = (t + 1)m+1
α+1 Θ(
x,s(t)
T
)
vx(x, t) = Vx
(
x,s(t)
T
)
s : [0,∞)→ [0,∞)
s = s(t) ←→ t = t̂(s)
θ(
x, t̂(τT ))
= (t̂(τT ) + 1)m+1
α+1 ΘT(
x, τ)
vx
(
x, t̂(τT ))
= V Tx
(
x, τ)
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Select s(t) = Tr
(t + 1)m+1
1+α
(ΘT ,ΣT ,ΓT , UT = V Tx ) satisfy
∂sU = Σxx,
1
T(βs + 1)Θs = ΣU −
m + 1
1 + αΘ,
1
T(βs + 1)Γs = U − Γ,
Σ = Θ−αΓmUn.
SendT →∞
r →∞so that T
r= O(1)
theory of relaxation
Chapman-Enskog expansion
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equilibria Σ = c U−α+m+n
1+α
Chapman-Enskog expansion gives
O( 1
T
)
∂sU = ∂xx
(
c U p)
O( 1
T 2
)
∂sU = ∂xx
(
c U p +λc2
T(βs + 1) U p−1∂xxU
p)
where c, β positive constants
p = −α+m+n1+α
p > 0 2nd order stable
p < 0 2nd order unstable (linearly ill-posed),4th order stable regularization (λ > 0)
of a backward parabolic equation
Comparison system vs effective equation
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X
Uvs
V x
0 0.2 0.4 0.6 0.8 1
1
1.002
1.004
1.006
1.008
1.01
Figure 1: Comparison of system (solid line) vs effective equation (dashedline) for T = 1000.
Conclusions
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■ Instability of uniform shear solution in adiabatic deformations
■ Onset of localization is captured by a forward-backward parabolicequation.
■ Collapse of diffusion across a formed shear band
■ Heat conduction has stabilizing effect; can diffuse or even supressshear bands
■ Numerical runs suggest metastable response for the heat conductingcase