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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