adiabatic shear bands · characteristics of bands ima - jul 09 - adiabatic shear bands – 4 / 37 a...

Adiabatic Shear Bands

Theocharis BaxevanisThodoros Katsaounis

University of Crete, Greece

Athanasios TzavarasUniversity of Maryland

andUniversity of Crete

Outline

IMA - Jul 09 - Adiabatic Shear Bands – 2 / 37

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


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


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?


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


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


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


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


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


■ 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


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


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


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


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


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


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


■ 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


■ ϕ(γ) = γ−m m > 0


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


■ ϕ(γ) = γ−m m > 0


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


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


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


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


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


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


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


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


■ 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


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


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


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)


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.


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


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 .


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


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


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)


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


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


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


■ 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

adiabatic shear bands · characteristics of bands ima - jul 09 - adiabatic shear bands – 4 / 37 a...

Documents