Plan Mechanical tests Rheological models Summary on rheology
Rheology
Georges Cailletaud
Centre des MatériauxMINES ParisTech/CNRS
October 2013
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Plan Mechanical tests Rheological models Summary on rheology
Contents
1 Mechanical testsStructuresRepresentative material elements
2 Rheological modelsBasic building bricksPlasticityViscoelasticityElastoviscoplasticity
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Plan Mechanical tests Rheological models Summary on rheology
Tests on a civil plane
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Plan Mechanical tests Rheological models Summary on rheology
Vibration of a wing
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Plan Mechanical tests Rheological models Summary on rheology
Biological structures (1/2)
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Biological structures (2/2)
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Plan Mechanical tests Rheological models Summary on rheology
Food industry
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Plan Mechanical tests Rheological models Summary on rheology
Testing machines
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Plan Mechanical tests Rheological models Summary on rheology
Tension test on a metallic specimen
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Mechanical tests
Basic tests
Time independent plasticityTension test, or hardening testCyclic load, or fatigue test
Time dependent plasticityTest at constant stress, or creep testTest at constant strain, or relaxation test
Other tests
Multiaxial loadTension–torsionInternal pressure
Bending tests
Crack propagation tests
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Plan Mechanical tests Rheological models Summary on rheology
Typical result on an aluminum alloy
For a stress σ0.2, it remains 0.2% residual strain after unloading
Stress to failure, σu
0.2% residual strainElastic slope
Tension curve
ε(mm/mm)
σ(M
Pa)
0.040.030.020.010
600
500
400
300
200
100
0
E=78000 MPa, σ0.2=430 MPa, σu=520 MPa Doc. Mines Paris-CDM, Evry
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Plan Mechanical tests Rheological models Summary on rheology
Typical result on an austenitic steel
Material exhibiting an important hardening : the yield stress increasesduring plastic flow
0.2% residual strainElastic slope
Tension curve
ε(mm/mm)
σ(M
Pa)
0.080.070.060.050.040.030.020.010
600
500
400
300
200
100
0
E=210000 MPa, σ0.2=180 MPa, σu=660 MPa Doc. ONERA-DMSE, Châtillon
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Plan Mechanical tests Rheological models Summary on rheology
Push–pull test on an aluminum alloy
Test under strain control ± 0.3%
Positive residual strain at zero stress
Negative stress at zero strain
ε(mm/mm)
σ(M
Pa)
0.0050.0030.001-0.001-0.003-0.005
300
200
100
0
-100
-200
-300
Doc. Mines Paris-CDM, Evry
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Plan Mechanical tests Rheological models Summary on rheology
Schematic models for the preceding results
σ
σ y
E
0 εa. Elastic–perfectly plastic
ε0
E
TE
σ
σy
b. Elastic–plastic (linear)
Elastoplastic modulus, ET = dσ/dε.
ET = 0 : elastic-perfectly plastic material
ET constant : linear plastic hardening
Et strain dependent in the general case
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Plan Mechanical tests Rheological models Summary on rheology
How does a plasticity model work ?
0 0’
A
B
ε
σ
Elastic regimeOA, O’B
Plastic flowAB
Residual strainOO’
Strain decomposition, ε = εe + εp ;
Yield domain, defined by a load function f
Hardening, defined by means of hardening variables,AI .
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Plan Mechanical tests Rheological models Summary on rheology
Result of a tension on a steel at hightemperature
Viscosity effect : Strain rate dependent behaviour
ε = 1.6 10−5s−1ε = 8.0 10−5s−1ε = 2.4 10−4s−1
725◦C
ε
σ(M
Pa)
0.10.080.060.040.020
80
60
40
20
0
Doc. Ecole des Mines, Nancy
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Plan Mechanical tests Rheological models Summary on rheology
Creep test on a tin–lead wire
Mines Paris-CDM, Evry
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Plan Mechanical tests Rheological models Summary on rheology
Creep on a cast iron
σ=25MPaσ=20MPaσ=16MPaσ=12MPa
t (s)
εp
10008006004002000
0.03
0.025
0.02
0.015
0.01
0.005
0
Doc. Mines Paris-CDM, Evry
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Plan Mechanical tests Rheological models Summary on rheology
Schematic representation of a creep curve
Primary creep , with hardening in the material
Secondary creep , steady state creep : εp is a power function ofthe applied stress
Tertiary creep , when damage mechanisms start
pε
III
III
t
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Plan Mechanical tests Rheological models Summary on rheology
Creep on a cast iron (2)
T=800◦CT=700◦CT=600◦CT=500◦C
σ (MPa)
εp(s−1
)
100101
0.001
0.0001
1e-05
1e-06
1e-07
1e-08
Doc. Mines Paris-CDM, Evry
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Plan Mechanical tests Rheological models Summary on rheology
Relaxation test
Constant strain during the test
During the test :ε = 0 = ε
p + σ/E
dεp =−dσ/E
The viscoplastic strain increases meanwhile stress decreases
The asymptotic stress may be zero (total relaxation) or not (partialrelaxation)
Partial relaxation if there is an internal stress or a threshold in thematerial
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Plan Mechanical tests Rheological models Summary on rheology
Schematic representation of a relaxation curve
The current point in stress space is obtained as the sum of a thresholdstress σs and of a viscous stress σv
The threshold stress represents the plastic behaviour that is reached forzero strain rate
σv
pε
σs
t
σ σ
E
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Plan Mechanical tests Rheological models Summary on rheology
Contents
1 Mechanical testsStructuresRepresentative material elements
2 Rheological modelsBasic building bricksPlasticityViscoelasticityElastoviscoplasticity
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Plan Mechanical tests Rheological models Summary on rheology
Building bricks for the material models
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Plan Mechanical tests Rheological models Summary on rheology
Various types of rheologies
Time independent plasticity
ε = εe + ε
p dεp = f (...)dσ
Elasto-viscoplasticity
ε = εe + ε
p dεp = f (...)dt
ViscoelasticityF(σ, σ,ε, ε) = 0
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Time independent plasticity
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Plan Mechanical tests Rheological models Summary on rheology
Elastic–perfectly plastic model
The elastic/plastic regime is defined by means ofa load function f (from stress space into R)
f (σ) = |σ|−σy
Elasticity domainif f < 0 ε = ε
e = σ/E
Elastic unloading
if f = 0 and f < 0 ε = εe = σ/E
Plastic flowif f = 0 and f = 0 ε = ε
p
The condition f = 0 is the consistency condition
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Plan Mechanical tests Rheological models Summary on rheology
Prager model
Loading function with two variables, σ and X
f (σ,X) = |σ−X |−σy with X = Hεp
Plastic flow if both conditions are verified f = 0 and f = 0.
∂f∂σ
σ+∂f∂X
X = 0
sign(σ−X) σ− sign(σ−X) X = 0 thus : σ = X
Plastic strain rate as a function of the stress rate
εp = σ/H
Plastic strain rate as a function of the total strain rate (once an elasticstrain is added)
εp =
EE +H
ε
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Plan Mechanical tests Rheological models Summary on rheology
Equation of onedimensional elastoplasticity
Elasticity domainif f (σ,Ai) < 0 ε = σ/E
Elastic unloading
if f (σ,Ai) = 0 and f (σ,Ai) < 0 ε = σ/E
Plastic flow
if f (σ,Ai) = 0 and f (σ,Ai) = 0 ε = σ/E + εp
The consistency condition writes :
f (σ,Ai) = 0
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Plan Mechanical tests Rheological models Summary on rheology
Illustration of the two hardening types
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Plan Mechanical tests Rheological models Summary on rheology
Isotropic hardening model
Loading function with two variables, σ and R
f (σ,R) = |σ|−R−σy
R depends on p, accumulated plastic strain : p = |εp|dR/dp = H thus R = Hp
Plastic flow iff f = 0 and f = 0
∂f∂σ
σ+∂f∂R
R = 0
sign(σ) σ− R = 0 thus sign(σ) σ−Hp
Plastic strain rate as a function of the stress rate
p = sign(σ) σ/H thus εp = σ/H
Classical modelsRamberg-Osgood : σ = σy +Kpm
Exponential rule : σ = σu +(σy −σu)exp(−bp)
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Plan Mechanical tests Rheological models Summary on rheology
Viscoelasticity
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Plan Mechanical tests Rheological models Summary on rheology
Elementary responses in viscoelasticity
Serie, Maxwell model : ε = σ/E0 +σ/η
Creep under a stress σ0 : ε = σ0/E0 +σ0 t /η
Relaxation for a strain ε0 : σ = E0ε0 exp[−t/τ]
Parallel, Voigt model : σ = Hε+ηε or ε = (σ−H ε)/η
Creep under a stress σ0 : ε = (σ0 /H)(1−exp[−t/τ′])
The constants τ = η/E0 and τ′ = η/H are in seconds, τ denoting the socalled le relaxation time of the Maxwell model
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More complex models
a. Kelvin–Voigt
(E0)
(H)
(η)
b. Zener
(η)(E2)
(E1)
Creep and relaxation responses
ε(t) = C(t)σ0 =
(1
E0+
1H
(1−exp[−t/τf ])
)σ0
σ(t) = E(t)ε0 =
(H
H +E0+
E0
H +E0exp[−t/τr ]
)E0ε0
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Elasto-viscoplasticity
Scheme of the model Tensile response
X = Hεvp
σv = ηεvp |σp|6 σy
σ = X +σv +σp
Elasticity domain, whose boundary is |σp|= σy
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Plan Mechanical tests Rheological models Summary on rheology
Model equations
Three regimes
(a) εvp =0 |σp|= |σ−Hε
vp| 6σy
(b) εvp >0 σp =σ−Hε
vp−ηεvp =σy
(c) εvp <0 σp =σ−Hε
vp−ηεvp = −σy
(a) interior or boundary of the elasticity domain (|σp| < σy )(b),(c) flow (|σp|= σy and |σp| = 0 )
One can summarize the three equations (with < x >= max(x ,0)) by
ηεvp = 〈|σ−X |−σy 〉 sign(σ−X)
or :
εvp =
< f >
ηsign(σ−X) , with f (σ,X) = |σ−X |−σy
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Creep with a Bingham model
t
σ σo y-
H
εvp
Viscoplastic strainversus time
σ
σ
Xo
y
σ
vpεEvolution in the planestress– vsicoplastic
strain
εvp =
σo−σy
H
(1−exp
(− t
τf
))with : τf = η/H
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Relaxation with a Bingham model
σ
H-E
vpε
σ
y
Relaxation
H
ε
Transitoire : OA = BC
Relaxation : AB
Effacementincomplet : CDO
A
B
DC
vp
Fading memory
σ = σyE
E +H
(1−exp
(− t
τr
))+
Eεo
E +H
(H +E exp
(− t
τr
))with : τr =
η
E +H
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Plan Mechanical tests Rheological models Summary on rheology
Ingredients for classical viscoplastic models
Bingham model
εvp =
< f >
ηsign(σ−X)
More generallyε
vp = φ(f )
φ(0) = 0 and φ monotonically increasing
εvp is zero if the current point is in the elasticity domain or on theboundary
εvp is non zero if the current point is outside from the elasticity domain
There are models with/without threshold, with/without hardening
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Plan Mechanical tests Rheological models Summary on rheology
Viscoplastic models without hardening
Models without threeshold : the elastic domain is reduced to the origin(σ = 0)
Norton model
εvp =
( |σ|K
)n
sign(σ)
Sellars–Tegart model
εvp = Ash
( |σ|K
)sign(σ)
Models with a thresholdPerzyna model
εvp =
⟨ |σ|−σy
K
⟩n
sign(σ) , εvp = ε0
⟨ |σ|σy−1
⟩n
sign(σ)
Georges Cailletaud | Rheology 40/44
Plan Mechanical tests Rheological models Summary on rheology
Viscoplastic models with hardening
The concept of additive hardening : The hardening comes from thevariables that represent the threshold (X and R)
εvp =
⟨ |σ−X |−R−σy
K
⟩n
sign(σ−X)
X stands for the internal stress (kinematical hardening)R +σy stands for the friction stress (isotropic hardening)σv is the viscous stress or drag stress
The concept of multiplicative hardening : one plays on viscous stress, forinstance :
εvp =
( |σ|K (εp)
)n
sign(σ) =
( |σ|K0|εp|m
)n
sign(σ)
strain hardening
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Plan Mechanical tests Rheological models Summary on rheology
For plasticity and viscoplasticity...
Elasticity defined by a loading function f < 0
Isotropic and kinematic variables
For plasticity :
Plastic flow defined by the consistency condition f = 0, f = 0
Plastic flow :dε
p = g(σ, . . .)dσ
For viscoplasticity :
Flow defined by the viscosity function if f > 0
Possible hardening on the viscous stress
Delayed viscoplastic flow
dεvp = g(σ, . . .)dt
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Plan Mechanical tests Rheological models Summary on rheology
Identification of the material parametersNorton model on tin–lead wires
0
0.02
0.04
0.06
0.08
0.1
0 1000 2000 3000 4000 5000
cree
p st
rain
time (s)
1534 g1320 g1150 g997 g720 g
0
2
4
6
8
10
12
14
0 5000 10000 15000 20000 25000
stre
ss (
MP
a)
time (s)
expsim
Creep test Relaxation ε=20%
Curves obtained with a Norton model
εp =
(σ
800
)2.3
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Plan Mechanical tests Rheological models Summary on rheology
Identification of the creep on salt
0
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0.008
0 0.5 1 1.5 2 2.5 3 3.5 4
stra
in
time (Ms)
expsim
Specimen Three level test (3, 6, 9 MPa)
Curves obtained with a Lemaitre model (strain hardening)
εp =
(σ
K
)n(εp + v0)
m
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