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Plan Mechanical tests Rheological models Summary on rheology

Rheology

Georges Cailletaud

Centre des MatériauxMINES ParisTech/CNRS

October 2013

Georges Cailletaud | Rheology 1/44

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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Plan Mechanical tests Rheological models Summary on rheology

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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Plan Mechanical tests Rheological models Summary on rheology

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

Georges Cailletaud | Rheology 11/44

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

Georges Cailletaud | Rheology 12/44

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

Georges Cailletaud | Rheology 13/44

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

Georges Cailletaud | Rheology 14/44

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 .

Georges Cailletaud | Rheology 15/44

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

Georges Cailletaud | Rheology 19/44

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

Georges Cailletaud | Rheology 20/44

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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Plan Mechanical tests Rheological models Summary on rheology

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

Georges Cailletaud | Rheology 29/44

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)

Georges Cailletaud | Rheology 31/44

Plan Mechanical tests Rheological models Summary on rheology

Viscoelasticity

Georges Cailletaud | Rheology 32/44

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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Plan Mechanical tests Rheological models Summary on rheology

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

Georges Cailletaud | Rheology 34/44

Plan Mechanical tests Rheological models Summary on rheology

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

Georges Cailletaud | Rheology 35/44

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

Georges Cailletaud | Rheology 36/44

Plan Mechanical tests Rheological models Summary on rheology

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

Georges Cailletaud | Rheology 37/44

Plan Mechanical tests Rheological models Summary on rheology

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

Georges Cailletaud | Rheology 38/44

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

Georges Cailletaud | Rheology 39/44

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

Georges Cailletaud | Rheology 41/44

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

Georges Cailletaud | Rheology 42/44

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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Georges Cailletaud | Rheology 43/44

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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Georges Cailletaud | Rheology 44/44