Georges Cailletaud Centre des Matériaux MINES …mms2.ensmp.fr/msi_paris/transparents/Georges_Cailletaud/...Isotropic : Tresca, von Mises An anisotropic criterion : Hill 5 Pressure

Plan Mechanical tests Rheological models Summary on rheology Multiaxial plasticity criterion Pressure dependent models Pressure independent models Synthesis on criteria

Rheology and multiaxial criteria

Georges Cailletaud

Centre des MatériauxMINES ParisTech/CNRS

Non Linear Computational MechanicsAthens MP06

Georges Cailletaud | Plasticity 1/82


Contents

1 Mechanical testsStructuresRepresentative material elements

2 Rheological modelsBasic building bricksPlasticityViscoelasticityElastoviscoplasticity

3 Multiaxial plasticity criterionMechanical testsMechanisms

4 Pressure dependent modelsIsotropic : Tresca, von MisesAn anisotropic criterion : Hill

5 Pressure independent models

6 Synthesis on criteria



Tests on a civil plane

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Vibration of a wing

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Biological structures (1/2)

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Biological structures (2/2)

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Food industry

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Testing machines

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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



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



Typical result on an austenitic steel

Material exhibiting an important hardening : the yield stress increasesduring plastic flow


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



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



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



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 .



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



Creep test on a tin–lead wire

Mines Paris-CDM, Evry



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




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



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




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



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



Contents









Building bricks for the material models



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



Time independent plasticity



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



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



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



Illustration of the two hardening types



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)



Viscoelasticity



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



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



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



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



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



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



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



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)



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



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



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

I try by myself on the site mms2.ensmp.fr O


http://mms2.ensmp.fr/mms_paris/experimental/exercices/f_crrel.pdf


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

I try by myself on the site mms2.ensmp.fr O


http://mms2.ensmp.fr/mms_paris/experimental/exercices/intro_salt.pdf


Contents









Biaxial loading path



Shear



From the lab to real world (1)


Tension curve

ε(mm/mm)

σ(M

Pa)

0.080.070.060.050.040.030.020.010

600

500

400

300

200

100

0





Tension curve

connueCourbe de traction

ε(mm/mm)

σ(M

Pa)

0.080.070.060.050.040.030.020.010

600

500

400

300

200

100

0

In most of the cases, the material is characterizedby a simple tension curve





Tension curve


ε(mm/mm)

σ(M

Pa)

0.080.070.060.050.040.030.020.010

600

500

400

300

200

100

0

212 53 10753 88 32

107 32 316

How can wetranspose ?

312312332211

Chargement reel complexe

t (s)

σ(M

Pa

)

6050403020100

400

350

300

250

200

150

100

50

0





Tension curve


ε(mm/mm)

σ(M

Pa)

0.080.070.060.050.040.030.020.010

600

500

400

300

200

100

0

212 53 10753 88 32

107 32 316


312312332211


t (s)

σ(M

Pa

)

6050403020100

400

350

300

250

200

150

100

50

0





Tension curve


ε(mm/mm)

σ(M

Pa)

0.080.070.060.050.040.030.020.010

600

500

400

300

200

100

0

212 53 10753 88 32

107 32 316


312312332211


t (s)

σ(M

Pa

)

6050403020100

400

350

300

250

200

150

100

50

0



How can one characterize multiaxial behaviour ?

Multiaxial mechanical tests

Research on the physical deformationmechanisms



Tension–torsion tests on tubes

Tension–torsion specimen

For a tube of length L, diameter2R and width e :

Strain measured by a gauge,or use of the relation betweenthe angle (β) and the strain(γ) :

β = γLR

Relation between themoment (M) and the shear(τ) :

M = 2πeR2τ0 0 0

0 0 σθz

0 σθz σzz

(rθz)



Biaxial testsBiaxial test on a cruciform specimen vinylester–glas fiber

More on the websiteSciences de l’Ingénieur, ENS Cachan


http://www.si.ens-cachan.fr/accueil_V2.php?page=affiche_ressource&id=131&page2=annexe&numannexe=4


Shear test

Double Arcanshear specimen

(rubber)

Doc. Centre des Matériaux, MINES ParisTech



Shear setup



Search of the yield surface in tension–shear

PhD Rousset, ENS Cachan



Initial surf. and after the first compression




Initial surface and square-shape loading




Shear on basalt



Summary of the experimental observations

Crystalline material, where deformation comes from shear (alloys, rocks)

Crystal network No volume change

Powders, geomaterials, damaged materials
















Critical variable ?







Critical variable ?



ShearDeviator





Critical variable ?



ShearDeviator


Deviator+ spherical

part



Slip systems in a monocrystal

PhD F. Hanriot (ENSMP-CDM, Evry)



Slip systems in a polycrystal

Clavel (ECP, Châtenay)



Rupture under dynamic loading



Schmid law

The deformation comes from slip on systems s defined by a plane ofnormal ns, and a shear direction ls, iif the resolved shear stress, τs

reaches a critical value τc

Projection of the stress vector on the slip direction. For a sungle crystalsubmitted to σ∼

τs = (σ∼ .n

s).ls

There is as many criteria linear in stress as the number of slip systems

f (σ∼) = τs− τc



Yield surfaces for polycrystals (Uniformelasticity)

Disrectionally Polycrystalsolidifiedmaterial

Compute yield surfaces


http://mms2.ensmp.fr/mms_paris/critere/animations/animations.php


Loading surfaces in tension–shear

001

σ11

σ 12

2001000-100-200

200

100

0

-100

-200

One cubic grain oriented along (001) axes




234001

σ11

σ 12

2001000-100-200

200

100

0

-100

-200

One grain oriented along (234)




2g234001

σ11

σ 12

2001000-100-200

200

100

0

-100

-200

One grain (001) and one grain (234)




10g2g

234001

σ11

σ 12

2001000-100-200

200

100

0

-100

-200

Ten randomly oriented grains




100g10g2g

234001

σ11

σ 12

2001000-100-200

200

100

0

-100

-200

Hundred randomly oriented grains




Tresca100g

10g2g

234001

σ11

σ 12

2001000-100-200

200

100

0

-100

-200

σ211 +4σ2

12 = σ2y , Tresca criterion



Contents









Characterization of the maximum shear

Stres tensor in the eigendirections :=

σ1 0 00 σ2 00 0 σ3

Stress vector for a normal n in the plan (x1–x2) (with θ = angle(x1,n) :

Tn = σ1 cos2θ+σ2 sin2

θ =σ1 +σ2

2+

σ1−σ2

2cos2θ

|Tt |=(T 2−T 2

n

)1/2=|σ1−σ2|

2sin2θ

Mohr circles : (Tn− σ1 +σ2

2

)2

+T 2t =

(σ1−σ2

2

)2

Max shear

|T maxt |= |σ1−σ2|

2



Tresca criterion

σ1σ2σ3Tn

Tt

Tmax

The maximum shear remains smaller than a critical value

Maxi,j |σi −σj |−σy = 0

σy is the elastic limit in tension

→WIKI


http://fr.wikipedia.org/wiki/Henri_Tresca


Representation of an isotropic material

- Invariants of the stress tensor :

I1 = trace(σ∼) =σii

I2 =(1/2) trace(σ∼)2 =(1/2)σijσji

I3 =(1/3) trace(σ∼)3 =(1/3)σijσjk σki

- Invariants of the deviator (s∼ = σ∼− (I1/3) I∼) :

J1 = trace(s∼) =0

J2 =(1/2) trace(s∼)2 =(1/2)sijsji

J3 =(1/3) trace(s∼)3 =(1/3)sijsjk ski

- One notes :

J = ((3/2)sijsji)0,5 =

((1/2)

((σ1−σ2)

2 +(σ2−σ3)2 +(σ3−σ1)

2))0,5= |σ|



Physical meaning of J

Sphere in the space of the deviatoric stresses

Octahedral shear stress :one a facet of normal (1,1,1), the stres vector has the followingcomponents : normal stress σoct and tangential stress τoct :

σoct = (1/3) I1 ; τoct = (√

2/3)J

The elastic distorsional energy (associated to the deviatoric part of σ∼and ε∼).

Wed =12

s∼ : e∼ =16µ

J2

Von Mises criterionf (σ∼) = J−σy

Note : formulated by Maxwell in 1865, and Huber in 1904 (→WIKI)


http://en.wikipedia.org/wiki/Von_Mises_yield_criterion


Von Mises contour in the deviatoric plane

CS

CI

TS

σ1

CS

TSσ2

CS

TS

σ3

TS stands for the points that are equiva-lent to simple tension, CS those that areequivalent to simple compression (forinstance a biaxial load, since a stressstate like σ1 = σ2 = σ is equivalent toσ3 =−σ), CI corresponds to shear

f (σ∼) = J−σy



Criteria without hydrostatic pressure

Von Mises criterion

f (σ∼) = J−σy

Tresca criterion

f (σ∼) = Maxi,j |σi −σj |−σy

Use of the second and third invariant

f (σ∼) = fct(J2,J3)



Comparaison Tresca–von Mises

In the tension–shear plane

− von Mises : f (σ,τ) =(σ

2 +3τ2)0,5−σy

−Tresca : f (σ,τ) =(σ

2 +4τ2)0,5−σy

In the plane of eigenstresses (σ1,σ2)

− von Mises : f (σ1,σ2) =(σ

21 +σ

22−σ1σ2

)0,5− σy

− Tresca : f (σ1,σ2) = σ2−σy if 0 6 σ1 6 σ2

f (σ1,σ2) = σ1−σy if 0 6 σ2 6 σ1

f (σ1,σ2) = σ1−σ2−σy if σ2 6 0 6 σ1

(symmetry with respect to axis σ1 = σ2)

In the deviatoric plane, von Mises = circle, Tresca = hexagon

Int the eigenstress space, cylindres of axis (1,1,1)



Comparisons Tresca–von Mises

σ12

σ11

τt

τm

σyσy

τm

τt-

-

-

a. In tension–shear (von Mises :τm = σy/

√3, Tresca : τt = σy/2)

σ1

σ2

σy

σy

σy

σy-

-

b. In biaxial tension



Anisotropic criteria

f (σ∼) = ((3/2)Hijkl sij skl)0,5−σy (or Hijkl σij σkl)

Hill’s criterion

In the orthotropy axes :

f (σ∼) =(F(σ11−σ22)2 +G(σ22−σ33)

2 +H(σ33−σ11)2

+2Lσ212 +2Mσ

223 +2Nσ

213)

0,5−σy

Transverse, 3 independent coefficients

Cubic symmetry cubique, one coefficient only



Contents









Drucker–Prager criterion

Linear combination of the first and second invariant (with 0 < α < 0.5)

f (σ∼) = (1−α)J +α I1−σy

Elastic yield in tension (σt ) and in compression (σc)

σt = σy σc =−σy/(1−2α)

σ1

2

3

σ

σ

In the eigenstress space

I1

J

σy

1−α

σy/α

In the plane I1− J



Mohr–Coulomb criterion

Combination of the tangential and normal stresses in the Mohr plane

|Tt |<− tan(φ)Tn +C

Could also be expressed as the combination of the sum and thedifference of the extremal stresses (σ3 6 σ2 6 σ1)

f (σ∼) = σ1−σ3 +(σ1 +σ3)sinφ−2C cosφ

f<0

σ 3 σ1

T

Tn

t

C cohesion, φ internal friction ofthe material

If C is zero and φ non zero,powder material

If φ is zero and C non zero,coherent material



Representation of Mohr-Coulomb’s criterion

σ

σσ

1

2

3

In the deviatoric plane, one et a regularhexagon

TS = 2√

6(C cosφ−p sinφ)/(3+ sinφ)

CS = 2√

6(−C cosφ+p sinφ)/(3−sinφ)

As a function of Kp and of the elasticitylimit in compression, Rp :

f (σ∼) = Kp σ1−σ3−Rp

Kp =1+ sinφ

1− sinφ= tan2

(π

4+

φ

2

)Rp =−2 cosφC

1− sinφ



Closed criteriaThe material cannot be infinitely strong in compression

Cap model, closes by one ellipse Drucker–Prager’s criterion

Cam–clay model has its limit curve defined by two ellipses in the plane(I1− J)

−I1

J Criticalline



Contents









Criteria, synthesis

The boundary of the initial elasticity domain is defined by a function fromthe stress space in R, that can be

Piecewise linear (Schmid, Tresca)Quadratic, or more

The elastic domain is convex

The criter can depend or not from the hydrostatic pressure


Documents

Georges Cailletaud Centre des Matériaux MINES …mms2.ensmp.fr/msi_paris/transparents/Georges_Cailletaud/...Isotropic : Tresca, von Mises An anisotropic criterion : Hill 5 Pressure