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Plan Mechanical tests Rheological models Plasticity criterion 3D plasticity

Rheology

Georges Cailletaud

Centre des MatériauxMINES ParisTech/CNRS

WEMESURF contact course, Paris 21-25 juin

Georges Cailletaud | Rheology 1/80


Contents

1 Mechanical testsTests on everythingResults on material elements

2 Rheological modelsBuilding bricksPlasticity

3 Plasticity criterionMultiaxial stress statesPlasticity mechanismsModels insensitive to hydrostatic pressureModels sensitive to hydrostatic pressureSynthesis on the criteria

4 3D plasticityThe ingredients of a 3D plasticity modelPlastic flowEvaluation of the flow directionsPerfectly plastic model



Contents







Tests on a civil plane

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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



Contents







Typical result on an aluminium 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

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, Chllon



Push–pull test on an aluminium 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 .



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



Contents







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)



Contents







Biaxial loading path



Shear



Tension–torsion tests on tubes

Tension–torsion specimen

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

Strain measured by a straingauge, or use of the relationbetween the angle (β) andthe 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 shear Arcan(rubber) specimen

Doc. Centre des Matériaux, MINES ParisTech



Search of the yield surface in tension–shear

PhD Rousset, ENS Cachan



Yield surfaceInitial and after the first compression

PhD Rousset, ENS Cachan



Shear on basalt



Contents







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 single crystal

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, iff the resolved shear stress, τs

reaches a critical value τc

Projection of the stress vector on the slip direction. For a single 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



Polycrystalline aggregate yield surfacesAssuming uniform elasticity

Directionally Polycrystalsolidifiedmaterial

Compute yield surfaces


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


Yield 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

Stress 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


Model ingredients for 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 :on a facet of normal (1,1,1), the normal component of the stress vectoris σoct and the tangential component is 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 criterion 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



Tresca versus 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

In the eigenstress space, cylinders of axis (1,1,1)



Comparisons of Tresca and von Mises criteria

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



An anisotropic criterion : Hill

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, 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 convexThe criter can depend or not from the hydrostatic pressure



Contents







From elastic to plastic behaviour

σ11

σ12

σ11

σ22

Tresca von Mises




σ11

σ12

σ11

σ22

von MisesElastic domain

Von Mises criterion




σ11

σ12

σ11

σ22

Elastic regimevon Mises

Elastic domainVon Mises criterion




σ11

σ12

σ11

σ22

Plasticityvon Mises

Elastic domainVon Mises criterion



3D plasticity

Strain decomposition

ε∼ = ε∼e + ε∼

th + ε∼p + ε∼

vp

Criteriondefined by the function f (σ∼ ,AI)

Flow rulesε∼

p = ...

Hardening rulesAI = ... later

? Today, model without hardening, e.g. :

f (σ∼) = J(σ∼)−σy



Contents







Principle of maximal power

Formulation due to Hill (1951) :The power of the real stress tensor σ∼ associated to realviscoplastic strain rate ε∼

p is larger than the power computed withany other admissible stress tensor σ∼

∗ (id est that does notovercome f = 0) associated with ε∼

p

(σ∼−σ∼∗) : ε∼

p > 0

The form F is generated to combine the function that must bemaximized and the constraint by means of a plastic multiplyier(non linear optimization problem)

F(σ∼) = σ∼ : ε∼p− λ f (σ∼)

Its derivative with respect to σ∼ must be zero to reach anextremum. The extremum is a max iff the function f is convex

ε∼p = λ

∂f∂σ∼







p

(σ∼−σ∼∗) : ε∼

p > 0

The solution of the plastic flow problem maximizes the plasticpower σ∼ : ε∼

p


F(σ∼) = σ∼ : ε∼p− λ f (σ∼)


ε∼p = λ

∂f∂σ∼







p

(σ∼−σ∼∗) : ε∼

p > 0

The solution of the plastic flow problem maximizes the plasticpower σ∼ : ε∼

p

One has to develop a maximization under constraint, since onemust have also f (σ∼) 6 0


F(σ∼) = σ∼ : ε∼p− λ f (σ∼)


ε∼p = λ

∂f∂σ∼







p

(σ∼−σ∼∗) : ε∼

p > 0


F(σ∼) = σ∼ : ε∼p− λ f (σ∼)


ε∼p = λ

∂f∂σ∼



Geometrical aspects of Hill’s principle

(σ∼−σ∼∗) : ε∼

p > 0

σ∼∗ on the yield surface, σ∼ in the domain, ε∼

p = 0∼Normality rule, by working in the tangent plane, (σ∼ on the surface),

k t∼∗ : ε∼

p > 0 and − k t∼∗ : ε∼

p > 0

so that t∼∗ : ε∼

p = 0

Sign of the multiplyier, by working on the interior normal, (σ∼ on thesurface), (σ∼−σ∼

∗) = k n∼ colinear to n∼ (k>0), and :

k n∼ : λn∼ > 0 thus λ > 0



Convexity of the yield surface

.ε. p

f<0σ

n

~

~

~

a. Illustration of the normality rule

*

~

~~σ n

σ

b. Convexity of f

Hill’s principle≡ (f convex and normality of the flow)



Contents







NOTE : computation of ∂J/∂σ∼

One will have to express∂J∂σ∼

to compute n∼The derivative s∼ with respect to σ∼ , writes with index notation :

Jijkl =12

(δik δjl + δilδjk )− 13

δijδkl

since :s∼ = J

≈: σ∼

Derivative of J with respect to σ∼ :

∂J∂σ∼

=∂J∂s∼

:∂s∼∂σ∼

=∂((3/2)s∼ : s∼)1/2

∂s∼: J≈

=32

s∼J

: J≈

=32

s∼J

Other solution :

J2 =32

sijsij thus 2J dJ = 3sij dsij = 3sij dσij

∂J∂σij

=32

sij

J∂J∂σ∼

=3s∼2J



Flow direction ssociated to von Mises’ criterion

Perfectly plastic model f (σ∼) = J(σ∼)−σy

εpij = λnij ε∼

p = λn∼

nij =∂J∂σij

=32

sij

Jn∼ =

∂J∂σ∼

=32

s∼J

One denotes by accumulated plastic strain, p, the length of the pathrepresenting plastic flow in the plastic strain space. At time t ,

p(t) =Z t

0p(τ)dτ p =

(23

εpij ε

pij

)1/2

=

(23

ε∼p : ε∼

p)1/2

With von Mises’s criterion

p =

(23

λ32

s∼J

: λ32

s∼J

)1/2

= λ



Von Mises’ model

Only one stress component is non zero

σ∼ =

σ 0 00 0 00 0 0

s∼ =2σ

3

1 0 00 −1/2 00 0 −1/2

Computation of the normal, with J = |σ|

n∼ =32

s∼J

=

1 0 00 −1/2 00 0 −1/2

sign(σ)

Components of the plastic strain rate

εp11 = λsign(σ) = p sign(σ) ε

p22 = ε

p33 =−ε

p11



Flow direction associated to Tresca’s criterion

Si σ1 > σ2 > σ3 : f (σ∼) = σ1−σ3−σy , there is a plastic flow in pureshear, with ε∼

p22 = 0

ε∼p = λ

1 0 00 0 00 0 −1

For simple tension , for example with σ1 > σ2 = σ3 = 0 :

f (σ∼) = σ1−σ2−σy or f (σ∼) = σ1−σ3−σy

ε∼p = λ

1 0 00 0 00 0 −1

+ µ

1 0 00 −1 00 0 0



Flow direction associated to Drucker–Pragercriterion

Expression of the criterion

f (σ∼) = (1−α)J(σ∼) + α trace(σ∼)−σy

The normal contains the identity tensor

n∼ =32

(1−α)s∼J

+ α I∼

Volume increase for any applied load :

trace(ε∼p) = λ trace n∼ = 3αλ



Contents







Perfectly plastic model

Flow to determine from f (σ∼) = 0 and f (σ∼) = 0

ε∼p = λ

∂f∂σ∼

= λn∼

Consistency condition written on f (σ∼) = J−σy

f =∂f∂σ∼

: σ∼ = n∼ : σ∼ = 0

During plastic flow, the current stress point can only stay around theyield surface. The plastic multiplyier cannot be determined in terms ofstress rate.



Perfectly plastic behaviour, strain control

Elasticityσ∼ = Λ

≈: (ε∼− ε∼

p) and n∼ : σ∼ = 0

Projection on the normal to the surface

n∼ : σ∼ = n∼ : Λ≈

: (ε∼− ε∼p) = n∼ : Λ

≈: ε∼−n∼ : Λ

≈: λn∼

Expression of the plastic multiplyier as a function of the strain rate

λ =n∼ : Λ

≈: ε∼

n∼ : Λ≈

: n∼Expression of the plastic strain rate

ε∼p = λn∼ =

n∼ : Λ≈

: ε∼n∼ : Λ

≈: n∼

n∼



The elastoplastic tensor L≈

ep

Stress rate σ∼ = Λ≈

: (ε∼− ε∼p)

By replacing the plastic strain rate

σ∼ = Λ≈

: ε∼−Λ≈

:

(n∼ : Λ

≈: ε∼

n∼ : Λ≈

: n∼n∼

)Since ΛijklnpqΛpqrs εrsnkl = ΛijklnklnpqΛpqrs εrs, it comes σ∼ = L

≈ep ε∼, with

L≈

ep = Λ≈−

(Λ≈

: n∼)⊗ (n∼ : Λ≈

)

n∼ : Λ≈

: n∼

Formally, this is an elastic type rule, but expressed in terms of rate



Isotropic elasticity and of von Mises’ criterion

Possible simplifications

Λijkl = λδijδkl + µ(δik δjl + δilδjk ) ; nij =32

sij

J

nij Λijkl = 2µnkl ; nij Λijkl nkl = 3µ ; nij Λijkl εkl = 2µnkl εkl

λ =23

n∼ : ε∼

Case of the onedimensional loading : during plastic flow, the plasticstrain rate is equal to the total strain rate, the normal is then the diagonal(1,−1/2,−1/2)sign(σ∼ ) . The total strain rate is the diagonal(ε

p11,−ε

p11/2,−ε

p11/2

), so that one get :

λ = εp11 sign(σ)


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