F7a: Deformation and failure phenomena: Elasticity ...ragnar/Volvo_course_home/lectures/Notes_F7.… · Some phenomena and models listed above will be considered in the course! Questions

F7. Characteristic behavior of solids

F7a: Deformation and failure phenomena: Elasticity, inelasticity, creep, fatigue.

à Choice of constitutive model: Issues to be considered

è Relevance ?

Physical effect of interest (deform., life time ...)

è Actual physical conditions ?

Temperature dependence

e.g. Enhanced creep in metals Reduced yield stressIncreased material ductility

Dependence on loading rate (strain rate):

increased strengthreduced ductility

è Accuracy ?

Application ? (building structure, microsystem component)

è Computational aspects - complexity, reliability, costs

Hand calculation, commercial code ?

› Material behavior is essentially determined by material microstructure

› Material behavior is represented by a constitutive model under given conditions - "a constitutive model is just a model".

à The constitutive problem

ü Different purposes and relevant models

è Consider some concepts intuitively: elasticity, viscoelasticity, plasticity, viscoplasicity ...

Elasticity (reversible, time independent)�ex. Hooke’s law, hyperelasticity: Neo-Hooke, Money-Rivlin�material: metals (small deformations), rubber

Figure 1

Viscoelasticity (irreversible, time dependent)�ex. Maxwell, Kelvin, Norton�material: polymers, second-ary creep in metals�

Figure 2

è Structural analysis under working load: Linear elasticity

è Analysis of damped vibrations: Viscoelasticity

è Calculation of limit load: Perfect plasticity

è Accurate calculation of permanent deformation after monotonic cyclic loading: Hardening elasto-plasticity

è Analysis of stationary creep and relaxation: Perfect elastoviscoplasticity

è Prediction of lifetime in high-cycle-fatigue: Damage coupled to elastic deformations

è Prediction of lifetime in high-cycle-fatigue: Damage coupled to plastic deformations

è Prediction of lifetime in creep and creep fatigue: Damage coupled to viscoplastic deformations

è Prediction of stability of a preexisting crack: Linear elasticity (singular stress field determined from sharpcracks)

è Prediction of strain localization in shear bands and incipient material failure: Softening plasticity (ordamage coupled to plastic deformation)

ü Basic question

Some phenomena and models listed above will be considered in the course!

Questions that should posed in regard to different models are:

- Is the model relevant for the current physical problem?

- Does the model produce sufficiently accurate predictions for the given purpose ?

- Is it possible to implement a robust numerical algorithm to obtain a truly operational algorithm ?

2 Notes_F7.nb

ü Approaches to constitutive modeling

-Phenomenological approach (considered here!)

Macroscopic (phenomenological) modeling: �

Figure 3

- Micro-structural processes represented as mean values of internal variables like plastic strain, damage.

- Constitutive equations based on macroscopic experiments.

Continuum idealization of stress, strain, etc,Assumed homogeneous elementary testsNote ! microstructure processes represented by “internal” continuum variables

-Micromechanics (fundamental) approach

Control volume on micro structural scalee.g. steel (grains) 10－6 － 10－4 me.g. concrete stones 10－2 mmicromechanics considerations via homogenization Ø macroscopical relation

Micromechanical modeling:

Figure 4

- Representative volume of microstructure modeled in detail by mechanical models (e.g. crystal-plastic-ity).

- Homogenization provides link to macroscopic level.�- Computationally demanding

-Statistical approach

Variation of size, shape etc. specimen for “same”stress and strainMathematical distribution of strength

ü "Typical" material behavior (metals and alloys)

Consider 1) Monotonic loading

Notes_F7.nb 3

Creep and relaxation

- Temperature dependence- Identifiable stages with time fl

Strain rate dependence

- Static (slow) loading- Dynamic (rapid) loading fl “Higher stiffness and strength for larger loading rate”!!

Figure 5

Consider 2) Cyclic loading

Cyclic loading and High--Cycle--Fatigue (HCF)

- Elastic deformation (macroscopically) degradation of elasticity close to failure(Note! “weak” theoretical basis at present)

Cyclic loading and Low--Cycle--Fatigue (LCF)

Consider 1D test:

Figure 6

Response in loading-unloading

4 Notes_F7.nb

Figure 7

Response in loading-unloading-loading-unloading (cyclic behavior)

Figure 8

Test modes: εa = const. or σa = const.

- Plastic deformation in each cycle fl "hysteresis loops"

Isotropic hardening

Figure 9

Kinematic hardening

Notes_F7.nb 5

Figure 10

- Stages of fatigue process, stress control fl

fl Shakedown (=stabilized cyclic curve) or ratcheting behavior

Figure 11

ü Characteristics of material "fluidity"

Creep: ε ≠ 0 when σ = 0

Figure 12

Relaxation: σ ≠ 0 when ε = 0

6 Notes_F7.nb

Figure 13

Stages of creep process (pertinent to stress controlled process):

- Transient (primary) ε decreasing

- Stationary (secondary) ε ∽constant

- Creep failure (tertiary) ε Ø ¶ when t → tR

Figure 14

F7b. Linear viscoelasticity

Duggafrågor1) For a static uniaxial bar problem at isothermal (θ = const. ) conditions, state principle of energyconservation (first law of thermodynamics). On the basis of this relationship, derive the relation e = σ ε ,where e is the internal energy, for the considered bar problem.

Notes_F7.nb 7

2) For the same bar problem as in T3, state the dissipation inequality ≥ 0 . In this context, state theexpression for a stored elastic (or free) energy ψ@ε, κD in the case of the linear visco-elastic Maxwellmodel. Derive also the expression for the uniaxial stress σ and the micro-stress K in the dash-pot. Inaddition, show that the dissipation is positive for the (linear) evolution of the visco-plastic strain.

3) On the basis of the established visco-plastic model derive the model behavior during the step load-ings: Creep behavior: σ@tD = σ0 H@tD and relaxation behavior ε@tD = ε0 H@tD . Assume zero temper-ature change: θ = 0 ⇒ εi = αθ = 0

à Constitutive relations: Introduction to thermodynamic basis

ü Uniaxial bar problem

Assume: 1) Isothermal (θ = const.) process of uniaxial bar!2) Dissipative material: function of strain ∈ (observable) and internal (hidden) variable κ (repr. irreversible processes in microstructure)

Figure 15

ü Consider thermodynamic relations

a) Principle of energy conservation

E + K = W H+QLE = ‡

Le x = Internal energy

K =12 ‡

Lρ u2 x = Kinetic energy = 0 in static case!

W = ‡LU u x + @σ uDx=0

x=L = Mechanical workHQ = Heat supply not consideredLNote! for considered barHσ uL' = σ' u + σ u' = −U u + σ ε ⇒ U u = σ ε − Hσ uL' ⇒

∴ W = ‡LHσ ε − Hσ uL'L x + @σ uDx=0

x=L = ‡Lσ ε x ⇒

Energy equation for static case:

8 Notes_F7.nb

E = W ⇒ ∴ e = σ ε

Note!

e = ‡ε0

ε

σ ε = actual strain energy experienced by material

b) Dissipation (or entropy) inequality

Introduce stored elastic energy: ψ@ε, κD = Helmholtz free elastic energy.Assume strain energy: e = e@εDDef.: Consider dissipation from the inequality

‡L

x = ‡LHe − ψL x ≥ 0 ≥ 0 pointwise

Express entropy inequality; consider dissiplation

:= e − ψ = σ ε −∂ψ∂ε

ε −∂ψ∂κ

κ ≥ 0 ⇒

∴ ≥ 0 σ =∂ψ∂ε

and = K κ ≥ 0 with K = −∂ψ∂κ

ü Dissipative materials - classes

1) Non-dissipative elastic material: =def. 0 fl

ψ = ψ@εD ⇒ = K κ := 0 ⇒ K, κ irrelevant!

e.g. linear thermo-elasticity with

ψ@ε; θD =12 E HεeL2 =

12 E Hε − α θL2 ⇒ σ = E Hε − α θL

2) Viscous dissipative material: ≥ 0 such that K depends on state as

K = f@ε, κ, κDe.g. visco-elasticity or visco-plasticty

ψ@ε; θD =12 E Hε − κ − α θL2 ⇒ σ = E Hε − κ − α θL; K = −

∂ψ∂κ

= σ ⇒

K = σ = µ κ ⇒ =1µ

σ2 ≥ 0 OK!

3) Inviscid (rate-independent) dissipative material: ≥ 0 so that K depends

K = f@ε, κD ε

e.g. elasto-plasticity.

Note! rate-insensitivity; s=another time scale !

ε =dεdt =

dεds

dsdt =

dεds s ⇒ K =

dKds s = f@ε, κD

dεds s ⇒

K = f@ε, κD εdKds = f@ε, κD

dεds

Notes_F7.nb 9

à Prototype model of viscoelastic material - Maxwell model

ü Model definition

Reological model:

Figure 16

Free energy:

ψ@ε, κ; θD =12 E Hε − κ − α θL2 ⇒ σ = K = E Hε − κ − α θL ⇒ K = −

∂ψ∂κ

= σ

Dissipation

= K κ = 9Assume : κ =Kµ, K = σ= =

1µ

σ2 ≥ 0 OK!

fl Evolution of deformation in damper:

κ =1µ

σ, µ > 0 viscosity parameterNote, relaxation time t∗ (used instead of µ ):

t∗ =def. µE ⇒ κ =

1t∗

σE =

1t∗ εe

ü Model behavior for step loading

Assume zero temperature change: θ = 0 ⇒ εi = αθ = 0Creep behavior: σ@tD = σ0 H@tD

Figure 17

Consider evolution rule, t ≥ 0 :

10 Notes_F7.nb

κ =1µ

σ0 , IC : κ@0D = 0Introduce relaxation time: t∗ =def. µ

EEstablish soltution by direct integration as

κ =1µ

σ0 ⇒ κ@tD = C@1D +σ0µ

t = C@1D +σ0E

tt∗

Inititial condition:

κ@0D = 0 ⇒ C@1D = 0Establish solution

σ0 = E Hε − κ@tDL ⇒ ε =σ0 Ht + t∗L

E t∗ = ε0 I tt∗ + 1M

Note! Initially at t = 0 , we get ε@0D = ε0 = σ0E fl Only the spring is activated! (No deformation in

dash-pot)

0.5 1 1.5 2 time t

0.5

1

1.5

2

2.5

3ε@tD×

σ0EStep−loading: creep, increasing t∗

Analysis: Creep behavior: σ@tD = σ0 H@tDRelaxation behavior: strain driven step ε@tD = ε0 H@tD

Figure 18

Consider evolution rule for damper

κ =1µ

σ, t ≥ 0, κ@0D = 0

Notes_F7.nb 11

Establish solution as:

κ@tD =1µ

σ with σ → E Hε0 − κ@tDL ⇒ κ@tD → − tt∗ C@1D + ε0

Establish solution with intitial value κ@0D = 0 and µ → E t∗

κ@tD = H1 − − tt∗ L ε0

σ = E Hε0 − κ@tDL = E ε0 − tt∗

0.5 1 1.5 2 time t

0.2

0.4

0.6

0.8

1

ε@tD×Eε0Step−loading: relaxation, increasing t∗

Analysis: Relaxation behavior: strain driven step ε@tD = ε0 H@tDà Linear (generalized) standard model

Consider generalization in reological model

Figure 19

Free energy:

ψ@ε, κ; θD =12 ‚

α=1

N

Eα Hε − κα − αθL2 ⇒

σ = ‚α=1

N

Eα Hε − κα − αθL = E∞ Hε − αθL − ‚α=1

N

Eα κα with E∞ = ‚α=1

N

Eα

Evolution of deformation in damper:

12 Notes_F7.nb

κα =1µα

Kα, Kα = Eα Hε − κα − αθLDissipation

= ‚α=1

N

Kα κα = ‚α=1

N 1µα

Kα2 ≥ 0 OK!

ü Three parameter model: Linear solid model

Figure 20

Consider special case: α = 2 where

σ = E∞ Hε − αθL − E1 κ1

κ1 = κ =1µ

K, K = E1 Hε − αθ − κL, µ = E1 t∗

κ2 := 0Relaxation behavior fl strain driven step ε@tD = ε0 H@tDwith α θ := 0

κ =1µ

K with K → E1 Hε − κL ⇒ κ@tD = ε0 + − tt∗ C@1D

Initial condition κ@0D = 0 fl

κ@tD = ε0 I1 − − tt∗ M ⇒ σ = E∞ ε0 − E1 κ@tD =. .. = E∞ J 1 + I − t

t∗ − 1M E1E∞

N ε0Note! κ@0D = 0 fl

σ = E¯@tD ε0, E¯@tD = E∞ J 1 + I − tt∗ − 1M E1

E∞N

Note: Maxwell E2 → 0 , E1 = E∞ → E fl E¯@tD = E − tt∗ (cf. previous case)

Notes_F7.nb 13

Figure 21

2000 4000 6000 8000 10000time t

0.2

0.4

0.6

0.8

1

σ@tD×E∞ε0Relaxation: 3−parameter

ü Summary: Linear elasticity

Consider σ = E¯@tD ε0 as linear relation at given times t fl Isochrones

Figure 22

0.2 0.4 0.6 0.8 1ε0E∞

0.2

0.4

0.6

0.8

1σ@tD Isocrones : Solid behavior

Figure 23

0.2 0.4 0.6 0.8 1ε0E∞

0.2

0.4

0.6

0.8

1σ@tDIsocrones : Fluid behavior HMaxwellL

Remove@"`∗"D

14 Notes_F7.nb