F7. Characteristic behavior of solids

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 stress Increased material ductility Dependence on loading rate (strain rate): increased strength reduced 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- Rivlinmaterial: metals (small deformations), rubber

2 Notes_F7.nb Figure Viscoelasticity (irreversible, time dependent)ex. Maxwell, Kelvin, Nortonmaterial: polymers, secondary 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 sharp cracks) è Prediction of strain localization in shear bands and incipient material failure: Softening plasticity (or damage 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?

Notes_F7.nb 3 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 tests Note! microstructure processes represented by internal continuum variables -Micromechanics (fundamental) approach Control volume on micro structural scale e.g. steel (grains) 0-6 - 0-4 m e.g. concrete stones 0-2 m micromechanics considerations via homogenization Ø macroscopical relation Micromechanical modeling: Figure 4 - Representative volume of microstructure modeled in detail by mechanical models (e.g. crystal-plasticity). - Homogenization provides link to macroscopic level.- Computationally demanding -Statistical approach Variation of size, shape etc. specimen for same stress and strain Mathematical distribution of strength ü "Typical" material behavior (metals and alloys) Consider ) Monotonic loading

4 Notes_F7.nb 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 D test: Response in loading-unloading Figure 6

Notes_F7.nb 5 Figure 7 Response in loading-unloading-loading-unloading (cyclic behavior) Figure 8 Test modes: ε a = const. or σ a = const. - Plastic deformation in each cycle fl "hysteresis loops" Isotropic hardening Kinematic hardening Figure 9

6 Notes_F7.nb Figure 0 - Stages of fatigue process, stress control fl fl Shakedown (=stabilized cyclic curve) or ratcheting behavior Figure ü Characteristics of material "fluidity" Creep: ε 0 when σ = 0 Relaxation: σ 0 when ε = 0 Figure 2

Notes_F7.nb 7 Figure 3 Stages of creep process (pertinent to stress controlled process): - Transient (primary) ε decreasing - Stationary (secondary) ε constant - Creep failure (tertiary) ε Ø when t t R Figure 4 F7b. Linear viscoelasticity Duggafrågor ) For a static uniaxial bar problem at isothermal (θ =const.) conditions, state principle of energy conservation (first law of thermodynamics). On the basis of this relationship, derive the relation e =σε, where e is the internal energy, for the considered bar problem.

8 Notes_F7.nb 2) For the same bar problem as in T3, state the dissipation inequality 0. In this context, state the expression for a stored elastic (or free) energy ψ@ε, D in the case of the linear visco-elastic Maxwell model. Derive also the expression for the uniaxial stress σ and the micro-stress K in the dash-pot. In addition, 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 loadings: Creep behavior: σ@td =σ 0 H@tD and relaxation behavior ε@td =ε 0 H@tD. Assume zero temperature change: θ =0 ε i =αθ=0 à Constitutive relations: Introduction to thermodynamic basis ü Uniaxial bar problem Assume: ) Isothermal (θ =const.) process of uniaxial bar! 2) Dissipative material: function of strain (observable) and internal (hidden) variable (repr. irreversible processes in microstructure) Figure 5 ü Consider thermodynamic relations a) Principle of energy conservation E + K = W H+Q L E = e x = Internal energy L K = 2 ρ u 2 x = Kinetic energy = 0 in static case! L W = Uu x + @σ u D x=l x=0 = Mechanical work L HQ = Heat supply not consideredl Note! for considered bar Hσ u L' =σ' u +σ u ' = U u +σ ε U u =σ ε Hσ u L' W = Hσ ε Hσ u L'L x + @σ u D x=l x=0 = σ ε x L L Energy equation for static case:

Notes_F7.nb 9 Note! E = W e =σ ε e = ε ε 0 σ ε = actual strain energy experienced by material b) Dissipation (or entropy) inequality Introduce stored elastic energy: ψ@ε, D = Helmholtz free elastic energy. Assume strain energy: e = e@εd Def.: Consider dissipation from the inequality x = He ψ L x 0 0 pointwise L L Express entropy inequality; consider dissiplation := e ψ =σε ψ ε ε ψ 0 0 σ = ψ ε and = K 0 with K = ψ ü Dissipative materials - classes ) Non-dissipative elastic material: def. = 0 fl ψ=ψ@εd = K := 0 K, irrelevant! e.g. linear thermo-elasticity with ψ@ε; θd = 2 E Hεe L 2 = 2 E Hε αθl2 σ=ehε αθl 2) Viscous dissipative material: 0 such that K depends on state as K = f@ε,, D e.g. visco-elasticity or visco-plasticty ψ@ε; θd = 2 E Hε αθl2 σ=ehε αθl; K = ψ =σ K =σ=µ = µ σ2 0OK! 3) Inviscid (rate-independent) dissipative material: 0 so that K depends K = f@ε, D ε e.g. elasto-plasticity. Note! rate-insensitivity; s=another time scale! ε dε = dt = dε ds ds dt = dε ds s K = dk ds s = f@ε, D dε ds s K = f@ε, D ε dk ds = f@ε, D dε ds

0 Notes_F7.nb à Prototype model of viscoelastic material - Maxwell model ü Model definition Reological model: Free energy: Figure 6 ψ@ε, ; θd = 2 E Hε αθl2 σ=k = E Hε αθl K = ψ =σ Dissipation = K = 9Assume : = K µ, K =σ= = µ σ2 0OK! fl Evolution of deformation in damper: = σ, µ>0 viscosity parameter µ Note, relaxation time t (used instead of µ): t = def. µ E = σ t E = t εe ü Model behavior for step loading Assume zero temperature change: θ =0 ε i =αθ=0 Creep behavior: σ@td =σ 0 H@tD Consider evolution rule, t 0: Figure 7

Notes_F7.nb = µ σ 0, IC:@0D = 0 Introduce relaxation time: t = def. µ E Establish soltution by direct integration as = µ σ 0 @td = C@D + σ 0 µ t = C@D + σ 0 E Inititial condition: t t @0D = 0 C@D = 0 Establish solution σ 0 = E Hε @tdl ε= σ 0 Ht + t L Et =ε 0 I t t + M Note! Initially at t = 0, we get ε@0d =ε 0 = σ 0 E fl Only the spring is activated! (No deformation in dash-pot) ε@td σ 0 Step loading: E creep, increasing t 3 2.5 2.5 0.5 0.5.5 2 time t Analysis: Creep behavior: σ@td =σ 0 H@tD Relaxation behavior: strain driven step ε@td =ε 0 H@tD Consider evolution rule for damper Figure 8 = µ σ, t 0, @0D = 0

2 Notes_F7.nb Establish solution as: @td = µ σ with σ E Hε 0 @tdl @td t t C@D +ε 0 Establish solution with intitial value @0D = 0 and µ Et @td = H t t L ε 0 σ= E Hε 0 @tdl = E ε 0 t t ε@td Eε 0 Step loading: relaxation, increasing t 0.8 0.6 0.4 0.2 0.5.5 2 time t Analysis: Relaxation behavior: strain driven step ε@td =ε 0 H@tD à Linear (generalized) standard model Consider generalization in reological model Free energy: Figure 9 N ψ@ε, ; θd = 2 E α Hε α αθl 2 α= N σ= E α Hε α αθl = E Hε αθl α= α= Evolution of deformation in damper: N E α α with E = N α= E α

Notes_F7.nb 3 α = K µ α, K α = E α Hε α αθl α Dissipation N α= N = K α α = K µ 2 α 0OK! α α= ü Three parameter model: Linear solid model Consider special case: α =2 where Figure 20 σ=e Hε αθl E = = µ K, K = E Hε αθ L, µ= E t 2 := 0 Relaxation behavior fl strain driven step ε@td =ε 0 H@tDwith αθ:= 0 = µ K with K E Hε L @td =ε 0 + t Initial condition @0D = 0 fl @td =ε 0 I t t Note! @0D = 0 fl t C@D M σ=e ε 0 E @td =...= E J + I t σ=ē@tdε 0, Ē@tD = E J + I t t M E N E Note: Maxwell E 2 0, E = E E fl Ē@tD = E t t (cf. previous case) t M E N ε0 E

4 Notes_F7.nb σ@td E ε 0 Relaxation: 3 parameter 0.8 0.6 0.4 0.2 2000 4000 6000 8000 0000 time t Figure 2 ü Summary: Linear elasticity Consider σ =Ē@tD ε 0 as linear relation at given times t fl Isochrones σ@td Isocrones : Solid behavior 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 Figure 22 ε 0 E Isocrones σ@td : Fluid behavior HMaxwellL 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 Figure 23 ε 0 E Remove@"` "D