fundamentals of computational thermo-elasticityjpamin/dyd/soki/termoelast.pdf · fundamentals of...

Thermo-elasticityExamples

Fundamentals of computational thermo-elasticity

J. Pamin

in cooperation with

J. Jaśkowiec, P. Pluciński, R. Putanowicz,

A. Stankiewicz, A. Wosatko

Cracow University of Technology, Department of Civil Engineering,Institute for Computational Civil Engineering

5 listopada 2015

J. Pamin & co. (Cracow University of Technology) Fundamentals of computational thermo-elasticity


Overview and assumptions

1 Thermo-elasticityThermodynamic backgroundTheory and algorithms

2 Examples

Assumptions

small displacements and strains

isotropic continuum

linear elasticity

static loading

nonstationary heat transport

coupling due to thermal expansion

selected material parameters temperature-dependent



Thermodynamic backgroundTheory and algorithms

Thermodynamics and mechanics

Thermodynamics provides restrictions on the form of constitutivemodels

State variables (observable and internal), e.g.

(ε, κ, θ) - strain tensor, internal variable vector, temperature, resp.

A thermodynamic system is reversible if thermodynamic potentialsdo not depend on internal variables

Laws of thermodynamics (for elementary material volume)

1 → e = σ : ε+ r −∇q (balance of internal energy)

2 → spro 0 (specific internal entropy production)




Dissipation

Definition of energy dissipation density

D = θspro = Dmech +Dther 0 (local + conductive)

D = σ : ε− (e − θs)− q · ∇θθ 0 (Clausius-Duhem inequality)

Helmholtz free energy ψ(ε, κ, θ)

e − θs = ψ

Dissipation inequality for isothermal conditions

D = σ : ε− ψ 0

ψ = ∂εψ : ε+ ∂κψ · κ

D = (σ − ∂εψ) : ε+K · κ 0

K = −∂κψ (thermodynamic forces)




Material models - isothermal elasticity

Hyperelasticityσ = σ(ε) and D = 0

Selected ψ is path-independent internal stress power W

ψ = W (ε) = σ : ε (stored elastic energy)

Work equals stored elastic energy

W |t1t0 =

∫ t1t0

W (ε)dt = ψ(ε1)− ψ(ε0)

ψ =∂ψ

∂ε: ε

D = σ : ε−∂ψ

∂ε: ε = 0 → σ =

∂ψ

∂ε

σ =∂2ψ

∂ε⊗ ∂ε: ε (tangent stiffness)




Non-isothermal conditions

Dissipation inequality for non-isothermal conditions

D = σ : ε− (ψ + s θ)− q·∇θθ 0

Thermo-elasticity

ψ =∂ψ

∂ε: ε+

∂ψ

∂θθ , s = −

∂ψ

∂θ, D = −q · ∇θ

θ

Thermo-elasto-plasticity

ψ = ψ(εe, κ, θ), εe = ε− εp

ψ =∂ψ

∂εe: εe +

∂ψ

∂κκ+

∂ψ

∂θθ

Dmech = σ : εp + K · κ 0

Dther = −q · ∇θθ 0




Temperature dependence

In principle all materialparameters (isotropy assumed)are temperature-dependent:– density ρ– Young modulus E– Poisson ratio ν– expansion coefficient α– heat conductivity k– heat capacity c

For elastic-plastic materialsadditionally:– yield stress σy– hardening modulus hp

Figures from (Ottosen and Ristinmaa, 2005)




Sources of nonlinearity

Arbitrary choice made for present derivation: E (θ), k(θ)

Hence, nonlinearity is due to temperature-dependence of Young modulusand heat conductivity, and possibly natural boundary conditions for thethermal subproblem (radiation).

For generalization refer e.g. to (Ottosen and Ristinmaa, 2005)




Thermoelasticity

Voigt’s notation:σ = [σx , σy , σz , τxy , τxz , τyz ]

T

ε = [εx , εy , εz , γxy , γxz , γyz ]T

Assumed b, t independent of temperature

Equilibrium

LTσ + b = 0 in V

σn = t on Stu = u on Su

Linear elasticity + thermal expansion (θ = T − T0)

ε = εe + εθ, σ = E(ε− εθ), ε = Lu

εθ = αθΠ, Π = [1 1 1 0 0 0]T, E = E(θ)




Momentum balance

Weak form at t + ∆t (u = u, vu = 0 on Su)∫V

(Lvu)T σ dV =

∫V

vTu b dV +

∫St

vTu t dS ∀vu

σt+∆t = σt + ∆σ

∆σ = limi→∞

∆σi , ∆σi+1 = ∆σi + dσ, σt+∆ti = σt + ∆σi∫

V

(Lvu)T dσ dV = Wt+∆text −Wt+∆t

int,i

Wt+∆tint,i =

∫V

(Lvu)T σt+∆ti dV




Momentum balance

Linearized weak form

dσ = E(dε− dεθ

)+ dE

(ε− εθ

)dE =

dEdθdθ, dεθ = αΠ dθ, sθ =

dEdθ

(ε− αθΠ)− αEΠ

dσ = E dε+ sθ dθ

dε = L du∫V

(Lvu)T E L du dV +

∫V

(Lvu)T sθ dθ dV = Wt+∆text −Wt+∆t

int,i




Heat transport

Assumed ρ, r , q independent of temperature

Heat transportc θ +∇Tq = r in V

qTn = q on Sq

θ = θ on Sθ

Fourier’s lawq = −Λ∇θ, Λ = Λ(θ)




Heat transport

Weak form at t + ∆t (θ = θ, vθ = 0 on Sθ)∫V

vθc θ dV −∫V

(∇vθ)T q dV =

∫V

vθr dV −∫Sq

vθq dS ∀vθ

θt+∆t = θt + ∆θ

Generalized midpoint rule(1− γ)θt + γ θt+∆t =

θt+∆t − θt

∆t

For γ = 1 backward Euler for time integration

θt+∆t =∆θ

∆t∫V

vθc∆θ

∆tdV −

∫V

(∇vθ)T qt+∆t dV = Qt+∆text




Heat transport

Weak form at t + ∆t (θ = θ, vθ = 0 on Sθ)

∆θ = limi→∞

∆θi , ∆θi+1 = ∆θi + dθ, θt+∆ti = θt + ∆θi

qt+∆t = qt + ∆q

∆q = limi→∞

∆qi , ∆qi+1 = ∆qi + dq, qt+∆ti = qt + ∆qi

Linearization:

dq = −dΛdθ∇θ dθ − Λ∇( dθ)




Heat transport

Linearized weak form

1∆t

∫V

vθc dθ dV +

∫V

(∇vθ)TdΛdθ∇θ dθ dV+

+

∫V

(∇vθ)T Λ∇( dθ) dV = Qt+∆text + Qt+∆t

int,i

Qt+∆tint,i =

∫V

(∇vθ)T qt+∆ti dV − 1

∆t

∫V

vθc∆θi dV




Discretization and coupling

u, vu using Nu, u = Nuu, Bu = LNu

θ, vθ using Nθ, θ = Nθθ, Bθ = ∇Nθ

Substitution leads to:[Ki Kθ,i

0 C + Hi

][d u

d θ

]=

[ft+∆text − ft+∆t

int,i

ht+∆text − ht+∆t

int,i

]where

Ki =

∫V

BTu EiBu dV , Kθ,i =

∫V

BTu sθ,iNθ dV

C =1

∆t

∫V

NTθ cNθ dV , Hi =

∫V

BTθ

(dΛdθ

∣∣∣∣i

∇θiNθ + ΛiBθ

)dV

ft+∆tint,i =

∫V

BTu σt+∆ti dV , ht+∆t

int,i =

∫V

NTθ c∆θi∆tdV −

∫V

BTθ qt+∆ti dV

Staggered algorithm possibleJ. Pamin & co. (Cracow University of Technology) Fundamentals of computational thermo-elasticity


Square configuration under thermal load

qn = 0

qn = 0

qn = 0

T (t)

t

T

1

Mechanics:

– density ρ = 2– Young modulus E = 20000– Poisson ratio ν = 0.2– yield stress σy = 300– hardening modulus ET = 2000

Heat flow:

– initial temperature T0 = 0– heat conductivity k = 10– heat capacity c = 1– expansion coefficient α = 0.01



Square configuration under thermal load (FEAP)Equivalent stress (elasticity)

Evolution of equivalent stress (FEAP)






Square configuration under thermal load (ANSYS)Equivalent stress (elasticity)

coarse mesh dense mesh

Caution: sensitivity of results to space and time discretization!



Square configuration under thermal load (ANSYS)Equivalent stress (plasticity)


Caution: locking in plasticity must be prevented!



Square configuration under thermal load (ANSYS)Plastic strain measure


Question: what if we need full coupling due to plastic dissipation andthermal softening?



FEMDK (Tochnog kernel): heat induced deformationProblem setup

0.9

1.0

1.0

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

convection

to environment

T

t

T(t) free to

move

fixed

fixed

T(0) = 0

T(2) = 1600

insulation

insulation

Material properties (SI Units): Steel AISI-304, density ρ = 8030.0, Youngmodulus E = 1.93e11, Poisson ratio ν = 0.29, yield stress σy = 24.3e6, linearexpansion coefficient α = 1.78e − 5, heat conductivity k = 16.3, convectioncoefficient hc = 10.0, environment temperature Tambient = 0.0.



FEMDK (Tochnog kernel): heat induced deformationDiscretisation



FEMDK (Tochnog kernel): heat induced deformationFinal temperature distribution



FEMDK (Tochnog kernel): heat induced deformationFinal displacement



FEMDK (Tochnog kernel): heat induced deformationFinal total strain distribution



FEMDK (Tochnog kernel): heat induced deformationFinal isotropic hardening parameter

Movie



FEMDK (Tochnog kernel): heat induced deformationEffective stress distribution in wall cross-section



Literature

G. A. Maugin.The Thermomechanics of Plasticity and Fracture.Cambridge University Press, Cambridge, 1992.

D.W. Nicholson.Finite element analysis. Theromechanics of solids.CRC Press, Boca Raton, 2008.

N.S. Ottosen and M. Ristinmaa.The Mechanics of Constitutive Modeling.Elsevier, 2005.

W. Prager.Introduction to Mechanics of Continua.Dover Publications, 1961.


fundamentals of computational thermo-elasticityjpamin/dyd/soki/termoelast.pdf · fundamentals of...

Documents