CONTINUUM MECHANICS I

Jirı Plesek

Institute of Thermomechancs

The Czech Academy of Sciences

Prague

Download text: http://www.it.cas.cz/cs/plesek/vyuka

Lecture 2

Contents

• Overview of continuum mechanics

kinematics

fundamental laws

(25 min)

• Uniaxial problem

dissipation inequality

notion of the free energy

concepts of material modelling

(35 min)

Kinematics

-

X1, x1

6X2, x2

*

r r-t = 0 t > 0

u

xX

Ω0

Ωt

X3, x3

F = Grad x = RU 7→ E := f (U)

J = det F

L = grad v = D + W

Mass conservation

For arbitrary volume

V0 ⊂ Ω0 → Vt ⊂ Ωt∫V0

ρ0 dV0 =

∫Vt

ρ dVt

Continuity equation

Jρ = ρ0 ⇒ ρ + ρ div v = 0

Balance of momentum

For arbitrary volumedp

dt=

∫Vt

b dVt +

∫St

t dSt

Equations of motion

divσ + b = ρv

Div P + B = ρ0u

P = JσF−T 1st Piola-Kirchhoff tensor

Balance of angular momentum

Moment of momentum

d

dt

∫Vt

ρx× v dVt =

∫Vt

x× b dVt +

∫St

x× t dSt

Symmetry of the Cauchy stress

σT = σ

Energy conservation (1/2)

Power balance

Q + W =d

dt(U + K)

Balance of mechanical energy

W =d

dtK +

∫Vt

σ :D dVt ⇒ Q +

∫Vt

σ :D dVt =d

dtU

Energy conservation (2/2)

Equation of heat conduction

κ− div h + σ :D = ρu

K − Div H + Σ:E = ρ0u

H = JF−1h Piola vector

Conjugate pair

Σ,E ⇐⇒∫Vt

σ :D dVt =

∫V0

Σ:E dV0

Example: 2nd PK stress, GL strain

Entropy inequality

Clausius-Duhem formulation

d

dtS ≥

∫Vt

κ

TdVt −

∫St

h · nT

dSt

Local dissipation inequality

−ρηT + σ :D− T−1h · gradT ≥ ρψ

−ρ0ηT + Σ:E− T−1H ·GradT ≥ ρ0ψ

ψ = u− Tη Helmholtz free energy

Contents

• Overview of continuum mechanics

kinematics

fundamental laws

(25 min)

• Uniaxial problem

dissipation inequality

notion of the free energy

concepts of material modelling

(35 min)

Tensile test

-F F

T

Nominal stress and small strain

σ =F

Aand ε =

∆l

l

A = the initial cross section

l = the initial length

First law

Internal energy

Heat power

Mechanical power

u = U/V

q = Q/V

w = W/V = (F/A)d

dt(∆l/l) = σε

q + w = u

Second law

Entropy η = S/V

Heat content inequality

η ≥ q

T

Dissipation inequality

η ≥ q

T

Dissipation inequality

T η ≥ q

Dissipation inequality

T η ≥ q = u− w

Dissipation inequality

T η ≥ q = u− σε

Dissipation inequality

T η ≥ q = u− σε

σε + T η ≥ u

Dissipation inequality

T η ≥ q = u− σεLegendre transform

Dissipation inequality

T η ≥ q = u− σεLegendre transform

T η = (Tη)· − T η

Dissipation inequality

T η ≥ q = u− σεLegendre transform

T η = (Tη)· − T η−ηT + σε ≥ u− (Tη)·

Dissipation inequality

T η ≥ q = u− σεLegendre transform

T η = (Tη)· − T η−ηT + σε ≥ u− (Tη)· = (u− Tη)·

Dissipation inequality

T η ≥ q = u− σεLegendre transform

T η = (Tη)· − T η−ηT + σε ≥ u− (Tη)· = (u− Tη︸ ︷︷ ︸

ψ

)·

Dissipation inequality

T η ≥ q = u− σεLegendre transform

T η = (Tη)· − T η−ηT + σε ≥ u− (Tη)· = (u− Tη︸ ︷︷ ︸

ψ

)·

ψ := u− Tη Helmholtz free energy

Dissipation inequality

T η ≥ q = u− σεLegendre transform

T η = (Tη)· − T η−ηT + σε ≥ u− (Tη)· = (u− Tη︸ ︷︷ ︸

ψ

)·

ψ := u− Tη Helmholtz free energy

−ηT + σε ≥ ψ

The Helmholtz free energy (1/2)

Isothermal process

∆w ≥ ψ2 − ψ1

Cyclic loading

1L→ 2

U→ 1

∆wL ≥ ψ2 − ψ1

∆wU ≥ ψ1 − ψ2

⇒ ∆wL ≥ ψ2 − ψ1 ≥ −∆wU

The Helmholtz free energy (1/2)

Isothermal process

∆w ≥ ψ2 − ψ1

Cyclic loading

1L→ 2

U→ 1

∆wL ≥ ψ2 − ψ1

∆wU ≥ ψ1 − ψ2

⇒ ∆wL ≥ ψ2 − ψ1 ≥ −∆wU

. . . assuming ∆wL > 0 and ∆wU < 0

The Helmholtz free energy (1/2)

Isothermal process

∆w ≥ ψ2 − ψ1

Cyclic loading

1L→ 2

U→ 1

∆wL ≥ ψ2 − ψ1

∆wU ≥ ψ1 − ψ2

⇒ |∆wL| ≥ ψ2 − ψ1 ≥ |∆wU |

. . . assuming ∆wL > 0 and ∆wU < 0

The Helmholtz free energy (1/2)

Isothermal process

∆w ≥ ψ2 − ψ1

Cyclic loading

1L→ 2

U→ 1

∆wL ≥ ψ2 − ψ1

∆wU ≥ ψ1 − ψ2

⇒ |∆wL| ≥ ψ2 − ψ1 ≥ |∆wU |

. . . assuming ∆wL > 0 and ∆wU < 0

ψ = strain energy

The Helmholtz free energy (2/2)

First law

∆q + ∆w = u2 − u1Isothermal process

∆w ' ψ2 − ψ1

Tensile test

∆q = ? ∆w

The Helmholtz free energy (2/2)

First law

∆q + ∆w = u2 − u1Isothermal process

∆w ' ψ2 − ψ1

Tensile test

∆q = 10×∆w

Thermoelasticity

Assumption

∃ψ(ε, T ) : ψ =∂ψ

∂εε +

∂ψ

∂TT

Dissipation inequality

−(η +

∂ψ

∂T

)T +

(σ − ∂ψ

∂ε

)ε ≥ 0

Thermoelasticity

Assumption

∃ψ(ε, T ) : ψ =∂ψ

∂εε +

∂ψ

∂TT

Dissipation inequality

−(η +

∂ψ

∂T

)T +

(σ − ∂ψ

∂ε

)ε ≥ 0

η = −∂ψ∂T

and σ =∂ψ

∂εcorollary: η =

q

T(ideal process)

Duhamel-Neumann’s law

Thermal expansion

ε = εe + α∆T ⇒ σ = E(ε− α∆T )

Duhamel-Neumann’s law

Thermal expansion

ε = εe + α∆T ⇒ σ = E(ε− α∆T )

Free energy

ψ(ε, T ) = 12Eε

2 − Eα∆Tε + f (T )

Entropy

η(ε, T ) = −∂ψ∂T

= Eαε− f ′(T )

Heat exchange

q = T η = αTEε− Tf ′′(T )T

Duhamel-Neumann’s law

Thermal expansion

ε = εe + α∆T ⇒ σ = E(ε− α∆T )

Free energy

ψ(ε, T ) = 12Eε

2 − Eα∆Tε + f (T )

Entropy

η(ε, T ) = −∂ψ∂T

= Eαε− f ′(T )

Heat exchange

q = T η = αTEε− Tf ′′(T )T

q = αTEε + cT . . . note that c = function(T)

Numerical example

q = αTEε + cT

Tensile test 100 MPa

∆q = αTσ = 10−5 × 300× 108 = 3× 105 J/m3

Mechanical work

∆w =σ2

2E=

(108)2

2× 1011= 0.25× 105 J/m3

Elasto-plasticity

Additive decomposition

ε = εe + εp

Assumption

ψ(εe) = ψ(ε− εp) ⇒ ψ =∂ψ

∂ε(ε− εp)

Dissipation inequality (σ − ∂ψ

∂ε

)ε +

∂ψ

∂εεp ≥ 0

σ =∂ψ

∂εand D = σεp ≥ 0

Recap

• The Helmholtz free energy (HFE) is the strain (stored) energy.

• HFE substantially differs from the internal energy.

• The dissipation inequality restricts the HFE structure.

• Thermoelasticity is the ideal process (η = q/T ).

• Specific heat capacity of solids is independent of stress.

• An elastic-plastic model introduces dissipation D > 0.