V. Constitutive equations Continuum mechanics -...

Continuum mechanicsV. Constitutive equations

Ales Janka

office Math [email protected]

http://perso.unifr.ch/ales.janka/mechanics

Mars 16, 2011, Universite de Fribourg

Ales Janka V. Constitutive equations

1. Constitutive equation: definition and basic axioms

Constitutive equation: relation between two physical quantitiesspecific to a material, e.g.:

τ ij = τ ij(u, {ek`}, {F k` },T )

Basic axioms

Axiom of causality

Axiom of determinism

Axiom of equipresence

Axiom of neighbourhood

Axiom of memory

Axiom of objectivity

Axiom of material invariance

Axiom of admissibility


1. Basic axioms: causality

Axiom of causality:Independent variables in the constitutive laws are:

Continuum position y i (x, t)

Temperature T

Dependent variables (responses) are e.g.:

Helmholtz free energy ϕ (thermodynamic potential, measureof the ”useful” work obtainable from a closed thermodynamicsystem)

Strain energy density Ψ

Stress tensor τ ij

Heat flux qi

Internal energy ε

Entropy S


1. Basic axioms: determinism and equipresence

Axiom of determinisimResponses of the constitutive functions at a material point x attime t are determined by the history of the motion and historyof the temperature of all points of the body.

Axiom of equipresenceIf an independent variable enters in one function of response, itshould be present in all constitutive laws (until the proof of thecontrary)


1. Basic axioms: neighbourhood

Axiom of neighbourhoodResponses at a point x are not much influenced by values ofindependent variables (temperature and displacement) at adistant point x.

Hypothesis: functions y(x, t) and T (x, t) are sufficiently smoothto be expanded into a Taylor series:

y i (x, t) = y i (x, t)+∂y i

∂x j

∣∣∣∣x,t

(x j−x j)+1

2

∂2y i

∂x j∂xk

∣∣∣∣x,t

(x j−x j)(xk−xk)+. . .

with negligible higher-order terms.Simple thermomechanical material: Taylor expansion termswith the second+higher derivatives are negligible:

τ(x, t) = T [y(x, t ′), y,x(x, t ′),T (x, t ′),T,x(x, t ′); x, t ′ ≤ t]

This class of material also called gradient continua.Ales Janka V. Constitutive equations

1. Basic axioms: memory

Axiom of memoryValues of constitutive variables from a distant past do not affectappreciably the values of constitutive laws now.

Smooth memory material: constitutive variables can beexpanded to Taylor series in time with negligible higher order terms

Fading memory: response functionals must smooth possiblediscontinuities in memory


1. Basic axioms: objectivity

Axiom of objectivity

Invariance of constitutive laws with respect to rigid body motionof the spatial frame of reference (spatial coordinates).

Simple consequence: constitutive laws depend of thedeformation gradient (or strain tensor) rather than y(x).


1. Basic axioms: material invariance

Axiom of material invariance

Invariance of constitutive laws with respect to certainsymmetries/transformations of the material frame of reference(material coordinates).

Symmetries in material properties due to crystallographicorientation.

Hemitropic continuum: invariant w.r.t. all rotationsIsotropic continuum: hemitropic + invariant to reflectionAnisotropic continuum: otherwise (can have some invarianceproperties, but not all)Homogeneous continuum: invariance w.r.t. shift of coord system


1. Basic axioms: admissibility

Axiom of admissibility

Consistence with respect to basic conservation laws (mass,momentum, energy), and the 2nd law of thermodynamics(entropy).

This axiom can also help to eliminate dependences on someconstitutive variables.


2. Constitutive laws for simple thermo-mechanical continua

Thermo-elastic continua: simple thermo-mechanical continuawith no memory.

By applying the basic axioms, all material properties depend onlyon the current values of deformation and temperature: hence alsofor Helmholtz free energy:

ϕ = ϕ({eij)},T )

For simplicity, consider only small deformations (Cauchy straintensor eij).

Are we able to say more about the form of the constitutivelaws in this case?



Axiom of admissibility: we need to be consistent withthermodynamical laws and basic equilibria.

Let us derive ϕ with respect to time

ϕ =∂ϕ

∂eijeij +

∂ϕ

∂TT

and substitute it into the dissipation inequality

ρ ϕ+ ρ T η − τ ij ∇jvi +qi

T∇iT ≤ 0.

We get

ρ∂ϕ

∂eijeij + ρ

∂ϕ

∂TT + ρ T η − τ ij ∇jvi +

qi

T∇iT ≤ 0.

NB. Due to the symmetry of τ ij = τ ji , we have

τ ij ∇jvi =1

2

(τ ij ∇jvi + τ ij ∇ivj

)= τ ij eij



Hence, the dissipation inequality looks now like:

eij

(ρ∂ϕ

∂eij− τ ij

)+ T ρ

(η +

∂ϕ

∂T

)+

qi

T∇iT ≤ 0.

This inequality must hold for any time-dependent process, ie.for any eij and T !

Hence there must be:

ρ∂ϕ

∂eij− τ ij = 0 ⇒ τ ij = ρ

∂ϕ

∂eij,

η +∂ϕ

∂T⇒ η = − ∂ϕ

∂T,

qi

T∇iT ≤ 0.


3. Simple thermo-mechanical continuum: large deformation

Small deformations: (Cauchy stress tensor)

τ ij = ρ∂ϕ

∂eij, η = − ∂ϕ

∂T,

qi

T∇iT ≤ 0

Large deformations: (2nd Piola-Kirchhoff)

T ij = ρ0∂ϕ

∂εij, η = − ∂ϕ

∂T,

qi

T∇iT ≤ 0

Define strain (or stored) energy density Ψ = ρ0 ϕ, then:

T ij =∂Ψ

∂εij, η = − 1

ρ0

∂Ψ

∂T,

qi

T∇iT ≤ 0

NB: T = ∂Ψ∂ε is a derivative of a scalar function with respect to a

tensor, see M2, Section 2.

Hyperelastic material: material for which T ij =∂Ψ

∂εijAles Janka V. Constitutive equations

4. Hooke’s law (Robert Hooke 1635–1703)

Neglect temperature: Taylor expansion of strain energy density:

Ψ(e) = Ψ0 + E ij eij +1

2E ijk` eij ek` + . . .

with material coefficients E ij , E ijk` called elastic tensors.

Suppose small deformations: take only the 3 first terms of theexpansion: then from τ ij = ∂Ψ

∂eijwe obtain the Hooke’s law (1660):

τ ij = E ij + E ijk`ek`

with the pre-stress E ij at initial configuration.If no pre-stress:

τ ij = E ijk`ek`


4.1. Hooke’s law: elastic tensor E ijk`

Elastic tensor E ijk`: 34 = 81 components depending only onmaterial coordinates

Possible reduction of degrees of freedom:

Symmetry of τ ij and ek` ⇒ symmetry of E ijk` within ij and k`:

E ijk` = E jik` = E ij`k = E ji`k

no. of components reduced to 36.

Taylor expansion of Ψ(e) ⇒ symmetry in pairs ij and k`:

E ijk` = E k`ij

no. of components thus reduced to 21.


4.2. Hooke’s law in deviator-form

Consider only small deformations. Physical meaning of Cauchy strain e:

Relative volume change:dV − dV0

dV0= e1

1 + e22 + e3

3 = tr(e) = e``

(cf. “1. Kinematics”, section 4.)

Define:Volumic dilatation e: volume-changing deformation component:

e =1

3tr(e) =

1

3e``

Strain deviator e: volume-preserving deformation component

eij = eij − e gij e ij = e i

j − e δij

changes only shape, not the volume, tr(e) = 0.

Hydrostatic tension s: forces opposed to volume change

s =1

3tr(τ) =

1

3τ jj

Stress deviator τ : forces opposed to shape change:

τ ij = τ ij − s g ij τ ij = τ i

j − s δij


4.2. Hooke’s law in deviator-form, shear and bulk moduli

Hooke’s law for isotropic materials (in deviator form):

τ ij = 2µ e i

j volume-preserving deformations

s = 3 K e volume-change

Material properties characterized only by 2 constants

shear modulus µ characterizes genuine shear

bulk modulus K characterizes (in)compressibility(incompressible material for K →∞)

Total strain tensor:

τ ij = τ i

j + s δij = 2µe ij + 3 K e δij

= 2µ

(e ij −

1

3e`` δ

ij

)+ K e`` δ

ij

= 2µ e ij +

(K − 2µ

3

)e`` δ

ij


4.3. Hooke’s law for isotropic materials: elastic tensor E ijk`

Total strain tensor:

τ ij = 2µ e ij +

(K − 2µ

3

)e`` g ij

= 2µ g ik g j` ek` +

(K − 2µ

3

)︸︷︷︸

=λ

g ij gk` ek`

= µ(g ik g j` + g i` g jk

)ek` + λ g ij gk` ek`

Here, λ and µ are the so called Lame coefficients, K =2µ+ 3λ

3.

The corresponding elastic tensor E ijk` is thus (τ ij = E ijk` ek`):

E ijk` = µ(g ik g j` + g i` g jk

)+ λ g ij gk`


4.4. Hooke’s law for isotropic materials: compliance Cijk`

The tensor-inverse of E ijk` is called compliance C :

τ ij = E ijk` ekl eij = Cijk` τk`.

Let us inverse the Hooke’s law (ie. express e as a function of τ):

τ = 2µ e + λ tr(e) Id

Take a trace:

tr(τ) = 2µ tr(e) + 3λ tr(e) = (2µ+ 3λ) tr(e)

Plug back tr(e) into the Hooke’s law above to get

τ = 2µe +λ

2µ+ 3λtr(τ) Id

Hence,

e =1

2µ

(τ − λ tr(τ)

2µ+3λId

)or eij =

1

2µ

(gik gj` − λ

2µ+3λgk` gij

)︸︷︷︸

Cijk`

τk`


4.5. Hooke’s law: Young’s modulus E , Poisson’s ratio ν

Material constants in Hooke’s law by analogy with linear springs:

In 1D, spring stiffness E =force F

relative elongation ε

Compare with a special case in 3D:

Mono-axial loadingSuppose τ ij = 0 for all i , j , except τ11 6= 0.Compliance-form of a 3D Hooke’s law gives:

e11 =1

2µ

(1− λ

2µ+ 3λ

)τ11 =

µ+ λ

µ (2µ+ 3λ)τ11

e22 = e33 = − λ

2µ (2µ+ 3λ)τ11

dudy

u(x,y)

(u)F

F

x

y

yyεyyτ (u)

yyτ (u)

u(x,y)

=


4.5. Hooke’s law: Young’s modulus E , Poisson’s ratio ν

Young’s modulus defined as apparent “1D spring stiffness” in thecase of mono-axial loading, ie:

E =τ11

e11=

µ+ λ

µ (2µ+ 3λ)

with τ11 and e11 from the mono-axial loading in cartesiancoordinates.

Poisson’s ratio measures transversal vs. axial elongation

ν = −e22

e11=

λ

2 (µ+ λ)

Relative volume change:

dV − dV0

dV0= e11 + e22 + e33 = (1− 2 ν) e11

ie. ν = 0.5 for incompressible materialsAles Janka V. Constitutive equations

4.6. Hooke’s law for isotropic materials: summary

Isotropic material characterized by two constants:

shear modulus µ and bulk modulus K ,

µ =E

2 (1 + ν)K =

1

3(2µ+ 3λ) =

1

3

E

1− 2ν

Lame’s coefficients µ and λ,

µ =E

2 (1 + ν)λ =

E ν

(1 + ν)(1− 2ν)= K − 2µ

3

Young’s modulus E and Poisson’s ratio ν,

E =µ (2µ+ 3λ)

µ+ λν =

λ

2(µ+ λ)


4.6. Hooke’s law for isotropic materials: summary

Corresponding form of Hooke’s law:

using shear modulus µ and bulk modulus K , in deviator form:

τ ij = 2µ e i

j , s = 3 K e

using Lame’s coefficients µ and λ:

τ ij = µ(g ik g j` + g i` g jk

)ek` + λ g ij gk` ek`

Or in global form

τ = 2µe +λ

2µ+ 3λtr(τ) Id

using Young’s modulus E and Poisson’s ratio ν:

τ ij =E

2(1 + ν)

(2ν

1− 2νg ij gk` + g ik g j` + g i` g jk

)ek`

large deformations: Saint Venant-Kirchhoff material

Ψ(ε) = µ tr(ε2) +λ

2

[tr(ε)

]2, T ij =

∂Ψ

∂εijAles Janka V. Constitutive equations

4.7. Hooke’s law: measuring stress-strain curve for steel

0 strain e11

stre

ss

2

13

3

4

uniform plasticelastic necking

B

A

5

1 Ultimate Strength

2 Yield Strength (elastic limit)

3 Rupture

4 Strain hardening region

5 Necking region

A 1st Piola-Kirchhoff stress σ =F

A0

B Euler stress τ =F

A


5. Linear thermo-elasticity: Duhamel-Neumann’s law

Consider also temperature: Taylor expansion of strain energy:

Ψ(e,T ) = Ψ0(T ) + E ij(T ) eij +1

2E ijk`(T ) eij ek` + . . .

= Ψ0(T ) +

[E ij(T0) +

∂E ij

∂T(T−T0) + . . .

]eij

+1

2

[E ijk`(T0) +

∂E ijk`

∂T(T−T0) + . . .

]eij ek` + . . .

Suppose |T − T0| << T0 and small deformations and neglect all(mixed) 3rd order terms and higher.

Duhamel-Neumann’s law: from τ ij = ∂Ψ∂eij

we obtain:

τ ij = E ijT0

+ E ijk` ek` − βij (T−T0)

with βij = −∂E ij

∂T . For isotropic materials βij = β g ij .

Usually, we take T0 with no pre-strain, E ijT0

= 0.


5. Linear thermo-elasticity: Duhamel-Neumann’s law

From the Hooke’s law, we can write:

τ ij = 2µ e i

j + λ δij e`` − βij (T−T0)

Let us derive the compliance-form e = e(τ ,T ):Index-contraction of the above gives

τ ii = (2µ+3λ) e``−βk

k (T−T0) ie. e`` =1

2µ+ 3λ

[τ ii + βk

k (T−T0)]

Substitute it back to the Duhamel-Neumann’s law to obtain

e ij =

1

2µ

[δik δ

`j −

λ

2µ+3λδij δ

`k

]τk` −

1

2µ

(λ

2µ+3λδij β

mm − βi

j

)︸︷︷︸αi

j ...thermal dilatation coeff

(T−T0)


6. Constitutive law for heat flux q: Fourier’s law

Suppose simple thermo-mechanical continuum, small deformations:

q = q(T ,∇T , e)

Use first-order Taylor expansion around a deformation-freeconfiguration at T0 to approximate:

qi = k i0 + k i

1 (T−T0) + k ij2 ∇jT + k ijk

3 ejk

with some coefficients k i0, k i

1 and k ijk3 .

This law must not contradict the 2nd law of thermodynamicsin particular there must be (cf. Section 2 and 3 above):

qi

T∇iT =

1

T

(k i

0 + k i1 (T−T0) + k ij

2 ∇jT + k ijk3 ejk

)∇iT ≤ 0

for any state of the continuum, ie. ∀ T > 0 and ∀ e. This issatisfied only if

k i0 = k i

1 ≡ 0 , k ijk3 ≡ 0 and −[k ij

2 ] is sym.positive definite


6. Constitutive law for heat flux q: Fourier’s law

Hence, we have derived the Fourier’s law for heat flux:

qi = −k ij ∇jT , q = k · ∇T

with the heat conductivity tensor k, k ij = −k ij2 , [k ij ] is a

symmetric positive definite matrix.

For isotropic materials: k = k Id .


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