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1.14 Elastic Materials

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1.14.1 General form

Elastic materials are defined as materials without memory. Consequently, the time derivatives

are dropped in Equation (1.382) and one obtains

S(X, t) = S[C(X, t ), ρ

−1(X, t), θ(X, t),∇0θ(X, t),X] (1.383)

48 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

and similarly (dropping the dependence),

Q = Q[C, ρ

−1, θ,∇0θ,X] (1.384)

ε = ε[C, ρ

−1, θ,∇0θ,X] (1.385)

η = η[C, ρ

−1, θ,∇0θ,X]. (1.386)

Since detC = J 2 and ρJ = ρ0 (balance of mass) the explicit dependence on ρ is dropped.

The balance of momentum, the balance of energy and the entropy inequality remain to be

satisfied. It is amazing that the entropy inequality, being an inequality, plays an extremely

important role in the derivation of the material laws. To see this, we first define the free

energy ψ(X, t) to simplify the calculations:

ψ := ε − θη. (1.387)

Since

˙ψ

= ˙ε − ˙ θη − θ˙η (1.388)

the entropy inequality now reads

−ρ0

( ˙ψ + ˙ θη) + 1

S : ˙E − 1

θ2Q· ∇0θ ≥ 0. (1.389)

Notice that because of Equation (1.385) and Equation (1.386), dropping the dependence on

−1

ψ = ψ[C, θ,∇0θ,X]. (1.390)

Accordingly,

˙ψ

= ∂ψ

∂C

: ˙C + ∂ψ

∂θ

˙ θ + ∂ψ

∂∇0θ

· ∇˙0θ. (1.391)

Substituting Equation (1.391) into Equation (1.389) and noting that ˙E = ˙C/2 yields

_−ρ0

∂ψ

∂C

+ 12

S_ : ˙C − ρ0

_∂ψ

∂θ

+ η

_ ˙ θ − ρ0

∂ψ

∂∇0θ

· ∇˙0θ − 1

θ2Q· ∇0θ ≥ 0. (1.392)

Since S, Q, ψ and η are not a function of ˙C nor ˙ θ nor

∇˙0θ, Equation (1.392) is linear in

˙C

, ˙ θ and ∇0

˙ θ. Hence, for Equation (1.392) to be valid for any ˙C , ˙ θ or ∇0

˙ θ, the coefficients

of these terms must be zero. Defining _ := ρ0ψ one obtains

S = 2ρ0

∂ψ

∂C

= 2

∂_

∂C

= ∂_

∂E

(1.393)

η = −∂ψ

∂θ

= − 1

ρ0

∂_

∂θ

(1.394)

∂ψ

∂∇0θ

= ∂_

∂∇0θ

= 0 (1.395)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 49

and the entropy inequality reduces to

−Q· ∇0θ ≥ 0. (1.396)

Consequently, for elastic materials there exists a function _(C, θ,X) such that S and η

can be obtained by partial differentiation. ε satisfies

ε(C, θ,X) = _

ρ0

+ θη. (1.397)

The only dependent variable that depends on ∇0θ is Q. Equation (1.396) requires that Q

is at least linear in ∇0θ, that is,

Q = −κ(C, θ,∇0θ,X) · ∇0θ. (1.398)

The entropy inequality has dictated the shape of nearly all variables! The only equations

left to satisfy are the balance of momentum and the balance of energy. Summarizing,

S = ∂_

∂E

(C, θ,X) (1.399)

η = − 1

ρ0

∂_

∂θ

(C, θ,X) (1.400)

ε = _

ρ0

+ θη (1.401)

Q = Q(C, θ,∇0θ,X). (1.402)

Elastic materials in this general form are also called hyperelastic materials. _ is sometimes

called the stored energy function (Ciarlet 1993).

1.14.2 Linear elastic materials

Special forms arise if we linearize S with respect to E and Q with respect to E and ∇0θ

(C and E are equivalent independent variables). To obtain a linear relation between S and

E, we expand _ about E = 0 and truncate the series after the quadratic terms:

_ ∼ _0(θ,X) + _KL(θ,X)EKL + 12

_KLMN(θ,X)EKLEMN, _E_ → 0 (1.403)

while Q is expanded at ∇0θ = 0,E = 0 and the linear terms are kept

Q ∼ −κK(θ,X) − κKL(θ,X)θ,L − κKLM(θ,X)ELM, _∇0θ_ → 0, _E_ → 0.

(1.404)

Because of the symmetry of E one finds

_KL = _LK (1.405)

_KLMN = _LKMN = _KLNM = _MNKL (1.406)

κKLM = κKML. (1.407)

50 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

Applying Equation (1.399) yields

SKL(θ,X) = _KL(θ,X) + _KLMN(θ,X)EMN. (1.408)

Physical observations and the second law of thermodynamics, cf Equations (1.396) and

(1.398), dictate that there is no heat flux if the temperature gradient is zero. This leads to

(see Equation (1.404))

κK = 0, κKLM = 0 (1.409)

and

QK = −κKL(θ,X)θ,L. (1.410)

The entropy inequality now amounts to

κKLθ,K θ,L ≥ 0 (1.411)

which means that the symmetric part of κKL must be positive definite. The physical meaning

of κKL is the conduction coefficient matrix.

The term _KL(θ,X) in Equation (1.408) contains the thermal stress. Let the temperature

θref represent a homogeneous temperature distribution without any thermal stresses.

Then one can write

_KL(θ,X) = γ KL(X) − βKL(θ,X)(θ − θref). (1.412)

γ KL are residual stresses from other sources and βkl is the compressive stress rise per unit

temperature increase if no expansion is allowed. Furthermore, we define αKL assuming

_KLMN to be invertible:

βKL(θ,X) =: _KLMN(θ,X)αMN(θ,X). (1.413)

Hence,

SKL(θ,X) = γ KL(X) − βKL(θ,X)(θ − θref) + _KLMN(θ,X)EMN

= γ KL(X) + _KLMN(θ,X) [EMN − αMN(θ,X)(θ − θref)] . (1.414)

The tensor α contains the expansion coefficients. Now, let us expand _0(θ,X) in Equation

(1.403) about θref:

_0(θ,X) = ρ0(X)ψ0(X) − ρ0(X)η0(X)(θ − θref) − ρ0(X)c(θ,X)

2θref

(θ − θref)2. (1.415)

Notice that the equality sign applies, since c(θ,X) in the last term may depend on θ.

Dropping the dependence on X in the notation and defining T := θ − θref yields

_ = ρ0ψ0 − ρ0η0T − ρ0c(θ)

2θref

T 2 + [γ KL − βKL(θ)T ]EKL + 1

_KLMN(θ)EKLEMN

(1.416)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 51

and

η = η0 + c(θ)T

θref

+ βKL(θ )

ρ0

EKL + c

(θ)T 2

2θref

+ βKL

(θ )

ρ0

TEKL + 1

_KLMN

(θ)EKLEMN.

(1.417)

The last three terms are due to the temperature dependence of the coefficients. Since

ρ0ε = _ + ρ0θη, one obtains

ρ0ε = ρ0ψ0 + ρ0η0θref + ρ0c(θ)

T + T 2

2θref

+ [γ KL + βKL(θ )θref]EKL + 1

_KLMN(θ)EKLEMN +

ρ0

θc

(θ)T 2

2θref

+ θβKL

(θ)T EKL + 1

ρ0θ_KLMN

(θ)EKLEMN. (1.418)

From Equation (1.418), it follows that c is the specific heat. Substituting the above equations

into the energy balance, Equation (1.355), is quite a tedious task. Generally, the derivative of

the coefficients with respect to the temperature can be neglected (the coefficients, however,

are still a function of temperature). Furthermore, discarding the quadratic T term leads to

ρ0ε = ρ0ψ0 + ρ0η0θref + ρ0c(θ)T + [γ KL + βKL(θ )θref]EKL + 12

_KLMN(θ)EKLEMN

(1.419)

and for the stress

SKL = [γ KL − βKL(θ)T ] + _KLMN(θ)EMN. (1.420)

Substitution into Equation (1.355) finally yields (after further linearization: θ ˙EKL ≈θref

˙E

KL)

ρ0c(θ) ˙ T + βKL(θ )θref

˙E

KL − (κKL(θ )θ,L);K − ρ0h = 0. (1.421)

This is the classical heat equation for linear elastic materials.

If in Equation (1.420)

γ KL =0 forK _= L

βKL =0 forK _= L

_KLMN =0 forK _= L and M = N (1.422)

one obtains SKL = 0,K _= L if EKL = 0,K _= L and vice versa. If this is true for arbitrary

orientations of the axes as in the case of isotropic materials, then the principal axes of E

are also principal axes of S. Indeed, take the principal axes of E as a local rectangular

coordinate system. This means that EKL = 0,K _= L and consequently SKL = 0,K _= L:

the shear stress is zero. Accordingly,

E =_

iENi ⊗ Ni (1.423)

S =_

iSNi ⊗ Ni . (1.424)

52 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

Furthermore, since

F =_

_ iCni ⊗ Ni (1.425)

and

σ = J

−1F · S · FT (1.426)

one finds

σ = J

−1 _

_ iCni ⊗ Ni_





jSNj ⊗ Nj





_ kCNk ⊗ nk

= J

−1_

iC iSni ⊗ ni (1.427)

which yields for the true principal stresses

λiσ = J

−1 iC iS = J

−1(2 iE + 1) iS. (1.428)

Since J, Ci > 0, the true stress and the second Piola–Kirchhoff stress have the same sign.

1.14.3 Isotropic linear elastic materials

For a linear elastic material, we found

_ = ρ0ψ0 − ρ0η0T − ρ0c(θ)

2θref

T 2 + [γ KL − βKL(θ)T ]EKL + 1

_KLMN(θ)EKLEMN

(1.429)

SKL = [γ KL − βKL(θ)T ] + _KLMN(θ)EMN (1.430)

QK = −κKL(θ )θ,L. (1.431)

Isotropy means that the material data are independent of the direction in the material frame

of reference. Hence, a transformation Q such that

_ = Q· X, QT ·Q = Q·QT = 1, detQ = ±1 (1.432)

must leave the constitutive equations invariant. Under such a transformation, second-order

and fourth-order tensors transform according to

_KL = γMNQK

MQL

N (1.433)

_KLMN = _PQRSQK

PQL

QQM

RQN

S. (1.434)

One can show that for this to be true for an arbitrary rotation, the tensors must satisfy

γ KL = γGKL (1.435)

_KLMN = λGKLGMN + μ(GKMGLN + GKNGLM) (1.436)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 53

and similarly for the other tensors

βKL = βGKL (1.437)

κKL = κGKL. (1.438)

Since

trE = GKLEKL (1.439)

is the trace of the tensor E, one finds

_(θ) = ρ0ψ0 − ρ0η0T − ρ0c(θ)

2θref

T 2 + [γ − β(θ)T ]trE + 12

λ(θ)(trE)2 + μ(θ)tr(E2)

(1.440)

SKL = [γ − β(θ)T ]GKL + λ(θ)(trE)GKL + 2μ(θ)EMNGKMGLN (1.441)

QK = −κ(θ)θ,LGKL. (1.442)

The energy equation reduces to

ρc(θ) ˙ T + β(θ)θref

˙E

− (GKLκ(θ)θ,L);K − ρ0h = 0. (1.443)

The kind of material described by Equations (1.440) to (1.443) is also called a St Venant–

Kirchhoff material.

The first and second invariant of a tensor E are defined by

I1E = trE (1.444)

I2E = 1

[I 2

− tr(E2)]. (1.445)

Consequently, the free energy _ can also be written as

_ = ρ0ψ0 − ρ0η0T − ρ0c(θ)

2θref

T 2 + [γ − β(θ)T ]I1E + 1

[λ(θ) + 2μ(θ)]I 2

− 2μ(θ)I2E.

(1.446)

λ(θ) and μ(θ) are called Lamґe’s constants, κ(θ) is the conduction coefficient, c(θ) is the

specific heat and β(θ) satisfies (substitute Equation (1.436) into Equation (1.413))

β(θ) = [3λ(θ) + 2μ(θ)]α(θ) (1.447)

where α(θ) is the isotropic expansion coefficient. The thermal stress now yields

SKL = −[3λ(θ) + 2μ(θ)]α(θ)TGKL. (1.448)

This stress is needed to suppress

EKL = α(θ)TGKL (1.449)

which is the strain resulting from the temperature change.

54 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

Finally, it should be noted that frequently other elastic constants are used instead of

the Lamґe’s constants λ and μ, the latter of which is also called the shear modulus. The

Poisson coefficient ν and Young’s modulus E satisfy

μ = E

2(1 + ν)

(1.450)

λ = νE

(1 + ν)(1 − 2ν)

(1.451)

which can be inverted to yield

ν = λ

2(λ + μ)

(1.452)

E = μ(3λ + 2μ)

λ + μ

. (1.453)

Another frequently used constant is the bulk modulus K. For linearized strains, it will be

proven in the next section that K is the ratio of the hydrostatic pressure p to the volume

reduction it produces. The following relations apply

K = λ + 2

μ (1.454)

ν = 3K − 2μ

6K + 2μ

. (1.455)

1.14.4 Linearizing the strains

So far, we consistently used the Lagrange strain tensor E. Equation (1.82) shows that E

is not linear in the displacement U. To obtain a truly linear theory, the quadratic terms in

E are dropped and one obtains the infinitesimal strains ˜E, Equation (1.88):

˜E

KL = 12

(UK;L + UL;K). (1.456)

Recall that the infinitesimal rotation is defined by

˜R

KL = 12

(UK;L − UL;K). (1.457)

Equation (1.90) has shown that EKL can only be replaced by ˜EKL if both the strain and

the rotations are small. The same applies to ekl and ˜ekl . Under the above assumptions, one

can write

EKL ∼ ˜EKL _˜E _, _˜R_ → 0 (1.458)

ekl ∼ ˜ekl _˜E_, _˜R_ → 0. (1.459)

To derive further simplifications we start from Equation (1.67):

xkgk

= (XL + UL)GL − o. (1.460)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 55

Taking the derivative with respect to K yields

,Kgk

= (δL

+ UL

;K)GL (1.461)

leading to

= (δL

+ ˜E L

+ ˜RL

K)GL · gk (1.462)

= (δL

+ ˜E L

+ ˜RL

K)g k

L (1.463)

where

g k

= GL · gk = gk · GL = gk

L. (1.464)

In a similar way, one arrives at

= (δl

− ˜el

− ˜rl

k)gK

l . (1.465)

For small strains and rotations, Equation (1.463) and Equation (1.465) reduce to

∼ gk

K, _˜E_, _˜R_ → 0 (1.466)

∼ gK

k, _˜E _, _˜R_ → 0. (1.467)

From Equation (1.352) and Equation (1.353), one finds

EKL = eklxk

,Kxl

,L (1.468)

which reduces by the use of Equations (1.458), (1.459), (1.466) and (1.467) to

˜E

KL ∼ ˜eklgk

Kgl

L, _˜E_, _˜R_ → 0. (1.469)

On the basis of Equation (1.457), a similar relationship applies to the infinitesimal rotation

˜R

KL ∼ ˜rklgk

Kgl

L, _˜E_, _˜R _ → 0. (1.470)

Furthermore, J = det xk

,K . Substituting Equation (1.463) and linearizing yields

J ∼ 1

3! eKLMeklm[gk

Kgl

Lgm

+ gk

Kgl

L( ˜E R

+ ˜RR

M)gm

+ gk

Kgm

M( ˜EQ

+ ˜RQ

L)gl

+ gl

Lgm

M( ˜E P

+ ˜RP

K)gm

P ], _˜E _, _˜R _ → 0 (1.471)

where eKLM and eklm are alternating symbols. This is equivalent to

J ∼ 1 + ˜E K

∼ 1 + ˜ek

k, _˜E_, _˜R_ → 0. (1.472)

Substituting Equation (1.463) and Equation (1.472) into the relationship between the Cauchy

stress and the second Piola–Kirchhoff stress leads to

σ kl = J

−1SKLxk

,Kxl

,L (1.473)

56 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

and linearizing yields

σ kl ∼ SKL[gk

Kgl

L(1− ˜em

m) + (˜ek

+ ˜rk

m)gm

Kgl

+ (˜el

+ ˜rl

m)gk

Kgm

L], _˜E _, _˜R_ → 0. (1.474)

The inverse of Equation (1.474) amounts to

SKL ∼ σ kl[gK

kgL

l(1 + ˜em

m) − gL

lgK

m(˜e m

+ ˜rm

− gK

kgL

m(˜em

+ ˜rm

l)], _˜E _, _˜R_ → 0. (1.475)

Substituting the above relations into Equation (1.420) yields a linearized expression for the

stress:

σ kl ∼ γ kl(1 − ˜em

m) + γ ml(˜e k

+ ˜rk

m) + γ km(˜el

+ ˜rl

− βklT + σ klmn˜emn, _˜E_, _˜R_ → 0 (1.476)

where

γ kl = γ KLgk

Kgl

L (1.477)

βkl = βKLgk

Kgl

L (1.478)

σ klmn = _KLMNgk

Kgl

Lgm

Mgn

N. (1.479)

In a similar way, by combining Equation (1.274) and Equation (1.410) one arrives at

qk = −J

−1κKLθ,lxk

,Kxl

,L. (1.480)

Linearizing yields

qk ∼ −κklθ,l , _˜E _, _˜R _ → 0 (1.481)

where

κkl = κKLgk

Kgl

L. (1.482)

Analogous considerations lead to

ρ0ε ∼ ρ0ψ0 + ρ0η0θref + ρ0c(θ)T + [γ kl + βklθref]˜ekl

+ 1

σ klmn˜ekl˜emn, _˜E_, _˜R_ → 0 (1.483)

η = η0 + cT

θref

+ βkl

ρ0

˜e

kl , _˜E _, _˜R_ → 0 (1.484)

_ = ρ0ψ0 − ρ0η0T − ρ0c

2θref

T 2 + [γ kl − βklT ]˜ekl

+ 1

σ klmn˜ekl˜emn, _˜E _, _˜R_ → 0. (1.485)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 57

The balance equations now read

σ kl

+ ρ(f l − ˙vl) = 0 (1.486)

ρ0c ˙ T + βklθref

˙˜ekl − (κklT,l);k − ρ0h = 0 (1.487)

κklT,kT,l ≥ 0. (1.488)

The derivation for isotropic materials runs along the same lines and yields for

_˜E _, _˜R _ → 0

σ kl ∼ γ (1 − ˜em

m)gkl − βT gkl + λ˜em

mgkl + 2(μ + γ )˜ekl (1.489)

qk ∼ −κT,lgkl (1.490)

ρ0ε ∼ ρ0ψ0 + ρ0η0θref + ρ0cT + [γ + βθref]˜em

+ 1

(λ + 2μ)I2

1˜e

− 2μI2˜e (1.491)

η = η0 + cT

θref

+ β

ρ0

I1˜e (1.492)

_ = ρ0ψ0 − ρ0η0T − ρ0c

2θref

T 2 + [γ − βT ]I1˜e

+ 1

(λ + 2μ)I2

1˜e

− 2μI2˜e (1.493)

σ kl

+ ρ(f l − ˙vl) = 0 (1.494)

ρ0c ˙ T + βθref

˙ I1˜e

− (κT,lgkl);k − ρ0h = 0 (1.495)

κT,kT,lgkl ≥ 0. (1.496)

For materials without residual stress and T = 0 Equation (1.489) reduces to

σ kl ∼ λ˜e m

mgkl + 2μ˜ekl . (1.497)

Hence,

σ k

∼ (3λ + 2μ)˜em

m. (1.498)

For a uniform pressure p we have

σ k

= 3p (1.499)

and (see Equation (1.472)),

˜e

∼ J − 1 ∼ dv − dV

, (1.500)

which is the volume change. Hence,

p = (λ + 2

μ)

dv − dV

(1.501)

from which Equation (1.454) follows.

Summarizing, in the small deformation theory, the strain tensors E and e are replaced by

their infinitesimal counterparts ˜E and ˜e . This is only justified for small strains together with

58 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

small rotations. Therefore, it is better to use the expression small deformation theory rather

than infinitesimal strain theory. Using the infinitesimal strains and rotations, the constitutive

equations and balance laws can be simplified. Notice that the term “infinitesimal” does not

apply to other quantities such as stresses. Equations (1.474) and (1.475) show that also in

the linear strain theory the second Piola–Kirchhoff and Cauchy stress both exist and are

generally different. The derived equations are valid in the spatial frame of reference.

1.14.5 Isotropic elastic materials

In this section, we start again from Equation (1.399) to Equation (1.402) and assume that

_ is isotropic in C but that the resulting stress S is not necessarily linear in E. This

covers the large family of so-called isotropic hyperelastic models such as neo–Hooke,

Mooney–Rivlin, Ogden and many others, used for materials such as rubber and hyperfoam.

Because of the isotropy, _ can only be a function of the invariants of C. These will be

denoted in the present context by I1, I2 and I3 (dropping the index C for convenience).

Accordingly,

_ = _(I1, I2, I3, θ,X) (1.502)

where

I1 = tr(C) (1.503)

I2 = 12

[I 2

− tr(C2)] (1.504)

I3 = detC. (1.505)

Consequently, Equation (1.399),

S = 2 _∂_

∂I1

(I1, I2, I3, θ,X)

∂I1

∂C

+ ∂_

∂I2

(I1, I2, I3, θ,X)

∂I2

∂C

+∂_

∂I3

(I1, I2, I3, θ,X)

∂I3

∂C

. (1.506)

Since

∂I1

∂CKL

= ∂CMNGMN

∂CKL

= GKL (1.507)

∂I2

∂CKL

= 1

∂

∂CKL

I 2

− CPQCMNGPNGQM

= 1

2I1GKL − CMNGKNGLM − CPQCPLCQK

= I1GKL − CMNGKNGLM (1.508)

∂I3

∂CKL

= cofactor(CKL) = cofactor(CLK) = I3(C

−1)KL (1.509)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 59

we obtain,

S = 2 _∂_

∂I1

(I1, I2, I3, θ,X)G_ + ∂_

∂I2

(I1, I2, I3, θ,X)(I1G_ − C_)

+∂_

∂I3

(I1, I2, I3, θ,X)I3C

−1_

. (1.510)

Here, the θ-dependence is not specified yet. Whether the function _(I1, I2, I3, θ,X) has

to satisfy specific requirements to make sense physically will be discussed in Chapter 4 on

hyperelastic materials. Since C_ and C

−1 have the same eigenvectors and the eigenvectors

of C_ are not modified by adding or subtracting a multiple of G_, Equation (1.510) shows

that S has the same eigenvectors as C_. Consequently, for an isotropic elastic material

the principal second Piola–Kirchhoff stress directions coincide with the principal stretch

directions.

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