ThermoElasticity notes

8/14/2019 ThermoElasticity notes

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Basic Thermoelasticity

Biswajit Banerjee

November 15, 2006

Contents

1 Governing Equations 1

1.1 Balance Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 The Clausius-Duhem Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Constitutive Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3.1 Other strain and stress measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3.2 Stress Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3.3 Helmholtz and Gibbs free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3.4 Specific Heats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Appendix 9

1 Governing Equations

The equations that govern the motion of a thermoelastic solid include the balance laws for mass, momentum, and

energy. Kinematic equations and constitutive relations are needed to complete the system of equations. Physical

restrictions on the form of the constitutive relations are imposed by an entropy inequality that expresses the second

law of thermodynamics in mathematical form.

The balance laws express the idea that the rate of change of a quantity (mass, momentum, energy) in a volume must

arise from three causes:

1. the physical quantity itself flows through the surface that bounds the volume,

2. there is a source of the physical quantity on the surface of the volume, or/and,

3. there is a source of the physical quantity inside the volume.

Let be the body (an open subset of Euclidean space) and let be its surface (the boundary of).

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Let the motion of material points in the body be described by the map

x = (X) = x(X) (1)

where X is the position of a point in the initial configuration and x is the location of the same point in the deformed

configuration. The deformation gradient (F) is given by

F =x

X=0 x . (2)

1.1 Balance Laws

Let f(x, t) be a physical quantity that is flowing through the body. Let g(x, t) be sources on the surface of the bodyand let h(x, t) be sources inside the body. Let n(x, t) be the outward unit normal to the surface . Let v(x, t) bethe velocity of the physical particles that carry the physical quantity that is flowing. Also, let the speed at which the

bounding surface is moving be un (in the direction n).

Then, balance laws can be expressed in the general form ([1])

d

dt

f(x, t) dV

=

f(x, t)[un(x, t) v(x, t) n(x, t)] dA +

g(x, t) dA +

h(x, t) dV . (3)

Note that the functions f(x, t), g(x, t), and h(x, t) can be scalar valued, vector valued, or tensor valued -depending on the physical quantity that the balance equation deals with.

It can be shown that the balance laws of mass, momentum, and energy can be written as (see Appendix):

+ v = 0 Balance of Mass

v b = 0 Balance of Linear Momentum

= T Balance of Angular Momentum e : (v) + q s = 0 Balance of Energy.

(4)

In the above equations (x, t) is the mass density (current), is the material time derivative of , v(x, t) is theparticle velocity, v is the material time derivative ofv, (x, t) is the Cauchy stress tensor, b(x, t) is the body forcedensity, e(x, t) is the internal energy per unit mass, e is the material time derivative ofe, q(x, t) is the heat fluxvector, and s(x, t) is an energy source per unit mass.

With respect to the reference configuration, the balance laws can be written as

det(F) 0 = 0 Balance of Mass

0 x 0 PT 0 b = 0 Balance of Linear MomentumF P = PT FT Balance of Angular Momentum

0 e PT : F+0 q 0 s = 0 Balance of Energy.

(5)

In the above, P is the first Piola-Kirchhoff stress tensor, and 0 is the mass density in the reference configuration.

The first Piola-Kirchhoff stress tensor is related to the Cauchy stress tensor by

P = det(F) F1 . (6)

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We can alternatively define the nominal stress tensor Nwhich is the transpose of the first Piola-Kirchhoff stress tensor such that

N := PT = det(F) (F1 )T = det(F) FT . (7)

Then the balance laws become

det(F) 0 = 0 Balance of Mass

0 x 0 N 0 b = 0 Balance of Linear Momentum

F NT = N FT Balance of Angular Momentum

0 e N : F+0 q 0 s = 0 Balance of Energy.

(8)

The gradient and divergence operators are defined such that

v =

3i,j=1

vi

xjei ej = vi,jei ej ; v =

3i=1

vi

xi= vi,i ; S=

3i,j=1

Sij

xjei = ij,j ei . (9)

where v is a vector field, BS is a second-order tensor field, and ei are the components of an orthonormal basis in

the current configuration. Also,

0 v =3

i,j=1

vi

XjEiEj = vi,jEiEj ; 0 v =

3i=1

vi

Xi= vi,i ; 0 S=

3i,j=1

Sij

XjEi = Sij,j Ei (10)

where v is a vector field, BS is a second-order tensor field, and Ei are the components of an orthonormal basis in

the reference configuration.

The contraction operation is defined as

A : B =3

i,j=1

Aij Bij = Aij Bij . (11)

1.2 The Clausius-Duhem Inequality

The Clausius-Duhem inequality can be used to express the second law of thermodynamics for elastic-plastic

materials. This inequality is a statement concerning the irreversibility of natural processes, especially when energy

dissipation is involved.

Just like in the balance laws in the previous section, we assume that there is a flux of a quantity, a source of the

quantity, and an internal density of the quantity per unit mass. The quantity of interest in this case is the entropy.

Thus, we assume that there is an entropy flux, an entropy source, and an internal entropy density per unit mass ()

in the region of interest.

Let be such a region and let be its boundary. Then the second law of thermodynamics states that the rate ofincrease of in this region is greater than or equal to the sum of that supplied to (as a flux or from internalsources) and the change of the internal entropy density due to material flowing in and out of the region.

Let move with a velocity un and let particles inside have velocities v. Let n be the unit outward normal tothe surface . Let be the density of matter in the region, q be the entropy flux at the surface, and r be theentropy source per unit mass. Then (see [1, 2]), the entropy inequality may be written as

d

dt

dV

(un v n) dA +

q dA +

r dV . (12)

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The scalar entropy flux can be related to the vector flux at the surface by the relation q = (x) n. Under theassumption of incrementally isothermal conditions (see [3] for a detailed discussion of the assumptions involved),

we have

(x) =q(x)

T; r =

s

T

where q is the heat flux vector, s is a energy source per unit mass, and T is the absolute temperature of a materialpoint at x at time t.

We then have the Clausius-Duhem inequality in integral form:

d

dt

dV

(un v n) dA

q n

TdA +

s

TdV . (13)

We can show that (see Appendix) the entropy inequality may be written in differential form as

q

T+ s

T. (14)

In terms of the Cauchy stress and the internal energy, the Clausius-Duhem inequality may be written as (see

Appendix)

(e T ) :v q T

T. (15)

1.3 Constitutive Relations

A set of constitutive equations is required to close to system of balance laws. For large deformation plasticity, we

have to define appropriate kinematic quantities and stress measures so that constitutive relations between them may

have a physical meaning.

Let the fundamental kinematic quantity be the deformation gradient (F) which is given by

F =x

X=0 x ; detF > 0 .

A thermoelastic material is one in which the internal energy (e) is a function only ofF and the specific entropy (),

that is

e = e(F, ) .

For a thermoelastic material, we can show that the entropy inequality can be written as (see Appendix)

e

T

+

eF

FT

: F+ q TT

0 . (16)

At this stage, we make the following constitutive assumptions:

1. Like the internal energy, we assume that and T are also functions only ofF and , i.e.,

= (F, ) ; T = T(F, ) .

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2. The heat flux q satisfies the thermal conductivity inequality and ifq is independent of and F, we have

q T 0 = ( T) T 0 = 0

i.e., the thermal conductivity is positive semidefinite.

Therefore, the entropy inequality may be written as

e

T

+

e

F FT

: F 0 .

Since and F are arbitrary, the entropy inequality will be satisfied if and only if

e

T = 0 = T =

e

and

e

F FT = 0 = =

e

F FT .

Therefore,

T =

e

and =

e

F F

T

. (17)

Given the above relations, the energy equation may expressed in terms of the specific entropy as (see Appendix)

T = q + s . (18)

Effect of a rigid body rotation of the internal energy:If a thermoelastic body is subjected to a rigid body rotation Q, then its internal energy should not change. After a rotation, the new deformation

gradient (F) is given byF= Q F .

Since the internal energy does not change, we must have

e = e(F, ) = e(F, ) .

Now, from the polar decomposition theorem, F= RUwhereRis the orthogonal rotation tensor (i.e., RRT

= RT

R= 1 ) andU is thesymmetric right stretch tensor. Therefore,

e(Q RU, ) = e(F, ) .

Now, we can choose any rotation Q. In particular, if we choose Q= RT, we have

e(RT RU, ) = e(1 U, ) = e(U, ) .

Therefore,

e(U, ) = e(F, ) .

This means that the internal energy depends only on the stretch Uand not on the orientation of the body.

1.3.1 Other strain and stress measures

The internal energy depends on F only through the stretch U. A strain measure that reflects this fact and also

vanishes in the reference configuration is the Green strain

E=1

2(FT F 1 ) =

1

2(U2 1 ) . (19)

Recall that the Cauchy stress is given by

= e

F FT .

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We can show that the Cauchy stress can be expressed in terms of the Green strain as (see Appendix)

= F e

E FT . (20)

Recall that the nominal stress tensor is defined as

N= detF ( FT) .

From the conservation of mass, we have 0 = detF. Hence,

N=0

FT . (21)

The nominal stress is unsymmetric. We can define a symmetric stress measure with respect to the reference

configuration call the second Piola-Kirchhoff stress tensor (S):

S := F1 N= P FT =0

F1 FT . (22)

In terms of the derivatives of the internal energy, we have

S=0

F1

F

e

E FT

FT = 0

e

E

and

N=0

F

e

E FT

FT = 0 F

e

E.

That is,

S= 0 eE

and N= 0 F eE

. (23)

1.3.2 Stress Power

The stress power per unit volume is given by :v. In terms of the stress measures in the referenceconfiguration, we have

:v =

F

e

E FT

: (F F1) .

Using the identity A : (B C) = (A CT) : B, we have

:v =

F eE

FT

FT

: F = F e

E

: F =

0N : F .

We can alternatively express the stress power in terms ofSand E. Taking the material time derivative ofEwe

have

E=1

2(FT F+ FT F) .

Therefore,

S : E=1

2[S : (FT F) + S : (FT F)] .

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Using the identities A : (B C) = (A CT) : B = (BT A) : Cand A : B = AT : BT and using the symmetryofS, we have

S : E=1

2[(S FT) : FT + (F S) : F] =

1

2[(F ST) : F+ (F S) : F] = (F S) : F .

Now, S= F1

N. Therefore, S : E= N : F. Hence, the stress power can be expressed as

:v = N : F = S : E . (24)

If we split the velocity gradient into symmetric and skew parts using

v = l = d+w

where d is the rate of deformation tensor and w is the spin tensor, we have

:v = : d+ : w = tr(T d) + tr(T w) = tr( d) + tr( w) .

Since is symmetric and w is skew, we have tr( w) = 0. Therefore, :v = tr( d). Hence, we may alsoexpress the stress power as

tr( d) = tr(NT F) = tr(S E) . (25)

1.3.3 Helmholtz and Gibbs free energy

Recall that

S= 0e

E.

Therefore,

e

E=

1

0S .

Also recall thate

= T .

Now, the internal energy e = e(E, ) is a function only of the Green strain and the specific entropy. Let us assume,that the above relations can be uniquely inverted locally at a material point so that we have

E= E(S, T) and = (S, T) .

Then the specific internal energy, the specific entropy, and the stress can also be expressed as functions ofSand T,

or Eand T, i.e.,

e = e(E, ) = e(S, T) = e(E, T) ; = (S, T) = (E, T) ; and S= S(E, T)

We can show that (see Appendix)

d

dt(e T ) = T +

1

0S : E or

d

dt= T +

1

0S : E . (26)

andd

dt(e T

1

0S : E) = T

1

0S : E or

dg

dt= T +

1

0S : E . (27)

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The difference between Cp and Cv can be expressed as

Cp Cv =1

0

S T

S

T

:

E

T. (38)

However, it is more common to express the above relation in terms of the elastic modulus tensor as (see Appendixfor proof)

Cp Cv =1

0S : E +

T

0E : C : E (39)

where the fourth-order tensor of elastic moduli is defined as

C :=S

E= 0

2

EE. (40)

For isotropic materials with a constant coefficient of thermal expansion that follow the St. Venant-Kirchhoff

material model, we can show that (see Appendix)

Cp Cv =1

0

tr(S) + 9 2 K T

.

2 Appendix

1. The integral

F(t) =

Zb(t)a(t)

f(x, t) dx

is a function of the parameter t. Show that the derivative ofF is given by

dF

dt=

d

dt

Zb(t)a(t)

f(x, t) dx

!=

Zb(t)a(t)

f(x, t)

tdx + f[b(t), t]

b(t)

t f[a(t), t]

a(t)

t.

This relation is also known as the Leibniz rule.

The following proof is taken from [4].

We have,

dF

dt= lim

t0

F(t + t) F(t)

t.

Now,

F(t + t) F(t)

t=

1

t

"Zb(t+t)a(t+t)

f(x, t + t) dx

Zb(t)a(t)

f(x, t) dx

#

1

t

Zb+ba+a

f(x, t + t) dx

Zba

f(x, t) dx

=

1

t

Za+aa f(x, t + t) dx +

Zb+ba f(x, t + t) dx

Zba f(x, t) dx

=

1

t

Za+aa

f(x, t + t) dx +

Zba

f(x, t + t) dx +

Zb+bb

f(x, t + t) dx

Zba

f(x, t) dx

=

Zba

f(x, t + t) f(x, t)

tdx +

1

t

Zb+bb

f(x, t + t) dx 1

t

Za+aa

f(x, t + t) dx .

Since f(x, t) is essentially constant over the infinitesimal intervals a < x < a + a and b < x < b + b, we may write

F(t + t) F(t)

t

Zba

f(x, t + t) f(x, t)

tdx + f(b, t + t)

b

t f(a, t + t)

a

t.

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Therefore,

d

dt

Z(t)

f(x, t) dV

!=

Z(t)

f(x, t)

t+ [f(x, t)] v(x, t) + f(x, t) v(x, t)

dV

or,

d

dt

Z(t)

f dV

!=

Z(t)

f

t+f v + f v

dV .

Using the identity

(vw) = v( w) +v w

we then have

d

dt

Z(t)

f dV

!=

Z(t)

f

t+ (fv)

dV .

Using the divergence theorem and the identity (ab) n = (b n)a we have

d

dt

Z(t)

f dV

!=

Z(t)

f

tdV +

Z(t)

(fv) n dV =

Z(t)

f

tdV +

Z(t)

(v n)f dV .

3. Show that the balance of mass can be expressed as:

+ v = 0

where (x, t) is the current mass density, is the material time derivative of, and v(x, t) is the velocity of physical particles in the body bounded by the surface .

Recall that the general equation for the balance of a physical quantity f(x, t) is given by

d

dt

Z

f(x, t) dV

=

Z


Z

g(x, t) dA +

Z

h(x, t) dV .

To derive the equation for the balance of mass, we assume that the physical quantity of interest is the mass density (x, t). Since mass is neithercreated or destroyed, the surface and interior sources are zero, i.e., g(x, t) = h(x, t) = 0 . Therefore, we have

d

dt

Z

(x, t) dV

=

Z

(x, t)[un(x, t) v(x, t) n(x, t)] dA .

Let us assume that the volume is a control volume (i.e., it does not change with time). Then the surface has a zero velocity (un = 0) and we get

Z

t

dV = Z

(v n) dA .

Using the divergence theorem Z v dV =

Z

v n dA

we get Z

tdV =

Z ( v) dV.

or, Z

t+ ( v)

dV = 0 .

Since is arbitrary, we must have

t+ ( v) = 0 .

Using the identity

( v) = v + v

we have

t+ v + v = 0 .

Now, the material time derivative of is defined as

=

t+ v .

Therefore,

+ v = 0 .

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4. Show that the balance of linear momentum can be expressed as:

v b = 0

where (x, t) is the mass density, v(x, t) is the velocity, (x, t) is the Cauchy stress, and b is the body force density.

Recall the general equation for the balance of a physical quantity

d

dt

Z

f(x, t) dV

=Z

f(x, t)[un(x, t) v(x, t) n(x, t)] dA +Z

g(x, t) dA +Z

h(x, t) dV .

In this case the physical quantity of interest is the momentum density, i.e., f(x, t) = (x, t) v(x, t). The source of momentum flux at the surface isthe surface traction, i.e., g(x, t) = t. The source of momentum inside the body is the body force, i.e., h(x, t) = (x, t) b(x, t). Therefore, we have

d

dt

Z

v dV

=

Z

v[un v n] dA +

Z

t dA +

Z

b dV .

The surface tractions are related to the Cauchy stress by

t = n .

Therefore,d

dt

Z

v dV

=

Z

v[un v n] dA +

Z n dA +

Z

b dV .

Let us assume that is an arbitrary fixed control volume. Then,Z

t( v) dV =

Z

v (v n) dA +

Z n dA +

Z

b dV .

Now, from the definition of the tensor product we have (for all vectors a)

(uv) a = (a v) u .

Therefore, Z

t( v) dV =

Z

(vv) n dA +

Z n dA +

Z

b dV .

Using the divergence theorem Z v dV =

Z

v n dA

we have Z

t( v) dV =

Z [ (vv)] dV +

Z dV +

Z

b dV

or, Z

t( v) + [( v)v] b

dV = 0 .

Since is arbitrary, we have

t( v) + [( v)v] b = 0 .

Using the identity

(uv) = ( v)u + (u) v

we get

tv +

v

t+ ( v)(v) +( v) v b = 0

or,

t+ v

v +

v

t+( v) v b = 0

Using the identity

( v) = v + v ()

we get

t+ v

v +

v

t+ [v + v ()] v b = 0

From the definition

(uv) a = (a v) u

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we have

[v ()] v = [v ()] v .

Hence,

t+ v

v +

v

t+ v v + [v ()] v b = 0

or,

t + v + v

v +

v

t + v v b = 0 .

The material time derivative of is defined as

=

t+ v .

Therefore,

[ + v]v + v

t+ v v b = 0 .

From the balance of mass, we have

+ v = 0 .

Therefore,

v

t+ v v b = 0 .

The material time derivative ofv is defined as

v =

v

t +v v .

Hence,

v b = 0 .

5. Show that the balance of angular momentum can be expressed as:

= T

We assume that there are no surface couples on or body couples in . Recall the general balance equation

d

dt

Z

f(x, t) dV

=

Z


Z

g(x, t) dA +

Z

h(x, t) dV .

In this case, the physical quantity to be conserved the angular momentum density, i.e., f = x ( v). The angular momentum source at the surfaceis then g = x t and the angular momentum source inside the body is h = x ( b). The angular momentum and moments are calculated with

respect to a fixed origin. Hence we have

d

dt

Zx ( v) dV

=

Z

[x ( v)][un v n] dA +

Z

x t dA +

Zx ( b) dV .

Assuming that is a control volume, we have

Zx

t( v)

dV =

Z

[x ( v)][v n] dA +

Z

x t dA +

Zx ( b) dV .

Using the definition of a tensor product we can write

[x ( v)][v n] = [[x ( v)]v] n .

Also, t = n. Therefore we have

Zx

t( v)

dV =

Z

[[x ( v)]v] n dA +Z

x ( n) dA +Zx ( b) dV .

Using the divergence theorem, we get

Zx

t( v)

dV =

Z [[x ( v)]v] dV +

Z

x ( n) dA +

Zx ( b) dV .

To convert the surface integral in the above equation into a volume integral, it is convenient to use index notation. Thus,

Z

x ( n) dA

i

=

Z

eijk xj kl nl dA =

Z

Ail nl dA =

ZA n dA

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where [ ]i represents the i-th component of the vector. Using the divergence theorem

ZA n dA =

Z A dV =

Z

Ail

xldV =

Z

xl(eijk xj kl) dV .

Differentiating,

ZA n dA =

Z

eijk jl kl + eijk xj kl

xl

dV =

Z

eijk kj + eijk xj kl

xl

dV =

Z

eijk kj + eijk xj [ ]l

dV .

Expressed in direct tensor notation, ZA n dA =

Z

h[E : T]i + [x ( )]i

idV

where E is the third-order permutation tensor. Therefore,

Z

x ( n) dA

i

==

Z

h[E : T]i + [x ( )]i

idV

or, Z

x ( n) dA ==

Z

hE : T + x ( )

idV .

The balance of angular momentum can then be written as

Zx

t( v)

dV =

Z [[x ( v)]v] dV +

Z

hE : T + x ( )

idV +

Zx ( b) dV .

Since is an arbitrary volume, we have

x

t( v)

= [[x ( v)]v] + E : T + x ( ) + x ( b)

or,

x

t( v) b

= [[x ( v)]v] + E : T .

Using the identity,

(uv) = ( v)u + (u) v

we get

[[x ( v)]v] = ( v)[x ( v)] + ([x ( v)]) v .

The second term on the right can be further simplified using index notation as follows.

[([x ( v)]) v]i = [([ (xv)]) v]i =

xl( eijk xj vk) vl

= eijk

xlxj vk vl +

xj

xlvk vl + xj

vk

xlvl

= (eijk xj vk)

xlvl

+ (eijk jl vk vl) + eijk xj

vk

xlvl

= [(xv)( v) + vv + x (v v)]i

= [(xv)( v) + x (v v)]i .

Therefore we can write

[[x ( v)]v] = ( v)(x v) + ( v)(xv) + x (v v) .

The balance of angular momentum then takes the form

x

t( v) b

= ( v)(x v) ( v)(xv) x (v v) + E : T

or,

x

t( v) + v v b

= ( v)(x v) ( v)(xv) + E : T

or,

x

v

t+

tv + v v b

= ( v)(x v) ( v)(xv) + E : T

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The material time derivative ofv is defined as

v =v

t+v v .

Therefore,

x [ v b] = x

tv + ( v)(x v) ( v)(xv) + E : T .

Also, from the conservation of linear momentum

v b = 0 .

Hence,

0 = x

tv + ( v)(x v) + ( v)(xv) E : T

=

t+ v + v

(xv) E : T .

The material time derivative of is defined as

=

t+ v .

Hence,

( + v)(xv) E : T = 0 .

From the balance of mass

+ v = 0 .

Therefore,E : T = 0 .

In index notation,

eijk kj = 0 .

Expanding out, we get

12 21 = 0 ; 23 32 = 0 ; 31 13 = 0 .

Hence,

= T

6. Show that the balance of energy can be expressed as:

e : (v) + q s = 0

where (x, t) is the mass density, e(x, t) is the internal energy per unit mass, (x, t) is the Cauchy stress, v(x, t) is the particle velocity, q is theheat flux vector, and s is the rate at which energy is generated by sources inside the volume (per unit mass).

Recall the general balance equation

d

dt

Z

f(x, t) dV

=

Z


Z

g(x, t) dA +

Z

h(x, t) dV .

In this case, the physical quantity to be conserved the total energy density which is the sum of the internal energy density and the kinetic energy

density, i.e., f = e + 1/2 |v v|. The energy source at the surface is a sum of the rate of work done by the applied tractions and the rate of heatleaving the volume (per unit area), i.e, g = v t q n where n is the outward unit normal to the surface. The energy source inside the body is thesum of the rate of work done by the body forces and the rate of energy generated by internal sources, i.e., h = v (b) + s.

Hence we have

d

dt Z e +1

2v v dV = Z e +

1

2v v (un v n) dA + Z(v t q n) dA + Z (v b + s) dV .

Let be a control volume that does not change with time. Then we get

Z

t

e +

1

2v v

dV =

Z

e +

1

2v v

(v n) dA +

Z

(v t q n) dA +

Z

(v b + s) dV .

Using the relation t = n, the identity v ( n) = (T v) n, and invoking the symmetry of the stress tensor, we get

Z

t

e +

1

2v v

dV =

Z

e +

1

2v v

(v n) dA +

Z

( v q) n dA +

Z

(v b + s) dV .

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We now apply the divergence theorem to the surface integrals to get

Z

t

e +

1

2v v

dV =

Z

e +

1

2v v

v

dA +

Z ( v) dA

Z q dA +

Z

(v b + s) dV .

Since is arbitrary, we have

t

e +1

2 v v

=

e +1

2 v vv

+ ( v) q + (v b + s) .

Expanding out the left hand side, we have

t

e +

1

2v v

=

t

e +

1

2v v

+

e

t+

1

2

t(v v)

=

t

e +

1

2v v

+

e

t+

v

t v .

For the first term on the right hand side, we use the identity ( v) = v + v to get

e +

1

2v v

v

=

e +

1

2v v

v +

e +

1

2v v

v

= e +1

2

v v v + e +1

2

v v v + e +1

2

v v v=

e +

1

2v v

v +

e +

1

2v v

v + e v +

1

2(v v) v

=

e +

1

2v v

v +

e +

1

2v v

v + e v + (vT v) v

=

e +

1

2v v

v +

e +

1

2v v

v + e v + (v v) v .

For the second term on the right we use the identity (ST v) = S :v + ( S) v and the symmetry of the Cauchy stress tensor to get

( v) = :v + ( ) v .

After collecting terms and rearranging, we get

t

+ v + ve + 12v v+ v

t+ v v b v+ e

t+e v+ :v+ q s = 0 .

Applying the balance of mass to the first term and the balance of linear momentum to the second term, and using the material time derivative of the

internal energy

e =e

t+e v

we get the final form of the balance of energy:

e :v + q s = 0 .

7. Show that the Clausius-Duhem inequality in integral form:

d

dt

Z

dV

Z

(un v n) dA

Z

q n

TdA +

Z

s

TdV .

can be written in differential form as

qT

!+ s

T.

Assume that is an arbitrary fixed control volume. Then un = 0 and the derivative can be take inside the integral to give

Z

t( ) dV

Z

(v n) dA

Z

q n

TdA +

Z

s

TdV .

Using the divergence theorem, we get

Z

t( ) dV

Z ( v) dV

Z

q

T

!dV +

Z

s

TdV .

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Since is arbitrary, we must have

t( ) ( v)

q

T

!+

s

T.

Expanding out

t +

t () v ( v)

q

T

!+

s

T

or,

t +

t v v ( v)

q

T

!+

s

T

or,

t+ v + v

+

t+ v

q

T

!+

s

T.

Now, the material time derivatives of and are given by

=

t+ v ; =

t+ v .

Therefore,

( + v) + q

T! +

s

T

.

From the conservation of mass + v = 0 . Hence,

q

T

!+

s

T.

8. Show that the Clausius-Duhem inequality

q

T

!+

s

T

can be expressed in terms of the internal energy as

(e T ) :v q T

T

.

Using the identity ( v) = v + v in the Clausius-Duhem inequality, we get

q

T

!+

s

Tor

1

T q q

1

T

!+

s

T.

Now, using index notation with respect to a Cartesian basis ej ,

1

T

!=

xj

`T1

ej =

`T2

Txj

ej = 1

T2T .

Hence,

1

T q +

1

T2q T +

s

Tor

1

T( q s) +

1

T2q T .

Recall the balance of energy

e :v + q s = 0 = e :v = ( q s) .

Therefore,

1

T( e :v) +

1

T2q T = T e :v +

q T

T.

Rearranging,

(e T ) :v q T

T.

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9. For thermoelastic materials, the internal energy is a function only of the deformation gradient and the temperature, i.e., e = e(F, T). Show that, forthermoelastic materials, the Clausius-Duhem inequality

(e T ) :v q T

T

can be expressed as

e

T

+

eF

FT

: F q TT

.

Since e = e(F, T), we have

e =e

F: F+

e

.

Therefore,

e

F: F+

e

T

:v

q T

Tor

e

T

+

e

F: F :v

q T

T.

Now,v = l = F F1. Therefore, using the identity A : (B C) = (A CT) : B, we have

:v = : (F F1) = ( FT) : F .

Hence,

e

T

+

e

F: F ( FT) : F

q T

T

or,

e

T

+

e

F FT

: F

q T

T.

10. Show that, for thermoelastic materials, the balance of energy

e :v + q s = 0 .

can be expressed as

T = q + s .

Since e = e(F, T), we have

e = eF

: F+ e

.

Plug into energy equation to get

e

F: F+

e

:v + q s = 0 .

Recall,e

= T and

e

F= FT .

Hence,

( FT) : F+ T :v + q s = 0 .

Now,v = l = F F1. Therefore, using the identity A : (B C) = (A CT) : B, we have

:v = : (F F1) = ( FT) : F .

Plugging into the energy equation, we have

:v + T :v + q s = 0

or,

T = q + s .

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11. Show that, for thermoelastic materials, the Cauchy stress can be expressed in terms of the Green strain as

= F e

E FT .

Recall that the Cauchy stress is given by

= e

F FT = ij =

e

Fik FTkj =

e

Fik Fjk .

The Green strain E = E(F) = E(U) and e = e(F, ) = e(U, ). Hence, using the chain rule,

e

F=

e

E:

E

F=

e

Fik=

e

Elm

Elm

Fik.

Now,

E =1

2(FT F 1) = Elm =

1

2(FTlp Fpm lm) =

1

2(Fpl Fpm lm) .

Taking the derivative with respect to F, we get

E

F=

1

2

FT

F F+ FT

F

F

=

Elm

Fik=

1

2

Fpl

FikFpm + Fpl

Fpm

Fik

.

Therefore,

=1

2

e

E:

FT

F F+FT

F

F

FT = ij =

1

2

e

Elm

Fpl

FikFpm + Fpl

Fpm

Fik

Fjk .

Recall,A

A

Aij

Akl= ik jl and

AT

A

Aji

Akl= jk il .

Therefore,

ij =1

2

e

Elm

`pi lk Fpm + Fpl pi mk

Fjk =

1

2

e

Elm(lk Fim + Fil mk)

Fjk

or,

ij =1

2

e

EkmFim +

e

ElkFil

Fjk = =

1

2

"F

e

E

T+F

e

E

# FT

or,

=1

2 F

"e

E

T+

e

E

# FT .

From the symmetry of the Cauchy stress, we have

= (F A) FT and T = F (F A)T = F AT FT and = T = A= AT .

Therefore,

e

E=

e

E

Tand we get

= F e

E FT .

12. For thermoelastic materials, the specific internal energy is given by

e = e(E, )

whereE is the Green strain and is the specific entropy. Show that

d

dt(e T ) = T +

1

0S : E and

d

dt(e T

1

0S : E) = T

1

0S : E

where 0 is the initial density, T is the absolute temperature, Sis the 2nd Piola-Kirchhoff stress, and a dot over a quantity indicates the material timederivative.

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20/25

Taking the material time derivative of the specific internal energy, we get

e =e

E: E+

e

.

Now, for thermoelastic materials,

T =e

and S= 0e

E

.

Therefore,

e =1

0S : E+ T . = e T =

1

0S : E .

Now,d

dt(T ) = T + T .

Therefore,

e d

dt(T ) + T =

1

0S : E =

d

dt(e T ) = T +

1

0S : E .

Also,

d

dt

1

0S : E

!=

1

0S : E+

1

0S : E .

Hence,

e d

dt(T ) + T =

d

dt

1

0S : E

!

1

0S : E =

d

dt

e T

1

0S : E

!= T

1

0S : E .

13. For thermoelastic materials, show that the following relations hold:

E=

1

0S(E, T) ;

T= (E, T) ;

g

S=

1

0E(S, T) ;

g

T= (S, T)

where (E, T) is the Helmholtz free energy and g(S, T) is the Gibbs free energy.

Also show thatS

T = 0

E andE

T = 0

S .

Recall that

(E, T) = e T = e(E, ) T .

and

g(S, T) = e + T +1

0S : E .

(Note that we can choose any functional dependence that we like, because the quantities e, , E are the actual quantities and not any particularfunctional relations).

The derivatives are

E=

e

E=

1

0S;

T= .

andg

S=

1

0

S

S: E =

1

0E ;

g

T= .

Hence,

E=

1

0S(E, T) ;

T= (E, T) ;

g

S=

1

0E(S, T) ;

g

T= (S, T)

From the above, we have

2

T E=

2

ET=

E=

1

0

S

T.

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and2g

T S=

2g

ST=

S=

1

0

E

T.

Hence,

S

T= 0

Eand

E

T= 0

S.

14. For thermoelastic materials, show that the following relations hold:

e(E, T)

T= T

T= T

2

T2

ande(S, T)

T= T

T+

1

0S :

E

T= T

2g

T2+S :

2g

ST.

Recall,

(E, T) = = e T = e(E, T) T (E, T)

and

g(S, T) = g = e + T +1

0S : E = e(S, T) + T (S, T) +

1

0S : E(S, T) .

Therefore,e(E, T)

T=

T+ (E, T) + T

T

ande(S, T)

T=

g

T+ (S, T) + T

T+

1

0S :

E

T.

Also, recall that

(E, T) =

T=

T=

2

T2,

(S, T) =g

T=

T=

2g

T2,

and

E(S, T) = 0g

S=

E

T= 0

2g

ST.

Hence,

e(E, T)

T= T

T= T

2

T2

and

e(S, T)

T= T

T+

1

0S :

E

T= T

2g

T2+S :

2g

ST.

15. For thermoelastic materials, show that the balance of energy equation

T = q + s

can be expressed as either

Cv T = ( T) + s +

0T

S

T: E

or

Cp

1

0S :

E

T

!T = ( T) + s

0T

E

T: S

where

Cv =e(E, T)

Tand Cp =

e(S, T)

T.

If the independent variables are E and T, then

= (E, T) = =

E: E+

TT .

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22/25

On the other hand, if we consider Sand T to be the independent variables

= (S, T) = =

S: S+

TT .

Since

E

= 1

0

S

T

;

T

=Cv

T

;

S

=1

0

E

T

; and

T

=1

TCp

1

0

S :E

T!

we have, either

= 1

0

S

T: E+

Cv

TT

or

=1

0

E

T: S+

1

T

Cp

1

0S :

E

T

!T .

The equation for balance of energy in terms of the specific entropy is

T = q + s .

Using the two forms of , we get two forms of the energy equation:

0T

S

T : E+ Cv T = q + s

and

0T

E

T: S+ Cp T

0S :

E

TT = q + s .

From Fouriers law of heat conduction

q = T .

Therefore,

0T

S

T: E+ Cv T = ( T) + s

and

0T

E

T: S+ Cp T

0S :

E

TT = ( T) + s .

Rearranging,

Cv T = ( T) + s +

0T

S

T: E

or,

Cp

1

0S :

E

T

!T = ( T) + s

0T

E

T: S.

16. For thermoelastic materials, show that the specific heats are related by the relation

Cp Cv =1

0

S T

S

T

!:

E

T.

Recall that

Cv :=e(E, T)

T= T

T

and

Cp :=e(S, T)

T= T

T+

1

0S :

E

T.

Therefore,

Cp Cv = T

T+

1

0S :

E

T T

T.

Also recall that

= (E, T) = (S, T) .

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23/25

Therefore, keeping Sconstant while differentiating, we have

T=

E:

E

T+

T.

Noting thatE = E(S, T), and plugging back into the equation for the difference between the two specific heats, we have

Cp Cv = T E

: ET

+ 10S : E

T.

Recalling that

E=

1

0

S

T

we get

Cp Cv =1

0

S T

S

T

!:

E

T.

17. For thermoelastic materials, show that the specific heats can also be related by the equations

Cp Cv =1

0

S :E

T

+E

T

: 2

EE

:E

T =

1

0

S :E

T

+T

0

E

T

: S

E

:E

T .

Recall that

S= 0

E= 0 f(E(S, T), T) .

Recall the chain rule which states that if

g(u, t) = f(x(u, t), y(u, t))

then, if we keep u fixed, the partial derivative ofg with respect to t is given by

g

t=

f

x

x

t+

f

y

y

t.

In our case,

u = S, t = T, g(S, T) = S, x(S, T) = E(S, T), y(S, T) = T, and f = 0 f.

Hence, we have

S= g(S, T) = f(E(S, T), T) = 0 f(E(S, T), T) .

Taking the derivative with respect to T keepingSconstant, we have

g

T=7

0

S

T= 0

2664 fE : ET + fT

71

T

T

3775

or,

0 =f

E:

E

T+

f

T.

Now,

f=

E=

f

E=

2

EEand

f

T=

2

T E.

Therefore,

0 =2

EE:

E

T+

2

T E=

E

E

:

E

T+

T

E

.

Again recall that,

E=

1

0S.

Plugging into the above, we get

0 =2

EE:

E

T+

1

0

S

T=

1

0

S

E:

E

T+

1

0

S

T.

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24/25

Therefore, we get the following relation for S/T:

S

T= 0

2

EE:

E

T=

S

E:

E

T.

Recall that

Cp Cv =1

0S T S

T : E

T

.

Plugging in the expressions for S/T we get:

Cp Cv =1

0

S+ T 0

2

EE:

E

T

:

E

T=

1

0

S+ T

S

E:

E

T

:

E

T.

Therefore,

Cp Cv =1

0S :

E

T+ T

2

EE:

E

T

:

E

T=

1

0S :

E

T+

T

0

S

E:

E

T

:

E

T.

Using the identity (A : B) : C= C : (A : B), we have

Cp Cv =1

0S :

E

T+ T

E

T:

2

EE:

E

T

=

1

0S :

E

T+

T

0

E

T:

S

E:

E

T

.

18. Consider an isotropic thermoelastic material that has a constant coefficient of thermal expansion and which follows the St-Venant Kirchhoff model,

i.e,

E = 1 and C = 1 1 + 2 I

where is the coefficient of thermal expansion and 3 = 3 K 2 where K, are the bulk and shear moduli, respectively.

Show that the specific heats related by the equation

Cp Cv =1

0

tr(S) + 9 2 K T .

Recall that,

Cp Cv =1

0S : E +

T

0E : C : E .

Plugging the expressions ofE and C into the above equation, we have

Cp Cv =1

0S : ( 1) +

T

0( 1) : ( 1 1 + 2 I) : ( 1)

=

0tr(S) +

2 T

01 : ( 1 1 + 2 I) : 1

=

0tr(S) +

2 T

01 : ( tr(1 ) 1 + 2 1)

=

0tr(S) +

2 T

0(3 tr(1) + 2 tr(1 ))

=

0tr(S) +

3 2 T

0(3 + 2)

= tr(S)

0+

9 2 K T

0.

Therefore,

Cp Cv =1

0

tr(S) + 9 2 K T .

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References

[1] T. W. Wright. The Physics and Mathematics of Adiabatic Shear Bands. Cambridge University Press,

Cambridge, UK, 2002.

[2] R. C. Batra. Elements of Continuum Mechanics. AIAA, Reston, VA., 2006.

[3] G. A. Maugin. The Thermomechanics of Nonlinear Irreversible Behaviors: An Introduction. World Scientific,

Singapore, 1999.

[4] M. D. Greenberg. Foundations of Applied Mathematics. Prentice-Hall Inc., Englewood Cliffs, New Jersey,

1978.

[5] T. Belytschko, W. K. Liu, and B. Moran. Nonlinear Finite Elements for Continua and Structures. John Wiley

and Sons, Ltd., New York, 2000.

[6] M. E. Gurtin. An Introduction to Continuum Mechanics. Academic Press, New York, 1981.

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