ACME 2011 Proceedings of the 19 UK National Conference of the ...

( , t)
t
v = Dx
Dt = ∂x
X ∂ but material points are not necessarily tracked in the case of the mapping
* *
*
χ ֏ x χ , t . Note the use of D
*
D t rather than the ordinary derivative d dt in equations (1) even
( )
though these are identical when applied to a function of t. Although the integrals ∫ Ω ρhdV and ∫ Ω ρdV
* *
are functions of t the derivative D D t is used to immediately relay the notion that Ω is a control
*
*
volume transported through v . Although v is present in equations (1) it cannot influence the value of
ψ as this would be physically meaningless. Typical transport equations are:
*
D *
D
t
*
D t
∫
∫
ρdV + ∫ ρ
∫
( v − v ) ⋅ ndΓ
= 0
*
Ω Γ
*
D *
D
t
∫
( v − v ) ⋅ ndΓ
= ∫ v ⋅ σ ⋅ ndΓ
− ∫ q ⋅ ndΓ
+ ∫ ρQdV
+ ∫
ρedV + ρe
ρv
⋅ bdV
Ω
Γ
*
D *
∫
( v − v ) ⋅ ndΓ
= ∫ σ ⋅ ndΓ
+ ∫
ρvdV + ρv
ρbdV
*
Ω
Γ
Γ
Ω
*
D *
−1
−1
∫ sdV + ∫ρs(
v − v ) ⋅ ndΓ
≥ −∫
T q ⋅ ndΓ
+ ∫ρT
*
D t Ω
Γ
Γ
Ω
ρ QdV
Γ
Γ
1
for transport of mass, energy, momentum, and entropy, where = u + v ⋅ v , u and s are internal energy,
Ω
e 2
and entropy per unit mass, q ⋅ n is heat flux, Q represents a heat source, σ is the Cauchy stress tensor
and b is a body force. The equality arising in equation (5) is as a consequence of irreversibly arising
from finite temperature differences and dissipation.
3 MECHANICAL ENERGY EQUATIONS AND DISSIPATION
A particular interest here is the splitting (if possible) of the energy equation (3) into thermal and
mechanical parts. To achieve this partition it is shown here that a necessary and sufficient condition is
that the solid possesses an entropy function that is dependent on temperature only. Consider for
demonstration purposes the deformation of a hyperelastic material.
Theorem 3.1 A hyperelastic material dependent on state variables temperature T and the independent
strain components E ij of the Green-Lagrange strain tensor E has internal energy in the form
T E
u u u
= + , where
T
u and
dependent on temperature only.
E
u are functions of T and ij
Ω
(2)
(3)
(4)
(5)
E , respectively, if and only if its entropy is
Proof 3.2: The central equation of thermodynamics for the solid is Tds = du − dwR
, where reversible
differential work dw
−1 −1
= ρ dω = ρ S: dE , and where strain energy ( ) E S : E
~
ω
E ij
= ∫ S d 1 = , E is
R 0 0 0
0 Eij 0
ij ij 2
the Green-Lagrange strain measure, which is work conjugate to the 2 nd Piola-Kirchoff stress S . Here 0 ρ
represents the density in a material reference configuration and by construction ρ σ : εɺ
= ρS:
Eɺ
, where
E ɺ represents material strain rate and εɺ is the Eulerian strain-rate tensor.
Consider internal energy of the form u( T,E ij ) with the central equation of thermodynamics
−1
( Tds dT − ∂u ∂ T) dT = ( ∂u ∂Eij − ρ 0 Sij ) dEij
under the assumption s = s( T)
and it immediately
follows that
2
∂ u ∂Eij∂ T = 0 hence
T E
u u u
= + is obtained on integration. Similarly, conversely,
assuming internal energy to be of the form
0
T E
u u u
= + gives
10