3.1 Hooke Model
65
ψ() =
1
2
E
2
.
(3.2)
Then the (energetic ˆ
=) total stress σ, which is conjugated to the total strain , follows
as
σ() = ∂ ψ() = E .
(3.3)
Note that the total stress σ applied to the rheological model (that enters the equilibrium condition) coincides identically with the energetic stress, σ
≡ σ. The specific
Hooke model is entirely energetic, thus the dissipation potential π = 0 is zero (as is
the corresponding dual dissipation potential π
∗
= 0). Consequently, the dissipative
stress σ
≡ 0 vanishes identically and no distinction is made between the total stress
σ and the energetic stress σ
.
The corresponding free enthalpy density ψ
∗ , as determined from a Legendre
transformation
ψ
∗
(σ) = max
σ σ −
1
2
E ||
2
(3.4)
then reads, with C := 1/E denoting the compliance,
ψ
∗
(σ) =
1
2
C σ
2
.
(3.5)
Accordingly, the respective constitutive law for the total strain follows as
(σ) = ∂ σ ψ
∗
(σ) = C σ.
(3.6)
Obviously, the expressions in Eqs. 3.3 and 3.6 are inverse relations. The free energy
and free enthalpy densities ψ = ψ() and ψ
∗
= ψ
∗
(σ) that are quadratic in and σ,
respectively, together with the resulting constitutive relations σ = σ() and = (σ)
that are linear in and σ, respectively, are displayed in Fig. 3.2.
The specific Hooke model is summarized in Table 3.1.