EN2210: Continuum Mechanics
Homework 5: Thermodynamics and Constitutive Equations
Solutions
School of Engineering
Brown University
1. Define the expended power of external forces acting on a deformable solid (which could be a sub-
volume within a larger body) by
d 1
Wexp = ∫ t ⋅ v + ∫ ρ b ⋅ v − dt ∫ 2 ρ v ⋅ v
S R R
Show that the expended power is zero for any rigid velocity field of the form
v (y , t=
) v 0 (t ) + ω(t ) × (y − y 0 )
where v 0 (t ), ω(t ) are vector valued functions of time (but independent of position)
It is helpful to re-write the time derivative of kinetic energy in terms of accelerations
d 1 d 1 dv dv
dt ∫ 2 dt ∫ 2
ρ v ⋅ v= ρ0 v ⋅ vdV = ∫ ρ0 ⋅ vdV = ∫ ρ ⋅ vdV
dt dt
R R0 R0 R
Hence
dv
Wexp = ∫ t ⋅ v + ∫ ρ b − dt ⋅ vdV
S R
It is also convenient to re-write the velocity field as
v (y , t ) = v 0 (t ) + W (y − y 0 )
where W is a skew tensor that has ω(t ) as its dual vector.
We can re-write
dv
Wexp = ∫ (n ⋅ σ) ⋅ vdA + ∫ ρ b − dt ⋅ vdV
S R
dv
= ∫ ∇ y ⋅ (σv )dV + ∫ ρ b − ⋅ vdV
dt
R R
dv
= ∫ σ : ∇y vdV + ∫ ∇y ⋅ σ + ρ b − dt ⋅ vdV
R R
Finally note that ∇ y v (y , t ) =
W and the stress is symmetric to satisfy angular momentum. Recall that the
contracted product of a symmetric and skew symmetric tensor is zero. The second integral vanishes from
balance of linear momentum.
[5 POINTS]
2. It is helpful to have versions of the first and second laws of thermodynamics in terms of quantities
defined on the reference configuration, including (but not limited to!) :
• Reference mass density ρ0
• Nominal stress and deformation gradient S, F
•
Material stress and Lagrange strain rate Σ, E
• Referential heat flux Θ = JF −1q
Give two equations for the first law, and give two expressions for the Clausius Duhem inequality.
We need to repeat the steps used to derive the first and second laws in the current configuration into the
reference configuration.
For the first law
d 1 dF ji d
∫ ∫ ∫
W =bi vi dV + σ ij ni v j dA =σ ij Dij dV +
dt 2
ρ∫v v
i i dV =Sij
V0 dt ∫ dV0 + ( KE )
dt
V A V V
∂Qi
Q= ∫ ∫
qdV − q ⋅ ndA= ∫ qJdV0 − ∫ Q ⋅ n0 dA0= ∫ Jq −
∂xi
dV
V A V0 A0 V0
Hence
dF ji
d d ∂Q
∫ + KE
ρ0ε dV0= Sij ∫dV0 + ( KE ) + Jq − i ∫ dV0
dt V dt dt ∂xi
V0 0 V0
∂ε dF ji ∂Qi
ρ0 = Sij − + Jq
∂t x =const dt ∂xi
Similarly in terms of material stress
∂ε dEij ∂Qi
ρ0 =
Σij − + Jq
∂t x =const dt ∂xi
The second law can be expressed as
dF ji
1 ∂θ ∂ψ ∂θ
Sij − Qi − ρ0 +s ≥0
dt θ ∂xi ∂t ∂t
dEij 1 ∂θ ∂ψ ∂θ
Sij − Qi − ρ0 +s
dt θ ∂xi ∂t ∂t
[5 POINTS]
3. Starting with the local form of the second law of thermodynamics and mass conservation
∂s ∂ (qi / q ) q ∂ρ ∂ρ vi
ρ + − ≥0 + 0
=
∂t x =const ∂yi q ∂t y ∂yi
(the symbols have their usual meaning), derive the statement of the second law for a control volume
∂ q ⋅n q
∂t∫ ∫
ρ sdV + ρ s ( v ⋅ n)dA + ∫
qq
dA − ∫
dV ≥ 0
R B B R
Note that we can write the first term as
∂s ∂s ∂s ∂ρ s ∂ρ ∂s
ρ =ρ +ρ vi = −s +ρ
∂t x ∂t y ∂yi ∂t y ∂t y ∂yi
Then, using mass conservation
∂ρ ∂ρ vi ∂ρ s ∂ρ ∂s ∂ρ s ∂ρ vi ∂s ∂ρ s ∂ρ svi
+ = 0⇒ −s +ρ = +s +ρ = +
∂t y ∂yi ∂t y ∂t y ∂yi ∂t y ∂yi ∂yi ∂t y ∂yi
Hence
∂ρ s ∂ρ svi ∂ (qi / q ) q
+ + − ≥0
∂t y ∂yi ∂yi q
Integrating this over a fixed spatial volume, noting that the time derivative can be taken outside the
integral, and applying the divergence theorem gives the required solution.
[5 POINTS]
4. State how the following quantities transform under a change of observer
(i) The spatial heat flux vector q
(ii) The referential heat flux vector Θ = JF −1q
T
(iv) The infinitesimal strain tensor ε = ∇ x u + ( ∇ x u ) / 2
(iv) The spatial gradient of a scalar function of position in a deformed solid g = ∇ yφ (y )
(v) The material gradient of a function of particle position in a solid G = ∇φ (x)
(i) q is a spatial vector therefore q* = Qq
= JF −1q ⇒ Θ=
(ii) Θ *
J *F*−1q=
*
JF −1QT Qq
= Θ
(iii) The displacement transforms as u* = Qu and the referential derivative is invariant so
( )
T T T
ε*= ∇ x u* + ∇ x u* / 2= Q∇ x u + ( ∇ x Qu ) / 2= Q∇ xu + ( ∇ xu ) QT / 2
(iv) g* =
∇ y*φ = ∇ yφ QT =∇
Q yφ = Qg
(v) The referential gradient is invariant by inspection, or else note
* * * T
G = gF ⇒ G = g F = gQ QF = G
[4 POINTS]
