3.5 Balance Laws
121
• The term σ v in (3.173) represents the rate of work (per unit volume) done by
equal and opposite stresses that deform the element at the rate
◦
λ = ˙
λ/λ = v [see
Eq. (3.91)]. In the absence of thermal effects (q = r = 0), this work is stored in
the body as internal energy per unit volume ρu, which is called the strain-energy
density.
Example 3.18 For small deformation, write the 1D form of the strain-energy
density for a solid in terms of the linear axial strain .
Solution
With thermal energy neglected, Eq. (3.173) becomes
ρ
du
dt
= σ
∂v
∂x
= σ
1
λ
dλ
dt
,
where λ = ∂x/∂X. For small strain, setting λ ≈ 1 + gives the approximation
ρ
du
dt
= σ
dd
dt
or
ρ du = σ dd.
If ρ is assumed to remain constant for small deformation, then substituting the 1D
Hooke’s law σ = EE and integrating yield
ρ u =
EE dd =
1
2 EE
2 ,
which is the strain energy per unit volume.
Energy Balance in 3D
Skipping the 2D case, we derive the energy equation in 3D using the approach
followed for the equations of motion. That is, we write the energy balance relation
for the entire body, convert area integrals to volume integrals, and extract the
appropriate differential equation.
Spatial Form In the current configuration, a body has mass density ρ, surface area
A, and volume V (Fig. 3.22). In terms of the velocity v, the total kinetic energy is
K =
1
2
V
v · v ρ dV ,
(3.174)
121
• The term σ v in (3.173) represents the rate of work (per unit volume) done by
equal and opposite stresses that deform the element at the rate
◦
λ = ˙
λ/λ = v [see
Eq. (3.91)]. In the absence of thermal effects (q = r = 0), this work is stored in
the body as internal energy per unit volume ρu, which is called the strain-energy
density.
Example 3.18 For small deformation, write the 1D form of the strain-energy
density for a solid in terms of the linear axial strain .
Solution
With thermal energy neglected, Eq. (3.173) becomes
ρ
du
dt
= σ
∂v
∂x
= σ
1
λ
dλ
dt
,
where λ = ∂x/∂X. For small strain, setting λ ≈ 1 + gives the approximation
ρ
du
dt
= σ
dd
dt
or
ρ du = σ dd.
If ρ is assumed to remain constant for small deformation, then substituting the 1D
Hooke’s law σ = EE and integrating yield
ρ u =
EE dd =
1
2 EE
2 ,
which is the strain energy per unit volume.
Energy Balance in 3D
Skipping the 2D case, we derive the energy equation in 3D using the approach
followed for the equations of motion. That is, we write the energy balance relation
for the entire body, convert area integrals to volume integrals, and extract the
appropriate differential equation.
Spatial Form In the current configuration, a body has mass density ρ, surface area
A, and volume V (Fig. 3.22). In terms of the velocity v, the total kinetic energy is
K =
1
2
V
v · v ρ dV ,
(3.174)
