The governing equations for a continuum (apart from the kinematic relations
and the constitutive laws) are
Balance of mass
Balance of linear momentum
Balance of angular momentum
Balance of energy
Entropy inequality
Two important results from calculus
edit
There are two important results from calculus that are useful when we
use or derive these governing equations. These are
The Gauss divergence theorem.
The Reynolds transport theorem
These are useful enough to bear repeating at this point in the context of
second order tensors.
The Gauss divergence theorem
edit
The divergence theorem relates volume integrals to surface integrals.
Let
Ω
{\displaystyle \Omega }
be a body and let
∂
Ω
{\displaystyle \partial \Omega }
be its boundary with outward unit normal
n
{\displaystyle \mathbf {n} }
. Let
A
(
x
)
{\displaystyle {\boldsymbol {A}}(\mathbf {x} )}
be a second order tensor valued function of
x
{\displaystyle \mathbf {x} }
. Then,
if
A
{\displaystyle {\boldsymbol {A}}}
is differentiable at least once (i.e.,
A
∈
C
0
{\displaystyle {\boldsymbol {A}}\in C^{0}}
) then
∫
Ω
∇
⋅
A
(
x
)
dV
=
∫
∂
Ω
A
T
(
x
)
⋅
n
dA
{\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\cdot {\boldsymbol {A}}(\mathbf {x} )~{\text{dV}}=\int _{\partial \Omega }{\boldsymbol {A}}^{T}(\mathbf {x} )\cdot \mathbf {n} ~{\text{dA}}}
It is often of interest to consider the situation where
A
{\displaystyle {\boldsymbol {A}}}
is not
continuously differentiable, i.e., when there are jumps in
A
{\displaystyle {\boldsymbol {A}}}
within the
body. Let
∂
Ω
int
{\displaystyle \partial \Omega _{\text{int}}}
represent the set of surfaces internal to the
body where there are jumps in
A
{\displaystyle {\boldsymbol {A}}}
. In that case, the divergence theorem is
written as
∫
Ω
∇
⋅
A
dV
=
∫
∂
Ω
A
T
⋅
n
dA
+
∫
∂
Ω
int
[
A
+
T
−
A
−
T
]
⋅
n
+
dA
{\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\cdot {\boldsymbol {A}}~{\text{dV}}=\int _{\partial \Omega }{\boldsymbol {A}}^{T}\cdot \mathbf {n} ~{\text{dA}}+\int _{\partial \Omega _{\text{int}}}\left[{\boldsymbol {A}}_{+}^{T}-{\boldsymbol {A}}_{-}^{T}\right]\cdot \mathbf {n} _{+}~{\text{dA}}}
where the subscripts
+
{\displaystyle +}
and
−
{\displaystyle -}
represents the values on the two sides of
the jump discontinuity with normal
n
+
{\displaystyle \mathbf {n} _{+}}
.
Reynold's transport theorem
edit
The transport theorem shows you how to
calculate the material time derivative of an integral. It is a
generalization of the Leibniz formula.
Let
Ω
{\displaystyle \Omega }
be a body in its current configuration and let
∂
Ω
{\displaystyle \partial \Omega }
be
its surface. Also, let
v
=
x
˙
{\displaystyle \mathbf {v} ={\dot {\mathbf {x} }}}
.
If
ϕ
(
x
,
t
)
{\displaystyle \phi (\mathbf {x} ,t)}
is a scalar valued function of
x
{\displaystyle \mathbf {x} }
and
t
{\displaystyle t}
then
D
D
t
[
∫
Ω
ϕ
(
x
,
t
)
dV
]
=
∫
Ω
∂
ϕ
∂
t
dV
+
∫
∂
Ω
(
v
⋅
n
)
ϕ
dA
=
∫
Ω
[
∂
ϕ
∂
t
+
∇
⋅
(
ϕ
v
)
]
dV
{\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\phi (\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\Omega }{\frac {\partial \phi }{\partial t}}~{\text{dV}}+\int _{\partial \Omega }(\mathbf {v} \cdot \mathbf {n} )~\phi ~{\text{dA}}=\int _{\Omega }\left[{\frac {\partial \phi }{\partial t}}+{\boldsymbol {\nabla }}\cdot \left(\phi ~\mathbf {v} \right)\right]~{\text{dV}}}
If
f
(
x
,
t
)
{\displaystyle \mathbf {f} (\mathbf {x} ,t)}
is a vector valued function of
x
{\displaystyle \mathbf {x} }
and
t
{\displaystyle t}
.
Then
D
D
t
[
∫
Ω
f
(
x
,
t
)
dV
]
=
∫
Ω
∂
f
∂
t
dV
+
∫
∂
Ω
(
v
⋅
n
)
f
dA
=
∫
Ω
[
∂
f
∂
t
+
∇
⋅
(
f
⊗
v
)
]
dV
{\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\mathbf {f} (\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\Omega }{\frac {\partial \mathbf {f} }{\partial t}}~{\text{dV}}+\int _{\partial \Omega }(\mathbf {v} \cdot \mathbf {n} )~\mathbf {f} ~{\text{dA}}=\int _{\Omega }\left[{\frac {\partial \mathbf {f} }{\partial t}}+{\boldsymbol {\nabla }}\cdot (\mathbf {f} \otimes \mathbf {v} )\right]~{\text{dV}}}
To refresh your memory, recall that the material time derivative is given by
D
ϕ
D
t
=
∂
ϕ
∂
t
+
∇
ϕ
⋅
v
D
f
D
t
=
∂
f
∂
t
+
∇
f
⋅
v
D
F
D
t
=
∂
F
∂
t
+
(
∇
F
)
T
⋅
v
{\displaystyle {\begin{aligned}{\cfrac {D\phi }{Dt}}&={\frac {\partial \phi }{\partial t}}+{\boldsymbol {\nabla }}\phi \cdot \mathbf {v} \\{\cfrac {D\mathbf {f} }{Dt}}&={\frac {\partial \mathbf {f} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {f} \cdot \mathbf {v} \\{\cfrac {D{\boldsymbol {F}}}{Dt}}&={\frac {\partial {\boldsymbol {F}}}{\partial t}}+({\boldsymbol {\nabla }}{\boldsymbol {F}})^{T}\cdot \mathbf {v} \end{aligned}}}
Conservation of mass
edit
For the situation where a body does not gain or lose mass, the balance of mass
is written as
D
ρ
D
t
+
ρ
∇
⋅
v
=
0
{\displaystyle {\cfrac {D\rho }{Dt}}+\rho ~{\boldsymbol {\nabla }}\cdot \mathbf {v} =0}
Sometimes, this equation is also written in conservative form as
∂
ρ
∂
t
+
∇
⋅
(
ρ
v
)
=
0
{\displaystyle {\frac {\partial \rho }{\partial t}}+{\boldsymbol {\nabla }}\cdot (\rho ~\mathbf {v} )=0}
If the material is incompressible then the density does not change with
time and we get
∇
⋅
v
=
0
{\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {v} =0}
For Lagrangian descriptions we can show that
ρ
J
=
ρ
0
{\displaystyle \rho ~J=\rho _{0}}
where
ρ
0
{\displaystyle \rho _{0}}
is the initial density.
Conservation of linear momentum
edit
The balance of linear momentum is essentially Newton's second applied to
continua. Newton's second law can be written as
D
p
D
t
=
f
{\displaystyle {\cfrac {D\mathbf {p} }{Dt}}=\mathbf {f} }
where the linear momentum
p
{\displaystyle \mathbf {p} }
is given by
p
(
t
)
=
∫
Ω
ρ
(
x
,
t
)
v
(
x
,
t
)
dV
{\displaystyle \mathbf {p} (t)=\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {v} (\mathbf {x} ,t)~{\text{dV}}}
and the total force
f
{\displaystyle \mathbf {f} }
is given by
f
(
t
)
=
∫
Ω
ρ
(
x
,
t
)
b
(
x
,
t
)
dV
+
∫
∂
Ω
t
(
x
,
t
)
dA
{\displaystyle \mathbf {f} (t)=\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {b} (\mathbf {x} ,t)~{\text{dV}}+\int _{\partial \Omega }\mathbf {t} (\mathbf {x} ,t)~{\text{dA}}}
where
ρ
{\displaystyle \rho }
is the density,
v
{\displaystyle \mathbf {v} }
is the spatial velocity,
b
{\displaystyle \mathbf {b} }
is
the body force and
t
{\displaystyle \mathbf {t} }
is the surface traction. Therefore, the balance
of linear momentum can be written as
D
D
t
[
∫
Ω
ρ
(
x
,
t
)
v
(
x
,
t
)
dV
]
=
∫
Ω
ρ
(
x
,
t
)
b
(
x
,
t
)
dV
+
∫
∂
Ω
t
(
x
,
t
)
dA
{\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {v} (\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {b} (\mathbf {x} ,t)~{\text{dV}}+\int _{\partial \Omega }\mathbf {t} (\mathbf {x} ,t)~{\text{dA}}}
Now using the transport theorem, we have
D
D
t
[
∫
Ω
ρ
v
dV
]
=
∫
Ω
[
ρ
D
v
D
t
+
(
D
ρ
D
t
+
ρ
∇
⋅
v
)
]
dV
{\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\rho ~\mathbf {v} ~{\text{dV}}\right]=\int _{\Omega }\left[\rho ~{\cfrac {D\mathbf {v} }{Dt}}+\left({\cfrac {D\rho }{Dt}}+\rho ~{\boldsymbol {\nabla }}\cdot \mathbf {v} \right)\right]~{\text{dV}}}
From the conservation of mass, the second term on the right hand side is zero
and we are left with
D
D
t
[
∫
Ω
ρ
(
x
,
t
)
v
(
x
,
t
)
dV
]
=
∫
Ω
ρ
D
v
D
t
dV
{\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\rho (\mathbf {x} ,t)~\mathbf {v} (\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\Omega }\rho ~{\cfrac {D\mathbf {v} }{Dt}}~{\text{dV}}}
Therefore, the balance of linear momentum can be written as
∫
Ω
ρ
D
v
D
t
dV
=
∫
Ω
ρ
b
dV
+
∫
∂
Ω
t
dA
{\displaystyle \int _{\Omega }\rho ~{\cfrac {D\mathbf {v} }{Dt}}~{\text{dV}}=\int _{\Omega }\rho ~\mathbf {b} ~{\text{dV}}+\int _{\partial \Omega }\mathbf {t} ~{\text{dA}}}
Using Cauchy's theorem (
t
=
σ
⋅
n
{\displaystyle \mathbf {t} ={\boldsymbol {\sigma }}\cdot \mathbf {n} }
) and the divergence theorem we
can show that the balance of linear momentum can be written as
∇
⋅
σ
+
ρ
b
=
ρ
D
v
D
t
=
ρ
a
{\displaystyle {\boldsymbol {\nabla }}\cdot {\boldsymbol {\sigma }}+\rho ~\mathbf {b} =\rho ~{\cfrac {Dv}{Dt}}=\rho ~\mathbf {a} }
In index notation,
σ
i
j
,
j
+
ρ
b
i
=
ρ
D
v
i
D
t
{\displaystyle \sigma _{ij,j}+\rho ~b_{i}=\rho ~{\cfrac {Dv_{i}}{Dt}}}
Conservation of angular momentum
edit
We also have to make sure that the moments are balanced. This requirement takes
the form of the conservation of angular momentum and can be written as
D
D
t
[
∫
Ω
x
×
(
ρ
v
)
dV
]
=
∫
Ω
x
×
(
ρ
b
)
dV
+
∫
∂
Ω
x
×
t
dA
{\displaystyle {\cfrac {D}{Dt}}\left[\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {v} )~{\text{dV}}\right]=\int _{\Omega }\mathbf {x} \times (\rho ~\mathbf {b} )~{\text{dV}}+\int _{\partial \Omega }\mathbf {x} \times \mathbf {t} ~{\text{dA}}}
We can show that this equation reduces down to the requirement that the
Cauchy stress is symmetric, i.e.,
σ
=
σ
T
{\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{T}}
Conservation of energy
edit
The balance of energy can be expressed as
ρ
D
e
D
t
=
σ
:
(
∇
v
)
−
∇
⋅
q
+
ρ
s
{\displaystyle \rho ~{\cfrac {De}{Dt}}={\boldsymbol {\sigma }}:({\boldsymbol {\nabla }}\mathbf {v} )-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s}
where
e
{\displaystyle e}
is the internal energy per unit mass,
q
{\displaystyle \mathbf {q} }
is the heat flux vector
and
s
{\displaystyle s}
is the heat source per unit volume.
We may also write this equation as
ρ
D
e
D
t
=
σ
:
d
−
∇
⋅
q
+
ρ
s
{\displaystyle \rho ~{\cfrac {De}{Dt}}={\boldsymbol {\sigma }}:{\boldsymbol {d}}-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s}
where
d
{\displaystyle \mathbf {d} }
, the rate of deformation tensor, is the symmetric part of the
velocity gradient. We can do this because the contraction of the skew
symmetric part of the velocity gradient with the symmetric Cauchy stress gives
us zero.
For purely mechanical problems,
q
=
0
{\displaystyle \mathbf {q} =0}
and
s
=
0
{\displaystyle s=0}
. So we can write
ρ
D
e
D
t
=
σ
:
d
{\displaystyle \rho ~{\cfrac {De}{Dt}}={\boldsymbol {\sigma }}:{\boldsymbol {d}}}
This shows that
σ
{\displaystyle {\boldsymbol {\sigma }}}
and
d
{\displaystyle {\boldsymbol {d}}}
are conjugate in power .
The entropy inequality is useful in determining which forms of the constitutive
equations are admissible. This inequality is also called the dissipation
inequality. In its Clausius-Duhem form, the inequality may written as
d
d
t
(
∫
Ω
ρ
η
dV
)
≥
−
∫
∂
Ω
ρ
η
v
⋅
n
dA
−
∫
∂
Ω
q
⋅
n
T
dA
+
∫
Ω
ρ
s
T
dV
.
{\displaystyle {\cfrac {d}{dt}}\left(\int _{\Omega }\rho ~\eta ~{\text{dV}}\right)\geq -\int _{\partial \Omega }\rho ~\eta ~\mathbf {v} \cdot \mathbf {n} ~{\text{dA}}-\int _{\partial \Omega }{\cfrac {\mathbf {q} \cdot \mathbf {n} }{T}}~{\text{dA}}+\int _{\Omega }{\cfrac {\rho ~s}{T}}~{\text{dV}}~.}
where
η
{\displaystyle \eta }
is the specific entropy (entropy per unit mass) and
T
{\displaystyle T}
is the
temperature.
In differential form the Clausius-Duhem inequality can be written as
ρ
η
˙
≥
−
∇
⋅
(
q
T
)
+
ρ
s
T
.
{\displaystyle \rho ~{\dot {\eta }}\geq -{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}~.}
In terms of the specific internal energy, the entropy inequality can be
expressed as
ρ
(
e
˙
−
T
η
˙
)
−
σ
:
∇
v
≤
−
q
⋅
∇
T
T
.
{\displaystyle \rho ~({\dot {e}}-T~{\dot {\eta }})-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.}
If we define the Helmholtz free energy (the energy that is available to do
mechanical work) as
ψ
=
e
−
η
T
{\displaystyle \psi =e-\eta ~T}
we can also write,
ρ
(
ψ
˙
+
T
η
˙
)
−
σ
:
d
≤
−
q
⋅
∇
T
T
.
{\displaystyle \rho ~({\dot {\psi }}+T~{\dot {\eta }})-{\boldsymbol {\sigma }}:{\boldsymbol {d}}\leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.}
The Lagrangian form of the governing equations can be obtained using the
relations between the various measures in the deformed and reference
configurations.
The Lagrangian form of the balance of mass is
ρ
(
X
,
t
)
J
(
X
,
t
)
=
ρ
0
(
X
)
{\displaystyle \rho (\mathbf {X} ,t)~J(\mathbf {X} ,t)=\rho _{0}(\mathbf {X} )}
Balance of linear momentum
edit
The Lagrangian form of the balance of linear momentum is
∇
∘
⋅
N
+
ρ
0
b
=
ρ
0
∂
v
(
X
,
t
)
∂
t
{\displaystyle {\boldsymbol {\nabla }}_{\circ }\cdot {\boldsymbol {N}}+\rho _{0}~\mathbf {b} =\rho _{0}~{\frac {\partial \mathbf {v} (\mathbf {X} ,t)}{\partial t}}}
where
N
{\displaystyle {\boldsymbol {N}}}
is the nominal stress and
∇
∘
⋅
(
∙
)
{\displaystyle {\boldsymbol {\nabla }}_{\circ }\cdot (\bullet )}
indicates that the
derivatives are with respect to
X
{\displaystyle \mathbf {X} }
. In terms of the first Piola-Kirchhoff
stress
P
{\displaystyle {\boldsymbol {P}}}
we can write
∇
∘
⋅
P
T
+
ρ
0
b
=
ρ
0
∂
v
(
X
,
t
)
∂
t
{\displaystyle {\boldsymbol {\nabla }}_{\circ }\cdot {\boldsymbol {P}}^{T}+\rho _{0}~\mathbf {b} =\rho _{0}~{\frac {\partial \mathbf {v} (\mathbf {X} ,t)}{\partial t}}}
In index notation,
P
j
i
,
j
+
ρ
0
b
i
=
ρ
0
v
˙
i
{\displaystyle P_{ji,j}+\rho _{0}~b_{i}=\rho _{0}{\dot {v}}_{i}}
Balance of angular momentum
edit
The balance of angular momentum in Lagrangian form is
N
T
⋅
F
T
=
F
⋅
N
{\displaystyle {\boldsymbol {N}}^{T}\cdot {\boldsymbol {F}}^{T}={\boldsymbol {F}}\cdot {\boldsymbol {N}}}
In terms of the first Piola-Kirchhoff stress
P
⋅
F
T
=
F
⋅
P
T
{\displaystyle {\boldsymbol {P}}\cdot {\boldsymbol {F}}^{T}={\boldsymbol {F}}\cdot {\boldsymbol {P}}^{T}}
In terms of the second Piola-Kirchhoff stress
S
=
S
T
{\displaystyle {\boldsymbol {S}}={\boldsymbol {S}}^{T}}
In the material frame, the balance of energy takes the form
ρ
0
∂
e
(
X
,
t
)
∂
t
=
F
˙
T
:
N
−
∇
∘
⋅
Q
(
X
,
t
)
+
ρ
0
s
{\displaystyle \rho _{0}~{\frac {\partial e(\mathbf {X} ,t)}{\partial t}}={\dot {\boldsymbol {F}}}^{T}:{\boldsymbol {N}}-{\boldsymbol {\nabla }}_{\circ }\cdot \mathbf {Q} (\mathbf {X} ,t)+\rho _{0}~s}
where
Q
{\displaystyle \mathbf {Q} }
is the heat flux per unit reference area.
In terms of the first Piola-Kirchhoff stress tensor we have
ρ
0
∂
e
(
X
,
t
)
∂
t
=
F
˙
T
:
P
T
−
∇
∘
⋅
Q
(
X
,
t
)
+
ρ
0
s
{\displaystyle \rho _{0}~{\frac {\partial e(\mathbf {X} ,t)}{\partial t}}={\dot {\boldsymbol {F}}}^{T}:{\boldsymbol {P}}^{T}-{\boldsymbol {\nabla }}_{\circ }\cdot \mathbf {Q} (\mathbf {X} ,t)+\rho _{0}~s}
The entropy inequality in Lagrangian form is
ρ
0
(
ψ
˙
+
T
η
˙
)
−
N
:
F
˙
≤
−
Q
⋅
∇
0
T
T
.
{\displaystyle \rho _{0}~({\dot {\psi }}+T~{\dot {\eta }})-{\boldsymbol {N}}:{\dot {\boldsymbol {F}}}\leq -{\cfrac {\mathbf {\mathbf {Q} } \cdot {\boldsymbol {\nabla }}_{0}T}{T}}~.}
In terms of the first P-K stress we have
ρ
0
(
ψ
˙
+
T
η
˙
)
−
P
T
:
F
˙
≤
−
Q
⋅
∇
0
T
T
.
{\displaystyle \rho _{0}~({\dot {\psi }}+T~{\dot {\eta }})-{\boldsymbol {P}}^{T}:{\dot {\boldsymbol {F}}}\leq -{\cfrac {\mathbf {\mathbf {Q} } \cdot {\boldsymbol {\nabla }}_{0}T}{T}}~.}