62
3 Continuum Mechanics and Nonlinear Elasticity
dφ
dt
=
∂φ
∂t
+ v i
∂φ
∂x i
,
(3.27)
and Eq. (3.26) becomes
d
dt
=
∂
∂t
+ v i
∂
∂x i
,
(3.28)
in which the summation convention applies.
The following examples illustrate the mechanics of taking time derivatives, as
well as some important points.
Example 3.3 In Cartesian coordinates, the motion of a continuum is described by
the velocity field
v = x 1 e 1 + x 1 e
−αt e 2 − x 2 e 3 ,
and the temperature distribution is
T =
x 1 + 2x 2 − x
2
3
e
−αt .
Both equations are expressed in spatial form, α is a constant, and all quantities are
taken as dimensionless for convenience. Determine the material time derivative of
the temperature field.
Solution
Setting φ = T in Eq. (3.25) gives
dT
dt
=
∂T
∂t
+ v · ∇T .
For this problem, the first term becomes
∂T
∂t
= −α(x 1 + 2x 2 − x
2
3 )e
−αt ,
and the temperature gradient is
∇T =
e 1
∂
∂x 1
+ e 2
∂
∂x 2
+ e 3
∂
∂x 3
x 1 + 2x 2 − x
2
3
e
−αt
= e
−αt (e 1 + 2e 2 − 2x 3 e 3 ) .
3 Continuum Mechanics and Nonlinear Elasticity
dφ
dt
=
∂φ
∂t
+ v i
∂φ
∂x i
,
(3.27)
and Eq. (3.26) becomes
d
dt
=
∂
∂t
+ v i
∂
∂x i
,
(3.28)
in which the summation convention applies.
The following examples illustrate the mechanics of taking time derivatives, as
well as some important points.
Example 3.3 In Cartesian coordinates, the motion of a continuum is described by
the velocity field
v = x 1 e 1 + x 1 e
−αt e 2 − x 2 e 3 ,
and the temperature distribution is
T =
x 1 + 2x 2 − x
2
3
e
−αt .
Both equations are expressed in spatial form, α is a constant, and all quantities are
taken as dimensionless for convenience. Determine the material time derivative of
the temperature field.
Solution
Setting φ = T in Eq. (3.25) gives
dT
dt
=
∂T
∂t
+ v · ∇T .
For this problem, the first term becomes
∂T
∂t
= −α(x 1 + 2x 2 − x
2
3 )e
−αt ,
and the temperature gradient is
∇T =
e 1
∂
∂x 1
+ e 2
∂
∂x 2
+ e 3
∂
∂x 3
x 1 + 2x 2 − x
2
3
e
−αt
= e
−αt (e 1 + 2e 2 − 2x 3 e 3 ) .