16.21 Techniques of Structural Analysis and
Design
Spring 2005
Unit #4 ­ Thermodynamics Principles
Raúl Radovitzky
February 22, 2005
First Law of Thermodynamics
�
d�
K +U =P +H
dt
(1)
where:
• K: kinetic energy
• U : internal energy
• P : Power of external forces
• H: hear exchange per unit time
1
K=
2
�
V
�
∂u ∂u
1
·
ρ
dV =
∂t ∂t
2
�
U=
ρ
V
∂ui ∂ui
dV
∂t ∂t
(2)
�
�0 dV =
ρU
V
U0 dV
V
1
(3)
�0 , U0 are the internal energy densities per unit mass and per unit
where U
volume, respectively.
�
�
∂u
∂u
P =
f·
dV + t ·
dS
(4)
∂t
∂t
V
S
In components:
�
P =
V
∂ui
fi
dV +
∂t
�
ti
S
∂ui
dS
∂t
Replacing ti = nj σji in this expression:
�
�
∂ui
∂ui
dV + nj σji
dS
P =
fi
∂t
∂t
S
V
Using Gauss’ Theorem:
�
�
∂ � ∂ui �
∂ui
dV +
σji
dV
P =
fi
∂t
∂t
V ∂xj
V
� ��
� ∂u
∂σji
∂ ∂ui �
i
=
+ fi
dV
+ σji
∂xj
∂t
∂xj ∂t
V
�
��
�
� �� �
2
∂ ui
∂ ∂ui
ρ 2 (why?)
∂t
∂t ∂xj
� � 2
�
∂ ∂ui
∂ ui ∂ui
dV
=
ρ 2
+ σji
∂xj
V
�∂t �� ∂t� � ∂t
�� �
1 ∂ � ∂ui �2
∂
σji �ji
2 ∂t ∂t
∂t
Notation:
Time derivatives:
∂( )
=(˙)
∂t
Examples:
•
∂ui
∂t
= u̇i ,
•
∂ 2 ui
∂t2
= üi
∂u
∂t
= u̇
2
(5)
(6)
(7)
•
∂�ij
∂t
= �˙ij
Spatial derivatives:
∂( )
= ( ),i
∂xi
Examples:
•
∂σji
∂xj
= σji,j
With this notation, the power of the external forces can be rewritten as:
�
�
d
1 ∂ui ∂ui
P =
ρ
dV +
σji �˙ji dV
dt V 2 ∂t ∂t
V
(8)
��
� � �� �
�
K
deformation power
where the “ρdV ” inside the first integral was included inside the time deriva­
tive since it is a constant due to conservation of mass. We conclude that part
of the power of the external forces goes into changing the kinetic energy of
the material and the rest into deforming the material. We call the latter the
deformation power and it represents the rate at which the stresses do work
on the deforming material.
Replacing in the first law, equation (1):
�
�
d�
d� �
K +U =
K +
σji �˙ji dV + H
(9)
dt
dt
V
After canceling the kinetic energy from both sides, the first law expresses the
fact that the internal energy of a deforming material can be changed either
by heating or by deforming the material:
d
dU
=
dt
dt
�
�
�0 dV =
ρU
V
σji �˙ji dV + H
(10)
V
In the isothermal case (H = 0):
� � �
�
∂ U0
− σij �˙ij dV = 0
ρ
∂t
V
3
(11)
or, in local form:
ρ
�0
∂U
= σij �˙ij
∂t
(12)
In ideal elasticity, we assume that all the work of deformation is converted
into internal energy, i.e., the internal energy density is a state function of the
deformation:
�0 = U
�0 (�ij )
U
(13)
�0
�0
∂U
∂U
=
�˙ij
∂t
∂�ij
(14)
Then:
Replace in first law, equation (12:
ρ
�0
∂U
�˙ij = σij �˙ij ⇒
∂�ij
(15)
�0
∂U
= σij
∂�ij
(16)
ρ
i.e., the stresses derive from a potential.
4