Yamaguchi H. Engineering Fluid Mechanics (Fluid Mechanics and Its Applications)
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30 1 Fundamentals in Continuum Mechanics
or alternatively
IJIT p
(1.6.13)
The stress tensor
IJ
, which is often termed as deviatoric stress tensor,
may include various contributions, depending upon the physical character
of the continuum medium, such as compressibility, viscoelastic nature, and
external (such as electromagnetic) field effects, likewise for
p
mentioned
above. A general expression of the total stress
T may be expressed by
*
ijij
*
ij
pT
WG
(1.6.14)
where
*
p and
*
ij
W
are the extended pressure and stress tensor respectively.
The mean of
ij
T is defined as
*
ii
*
iim
pTT
W
3
1
3
1
(1.6.15)
and the deviatoric stress
ij
W
is defined as
ijmijij
TT
G
W
(1.6.16)
In viscous, incompressible Newtonian fluid, i.e 0
*
ii
W
and pp
*
, the
mean stress is equal to the pressure
p
as
pT
m
(1.6.17)
This fluid is sometimes called the perfect fluid.
Exercise
Exercise 1.1 Dyadic Product u
u
is called the gradient of vector u and is sometimes written xu ww or
grad
u
. u is the second order tensor in the Cartesian coordinates system.
Show
u
as the dyadic product, using suffix notation of tensor with unit
Ans.
i
j
jijj
i
i
x
u
u
x
w
w
¸
¸
¹
·
¨
¨
©
§
w
w
eeeeu
ˆˆˆˆ
(1)
dyads
ji
ee
ˆˆ
.
Exercise 31
Equation (1) shows that
u
is a second order tensor whose ij compo-
nents are
ij
uu ww .
Exercise 1.2 Convective Term
In Eq. (1.1.7), the term
uuuu is called the convective term in
fluid mechanics. Using the vector identities in Appendix B-5 (B.5-6), re-
duce Eq. (1.1.9).
uvvuuvvuvu u
uuu
(1)
Ans.
Set u
v
, which gives
uuuuu
2
uu
2
1
(2)
When the vorticity vector
uȦ u
, Eq. (2) becomes
Ȧuuuu
2
u
¸
¹
·
¨
©
§
2
1
(3)
If the velocity field
u is irrotational, i.e. 0 u uȦ , u has a scalar po-
tential
I
such that
I
u
(4)
and with the scalar potential the convective term will be written
¸
¹
·
¨
©
§
2
uu
I
2
1
(5)
Exercise 1.3 Euler’s Relation
Proof the Euler Relation given by Eq. (1.1.22)
u J
Dt
DJ
(1)
where
32 1 Fundamentals in Continuum Mechanics
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
321
]]]
]]]
]]]
]]]
H
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
xxx
xxx
xxx
xxx
J
kji
ijk
(2)
Ans.
We firstly write the following relation, for the velocity gradient
j
ii
jj
i
u
Dt
Dxx
Dt
D
]]]
w
w
w
w
w
w
(3)
With this relation, we are able to set an expression of the time change of
the Jacobian as follows
kji
ijk
kji
ijk
kji
ijk
uxxxuxxxu
Dt
DJ
]]]
H
]]]
H
]]]
H
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
321321321
(4)
While Eq. (3) may be formulated by chain rule
j
i
j
i
j
i
k
i
j
k
j
i
x
x
ux
x
ux
x
u
x
uxu
]]]]]
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
3
3
2
2
1
1
(5)
Thus, finally Eq. (4) can be reduced to give the required form
kji
ijk
kji
ijk
kji
ijk
kji
ijk
kji
ijk
kji
ijk
kji
ijk
kji
ijk
kji
ijk
k
l
lji
ijk
kj
l
li
ijk
kji
l
l
ijk
xxx
x
uxxx
x
uxxx
x
u
xxx
x
uxxx
x
uxxx
x
u
xxx
x
uxxx
x
uxxx
x
u
x
x
uxxxx
x
uxxxx
x
u
Dt
DJ
]]]
H
]]]
H
]]]
H
]]]
H
]]]
H
]]]
H
]]]
H
]]]
H
]]]
H
]]]
H
]]]
H
]]]
H
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
321
3
3321
2
3321
1
3
321
3
2321
2
2321
1
2
323
3
1322
2
1321
1
1
321321321
u
w
w
w
w
w
w
J
x
u
J
x
u
J
x
u
J
3
3
2
2
1
1
(6)
Exercise 33
Exercise 1.4 Pseudovector
If Ȧ is a pseudovector of a second order tensor S , examine the nature of
Ȧ in consideration of Eq. (1.1.28).
Sİ Ȧ
(1)
Ans.
Decompose S into the symmetric part and skew-symmetric part as
as
2
1
2
1
SS
SSSSS
TT
(2)
where the skew-symmetric part is written by components
»
»
»
¼
º
«
«
«
¬
ª
»
»
»
¼
º
«
«
«
¬
ª
0
0
0
2
1
0
0
0
2
1
12
13
23
23321331
32231221
31132112
a
ZZ
ZZ
ZZ
ᇫᇫ
ssss
ssss
ssss
S
(3)
Namely vector
Ȧ is expressed with components
1
Z
,
2
Z
and
3
Z
as
332211
Z
Z
Z
eeeȦ
ˆˆˆ
(4)
where
Ȧ is obtained by Sİ (see Eq. (1.1.26) for example), this implies
that components of the skew-symmetric part of the second order tensor are
composed of components of the pseudovector. If S is assumed to be a ve-
locity gradient
u , the vorticity vector
Ȧ
is derived from components of
the spin tensor, which is the skew-symmetric part of the tensor
u
.
Exercise 1.5 Material Objectivity
The upper convective Maxwell model constitutive equation (a linear vis-
coelastic model, in Chapter 7) can be written by
eIJIJ
0
2
KO
(1)
34 1 Fundamentals in Continuum Mechanics
where IJ is a second order tensor,
O
and
0
K
are constant and e is the rate
of deformation tensor (see Eq. (1.1.16)). Show that Eq. (1) satisfy the ma-
terial objectivity.
Ans.
The material objectivity has to be met by the linear transformation for
tensor A as follows
T
tt QAQA
(2)
where
tQ is a rotation tensor, defined in Eq. (1.2.1). Therefore, the con-
stitutive Equation of Eq. (1) has to be invariant by the transformation of Eq.
(2), that is
eIJIJ
0
2
KO
(3)
where
T
QIJQIJ
(4)
T
QIJQIJ
(5)
and
T
QeQe
(6)
knowing that scalar constants are frame invariants.
In order to verify the invariance of Eq. (1), take the transformation to the
model equation
T
QeIJIJQ
¸
¹
·
¨
©
§
0
2
KO
(7)
Since Eq. (7) is linear, we can write
TTT
QeQQIJQQIJQ
¸
¹
·
¨
©
§
0
2
KO
(8)
Thus we can recover the given equation by knowing Eqs. (4), (5) and (6)
eIJIJ
0
2
KO
(9)
'
'' '
'
'
'
'' '
Exercise 35
In deriving Eq. (9), we used the principal of frame invariance (the ma-
terial objectivity) for the upper convective derivative given in Eq. (1.3.16).
Exercise 1.6 Reynolds’ Transport Theorem
Give a physical picture of the Reynolds’ transport theorem, considering the
rate of change of a certain quantity
F
of matter moving through a control
volume, as depicted in Fig. 1.10,
Ans.
Consider a control volume of region A , which contains a quantity of
matter at some time
t , indicated by the solid line. At some time later
time
tt , the boundary of the system has a new physical location as
shown by the dotted line, at which the control volume occupies regions B
and A minus C . The increment of the matter is written
^`
tmttmttmttmm
ABCA
''' '
(1)
Fig. 1.10 System of moving control volume and fixed control volume
Taking differentiation to Eq. (1) with respect to time t and after rear-
rangement, we can write
t
ttmttm
t
tmttm
t
m
CBAA
(2)
'
'
'' ''
'''
36 1 Fundamentals in Continuum Mechanics
The rate of change of
m
is calculated by taking the limit of Eq. (2) as
0
ot . The first term on the right hand side of Eq. (2) thus becomes
³
w
w
w
w
'
'
o'
V
AA
AA
t
FdV
t
mm
tt
tmttm
0
lim
(3)
Since region
A is fixed in the coordinates, we can write
³³
w
w
w
w
V
V
dV
t
F
FdV
t
(4)
In a similar manner as the second term becomes
CB
CB
t
mm
t
ttmttm
o0
lim
(5)
where
B
m
is the rate of change of m through surface area
1
S from control
volume (region
A ), and which is expressed by surface integral
³
1
S
B
dFm Su
(6)
where uF is the flux of
F
(
F
is transported through the surface of the
control volume by stream of flow with velocity
u ).
Similarly
C
m is expressed
³
2
S
C
dFm Su
(7)
where the minus sign means the inward to the surface of the control vol-
ume. With Eqs. (6) and (7). Equation (5) becomes
³
S
CB
dFmm Su
(8)
Therefore, the rate of change of m is altogether written
³³³
w
w
'
'
o'
SV
V
t
dFdV
t
F
FdV
Dt
D
t
m
Su
0
lim
(9)
This is a Lagrangian-to-Eulerian description of the rate of change of an
extensive integral quantity given by Eq. (1.5.10).
'
''
'
'
Exercise 37
Exercise 1.7 Principal Axes of Stress
The pressure
p
defined in Eq. (1.6.17) is the mean of the normal stresses,
that is, one third of the trace of the total stress tensor
T . It appears that
p
is meant to be the mean of the principal stresses. Give the definition of the
principal stresses and its direction of the principal axis.
Ans .
When the direction of a stress vector
n
t is equal to that of an unit
normal vector
n
ˆ
, if
n
t is derived from a stress tensor T by Cauchy’s
stress formula, the direction of n
ˆ
is called the direction of the principal
axes of stress and the stresses are the principal stresses.
Thus, in case of
n
ˆ
being parallel with the principal axes, we can write
nnt
ˆ
T
ˆ
O
n
(1)
where
O
is a scalar quantity. Equation (1) gives a relationship written as
0 IT
ˆ
O
n
(2)
and Eq. (2) has to satisfy
0 IT
O
(3)
for the condition of
0zn
ˆ
. Equation (3) is called the characteristic equa-
T . Roots of Eq. (3) give eigenvalues, which are the
The perfect fluid given by Eq. (1.6.17) is an isotropic fluid in a sense
that a simple direct stress acting in it does not produce a shearing deforma-
tion. In the functional relation between stress and deformation must be in-
dependent of the orientation of the coordinates system.
The component form of Eq. (3) is written by
0
333231
232221
131211
O
O
O
TTT
TTT
TTT
(4)
with which we have a third order polynomial equation for
O
as follows
0
32
2
1
3
III
OOO
(5)
tion of stress tensor
principal stresses for the principal axes.
38 1 Fundamentals in Continuum Mechanics
Since
O
is independent of choice of the coordinates system, the coeffi-
cients of
1
I ,
2
I and
3
I in Eq. (5) are also independent of the choice of the
coordinates. Therefore,
1
I ,
2
I and
3
I are frame invariants. Equation (5) is
also true for other tensors such as the rate of deformation tensor, which is
discussed in more detail in Chapter 7,
1
I ,
2
I and
3
I are respectively given by
T
r
tI
1
(6)
>@
>@
2
22
2
2
1
2
1
TTT:ȉT
rrr
tttI
(7)
if
T is symmetric, i.e.
T:TT
2
r
t ,
and
Tdet
3
I
(8)
Note that if
1
V
,
2
V
and
3
V
are the principal stresses,
1
I ,
2
I and
3
I are re-
spectively given by
321
3
1
VVVVV
¦
m
mii
I
(9)
1332212
2
1
VVVVVVVVVV
jiijjjii
I
(10)
and
3213
det
VVVV
ij
I
(11)
Problems
1-1. Show that
T
QQ
is a skew-symmetric tensor, and whose pseudovec-
tor is .
Ans.
»
»
»
»
»
»
»
»
¼
º
«
«
«
«
«
«
«
«
¬
ª
u
u
rur
rrrrr
rrr
where
fixedforrotationrigid
thatso
0
1
0
00
T
T
TTT
TTT
tt
QQQQQ
)(Q
QQQQQQ
QQQQI,QQ
:
:
:
:
Problems 39
1-2. If
Q is a rotation tensor, show that Q is a unitary matrix, which satis-
fies
1T
QQ . Consider the two dimensional axis rotation by
ș
where the frame is transferred x' – y' to
x
– y .
Ans.
»
»
»
»
»
»
»
¼
º
«
«
«
«
«
«
«
¬
ª
?
c
¸
¸
¹
·
¨
¨
©
§
c
c
¸
¸
¹
·
¨
¨
©
§
¸
¸
¹
·
¨
¨
©
§
1
cossin
sincos
QQ
Q
T
y
x
y
x
xx
TT
TT
1-3. Proof for Cartesian vectors
u and
v
, that
uvvuvuvvu uu
Ans.
»
»
»
»
»
»
»
»
»
»
»
»
¼
º
«
«
«
«
«
«
«
«
«
«
«
«
¬
ª
¸
¸
¹
·
¨
¨
©
§
w
w
¸
¸
¹
·
¨
¨
©
§
w
w
¸
¸
¹
·
¨
¨
©
§
w
w
¸
¸
¹
·
¨
¨
©
§
w
w
w
w
¸
¸
¹
·
¨
¨
©
§
w
w
w
w
¸
¸
¹
·
¨
¨
©
§
w
w
°
¿
°
¾
½
°
¯
°
®
w
w
¸
¸
¹
·
¨
¨
©
§
w
w
w
w
w
w
w
w
j
j
ij
j
ii
j
ji
j
i
i
j
j
ij
j
j
j
i
j
i
j
m
j
l
m
l
j
jlimjmil
ml
j
klmkijml
j
klmijk
mlklm
j
ijk
u
x
vv
x
uv
x
uu
x
v
v
x
uvu
x
v
x
uvu
x
v
x
uvu
x
vu
x
vu
x
vu
x
GGGG
HHHHHH
1-4. When a scalar function
t,p x differs form a material surface, but
tp ,x
moves with a velocity v different from the stream velocity u ,
show that
p
dt
dp
¸
¹
·
¨
©
§
nvu
ˆ
Ans.
»
»
»
»
»
»
»
»
»
»
»
¼
º
«
«
«
«
«
«
«
«
«
«
«
¬
ª
w
w
w
w
pp
dt
dp
t
p
p
t
p
dt
dp
txp
nvuvu
vuw
w
ˆ
,
,
0pointmaterialwithnotand
,velocityrelativewithmovebut
surface,is the materialIf
u