Geology Reference
In-Depth Information
 
e Ω
i
t
Mz Cz Kz
+
+
=
p
(1.79)
But from Equation (1.82),
C
ˆ
ˆ
∂
∂
K z
∂
∂
z
∂
∂
z
= − ∂
∂
K z
Separating the real and imaginary parts, we
obtain
ˆ
ˆ
+
K
C
=
⇒
K
C
0
C
C
x
x
x
x
e
e
e
e
(1.87)
Bz
−
Ω
Ω
Cz
=
p
C
S
c
(1.80)
which can be used in Equation (1.86) to yield
Bz
+
Cz
=
0
S
C
ˆ
∂
∂
z
∂
∂
z
∂
∂
c
x
∂
∂
K z
1
2
1
2
1
2
ˆ
T
C
T
C
T
=
p
=
z K
= −
z
where B = K - Ω 2 M . Calculating z S from the
second equation in (1.80) yields
C
C
C
C
x
x
x
e
e
e
e
(1.88)
z
= −
Ω
B
−
1
Cz
(1.81)
ˆ / K x e . Before
We thus need to calculate ∂
∂
S
C
proceeding we note that
After substituting in the first equation in
(1.80), we get
∂
∂
B
−
1
∂
∂
B
∂
∂
B
−
1
∂
∂
B B
B B I
−
1
= ⇒
B B
+
−
1
=
⇒
= −
B
−
1
−
1
0
x
x
x
x
e
e
e
e
(1.89)
Kz
p
=
(1.82)
C
in which we used the fact that B is symmetric,
i.e. B T = B . Differentiating Equation (1.83) and
making use of Equation (1.89) and symmetry of
B and C , we obtain
where
K B
−
= +
Ω 2
CB C
1
(1.83)
K
∂
∂
= ∂
∂
B
∂
∂
B
∂
∂
C B C
The displacement vector u is the real part of z , so
−
−
−
−
Ω
2
CB
1
B C
1
+
Ω
2
1
2
x
x
x
x
e
e
e
e
(1.90)
(
) = (
(
)
) − (
)
i
Ω
t
u
= ℜ
z
+
i
z
e
cos
Ω
t
z
sin
Ω
t
z
C
S
C
S
(1.84)
Pre- and post-multiplying by z C and using
Equation (1.81), we achieve
Using this and p = p C cosΩ t in Equation (1.66),
we can write
ˆ
∂
∂
K z
∂
∂
B z
∂
∂
B
∂
∂
C z
z
T
z
T
z
T
z
z
T
=
−
−
2 Ω
C
C
C
C
S
S
C
S
x
x
x
x
Ω
1
2
2
π
/
Ω
(
) =
e
e
e
e
∫
T
2
T
T
c
=
p z
cos
Ω
t
−
p z
cos
Ω
t
sin
Ω
t dt
p z
(1.91)
C
C
C
s
C
C
2
π
0
(1.85)
Substituting in Equation (1.88) and replacing
B by K - Ω 2 M , we obtain
Differentiating with respect to design variable
x e , we obtain
c
x
1
2
K
M z
K
M
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=
z
T
−
Ω
2
−
z
T
−
Ω
2
z
∂
∂
c
x
∂
∂
z
1
2 p
S
S
C
C
x
x
x
x
=
T
C
(1.86)
e
e
e
e
e
C
x
∂
∂
C z
e
e
T
Ω
+
z
C
S
x
e
(1.92)
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