Biomedical Engineering Reference
In-Depth Information
a result that follows from (
8.50
) with the application of (
8.40
). When (
8.51
)is
dotted with
U
and (
8.40
) again employed, it follows that the undrained elastic
coefficients are also incompressible in the case of assumed incompressibility of the
matrix material and the fluid, as one would anticipate:
U S
u
¼ U S
d
K
Reff
ðU S
d
U S
d
UÞ¼U S
d
S
d
U ¼
0
:
(8.52)
In the isotropic compressible case (
8.49
) reduces to formulas for the undrained
bulk modulus
K
u
(where
K
u
1
S
u
¼ðU
U
u
Þ
) and the undrained Poisson's ratio
n
d
,
K
d
,
K
f
,
K
m
, and
in terms of
n
f
, thus
2
K
f
K
d
K
m
ð
1
ð
=
ÞÞ
K
u
K
d
¼
þ
and
K
f
K
m
K
d
K
m
=
ð
1
ð
=
ÞÞ f þ f
d
d
K
d
K
m
3
n
þ
S
ð
1
2
n
Þð
1
ð
=
ÞÞ
u
n
¼
:
(8.53)
K
d
K
m
3
S
ð
1
2
n
d
Þð
1
ð
=
ÞÞ
K
u
u
2,
consistent with the general result for incompressibility for the undrained constant
set. It follows that
E
u
In the isotropic incompressible case (
8.53
) reduces to 1
=
¼
0 and
n
¼
1
=
¼
3
G
.
Problems
8.6.1. Expand the second equality in (
8.50
) as a six-by-six matrix equation.
8.6.2. Expand the second equality in (
8.50
) as a six-by-six matrix equation for the
special case of transversely isotropic symmetry using the technical elastic
constants, i.e., Young's moduli and Poisson's ratios.
8.6.3. Show that (
8.53
) are a consequence of the two equations (7.20). Do this by
working backward; start with (
8.53
) and substitute
K
d
and
G
d
from (
8.8
),
(
K
d
/K
m
)] and employ the relations between the elastic isotropic
constants listed in Table 7.2. The mechanics of solving this problem is
straightforward algebraic substitution. However, it can become a task if
one is not careful to keep the algebraic objective in view. Try using a
symbolic algebra program.
a ¼
[1
8.7 Expressions of Mass and Momentum Conservation
The conservation of mass is expressed by the equation of continuity,
@r
@
t
þrðr
v
Þ¼
:
ð
:
Þ
0
3
6
repeated
The form of the mass conservation equation (3.6) is altered to apply to the pore
fluid volume by first replacing
r
by
fr
f
in (3.6) and then dividing the equation
through by
r
fo
, thus
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