Civil Engineering Reference
In-Depth Information
∂
u
t
u
−
u
0
,
k
+
1
0
,
k
=
δ
∆
t
∂
∂
2
u
c
h
v
=
(
u
+
u
−
2
u
)
2
,
k
4
,
k
0
,
k
y
2
2
∂
∂
u
x
2
c
h
v
=
(
u
+
u
−
2
u
)
1
,
k
3
,
k
0
,
k
2
2
Hence the explicit finite difference equation is:
u
+
=
r u
(
+
u
+
u
+
u
)
+
u
0
1 4
(
−
r
)
0
,
k
1
1
,
k
2
,
k
3
,
k
4
,
k
,
k
where
c
∆
2
t
v
r
=
h
The schematic form of this equation is illustrated in Fig.
12.16b
.
Impermeable boundary condition
Impermeable boundaries are treated as for the one-dimensional case.
12.14.2 Three-dimensional consolidation
For instances of radial symmetry the differential equation can be expressed in polar co-ordinates:
∂
∂
u
R
2
1
∂
∂
u
R
+
∂
∂
2
u
=
∂
u
t
c
+
v
2
R
z
2
∂
then
∂
∂
u
t
u
−
u
0
,
k
+
1
0
,
k
=
∆
t
∂
∂
u
z
2
u
+
u
−
2
u
2
,
k
4
,
k
0
,
k
=
2
∆
z
2
u
R
2
u
u
2
u
∂
∂
+
−
1
,
k
3
,
k
0
,
k
=
2
∆
R
2
1
∂
∂
u
R
1
u
−
u
R
3
,
k
1
,
k
=
R
R
2
∆
If we put
Δ
z
=
Δ
R
=
h the finite difference equation becomes:
h
R
+
h
2
u
+
=
r u
(
+
u
)
+
u
(
1 4
− +
r
)
ru
1
−
ru
1
+
0
,
k
1
2
,
k
4
,
k
0
,
k
1
,
k
3
,
k
2
where
c
∆
2
t
v
r
=
h
At the origin, where R
=
0:
1
∂
→
∂
∂
u
R
u
R
2
R
∂
2
