Soil Mechanics A

1
Numerical Solution of the 1-D Consolidation
Equation
2
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3
The 1-D Consolidation Equation
For homogeneous soil the equation governing one
dimensional consolidation is given by
Analytical solutions are only available for very
simple initial and boundary conditions. Approxima
te numerical solutions are required for other
cases.
4
Setting up the numerical solution grid
q
1
Soil Layer
2
3
4
5
Setting up the numerical solution grid
q
t 0
t t1
t t2
t
1
1
Soil Layer
2
2
3
3
4
4
z
6
Assumed excess pore pressure variation
u uA
A
u uB
B
u
z
C
u uC
t tn
7
Finite difference formulae
Assume
2
u
a
a
z
a
z



1
2
3
8
Finite difference formulae
Assume
2
u
a
a
z
a
z



1
2
3
2
u
a
a
z
a
z

-

D
D
A
1
2
3
Then
u
a

B
1
2
u
a
a
z
a
z



D
D
C
1
2
3
9
Finite difference formulae
Assume
2
u
a
a
z
a
z



1
2
3
2
u
a
a
z
a
z

-

D
D
A
1
2
3
Then
u
a

B
1
2
u
a
a
z
a
z



D
D
C
1
2
3
a
u

B
1
u
u
-
A
C
a

Hence
2
z
2
D
u
u
u
2

-
A
C
B
a

3
2
z
2
D
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Finite difference formulae

u
u
-

u
é
ù
A
C

ê
ú

z
z
2
D
(1 d)
ë
û
B
é
ù
2
u
u
u

-
2

u
A
C
B

ê
ú
2
2

z
z
D
ë
û
B
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Finite difference form of consolidation equation
12
Finite difference form of consolidation equation

13
Finite difference form of consolidation equation

14
Finite difference form of consolidation equation
where
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Finite difference form of consolidation equation
To proceed an approximate numerical solution of
the integral is needed. This is as follows
16
Finite difference form of consolidation equation
To proceed an approximate numerical solution of
the integral is needed. This is as follows
F(t)
Error in approximation
Area F(t) Dt
t
t
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Finite difference form of consolidation equation
(2d)
where
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Finite difference form of consolidation equation
D
u
q
u
t
u
t
u
t



-
b

(
)
(
)
(
)
2
u
(tDt)
(t)

(2e)
B
A
C
B
B
B

• This equation allows the value of the excess pore
  pressure at time tDt to be calculated from those
  at time t.
• This equation can be used for each point (depth)
  in the solution grid.
• To obtain solutions appropriate boundary
  conditions and initial conditions must be
  specified for the solution grid

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Finite difference approximation of a drainage
boundary
A drainage boundary is one where the excess pore
pressure is zero, either because it is the soil
surface, or because the adjacent layer is very
permeable i.e. uB 0
.
B
Drainage Boundary u 0
Saturated soil
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Finite difference approximation of an impermeable
boundary
Saturated soil
.
A
.
B
.
Impermeable layer
C

u
At an impermeable boundary

0

z
thus
and hence
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Example Clay layer subjected to a surcharge
loading
q 64 kPa
4 sub-layers cv 2 m2/year
4m
Impermeable bedrock
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Example Clay layer subjected to a surcharge
loading
Numerical Solution t(years)
0 0.25 0.5 0.75 1 1.25 1.5 z 0 64
0 0 0 0 0 0 z 1m 64 64 32 32 24 24 20 z
2m 64 64 64 48 48 40 40 z 3m 64 64 64 64 56 56
48 z 4m 64 64 64 64 64 56 56 dummy 64 64 64 6
4 56 56 48
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Selection of b
There is an important restriction on b to ensure
stability of the solution, which is
24
Selection of b
There is an important restriction on b to ensure
stability of the solution, which is
It is convienient to choose b 0.5 because this
simplifies the finite difference equation
25
Selection of b
There is an important restriction on b to ensure
stability of the solution, which is
It is convienient to choose b 0.5 because this
simplifies the finite difference equation
26
Selection of b
There is an important restriction on b to ensure
stability of the solution, which is
It is convienient to choose b 0.5 because this
simplifies the finite difference equation
or
D
D
u
t
t
u
t
q
t
u
t
u
t





-
b
(
)
(
)
u
(
)
(
)
(
)
2
B
B
B
A
C
B
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Selection of b
There is an important restriction on b to ensure
stability of the solution, which is
It is convienient to choose b 0.5 because this
simplifies the finite difference equation
or
D
D
u
t
t
u
t
q
t
u
t
u
t





-
b
(
)
(
)
u
(
)
(
)
(
)
2
B
B
B
A
C
B
1
1
if b

D
D
u
t
t
u
t
q
t
u
t
u
t





-
(
)
(
)
u
(
)
(
)
(
)
2
B
B
B
A
C
B
2
2
28
Selection of b
There is an important restriction on b to ensure
stability of the solution, which is
It is convienient to choose b 0.5 because this
simplifies the finite difference equation
or
D
D
u
t
t
u
t
q
t
u
t
u
t





-
b
(
)
(
)
u
(
)
(
)
(
)
2
B
B
B
A
C
B
1
1
if b

D
D
u
t
t
u
t
q
t
u
t
u
t





-
(
)
(
)
u
(
)
(
)
(
)
2
B
B
B
A
C
B
2
2
1
D
q
t
u
t



u
(
)
(
)
D
u
t
t

(
)
(4)
B
A
C
B
2
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Selection of b

• Before commencing the solution the values of b,
  Dt, Dz have to be chosen. Two of the three
  parameters may be chosen which then fix the
  other. There are restrictions on all the
  parameters.
• b must be less than or equal to 0.5.
• Dz must be chosen so that there is an integer
  number of soil layers.
• Dt must be chosen so that the solution is
  obtained at the correct time, and so that changes
  in load are modelled correctly (i.e. occur at the
  correct time).

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Example 1 Step 1 - Determine Dz
For example 1 1. Choose 4 soil layers - Hence Dz
1m
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Example 1
q 64 kPa
4 sub-layers cv 2 m2/year
4m
Impermeable bedrock
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Example 1 Step 2 - Determine Dt
For example 1 1. Choose 4 soil layers - Hence Dz
1m 2. Choose b 1/2
33
Example 1 Step 2 - Determine Dt
For example 1 1. Choose 4 soil layers - Hence Dz
1m 2. Choose b 1/2 3. Determine Dt from
t
D
2
1
thus

2
2
1
and
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Example 1 Step 2 - Determine Dt
For example 1 1. Choose 4 soil layers - Hence Dz
1m 2. Choose b 1/2 3. Determine Dt from
t
D
2
1
thus

2
2
1
and
4. Check Dt OK, if not select Dt and determine b.
35
Example 1 Step 3 - Initial excess pore pressures
q
64 kPa
t
There can be no instantaneous volume change, and
for 1-D conditions a sudden change in q gives an
equal change in u, the excess pore pressure.
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Example 1 Step 3 - Initial excess pore pressures
Commencing the Analysis
t(years) 0 z 0 64 z
1m 64 z 2m 64 z 3m 64 z 4m 64
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Example 1 Step 4 - Introduce dummy node for
impermeable base
Commencing the Analysis
t(years) 0 z 0 64 z
1m 64 z 2m 64 z 3m 64 z
4m 64 dummy
t(years) 0 z 0 64 z
1m 64 z 2m 64 z 3m 64 z
4m 64 dummy 64
38
Example 1 Step 5 - Use finite difference
equation to march forward solution
Consider a particular time step
t (years) 0.75 1 q(kPa) 64 64 z0 0 0 z
1m 32 24 z2m 48 48 z3m 64 56 z4m
64 64 dummy 64 56
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Initial State t (years) 0.75 1 q(kPa) 64 64
z0 0 z1m 32 z2m 48 z3m 64
z4m 64 dummy 64
40
Incorporation of upper boundary condition t
(years) 0.75 1 q(kPa) 64 64 z0 0 0 z1m
32 z2m 48 z3m 64 z4m 64 dumm
y 64
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Solution at 1m t years 0.75 1 q(kPa) 64 64 z
0 0 0 z 1m 32 24 z 2m 48 z
3m 64 z 4m 64 dummy 64
24 (0 48)/2
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Solution at 2m t (years) 0.75 1 q(kPa) 64 64
z 0 0 0 z 1m 32 24 z 2m 48 48 z
3m 64 z 4m 64 dummy 64
43
Solution at 3m t (years) 0.75 1 q(kPa) 64 64
z 0 0 0 z 1m 32 24 z 2m 48 48 z
3m 64 56 z 4m 64 dummy 64
44
Solution at 4m t years 0.75 1 q(kPa) 64 64 z
0 0 0 z1m 32 24 z2m 48 48 z3m 64 5
6 z4m 64 64 dummy 64
64 (64 64)/2
45
Incorporation of base boundary condition t
(years) 0.75 1 q(kPa) 64 64 z0 0 0 z1m
32 24 z2m 48 48 z3m 64 56 z4m 64 64
dummy 64 56
46
Calculation of Settlement
H
S(t)
dz

e
ò
v
0
H
m
q
u
dz

-
(
)
ò
v
0
H
m
q(t) H
m
udz

-
ò
v
v
0
An approximate numerical approach is required to
evaluate the integral.
47
Calculation of Settlement
u0
Dz
u1
Dz
H
Trapezoidal rule
u2
Dz
u3
Dz
u4
u
u

H
(
top
bot
)
m
z
u




D
udz
å
ò
v
rest
2
0
48
Example 1 Calculation of Settlement
)
t 0.75 1 q(kPa) 64 64 z 0 0 0 z
1m 32 24 z 2m 48 48 z 3m 64 56 z
4m 64 64 dummy 64 56 settlement(mm) 24 28.8.
S

x
x
0
0003
64
4
.

0
64
-
x
0
0003
.
(
2



x
24
48
56
1
)
28.8
49
Example 2 Time step 2 months
c
t
D
v
b

2
z
D
x
2
2
12
(
/
)

2
1
1

3
50
Example 2 Time step 2 months
u
t
t
u
t
u
t
u
t
u
t
(
)
(
)

(
)
(
)
(
)




-
D
b
2
B
B
A
C
b
.
.

.
.42
.



x

-
x
59
26
0
3333
48
20
62
2
59
26
.

56
63
t(mths) 0.00 2.00 4.00 6.00 8.00 10.00 12.00 se
ttlement(mm) 0.00 9.60 16.00 20.27 23.82 26.90 29.
67 q(kPa) 64.00 64.00 64.00 64.00 64.00 64.00 64.
00 z 0 64.00 0.00 0.00 0.00 0.00 0.00 0.00 z
1m 64.00 64.00 42.67 35.56 30.81 27.65 25.28 z
2m 64.00 64.00 64.00 56.89 52.15 48.20 45.04 z
3m 64.00 64.00 64.00 64.00 61.63 59.26 56.63 z
4m 64.00 64.00 64.00 64.00 64.00 62.42 60.31 dum
my 64.00 64.00 64.00 64.00 61.63 59.26 56.63
51
Example 3 Variable loading
q
120 kPa
t
12 months
Choose b 0.5 Then Dt 0.25 years and Dq 30
kPa (for t lt 12 months) then Dq 0 kPa
52
Example 3 Variable loading
t(years)
0
0.25
0.5
0.75
1
1.25
1.5
settlement(mm
0
4.5
13.5
24.75
38.25
48.938
56.813
)
q(kPa)
0
30
60
90
120
120
120
z0
0
0
0
0
0
0
0
z1m
0
30
45
60
71.25
52.5
46.875
z2m
0
30
60
82.5
105
93.75
82.5
z3m
0
30
60
90
116.25
112.5
105
z4m
0
30
60
90
120
116.25
112.5
dummy
0
30
60
90
116.25
112.5
105
53
Example 4 Abrupt change of load
q
t
1
1.25
0.75
1.00
t
28.8
37.2
24
28.8
settlement(mm)
96
96
64.00
64.00
q(kPa)
32
0
0.00
0.00
z0
56
56
32.00
24.00
z1m
80
72
48.00
48.00
z2m
88
88
64.00
56.00
z3m
96
88
64.00
64.00
z4m
88
88
64.00
56.00
dummy
54
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