Example 15.4 A fill is placed at an initial saturation of 80% (So) and initial porosity of37.5% (no). The average unit weight of the compacted fill is 20,000 N/m3. The dam is raised to a height of 15 m in the first stage after which there is a gap during which one-third of pore pressure will have dissipated. In the next stage, the dam is raised to a height of 35 m. Plot u v/s σ. The variation of ∆V/Vo v/s σ′ is as follows:
∆V/Vo(%)
1
2
3
4
4.5
4.75
σ′ (N/cm2)
5.0
12.5
23
40
54.5
66.0
Solution: During the first stage:
F
∆V I
G
J po
∆u = –
H
VoK
F
∆V I
G
J
+ no (1 − So + SoH)
H
VoK
with po = atmosheric pressure = 10.3 N/cm2 and H = 0.02
F
∆V I
F
∆V I
10.3 G
J
10.3 G
J
H
VoK
H
VoK
= –
= –
.
∆V
F
∆V I
+ 0.375 (1
− 0.80 + 0.80 × 0.02)
0.0798 + G
J
Vo
H
VoK
Using this equation and the curve
∆V
v/s σ′, the following table can be prepared.
V
o
∆V/Vo(%)
∆u
σ′
σ
1
1.48
5.0
6.48
2
3.45
12.5
15.95
3
6.21
23.0
29.21
4
10.35
40.0
50.35
4.5
13.32
54.5
67.82
4.75
15.15
66.0
81.15
520 IRRIGATION AND WATER RESOURCES ENGINEERING
At the end of the first stage of construction (1 on Fig. 15.23)
= γ H = 20000 × 15 N/m2 = 300000 N/m2
= 30 N/cm2
For
σ = 30 N/cm2
∆V
= 3.06%
from graph
V
o
∴
σ′ = 23.3 N/cm2
from graph
and
u = 6.7 N/cm2
from graph
After one-third of pore pressure dissipation ( = 2.23 N/cm2)