HOMEWORK SOLUTION
HW #8
GEOTECHNICAL ENGINEERING I
(Question no. 7.4, 7.7, 7.8, 7.12, ..Al-Khafaji textbook)
7.4 (a) See figure 1. and figure 2.
|
Pressure (kPa) |
Void Ratio |
|
20 |
0.853 |
|
40 |
0.848 |
|
80 |
0.838 |
|
160 |
0.82 |
|
320 |
0.778 |
|
640 |
0.689 |
|
1280 |
0.59 |
|
320 |
0.619 |
|
80 |
0.654 |
|
20 |
0.69 |
|
0 |
0.79 |
Figure 1.
Figure 2.
(b) From figure 2.
Compression Index = 
= 0.332
Recompression Index = 
= 0.057
(c) From figure 2. (according to Cassagrande)
The past pressure (max) = 250 kPa
7.7 (a) See figure 3
d0 = 3.5 mm
d100 = 4.36 mm
|
Time (min) |
Dial Reading (mm) |
|
0 |
3.394 |
|
0.1 |
3.493 |
|
0.25 |
3.529 |
|
0.5 |
3.573 |
|
1 |
3.638 |
|
2 |
3.731 |
|
4 |
3.872 |
|
8 |
4.048 |
|
15 |
4.192 |
|
30 |
4.312 |
|
70 |
4.415 |
|
140 |
4.47 |
|
260 |
4.506 |
|
455 |
4.54 |
|
1440 |
4.609 |
Figure 3.
(b) T50 = 0.197
ds = 3.44 mm (from graph)
H = Thickness/2=19.05/2mm
= 9.525mm (permitted drainage top and bottom)
t50 = corresponds with
(
)
= 4.5 min
so, Cv
= 
(c) Cv = 3.97 mm2/min = 3.97x 10-6 m2/min
av = 
gw = 9800 N/m3
e = (e1+e2)/2 = (0.845+0.712)/2 = 0.7785
So, k =
=
= 175.10-6 N/min = 2.9 . 10-4 cm/s
(d) Compression Index = 
7.8 Sample thickness = 25.4 mm
t consolidation 50% = 10 min
Drained at top and bottom , Hlab = 25.4/2=12.7 mm
Clay stratum = 7 m thick (covered by layer of sand and underlain by impervious bedrock)
So, Hfield = 7 m
In the lab
T50 = 0.197
T50
= 
Cv = 
In the field
t50 field = 
7.12 Clay layer thickness = 3.5 m (drain at top and bottom)
So, Hfield = 3.5/2 = 1.75 m
Load = 100 kPa
T30 = 0.08
Assume Cv = 2.5 m2/year = 0.007 m2/day
t30 =
Note: The result depends on Cv assumed.