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

T30 = 0.08

Assume            Cv = 2.5 m2/year = 0.007 m2/day

t30 =

Note: The result depends on Cv assumed.