Geotechnical Engineering

DHARM

COMPOSITION OF SOIL TERMINOLOGY AND DEFINITIONS 25

whence G = 2.74
Specific gravity of soil particles = 2.74
Void ratio = 0.393 × 2.74 = 1.08.
Example 2.8: The mass specific gravity of a fully saturated specimen of clay having a water
content of 30.5% is 1.96. On oven drying, the mass specific gravity drops to 1.60. Calculate the
specific gravity of clay. (S.V.U.—B.E.(R.R.)—Nov. 1972)

Saturated clay

Water content, w = 30.5%

Mass specific gravity, Gm = 1.96

∴γsat = Gm .γw = 1.96 γw

On oven-drying, Gm = 1.60

∴ γd = Gm.γw = 1.60γw

γsat = 1.96.γw =

()

()

Ge

e

+ w

+

γ

1

...(i)

γd = 1.60.γw =

G

e

. w
()

γ

1 +

...(ii)

For a saturated soil, e = wG

∴ e = 0.305G

From (i),

1.96 = (.)

(. )

.

(. )

GG

G

G

G

+

+

=

+

0 305

1 0 305

1 305

1 0 305

⇒ 1.96 + 0.598G = 1.305G

⇒ G = 1 960

0 707

.

.

= 2.77

From (ii),

1.60 = G/(1 + e)

⇒ G = (1 + 0.305G) 1.6

⇒ G = 1.6 + 0.485G

⇒ 0.512G = 1.6

⇒ G = 1.6/0.512 = 3.123

The latter part should not have been given (additional and inconsistent data).

Example 2.9: A sample of clay taken from a natural stratum was found to be partially satu-
rated and when tested in the laboratory gave the following results. Compute the degree of
saturation. Specific gravity of soil particles = 2.6 ; wet weight of sample = 2.50 N; dry weight of
sample = 210 N ; and volume of sample = 150 cm^3. (S.V.U.—B.E.(R.R.)—Nov., 1974)

Specific gravity of soil particles, G = 2.60

Wet weight, W = 2.50 N;

Volume, V = 150 cm^3

Dry weight, Wd = 2.10 N