Geotechnical Engineering

DHARM

COMPOSITION OF SOIL TERMINOLOGY AND DEFINITIONS 21

Water content, w =

weight of water

weight of solids

=

Se

G

w

w

γ

γ

= S.e/G

or w.G = S.e

γ =

total weight

total volume

=

+

+

=

+

+

(.)

()

()

()

GSe

e

Gw

w e

w

1

1

1

γ

γ

γd = weight of solids

total volume

=

+

G

e

. m
()

γ
1
and so on.

The reader may, in a similar manner, prove the other relationships also.

2.4 Illustrative Examples

Example 2.1: One cubic metre of wet soil weighs 19.80 kN. If the specific gravity of soil parti-
cles is 2.70 and water content is 11%, find the void ratio, dry density and degree of saturation.

(S.V.U.—B.E.(R.R.)—Nov. 1975)

Bulk unit weight, = 19.80 kN/m^3

Water content, w = 11% = 0.11

Dry unit weight, γd =

γ

()

.

1 (.)

19 80

+ 1011

=

w +

kN/m^3 = 17.84 kN/m^3

Specific gravity of soil particles G = 2.70

γd = G

e

.γw

1 +

Unit weight of water, γw = 9.81 kN/m^3

∴ 17.84 =

270 981

1

..

()

×

+e

(1 + e) = 270 981

17 84

..

.

×

= 1.485

Void ratio, e = 0.485

Degree of Saturation, S = wG/e

∴ S =

011 270

0 485

..

.

×

= 0.6124

∴ Degree of Saturation = 61.24%.

Example 2.2: Determine the (i) Water content, (ii) Dry density, (iii) Bulk density, (iv) Void
ratio and (v) Degree of saturation from the following data :

Sample size 3.81 cm dia. × 7.62 cm ht.

Wet weight = 1.668 N

Oven-dry weight = 1.400 N

Specific gravity = 2.7 (S.V.U.—B. Tech. (Part-time)—June, 1981)

Wet weight, W = 1.668 N

Oven-dry weight, Wd = 1.400 N