Geotechnical Engineering

DHARM

COMPOSITION OF SOIL TERMINOLOGY AND DEFINITIONS 23

Porosity, n = 35%

Void ratio, e = n/(1 – n) = 0.35/0.65 = 0.54

Specific gravity of soil particles = 2.7

Dry unit weight, γd =

G

e

. w
()

γ

1 +

= 27 981

154

..

.

×

kN/m^3 = 17.20 kN/m^3

Saturated unit weight, γsat =

()

()

Ge.

e w

+

1 +

γ

= (.. )

.

270 054

154

+ × 9.81 kN/m 3

= 20.64 kN/m^3

Submerged unit weight, γ′ = γsat – γw

= (20.64 – 9.81) kN/m^3

= 10.83 kN/m^3.

Example 2.5: (i) A dry soil has a void ratio of 0.65 and its grain specific gravity is = 2.80. What
is its unit weight?

(ii) Water is added to the sample so that its degree of saturation is 60% without any

change in void ratio. Determine the water content and unit weight.

(iii) The sample is next placed below water. Determine the true unit weight (not consid-

ering buoyancy) if the degree of saturation is 95% and 100% respectively.

(S.V.U.—B.E.(R.R.)—Feb, 1976)

(i)Dry Soil

Void ratio, e = 0.65

Grain specific gravity, G = 2.80

Unit weight, γd = G

e

. w
()

..

.

γ

1

280 98

+ 165

= × kN/m^3 = 16.65 kN/m^3.

(ii)Partial Saturation of the Soil

Degree of saturation, S = 60%

Since the void ratio remained unchanged, e = 0.65

Water content, w = Se

G

...

.

=060 065×

280

= 0.1393

= 13.93%

Unit weight =

()

()

.

(...)

.

.

GSe

e w

+

+

=

+×

1

280 060 065

165

γ 981 kN/m^3

= 18.97 kN/m^3.

(iii)Sample below Water

High degree of saturation S = 95%

Unit weight =

()

()

.

(... )

.

.

GSe

e w

+

+

=

+×

1

280 095 065

165

γ 981 kN/m^3

= 20.32 kN/m^3