Classification and Physical Properties of Soils - Elements of Soil Mechanics - page 27

Solution:

2

π 0 1
4

× .

Volume of sample = × 0 2 . = 0 0016 . m

3

M
V

3 15
0 0016

.

ρ b = = = 1969 kg/m

3

= 1 97 . Mg/m

3

.

3 15 2 82
2 82

. − .

w = = 11 7 . %

.

ρ 1 97

1 117

.
.

b

ρ = = = 1 76 . Mg/m

3

d

1 w

+

γ = × ρ 9 81 17 3 . = . kN/m

3

d d

Relationship between w, γ d and γ

W W
V

w s

γ = (1)

W
V

γ d = (2)

W
W

w

w = (3)

s

From Equation (3) W w = wW s and, substituting in Equation (1) ,

W
V

s

γ = ( 1 + w )

i.e.

γ

γ =

d

1 + w

Thus to find the dry unit weight from the bulk unit weight, divide the latter by (1 + w) where w is the
water content expressed as a decimal.

Relationship between e, w and G s for a saturated soil

W
W

V
V G

γ V

V G

e
G

w w w v

w ( V V if the soil is saturated )

= = = = = w v

γ s s w s s s s

i.e.

e wG s =