Geotechnical Engineering

DHARM

84 GEOTECHNICAL ENGINEERING

Difference in the volume of water at LL and SL = (30 – 16.6) cm^3 = 13.4 cm^3

Corresponding difference in weight of water = 0.134 N

But this is (0.75 – 0.25) Wd or 0.50 Wd from Fig. 3.23.

∴ 0.50 Wd = 0.134

Wd = 0.268 N

Weight of water at SL = 0.25 Wd = 0.25 × 0.268 = 0.067 N

∴ Volume of water at SL = 6.7 cm^3

Volume of solids, Vs = Total volume at SL – volume of water at SL.

= (16.6 – 6.7) cm^3 = 9.9 cm^3.

Weight of solids, Wd = 0.268 N

∴ γs =

0 268

99

.

.

N/cm^3 = 0.027 N/cm^3 = 27 kN/m^3

∴ Specific gravity of solids =

γ

γ

s

w

=^27

981.

= 2.71

Shrinkage ratio, R =

W

V

d

d

=

26 8

16 6

.

.

= 1.61

Volumetric shrinkage, Vs = R(wi – ws) = R(wL – ws), here

= 1.61 (75 – 25) = 80.5%.

Example 3.19: The mass specific gravity of a saturated specimen of clay is 1.84 when the
water content is 38%. On oven drying the mass specific gravity falls to 1.70. Determine the
specific gravity of solids and shrinkage limit of the clay.

For a saturated soil,

e = w.G

∴ e = 0.38 G

Mass specific gravity in the saturated condition

=

γ

γ

sat

w

Ge

e

GG

G

= +

+

= +

+

()

()

(.)

1 (.)

038

1038

∴ 1.84 =

138

1038

.

.

G

+ G

whence G = 2.71

∴ Specific gravity of the solids = 2.71

By Eq. 3.50, the shrinkage limit is given by

ws =

γ

γ

w

d G

−

F

HG

I

KJ

×

1

100

where γd = dry unit weight in dry state.

But, γd = (mass specific gravity in the dry state) γw

= 1.70 γw