Geotechnical Engineering

DHARM

80 GEOTECHNICAL ENGINEERING

the original soil was 50, what is the percentage of particles finer computed above in the entire

soil sample?

By Eqs. 3.21 and 3.22,

D = K Ht/

where K =

3

1

μ

γ

w

w()G−

∴ K =

3

1

×

×−

0.001

9.8 ()2.72

= 0.0133 mm

min

cm

D = 0.0133^10

10

mm = 0.0133 mm

This is the largest size remaining at the sampling depth.

By Eq. 3.26,

Nf =

W

W

V

V

p

sp

F

HG

I

KJ

F

HG

I

KJ

× 100

Here Wp = 0.0032 N; Ws = 0.5 N; V = 500 ml; Vp = 10 ml.

∴ Nf =

0.0032

050

500

. 10

× × 100 = 32%

Thus, the percentage of particles finer than 0.0133 mm is 32.

By Eq. 3.28, N = Nf(Wf/W)

Here Wf/W is given as 0.50.

∴ N = 32 × 0.50 = 16

Hence, this percentage is 16, based on the entire sample of soil.

Example 3.12: In a hydrometer analysis, the corrected hydrometer reading in a 1000 ml
uniform soil suspension at the start of sedimentation was 28. After a lapse of 30 minutes, the
corrected hydrometer reading was 12 and the corresponding effective depth 10.5 cm. The spe-
cific gravity of the solids was 2.70. Assuming the viscosity and unit weight of water at the
temperature of the test as 0.001 N-s/m^2 and 9.8 kN/m^3 respectively, determine the weight of
solids mixed in the suspension, the effective diameter corresponding to the 30-min. reading
and the percentage of particles finer than this size.

The corrected hydrometer reading initially, Rhi = 28

∴γi = 0.01028 N/cm^3

But, by Eq. 3.29,

γi = γw + [G – 1)/G] W/V

Substituting,

0.01028 =^001

27 1

2 7 1000

. (. )
.

+

−

× W

whence W =

0 028 2 7

17

..

.

×

N = 0.445 N