Geotechnical Engineering

(Jeff_L) #1
DHARM

82 GEOTECHNICAL ENGINEERING

∴ Liquid limit, wL = Ip + wp = 30 + 25 = 55%.
By Eq. 3.40,

Liquidity index, IL =

()ww
I

p
p


where w is the natural moisture content.

∴ Liquidity index, IL =

()34 25
30


= 0.30
By Eq. 3.39,

Consistency index, Ic =

()ww
I

L
p


∴ Consistency index, Ic =

()55 34
30


= 0.70
The consistency of the soil may be described as ‘medium soft’ or ‘medium stiff’.
Example 3.15: A fine grained soil is found to have a liquid limit of 90% and a plasticity index
of 50. The natural water content is 28%. Determine the liquidity index and indicate the prob-
able consistency of the natural soil.


Liquid limit, wL = 90%
Plasticity index, Ip = 50
By Eq. 3.37, Ip = wL – wp
∴ Plastic limit, wp = wL – Ip = 90 – 50 = 40%
The natural water content, w = 28%
Liquidity index, IL, by Eq. 3.40, is given by

IL =

ww
I

p
p


=

28 40
50

12
50


=− = – 0.24 (negative)

Since the liquidity index is negative, the soil is in the semi-solid state of consistency and
is stiff; this fact can be inferred directly from the observation that the natural moisture con-
tent is less than the plastic limit of the soil.
Example 3.16: A clay sample has void ratio of 0.50 in the dry condition. The grain specific
gravity has been determined as 2.72. What will be the shrinkage limit of this clay?


The void ratio in the dry condition also will be the void ratio of the soil even at the
shrinkage limit: but the soil has to be saturated at this limit.
For a saturated soil,
e = wG
or w = e/G


∴ ws = e/G = 0.50/2.72 = 18.4%,
Hence the shrinkage limit for this soil is 18.4%.


Example 3.17: The following are the data obtained in a shrinkage limit test:
Initial weight of saturated soil = 0.956 N
Initial volume of the saturated soil = 68.5 cm^3

Free download pdf