Question

A fine-grained soil has a liquid limit of 200%, determined from the Casagrande cup method. The plastic limit was measured by rolling threads of soil to be 45%. The water content of the soil is determined to be 60% in the field by oven-drying a field sample. The clay content is 63%, determined via sieve and hydrometer testing and the plasticity chart. Determine: (a) Plasticity index, liquidity index, and the activity. (b) What state is the clay (liquid, plastic, semi-solid, etc.) and why

Answer 1

Answer:

a)

Plasticity index $I_P$  = 155%

Liquidity index $I_L$ = 0.09677

Activity A = 2.4603

b)

Consistency index of the clay = 0.9032

Since Liquidity index of soil( 0.09677) is between 0 and 0.25 and Consistency index (0.9032 ) is between 0.75 and 1; Then, The Clay is in Plastic State

Explanation:

Given that;

Liquid Limit $W_{L}$ = 200%

Plastic Limit $W_{P}$ = 45%

Natural Water Content $W_N$ = 60%

Clay Content $C_C$ = 63%;

(a) Plasticity index, liquidity index, and the activity.

Plasticity index $I_P$ = Liquid Limit $W_{L}$ - Plastic Limit $W_{P}$

Plasticity index $I_P$ = 200% - 45%

Plasticity index $I_P$  = 155%

Liquidity index $I_L$ = [ (Natural Water Content $W_N$ - Plastic Limit $W_{P}$ = 45%) /  Plasticity index $I_P$]

so;

Liquidity index $I_L$ = ( 60 - 45)/155

Liquidity index $I_L$ = 15 / 155

Liquidity index $I_L$ = 0.09677

the activity A is;

A = Plasticity index $I_P$  / Clay Content $C_C$

Activity A = 155 / 63

Activity A = 2.4603

b) What state is the clay;

To get the state in which is in, we use the consistency index of Soil. which is given as follows;

$I_C$ = [( Liquid Limit $W_{L}$ - Natural Water Content $W_N$ ) / Plasticity index $I_P$]

we substitute

$I_C$  = ( 200 - 60) / 155

$I_C$  = 140 / 155

$I_C$  = 0.9032

Consistency index of the clay = 0.9032

Since Liquidity index of soil( 0.09677) is between 0 and 0.25 and Consistency index (0.9032 ) is between 0.75 and 1; Then, The Clay is in Plastic State