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

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

Step-by-step 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