Question

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2.70. Determine the volume of the sample when the water content is 10%.

Answer 1

`79 f t^(3)` is the volume of the sample when the water content is 10%.

Step-by-step explanation:

Given Data:

`V_(1)=100\ \mathrm{ft}^(3)`

First has a natural water content of 25% =
`(25)/(100)` = 0.25

Shrinkage limit,
`w_(1)=12 \%=(12)/(100)=0.12`

`G_(s)=2.70`

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

`V \propto[1+e]`

`(V_(2))/(V_(1))=(1+e_(2))/(1+e_(1))` ------> eq 1

`e_(1)=(w_(1) * G_(s))/(S_(r))`

The above equation is at
`S_(r)=1`,

`e_(1)=w_(1) * G_(s)`

Applying the given values, we get

`e_(1)=0.25 * 2.70=0.675`

Shrinkage limit is lowest water content

`e_(2)=w_(2) * G_(s)`

Applying the given values, we get

`e_(2)=0.12 * 2.70=0.324`

Applying the found values in eq 1, we get

`(V_(2))/(100)=(1+0.324)/(1+0.675)=(1.324)/(1.675)=0.7904`

`V_(2)=0.7904 * 100=79\ \mathrm{ft}^(3)`