Question

The unit weight of a soil is 96 lb/ft3 . The moisture content if this soil is 17% when the degree of saturation is 60%. Determine: a. Void ratio b. Specific gravity of solids c. Saturated unit weight.

Answer 1

To solve this problem we will begin by defining the variables that are sought to be found. The proper order with the data provided is to start determining the void time, to later determine the severity of solids and end with the saturated unit weight. Let's start defining the variables

The unit weight of soil can be written as,

$\gamma = \frac{G_s\gamma_w(1+w)}{1+e}$

Here,

$G_s =$  Specific gravity of soil

w = Unit Weight of water

e = Void ratio, which also can be defined as,

Void ratio

$e = \frac{wG_s}{S}$

Replacing,

$\gamma = \frac{G_s\gamma_w(1+w)}{1+\frac{wG_s}{S}}$

PART B) Now we have that,

$96 = \frac{G_s(62.4)(1+0.17)}{1+\frac{(0.17)G_s}{0.6}}$

So the specific gravity of the soil is

$G_s = 2.09$

PART A) Calculate the void ratio with this information,

$e = \frac{wG_s}{S}$

$e = \frac{(0.17)(2.09)}{0.60}$

$e = 0.5921$

PART C) Finally we can calculate the saturated unit weight

$\gamma_{sat} = \frac{(G+e)\gamma_w}{1+e}$

$\gamma_{sat} = \frac{(2.09+0.5921)(62.4)}{1+0.5921}$

$\gamma_{sat} = 105.12lb/ft^3$