Question

A 0.320 ft3 sample of a certain soil has a weight of 38.9 lb., moisture content of 19.2%, and specific gravity of solids of 2.67. Find the void ratio, degree of saturation, and saturated unit weight.

Answer 1

e = 0.6342

S = 0.808

γsat = 126.207 lb/ft³

Explanation:

Given

VT = 0.320 ft³

WT = 38.9 lb

w = 19.2% = 0.192

Gs = 2.67

then we apply

γ = WT / VT

γ = 38.9 lb / 0.320 ft³

γ = 121.5625 lb/ft³

then we get γdry as follows

γdry = γ / (1 + w)

γdry = 121.5625 lb/ft³ / (1 + 0.192)

γdry = 101.982 lb/ft³

the void ratio (e) can be obtained applying this equation

γdry = Gs*γw / (1 + e)

101.982 = 2.67*62.42 / (1 + e)

⇒ e = 0.6342

We can get the degree of saturation (S) as follows

S*e = Gs*w

S = Gs*w / e

S = 2.67*0.192 / 0.6342

S = 0.808

The saturated unit weight (γsat) will be obtained applying this formula

γsat = (Gs + e)*γw / (1 + e)

γsat = (2.67 + 0.6342)*62.42 lb/ft³/ (1 + 0.6342)

γsat = 126.207 lb/ft³