Question

A soil specimen was tested to have a moisture content of 32%, a void ratio of 0.95, and a specific gravity of soil solids of 2.75. Determine:
a. the degree of satruation
b. porosity
c. dry unit weight

Answer 1

Answer:

a. 0.9263

b. 0.4872

c. 13.83kN/m$^{3}$

Explanation:

moisture content (ω) = 0.32

void ratio (e) = 0.95

specific gravity ($G_{s}$) = 2.75

the degree of satruation (S) = $\frac{w . G_{s} }{e}$ =0.32×2.75/0.95 = 0.9263

b. porosity (n) = $\frac{e}{e + 1}$ = 0.95/(0.95 + 1)= 0.4872

c. dry unit weight (γ$_{d}$) = $\frac{G_{s} . V_{w} }{1 + e}$

taking specific unit weight of water (V$_{w}$)= 9.81kN/m$^{3}$

γ$_{d}$ = 2.75 × 1000/(1 + 0.95) = 13.83kN/m$^{3}$