Question

3. An undisturbed soil sample has the following data (Source - Lu & Evett): 1. Void ratio = 0.78 2. Water content -12% 3. Specific gravity of solids - 2.68 Determine: 1. Wet unit weight (in g/ce) 2. Dry unit weight (in g/ce) 3. Degree of saturation 4. Porosity

Answer 1

Answer:

Wet unit weight = 1.94 g/cc

Dry unit weight =1.05 g/cc

density of saturation = 0.41

porosity = 0.43

Explanation:

given data

Void ratio = 0.78

Water content -12%

Specific gravity of solids - 2.68

to find out

Wet unit weight and Dry unit weight and Degree of saturation and Porosity

solution

we know that Wet unit weight formula is

Wet unit weight = $\frac{Gs + e}{1+e}$ density of water

here Gs is specific gravity of soild and e is void ratio and density of water is 1

so Wet unit weight = $\frac{2.68 + 0.78}{1+0.78}$

Wet unit weight = 1.94 g/cc

and

Dry unit weight = $\frac{Gs - e}{1+e}$ density of water

Dry unit weight = $\frac{2.65 - 0.78}{1+0.78}$

Dry unit weight =1.05 g/cc

and

density of saturation is

density of saturation = Wc × $\frac{Gs}{e}$

her Wc is water content

so density of saturation = 0.12 × $\frac{2.68}{0.78}$

density of saturation = 0.41

and

porosity is

porosity = $\frac{e}{1+e}$  

porosity =  $\frac{0.78}{1+0.78}$  

porosity = 0.43