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

Wet unit weight = 1.94 g/cc

Dry unit weight =1.05 g/cc

density of saturation = 0.41

porosity = 0.43

Step-by-step 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 =
`(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 =
`(2.68 + 0.78)/(1+0.78)`

Wet unit weight = 1.94 g/cc

and

Dry unit weight =
`(Gs - e)/(1+e)` density of water

Dry unit weight =
`(2.65 - 0.78)/(1+0.78)`

Dry unit weight =1.05 g/cc

and

density of saturation is

density of saturation = Wc ×
`(Gs)/(e)`

her Wc is water content

so density of saturation = 0.12 ×
`(2.68)/(0.78)`

density of saturation = 0.41

and

porosity is

porosity =
`(e)/(1+e)`

porosity =
`(0.78)/(1+0.78)`

porosity = 0.43