Question

A dry sample of sand is placed in a container having a volume of 0.3 ft3. The dry weight of the sample is 31 lb. Water is carefully added to the container so not to disturb the condition of the sand. When the container is filled, the combined weight of soil plus water is 38.2 lb. Calculate: a) the void ratio of the soil in the container, b) the specific gravity of the soil particles, c) the wet and the dry unit weights specifying their units.

Answer 1

Answer:

Explanation:

Given:

Volume of dry soil = 0.3 ft^3

= 0.0085 m^3

Mass of dry soil = 31 lb

= 14.08 kg

Mass of dry soil + water = 38.2 lb

= 17.34 kg

Mass of water = 17.34 - 14.08

= 3.26 kg

A.

Density of water = 1000 kg/m^3

Volume of water = 0.0033 m^3

Porosity, e = Vv/Vs

Where,

Vv = volume of void-space (air and water)

Vs = volume of solids(dry soil)

= 0.0033/0.0085

= 0.39

B.

SG = density of dry soil/density of water

= (14.08/0.0085)/1000

= 1.656

Answer 2

The added water is taken as occupying the voids in the soil placed in the container.

The correct responses are;

• a) The void ratio of the soil is approximately 0.385

• b) The specific gravity of the soil is approximately 1.656

• c) The wet unit weight is $\underline{127.\overline 3}$ lb/ft.³, the dry unit weight is $\displaystyle \underline{103. \overline 3}$ lb/ft.³

Reasons:

Volume of the container, V = 0.3 ft.³

Dry weight of the sample = 31 lb

Combined weight of soil + water = 38.2 lb

a) To find the void ratio of the soil in the container

Solution;

Density of water, ρ = 62.4 lb/ft.³

$\displaystyle Volume \ of \ water \ added = \frac{Mass \ of \ water}{Density \ of \ water} = \frac{38.2 \ lb - 31 \ lb}{62.4 \ ft.^3} = \frac{3}{26} \ lb/ft.^3$

$\displaystyle Void \ ratio = \mathbf{\frac{Volume \ of \ voids, \ V_v}{Volume \ of \ the \ dry \ solids, \ V_s}}$

Volume of voids, $V_v$ = Volume of water added = $\displaystyle \frac{3}{26} \ lb/ft.^3$

Volume of the dry solid, $V_s$ = Volume of the container = 0.3 ft.³

$\displaystyle Void \ ratio, \ e = \frac{\frac{3}{26} }{0.03} = \frac{5}{13} \approx 0.385$

b) Specific gravity of the soil, $\mathbf{G_s}$, is found as follows;

$\displaystyle G_s = \mathbf{ \frac{M_s}{V_s \times \rho_w}}$

Which gives;

$\displaystyle G_s = \frac{31}{0.3 \times 62.4} = \frac{775}{468} \approx 1.656$

The specific gravity of the soil, $G_s$ ≈ 1.656

c) When all the void spaces are filled with water, the wet unit weight is given by the formula;

$\displaystyle \gamma_t = \gamma_{wet} = \frac{W_s + W_w}{V} = \frac{38.2 \ lb}{0.3 \ ft.^3} = 127.\overline 3 \ lb/ft.^3$

The wet unit weight, $\gamma_{wet}$ = $\underline{127.\overline 3 \ lb/ft.^3}$

The dry unit weight, $\gamma _d$, is obtained as follows;

$\displaystyle \gamma _d = \frac{W_s}{V} = \frac{31 \ lb }{0.3 \ ft.^3} = 103. \overline 3 \ lb/ft.^3$

The dry unit weight, $\gamma _d$ =$\underline{103. \overline 3 \ lb/ft.^3}$

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