Math  /  Data & Statistics

Question4. One cubic meter of total soil weighs 19.8 kN . if specific gravity of the soil particle is 2.1 and moisture content is 0.11 , find the void retio, dry density and degree of saturation. (3)

Studdy Solution

STEP 1

1. The unit weight of the soil is given as 19.8kN/m3 19.8 \, \text{kN/m}^3 .
2. The specific gravity (Gs G_s ) of the soil particles is 2.1 2.1 .
3. The moisture content (w w ) is 0.11 0.11 .
4. The density of water (ρw \rho_w ) is 1g/cm3 1 \, \text{g/cm}^3 or 1000kg/m3 1000 \, \text{kg/m}^3 .
5. The relationship between unit weight, density, and gravity is γ=ρg \gamma = \rho \cdot g , where g=9.81m/s2 g = 9.81 \, \text{m/s}^2 .

STEP 2

1. Calculate the dry density of the soil.
2. Calculate the void ratio.
3. Calculate the degree of saturation.

STEP 3

Calculate the dry density (ρd \rho_d ) using the formula:
ρd=γg(1+w)\rho_d = \frac{\gamma}{g(1 + w)}
Substitute the given values:
ρd=19.8kN/m39.81m/s2×(1+0.11)\rho_d = \frac{19.8 \, \text{kN/m}^3}{9.81 \, \text{m/s}^2 \times (1 + 0.11)}
Calculate:
ρd=19.89.81×1.111.81g/cm3\rho_d = \frac{19.8}{9.81 \times 1.11} \approx 1.81 \, \text{g/cm}^3

STEP 4

Calculate the void ratio (e e ) using the formula:
e=Gsρwρd1e = \frac{G_s \cdot \rho_w}{\rho_d} - 1
Substitute the given values:
e=2.1×11.811e = \frac{2.1 \times 1}{1.81} - 1
Calculate:
e0.16e \approx 0.16

STEP 5

Calculate the degree of saturation (Sr S_r ) using the formula:
Sr=wGseS_r = \frac{w \cdot G_s}{e}
Substitute the given values:
Sr=0.11×2.10.16S_r = \frac{0.11 \times 2.1}{0.16}
Calculate:
Sr1.44or144%S_r \approx 1.44 \, \text{or} \, 144\%
Since Sr S_r cannot exceed 100%, this indicates an error in assumptions or calculations.
The calculated values are: - Dry density: 1.81g/cm3 1.81 \, \text{g/cm}^3 - Void ratio: 0.16 0.16 - Degree of saturation: Calculation suggests 144% 144\% , which is not possible, indicating a need to re-evaluate assumptions or inputs.

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