Answer :
The allowable gross load on a 2m x 2m foundation embedded at a depth of 1.5m can be determined using the general bearing capacity equation. The given properties of the soil are: angle of internal friction (ϕ) = 30°, cohesion (c) = 2 kPa, and unit weight (γ) = 17 kN/m³. Using the provided values, the allowable gross load is calculated to be 803.9 kPa.
The allowable gross load on a foundation can be determined using the general bearing capacity equation. The equation is given by Qa = cNc+qNq+0.5ρBNγ.
Where:
c is the cohesion of the soil
q is the overburden pressure at the foundation depth
ρ is the unit weight of soil
B is the width of the foundation
Nc, Nq, and Nγ are bearing capacity factors
First, we need to calculate the overburden pressure at the foundation depth:
q = γ × depth = 17 kN/m³ × 1.5m = 25.5 kPa
Next, we can plug in the values into the general bearing capacity equation:
Qa = 2 kPa × Nc + 25.5 kPa × Nq + 0.5 × 17 kN/m³ × 2m × Nγ
We need to determine the bearing capacity factors for this particular soil condition. Since the soil properties are provided, we can use Terzaghi's bearing capacity factor charts to find the values for the angle of internal friction (ϕ) = 30°:
Nc = 9.4
Nq = 9.4
Nγ = 16.1
Finally, we can substitute these values into the equation and calculate the allowable gross load:
Qa = 2 kPa × 9.4 + 25.5 kPa × 9.4 + 0.5 × 17 kN/m³ × 2m × 16.1
Qa = 18.8 kPa + 239.7 kPa + 545.4 kPa = 803.9 kPa
Therefore, the allowable gross load on the foundation is 803.9 kPa.
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