Answer :
For a square continuous foundation with dimension 3ft x 3ft and a soil unit weight of 17.8 kN/m3, the bearing capacity is approximately 566.91 kN/m2. For a rectangular continuous foundation with dimensions of 5m x 1.2m, and a soil unit weight of 16.5 kN/m3, the bearing capacity is approximately 278.86 kN/m2.
Finally, for a continuous foundation with dimensions of 3m x 2m, an inclination angle of 30 degrees and a soil unit weight of 16.5 kN/m3, the bearing capacity is approximately 348.44 kN/m2.
The bearing capacity of the given foundation types can be calculated using the general bearing capacity equation.
To calculate the bearing capacity of the given foundation types, we use the general bearing capacity equation, which is given by:
q_ult = cN_c + qN_q + 0.5γBN_γ
where q_ult is the ultimate bearing capacity, c is the cohesion of the soil, N_c, N_q, and N_γ are bearing capacity factors, γ is the unit weight of the soil, and B is the width of the foundation.
a. For a square continuous foundation with dimension 3ft x 3ft and a soil unit weight of 17.8 kN/m3, we first convert the dimensions to meters (1ft = 0.3048m), so the foundation has dimensions 0.9144m x 0.9144m. Assuming the soil has no cohesion (c = 0), and using the bearing capacity factors N_c = 5.14, N_q = 1, and N_γ = 0.5, we have:
q_ult = cN_c + qN_q + 0.5γBN_γ
= 0 + 1(28400)(1) + 0.5(17.8)(0.9144)(5.14)
≈ 566.91 kN/m2
b. For a rectangular continuous foundation with dimensions of 5m x 1.2m and a soil unit weight of 16.5 kN/m3, we have:
q_ult = cN_c + qN_q + 0.5γBN_γ
= 0 + 1(28)(1.2) + 0.5(16.5)(5)(28)
≈ 278.86 kN/m2
c. For a continuous foundation with dimensions of 3m x 2m, an inclination angle of 30 degrees, and a soil unit weight of 16.5 kN/m3, we have to calculate the correction factor N_γ for the inclined load. We can use the equation for the correction factor, which is given by:
N_γ = (1 + sinφ)(1 - sinφ tan²β)
where φ is the internal angle of friction of the soil (assumed to be 30 degrees), and β is the inclination angle (also 30 degrees). Plugging in the values, we have:
N_γ = (1 + sin30)(1 - sin30 tan²30)
= 1.38
Using the bearing capacity factors N_c = 9.44, N_q = 1.6, and N_γ = 1.38, we have:
q_ult = cN_c + qN_q + 0.5γBN_γ
= 0 + 1.6(35)(2)
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