The soil angle of internal friction (φ) can be determined using the formula:
tan(φ) = (σ1 - σ3) / 2σ1
Where:
- σ1 = major principal stress (395.8 kN/m?)
- σ3 = minor principal stress (144.00 kN/m?)
Plugging in the values:
tan(φ) = (395.8 - 144.00) / (2 * 395.8)
tan(φ) = 251.8 / 791.6
φ = tan^(-1)(0.3179)
φ ≈ 17.95 degrees
Therefore, the soil angle of internal friction is approximately 17.95 degrees.
(b) A triaxial test was performed on dry cohesionless soil under a confining pressure of 144.00 kN/m?.
If the sample fails when the deviator stress reaches 395.8 kN/m?, determine the soil angle of internal fraction
1 answer