Drained Triaxial Compression of Saturated Sand A saturated, cohesionless sand specimen (c′ = 0) fails in a drained triaxial compression test under deviator stress q = (σ1 − σ3) = 3 kgf/cm^2 with cell (confining) pressure σ3 = 1 kgf/cm^2. Estimate the effective friction angle φ′ of the sand.

Introduction / Context:In drained triaxial compression of clean sands, the Mohr–Coulomb strength can be represented by σ1/σ3 = (1 + sin φ′) / (1 − sin φ′) when cohesion c′ ≈ 0. Given measured deviator stress and confining pressure, φ′ can be back-calculated, an essential step in geotechnical design for bearing capacity, slope stability, and earth pressure problems.

Given Data / Assumptions:

• Confining pressure: σ3 = 1 kgf/cm^2.
• Deviator stress at failure: q = σ1 − σ3 = 3 kgf/cm^2.
• Therefore, σ1 = 4 kgf/cm^2.
• Material: cohesionless sand ⇒ c′ = 0.

Concept / Approach:

For c′ = 0, the triaxial compression failure condition is σ1/σ3 = (1 + sin φ′) / (1 − sin φ′). Solving for sin φ′ with the measured stress ratio yields the effective friction angle φ′.

Step-by-Step Solution:

Verification / Alternative check:

Mohr’s circle at failure for these principal stresses will be tangent to a line at angle 45° + φ′/2 to the σ-axis; using φ′ ≈ 37° gives consistent geometry and shear stress at failure.

Why Other Options Are Wrong:

• 45° and 53°: Reflect much stronger sands or dense states not supported by the measured stress ratio of 4.
• 20°: Far too low for clean sand under drained loading with the stated failure stresses.

Common Pitfalls:

Using total stresses when pore pressures are significant (here test is drained and sand is cohesionless); mixing up q with σ1; rounding errors when converting to degrees.

Final Answer:

37°