Direct shear test on cohesionless soil (c = 0) — find angle of internal friction φ (Given σ_n = 200 kN/m^2; failure shear stress τ_f = 100 kN/m^2.)

Introduction / Context:
In a direct shear test on a cohesionless soil, the Mohr–Coulomb failure criterion reduces to τ = σ * tan φ because cohesion c = 0. The test provides τ at failure under a known normal stress σ, allowing the friction angle φ to be computed directly.

Given Data / Assumptions:

• Normal stress σ_n = 200 kN/m^2.
• Failure shear stress τ_f = 100 kN/m^2.
• Soil is cohesionless: c = 0.
• Plane strain conditions as per direct shear test.

Concept / Approach:
Mohr–Coulomb: τ_f = c + σ_n * tan φ → with c = 0, tan φ = τ_f / σ_n. Compute tan φ and then take arctangent to obtain φ in degrees.

Step-by-Step Solution:
1) tan φ = τ_f / σ_n = 100 / 200 = 0.5.2) φ = arctan(0.5) ≈ 26.565°.3) Rounded to one decimal place → 26.6°.4) Select the option 26.6.

Verification / Alternative check:
Sanity check: tan 30° = 0.577 > 0.5, so φ must be slightly below 30°, consistent with 26.6°.

Why Other Options Are Wrong:
29.5° and 30.0° correspond to tan φ close to 0.565–0.577 (too high); 32.6° is higher still.

Common Pitfalls:
Using wrong stress units (kPa vs kN/m^2 are equivalent); confusing cohesion intercept; rounding φ prematurely leading to an incorrect option match.

Final Answer:
26.6