Axially loaded pile in clay – compute skin friction capacity A single pile of diameter 0.50 m and length 10 m is embedded in saturated clay. Undrained strength parameters: cohesion cu = 60 kN/m², angle of internal friction φ = 0. Adhesion factor α = 0.6. Assuming only shaft resistance in undrained conditions, compute the skin friction capacity (in kN).

Introduction:For piles in saturated clays under short-term (undrained) loading, skin friction is commonly estimated using the α-method: unit shaft resistance fs = α * cu. The total shaft capacity equals fs times the pile surface area embedded in clay. This problem checks correct use of the α-method and geometric calculations.

Given Data / Assumptions:

• Diameter d = 0.50 m; embedded length L = 10 m.
• Undrained shear strength cu = 60 kN/m²; φ = 0 (undrained).
• Adhesion factor α = 0.6 ⇒ adhesion ca = α * cu.
• End bearing ignored; uniform properties along length.

Concept / Approach:

Compute the pile perimeter and multiply by length to get the surface area. Then apply fs = α * cu and multiply to obtain total skin friction. Units must remain consistent (kN/m² for stress, m² for area) to yield kN.

Step-by-Step Solution:

Verification / Alternative check:

Reasonableness: For medium-stiff clay and a 0.5 m pile, a few hundred kN of shaft capacity is typical; the magnitude aligns with experience.

Why Other Options Are Wrong:

671 assumes larger α or includes end bearing; 283 and 106 underuse parameters (e.g., half-perimeter or wrong cu). 412 reflects partial length or reduced α.

Common Pitfalls:

Mixing cu and ca; forgetting π in perimeter; using diameter instead of perimeter for area.

Final Answer:

565