Pile Foundations MCQ with answer and solution

Concept

Consolidation Settlement of Pile Groups

Formula

Primary consolidation settlement: S = (Δσ × H) / M, where S is settlement, Δσ is stress increase, H is clay layer thickness, and M is constrained modulus. Distributed load area: A = (B + z)², where B is the width of the pile group and z is the depth below the equivalent raft foundation level. Working load: Q = Δσ × A

Step-by-step solution

1. Find remaining clay thickness below equivalent raft at 2/3 pile length: H = 10 - 6 = 4 m. 2. Calculate stress increase using Δσ = (S × M) / H = (0.4 × 10,000) / 4 = 1000 kPa. 3. Calculate effective area at mid-depth (z = 2 m): B = 2(0.25) + 0.5 = 1.0 m, A = (1.0 + 2)² = 9 m². 4. Calculate working load: Q = Δσ × A = 1000 × 9 = 9000 kN.

Explanation

The correct working load is 9000 kN. For a pile group, the load is assumed to act at 2/3 of the pile length (6 m deep), leaving 4 m of clay below it.1. Calculate average stress increase: Stress Increase = (Settlement × Constrained Modulus) / Clay Thickness \= (0.4 m × 10,000 kPa) / 4 m = 1000 kPa 2. Calculate effective area at the middle of the layer (2 m below load depth): Width of group B = 2 × spacing + diameter = 2(0.25 m) + 0.5 m = 1.0 m Distributed Area = (B + z)² = (1.0 m + 2 m)² = 9 m² 3. Calculate working load: Load = Stress × Area = 1000 kPa × 9 m² = 9000 kN

Shortcut

For pile group consolidation in clay, remember the 2/3 length rule for equivalent raft depth and 2:1 stress dispersion. Use Δσ = (S × M) / H_clay to quickly find stress increase, then multiply by effective area A = (B + z)² to get working load Q = Δσ × A.

Common mistake

A common mistake is using the full length of the clay layer (10 m) or the full pile length (9 m) instead of the remaining clay thickness below the equivalent raft level at 2/3 depth (4 m) when calculating the stress increase and consolidation settlement.