Compaction, Stress Distribution and Consolidation

When a soil is loaded it compresses. Compaction removes air quickly by mechanical effort; consolidation squeezes water out slowly under sustained load. Between them sits the question of how much stress actually reaches each depth.

Compaction

• Standard Proctor test: 2.6 kg rammer, 310 mm drop, 3 layers × 25 blows. Modified Proctor uses more energy (4.9 kg, 450 mm, 5 layers).
• Dry density rises with water content up to the optimum moisture content (OMC), giving the maximum dry density, then falls.
• More compactive effort → higher maximum dry density and lower OMC.
• Zero air voids line: γ_d = Gγ_w/(1 + wG), the upper limit for any compaction curve.
• Field control: sand replacement or core cutter tests; relative compaction = field γ_d/lab max γ_d.

Stress distribution

• Boussinesq (point load Q, depth z, radial distance r): σ_z = (3Q/(2πz²)) × [1/(1 + (r/z)²)]^(5/2). Directly below the load: σ_z = 0.4775 Q/z².
• Westergaard (layered soils): directly below, σ_z = 0.3183 Q/z².
• 2:1 method: the load spreads at 2 vertical to 1 horizontal: σ_z = Q/((B + z)(L + z)).
• Pressure bulb: the 0.1q contour reaches about 2B below a square footing.

Consolidation settlement

Normally consolidated clay (present stress is the highest it has ever had):

S = (C_c H/(1 + e₀)) log₁₀((σ'₀ + Δσ)/σ'₀)

Alternatively, S = m_v Δσ H, with m_v = a_v/(1 + e₀). The over-consolidation ratio OCR = σ'_c/σ'₀; over-consolidated clays compress far less (use the recompression index C_r).

Empirical: C_c = 0.009(LL − 10) (Skempton, undisturbed clays).

Rate of consolidation (Terzaghi)

• Time factor T_v = c_v t/d², with d = drainage path: H/2 for double drainage, H for single drainage.
• For U < 60%: T_v = (π/4)U². For U = 50%, T_v = 0.197; for U = 90%, T_v = 0.848.
• Time for the same degree of consolidation scales with d²: double drainage is 4 times faster than single drainage.
• Secondary compression (creep) continues after primary consolidation ends.

Worked example

A 4 m clay layer (C_c = 0.3, e₀ = 1.0) has σ'₀ = 100 kPa at mid-depth and gets Δσ = 100 kPa.

• S = 0.3 × 4/2 × log₁₀(2) = 0.6 × 0.301 = 0.181 m.

Practice