A 3 m thick normally consolidated clay layer (γsat = 19 kN/m³, Cc = 0.15, e₀ = 1) lies below 6 m of sand (γ = 18 kN/m³). The water table is at the top of the clay; use γw = 9.81 kN/m³. A wide fill raises the vertical stress throughout the clay by 80 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.
Consolidation settlement worksheet with answers
A 5 m thick normally consolidated clay layer (γsat = 17.5 kN/m³, Cc = 0.3, e₀ = 0.65) lies below 4 m of sand (γ = 17 kN/m³). The water table is at the top of the clay; use γw = 9.81 kN/m³. A wide fill raises the vertical stress throughout the clay by 95 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.
A 7 m thick overconsolidated clay layer has Cc = 0.2, Cr = 0.05 and e₀ = 0.6. At its mid-depth the effective vertical stress is 45 kPa and the overconsolidation ratio is OCR = 3. A new building adds 130 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
A 7.5 m thick overconsolidated clay layer has Cc = 0.2, Cr = 0.06 and e₀ = 1.1. At its mid-depth the effective vertical stress is 90 kPa and the preconsolidation pressure is 130 kPa. A new building adds 25 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
In a laboratory oedometer test, a clay specimen 20 mm thick, drained at the top and bottom, reaches 50% consolidation in 15 min. In the field the same clay forms a layer 5 m thick. It lies on impervious rock, with sand above it, so water drains out only at the top. How many days t does the field layer take to reach 50% consolidation?
A clay layer 6.5 m thick has a coefficient of consolidation cv = 0.5 m²/year. It lies between two sand layers, so water drains out at the top and the bottom. For 50% average consolidation the time factor is Tv = 0.197. How many years t does the layer take to reach 50% consolidation?
Answer key
- 1. S = 49.34 mm
- 2. S = 290.9 mm
- 3. S = 203 mm
- 4. S = 22.81 mm
- 5. t = 2604 days
- 6. t = 4.162 years
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