A 6 m thick normally consolidated clay layer (γsat = 19.5 kN/m³, Cc = 0.5, e₀ = 0.85) lies below 3.5 m of sand (γ = 17.5 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 130 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 6 m thick normally consolidated clay layer (γsat = 19 kN/m³, Cc = 0.35, e₀ = 1.1) lies below 5.5 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 75 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.
A 4.5 m thick overconsolidated clay layer has Cc = 0.4, Cr = 0.02 and e₀ = 1.15. At its mid-depth the effective vertical stress is 50 kPa and the preconsolidation pressure is 170 kPa. A new building adds 80 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
A 3.5 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.06 and e₀ = 0.95. At its mid-depth the effective vertical stress is 120 kPa and the overconsolidation ratio is OCR = 1.5. A new building adds 100 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
In a laboratory oedometer test, a clay specimen 25 mm thick, drained at the top and bottom, reaches 50% consolidation in 4 min. In the field the same clay forms a layer 6 m thick. It lies between two sand layers, so water drains out at the top and the bottom. How many days t does the field layer take to reach 50% consolidation?
A clay layer 3.5 m thick has a coefficient of consolidation cv = 4 m²/year. It lies between two sand layers, so water drains out at the top and the bottom. For 90% average consolidation the time factor is Tv = 0.848. How many years t does the layer take to reach 90% consolidation?
Answer key
- 1. S = 628 mm
- 2. S = 209.4 mm
- 3. S = 17.37 mm
- 4. S = 58.07 mm
- 5. t = 160 days
- 6. t = 0.6492 years
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