An 8 m thick normally consolidated clay layer (γsat = 19 kN/m³, Cc = 0.25, e₀ = 1.3) lies below 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 65 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 layer of normally consolidated clay (Cc = 0.5, e₀ = 1.35) has an effective vertical stress of 115 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 145 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.
A 6.5 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.06 and e₀ = 1. At its mid-depth the effective vertical stress is 60 kPa and the overconsolidation ratio is OCR = 2.5. A new building adds 200 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
A 7 m thick overconsolidated clay layer has Cc = 0.35, Cr = 0.04 and e₀ = 0.8. At its mid-depth the effective vertical stress is 55 kPa and the preconsolidation pressure is 175 kPa. A new building adds 85 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 15 min. In the field the same clay forms a layer 2.5 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?
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 9 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?
Answer key
- 1. S = 161.5 mm
- 2. S = 376.9 mm
- 3. S = 310.5 mm
- 4. S = 63.12 mm
- 5. t = 104.2 days
- 6. t = 360 days
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