Consolidation settlement worksheet with answers

Free printable Consolidation settlement worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
Δσ = 65 kPasandγ = 19 kN/m³2.5 mclayγsat​ = 18.5 kN/m³Cc​ = 0.4, e0​ = 1.2510 mwater tableX

A 10 m thick normally consolidated clay layer (γsat = 18.5 kN/m³, Cc = 0.4, e₀ = 1.25) lies below 2.5 m of sand (γ = 19 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.

sand: Hs = 2.5 m, γ = 19 kN/m³clay: H = 10 m, γsat = 18.5 kN/m³γw = 9.81 kN/m³Cc = 0.4e0 = 1.25Δσ = 65 kPa
S = mm
2.Easy
Δσ = 105 kPasandclayCc​ = 0.5e0​ = 0.85σ′0​ = 130 kPa2.5 mwater table

A 2.5 m thick layer of normally consolidated clay (Cc = 0.5, e₀ = 0.85) has an effective vertical stress of 130 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 105 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 2.5 mCc = 0.5e0 = 0.85σ′0 = 130 kPaΔσ = 105 kPa
S = mm
3.Medium
Δσ = 25 kPasandclayCc​ = 0.3, Cr​ = 0.02e0​ = 0.65σ′0​ = 45 kPaσ′p​ = 85 kPa2 mwater table

A 2 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.02 and e₀ = 0.65. At its mid-depth the effective vertical stress is 45 kPa and the preconsolidation pressure is 85 kPa. A new building adds 25 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 2 mCc = 0.3Cr = 0.02e0 = 0.65σ′0 = 45 kPaσ′p = 85 kPaΔσ = 25 kPa
S = mm
4.Medium
Δσ = 20 kPasandclayCc​ = 0.3, Cr​ = 0.05e0​ = 1.05σ′0​ = 45 kPaσ′p​ = 75 kPa6.5 mwater table

A 6.5 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.05 and e₀ = 1.05. At its mid-depth the effective vertical stress is 45 kPa and the preconsolidation pressure is 75 kPa. A new building adds 20 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 6.5 mCc = 0.3Cr = 0.05e0 = 1.05σ′0 = 45 kPaσ′p = 75 kPaΔσ = 20 kPa
S = mm
5.Hard
sandclayH = 6.5 m6.5 mimpervious rockwater table

In a laboratory oedometer test, a clay specimen 25 mm thick, drained at the top and bottom, reaches 50% consolidation in 10 min. In the field the same clay forms a layer 6.5 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?

hlab = 25 mmtlab = 10 minH = 6.5 mfield: single drainage
t = days
6.Hard
sandclayH = 2 mcv​ = 1.2 m²/year2 msandwater table

A clay layer 2 m thick has a coefficient of consolidation cv = 1.2 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?

H = 2 mcv = 1.2 m²/yearU = 50%Tv = 0.197double drainage
t = years

Answer key

  1. 1. S = 416.3 mm
  2. 2. S = 173.7 mm
  3. 3. S = 4.652 mm
  4. 4. S = 25.32 mm
  5. 5. t = 1878 days
  6. 6. t = 0.1642 years

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