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 = 85 kPasandclayCc​ = 0.25e0​ = 0.6σ′0​ = 130 kPa6 mwater table

A 6 m thick layer of normally consolidated clay (Cc = 0.25, e₀ = 0.6) has an effective vertical stress of 130 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 85 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 6 mCc = 0.25e0 = 0.6σ′0 = 130 kPaΔσ = 85 kPa
S = mm
2.Easy
Δσ = 130 kPasandγ = 19 kN/m³2.5 mclayγsat​ = 18.5 kN/m³Cc​ = 0.4, e0​ = 1.053 mwater tableX

A 3 m thick normally consolidated clay layer (γsat = 18.5 kN/m³, Cc = 0.4, e₀ = 1.05) 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 130 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 = 3 m, γsat = 18.5 kN/m³γw = 9.81 kN/m³Cc = 0.4e0 = 1.05Δσ = 130 kPa
S = mm
3.Medium
Δσ = 100 kPasandclayCc​ = 0.3, Cr​ = 0.03e0​ = 0.8σ′0​ = 50 kPaOCR = 2.52 mwater table

A 2 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.03 and e₀ = 0.8. At its mid-depth the effective vertical stress is 50 kPa and the overconsolidation ratio is OCR = 2.5. A new building adds 100 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 2 mCc = 0.3Cr = 0.03e0 = 0.8σ′0 = 50 kPaOCR = 2.5Δσ = 100 kPa
S = mm
4.Medium
Δσ = 160 kPasandclayCc​ = 0.25, Cr​ = 0.05e0​ = 0.9σ′0​ = 45 kPaσ′p​ = 135 kPa4 mwater table

A 4 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.05 and e₀ = 0.9. At its mid-depth the effective vertical stress is 45 kPa and the preconsolidation pressure is 135 kPa. A new building adds 160 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4 mCc = 0.25Cr = 0.05e0 = 0.9σ′0 = 45 kPaσ′p = 135 kPaΔσ = 160 kPa
S = mm
5.Hard
sandclayH = 10 m10 mimpervious rockwater table

In a laboratory oedometer test, a clay specimen 20 mm thick, drained at the top and bottom, reaches 50% consolidation in 5 min. In the field the same clay forms a layer 10 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 = 20 mmtlab = 5 minH = 10 mfield: single drainage
t = days
6.Hard
sandclayH = 9.5 mcv​ = 1.5 m²/year9.5 mimpervious rockwater table

A clay layer 9.5 m thick has a coefficient of consolidation cv = 1.5 m²/year. It lies on impervious rock, with sand above it, so water drains out only at the top. For 60% average consolidation the time factor is Tv = 0.287. How many years t does the layer take to reach 60% consolidation?

H = 9.5 mcv = 1.5 m²/yearU = 60%Tv = 0.287single drainage
t = years

Answer key

  1. 1. S = 204.8 mm
  2. 2. S = 291.5 mm
  3. 3. S = 39.66 mm
  4. 4. S = 145.7 mm
  5. 5. t = 3472 days
  6. 6. t = 17.27 years

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