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 kPasandclayCc​ = 0.45e0​ = 1.1σ′0​ = 145 kPa6 mwater table

A 6 m thick layer of normally consolidated clay (Cc = 0.45, e₀ = 1.1) has an effective vertical stress of 145 kPa at its mid-depth. A wide fill raises the vertical stress in 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.

H = 6 mCc = 0.45e0 = 1.1σ′0 = 145 kPaΔσ = 65 kPa
S = mm
2.Easy
Δσ = 150 kPasandγ = 18 kN/m³5 mclayγsat​ = 17.5 kN/m³Cc​ = 0.35, e0​ = 1.1510 mwater tableX

A 10 m thick normally consolidated clay layer (γsat = 17.5 kN/m³, Cc = 0.35, e₀ = 1.15) lies below 5 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 150 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

sand: Hs = 5 m, γ = 18 kN/m³clay: H = 10 m, γsat = 17.5 kN/m³γw = 9.81 kN/m³Cc = 0.35e0 = 1.15Δσ = 150 kPa
S = mm
3.Medium
Δσ = 90 kPasandclayCc​ = 0.2, Cr​ = 0.06e0​ = 0.95σ′0​ = 65 kPaOCR = 1.57.5 mwater table

A 7.5 m thick overconsolidated clay layer has Cc = 0.2, Cr = 0.06 and e₀ = 0.95. At its mid-depth the effective vertical stress is 65 kPa and the overconsolidation ratio is OCR = 1.5. A new building adds 90 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7.5 mCc = 0.2Cr = 0.06e0 = 0.95σ′0 = 65 kPaOCR = 1.5Δσ = 90 kPa
S = mm
4.Medium
Δσ = 150 kPasandclayCc​ = 0.45, Cr​ = 0.06e0​ = 0.75σ′0​ = 115 kPaσ′p​ = 145 kPa3 mwater table

A 3 m thick overconsolidated clay layer has Cc = 0.45, Cr = 0.06 and e₀ = 0.75. At its mid-depth the effective vertical stress is 115 kPa and the preconsolidation pressure is 145 kPa. A new building adds 150 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 3 mCc = 0.45Cr = 0.06e0 = 0.75σ′0 = 115 kPaσ′p = 145 kPaΔσ = 150 kPa
S = mm
5.Hard
sandclayH = 9.5 m9.5 mimpervious rockwater table

In a laboratory oedometer test, a clay specimen 25 mm thick, drained at the top and bottom, reaches 50% consolidation in 4 min. In the field the same clay forms a layer 9.5 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 = 4 minH = 9.5 mfield: single drainage
t = days
6.Hard
sandclayH = 6.5 mcv​ = 4 m²/year6.5 mimpervious rockwater table

A clay layer 6.5 m thick has a coefficient of consolidation cv = 4 m²/year. It lies on impervious rock, with sand above it, so water drains out only at the top. For 80% average consolidation the time factor is Tv = 0.567. How many years t does the layer take to reach 80% consolidation?

H = 6.5 mcv = 4 m²/yearU = 80%Tv = 0.567single drainage
t = years

Answer key

  1. 1. S = 206.8 mm
  2. 2. S = 547 mm
  3. 3. S = 195.5 mm
  4. 4. S = 212.4 mm
  5. 5. t = 1604 days
  6. 6. t = 5.989 years

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