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 = 80 kPasandclayCc​ = 0.6e0​ = 0.65σ′0​ = 70 kPa9.5 mwater table

A 9.5 m thick layer of normally consolidated clay (Cc = 0.6, e₀ = 0.65) has an effective vertical stress of 70 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 80 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 9.5 mCc = 0.6e0 = 0.65σ′0 = 70 kPaΔσ = 80 kPa
S = mm
2.Easy
Δσ = 45 kPasandclayCc​ = 0.35e0​ = 0.9σ′0​ = 110 kPa10 mwater table

A 10 m thick layer of normally consolidated clay (Cc = 0.35, e₀ = 0.9) has an effective vertical stress of 110 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 45 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 10 mCc = 0.35e0 = 0.9σ′0 = 110 kPaΔσ = 45 kPa
S = mm
3.Medium
Δσ = 110 kPasandclayCc​ = 0.35, Cr​ = 0.04e0​ = 0.7σ′0​ = 90 kPaOCR = 24 mwater table

A 4 m thick overconsolidated clay layer has Cc = 0.35, Cr = 0.04 and e₀ = 0.7. At its mid-depth the effective vertical stress is 90 kPa and the overconsolidation ratio is OCR = 2. A new building adds 110 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4 mCc = 0.35Cr = 0.04e0 = 0.7σ′0 = 90 kPaOCR = 2Δσ = 110 kPa
S = mm
4.Medium
Δσ = 220 kPasandclayCc​ = 0.4, Cr​ = 0.05e0​ = 0.6σ′0​ = 95 kPaσ′p​ = 185 kPa3.5 mwater table

A 3.5 m thick overconsolidated clay layer has Cc = 0.4, Cr = 0.05 and e₀ = 0.6. At its mid-depth the effective vertical stress is 95 kPa and the preconsolidation pressure is 185 kPa. A new building adds 220 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 3.5 mCc = 0.4Cr = 0.05e0 = 0.6σ′0 = 95 kPaσ′p = 185 kPaΔσ = 220 kPa
S = mm
5.Hard
sandclayH = 5 m5 mimpervious rockwater table

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 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 = 15 minH = 5 mfield: single drainage
t = days
6.Hard
sandclayH = 5.5 mcv​ = 0.8 m²/year5.5 mimpervious rockwater table

A clay layer 5.5 m thick has a coefficient of consolidation cv = 0.8 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 = 5.5 mcv = 0.8 m²/yearU = 80%Tv = 0.567single drainage
t = years

Answer key

  1. 1. S = 1143 mm
  2. 2. S = 274.4 mm
  3. 3. S = 66.01 mm
  4. 4. S = 233.9 mm
  5. 5. t = 1667 days
  6. 6. t = 21.44 years

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