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 = 140 kPasandclayCc​ = 0.35e0​ = 1.2σ′0​ = 50 kPa2.5 mwater table

A 2.5 m thick layer of normally consolidated clay (Cc = 0.35, e₀ = 1.2) has an effective vertical stress of 50 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 140 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 = 0.35e0 = 1.2σ′0 = 50 kPaΔσ = 140 kPa
S = mm
2.Easy
Δσ = 20 kPasandclayCc​ = 0.6e0​ = 0.7σ′0​ = 60 kPa4 mwater table

A 4 m thick layer of normally consolidated clay (Cc = 0.6, e₀ = 0.7) has an effective vertical stress of 60 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 20 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 4 mCc = 0.6e0 = 0.7σ′0 = 60 kPaΔσ = 20 kPa
S = mm
3.Medium
Δσ = 160 kPasandclayCc​ = 0.5, Cr​ = 0.02e0​ = 0.9σ′0​ = 50 kPaσ′p​ = 160 kPa3 mwater table

A 3 m thick overconsolidated clay layer has Cc = 0.5, Cr = 0.02 and e₀ = 0.9. At its mid-depth the effective vertical stress is 50 kPa and the preconsolidation pressure is 160 kPa. A new building adds 160 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 3 mCc = 0.5Cr = 0.02e0 = 0.9σ′0 = 50 kPaσ′p = 160 kPaΔσ = 160 kPa
S = mm
4.Medium
Δσ = 190 kPasandclayCc​ = 0.35, Cr​ = 0.05e0​ = 0.75σ′0​ = 60 kPaOCR = 2.56.5 mwater table

A 6.5 m thick overconsolidated clay layer has Cc = 0.35, Cr = 0.05 and e₀ = 0.75. At its mid-depth the effective vertical stress is 60 kPa and the overconsolidation ratio is OCR = 2.5. A new building adds 190 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 6.5 mCc = 0.35Cr = 0.05e0 = 0.75σ′0 = 60 kPaOCR = 2.5Δσ = 190 kPa
S = mm
5.Hard
sandclayH = 5 mcv​ = 5 m²/year5 mimpervious rockwater table

A clay layer 5 m thick has a coefficient of consolidation cv = 5 m²/year. It lies on impervious rock, with sand above it, so water drains out only at the top. For 90% average consolidation the time factor is Tv = 0.848. How many years t does the layer take to reach 90% consolidation?

H = 5 mcv = 5 m²/yearU = 90%Tv = 0.848single drainage
t = years
6.Hard
sandclayH = 7.5 m7.5 msandwater table

In a laboratory oedometer test, a clay specimen 19 mm thick, drained at the top and bottom, reaches 50% consolidation in 4 min. In the field the same clay forms a layer 7.5 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?

hlab = 19 mmtlab = 4 minH = 7.5 mfield: double drainage
t = days

Answer key

  1. 1. S = 230.6 mm
  2. 2. S = 176.4 mm
  3. 3. S = 109.2 mm
  4. 4. S = 362.3 mm
  5. 5. t = 4.24 years
  6. 6. t = 432.8 days

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