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 = 90 kPasandclayCc​ = 0.6e0​ = 1.15σ′0​ = 120 kPa8 mwater table

An 8 m thick layer of normally consolidated clay (Cc = 0.6, e₀ = 1.15) has an effective vertical stress of 120 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 90 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 8 mCc = 0.6e0 = 1.15σ′0 = 120 kPaΔσ = 90 kPa
S = mm
2.Easy
Δσ = 120 kPasandclayCc​ = 0.35e0​ = 1.25σ′0​ = 45 kPa6.5 mwater table

A 6.5 m thick layer of normally consolidated clay (Cc = 0.35, e₀ = 1.25) has an effective vertical stress of 45 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 120 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 6.5 mCc = 0.35e0 = 1.25σ′0 = 45 kPaΔσ = 120 kPa
S = mm
3.Medium
Δσ = 130 kPasandclayCc​ = 0.3, Cr​ = 0.06e0​ = 0.65σ′0​ = 65 kPaσ′p​ = 145 kPa4.5 mwater table

A 4.5 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.06 and e₀ = 0.65. At its mid-depth the effective vertical stress is 65 kPa and the preconsolidation pressure is 145 kPa. A new building adds 130 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4.5 mCc = 0.3Cr = 0.06e0 = 0.65σ′0 = 65 kPaσ′p = 145 kPaΔσ = 130 kPa
S = mm
4.Medium
Δσ = 60 kPasandclayCc​ = 0.25, Cr​ = 0.06e0​ = 0.85σ′0​ = 95 kPaσ′p​ = 145 kPa7.5 mwater table

A 7.5 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.06 and e₀ = 0.85. At its mid-depth the effective vertical stress is 95 kPa and the preconsolidation pressure is 145 kPa. A new building adds 60 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7.5 mCc = 0.25Cr = 0.06e0 = 0.85σ′0 = 95 kPaσ′p = 145 kPaΔσ = 60 kPa
S = mm
5.Hard
sandclayH = 8 m8 msandwater table

In a laboratory oedometer test, a clay specimen 19 mm thick, drained at the top and bottom, reaches 50% consolidation in 15 min. In the field the same clay forms a layer 8 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 = 15 minH = 8 mfield: double drainage
t = days
6.Hard
sandclayH = 3 mcv​ = 4 m²/year3 msandwater table

A clay layer 3 m thick has a coefficient of consolidation cv = 4 m²/year. It lies between two sand layers, so water drains out at the top and the bottom. For 80% average consolidation the time factor is Tv = 0.567. How many years t does the layer take to reach 80% consolidation?

H = 3 mcv = 4 m²/yearU = 80%Tv = 0.567double drainage
t = years

Answer key

  1. 1. S = 542.6 mm
  2. 2. S = 570.5 mm
  3. 3. S = 162.3 mm
  4. 4. S = 74.03 mm
  5. 5. t = 1847 days
  6. 6. t = 0.3189 years

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