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​ = 0.45e0​ = 1.4σ′0​ = 120 kPa10 mwater table

A 10 m thick layer of normally consolidated clay (Cc = 0.45, e₀ = 1.4) has an effective vertical stress of 120 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 = 10 mCc = 0.45e0 = 1.4σ′0 = 120 kPaΔσ = 85 kPa
S = mm
2.Easy
Δσ = 110 kPasandclayCc​ = 0.15e0​ = 1.15σ′0​ = 120 kPa3.5 mwater table

A 3.5 m thick layer of normally consolidated clay (Cc = 0.15, 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 110 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 3.5 mCc = 0.15e0 = 1.15σ′0 = 120 kPaΔσ = 110 kPa
S = mm
3.Medium
Δσ = 90 kPasandclayCc​ = 0.45, Cr​ = 0.05e0​ = 1.2σ′0​ = 55 kPaσ′p​ = 165 kPa3 mwater table

A 3 m thick overconsolidated clay layer has Cc = 0.45, Cr = 0.05 and e₀ = 1.2. At its mid-depth the effective vertical stress is 55 kPa and the preconsolidation pressure is 165 kPa. A new building adds 90 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 3 mCc = 0.45Cr = 0.05e0 = 1.2σ′0 = 55 kPaσ′p = 165 kPaΔσ = 90 kPa
S = mm
4.Medium
Δσ = 120 kPasandclayCc​ = 0.5, Cr​ = 0.05e0​ = 1.2σ′0​ = 75 kPaOCR = 34 mwater table

A 4 m thick overconsolidated clay layer has Cc = 0.5, Cr = 0.05 and e₀ = 1.2. At its mid-depth the effective vertical stress is 75 kPa and the overconsolidation ratio is OCR = 3. A new building adds 120 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4 mCc = 0.5Cr = 0.05e0 = 1.2σ′0 = 75 kPaOCR = 3Δσ = 120 kPa
S = mm
5.Hard
sandclayH = 4 mcv​ = 4 m²/year4 msandwater table

A clay layer 4 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 50% average consolidation the time factor is Tv = 0.197. How many years t does the layer take to reach 50% consolidation?

H = 4 mcv = 4 m²/yearU = 50%Tv = 0.197double drainage
t = years
6.Hard
sandclayH = 4 m4 msandwater table

In a laboratory oedometer test, a clay specimen 20 mm thick, drained at the top and bottom, reaches 50% consolidation in 4 min. In the field the same clay forms a layer 4 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 = 20 mmtlab = 4 minH = 4 mfield: double drainage
t = days

Answer key

  1. 1. S = 436.1 mm
  2. 2. S = 68.99 mm
  3. 3. S = 28.7 mm
  4. 4. S = 37.72 mm
  5. 5. t = 0.197 years
  6. 6. t = 111.1 days

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