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 kPasandγ = 17 kN/m³5 mclayγsat​ = 19 kN/m³Cc​ = 0.25, e0​ = 1.38 mwater tableX

An 8 m thick normally consolidated clay layer (γsat = 19 kN/m³, Cc = 0.25, e₀ = 1.3) lies below 5 m of sand (γ = 17 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 65 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 = 17 kN/m³clay: H = 8 m, γsat = 19 kN/m³γw = 9.81 kN/m³Cc = 0.25e0 = 1.3Δσ = 65 kPa
S = mm
2.Easy
Δσ = 145 kPasandclayCc​ = 0.5e0​ = 1.35σ′0​ = 115 kPa5 mwater table

A 5 m thick layer of normally consolidated clay (Cc = 0.5, e₀ = 1.35) has an effective vertical stress of 115 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 145 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 5 mCc = 0.5e0 = 1.35σ′0 = 115 kPaΔσ = 145 kPa
S = mm
3.Medium
Δσ = 200 kPasandclayCc​ = 0.3, Cr​ = 0.06e0​ = 1σ′0​ = 60 kPaOCR = 2.56.5 mwater table

A 6.5 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.06 and e₀ = 1. At its mid-depth the effective vertical stress is 60 kPa and the overconsolidation ratio is OCR = 2.5. A new building adds 200 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 6.5 mCc = 0.3Cr = 0.06e0 = 1σ′0 = 60 kPaOCR = 2.5Δσ = 200 kPa
S = mm
4.Medium
Δσ = 85 kPasandclayCc​ = 0.35, Cr​ = 0.04e0​ = 0.8σ′0​ = 55 kPaσ′p​ = 175 kPa7 mwater table

A 7 m thick overconsolidated clay layer has Cc = 0.35, Cr = 0.04 and e₀ = 0.8. At its mid-depth the effective vertical stress is 55 kPa and the preconsolidation pressure is 175 kPa. A new building adds 85 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7 mCc = 0.35Cr = 0.04e0 = 0.8σ′0 = 55 kPaσ′p = 175 kPaΔσ = 85 kPa
S = mm
5.Hard
sandclayH = 2.5 m2.5 msandwater 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 2.5 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 = 25 mmtlab = 15 minH = 2.5 mfield: double drainage
t = days
6.Hard
sandclayH = 9 m9 msandwater 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 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 = 25 mmtlab = 4 minH = 9 mfield: double drainage
t = days

Answer key

  1. 1. S = 161.5 mm
  2. 2. S = 376.9 mm
  3. 3. S = 310.5 mm
  4. 4. S = 63.12 mm
  5. 5. t = 104.2 days
  6. 6. t = 360 days

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