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 = 55 kPasandγ = 19 kN/m³3.5 mclayγsat​ = 19 kN/m³Cc​ = 0.15, e0​ = 1.15 mwater tableX

A 5 m thick normally consolidated clay layer (γsat = 19 kN/m³, Cc = 0.15, e₀ = 1.1) lies below 3.5 m of sand (γ = 19 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 55 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

sand: Hs = 3.5 m, γ = 19 kN/m³clay: H = 5 m, γsat = 19 kN/m³γw = 9.81 kN/m³Cc = 0.15e0 = 1.1Δσ = 55 kPa
S = mm
2.Easy
Δσ = 110 kPasandclayCc​ = 0.15e0​ = 1.2σ′0​ = 60 kPa2 mwater table

A 2 m thick layer of normally consolidated clay (Cc = 0.15, e₀ = 1.2) has an effective vertical stress of 60 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 = 2 mCc = 0.15e0 = 1.2σ′0 = 60 kPaΔσ = 110 kPa
S = mm
3.Medium
Δσ = 190 kPasandclayCc​ = 0.2, Cr​ = 0.03e0​ = 0.75σ′0​ = 80 kPaOCR = 25 mwater table

A 5 m thick overconsolidated clay layer has Cc = 0.2, Cr = 0.03 and e₀ = 0.75. At its mid-depth the effective vertical stress is 80 kPa and the overconsolidation ratio is OCR = 2. A new building adds 190 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 5 mCc = 0.2Cr = 0.03e0 = 0.75σ′0 = 80 kPaOCR = 2Δσ = 190 kPa
S = mm
4.Medium
Δσ = 50 kPasandclayCc​ = 0.3, Cr​ = 0.04e0​ = 1.1σ′0​ = 100 kPaσ′p​ = 200 kPa3 mwater table

A 3 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.04 and e₀ = 1.1. At its mid-depth the effective vertical stress is 100 kPa and the preconsolidation pressure is 200 kPa. A new building adds 50 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 3 mCc = 0.3Cr = 0.04e0 = 1.1σ′0 = 100 kPaσ′p = 200 kPaΔσ = 50 kPa
S = mm
5.Hard
sandclayH = 5.5 m5.5 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 5.5 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 = 15 minH = 5.5 mfield: double drainage
t = days
6.Hard
sandclayH = 7 mcv​ = 1.2 m²/year7 mimpervious rockwater table

A clay layer 7 m thick has a coefficient of consolidation cv = 1.2 m²/year. It lies on impervious rock, with sand above it, so water drains out only at the top. For 70% average consolidation the time factor is Tv = 0.403. How many years t does the layer take to reach 70% consolidation?

H = 7 mcv = 1.2 m²/yearU = 70%Tv = 0.403single drainage
t = years

Answer key

  1. 1. S = 74.32 mm
  2. 2. S = 61.68 mm
  3. 3. S = 155.7 mm
  4. 4. S = 10.06 mm
  5. 5. t = 872.9 days
  6. 6. t = 16.46 years

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