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 = 40 kPasandclayCc​ = 0.35e0​ = 0.65σ′0​ = 60 kPa3 mwater table

A 3 m thick layer of normally consolidated clay (Cc = 0.35, e₀ = 0.65) has an effective vertical stress of 60 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 40 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 3 mCc = 0.35e0 = 0.65σ′0 = 60 kPaΔσ = 40 kPa
S = mm
2.Easy
Δσ = 135 kPasandγ = 18.5 kN/m³4.5 mclayγsat​ = 18.5 kN/m³Cc​ = 0.6, e0​ = 0.853 mwater tableX

A 3 m thick normally consolidated clay layer (γsat = 18.5 kN/m³, Cc = 0.6, e₀ = 0.85) lies below 4.5 m of sand (γ = 18.5 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 135 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

sand: Hs = 4.5 m, γ = 18.5 kN/m³clay: H = 3 m, γsat = 18.5 kN/m³γw = 9.81 kN/m³Cc = 0.6e0 = 0.85Δσ = 135 kPa
S = mm
3.Medium
Δσ = 50 kPasandclayCc​ = 0.25, Cr​ = 0.02e0​ = 0.95σ′0​ = 40 kPaOCR = 2.57.5 mwater table

A 7.5 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.02 and e₀ = 0.95. At its mid-depth the effective vertical stress is 40 kPa and the overconsolidation ratio is OCR = 2.5. A new building adds 50 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7.5 mCc = 0.25Cr = 0.02e0 = 0.95σ′0 = 40 kPaOCR = 2.5Δσ = 50 kPa
S = mm
4.Medium
Δσ = 110 kPasandclayCc​ = 0.25, Cr​ = 0.02e0​ = 0.75σ′0​ = 100 kPaσ′p​ = 190 kPa3.5 mwater table

A 3.5 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.02 and e₀ = 0.75. At its mid-depth the effective vertical stress is 100 kPa and the preconsolidation pressure is 190 kPa. A new building adds 110 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 3.5 mCc = 0.25Cr = 0.02e0 = 0.75σ′0 = 100 kPaσ′p = 190 kPaΔσ = 110 kPa
S = mm
5.Hard
sandclayH = 7 m7 mimpervious rockwater table

In a laboratory oedometer test, a clay specimen 20 mm thick, drained at the top and bottom, reaches 50% consolidation in 10 min. In the field the same clay forms a layer 7 m thick. It lies on impervious rock, with sand above it, so water drains out only at the top. How many days t does the field layer take to reach 50% consolidation?

hlab = 20 mmtlab = 10 minH = 7 mfield: single drainage
t = days
6.Hard
sandclayH = 6 mcv​ = 1.2 m²/year6 mimpervious rockwater table

A clay layer 6 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 = 6 mcv = 1.2 m²/yearU = 70%Tv = 0.403single drainage
t = years

Answer key

  1. 1. S = 141.2 mm
  2. 2. S = 370.3 mm
  3. 3. S = 27.09 mm
  4. 4. S = 32.88 mm
  5. 5. t = 3403 days
  6. 6. t = 12.09 years

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