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 = 95 kPasandγ = 18 kN/m³3 mclayγsat​ = 19.5 kN/m³Cc​ = 0.2, e0​ = 0.9510 mwater tableX

A 10 m thick normally consolidated clay layer (γsat = 19.5 kN/m³, Cc = 0.2, e₀ = 0.95) lies below 3 m of sand (γ = 18 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 95 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 m, γ = 18 kN/m³clay: H = 10 m, γsat = 19.5 kN/m³γw = 9.81 kN/m³Cc = 0.2e0 = 0.95Δσ = 95 kPa
S = mm
2.Easy
Δσ = 45 kPasandclayCc​ = 0.3e0​ = 1.1σ′0​ = 110 kPa8 mwater table

An 8 m thick layer of normally consolidated clay (Cc = 0.3, e₀ = 1.1) has an effective vertical stress of 110 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 45 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 = 0.3e0 = 1.1σ′0 = 110 kPaΔσ = 45 kPa
S = mm
3.Medium
Δσ = 180 kPasandclayCc​ = 0.4, Cr​ = 0.06e0​ = 1.15σ′0​ = 85 kPaσ′p​ = 145 kPa2.5 mwater table

A 2.5 m thick overconsolidated clay layer has Cc = 0.4, Cr = 0.06 and e₀ = 1.15. At its mid-depth the effective vertical stress is 85 kPa and the preconsolidation pressure is 145 kPa. A new building adds 180 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 2.5 mCc = 0.4Cr = 0.06e0 = 1.15σ′0 = 85 kPaσ′p = 145 kPaΔσ = 180 kPa
S = mm
4.Medium
Δσ = 200 kPasandclayCc​ = 0.3, Cr​ = 0.06e0​ = 1.15σ′0​ = 85 kPaOCR = 27.5 mwater table

A 7.5 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.06 and e₀ = 1.15. At its mid-depth the effective vertical stress is 85 kPa and the overconsolidation ratio is OCR = 2. A new building adds 200 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7.5 mCc = 0.3Cr = 0.06e0 = 1.15σ′0 = 85 kPaOCR = 2Δσ = 200 kPa
S = mm
5.Hard
sandclayH = 5 mcv​ = 0.8 m²/year5 mimpervious rockwater table

A clay layer 5 m thick has a coefficient of consolidation cv = 0.8 m²/year. It lies on impervious rock, with sand above it, so water drains out only at the top. For 50% average consolidation the time factor is Tv = 0.197. How many years t does the layer take to reach 50% consolidation?

H = 5 mcv = 0.8 m²/yearU = 50%Tv = 0.197single drainage
t = years
6.Hard
sandclayH = 4 m4 msandwater table

In a laboratory oedometer test, a clay specimen 19 mm thick, drained at the top and bottom, reaches 50% consolidation in 6 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 = 19 mmtlab = 6 minH = 4 mfield: double drainage
t = days

Answer key

  1. 1. S = 292.3 mm
  2. 2. S = 170.2 mm
  3. 3. S = 138 mm
  4. 4. S = 297.8 mm
  5. 5. t = 6.156 years
  6. 6. t = 184.7 days

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