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 = 75 kPasandclayCc​ = 0.5e0​ = 0.75σ′0​ = 145 kPa2.5 mwater table

A 2.5 m thick layer of normally consolidated clay (Cc = 0.5, e₀ = 0.75) has an effective vertical stress of 145 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 75 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 2.5 mCc = 0.5e0 = 0.75σ′0 = 145 kPaΔσ = 75 kPa
S = mm
2.Easy
Δσ = 100 kPasandγ = 17 kN/m³4 mclayγsat​ = 17.5 kN/m³Cc​ = 0.3, e0​ = 0.63 mwater tableX

A 3 m thick normally consolidated clay layer (γsat = 17.5 kN/m³, Cc = 0.3, e₀ = 0.6) lies below 4 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 100 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 m, γ = 17 kN/m³clay: H = 3 m, γsat = 17.5 kN/m³γw = 9.81 kN/m³Cc = 0.3e0 = 0.6Δσ = 100 kPa
S = mm
3.Medium
Δσ = 25 kPasandclayCc​ = 0.3, Cr​ = 0.03e0​ = 0.7σ′0​ = 60 kPaσ′p​ = 170 kPa7.5 mwater table

A 7.5 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.03 and e₀ = 0.7. At its mid-depth the effective vertical stress is 60 kPa and the preconsolidation pressure is 170 kPa. A new building adds 25 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7.5 mCc = 0.3Cr = 0.03e0 = 0.7σ′0 = 60 kPaσ′p = 170 kPaΔσ = 25 kPa
S = mm
4.Medium
Δσ = 20 kPasandclayCc​ = 0.35, Cr​ = 0.05e0​ = 0.85σ′0​ = 115 kPaOCR = 1.54.5 mwater table

A 4.5 m thick overconsolidated clay layer has Cc = 0.35, Cr = 0.05 and e₀ = 0.85. At its mid-depth the effective vertical stress is 115 kPa and the overconsolidation ratio is OCR = 1.5. A new building adds 20 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4.5 mCc = 0.35Cr = 0.05e0 = 0.85σ′0 = 115 kPaOCR = 1.5Δσ = 20 kPa
S = mm
5.Hard
sandclayH = 5 mcv​ = 1.5 m²/year5 msandwater table

A clay layer 5 m thick has a coefficient of consolidation cv = 1.5 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 = 5 mcv = 1.5 m²/yearU = 50%Tv = 0.197double drainage
t = years
6.Hard
sandclayH = 9.5 m9.5 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 9.5 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 = 6 minH = 9.5 mfield: double drainage
t = days

Answer key

  1. 1. S = 129.3 mm
  2. 2. S = 198.9 mm
  3. 3. S = 20.02 mm
  4. 4. S = 8.469 mm
  5. 5. t = 0.8208 years
  6. 6. t = 1042 days

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