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 kPasandclayCc​ = 0.5e0​ = 0.9σ′0​ = 150 kPa7.5 mwater table

A 7.5 m thick layer of normally consolidated clay (Cc = 0.5, e₀ = 0.9) has an effective vertical stress of 150 kPa at its mid-depth. A wide fill raises the vertical stress in 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.

H = 7.5 mCc = 0.5e0 = 0.9σ′0 = 150 kPaΔσ = 55 kPa
S = mm
2.Easy
Δσ = 90 kPasandγ = 17.5 kN/m³3 mclayγsat​ = 18 kN/m³Cc​ = 0.3, e0​ = 0.84 mwater tableX

A 4 m thick normally consolidated clay layer (γsat = 18 kN/m³, Cc = 0.3, e₀ = 0.8) lies below 3 m of sand (γ = 17.5 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 90 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 = 17.5 kN/m³clay: H = 4 m, γsat = 18 kN/m³γw = 9.81 kN/m³Cc = 0.3e0 = 0.8Δσ = 90 kPa
S = mm
3.Medium
Δσ = 60 kPasandclayCc​ = 0.35, Cr​ = 0.05e0​ = 0.6σ′0​ = 60 kPaOCR = 32 mwater table

A 2 m thick overconsolidated clay layer has Cc = 0.35, Cr = 0.05 and e₀ = 0.6. At its mid-depth the effective vertical stress is 60 kPa and the overconsolidation ratio is OCR = 3. A new building adds 60 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 2 mCc = 0.35Cr = 0.05e0 = 0.6σ′0 = 60 kPaOCR = 3Δσ = 60 kPa
S = mm
4.Medium
Δσ = 150 kPasandclayCc​ = 0.4, Cr​ = 0.03e0​ = 0.8σ′0​ = 110 kPaOCR = 36 mwater table

A 6 m thick overconsolidated clay layer has Cc = 0.4, Cr = 0.03 and e₀ = 0.8. At its mid-depth the effective vertical stress is 110 kPa and the overconsolidation ratio is OCR = 3. A new building adds 150 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 6 mCc = 0.4Cr = 0.03e0 = 0.8σ′0 = 110 kPaOCR = 3Δσ = 150 kPa
S = mm
5.Hard
sandclayH = 3.5 mcv​ = 3 m²/year3.5 mimpervious rockwater table

A clay layer 3.5 m thick has a coefficient of consolidation cv = 3 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 = 3.5 mcv = 3 m²/yearU = 50%Tv = 0.197single drainage
t = years
6.Hard
sandclayH = 7 mcv​ = 5 m²/year7 msandwater table

A clay layer 7 m thick has a coefficient of consolidation cv = 5 m²/year. It lies between two sand layers, so water drains out at the top and the bottom. For 90% average consolidation the time factor is Tv = 0.848. How many years t does the layer take to reach 90% consolidation?

H = 7 mcv = 5 m²/yearU = 90%Tv = 0.848double drainage
t = years

Answer key

  1. 1. S = 267.8 mm
  2. 2. S = 242 mm
  3. 3. S = 18.81 mm
  4. 4. S = 37.36 mm
  5. 5. t = 0.8044 years
  6. 6. t = 2.078 years

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