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 = 120 kPasandclayCc​ = 0.2e0​ = 0.8σ′0​ = 60 kPa10 mwater table

A 10 m thick layer of normally consolidated clay (Cc = 0.2, e₀ = 0.8) has an effective vertical stress of 60 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 120 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 10 mCc = 0.2e0 = 0.8σ′0 = 60 kPaΔσ = 120 kPa
S = mm
2.Easy
Δσ = 140 kPasandγ = 18.5 kN/m³6 mclayγsat​ = 19.5 kN/m³Cc​ = 0.3, e0​ = 1.256 mwater tableX

A 6 m thick normally consolidated clay layer (γsat = 19.5 kN/m³, Cc = 0.3, e₀ = 1.25) lies below 6 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 140 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

sand: Hs = 6 m, γ = 18.5 kN/m³clay: H = 6 m, γsat = 19.5 kN/m³γw = 9.81 kN/m³Cc = 0.3e0 = 1.25Δσ = 140 kPa
S = mm
3.Medium
Δσ = 120 kPasandclayCc​ = 0.4, Cr​ = 0.04e0​ = 1.2σ′0​ = 40 kPaσ′p​ = 90 kPa4 mwater table

A 4 m thick overconsolidated clay layer has Cc = 0.4, Cr = 0.04 and e₀ = 1.2. At its mid-depth the effective vertical stress is 40 kPa and the preconsolidation pressure is 90 kPa. A new building adds 120 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4 mCc = 0.4Cr = 0.04e0 = 1.2σ′0 = 40 kPaσ′p = 90 kPaΔσ = 120 kPa
S = mm
4.Medium
Δσ = 120 kPasandclayCc​ = 0.25, Cr​ = 0.06e0​ = 0.7σ′0​ = 45 kPaσ′p​ = 155 kPa4.5 mwater table

A 4.5 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.06 and e₀ = 0.7. At its mid-depth the effective vertical stress is 45 kPa and the preconsolidation pressure is 155 kPa. A new building adds 120 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4.5 mCc = 0.25Cr = 0.06e0 = 0.7σ′0 = 45 kPaσ′p = 155 kPaΔσ = 120 kPa
S = mm
5.Hard
sandclayH = 10 mcv​ = 0.5 m²/year10 msandwater table

A clay layer 10 m thick has a coefficient of consolidation cv = 0.5 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 = 10 mcv = 0.5 m²/yearU = 90%Tv = 0.848double drainage
t = years
6.Hard
sandclayH = 4.5 mcv​ = 5 m²/year4.5 mimpervious rockwater table

A clay layer 4.5 m thick has a coefficient of consolidation cv = 5 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 = 4.5 mcv = 5 m²/yearU = 50%Tv = 0.197single drainage
t = years

Answer key

  1. 1. S = 530.1 mm
  2. 2. S = 240.7 mm
  3. 3. S = 207.3 mm
  4. 4. S = 103.3 mm
  5. 5. t = 42.4 years
  6. 6. t = 0.7979 years

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