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 = 140 kPasandγ = 17 kN/m³5 mclayγsat​ = 18.5 kN/m³Cc​ = 0.25, e0​ = 16 mwater tableX

A 6 m thick normally consolidated clay layer (γsat = 18.5 kN/m³, Cc = 0.25, e₀ = 1) lies below 5 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 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 = 5 m, γ = 17 kN/m³clay: H = 6 m, γsat = 18.5 kN/m³γw = 9.81 kN/m³Cc = 0.25e0 = 1Δσ = 140 kPa
S = mm
2.Easy
Δσ = 105 kPasandclayCc​ = 0.25e0​ = 1σ′0​ = 45 kPa3.5 mwater table

A 3.5 m thick layer of normally consolidated clay (Cc = 0.25, e₀ = 1) has an effective vertical stress of 45 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 105 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 3.5 mCc = 0.25e0 = 1σ′0 = 45 kPaΔσ = 105 kPa
S = mm
3.Medium
Δσ = 85 kPasandclayCc​ = 0.5, Cr​ = 0.06e0​ = 1.05σ′0​ = 110 kPaσ′p​ = 210 kPa3 mwater table

A 3 m thick overconsolidated clay layer has Cc = 0.5, Cr = 0.06 and e₀ = 1.05. At its mid-depth the effective vertical stress is 110 kPa and the preconsolidation pressure is 210 kPa. A new building adds 85 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 3 mCc = 0.5Cr = 0.06e0 = 1.05σ′0 = 110 kPaσ′p = 210 kPaΔσ = 85 kPa
S = mm
4.Medium
Δσ = 150 kPasandclayCc​ = 0.5, Cr​ = 0.06e0​ = 0.9σ′0​ = 110 kPaσ′p​ = 150 kPa7 mwater table

A 7 m thick overconsolidated clay layer has Cc = 0.5, Cr = 0.06 and e₀ = 0.9. At its mid-depth the effective vertical stress is 110 kPa and the preconsolidation pressure is 150 kPa. A new building adds 150 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7 mCc = 0.5Cr = 0.06e0 = 0.9σ′0 = 110 kPaσ′p = 150 kPaΔσ = 150 kPa
S = mm
5.Hard
sandclayH = 7 mcv​ = 4 m²/year7 mimpervious rockwater table

A clay layer 7 m thick has a coefficient of consolidation cv = 4 m²/year. It lies on impervious rock, with sand above it, so water drains out only at the top. 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 = 4 m²/yearU = 90%Tv = 0.848single drainage
t = years
6.Hard
sandclayH = 7 m7 msandwater table

In a laboratory oedometer test, a clay specimen 20 mm thick, drained at the top and bottom, reaches 50% consolidation in 8 min. In the field the same clay forms a layer 7 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 = 20 mmtlab = 8 minH = 7 mfield: double drainage
t = days

Answer key

  1. 1. S = 265.6 mm
  2. 2. S = 228.8 mm
  3. 3. S = 21.83 mm
  4. 4. S = 469.8 mm
  5. 5. t = 10.39 years
  6. 6. t = 680.6 days

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