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 = 125 kPasandclayCc​ = 0.45e0​ = 0.7σ′0​ = 55 kPa5.5 mwater table

A 5.5 m thick layer of normally consolidated clay (Cc = 0.45, e₀ = 0.7) has an effective vertical stress of 55 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 125 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 5.5 mCc = 0.45e0 = 0.7σ′0 = 55 kPaΔσ = 125 kPa
S = mm
2.Easy
Δσ = 105 kPasandγ = 17.5 kN/m³5.5 mclayγsat​ = 18 kN/m³Cc​ = 0.15, e0​ = 0.810 mwater tableX

A 10 m thick normally consolidated clay layer (γsat = 18 kN/m³, Cc = 0.15, e₀ = 0.8) lies below 5.5 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 105 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

sand: Hs = 5.5 m, γ = 17.5 kN/m³clay: H = 10 m, γsat = 18 kN/m³γw = 9.81 kN/m³Cc = 0.15e0 = 0.8Δσ = 105 kPa
S = mm
3.Medium
Δσ = 15 kPasandclayCc​ = 0.25, Cr​ = 0.04e0​ = 1σ′0​ = 60 kPaσ′p​ = 150 kPa4 mwater table

A 4 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.04 and e₀ = 1. At its mid-depth the effective vertical stress is 60 kPa and the preconsolidation pressure is 150 kPa. A new building adds 15 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4 mCc = 0.25Cr = 0.04e0 = 1σ′0 = 60 kPaσ′p = 150 kPaΔσ = 15 kPa
S = mm
4.Medium
Δσ = 220 kPasandclayCc​ = 0.35, Cr​ = 0.06e0​ = 0.65σ′0​ = 55 kPaσ′p​ = 165 kPa7.5 mwater table

A 7.5 m thick overconsolidated clay layer has Cc = 0.35, Cr = 0.06 and e₀ = 0.65. At its mid-depth the effective vertical stress is 55 kPa and the preconsolidation pressure is 165 kPa. A new building adds 220 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7.5 mCc = 0.35Cr = 0.06e0 = 0.65σ′0 = 55 kPaσ′p = 165 kPaΔσ = 220 kPa
S = mm
5.Hard
sandclayH = 2 mcv​ = 1.5 m²/year2 mimpervious rockwater table

A clay layer 2 m thick has a coefficient of consolidation cv = 1.5 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 = 2 mcv = 1.5 m²/yearU = 50%Tv = 0.197single drainage
t = years
6.Hard
sandclayH = 9 m9 msandwater table

In a laboratory oedometer test, a clay specimen 20 mm thick, drained at the top and bottom, reaches 50% consolidation in 5 min. In the field the same clay forms a layer 9 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 = 5 minH = 9 mfield: double drainage
t = days

Answer key

  1. 1. S = 749.6 mm
  2. 2. S = 205.7 mm
  3. 3. S = 7.753 mm
  4. 4. S = 483.1 mm
  5. 5. t = 0.5253 years
  6. 6. t = 703.1 days

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