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 = 45 kPasandγ = 18 kN/m³3 mclayγsat​ = 19.5 kN/m³Cc​ = 0.45, e0​ = 0.959 mwater tableX

A 9 m thick normally consolidated clay layer (γsat = 19.5 kN/m³, Cc = 0.45, e₀ = 0.95) lies below 3 m of sand (γ = 18 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 45 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 = 18 kN/m³clay: H = 9 m, γsat = 19.5 kN/m³γw = 9.81 kN/m³Cc = 0.45e0 = 0.95Δσ = 45 kPa
S = mm
2.Easy
Δσ = 85 kPasandclayCc​ = 0.6e0​ = 0.7σ′0​ = 55 kPa9.5 mwater table

A 9.5 m thick layer of normally consolidated clay (Cc = 0.6, 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 85 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 9.5 mCc = 0.6e0 = 0.7σ′0 = 55 kPaΔσ = 85 kPa
S = mm
3.Medium
Δσ = 180 kPasandclayCc​ = 0.45, Cr​ = 0.06e0​ = 0.85σ′0​ = 70 kPaσ′p​ = 150 kPa7.5 mwater table

A 7.5 m thick overconsolidated clay layer has Cc = 0.45, Cr = 0.06 and e₀ = 0.85. At its mid-depth the effective vertical stress is 70 kPa and the preconsolidation pressure is 150 kPa. A new building adds 180 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7.5 mCc = 0.45Cr = 0.06e0 = 0.85σ′0 = 70 kPaσ′p = 150 kPaΔσ = 180 kPa
S = mm
4.Medium
Δσ = 170 kPasandclayCc​ = 0.3, Cr​ = 0.04e0​ = 1.1σ′0​ = 45 kPaσ′p​ = 125 kPa4.5 mwater table

A 4.5 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.04 and e₀ = 1.1. At its mid-depth the effective vertical stress is 45 kPa and the preconsolidation pressure is 125 kPa. A new building adds 170 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4.5 mCc = 0.3Cr = 0.04e0 = 1.1σ′0 = 45 kPaσ′p = 125 kPaΔσ = 170 kPa
S = mm
5.Hard
sandclayH = 6 m6 msandwater table

In a laboratory oedometer test, a clay specimen 19 mm thick, drained at the top and bottom, reaches 50% consolidation in 4 min. In the field the same clay forms a layer 6 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 = 4 minH = 6 mfield: double drainage
t = days
6.Hard
sandclayH = 9.5 mcv​ = 1 m²/year9.5 msandwater table

A clay layer 9.5 m thick has a coefficient of consolidation cv = 1 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 = 9.5 mcv = 1 m²/yearU = 50%Tv = 0.197double drainage
t = years

Answer key

  1. 1. S = 342 mm
  2. 2. S = 1361 mm
  3. 3. S = 485.2 mm
  4. 4. S = 189.4 mm
  5. 5. t = 277 days
  6. 6. t = 4.445 years

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