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 = 110 kPasandclayCc​ = 0.6e0​ = 1.35σ′0​ = 100 kPa3.5 mwater table

A 3.5 m thick layer of normally consolidated clay (Cc = 0.6, e₀ = 1.35) has an effective vertical stress of 100 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 110 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 = 0.6e0 = 1.35σ′0 = 100 kPaΔσ = 110 kPa
S = mm
2.Easy
Δσ = 135 kPasandγ = 17.5 kN/m³4 mclayγsat​ = 17.5 kN/m³Cc​ = 0.4, e0​ = 1.153 mwater tableX

A 3 m thick normally consolidated clay layer (γsat = 17.5 kN/m³, Cc = 0.4, e₀ = 1.15) lies below 4 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 135 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

sand: Hs = 4 m, γ = 17.5 kN/m³clay: H = 3 m, γsat = 17.5 kN/m³γw = 9.81 kN/m³Cc = 0.4e0 = 1.15Δσ = 135 kPa
S = mm
3.Medium
Δσ = 80 kPasandclayCc​ = 0.45, Cr​ = 0.02e0​ = 0.65σ′0​ = 95 kPaσ′p​ = 215 kPa6 mwater table

A 6 m thick overconsolidated clay layer has Cc = 0.45, Cr = 0.02 and e₀ = 0.65. At its mid-depth the effective vertical stress is 95 kPa and the preconsolidation pressure is 215 kPa. A new building adds 80 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 6 mCc = 0.45Cr = 0.02e0 = 0.65σ′0 = 95 kPaσ′p = 215 kPaΔσ = 80 kPa
S = mm
4.Medium
Δσ = 180 kPasandclayCc​ = 0.5, Cr​ = 0.02e0​ = 1.2σ′0​ = 120 kPaσ′p​ = 150 kPa4 mwater table

A 4 m thick overconsolidated clay layer has Cc = 0.5, Cr = 0.02 and e₀ = 1.2. At its mid-depth the effective vertical stress is 120 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 = 4 mCc = 0.5Cr = 0.02e0 = 1.2σ′0 = 120 kPaσ′p = 150 kPaΔσ = 180 kPa
S = mm
5.Hard
sandclayH = 6.5 mcv​ = 1 m²/year6.5 msandwater table

A clay layer 6.5 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 = 6.5 mcv = 1 m²/yearU = 50%Tv = 0.197double drainage
t = years
6.Hard
sandclayH = 5.5 m5.5 mimpervious rockwater table

In a laboratory oedometer test, a clay specimen 25 mm thick, drained at the top and bottom, reaches 50% consolidation in 10 min. In the field the same clay forms a layer 5.5 m thick. It lies on impervious rock, with sand above it, so water drains out only at the top. How many days t does the field layer take to reach 50% consolidation?

hlab = 25 mmtlab = 10 minH = 5.5 mfield: single drainage
t = days

Answer key

  1. 1. S = 287.9 mm
  2. 2. S = 236.8 mm
  3. 3. S = 19.3 mm
  4. 4. S = 277.2 mm
  5. 5. t = 2.081 years
  6. 6. t = 1344 days

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