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 = 135 kPasandγ = 18.5 kN/m³5.5 mclayγsat​ = 17.5 kN/m³Cc​ = 0.25, e0​ = 0.93 mwater tableX

A 3 m thick normally consolidated clay layer (γsat = 17.5 kN/m³, Cc = 0.25, e₀ = 0.9) lies below 5.5 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 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 = 5.5 m, γ = 18.5 kN/m³clay: H = 3 m, γsat = 17.5 kN/m³γw = 9.81 kN/m³Cc = 0.25e0 = 0.9Δσ = 135 kPa
S = mm
2.Easy
Δσ = 25 kPasandclayCc​ = 0.2e0​ = 1σ′0​ = 105 kPa9.5 mwater table

A 9.5 m thick layer of normally consolidated clay (Cc = 0.2, e₀ = 1) has an effective vertical stress of 105 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 25 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 = 0.2e0 = 1σ′0 = 105 kPaΔσ = 25 kPa
S = mm
3.Medium
Δσ = 110 kPasandclayCc​ = 0.4, Cr​ = 0.02e0​ = 1σ′0​ = 115 kPaσ′p​ = 145 kPa2 mwater table

A 2 m thick overconsolidated clay layer has Cc = 0.4, Cr = 0.02 and e₀ = 1. At its mid-depth the effective vertical stress is 115 kPa and the preconsolidation pressure is 145 kPa. A new building adds 110 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 2 mCc = 0.4Cr = 0.02e0 = 1σ′0 = 115 kPaσ′p = 145 kPaΔσ = 110 kPa
S = mm
4.Medium
Δσ = 110 kPasandclayCc​ = 0.2, Cr​ = 0.05e0​ = 0.95σ′0​ = 40 kPaσ′p​ = 130 kPa5.5 mwater table

A 5.5 m thick overconsolidated clay layer has Cc = 0.2, Cr = 0.05 and e₀ = 0.95. At its mid-depth the effective vertical stress is 40 kPa and the preconsolidation pressure is 130 kPa. A new building adds 110 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 5.5 mCc = 0.2Cr = 0.05e0 = 0.95σ′0 = 40 kPaσ′p = 130 kPaΔσ = 110 kPa
S = mm
5.Hard
sandclayH = 7 m7 mimpervious rockwater table

In a laboratory oedometer test, a clay specimen 25 mm thick, drained at the top and bottom, reaches 50% consolidation in 5 min. In the field the same clay forms a layer 7 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 = 5 minH = 7 mfield: single drainage
t = days
6.Hard
sandclayH = 5 mcv​ = 0.8 m²/year5 msandwater table

A clay layer 5 m thick has a coefficient of consolidation cv = 0.8 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 = 5 mcv = 0.8 m²/yearU = 50%Tv = 0.197double drainage
t = years

Answer key

  1. 1. S = 134.5 mm
  2. 2. S = 88.12 mm
  3. 3. S = 78.34 mm
  4. 4. S = 107.2 mm
  5. 5. t = 1089 days
  6. 6. t = 1.539 years

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