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​ = 0.3e0​ = 1.2σ′0​ = 50 kPa7 mwater table

A 7 m thick layer of normally consolidated clay (Cc = 0.3, e₀ = 1.2) has an effective vertical stress of 50 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 = 7 mCc = 0.3e0 = 1.2σ′0 = 50 kPaΔσ = 125 kPa
S = mm
2.Easy
Δσ = 65 kPasandclayCc​ = 0.15e0​ = 0.95σ′0​ = 105 kPa9 mwater table

A 9 m thick layer of normally consolidated clay (Cc = 0.15, e₀ = 0.95) has an effective vertical stress of 105 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 65 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 9 mCc = 0.15e0 = 0.95σ′0 = 105 kPaΔσ = 65 kPa
S = mm
3.Medium
Δσ = 170 kPasandclayCc​ = 0.4, Cr​ = 0.06e0​ = 0.95σ′0​ = 90 kPaσ′p​ = 200 kPa7 mwater table

A 7 m thick overconsolidated clay layer has Cc = 0.4, Cr = 0.06 and e₀ = 0.95. At its mid-depth the effective vertical stress is 90 kPa and the preconsolidation pressure is 200 kPa. A new building adds 170 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 7 mCc = 0.4Cr = 0.06e0 = 0.95σ′0 = 90 kPaσ′p = 200 kPaΔσ = 170 kPa
S = mm
4.Medium
Δσ = 15 kPasandclayCc​ = 0.2, Cr​ = 0.06e0​ = 0.6σ′0​ = 120 kPaσ′p​ = 150 kPa3.5 mwater table

A 3.5 m thick overconsolidated clay layer has Cc = 0.2, Cr = 0.06 and e₀ = 0.6. At its mid-depth the effective vertical stress is 120 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 = 3.5 mCc = 0.2Cr = 0.06e0 = 0.6σ′0 = 120 kPaσ′p = 150 kPaΔσ = 15 kPa
S = mm
5.Hard
sandclayH = 7.5 m7.5 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 7.5 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 = 7.5 mfield: double drainage
t = days
6.Hard
sandclayH = 6 mcv​ = 1.5 m²/year6 msandwater table

A clay layer 6 m thick has a coefficient of consolidation cv = 1.5 m²/year. It lies between two sand layers, so water drains out at the top and the bottom. For 80% average consolidation the time factor is Tv = 0.567. How many years t does the layer take to reach 80% consolidation?

H = 6 mcv = 1.5 m²/yearU = 80%Tv = 0.567double drainage
t = years

Answer key

  1. 1. S = 519.3 mm
  2. 2. S = 144.9 mm
  3. 3. S = 238.3 mm
  4. 4. S = 6.714 mm
  5. 5. t = 488.3 days
  6. 6. t = 3.402 years

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