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 = 105 kPasandclayCc​ = 0.4e0​ = 0.9σ′0​ = 125 kPa8.5 mwater table

An 8.5 m thick layer of normally consolidated clay (Cc = 0.4, e₀ = 0.9) has an effective vertical stress of 125 kPa at its mid-depth. A wide fill raises the vertical stress in 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.

H = 8.5 mCc = 0.4e0 = 0.9σ′0 = 125 kPaΔσ = 105 kPa
S = mm
2.Easy
Δσ = 70 kPasandclayCc​ = 0.15e0​ = 1.15σ′0​ = 70 kPa8.5 mwater table

An 8.5 m thick layer of normally consolidated clay (Cc = 0.15, e₀ = 1.15) has an effective vertical stress of 70 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 70 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 8.5 mCc = 0.15e0 = 1.15σ′0 = 70 kPaΔσ = 70 kPa
S = mm
3.Medium
Δσ = 50 kPasandclayCc​ = 0.25, Cr​ = 0.04e0​ = 0.6σ′0​ = 75 kPaOCR = 1.56 mwater table

A 6 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.04 and e₀ = 0.6. At its mid-depth the effective vertical stress is 75 kPa and the overconsolidation ratio is OCR = 1.5. A new building adds 50 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 6 mCc = 0.25Cr = 0.04e0 = 0.6σ′0 = 75 kPaOCR = 1.5Δσ = 50 kPa
S = mm
4.Medium
Δσ = 90 kPasandclayCc​ = 0.45, Cr​ = 0.04e0​ = 0.85σ′0​ = 45 kPaOCR = 24.5 mwater table

A 4.5 m thick overconsolidated clay layer has Cc = 0.45, Cr = 0.04 and e₀ = 0.85. At its mid-depth the effective vertical stress is 45 kPa and the overconsolidation ratio is OCR = 2. A new building adds 90 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 4.5 mCc = 0.45Cr = 0.04e0 = 0.85σ′0 = 45 kPaOCR = 2Δσ = 90 kPa
S = mm
5.Hard
sandclayH = 2.5 m2.5 mimpervious rockwater table

In a laboratory oedometer test, a clay specimen 20 mm thick, drained at the top and bottom, reaches 50% consolidation in 8 min. In the field the same clay forms a layer 2.5 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 = 20 mmtlab = 8 minH = 2.5 mfield: single drainage
t = days
6.Hard
sandclayH = 6 mcv​ = 0.8 m²/year6 msandwater table

A clay layer 6 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 = 6 mcv = 0.8 m²/yearU = 50%Tv = 0.197double drainage
t = years

Answer key

  1. 1. S = 473.9 mm
  2. 2. S = 178.5 mm
  3. 3. S = 69.31 mm
  4. 4. S = 222 mm
  5. 5. t = 347.2 days
  6. 6. t = 2.216 years

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