Consolidation settlement: practice problems with step-by-step solutions
Free Consolidation settlement practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: normally consolidated clay, overconsolidated clay, time for consolidation.
Free practice problemEasy
A 7.5 m thick layer of normally consolidated clay (Cc = 0.5, e₀ = 1.15) has an effective vertical stress of 135 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.
When you build on a clay layer, the extra load squeezes the clay. Water is pushed out of the pores and the grains move closer together. The ground surface sinks: this is consolidation settlement.
Clay has very small pores, so the water leaves slowly. A building on clay can keep settling for years.
The amount of pore space is measured by the void ratioe: the volume of the pores divided by the volume of the solid grains. A clay with e = 1 has as much pore volume as grain volume.
When the clay is squeezed, e goes down. The layer becomes thinner by the same fraction of its total volume:
S = H · Δe / (1 + e0)
Here H is the layer thickness, e0 the starting void ratio and Δe the drop in void ratio. The 1 + e0 is the total volume per unit volume of grains.
Self-check Why does a clay layer take years to settle, while sand settles almost at once?
Clay grains are heavier than sand grains
Clay has no water in its pores
The load on clay grows slowly over the years
Water has to flow out of very small pores, which takes time
Answer: Water has to flow out of very small pores, which takes time. Settlement needs water to leave the pores. In clay the pores are tiny, so the water flows out slowly. In sand it escapes almost at once.
2Normally and overconsolidated clay
Tests show that e drops in a straight line when you plot it against the logarithm of the effective stress σ′. The slope of that line is the compression indexCc.
Some clays were loaded more heavily in the past, for example by a glacier or soil that has since eroded. The largest effective stress a clay has ever carried is its preconsolidation pressureσ′p.
Below σ′p the clay only recompresses along a flat line with slope Cr, the recompression index. Cr is small, often Cc/10.
Above σ′p the clay follows the steep line with slope Cc.
A clay now at σ′0 = σ′p is normally consolidated (NC). A clay with σ′0 < σ′p is overconsolidated (OC). The overconsolidation ratio is OCR = σ′p/σ′0.
With σ′f = σ′0 + Δσ the final effective stress, the settlement is:
All stresses are taken at the middle of the clay layer, which stands for the average of the whole layer.
Self-check A clay has σ′₀ = 60 kPa and σ′p = 120 kPa. A load raises the stress to σ′f = 100 kPa. Which index do you use?
Cc only
Cr only
Cr up to 120 kPa, then Cc
Neither: the clay does not settle
Answer: Cr only. The final stress of 100 kPa stays below σ′p = 120 kPa, so the clay only recompresses along the flat Cr line. Cc would be needed only above 120 kPa.
3How long it takes
The time to reach a certain part of the final settlement depends on how far the water has to travel. That distance is the drainage pathHdr.
Sand (or another free-draining layer) above and below the clay: water leaves at both faces, so Hdr = H/2. This is double drainage.
Rock or another impervious layer below: water can only leave at the top, so Hdr = H. This is single drainage.
t = Tv · Hdr² / cv
cv is the coefficient of consolidation of the clay, for example in m²/year. Tv is the time factor for the degree of consolidation U you want: 0.197 for 50%, 0.403 for 70%, 0.848 for 90%.
Example: a 4 m clay layer on rock, cv = 2 m²/year, U = 50%. t = 0.197 · 4²/2 = 1.576 years. With sand below it, Hdr = 2 m and t = 0.394 years: four times faster.
Self-check A clay layer 6 m thick lies between two sand layers. What is the drainage path Hdr?
1.5 m
12 m
3 m
6 m
Answer: 3 m. Water leaves at the top and the bottom, so the longest path is half the layer: 6/2 = 3 m. Using the full 6 m is right only when one face is closed.
4The method step by step
Find the initial effective stress σ′0 at the middle of the clay (given, or from the soil profile).
Add the extra stress: σ′f = σ′0 + Δσ.
Decide the clay type. If σ′p or OCR is given, compare σ′f with σ′p = OCR·σ′0.
Pick the matching formula and use log₁₀, not ln.
Give S in the unit asked: metres times 1000 gives millimetres.
For a time question: find Hdr from the drainage at the top and bottom, then t = Tv·Hdr²/cv.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A 4 m thick layer of normally consolidated clay (Cc = 0.3, e0 = 1.0) has σ′0 = 100 kPa at its mid-depth. A wide fill adds Δσ = 100 kPa. Find the primary consolidation settlement S in mm.
1
Look at the layer
The clay is normally consolidated, so the whole stress increase follows the Cc line.
2
Final effective stress
σ′f = σ′0 + Δσ = 100 + 100 = 200 kPa
3
The logarithm
log₁₀(σ′f/σ′0) = log₁₀(200/100) = log₁₀ 2 = 0.301
4
The layer factor
Cc·H/(1 + e0) = 0.3 · 4/(1 + 1.0) = 0.6 m
5
Settlement
S = 0.6 · 0.301 = 0.1806 m = 180.6 mm
6
Check the answer
The void ratio drops by Δe = Cc·0.301 = 0.0903. Then S = H·Δe/(1 + e0) = 4 · 0.0903/2 = 0.1806 m. That is 4.5% of the layer, a normal value for soft clay.
Answer
S ≈ 181 mm (0.1806 m).
6Common mistakes
Using ln instead of log₁₀. The answer is then 2.3 times too big.
Forgetting to divide by 1 + e0.
Writing log₁₀(Δσ/σ′0) instead of log₁₀((σ′0 + Δσ)/σ′0).
Using Cc for an overconsolidated clay that stays below σ′p.
Taking Hdr = H/2 when the clay lies on rock, or Hdr = H when both faces drain.
Forgetting to square Hdr in the time formula.
Self-check NC clay: Cc = 0.25, H = 4 m, e₀ = 1.0, σ′₀ = 50 kPa, Δσ = 50 kPa. What is S?
0 mm
150.5 mm
301.0 mm
346.6 mm
Answer: 150.5 mm. S = 0.25·4/2 · log₁₀(100/50) = 0.5 · 0.301 = 0.1505 m. Using ln gives 346.6 mm, leaving out 1 + e₀ gives 301 mm, and log₁₀(Δσ/σ′₀) = log₁₀ 1 gives 0.
7How exam questions look
Easy: normally consolidated clay. You get H, Cc, e0, σ′0 and Δσ, as in the example. Sometimes you first find σ′0 from a soil profile.
Medium: overconsolidated clay with Cr, Cc and σ′p or OCR. Compare σ′f with σ′p first; split the log at σ′p if you pass it.
Hard: time for consolidation. Decide single or double drainage, then t = Tv·Hdr²/cv.
Worked OC example: H = 2 m, e0 = 1, Cr = 0.04, Cc = 0.4, σ′0 = 50 kPa, σ′p = 100 kPa, Δσ = 100 kPa. Then σ′f = 150 kPa passes σ′p.
S = 2/2 · [0.04·log₁₀(100/50) + 0.4·log₁₀(150/100)] = 0.0120 + 0.0704 = 0.0825 m
A 7 m thick normally consolidated clay layer (γsat = 18.5 kN/m³, Cc = 0.2, e₀ = 0.9) lies below 4 m of sand (γ = 19 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 70 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 = 19 kN/m³clay: H = 7 m, γsat = 18.5 kN/m³γw = 9.81 kN/m³Cc = 0.2e0 = 0.9Δσ = 70 kPa
Find S(mm)
Show hints
Normally consolidated: S = Cc·H/(1 + e₀)·log₁₀(σ′f/σ′₀).
At the middle of the clay: σ′₀ = γ·Hs + (γsat − γw)·H/2.
Cc·H/(1 + e0) = 0.7368 m.
Show the answer and the step-by-step solution
Primary consolidation of a clay layer of thickness H, using the effective stress at the middle of the layer: σ′f = σ′0 + Δσ. Normally consolidated clay: S = Cc·H/(1 + e0)·log₁₀(σ′f/σ′0). Overconsolidated clay recompresses along Cr up to the preconsolidation pressure σ′p and follows Cc only beyond it. Time: t = Tv·Hdr²/cv, where the drainage path Hdr is H/2 when water can leave at the top and bottom, and H when only one side drains.
Effective stress at the middle of the clay
The sand lies above the water table; below it, the clay is submerged, so use γsat − γw.
σ′0 = 19·4 + (18.5 − 9.81)·3.5 = 106.4 kPa
Final effective stress
σ′f = 106.4 + 70 = 176.4 kPa
Settlement
S = 0.2·7/(1 + 0.9)·log₁₀(176.4/106.4) = 0.7368·0.2195 = 0.1618 m = 161.8 mm
Answer
S = 161.8 mm
Example 2Medium
A 2 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 50 kPa and the preconsolidation pressure is 150 kPa. A new building adds 230 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
Compare the final stress σ′₀ + Δσ with the preconsolidation pressure.
σ′f = 50 + 230 = 280 kPa > σ′p = 150 kPa.
Split the stress range at σ′p: Cr from σ′₀ to σ′p, then Cc from σ′p to σ′f.
Show the answer and the step-by-step solution
Primary consolidation of a clay layer of thickness H, using the effective stress at the middle of the layer: σ′f = σ′0 + Δσ. Normally consolidated clay: S = Cc·H/(1 + e0)·log₁₀(σ′f/σ′0). Overconsolidated clay recompresses along Cr up to the preconsolidation pressure σ′p and follows Cc only beyond it. Time: t = Tv·Hdr²/cv, where the drainage path Hdr is H/2 when water can leave at the top and bottom, and H when only one side drains.
Compare with the preconsolidation pressure
σ′f = 50 + 230 = 280 kPa > 150 kPa
Settlement
Recompression up to σ′p, virgin compression beyond it.
S = 0.02·2/(1 + 1.2)·log₁₀(150/50) + 0.5·2/(1 + 1.2)·log₁₀(280/150)
S = 8.675 + 123.2 = 131.9 mm
Answer
S = 131.9 mm
Example 3Hard
A clay layer 9.5 m thick has a coefficient of consolidation cv = 3 m²/year. It lies between two sand layers, so water drains out at the top and the bottom. For 60% average consolidation the time factor is Tv = 0.287. How many years t does the layer take to reach 60% consolidation?
Two drained faces: the longest path a water particle travels is half the layer.
Hdr = 4.75 m.
Show the answer and the step-by-step solution
Primary consolidation of a clay layer of thickness H, using the effective stress at the middle of the layer: σ′f = σ′0 + Δσ. Normally consolidated clay: S = Cc·H/(1 + e0)·log₁₀(σ′f/σ′0). Overconsolidated clay recompresses along Cr up to the preconsolidation pressure σ′p and follows Cc only beyond it. Time: t = Tv·Hdr²/cv, where the drainage path Hdr is H/2 when water can leave at the top and bottom, and H when only one side drains.
Drainage path
Hdr = H/2 = 9.5/2 = 4.75 m
Time
t = Tv·Hdr²/cv = 0.287·4.75²/3 = 2.158 years
Answer
t = 2.158 years
Questions and answers
What is Consolidation settlement?
When you build on a clay layer, the extra load squeezes the clay. Water is pushed out of the pores and the grains move closer together. The ground surface sinks: this is consolidation settlement.
Clay has very small pores, so the water leaves slowly. A building on clay can keep settling for years.
How do I solve Consolidation settlement problems step by step?
Primary consolidation of a clay layer of thickness H, using the effective stress at the middle of the layer: σ′f = σ′0 + Δσ. Normally consolidated clay: S = Cc·H/(1 + e0)·log₁₀(σ′f/σ′0). Overconsolidated clay recompresses along Cr up to the preconsolidation pressure σ′p and follows Cc only beyond it. Time: t = Tv·Hdr²/cv, where the drainage path Hdr is H/2 when water can leave at the top and bottom, and H when only one side drains.
What are common mistakes in Consolidation settlement?
Use the base-10 logarithm, log₁₀, not the natural logarithm ln.
Divide by 1 + e₀, not just multiply by Cc·H.
The log uses the final over the initial stress: (σ′₀ + Δσ)/σ′₀, not Δσ/σ′₀.
What do the Consolidation settlement exercises cover?
Easy: normally consolidated clay
Medium: overconsolidated clay
Hard: time for consolidation
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.