Effective stress in soil
- Total stress
- σ = Σγi·hi
- Pore water pressure
- u = γw·(z − zw)
- Effective stress
- σ′ = σ − u
- Water
- γw = 9.81 kN/m³
Geotechnical Engineering: cram pack
Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.
1Formulas and methods
Consolidation settlement
- Normally consolidated
- S = Cc·H/(1 + e0)·log₁₀(σ′f/σ′0)
- Final stress
- σ′f = σ′0 + Δσ (at mid-depth of the layer)
- Overconsolidated, σ′f ≤ σ′p
- S = Cr·H/(1 + e0)·log₁₀(σ′f/σ′0)
- Overconsolidated, σ′f > σ′p
- S = H/(1 + e0)·[Cr·log₁₀(σ′p/σ′0) + Cc·log₁₀(σ′f/σ′p)]
- Overconsolidation ratio
- OCR = σ′p/σ′0
- Time for consolidation
- t = Tv·Hdr²/cv
- Drainage path
- Hdr = H/2 (double drainage), H (single drainage)
- Time factors
- U 50% → 0.197, 60% → 0.287, 70% → 0.403, 80% → 0.567, 90% → 0.848
Hydrostatic force
- Pressure at depth
- p = ρgh
- Force on a vertical wall
- F = ρg·hc·A
- Centre of pressure
- yp = yc + Ic/(ycA)
- Rectangle
- Ic = bH³/12
2Practice problems
1.Effective stress in soilMedium
For the soil profile shown, find the vertical effective stress σ′ at point X.
z = 6 mzw = 4 mγw = 9.81 kN/m³sand: γd = 15 kN/m³sand: γsat = 19.5 kN/m³
σ′ = kPa
2.Effective stress in soilHard
For the soil profile shown, find the vertical effective stress σ′ at point X.
z = 9.5 mzw = 2 mγw = 9.81 kN/m³sand: γd = 16.5 kN/m³sand: γsat = 18.5 kN/m³clay: γsat = 17 kN/m³
σ′ = kPa
3.Consolidation settlementMedium
A 7 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.04 and e₀ = 1.1. At its mid-depth the effective vertical stress is 65 kPa and the preconsolidation pressure is 155 kPa. A new building adds 210 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
H = 7 mCc = 0.25Cr = 0.04e0 = 1.1σ′0 = 65 kPaσ′p = 155 kPaΔσ = 210 kPa
S = mm
4.Consolidation settlementHard
In a laboratory oedometer test, a clay specimen 25 mm thick, drained at the top and bottom, reaches 50% consolidation in 15 min. In the field the same clay forms a layer 9 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 = 15 minH = 9 mfield: single drainage
t = days
5.Hydrostatic forceMedium
A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 2 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
6.Hydrostatic forceHard
A vertical rectangular gate 3 m wide sits in the wall of a tank of seawater, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 2.5 mb = 3 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3Solutions
1.Effective stress in soil
Answer: σ′ = 79.38 kPa
Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).
Total stress
Add the weight of each soil band above X. Above the water table use the dry unit weight, below it the saturated one.
σ = γd·zw + γsat·(z − zw) = 15·4 + 19.5·2 = 99 kPa
Pore water pressure
Water pressure grows with depth below the water table.
u = γw·(z − zw) = 9.81·2 = 19.62 kPa
Effective stress
σ′ = σ − u = 99 − 19.62 = 79.38 kPa
Answer
σ′ = 79.38 kPa
2.Effective stress in soil
Answer: σ′ = 92.92 kPa
Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).
Total stress
Add the weight of each soil band above X. Above the water table use the dry unit weight, below it the saturated one.
σ = γd·2 + γsat,sand·4 + γsat,clay·3.5
σ = 16.5·2 + 18.5·4 + 17·3.5 = 166.5 kPa
Pore water pressure
Water pressure grows with depth below the water table.
u = γw·(z − zw) = 9.81·7.5 = 73.58 kPa
Effective stress
σ′ = σ − u = 166.5 − 73.58 = 92.92 kPa
Answer
σ′ = 92.92 kPa
3.Consolidation settlement
Answer: S = 257.8 mm
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 = 65 + 210 = 275 kPa > 155 kPa
Settlement
Recompression up to σ′p, virgin compression beyond it.
S = 0.04·7/(1 + 1.1)·log₁₀(155/65) + 0.25·7/(1 + 1.1)·log₁₀(275/155)
S = 50.32 + 207.5 = 257.8 mm
Answer
S = 257.8 mm
4.Consolidation settlement
Answer: t = 5400 days
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 paths
Hdr,lab = 25/2 = 12.5 mm = 0.0125 m
Hdr,field = 9 m
Scale the time
t = Tv·Hdr²/cv with the same T_v and cv in the lab and the field.
tfield = 15·(9/0.0125)² = 7.776 × 10⁶ min
tfield = 7.776 × 10⁶/1440 = 5400 days
Answer
t = 5400 days
5.Hydrostatic force
Answer: F = 147.2 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 1.5 + 2/2 = 2.5 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·6 = 147.2 kN
Answer
F = 147.2 kN
6.Hydrostatic force
Answer: F = 395.9 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 4 + 2.5/2 = 5.25 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·5.25·7.5 = 395.9 kN
Answer
F = 395.9 kN
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