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 = 4 mzw = 2 mγw = 9.81 kN/m³sand: γd = 15.5 kN/m³sand: γsat = 20 kN/m³
σ′ = kPa
2.Effective stress in soilHard
For the soil profile shown, find the total vertical stress σ at point X.
z = 6.5 mzw = 1 mγw = 9.81 kN/m³sand: γd = 15 kN/m³sand: γsat = 20.5 kN/m³clay: γsat = 17.5 kN/m³
σ = kPa
3.Consolidation settlementMedium
A 3.5 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.02 and e₀ = 0.6. At its mid-depth the effective vertical stress is 95 kPa and the overconsolidation ratio is OCR = 2.5. A new building adds 50 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
H = 3.5 mCc = 0.3Cr = 0.02e0 = 0.6σ′0 = 95 kPaOCR = 2.5Δσ = 50 kPa
S = mm
4.Consolidation settlementHard
A clay layer 3.5 m thick has a coefficient of consolidation cv = 0.5 m²/year. It lies on impervious rock, with sand above it, so water drains out only at the top. For 50% average consolidation the time factor is Tv = 0.197. How many years t does the layer take to reach 50% consolidation?
H = 3.5 mcv = 0.5 m²/yearU = 50%Tv = 0.197single drainage
t = years
5.Hydrostatic forceMedium
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 2.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
6.Hydrostatic forceHard
A vertical rectangular gate 1.5 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 = 1.5 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3Solutions
1.Effective stress in soil
Answer: σ′ = 51.38 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) = 15.5·2 + 20·2 = 71 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 = 71 − 19.62 = 51.38 kPa
Answer
σ′ = 51.38 kPa
2.Effective stress in soil
Answer: σ = 117.3 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·1 + γsat,sand·2 + γsat,clay·3.5
σ = 15·1 + 20.5·2 + 17.5·3.5 = 117.3 kPa
Answer
σ = 117.3 kPa
3.Consolidation settlement
Answer: S = 8.034 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.
Preconsolidation pressure
σ′p = OCR·σ′0 = 2.5·95 = 237.5 kPa
Compare with the preconsolidation pressure
σ′f = 95 + 50 = 145 kPa ≤ 237.5 kPa
Settlement
The clay stays on its recompression line.
S = Cr·H/(1 + e0)·log₁₀(σ′f/σ′0) = 0.02·3.5/(1 + 0.6)·log₁₀(145/95) = 8.034 mm
Answer
S = 8.034 mm
4.Consolidation settlement
Answer: t = 4.827 years
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 = 3.5 m
Time
t = Tv·Hdr²/cv = 0.197·3.5²/0.5 = 4.827 years
Answer
t = 4.827 years
5.Hydrostatic force
Answer: F = 156.3 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 = 3 + 2.5/2 = 4.25 m
A = H·b = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.25·3.75 = 156.3 kN
Answer
F = 156.3 kN
6.Hydrostatic force
Answer: F = 198 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 = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·5.25·3.75 = 198 kN
Answer
F = 198 kN
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