CivilSolve

Geotechnical

Vertical Effective Stress Profile

Total stress σ, hydrostatic pore pressure u, and effective stress σ′ = σ − u at chosen depths in a layered soil profile with a water table (which may be at depth, at the surface, or ponded above it). Emits the σ/u/σ′ vs depth diagram. Use for: 'find the effective stress at depth', stress profiles for consolidation or bearing problems. Not for: seepage-modified pore pressures or excess pore pressure during consolidation.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
layersSoil layers from the ground surface downward
layers 1
depths of interestDepths to report σ, u, σ′ at; defaults to every layer boundary plus the water table

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SI single layer, WT at 2 m: σ/u/σ′ at 5 m = 96 / 29.43 / 66.57 kPa

Given

layers
  1. 1.thickness 5 mmoist unit weight 18 kN/m^3saturated unit weight 20 kN/m^3
water table depth
2 m

Assumptions

  • Pore pressure is hydrostatic below the water table; capillary rise above it is excluded (u = 0 above the water table).
  • No seepage or excess pore pressure — the groundwater is static.
  • Unit weight of water taken as 9.81 kN/m³ (62.4 pcf differs by under 0.1%).

Solution steps

  1. Set up the stress column

    The profile is split into constant-unit-weight segments at layer boundaries and at the water table: moist unit weight above the water table, saturated below.

    0 m–2 m: γ = 18 kN/m^3; 2 m–5 m (submerged): γ = 20 kN/m^3

  2. Total vertical stress

    σ at a depth is the full weight of everything above it — soil (moist or saturated) and any ponded water — per unit area.

    σ = Σ γi·Δzi (+ γ_w · ponded depth)

  3. Hydrostatic pore pressure

    Below the water table the pore water is continuous and static, so u grows linearly with depth below it; above the water table u is taken as zero.

    u = γ_w·(z − z_wt) for z below the water table, else u = 0

  4. Stresses at z = 0 m

    At the ground surface only ponded water (if any) contributes — and it adds equally to σ and u.

    σ′ = σ − u

    at z = 0 m: σ = 0 kPa, u = 0 kPa, σ′ = 0 kPa − 0 kPa = 0 kPa

    = 0 kPa

  5. Stresses at z = 2 m

    Effective stress is what the soil skeleton actually carries: total stress minus pore pressure.

    σ′ = σ − u

    at z = 2 m: σ = 36 kPa, u = 0 kPa, σ′ = 36 kPa − 0 kPa = 36 kPa

    = 36 kPa

  6. Stresses at z = 5 m

    Effective stress is what the soil skeleton actually carries: total stress minus pore pressure.

    σ′ = σ − u

    at z = 5 m: σ = 96 kPa, u = 29.4 kPa, σ′ = 96 kPa − 29.4 kPa = 66.6 kPa

    = 66.6 kPa

Stress profileStress profile500096Stress (kPa)Depth z (m)water table (0, 2)

Results

Total stress σ at z = 0 m

0kPa

Pore pressure u at z = 0 m

0kPa

Effective stress σ′ at z = 0 m

0kPa

Total stress σ at z = 2 m

36kPa

Pore pressure u at z = 2 m

0kPa

Effective stress σ′ at z = 2 m

36kPa

Total stress σ at z = 5 m

96kPa

Pore pressure u at z = 5 m

29.4kPa

Effective stress σ′ at z = 5 m

66.6kPa

Where this answer was checked
source
Terzaghi effective stress, closed form (NCEES FE Reference Handbook) — σ = Σγh; u = γw·(z − zwt); σ′ = σ − u
verified by
hand-recomputed
derivation
Depths default to [0, 2 (WT), 5]. At 2 m: σ = 2·18 = 36 kPa, u = 0, σ′ = 36. At 5 m: σ = 2·18 + 3·20 = 96 kPa; u = 3·9.81 = 29.43 kPa; σ′ = 96 − 29.43 = 66.57 kPa.

Example 2

SI two layers, WT at the interface: at 7 m σ′ = 129 − 39.24 = 89.76 kPa

Given

layers
  1. 1.name sandthickness 3 mmoist unit weight 17 kN/m^3
  2. 2.name claythickness 4 msaturated unit weight 19.5 kN/m^3
water table depth
3 m

Assumptions

  • Pore pressure is hydrostatic below the water table; capillary rise above it is excluded (u = 0 above the water table).
  • No seepage or excess pore pressure — the groundwater is static.
  • Unit weight of water taken as 9.81 kN/m³ (62.4 pcf differs by under 0.1%).

Solution steps

  1. Set up the stress column

    The profile is split into constant-unit-weight segments at layer boundaries and at the water table: moist unit weight above the water table, saturated below.

    0 m–3 m: γ = 17 kN/m^3; 3 m–7 m (submerged): γ = 19.5 kN/m^3

  2. Total vertical stress

    σ at a depth is the full weight of everything above it — soil (moist or saturated) and any ponded water — per unit area.

    σ = Σ γi·Δzi (+ γ_w · ponded depth)

  3. Hydrostatic pore pressure

    Below the water table the pore water is continuous and static, so u grows linearly with depth below it; above the water table u is taken as zero.

    u = γ_w·(z − z_wt) for z below the water table, else u = 0

  4. Stresses at z = 0 m

    At the ground surface only ponded water (if any) contributes — and it adds equally to σ and u.

    σ′ = σ − u

    at z = 0 m: σ = 0 kPa, u = 0 kPa, σ′ = 0 kPa − 0 kPa = 0 kPa

    = 0 kPa

  5. Stresses at z = 3 m

    Effective stress is what the soil skeleton actually carries: total stress minus pore pressure.

    σ′ = σ − u

    at z = 3 m: σ = 51 kPa, u = 0 kPa, σ′ = 51 kPa − 0 kPa = 51 kPa

    = 51 kPa

  6. Stresses at z = 7 m

    Effective stress is what the soil skeleton actually carries: total stress minus pore pressure.

    σ′ = σ − u

    at z = 7 m: σ = 129 kPa, u = 39.2 kPa, σ′ = 129 kPa − 39.2 kPa = 89.8 kPa

    = 89.8 kPa

Stress profileStress profile7000129Stress (kPa)Depth z (m)water table (0, 3)

Results

Total stress σ at z = 0 m

0kPa

Pore pressure u at z = 0 m

0kPa

Effective stress σ′ at z = 0 m

0kPa

Total stress σ at z = 3 m

51kPa

Pore pressure u at z = 3 m

0kPa

Effective stress σ′ at z = 3 m

51kPa

Total stress σ at z = 7 m

129kPa

Pore pressure u at z = 7 m

39.2kPa

Effective stress σ′ at z = 7 m

89.8kPa

Where this answer was checked
source
Terzaghi effective stress, closed form (NCEES FE Reference Handbook) — Sand 3 m at γ = 17 above the WT; clay 4 m at γsat = 19.5 below
verified by
hand-recomputed
derivation
Depths default to [0, 3 (interface = WT), 7]. At 3 m: σ = 3·17 = 51 kPa, u = 0, σ′ = 51. At 7 m: σ = 51 + 4·19.5 = 51 + 78 = 129 kPa; u = 4·9.81 = 39.24 kPa; σ′ = 129 − 39.24 = 89.76 kPa.

Example 3

US sand column, WT at 10 ft: at 20 ft σ = 2350 psf, u ≈ 624 psf, σ′ ≈ 1726 psf

Given

layers
  1. 1.name moist sandthickness 10 ftmoist unit weight 110 pcf
  2. 2.name saturated sandthickness 10 ftsaturated unit weight 125 pcf
water table depth
10 ft

Assumptions

  • Pore pressure is hydrostatic below the water table; capillary rise above it is excluded (u = 0 above the water table).
  • No seepage or excess pore pressure — the groundwater is static.
  • Unit weight of water taken as 9.81 kN/m³ (62.4 pcf differs by under 0.1%).

Solution steps

  1. Set up the stress column

    The profile is split into constant-unit-weight segments at layer boundaries and at the water table: moist unit weight above the water table, saturated below.

    0 ft–10 ft: γ = 110 pcf; 10 ft–20 ft (submerged): γ = 125 pcf

  2. Total vertical stress

    σ at a depth is the full weight of everything above it — soil (moist or saturated) and any ponded water — per unit area.

    σ = Σ γi·Δzi (+ γ_w · ponded depth)

  3. Hydrostatic pore pressure

    Below the water table the pore water is continuous and static, so u grows linearly with depth below it; above the water table u is taken as zero.

    u = γ_w·(z − z_wt) for z below the water table, else u = 0

  4. Stresses at z = 0 ft

    At the ground surface only ponded water (if any) contributes — and it adds equally to σ and u.

    σ′ = σ − u

    at z = 0 ft: σ = 0 psf, u = 0 psf, σ′ = 0 psf − 0 psf = 0 psf

    = 0 psf

  5. Stresses at z = 10 ft

    Effective stress is what the soil skeleton actually carries: total stress minus pore pressure.

    σ′ = σ − u

    at z = 10 ft: σ = 1100 psf, u = 0 psf, σ′ = 1100 psf − 0 psf = 1100 psf

    = 1100 psf

  6. Stresses at z = 20 ft

    Effective stress is what the soil skeleton actually carries: total stress minus pore pressure.

    σ′ = σ − u

    at z = 20 ft: σ = 2350 psf, u = 624 psf, σ′ = 2350 psf − 624 psf = 1730 psf

    = 1730 psf

Stress profileStress profile200002350Stress (psf)Depth z (ft)water table (0, 10)

Results

Total stress σ at z = 0 ft

0psf

Pore pressure u at z = 0 ft

0psf

Effective stress σ′ at z = 0 ft

0psf

Total stress σ at z = 10 ft

1100psf

Pore pressure u at z = 10 ft

0psf

Effective stress σ′ at z = 10 ft

1100psf

Total stress σ at z = 20 ft

2350psf

Pore pressure u at z = 20 ft

624psf

Effective stress σ′ at z = 20 ft

1730psf

Where this answer was checked
source
Terzaghi effective stress, closed form with US units (γw ≈ 62.4 pcf) — 10 ft moist sand at 110 pcf over 10 ft saturated sand at 125 pcf
verified by
hand-recomputed
derivation
At 20 ft: σ = 10·110 + 10·125 = 1100 + 1250 = 2350 psf; u = 62.4·10 = 624 psf; σ′ = 2350 − 624 = 1726 psf. Engine γw = 9.81 kN/m³ = 62.45 pcf gives u = 624.5 psf, σ′ = 1725.5 psf — within 0.1% of the 62.4-pcf hand values.