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CivilSolve

Geotechnical

Vertical Effective Stress Profile

Calculate total stress, pore water pressure, and effective stress with depth in layered soil. Set the water table and view the stress profile.

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.