Calculate total stress, pore water pressure, and effective stress with depth in layered soil. Set the water table and view the stress profile.
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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.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
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.
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)
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
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
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
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
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.name sandthickness 3 mmoist unit weight 17 kN/m^3
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
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.
US sand column, WT at 10 ft: at 20 ft σ = 2350 psf, u ≈ 624 psf, σ′ ≈ 1726 psf
Given
layers
1.name moist sandthickness 10 ftmoist unit weight 110 pcf
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
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.