Total vertical stress is carried by both the soil skeleton and pore water. Effective stress is the part carried by soil grains, and it controls strength and compression behavior.
The effective unit weight below the water table is the saturated unit weight minus the unit weight of water. It is also called submerged or buoyant unit weight.
Above the water table, capillary water creates negative pore pressure. This suction can increase effective stress even when the soil is above the water table.
Typical capillary rise ranges: coarse sand 0.1 to 0.2 m, fine sand 0.3 to 1.2 m, silt 1.3 to 7.5 m, and clay 7.5 to 23 m.
Upward seepage increases pore pressure and reduces effective stress. Downward seepage reduces pore pressure and increases effective stress.
In a deposit of fine sand, the water table is 3.5 m below the surface, but sand to a height of 1.0 m above the water table is saturated by capillary water. Above this height, the sand may be assumed dry. The saturated and dry unit weights are 20 kN/m3 and 16 kN/m3. Calculate effective vertical stress 8 m below the surface.
Answer: 105.86 kPa.
A soil profile is shown in the reference figure with unit weights of soil layers. The groundwater table is 4 m below the ground surface. Determine how high the groundwater table should rise so that effective stress at the sand-clay interface is 110 kPa.
The figure contains the layer thicknesses and unit weights needed for the numerical solution. Use this setup:
Solve for the required water pressure head $h_w$, then compare it with the original water table depth to get the water table rise.
For the given soil profile in the reference figure, calculate the saturated unit weight of soil 1, total vertical stress at A, and effective vertical stress at B.
The profile dimensions and soil properties are embedded in the figure. Use these equations in order:
A clay layer 5 m thick rests beneath submerged sand 8 m thick. The top of sand is 3 m below the water surface. Saturated unit weights are 25 kN/m3 for sand and 20 kN/m3 for clay. Evaluate total vertical pressure, pore water pressure, and effective stress at mid-height of the clay.
Answer: $\sigma_v=279.43$ kPa, $u=132.44$ kPa, and $\sigma'_v=146.99$ kPa.
A 7.6 m clay layer is overlain by 15.2 m of sand. Initially the water table is at ground level. The saturated unit weights of sand and clay are 21.2 kN/m3 and 18.8 kN/m3, and the dry unit weight of sand is 18.2 kN/m3. When the water table lowers by 7.6 m, the sand above the table has $S=20\%$. Determine effective pressure at mid-height of the clay before and after lowering, and when the sand has no water table but remains at the same saturation.
Answer: 207.29 kPa before lowering, 263.61 kPa after lowering, and 319.92 kPa for the stated no-water-in-sand condition.
A granular soil layer has upward seepage caused by water supplied through a valve at the bottom of a tank. Determine the head $h$ that causes boiling.
Boiling occurs when the upward hydraulic gradient equals the critical hydraulic gradient.
A tank has upward seepage through granular soil. From the reference solution, at point C the total stress is due to 0.7 m of water and 1.2 m of saturated soil with $\gamma_{sat}=20$ kN/m3, and pore pressure head is 2.65 m.
Answer: 4.87 kPa.
For upward flow through sand, compute the seepage force per unit volume from the hydraulic gradient.
Use the head loss and soil thickness from the figure, then substitute into $f_s=i\gamma_w$.
A granular soil deposit is 4 m thick and is underlain by clay 5 m thick. The water table is at the sand-clay interface. Unit weight of sand is 16.6 kN/m3 and saturated unit weight of clay is 17.8 kN/m3. If water is 3 m above the soil surface, find effective stress at the bottom of clay.
Answer: 67.11 kPa.
A uniform soil deposit has $e=0.60$ and $G_s=2.65$. The natural water table is at depth $h$ below ground surface. Neutral pressure at 6 m depth is 34.34 kPa. Due to capillary moisture, average degree of saturation above the water table is 50 percent. Compute $h$.
Answer: The water table is 2.5 m below ground surface.
At an offshore location, the soil surface is 9 m below the water surface. The saturated unit weight of soil is 18.56 kN/m3. Determine effective stress at a depth of 16 m below the water surface.
Answer: 61.25 kPa.
Total stress is simply the weight of all material above a point divided by the area. Pore water pressure is the pressure carried by water in the voids. Effective stress is the difference, and it is the pressure actually transmitted through the grain-to-grain contacts. Because soil deformation and shear failure depend on grain contact forces, effective stress governs settlement, strength, and stability — not total stress.
When a saturated clay is suddenly loaded, the load is initially carried almost entirely by pore water because the clay cannot drain instantly. As pore pressure dissipates over time, effective stress increases and settlement occurs. This process is called primary consolidation. Engineers must track effective stress, not total stress, when predicting long-term behavior.
A soil profile has three layers. Layer 1 is 2 m of dry sand with unit weight 16 kN/m3. Layer 2 is 4 m of saturated sand with unit weight 20 kN/m3, and the water table is at the top of this layer. Layer 3 is 3 m of saturated clay with unit weight 19 kN/m3. Compute the total vertical stress, pore pressure, and effective vertical stress at the base of each layer.
At the base of Layer 1 (z = 2 m):
At the base of Layer 2 (z = 6 m):
At the base of Layer 3 (z = 9 m):
Answer: Layer 1 base: $\sigma'_v=32$ kPa; Layer 2 base: $\sigma'_v=72.76$ kPa; Layer 3 base: $\sigma'_v=100.33$ kPa.
A 3 m thick layer of sand is fully saturated ($\gamma_{sat}=20$ kN/m3). Upward seepage occurs with a head difference of 1.2 m over the 3 m thickness. Compute the effective stress at the bottom of the sand layer with and without seepage.
Without seepage (submerged condition):
With upward seepage:
The upward seepage force has reduced the effective stress by about 39 percent.
Answer: Without seepage: 30.57 kPa; with upward seepage: 18.80 kPa.
A 4 m thick layer of sand is submerged with $\gamma_{sat}=19.5$ kN/m3. Downward seepage occurs with a hydraulic gradient of 0.25. Compute the effective stress at depths of 2 m and 4 m below the top of the sand layer.
At $z=2$ m with downward seepage:
At $z=4$ m:
Answer: $\sigma'_v=24.29$ kPa at 2 m depth and $\sigma'_v=48.57$ kPa at 4 m depth.
A profile consists of 5 m of clay ($\gamma_{sat}=18$ kN/m3) overlying a confined sand aquifer. The piezometric head in the sand aquifer rises to 3 m above the ground surface. Compute the effective stress at the base of the clay. Then determine whether artesian conditions could cause heave of the clay layer above the aquifer.
Pore pressure at the base of clay equals artesian head times unit weight of water:
For heave, check if artesian uplift exceeds the weight of the clay:
Effective stress is positive (11.52 kPa), so heave does not occur, though the margin is small.
Answer: $\sigma'_v=11.52$ kPa at the base of clay; no heave since $\sigma'_v > 0$.