Uniform sand, water table 2 m, capillary

Given

Uniform sand, water table (WT) at 2 m depth, $\gamma = 18$ kN/m³, $\gamma_w = 9.81$ kN/m³. Depth of interest = 5 m.

(i) Dry sand above the WT

• Total stress: $\sigma = 18 \times 5 = 90.0$ kPa
• Pore pressure at 5 m: $u = 9.81 \times (5-2) = 29.43$ kPa

$$\sigma' = 90.0 - 29.43 = 60.57\ \text{kPa}$$

Variation with depth (linear):

(ii) Sand above the WT saturated by capillary action

The pore water in the 2 m above the WT is under suction: $u = -\gamma_w z_c$, so $u = -19.62$ kPa at the ground surface (top of the capillary zone) and 0 at the WT. The soil weight is still taken as 18 kN/m³ (no other unit weight is given), so the total stress is unchanged.

At 5 m: $\sigma = 90.0$ kPa, $u = 29.43$ kPa, so

$$\sigma' = 90.0 - 29.43 = 60.57\ \text{kPa}$$

The effective stress at 5 m is the same as in (i): capillary suction changes the stress only within the capillary zone (and below it, if the saturated weight were larger).

If the saturated unit weight were higher than 18 kN/m³ in the capillary zone, the total and effective stress below would increase by $(\gamma_{sat}-18)\times 2$.

Plot

 Case (i)               Case (ii)
 σ' at 0 m = 0          σ' at 0 m = 19.6  (σ'-line shifts right
                                           above the WT)
 σ' at 2 m = 36         σ' at 2 m = 36
 σ' at 5 m = 60.6       σ' at 5 m = 60.6

Answer: (i) $\sigma' = 60.57$ kPa. (ii) $\sigma' = 60.57$ kPa at 5 m; the effective stress is increased only in the top 2 m (by $\gamma_w z_c$, up to 19.62 kPa at the surface).