Two-layer profile stress diagram, 6 m sand

Given and assumptions

• Profile: clay 0 to 4 m, fine sand 4 m to 10 m. Water table (WT) at 2 m depth, capillary rise 1 m above the WT (saturated zone from 1 m to 2 m depth).
• Clay: $\gamma = 18$ kN/m³ above the WT, $\gamma_{sat} = 22$ kN/m³ below. The capillary zone is saturated, so 22 kN/m³ is used from 1 m to 2 m.
• Sand: $n = 0.44$, $G = 2.65$, $\gamma_w = 9.81$ kN/m³. The soil above the capillary zone is taken as having $u = 0$.

$$e = \frac{n}{1-n} = \frac{0.44}{0.56} = 0.786, \qquad \gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{(2.65+0.786)\times 9.81}{1.786} = 18.87\ \text{kN/m}^3$$

Stresses (kPa)

$$\sigma = \sum \gamma h, \qquad u = \gamma_w \times (\text{depth below WT}) \text{ (negative in capillary zone)}, \qquad \sigma' = \sigma - u$$

Diagram

 Depth  total σ        u                 σ'
  0 m   0              0                 0
  1 m   18       0 -> -9.81 (suction)    18 -> 27.8
  2 m   40       0 (WT)                  40
  4 m   84       19.6                    64.4
 10 m   197.3    78.5                    118.8
       (linear between points; u is negative
        only between 1 m and 2 m)

Plot $\sigma$, $u$ and $\sigma'$ against depth, joining the points with straight lines. $\sigma$ changes slope at 1 m and 4 m. At 1 m, $u$ jumps from 0 to $-9.81$ kPa, so $\sigma'$ jumps from 18 to 27.81 kPa; $\sigma'$ then changes slope again at 4 m. $u$ rises linearly from $-9.81$ kPa at 1 m to 0 at 2 m (the negative triangle above the WT) and then to 78.48 kPa at 10 m.

Answer: At the bottom (10 m): $\sigma = 197.25$, $u = 78.48$, $\sigma' = 118.77$ kPa. Capillarity gives a gain of $9.81$ kPa in effective stress at 1 m.