CivilGeotechnicaleasyFE ExamCoursework

Effective Stress Below the Water Table

A uniform soil deposit has total unit weight $\gamma = 18\ \text{kN/m}^3$ throughout. The water table sits $2.0\ \text{m}$ below ground surface.

Take $\gamma_w = 9.81\ \text{kN/m}^3$ .

Find the vertical effective stress at a depth of $5.0\ \text{m}$ .

Given

γ	18 kN/m^3	Total unit weight
z	5 m	Depth of interest
zw	2 m	Depth to water table

Hint 1

Effective stress is total stress minus pore water pressure.

Hint 2

Total stress uses the full 5 m of soil above the point.

Hint 3

Pore pressure is hydrostatic measured down from the water table — 3 m, not 5 m.

Worked solution — try the problem first

Terzaghi's principle: $\sigma' = \sigma - u$ .

Total vertical stress at 5 m — the full weight of soil above, regardless of the water table:

$\sigma = \gamma z = 18 \times 5.0 = 90.0\ \text{kPa}$

Pore water pressure. Hydrostatic below the water table, measured from the water table down — a depth of $5.0 - 2.0 = 3.0$ m:

$u = \gamma_w h_w = 9.81 \times 3.0 = 29.4\ \text{kPa}$

Effective stress:

$\sigma' = 90.0 - 29.4 = 60.6\ \text{kPa}$

$\boxed{\sigma' = 60.6\ \text{kPa}}$

Effective stress governs shear strength and settlement — which is why lowering a water table (raising $\sigma'$ ) causes ground to settle, and why raising one can trigger a slope failure.

Concepts:Effective stressPore water pressureTerzaghi's principle

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