Effective vertical stress in a clay layer over an artesian sand: 16-Civ-A4 Geotechnical Materials and Analysis, December 2015, Q3

Question 3: Effective vertical stress in a clay layer over an artesian sand (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Profile from ground level (GL): upper sand 0–4 m, clay 4–8 m, lower (artesian) sand 8–12 m. Water table 2 m below GL; lower sand piezometric surface 4 m above GL.

Given data
Quantity	Value
Upper-sand unit weight, above WT	16.5 kN/m³
Sand saturated unit weight	19 kN/m³
Clay saturated unit weight	20 kN/m³
Water table depth	2 m below GL
Artesian piezometric surface	4 m above GL
Unit weight of water, $\gamma_w$	9.81 kN/m³

Find. The effective vertical stress $\sigma'_v$ at the top of the clay (depth 4 m) and at the bottom of the clay (depth 8 m).

Figure 3.1 — layered profile: water table 2 m down, lower sand artesian (piezometric surface 4 m above GL).

Approach. Build the total vertical stress by summing $\gamma\,z$ layer by layer, obtain the pore pressure separately (hydrostatic from the water table in the upper sand; from the artesian piezometric surface at the base of the clay), and subtract: $\sigma'_v = \sigma_v - u$.

Total stress at the top of the clay (z = 4 m). The upper sand is moist over the top 2 m and saturated over the next 2 m: $$\sigma_v = (2)(16.5) + (2)(19) = 33 + 38 = 71\ \text{kPa}.$$
Pore pressure at the top of the clay. The upper sand is in free hydrostatic communication with the water table 2 m above this level: $$u = (4-2)\,\gamma_w = (2)(9.81) = 19.62\ \text{kPa}.$$
Effective stress at the top of the clay. $$\sigma'_{v,\text{top}} = 71 - 19.62 = \boxed{51.4\ \text{kPa}}.$$
Total stress at the base of the clay (z = 8 m). Add the 4 m clay column: $$\sigma_v = 71 + (4)(20) = 71 + 80 = 151\ \text{kPa}.$$
Pore pressure at the base of the clay. Here the clay is in contact with the artesian lower sand, whose piezometric surface stands 4 m above GL. The pressure head at this point is the height of the piezometric surface above it, $8 + 4 = 12$ m: $$u = (12)(9.81) = 117.72\ \text{kPa}.$$
Effective stress at the base of the clay. $$\sigma'_{v,\text{base}} = 151 - 117.72 = \boxed{33.3\ \text{kPa}}.$$

Final results — Question 3
Location	$\sigma_v$ (kPa)	$u$ (kPa)	$\sigma'_v$ (kPa)
Top of clay (4 m)	71.0	19.6	51.4
Base of clay (8 m)	151.0	117.7	33.3

Note that the effective stress decreases from 51.4 kPa at the top of the clay to 33.3 kPa at its base, even though the total stress rises by 80 kPa. The artesian head drives an upward pore-pressure gradient across the clay, so the clay near the lower sand is the most vulnerable to loss of effective stress (and to heave if the artesian head were to rise).

Topic: Effective-stress profile through a confined clay under artesian pressure.

Key relations: $\sigma_v=\sum\gamma_i z_i$; hydrostatic $u=\gamma_w h_p$ with the pressure head $h_p$ measured from the relevant piezometric surface; $\sigma'_v=\sigma_v-u$ (Terzaghi).

Why this works: Total stress depends only on the weight of everything above the point, but pore pressure depends on the hydraulic boundary controlling that stratum. The upper sand is controlled by the water table; the lower sand is controlled by its artesian piezometric surface. Using the correct head for each is the whole of the problem.

Common pitfall: Taking the pore pressure at the base of the clay as hydrostatic from the water table (which would give $u=(8-2)(9.81)=58.9$ kPa) instead of from the artesian surface — this overstates $\sigma'_v$ by 59 kPa and hides the heave risk.

Source: Das & Sobhan, Principles of Geotechnical Engineering, ch. 9 (effective stress, seepage and artesian conditions).

Question 3 of 6: Effective vertical stress in a clay layer over an artesian sand

Question 3: Effective vertical stress in a clay layer over an artesian sand (10 marks)