07-Str-A3 · December 2015

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Engineers Canada / PEO National Examinations, December 2015 — 07-Str-A3 Geotechnical Materials and Analysis. Closed book, three hours, 100 marks, one approved Casio or Sharp calculator, drawing instruments required. All six questions are compulsory and are weighted 20 / 10 / 10 / 20 / 20 / 20. The paper carries its own appendix: a formula sheet, the rectangular-loading m–n influence chart and a Newmark influence chart whose influence value is printed as $I_N = 0.005$ (200 elements). Values quoted below are taken from those sheets.

Reference texts.

B. M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. — Ch. 6 (compaction), Ch. 7 (permeability), Ch. 9 (stresses in a soil mass), Ch. 10 (consolidation), Ch. 12 (shear strength).
R. F. Craig / J. A. Knappett, Craig’s Soil Mechanics, 8th ed. — Ch. 3 (effective stress, artesian profiles), Ch. 4 (pore pressure coefficients $A$ and $B$), Ch. 5 (shear strength and the choice of test), Ch. 6 (stress distribution).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — Ch. 4 (site investigation and sampling categories), Ch. 6 (excavations and base stability). The Canadian reference for practice, sampling classes and terminology.
K. Terzaghi, R. B. Peck & G. Mesri, Soil Mechanics in Engineering Practice, 3rd ed. — Art. 15–17 (consolidation and seepage), Art. 19–20 (shear strength of clays).

Check — three readings of the printed paper, carried as stated. (1) Question 1 is headed “(4 x 5 = 20 marks)” but prints five lettered parts (i)–(v); all five are answered and the header total of 20 marks is kept. (2) The section figure for Question 5 is captioned “Figure 2” on page 4 although the stem of Question 5 calls it Figure 3 — the same drawing, mislabelled in the source. (3) On that figure the “3.0 m” dimension is drawn with its arrowhead at the top of the sand, but at the figure’s own vertical scale (calibrated on the 2 m of clay left beneath the cut) the dimensioned distance is $3.1$ m from the piezometric surface down to the base of the cut, and $5.1$ m down to the sand. The piezometric surface is therefore taken as $3.0$ m above the base of the cut, i.e. $5.0$ m above point $A$; this is also the only reading for which a positive depth of water $h$ exists, and it reproduces the depth of water drawn in the figure. The alternative reading is worked through and dismissed in Question 5.

Question 3: Effective 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.

Quantity	Value
Upper sand, ground level to 4 m depth	$\gamma = 16.5\ \text{kN/m}^{3}$ above the water table, $\gamma_{sat} = 19\ \text{kN/m}^{3}$ below
Clay, 4 m to 8 m depth	$\gamma_{sat} = 20\ \text{kN/m}^{3}$
Lower sand, 8 m to 12 m depth	$\gamma_{sat} = 19\ \text{kN/m}^{3}$, artesian
Water table	2 m below ground level, in the upper sand
Piezometric surface of the lower sand	4 m above ground level
Unit weight of water	$\gamma_w = 9.81\ \text{kN/m}^{3}$

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

The profile. Two independent pore-pressure regimes meet across the clay: hydrostatic below a water table 2 m down in the upper sand, and artesian in the lower sand with its piezometric surface 4 m above ground level. Pore pressure is therefore not hydrostatic through the clay.

Approach. Total vertical stress is the accumulated weight of everything above the level in question; pore pressure at each of the two levels is read from the water regime of the layer that controls it — the water table for the top of the clay, the artesian piezometric surface for the bottom — and Terzaghi’s $\sigma' = \sigma - u$ is applied at each level separately.

Total vertical stress at the top of the clay. The 4 m of upper sand is moist over the top 2 m and saturated over the lower 2 m, so $$\sigma_{v,4} = 2(16.5) + 2(19) = 33.0 + 38.0 = \boxed{71.0\ \text{kPa}} .$$
Pore pressure at the top of the clay. This level sits inside the upper sand’s hydrostatic regime, whose water table is 2 m below ground level, so the pressure head at 4 m depth is $4 - 2 = 2$ m of water: $$u_4 = 2 \times 9.81 = 19.62\ \text{kPa} .$$
Effective stress at the top of the clay. Applying Terzaghi’s principle from the formula sheet, $\sigma' = \sigma - u$, $$\sigma_{v,4}' = 71.0 - 19.62 = \boxed{51.38\ \text{kPa}} .$$
Total vertical stress at the bottom of the clay. Adding the weight of the 4 m of saturated clay to the value from step 1, $$\sigma_{v,8} = 71.0 + 4(20) = 71.0 + 80.0 = \boxed{151.0\ \text{kPa}} .$$
Pore pressure at the bottom of the clay. This level is the top of the artesian sand, so its pore pressure is governed by that layer’s piezometric surface, which stands 4 m above ground level. The pressure head is measured from the level in question up to the piezometric surface, a total of $8 + 4 = 12$ m of water: $$u_8 = 12 \times 9.81 = 117.72\ \text{kPa} .$$ Note that this is emphatically not the hydrostatic value that a continuation of the upper water table would give, $(8-2)(9.81) = 58.9$ kPa. Using the water table here is the single commonest error in this question.
Effective stress at the bottom of the clay. Subtracting again, $$\sigma_{v,8}' = 151.0 - 117.72 = \boxed{33.28\ \text{kPa}} .$$
Check the result for physical sense. The effective stress has decreased from 51.38 kPa to 33.28 kPa across the clay even though 80 kPa of total stress was added, because the pore pressure rose by 98.1 kPa over the same 4 m. The head difference driving that is $12 - 2 = 10$ m across a 4 m layer, an upward hydraulic gradient of $i = 10/4 = 2.5$. Steady upward seepage through the clay is therefore taking place, which is what reduces effective stress with depth. The clay itself is still in place: its base carries a total stress of 151.0 kPa against an uplift pressure of 117.72 kPa, a factor of safety against uplift of $151.0/117.72 = 1.28$. Excavating in the upper sand, or thinning the clay, would erode that margin quickly — the same mechanism Question 5 asks about explicitly.

Level	$\sigma_v$ (kPa)	$u$ (kPa)	$\sigma_v'$ (kPa)
Top of clay, 4 m depth	71.0	19.62	51.38
Bottom of clay, 8 m depth	151.0	117.72	33.28
Upward hydraulic gradient through the clay	$i = 2.5$
Factor of safety of the clay against uplift	$151.0/117.72 = 1.28$

Topic. Effective stress where two pore-pressure regimes are separated by an aquitard.

Key relations. $\sigma_v = \sum \gamma_i h_i$; $u = \gamma_w h_p$ with $h_p$ the height of the relevant piezometric surface above the point; $\sigma' = \sigma - u$; hydraulic gradient $i = \Delta h / L$.

Why this works. Pore pressure is a property of the water, not of the soil column, and it is continuous only where the water is. A clay layer of low permeability lets two quite different heads coexist a few metres apart: hydrostatic above it, artesian below it, with a linear head loss across the clay itself. The correct pressure at any level is therefore read from the piezometric surface of whichever body of water is connected to that level, which is why the two ends of the same clay layer are treated with different water tables. Total stress, by contrast, is simply weight and never cares about the water regime.

Common pitfall. Drawing one hydrostatic pore-pressure line from the water table straight through the profile. That would give $u_8 = 58.9$ kPa and $\sigma_{v,8}' = 92.1$ kPa — nearly three times the true value, and it would hide the fact that the clay is close to floating. The second trap is measuring the artesian head from ground level instead of from the point, losing the 8 m.

Source. Craig’s Soil Mechanics 8th ed., §3.2–3.4 and Example 3.2 (this profile is the textbook case); Das & Sobhan 9th ed., §9.4–9.6.

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