07-Str-A3 · December 2015

Question 5 of 6: Water needed in a cut to prevent base uplift

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 5: Water needed in a cut to prevent base uplift (Value: 20 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
Clay stratum, ground level to the sand	7 m thick, $\rho_{sat} = 1900\ \text{kg/m}^{3}$
Depth of the cut	$H = 5$ m
Clay remaining beneath the base of the cut	$7 - 5 = 2$ m
Underlying sand	Artesian, $\rho_{sat} = 1800\ \text{kg/m}^{3}$
Piezometric surface of the sand	3.0 m above the base of the cut, i.e. 5.0 m above point $A$
Unit weight of water	$\gamma_w = 9.81\ \text{kN/m}^{3}$

Find. The depth of water $h$ that must be maintained in the cut so that the clay plug beneath it is not lifted off the sand.

The section. The 2 m of clay left beneath the base of the cut acts as a plug held down by its own weight and by any water ponded on it, and pushed up by the artesian pressure in the sand at point A.

Approach. The mechanism is uplift, not shear: the clay plug beneath the cut has low permeability, so the artesian pressure acts on its underside as a boundary pressure. The plug stays in place while the total vertical stress it and the ponded water deliver at point $A$ is at least equal to that pressure, which gives a direct equation for $h$.

Convert the density of the clay into a unit weight. The figure gives a density, and stresses need a unit weight, so $$\gamma_{clay} = \rho_{sat}\,g = 1900 \times 9.81 \times 10^{-3} = 18.639\ \text{kN/m}^{3} .$$ The density of the sand, $1800\ \text{kg/m}^{3}$, plays no part: the sand lies below the level being checked and its weight neither helps nor hinders.
Write the uplift pressure at point $A$. Point $A$ is on the top of the artesian sand, whose piezometric surface stands 3.0 m above the base of the cut and therefore $3.0 + 2.0 = 5.0$ m above $A$ itself. The pressure trying to lift the plug is $$u_A = \gamma_w h_p = 9.81 \times 5.0 = 49.05\ \text{kPa} .$$
Write the total vertical stress at $A$ holding the plug down. Two things press on $A$: the 2 m of saturated clay left beneath the cut, and the water ponded to a depth $h$ on the base of the cut. Nothing else remains, because the excavation has removed the upper 5 m. Hence $$\sigma_A = \gamma_{clay}(7 - 5) + \gamma_w h = 2(18.639) + 9.81h = 37.278 + 9.81h\ \ \text{kPa} .$$
Impose the limiting condition and solve for $h$. Stability is lost the instant the uplift pressure equals the total stress, that is when the effective stress at the base of the plug falls to zero, so the limiting case is $\sigma_A = u_A$: $$37.278 + 9.81h = 49.05 \quad\Longrightarrow\quad h = \frac{49.05 - 37.278}{9.81} = \boxed{1.20\ \text{m}} .$$
Read the same result in closed form as a check. Dividing the limiting equation through by $\gamma_w$ turns it into a statement about heads and densities alone, $$h = h_p - T\,\frac{\rho_{clay}}{\rho_w} = 5.0 - 2.0\left(\frac{1900}{1000}\right) = 5.0 - 3.8 = 1.20\ \text{m} ,$$ which is independent of the value taken for $g$ and confirms the arithmetic exactly. In words, each metre of clay is worth 1.9 m of water, so 2 m of clay balances 3.8 m of the 5 m of artesian head and the remaining 1.2 m must be supplied by ponding.
Quantify how unsafe the empty cut is. With no water in the cut the factor of safety against uplift is $$F = \frac{\sigma_A}{u_A} = \frac{37.278}{49.05} = 0.76 ,$$ so the base would fail as soon as the excavation reached 5 m. At $h = 1.20$ m the factor of safety is exactly $1.00$, which is a limiting state and not a design value.
Convert the limiting depth into a construction recommendation. A margin is needed because the clay thickness is never uniform, the piezometric level in the sand varies seasonally, and the plug may be locally thinned by a soft zone or a buried channel. Requiring $F = 1.2$ gives $$h = \frac{1.2(49.05) - 37.278}{9.81} = 2.20\ \text{m} ,$$ while a more modest ponded depth of 1.5 m already gives $F = 1.06$. The practical answer is therefore: maintain at least 1.2 m of water at all times to keep the base stable, and pond 1.5 m to 2.2 m in service, with standpipe piezometers installed in the sand and monitored daily. The alternative, if the cut must be worked dry, is to relieve the artesian pressure by pumping from deep wells in the sand until the piezometric surface has been drawn down below the level that 2 m of clay alone can hold, namely 3.8 m above $A$.

Check — the alternative reading of the “3.0 m” dimension. The dimension line in the source is drawn with its lower arrowhead on the top of the sand rather than on the base of the cut. Read that way, the artesian head at $A$ would be 3.0 m, $u_A = 29.43$ kPa, and the 2 m of clay alone would supply 37.28 kPa: the base would then be stable with the cut empty ($F = 1.27$) and the required $h$ would come out negative, at $-0.80$ m. Three things rule that reading out. The dimensioned length scales at 4.4 to 5.1 m against the figure’s own 2 m of clay beneath the cut, not 3.0 m; the distance from the piezometric surface to the base of the cut scales at 2.4 to 3.1 m, which does match; and the figure draws a positive depth $h$ of roughly 1.2 to 1.4 m, which is what the reading adopted here predicts. The solution above is worked on that basis, and the method is identical under either reading — only the head substituted in step 2 changes.

Quantity	Value
Unit weight of the clay	18.639 kN/m$^3$
Thickness of clay beneath the cut	2.0 m
Artesian head at $A$	5.0 m of water, $u_A = 49.05$ kPa
Total stress at $A$ with an empty cut	37.278 kPa, $F = 0.76$
Required depth of water for $F = 1.0$	$h = 1.20$ m
Depth of water for $F = 1.2$	2.20 m
Drawdown alternative	Relieve the sand to below 3.8 m of head above $A$

Topic. Base heave of an excavation over an artesian aquifer.

Key relations. Limiting condition $\sigma_A = u_A$, i.e. $\sigma_A' = 0$; $\sigma_A = \gamma_{clay}T + \gamma_w h$; $u_A = \gamma_w h_p$; closed form $h = h_p - T\rho_{clay}/\rho_w$; factor of safety $F = \sigma_A/u_A$.

Why this works. The clay plug is effectively impermeable over the duration of the works, so the water in the sand cannot seep into it and instead presses on its underside as a uniform boundary pressure. The plug is then a slab in a straightforward force balance: weight plus surcharge down, uplift pressure up. Because both sides of that balance are pressures at the same level, the condition is identical to requiring the effective stress at the base of the clay to remain positive, which is the general statement of the criterion. The shear strength of the clay does not enter for a wide excavation, which is why the stiff clay described in the stem gets no strength parameter at all.

Common pitfall. Including the 5 m of excavated clay in $\sigma_A$, which gives an unsafe result of about 130 kPa of restraint that no longer exists. A second, subtler error is using the buoyant unit weight of the clay: the plug is being weighed as a total force against a total pressure, so the saturated unit weight is the correct one. A third is forgetting that ponded water is the cheapest possible ballast — a metre of water is worth more than half a metre of clay and can be added or removed in hours.

Source. Das & Sobhan, Principles of Geotechnical Engineering 9th ed., §9.7 (uplift and heave beneath excavations); CFEM 4th ed., Ch. 6 (base stability and dewatering of excavations); Terzaghi, Peck & Mesri, Art. 47.

← Question 4 Next: Question 6 →