Water level required to prevent bottom heave of a cut over artesian sand: 16-Civ-A4 Geotechnical Materials and Analysis, December 2015, Q5

Question 5: Water level required to prevent bottom heave of a cut over artesian sand (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 (read from Figure 3). Stiff saturated clay, $\rho_{sat} = 1900$ kg/m³, 7 m thick from ground level down to the top of the sand; cut depth $H = 5$ m; sand layer 2 m thick, $\rho_{sat} = 1800$ kg/m³, under artesian pressure. A standpipe tapping the sand shows water rising 3.0 m above the top of the sand. Point A lies at the clay–sand contact beneath the floor of the cut. Water stands to height $h$ in the cut.

Given data
Quantity	Value
Clay saturated density	1900 kg/m³
Clay thickness (ground level to sand)	7 m
Depth of cut, $H$	5 m
Standpipe water level above the top of the sand (point A)	3.0 m
Sand thickness, density	2 m, 1800 kg/m³
Unit weight of water, $\gamma_w$	9.81 kN/m³

Find. The height of water $h$ that must stand in the cut so that the clay beneath the cut floor does not heave (bottom blow-out).

Figure 3 — 5 m cut in 7 m of saturated clay over 2 m of artesian sand; the standpipe level is 3.0 m above the sand, and heave is checked at A.

Approach. Heave occurs when the upward water pressure of the artesian sand acting on the base of the clay left under the cut (point A) equals the downward total stress there, which is the weight of that clay plus the water ponded in the cut. Setting the two equal gives the limiting condition, factor of safety $F = 1$, which is solved for $h$.

Clay unit weight. $$\gamma_{clay} = \rho g = \dfrac{(1900)(9.81)}{1000} = 18.64\ \text{kN/m}^3.$$
Thickness of clay left beneath the cut. $$D = 7 - H = 7 - 5 = 2.0\ \text{m}.$$
Uplift pressure at A. The standpipe water level stands 3.0 m above A, so the pressure head at A is 3.0 m: $$u_A = (3.0)(9.81) = 29.43\ \text{kPa}.$$
Downward total stress at A under the cut. The 2.0 m clay plug plus $h$ of water: $$\sigma_A = (2.0)(18.64) + h\,\gamma_w = 37.28 + 9.81\,h\ \ (\text{kPa}).$$
Limiting stability ($F = 1$). $$37.28 + 9.81\,h = 29.43 \;\Rightarrow\; h = \dfrac{29.43 - 37.28}{9.81} = -0.80\ \text{m}.$$ The required height is negative: the weight of the clay plug alone already exceeds the artesian uplift, so no water is needed in the cut ($h \ge 0$ satisfies stability). Even with the cut completely dry, $$F = \dfrac{37.28}{29.43} = \boxed{1.27} > 1.$$
With a design margin. If a factor of safety of 1.5 against uplift is required, $1.5(29.43) = 37.28 + 9.81\,h$, which gives $$h = \dfrac{44.15 - 37.28}{9.81} = \boxed{0.70\ \text{m}}.$$ So about 0.7 m of water would be kept in the cut (or the artesian head relieved with pressure-relief wells) if $F = 1.5$ is specified.

Final results — Question 5
Quantity	Value
Clay plug beneath cut, $D$	2.0 m
Uplift pressure at A, $u_A$	29.43 kPa
Weight of clay plug	37.28 kPa
$h$ for $F=1$	−0.80 m ⇒ no water required
$F$ with the cut dry	1.27
$h$ for $F=1.5$	0.70 m

Check: The figure is not drawn to scale. The answer uses the printed dimensions: 7 m of clay, a 5 m cut and a standpipe level 3.0 m above the top of the sand. Measured off the drawing, the standpipe level would sit about 4.4 m above the sand; with that head $u_A = 43.2$ kPa and the limiting height would be $h = 0.60$ m. The printed labels govern.

Topic: Base (bottom) heave of an excavation floored by clay over an artesian aquifer.

Key relations: Heave factor of safety $F = \dfrac{\text{downward total stress at base}}{\text{artesian uplift}} = \dfrac{D\,\gamma_{clay} + h\,\gamma_w}{\gamma_w\,h_{p}}$ with $D$ the clay thickness left below the cut and $h_p$ the pressure head at the base of the clay.

Why this works: The thin clay plug has negligible shear resistance around its perimeter relative to the uplift, so stability is governed by a simple vertical pressure balance at its base. Ponding water in the cut adds downward stress, which is exactly the counterweight needed if the clay alone cannot resist the artesian pressure.

Common pitfall: Using the full clay thickness instead of the clay left below the cut floor, measuring the piezometric head from ground level instead of from the level the standpipe dimension is drawn to, or using the buoyant instead of the total clay weight.

Source: Das & Sobhan, Principles of Geotechnical Engineering, ch. 9 (effective stress, heave in clay due to artesian pressure); Knappett & Craig, Craig’s Soil Mechanics, ch. 3.

Question 5 of 6: Water level required to prevent bottom heave of a cut over artesian sand

Question 5: Water level required to prevent bottom heave of a cut over artesian sand (20 marks)