16-Civ-B3 Geotechnical Design · May 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. EGBC / Engineers Canada National Examination 16-Civ-B3 Geotechnical Design, May 2018. Three hours, open book, any non-communicating calculator. Section A — five discussion questions of 7 marks each, answer any four. Section B — four design questions of 24 marks each, answer any three. Examinable total $4\times 7 + 3\times 24 = 100$ marks. Page 1 Note 6 requires the candidate to identify clearly the source of every design chart and assumed value used, so the provenance of each correlation is named where it is used, not only in the concept notes. All nine questions are solved below, because the set is a study resource rather than a three-hour sitting.
Reference texts for this subject. B. M. Das, Principles of Foundation Engineering, 9th ed. (Cengage) — Ch. 3 (subsurface exploration and SPT corrections), Ch. 4 (bearing capacity), Ch. 5 (settlement, Schmertmann), Ch. 8 (retaining walls), Ch. 11–12 (pile foundations and drilled shafts); B. M. Das, Principles of Geotechnical Engineering, 9th ed. — Ch. 8 (shear strength), Ch. 15 (slope stability); R. D. Holtz, W. D. Kovacs & T. C. Sheahan, An Introduction to Geotechnical Engineering, 2nd ed.; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the Canadian design authority for the factors of safety and serviceability limits quoted here; L. C. Reese & M. W. O'Neill, Drilled Shafts: Construction Procedures and Design Methods (FHWA-HI-88-042).
Check — figure readings. Two dimensions are read from the drawings, as follows. (1) In Figure 2 the “1 m” dimension is the height of the bell: its arrows point inward at the flare, and scaling against the 4 m dimension on the same figure puts the bell base exactly on the 12 m line. The pile is therefore $L = 12$ m long with the bell top at 11 m, not 11 m long with its base floating 1 m clear of the layer base. (2) In Figure 3 the “0.35 m” label carries extension lines from the top and bottom corners of the base slab, so it is the base thickness; the “0.5 m” at the left is measured to the base underside, so only 0.15 m of soil covers the toe. The toe projection is not dimensioned and follows from the printed values as $4.0 - 0.3 - 2.0 = 1.7$ m. Figure 3 is not drawn to scale — its toe is drawn about half its dimensioned length — so the printed numbers govern, as page 1 Note 1 anticipates.
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.
The answer has two parts: a clay possesses a cohesion intercept in the short term, and cohesion — unlike friction — does not need a confining stress to mobilise, so it can hold a vertical face. The mathematics of the critical height then shows how deep that face may be, and the mechanics of pore-pressure dissipation shows why the same face collapses later.
The undrained condition. A clay has a permeability of order $10^{-9}$ m/s, so the time for the pore pressures around an excavation to come to equilibrium, $t \sim T\,H^2/c_v$, runs to months or years for a cut of a few metres. Immediately after digging, therefore, the clay is undrained. Removing the soil reduces the mean total stress, which generates a negative excess pore pressure; the effective stresses on any potential failure surface are temporarily raised, and the mobilised strength is the undrained strength $c_u$ with $\phi_u = 0$. In a partly saturated clay above the water table, matric suction adds a further apparent cohesion of the same kind.
The critical height. Consider a vertical face of height $H$ in a clay of unit weight $\gamma$. In the Rankine active state the horizontal effective stress on the face is
$$\sigma_a = \gamma z\,K_a - 2c_u\sqrt{K_a}, \qquad K_a = \tan^2\!\left(45^{\circ}-\frac{\phi_u}{2}\right) = 1 \ \ \text{for}\ \ \phi_u = 0 .$$
The pressure is therefore $\sigma_a = \gamma z - 2c_u$, which is negative — that is, tensile — down to the depth
$$z_0 = \frac{2c_u}{\gamma}\, ,$$
and positive below it. Because soil cannot carry tension, a tension crack opens to $z_0$; but over that same depth the theoretical push on the face is negative, so no support is required. The net active thrust on a face of height $H$ is the area of the pressure diagram, $P_a = \tfrac12\gamma H^2 - 2c_u H$, and this vanishes when
$$\boxed{\,H_c = \frac{4c_u}{\gamma}\,}$$
which is the classical critical height of an unsupported vertical cut. Taylor's stability chart gives the slightly lower value $H_c = c_u/(0.261\gamma) = 3.83\,c_u/\gamma$ for a $90^{\circ}$ slope with $\phi_u = 0$, because it searches for the true critical circular arc rather than assuming a plane wedge; and if the tension crack is assumed to be full of water the value drops further, to about $2.67\,c_u/\gamma$.
What the numbers mean on site. Take a firm clay with $c_u = 40$ kPa and $\gamma = 18$ kN/m3. Then $H_c = 4(40)/18 = 8.9$ m by the Rankine expression and $3.83(40)/18 = 8.5$ m by Taylor; with a water-filled tension crack, 5.9 m. Applying a factor of safety of 1.5 to the strength gives a working unsupported depth of about $8.5/1.5 \approx 5.7$ m — and every Canadian provincial occupational health and safety regulation caps unsupported excavation depth far below that (commonly 1.2 m before shoring or sloping is required), precisely because the calculation depends on a transient strength.
Why it does not last. Four processes destroy the temporary stability, all of them time dependent. The negative excess pore pressures dissipate as water is drawn toward the face, so effective stresses and strength fall toward their drained values $c'$, $\phi'$ — and for a normally consolidated clay $c'$ is essentially zero, so the vertical face has no long-term resistance at all. Surface water enters the tension crack and adds a hydrostatic thrust $\tfrac12\gamma_w z_0^2$ at the worst possible lever arm. The exposed face swells and softens, and desiccation cracking in dry weather followed by wetting destroys the fabric. Finally, in a fissured clay the mass strength is well below the intact $c_u$, and progressive failure along fissures can bring the face down long before consolidation is complete. This is why the temporary stability of a steep cut in clay is treated as a construction convenience, never as a design condition.
