16-Civ-B3 Geotechnical Design · December 2019
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Examinations, December 2019 — 16-Civ-B3 Geotechnical Design. Three hours, open book, any non-communicating calculator. Section A holds five discussion questions worth 7 marks each (answer any four); Section B holds four design questions worth 24 marks each (answer any three). The examinable total is therefore 4 × 7 + 3 × 24 = 100 marks. Page-1 Note 3 sets the answer-any-four / any-three rule, and Note 6 requires the candidate to name the source of every design chart and of every assumed value — so every chart read, correlation and assumption below is attributed where it is used. All nine questions are solved here, because the set is a study resource rather than a timed sitting.
Reference texts. B. M. Das, Principles of Foundation Engineering, 9th ed. (bearing capacity, elastic settlement, retaining walls, drilled shafts); B. M. Das, Principles of Geotechnical Engineering, 9th ed. (shear strength, lateral earth pressure, slope stability); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. (Canadian practice, factors of safety, in-situ testing); R. F. Craig, Craigʹs Soil Mechanics, 9th ed. (effective stress, undrained strength); D. P. Coduto, Foundation Design: Principles and Practices, 2nd ed. (shallow-foundation design, settlement serviceability).
Check — conventions used throughout this paper. Unit weights printed on the figures are treated as bulk (saturated below any water table); effective unit weights use γw = 9.81 kN/m3. Reinforced concrete is taken at γc = 24 kN/m3 (CFEM 4th ed.; the exam gives no value), and Question 8 shows that the conclusion is unchanged anywhere in the 23–25 kN/m3 range. Where the paper omits a number the solution needs, the assumption is stated at the point of use and its influence on the answer is quantified, as page-1 Notes 1 and 7 invite.
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.
A cut excavated quickly in a saturated clay is a total-stress problem. The clay has too low a permeability to drain during the few days the face stands open, so the appropriate strength is the undrained shear strength cu mobilised at constant volume, and the appropriate friction angle is φu = 0. That single substitution is what makes the derivation short, because it collapses the Rankine active coefficient to unity.
Derivation. The Rankine active pressure on a vertical, frictionless face in a soil with cohesion is the expression drawn in Figure 1 — the triangular term driven by vertical stress, less the uniform term contributed by cohesion:
$$\begin{aligned} \sigma_a &= \sigma_v^{\prime} K_a - 2c^{\prime}\sqrt{K_a} \\ K_a &= \tan^2\!\left(45^{\circ} - \frac{\varphi^{\prime}}{2}\right) \end{aligned}$$Setting φu = 0 gives Ka = 1 and replaces cʹ by cu, so with σv = γz the pressure at depth z becomes the simple linear function that the third panel of Figure 1 shows:
$$\sigma_a(z) = \gamma z - 2c_u$$This is negative — tensile — near the surface, and clay cannot sustain tension against a free face. The depth at which it changes sign defines the tension crack:
$$\gamma z_c - 2c_u = 0 \;\Longrightarrow\; z_c = \frac{2c_u}{\gamma}$$The cut stands unsupported for as long as the net thrust it must resist is zero, that is, for as long as the tensile area above zc exactly cancels the compressive area below it. Integrating the pressure over the full face height H and setting the total to zero:
$$P_a = \int_0^{H}\left(\gamma z - 2c_u\right)\mathrm{d}z = \tfrac{1}{2}\gamma H^{2} - 2c_u H = 0$$Dividing through by H and solving gives the critical height, which is exactly twice the tension-crack depth:
$$\boxed{H_c = \frac{4c_u}{\gamma} = 2z_c}$$Given. The numerical part of the question asks about a cut H = 8.0 m deep in a saturated clay of undrained shear strength cu = 30 kPa. The paper does not state a unit weight, so a saturated bulk unit weight γ = 18 kN/m3 is adopted, which is representative of a soft to firm saturated clay (CFEM 4th ed., Table 3.1).
Find. Whether the 8 m cut will stand unsupported, and if not, by how much it is over-deep.
Approach. Evaluate the boxed critical height for the given strength, compare it with the required 8 m, and then apply a factor of safety to convert the theoretical limit into a depth that could actually be left open.
Answer to the comment. Support is unequivocally required. The 8 m cut exceeds the theoretical critical height of 6.67 m, so it is not a question of an inadequate margin — the face is beyond collapse as drawn. In practice the excavation must either be shored over its full depth (soldier piles and lagging, sheet piling, or a trench box for a narrow cut), or benched and battered back to a stable geometry, or dewatered and cut in stages with the upper 4 m or so free-standing and the remainder supported. The tension crack of 3.33 m is a second, independent reason for support: once it fills with surface water it applies a hydrostatic thrust of ½γwzc2 ≈ 55 kN/m that the derivation above assumes is absent, which is why a clay cut that has stood for a week can fail during the first heavy rain.
| Quantity | Value |
|---|---|
| Critical height (wedge derivation) | Hc = 4cu/γ = 6.67 m |
| Tension-crack depth | zc = 3.33 m |
| Critical height (Taylor, Ns = 0.261) | 6.39 m |
| Depth safe to leave unsupported (FS = 1.5) | 4.44 m |
| Verdict on the 8 m cut | support required over the full depth |
Check — sensitivity to the assumed unit weight. The unit weight is assumed, not given, so it is worth showing the conclusion does not depend on it: Hc = 6.67 m at 18 kN/m3, 6.32 m at 19 kN/m3 and 6.00 m at 20 kN/m3. Every value in the plausible range for a saturated clay falls short of 8 m, so the answer — support is required — is invariant. Note also that the expression is a short-term result: as the clay drains, cʹ falls towards zero and the long-term stable vertical height tends to zero, so an unsupported clay face has no permanent stability whatever its depth.
