16-Civ-B3 Geotechnical Design · May 2017

Question 4 of 9: Why a steep excavation in clay stands unsupported for a short time

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

Notes on this paper

Paper format. National Examinations, May 2017 — 16-Civ-B3 Geotechnical Design; three hours, open book, any non-communicating calculator. Section A holds five discussion questions worth 7 marks each of which four are marked; Section B holds four design questions worth 24 marks each of which three are marked, so the examinable total is 4 × 7 + 3 × 24 = 100 marks. All nine questions are worked below, because the set is a study resource rather than a marked script.

Reference texts. B. M. Das, Principles of Foundation Engineering, 7th–9th ed. (Cengage); B. M. Das, Principles of Geotechnical Engineering; R. F. Craig, Craig's Soil Mechanics, 8th ed. (Knappett & Craig); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed.; J. E. Bowles, Foundation Analysis and Design; ASTM D1586 (SPT) and D5778 (CPT).

Source of charts and assumed values (page 1, Note 6). Every design coefficient used below is named where it is used: Terzaghi bearing-capacity factors from Das, Principles of Foundation Engineering, Table 3.1 (values computed by Kumbhojkar, 1993); Vesic/Reissner factors and the shape, depth and inclination factors from Das Table 3.4 and Eqs. (3.19)–(3.26); drilled-shaft adhesion factor alpha* = 0.55 from Reese & O'Neill (1989) as tabulated by Das, Chapter 12; bearing factor Nc* = 9 from Skempton (1951); earth-pressure coefficient for downdrag K' = 1 − sin(phi') from Das, Chapter 11; Janbu's bearing-capacity number for the pile point from Das Eq. (11.33). Assumed values (adhesion ratio, pile spacing, rigidity index) are stated in a callout beside the step that uses them.

Question 4: Why a steep excavation in clay stands unsupported for a short time (7 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.

The short answer is that a saturated clay, loaded or unloaded quickly, has a shear strength that is independent of the confining stress removed by the excavation, and Rankine's active-pressure distribution for such a material is negative over the upper part of the cut. Over a finite depth the negative and positive parts of the pressure diagram cancel, so the net thrust that a support system would have to resist is zero and no support is required. That state is temporary because it depends on pore pressures that are out of equilibrium, and equilibrium is restored by seepage.

Undrained active pressure on an unsupported vertical cut in clay: the tension zone of depth z₀ exactly cancels the positive block when H = H_c.

The lateral-earth-pressure argument. For a saturated clay sheared without drainage the failure envelope in total stress is horizontal: tau_f = c_u with phi_u = 0. Rankine's active earth-pressure coefficient is then

$$K_a=\tan^{2}\!\left(45^\circ-\frac{\phi_u}{2}\right)=\tan^{2}45^\circ=1$$

and the active pressure on a vertical, smooth, frictionless plane at depth z behind the face is

$$\sigma_a=K_a\,\gamma z-2c_u\sqrt{K_a}=\gamma z-2c_u$$

This expression is negative from the surface down to the depth at which it changes sign, the theoretical tension-crack depth

$$z_0=\frac{2c_u}{\gamma}$$

Soil cannot sustain tension against a vertical face, so over that depth the clay simply stands by itself; below it the pressure grows linearly. Integrating the whole diagram from the surface to a depth H gives the net active thrust

$$P_a=\tfrac{1}{2}\gamma H^{2}-2c_u H$$

and setting the thrust to zero gives the depth at which the cut needs no support at all — the critical or unsupported height,

$$\boxed{H_c=\frac{4c_u}{\gamma}}$$

Taking a typical firm Toronto-area clay with c_u = 50 kPa and gamma = 19 kN/m³, the tension zone extends to z₀ = 5.26 m and the theoretical unsupported height is H_c = 10.53 m. Taylor's stability-number solution for a vertical slope in a phi_u = 0 material, which uses a circular arc rather than a plane and accounts for the tension crack, gives the slightly lower value H_c = 3.85 c_u/gamma = 10.13 m. Applying the customary factor of safety of 1.5 to the critical height leaves a working unsupported depth of about 7.0 m — which is why a trench a few metres deep in a stiff clay routinely stands open while a trench of the same depth in a clean sand collapses the moment the excavator withdraws.

Why the strength is available at all, and why it does not last. Excavation is an unloading process. Removing the soil above a point reduces the total mean stress there, and in a saturated clay of low permeability that reduction is taken almost entirely by the pore water: the pore pressure falls, often below the pre-existing equilibrium value and sometimes below atmospheric. Since sigma' = sigma − u, a fall in u at nearly constant total stress means the effective stress, and therefore the frictional component of strength, is temporarily higher than the long-term value. The clay is, in effect, borrowing strength from suction in its pore water. That is what the undrained strength c_u represents in a total-stress analysis.

The borrowed strength is repaid. Water flows towards the zones of low pore pressure from the surrounding ground and from surface infiltration, the negative excess pore pressures dissipate, the effective stresses fall to their long-term values, and the operative strength falls from c_u to the drained envelope c' + sigma' tan(phi'). For a normally consolidated or lightly overconsolidated clay c' is essentially zero, so in the long term a vertical face has no strength at all at zero effective confining stress and must fail. The time available is governed by consolidation: with the characteristic time t proportional to H²/c_v, a stiff clay of coefficient of consolidation around 1 m²/yr and a 5 m cut has a characteristic swelling time of years, whereas a soft silty clay may soften in weeks. Three further mechanisms shorten the safe period in practice: a tension crack that fills with rainwater applies a hydrostatic thrust of one half gamma_w z₀² exactly where the analysis assumed zero pressure; fissures in an overconsolidated clay let water in far faster than the intact permeability suggests and reduce the mass strength below the laboratory value; and desiccation, vibration and surcharge from plant near the crest all degrade the crest zone.

Check — practice, not theory, governs the site. The calculation above explains why the face stands; it is not an authorisation to leave it unsupported. Canadian occupational health and safety law sets the requirement independently of the analysis: for example the BC Occupational Health and Safety Regulation, Part 20, requires an excavation deeper than 1.2 m in which a worker will work to be sloped, benched or supported unless it is in stable rock or a professional engineer has certified the design, and the equivalent provisions of the Ontario Regulation 213/91 for construction projects classify soil types and prescribe the allowable unsupported geometry. The engineering answer is that the cut stands because of temporary undrained strength; the regulatory answer is that temporary strength is not a design basis for worker safety.

Topic. Undrained (phi = 0) lateral earth pressure and the temporary nature of excavation stability in clay.

Key relations. K_a = tan²(45° − phi/2), equal to 1 when phi_u = 0; sigma_a = gamma z − 2c_u; tension-crack depth z₀ = 2c_u/gamma; net thrust P_a = 0.5 gamma H² − 2 c_u H; critical height H_c = 4c_u/gamma, or 3.85 c_u/gamma from Taylor's circular-arc solution; effective stress sigma' = sigma − u; swelling time t proportional to H²/c_v.

Why this works. In a total-stress analysis the clay is treated as a purely cohesive material, so the active pressure diagram is the self-weight line gamma z shifted down by the constant 2c_u. Because the shift is constant and the self-weight term grows linearly, there is always a depth over which the diagram is negative and a depth at which the areas above and below the sign change are equal. That second depth is the unsupported height. The physical content of the constant 2c_u is the suction the clay holds in its pore water after unloading, which is why the result is valid only while that suction persists.

Common pitfall. Treating H_c as a design depth rather than a failure depth, and ignoring the tension crack. If the crack fills with water the thrust is no longer zero at H = H_c and the face fails at a smaller height; the usual design rule is to apply a factor of safety of at least 1.5 to H_c and to assume the crack is water-filled unless it is demonstrably protected.

Source. Craig's Soil Mechanics, 8th ed., Sections 6.2–6.3 (Rankine active state, tension cracks) and Chapter 12 (Taylor's stability numbers); Das, Principles of Geotechnical Engineering, Sections 13.4 and 15.5.

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