16-Civ-B3 Geotechnical Design · December 2015

Question 3 of 9: The Rationale of the "Phi Equals Zero" Concept

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

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada National Examinations, December 2015 — 98-Civ-B3 Geotechnical Design. Three hours, OPEN BOOK, any non-communicating calculator. Section A carries five discussion questions of 7 marks each (answer any four); Section B carries four design questions of 24 marks each (answer any three); 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 timed attempt.

Reference texts (98-Civ-B3 / 16-Civ-B3 Geotechnical Design).

B. M. Das, Principles of Foundation Engineering, 9th ed. — subsurface exploration (Ch. 2), bearing capacity (Ch. 3), settlement of shallow foundations (Ch. 5), stress increase in a soil mass (Ch. 6), retaining walls (Ch. 8), pile foundations (Ch. 11).
B. M. Das, Principles of Geotechnical Engineering, 9th ed. — consolidation (Ch. 11), shear strength (Ch. 12), lateral earth pressure (Ch. 13), slope stability (Ch. 15).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for site investigation, in-situ testing, bearing resistance, deep foundations, earth-retaining structures and expansive soils.
R. F. Craig, Craig's Soil Mechanics, 9th ed. — effective stress, earth pressure theory and slope stability.
D. P. Coduto, Foundation Design: Principles and Practices, 3rd ed. — in-situ testing correlations and settlement of shallow foundations.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — shaft adhesion in clay, bored and augered piles, load testing.

Sources of design charts and assumed values (page-1 Note 6). Note 6 of this paper requires the candidate to identify the source of every design chart used and of every value assumed in the absence of data. They are named where used and collected here:

Q6 — Rankine active coefficient for a sloping backfill, Das, Principles of Foundation Engineering, Eq. (8.5); base friction and adhesion mobilisation factors $k_1 = k_2 = \tfrac{2}{3}$ after Das §8.5; Rankine passive coefficient $K_p = \tan^2(45^{\circ} + \phi'_2/2)$. Assumed: stem height $H = 10.0$ m (the exam omits it — see the callout in Q6); reinforced concrete $\gamma_c = 24$ kN/m$^3$ (CFEM §4; CSA A23.3 normal-density concrete); the backfill is fully drained so no water force acts.
Q7 — overburden correction $C_N$ after Liao & Whitman (1986); $\phi'$ from $(N_1)_{60}$ after Peck, Hanson & Thornburn (1974) as fitted by Wolff (1989), cross-checked against Hatanaka & Uchida (1996); bearing capacity factors from Das Table 3.3 (Prandtl–Reissner $N_q$, Vesic $N_{\gamma} = 2(N_q+1)\tan\phi'$), shape factors after De Beer (1970) and depth factors after Hansen (1970), Das Table 3.4; settlement from Meyerhof's (1965) SPT expression, Das Eq. (5.42), cross-checked by Schmertmann's strain-influence method with $E_s = 500(N_{60}+15)$ kPa. Assumed: founding depth $D_f = 2.0$ m; the sand is uniform to at least $2B$ below the base; tolerable settlement 25 mm.
Q8 — compression index from Terzaghi & Peck (1967), $C_c = 0.009(LL-10)$; stress increase by the 2:1 method (Das §6.2) with a Boussinesq rectangular-area cross-check (Das Table 6.6); Simpson weighting of $\Delta\sigma'$ prescribed on the exam paper itself. Assumed: $\gamma_w = 9.81$ kN/m$^3$; the clay is saturated so $e_0 = wG_s$; the sand layers are incompressible relative to the clay.
Q9 — undrained ($\phi_u = 0$) mass procedure, Das, Principles of Geotechnical Engineering, §15.5; drained comparison by the ordinary method of slices and Bishop's simplified method, Das §15.11–15.12, and by the infinite-slope criterion, Das Eq. (15.10). Assumed: no external water force and no seismic loading; the sliding mass is homogeneous.

Section A — discussion questions (7 marks each; answer any four)

Question 3: The Rationale of the "Phi Equals Zero" Concept (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 $\phi_u = 0$ concept is the statement that a saturated clay loaded rapidly enough that no water can leave it behaves as a purely cohesive material whose shear strength is a constant $c_u$, independent of the applied total normal stress. It is not a claim about the soil's true friction; a clay is frictional at the particle level and its drained envelope has $\phi'$ of the order of 20 to 30 degrees. It is a claim about what happens when the volume of a saturated soil is prevented from changing.

The reasoning runs through effective stress. Shear resistance depends on the effective normal stress on the failure plane, $\tau_f = c' + \sigma'\tan\phi'$. If a saturated specimen is loaded so quickly that the pore water cannot drain, the specimen cannot change volume, so any increase in total normal stress on the potential failure plane is carried entirely by the pore water. The pore pressure rises by exactly the amount that the total stress rose, the effective stress is unchanged, and therefore the shear strength is unchanged. Plotting the results of several undrained triaxial tests on identical saturated specimens at different cell pressures gives a set of Mohr circles of the same diameter at different positions, and their common tangent is horizontal: an apparent envelope with $\phi_u = 0$ and an intercept $c_u = q_u/2$, where $q_u$ is the unconfined compressive strength. The magnitude of $c_u$ is not a material constant either — it is fixed by the effective stress that existed in the ground before loading, which is why $c_u$ increases with depth in a normally consolidated deposit and why a clay that has been allowed to consolidate under a preload comes back with a higher $c_u$.

The design rationale that follows is a matter of timing. In a clay of low permeability, construction is fast and drainage is slow, so at the end of construction the soil is essentially in the undrained state and the pore pressures induced by the new load are still present. If the loading is a net increase in stress — an embankment, a footing, a tank — that undrained condition is the most critical one, because the subsequent dissipation of positive excess pore pressure raises effective stress and makes the ground stronger with time. The engineer can therefore design the end-of-construction case with a total-stress analysis in which $\phi_u = 0$, using only $c_u$ and total unit weights, and needs no estimate of the pore pressures at all. That is an enormous simplification: bearing capacity collapses to $q_u = 5.14\,c_u\,F_{cs}F_{cd} + q$, and a circular slip analysis collapses to a single moment equation with no slice bookkeeping, because the mobilised strength is constant around the arc.

A concrete example from practice is the design of a raft or spread footing for a low-rise building on the soft to firm marine clays of the Fraser delta or the Champlain Sea clays of the St. Lawrence lowlands. The site investigation returns a $c_u$ profile from field vane tests corrected for plasticity by Bjerrum's factor, together with unconfined compression and unconsolidated-undrained triaxial results. The undrained bearing capacity of the footing is then computed directly from $c_u$ with $N_c = 5.14$, a factor of safety of 3 is applied, and the result governs the plan area. Long-term consolidation settlement is treated as a separate, entirely independent calculation using $C_c$, $e_0$ and the stress increase; there is no attempt to combine the two. A second familiar example is the short-term stability of a temporary excavation or a cut slope in clay, where the undrained analysis gives the safe unsupported height or the end-of-excavation factor of safety — but here the sign of the pore-pressure change is reversed, and that reversal is the essential caveat on the whole method.

Topic. Total-stress (undrained) analysis of saturated clays.

Key relations. $\tau_f = c' + \sigma'\tan\phi'$ (effective stress); under undrained conditions on a saturated soil $\Delta u = \Delta\sigma$ so $\Delta\sigma' = 0$ and $\tau_f = c_u$ with $\phi_u = 0$; $c_u = q_u/2$; $q_u = 5.14\,c_u\,F_{cs}F_{cd} + q$ for a strip footing.

Why this works. Saturated water is effectively incompressible relative to the soil skeleton. Prevent drainage and you prevent volume change; prevent volume change and the skeleton cannot take up any new load, so the pore water takes all of it. Effective stress therefore does not move, and neither does strength. The $\phi_u = 0$ envelope is a consequence of that constraint rather than a property of the clay.

Common pitfall. Applying the undrained case as if it were always the critical one. For loading problems (footings, fills) the undrained case is critical and the long-term case improves. For unloading problems (excavations, cut slopes, the upstream face of a dam on rapid drawdown) the excess pore pressures are negative, the ground is temporarily stronger than it will be, and the drained long-term case governs. A cut slope checked only with $c_u$ can pass comfortably and fail years later, which is precisely the situation examined in Question 9 of this paper.

Source. Das, Principles of Geotechnical Engineering, 9th ed., §12.6–12.9; Craig's Soil Mechanics, 9th ed., §5.4; CFEM 4th ed., §3 and §10.

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