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16-Civ-B3 Geotechnical Design · December 2013

Question 3 of 9: Explaining active failure behind a yielding wall with Mohr circles

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

Notes on this paper

Paper format. National Examinations, December 2013 — 98-Civ-B3 Geotechnical Design. Three hours, open book, any non-communicating calculator. Section A holds five 7-mark discussion questions (answer any four); Section B holds four 24-mark design questions (answer any three), so the examinable total is 4 × 7 + 3 × 24 = 100 marks. Every one of the nine questions is answered here, because the set is a study resource rather than a sitting.

Reference texts. B. M. Das, Principles of Foundation Engineering (9th ed.) and Principles of Geotechnical Engineering (9th ed.); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed.) — the governing Canadian reference for foundation practice; R. F. Craig, Craig's Soil Mechanics (9th ed.); D. P. Coduto, Foundation Design: Principles and Practices (3rd ed.).

Sources of design charts and assumed values (paper Note 6). The paper requires every chart and assumed value to be identified. Bearing-capacity factors are Terzaghi’s (Das, Foundation Engineering, Table 3.1, with Nγ after Kumbhojkar 1993); the pile end-bearing factor Nq* is read from Meyerhof’s chart (Das Fig. 11.14) and cross-checked against Janbu’s closed-form expression; the adhesion factor α is Das Table 11.5 (Terzaghi, Peck & Mesri); the strain-influence distribution is Schmertmann, Hartman & Brown (1978). Assumed values — specific gravity of solids Gs = 2.65 for the Question 8 backfill, base friction δ = ⅔φ′ and base adhesion ca = ⅔c′ (CFEM §24), and a driving-parameter value K = 1.4K0 for a high-displacement driven pile (Das §11.11) — are flagged where they are used.
Check — Question 6 text and figure disagree. The printed text of Question 6 states a 1 m × 1 m square footing with γ = 20 kN/m³ and φ = 36°, while Figure 3 is drawn for a 2.5 m footing with γ = 18 kN/m³ and φ′ = 38°. The question text governs (it is the instruction to the candidate); the figure is used only for the embedment Df = 1.5 m and for the CPT modulus profile, which the text does not restate. The alternative reading is noted at the end of the answer.

Question 3: Explaining active failure behind a yielding wall with Mohr circles (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.

Consider the soil element at depth $z$ behind the wall. Before the wall moves, the backfill is in the at-rest condition: the vertical effective stress is $\sigma^{\prime}_0 = \gamma z$ and the horizontal effective stress is $\sigma^{\prime}_h = K_0 \sigma^{\prime}_0$ with $K_0 = 1 - \sin\phi^{\prime}$ for a normally consolidated granular fill. Both are principal stresses because the wall face is smooth and vertical, so on the Mohr diagram the state is a circle whose diameter runs from $K_0\sigma^{\prime}_0$ to $\sigma^{\prime}_0$. That circle sits well inside the Mohr–Coulomb envelope $\tau = c^{\prime} + \sigma^{\prime}\tan\phi^{\prime}$, so no element is at failure and the soil is elastic.

σ′τMohr–Coulomb envelope τ = σ′ tan φ′σ′₋ (constant)K₀σ′₋Kₐσ′₋wall yields outwards (Δx)the circle grows until it touches the envelope
Figure Q3 — the stress path of an element behind a wall that yields outwards. $\sigma^{\prime}_v$ is held constant while $\sigma^{\prime}_h$ falls from $K_0\sigma^{\prime}_v$ to $K_a\sigma^{\prime}_v$; the circle expands leftwards until it touches the failure envelope.

Now let the wall rotate outwards about its base through the movement $\Delta x$ shown in the figure. The soil above the toe is free to expand horizontally, but nothing has changed the weight of overburden, so the vertical stress $\sigma^{\prime}_0$ stays constant while the horizontal stress falls. On the Mohr diagram the right-hand end of the diameter is pinned and the left-hand end travels towards the origin, so the circle grows steadily larger. Its radius, which is the maximum shear stress in the element, increases with every increment of wall movement.

The circle can only grow until it becomes tangent to the Mohr–Coulomb envelope. At tangency the soil has mobilised its full shear strength on one pair of planes and cannot sustain any further reduction of $\sigma^{\prime}_h$: this is the Rankine active state, and the horizontal stress has reached its minimum possible value

$$\begin{aligned}\sigma^{\prime}_a &= K_a \sigma^{\prime}_0 - 2c^{\prime}\sqrt{K_a} \\ K_a &= \tan^{2}\!\left(45^{\circ} - \tfrac{\phi^{\prime}}{2}\right).\end{aligned}$$

The geometry of the tangent point is what produces the failure planes drawn in Figure 2. On a Mohr circle, the pole construction places the plane of failure at an angle $45^{\circ} + \phi^{\prime}/2$ to the plane on which the major principal stress acts. Here the major principal stress is vertical, so the slip planes rise at $45^{\circ} + \phi^{\prime}/2$ from the horizontal — equivalently at $45^{\circ} - \phi^{\prime}/2$ from the vertical, which is exactly how the figure labels them. They occur as a conjugate pair, and the wedge bounded by the wall, the ground surface and the plane through the heel is the classical Rankine active wedge that slides down and towards the wall as the wall yields.

Two consequences follow directly from the diagram. First, the thrust on the wall decreases as the wall moves out, from the at-rest value $\tfrac12 K_0\gamma H^2$ to the active value $\tfrac12 K_a\gamma H^2$, which for $\phi^{\prime} = 30^{\circ}$ is a fall of about a third; designing a wall that is free to yield for at-rest pressure is therefore uneconomic, while designing a braced or restrained wall for active pressure is unsafe. Second, the movement needed to reach the active state is small — of order $0.001H$ in dense sand and $0.004H$ in loose sand — so for an ordinary cantilever wall founded on compressible soil the active condition is reached essentially on backfilling. If the wall were instead pushed into the fill, the left-hand end of the diameter would sweep past $\sigma^{\prime}_0$ and the circle would grow towards the envelope on the other side, giving the passive state with slip planes at $45^{\circ} - \phi^{\prime}/2$ to the horizontal and a much larger required movement.