07-Str-A3 · December 2014

Question 2 of 6: The three limiting lateral earth pressures

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

Notes on this paper

Paper format: PEO / Engineers Canada National Examinations, December 2014, 07-Str-A3 Geotechnical Materials and Analysis. Three hours, closed book, drawing instruments required, one approved Casio or Sharp calculator. All charts and equations are supplied at the back of the paper (the m–n influence chart on page 8, a Newmark chart with $I_N = 0.005$ on page 9, and a two-page formula sheet). Total value 100 marks over six compulsory questions (20 + 10 + 10 + 20 + 20 + 20). Every question and every sub-part is answered in full below.

Reference texts: Das, Principles of Geotechnical Engineering, 9th ed. (Ch. 6 compaction, Ch. 7 permeability, Ch. 8 seepage and flow nets, Ch. 10 stresses in a soil mass, Ch. 11 consolidation, Ch. 12 shear strength); Knappett & Craig, Craig’s Soil Mechanics, 8th ed. (Ch. 2 seepage and flow-net construction, Ch. 5 shear strength, Ch. 11 lateral earth pressure); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed. (compacted-fill fabric and Proctor behaviour by USCS group); Harr, Groundwater and Seepage, Ch. 4 (closed-form conformal solutions for a single cut-off); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for earth pressures, seepage control and settlement; Canadian Dam Association, Dam Safety Guidelines (2013, rev. 2019); ASTM D698 / D1557 (Proctor compaction), ASTM D4767 (consolidated-undrained triaxial with pore-pressure measurement), ASTM D7181 (consolidated-drained triaxial).

Check — two printing slips in the source, carried exactly as printed. (1) The Question 1 header reads “(4 × 5 = 20 marks)” but five statements (i)–(v) are printed; this solution treats the question as 5 × 4 = 20 marks and answers all five parts. (2) Question 5(a) instructs the candidate to “draw on Figure 4”, while the only section supplied is labelled Figure 3 — the blank grid sheets on pages 6 and 7 are the intended drawing space. Neither slip changes the 100-mark total.

Question 2: The three limiting lateral earth pressures (10 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 three states are not three different soils; they are the same soil element at the same vertical effective stress, distinguished only by how much the wall has been allowed to move. The vertical effective stress $\sigma'_v = \gamma' z$ is fixed by the depth of overburden and does not change. The horizontal effective stress is whatever the wall permits, and the coefficient $K = \sigma'_h / \sigma'_v$ is the bookkeeping device that records it.

Figure 2.1 — the same soil element behind the same wall in the three limiting states. Only the wall movement differs; the vertical stress is unchanged.

At rest. If the wall is rigid enough and restrained enough that no lateral strain occurs at all — a basement wall propped by the floor slabs, a bridge abutment locked in by the deck, an unyielding culvert — the ground stays in the state it consolidated into. For a normally consolidated soil Jaky’s expression $K_0 = 1 - \sin\phi'$ is the standard estimate, raised for over-consolidated ground to $K_0 = (1 - \sin\phi')\,\mathrm{OCR}^{\sin\phi'}$, which can exceed unity in a heavily over-consolidated glacial till. No shear failure has occurred, so the Mohr circle sits entirely inside the failure envelope.

Active. Allow the wall to move away from the retained soil and the horizontal stress falls while the vertical stress stays put. The circle grows to the left until it touches the envelope, at which point the soil is failing and the horizontal stress can drop no further. For a smooth vertical wall retaining level ground the Rankine value is

$$K_a \;=\; \frac{1 - \sin\phi'}{1 + \sin\phi'} \;=\; \tan^{2}\!\left(45^\circ - \frac{\phi'}{2}\right)$$

and the failure planes form at $45^\circ + \phi'/2$ to the horizontal. The movement needed is small — of order $0.001H$ in a dense sand and $0.004H$ in a loose sand — which is why almost every free-standing retaining wall in service is at or near the active state.

Passive. Push the wall into the soil and the horizontal stress rises past the vertical, the two principal stresses exchange roles, and the circle grows to the right until it again touches the envelope. Now

$$K_p \;=\; \frac{1 + \sin\phi'}{1 - \sin\phi'} \;=\; \tan^{2}\!\left(45^\circ + \frac{\phi'}{2}\right) \;=\; \frac{1}{K_a}$$

with failure planes at $45^\circ - \phi'/2$. The critical practical point is that passive resistance needs one to two orders of magnitude more movement to develop — roughly $0.01H$ to $0.05H$ — so full passive pressure in front of a sheet pile or a footing key cannot be relied upon unless that displacement is tolerable. The CFEM accordingly recommends applying a substantial reduction factor to $K_p$ in design.

Figure 2.2 — the three Mohr circles for one element, drawn for $\sigma'_v = 100$ kPa in a $c' = 0$, $\phi' = 30^\circ$ soil. The at-rest circle does not reach the envelope; the active and passive circles are both tangent to it, with $\sigma'_v$ acting as the major principal stress in the active case and as the minor principal stress in the passive case.

Question 2 — the three states summarised, illustrated for $\phi' = 30^\circ$, $c' = 0$, $\sigma'_v = 100$ kPa
State	Wall movement	Coefficient	Value	$\sigma'_h$ (kPa)	Mohr circle
At rest	None	$K_0 = 1 - \sin\phi'$	0.500	50.0	Inside the envelope — no failure
Active	Away from soil, $\approx 0.001H$	$K_a = \tan^{2}(45^\circ - \phi'/2)$	0.333	33.3	Tangent; $\sigma'_1 = \sigma'_v$
Passive	Into soil, $\approx 0.01H$–$0.05H$	$K_p = \tan^{2}(45^\circ + \phi'/2)$	3.000	300.0	Tangent; $\sigma'_1 = \sigma'_h$

Topic. Rankine limiting states and their representation as Mohr circles on a fixed failure envelope.

Key relations. $K_0 = 1 - \sin\phi'$ (Jaky, NC) and $K_0 = (1-\sin\phi')\,\mathrm{OCR}^{\sin\phi'}$ (OC); $K_a = \tan^{2}(45^\circ - \phi'/2)$; $K_p = \tan^{2}(45^\circ + \phi'/2) = 1/K_a$; $\sigma'_1 = \sigma'_3\tan^{2}(45^\circ + \phi'/2) + 2c'\tan(45^\circ + \phi'/2)$ from the formula sheet, which reduces to both Rankine coefficients when $c' = 0$.

Why this works. The failure envelope belongs to the soil and never moves. What moves is the circle, and the only thing that moves it is the wall. Holding $\sigma'_v$ fixed and letting $\sigma'_h$ shrink drives the circle left until it is tangent (active); letting $\sigma'_h$ grow drives it right until it is tangent again (passive). Everything in between is a stable, non-failing state, of which the at-rest condition is the particular one reached with zero lateral strain. Because the passive circle must grow from $\sigma'_h = \sigma'_v$ all the way to $K_p\sigma'_v$, it needs far more strain than the active circle, which only has to shrink a little — that asymmetry is the whole reason passive resistance is treated so cautiously in design.

Common pitfall. Drawing the passive circle with $\sigma'_v$ still as the major principal stress; the principal directions swap over, and the failure planes rotate with them. A close second is applying $K$ to a total vertical stress: the coefficients act on effective stress, and any water pressure must be added separately as a full hydrostatic thrust.

Source. Knappett & Craig, Craig’s Soil Mechanics, 8th ed., Ch. 11; Das, Principles of Geotechnical Engineering, 9th ed., Ch. 13; Canadian Geotechnical Society, CFEM, 4th ed., Ch. 27 (earth pressures and their mobilisation).

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