07-Str-A3 · May 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. Engineers Canada / PEO National Examinations, May 2016 — 07-Str-A3 Geotechnical Materials and Analysis. Closed book, three hours, 100 marks, one approved Casio or Sharp calculator, drawing instruments required. All six questions are compulsory and are weighted 20 / 10 / 10 / 20 / 20 / 20. The paper carries its own appendix (pages 8–11): a formula sheet, the rectangular-loading m–n influence chart and a Newmark influence chart whose influence value is printed on the formula sheet as $\sigma_z = 0.005\,N q$, i.e. $I_N = 0.005$ over 200 elements. Values quoted below are taken from those sheets.
Reference texts.
Check — three readings of the printed paper, carried as stated.
(1) Question 1 is headed “(4 x 5 = 20 marks)” but prints five lettered parts (i)–(v). All five are answered and the header total of 20 marks is kept as printed; the mismatch is a printing slip in the source, not a missing part.
(2) Figure 1 (part (ii)) prints no dimension for the depth of the permeable stratum. The dam base is dimensioned $L = 15$ m and the section is drawn to a single horizontal scale, so the stratum has been scaled off the drawing at $T \approx 5.6$ m ($T/L = 0.37$). The answer to (ii) does not depend on that number — it is read from the printed flow net — but the independent numerical check below does, and the sensitivity is stated where it is used.
(3) In Figure 4 the stem of Question 5(b) says point $A$ is at the “back of the piling”, while the dot is drawn on the upstream (headwater) face of the sheet pile, 1 m below ground. The drawn face is answered, and the value on the downstream face at the same elevation is reported alongside it.
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 and not three different depths — they are the same element of soil at the same depth, with the same vertical effective stress $\sigma_v'$, responding to three different amounts of wall movement. That is the single idea the question is testing, and the figure below is drawn to make it: all three Mohr circles share the same right-hand or left-hand point at $\sigma_v' = 100$ kPa, and only the horizontal stress changes.
At rest. If the wall does not move at all, the soil behind it undergoes no lateral strain. The horizontal effective stress is then whatever the deposition history left in place, and the soil is nowhere near failure — its Mohr circle is small and sits well inside the envelope. For a normally consolidated soil Jaky’s expression is used,
$$K_0 = 1 - \sin\phi', \qquad \sigma_h' = K_0\,\sigma_v',$$which for $\phi' = 30^\circ$ gives $K_0 = 0.50$ and $\sigma_h' = 50$ kPa against $\sigma_v' = 100$ kPa. An over-consolidated soil retains locked-in horizontal stress and takes $K_0 \approx (1-\sin\phi')\,\text{OCR}^{\sin\phi'}$, which can exceed unity in a heavily over-consolidated clay. The at-rest condition governs basement walls, braced excavations, bridge abutments on piles and any other structure stiff enough that the soil cannot strain laterally.
Active. If the wall yields away from the soil, the retained mass stretches horizontally. The vertical stress is unchanged, so $\sigma_h'$ falls, the Mohr circle grows to the left, and at a small movement it touches the failure envelope. Beyond that the soil cannot shed any more horizontal stress: it is at failure, and the wedge behind the wall is on the point of sliding down and forward. The limiting value is Rankine’s
$$K_a = \tan^2\!\left(45^\circ - \frac{\phi'}{2}\right) = \frac{1-\sin\phi'}{1+\sin\phi'}, \qquad \sigma_a' = K_a\sigma_v' - 2c'\sqrt{K_a},$$giving $K_a = 0.333$ and $\sigma_a' = 33.3$ kPa for the same element. In the active state the vertical stress is the major principal stress, and the failure planes rise at $45^\circ + \phi'/2$ from the horizontal. The movement required is small — of order $0.001H$ for a dense sand and $0.004H$ for a loose one — which is why almost every free cantilever or gravity wall is designed for active pressure.
Passive. If instead the wall is pushed into the soil, the mass is compressed horizontally. Now $\sigma_h'$ rises, and the circle first shrinks to a point (when $\sigma_h' = \sigma_v'$, the isotropic state), then grows again on the other side until it touches the envelope with the horizontal stress as the major principal stress. The limiting value is
$$K_p = \tan^2\!\left(45^\circ + \frac{\phi'}{2}\right) = \frac{1+\sin\phi'}{1-\sin\phi'}, \qquad \sigma_p' = K_p\sigma_v' + 2c'\sqrt{K_p},$$which for $\phi' = 30^\circ$ gives $K_p = 3.0$ and $\sigma_p' = 300$ kPa — nine times the active value. The failure planes are flatter, at $45^\circ - \phi'/2$. The movement needed to mobilise full passive resistance is an order of magnitude larger than for the active case, typically 2–5 % of the wall height, which is why designers rarely take the full $K_p$: the passive resistance in front of a sheet pile toe is normally divided by a factor of 1.5–2, or the wall is checked at a strain compatible with the active side.
| Condition | Wall movement | Coefficient ($\phi' = 30^\circ$) | $\sigma_h'$ at $\sigma_v' = 100$ kPa | State of the circle |
|---|---|---|---|---|
| At rest | none | $K_0 = 1-\sin\phi' = 0.50$ | 50.0 kPa | inside the envelope |
| Active | away from soil, $\approx 0.001H$ | $K_a = \tan^2(45^\circ-\phi'/2) = 0.333$ | 33.3 kPa | tangent, $\sigma_v'$ major |
| Passive | into soil, $\approx 0.02$–$0.05H$ | $K_p = \tan^2(45^\circ+\phi'/2) = 3.00$ | 300 kPa | tangent, $\sigma_h'$ major |
The ordering $K_a < K_0 < K_p$ is not a coincidence of the numbers chosen: the active and passive values are the two extremes the soil can reach without failing, and the at-rest value is necessarily somewhere between them.
