07-Str-A3 · December 2015

Question 6 of 6: Undrained strength beneath a staged embankment

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

Notes on this paper

Paper format. Engineers Canada / PEO National Examinations, December 2015 — 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: a formula sheet, the rectangular-loading m–n influence chart and a Newmark influence chart whose influence value is printed as $I_N = 0.005$ (200 elements). Values quoted below are taken from those sheets.

Reference texts.

B. M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. — Ch. 6 (compaction), Ch. 7 (permeability), Ch. 9 (stresses in a soil mass), Ch. 10 (consolidation), Ch. 12 (shear strength).
R. F. Craig / J. A. Knappett, Craig’s Soil Mechanics, 8th ed. — Ch. 3 (effective stress, artesian profiles), Ch. 4 (pore pressure coefficients $A$ and $B$), Ch. 5 (shear strength and the choice of test), Ch. 6 (stress distribution).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — Ch. 4 (site investigation and sampling categories), Ch. 6 (excavations and base stability). The Canadian reference for practice, sampling classes and terminology.
K. Terzaghi, R. B. Peck & G. Mesri, Soil Mechanics in Engineering Practice, 3rd ed. — Art. 15–17 (consolidation and seepage), Art. 19–20 (shear strength of clays).

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. (2) The section figure for Question 5 is captioned “Figure 2” on page 4 although the stem of Question 5 calls it Figure 3 — the same drawing, mislabelled in the source. (3) On that figure the “3.0 m” dimension is drawn with its arrowhead at the top of the sand, but at the figure’s own vertical scale (calibrated on the 2 m of clay left beneath the cut) the dimensioned distance is $3.1$ m from the piezometric surface down to the base of the cut, and $5.1$ m down to the sand. The piezometric surface is therefore taken as $3.0$ m above the base of the cut, i.e. $5.0$ m above point $A$; this is also the only reading for which a positive depth of water $h$ exists, and it reproduces the depth of water drawn in the figure. The alternative reading is worked through and dismissed in Question 5.

Question 6: Undrained strength beneath a staged embankment (Value: 20 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.

Given.

Quantity	Value
Density of the fill	$\rho = 1.7\ \text{Mg/m}^{3}$, so $\gamma_{fill} = 16.677\ \text{kN/m}^{3}$
Foundation soil, effective strength parameters	$c' = 50\ \text{kPa}$, $\phi' = 21^{\circ}$
Foundation soil density	$\rho = 1.6\ \text{Mg/m}^{3}$
Pore pressure parameters	$A = 0.2$, $B = 0.98$
Fill height, before and after this stage	3 m raised to 6 m
Lateral stress increment	$\Delta\sigma_3 = \Delta\sigma_1/2$
Drainage during the stage	None — dissipation negligible

Find. The shear strength available immediately beneath the centre line of the embankment at the moment the fill reaches 6 m, and whether the foundation soil is normally or over-consolidated.

Left: the element considered, at original ground level beneath the centre line. Right: the effective-stress Mohr circles before and after the 3 m lift. The lift moves the major principal effective stress out and drags the minor one back, so the circle grows and shifts towards the envelope.

Approach. The element is taken to have consolidated fully under the first 3 m of fill, so its effective stresses at the start of the stage are known. The 3 m lift is applied undrained; Skempton’s expression converts the total stress increments into an excess pore pressure, the effective stresses follow by subtraction, and the Mohr–Coulomb envelope then gives the strength available at that stress state.

Convert the fill density into a unit weight and find the vertical stress increment. With $\gamma_{fill} = 1.7 \times 9.81 = 16.677\ \text{kN/m}^{3}$, raising the fill by $6 - 3 = 3$ m applies $$\Delta\sigma_1 = \gamma_{fill}\,\Delta H = 16.677 \times 3 = 50.03\ \text{kPa} .$$ Beneath the centre line the vertical direction is the major principal direction, which is why this increment is written $\Delta\sigma_1$.
Apply the stated lateral condition. The question fixes the horizontal increment at half the vertical one, so $$\Delta\sigma_3 = \frac{\Delta\sigma_1}{2} = \frac{50.03}{2} = 25.02\ \text{kPa} .$$ The deviator increment is therefore $\Delta\sigma_1 - \Delta\sigma_3 = 25.02\ \text{kPa}$ as well.
Compute the excess pore pressure from Skempton’s equation. Using the expression printed in the hint with $A = 0.2$ and $B = 0.98$, $$\Delta u = B\left[\Delta\sigma_3 + A(\Delta\sigma_1 - \Delta\sigma_3)\right] = 0.98\left[25.02 + 0.2(25.02)\right] = 0.98(30.02) = \boxed{29.42\ \text{kPa}} .$$ The isotropic part contributes 25.02 kPa of that and the deviatoric part only 5.00 kPa, because $A$ is small.
Find the change in each effective stress. Subtracting the excess pore pressure from each total increment, $$\begin{aligned} \Delta\sigma_1' &= 50.03 - 29.42 = +20.61\ \text{kPa}, \\ \Delta\sigma_3' &= 25.02 - 29.42 = -4.40\ \text{kPa} . \end{aligned}$$ The lift therefore reduces the minor effective stress even while it raises the major one. That is the whole danger of rapid filling: the Mohr circle grows in radius and its left-hand end moves the wrong way.
Establish the effective stress state at the end of the lift. Before the lift the element had consolidated under 3 m of fill, so $\sigma_1' = 16.677(3) = 50.03$ kPa and, on the same lateral rule, $\sigma_3' = 25.02$ kPa. Adding the changes from step 4, $$\begin{aligned} \sigma_1' &= 50.03 + 20.61 = 70.64\ \text{kPa}, \\ \sigma_3' &= 25.02 - 4.40 = 20.61\ \text{kPa} . \end{aligned}$$ Both remain positive, which is the check that the staged reading of the question is the admissible one: applying the whole 6 m in a single undrained lift would give $\Delta u = 58.84$ kPa and a minor effective stress of $-8.8$ kPa, which no soil can sustain.
Evaluate the shear strength at that state. The Mohr–Coulomb criterion in effective stresses is $\tau_f = c' + \sigma'\tan\phi'$. Beneath the centre line of a wide embankment the potential slip surface through this element is close to horizontal, so the relevant normal effective stress is the vertical one just found, and $$\tau_f = 50 + 70.64\tan 21^{\circ} = 50 + 70.64(0.3839) = 50 + 27.12 = \boxed{77.1\ \text{kPa}} .$$ This is the undrained strength in the sense the question asks for: the strength available at the instant the fill reaches 6 m, before any dissipation has occurred.
Compare it with the shear stress actually mobilised. The maximum shear stress on the element is the radius of its Mohr circle, $$t = \frac{\sigma_1' - \sigma_3'}{2} = \frac{70.64 - 20.61}{2} = 25.02\ \text{kPa},$$ centred at $s' = (70.64 + 20.61)/2 = 45.63$ kPa. The envelope stands well clear of that circle, so the element is not close to failure at this stage; the margin would close if further lifts were placed without allowing dissipation, and it is that progression, not this single stage, that staged construction is designed to control.
Answer the stress-history question. The soil is over-consolidated, on two independent grounds. The first is the cohesion intercept: a genuinely normally consolidated soil has $c' \approx 0$ because it has never experienced a higher effective stress, and its effective-stress envelope passes through the origin. A value of $c' = 50\ \text{kPa}$ is far too large to be a fitting artefact and points to a soil unloaded from a higher past pressure, or cemented, or desiccated. The second is the pore pressure parameter: $A = 0.2$ lies well below the $0.5$ to $1.0$ band typical of a normally consolidated clay and inside the range expected of a lightly over-consolidated one, meaning the soil has only a weak tendency to contract on shearing. Both observations say the same thing, and they are consistent with the modest $\phi' = 21^{\circ}$ of a stiff plastic clay.

Check — why the strength is quoted on the plane and not as a circle radius. An alternative definition of undrained strength is the radius of the Mohr circle at failure, $s_u = c'\cos\phi' + s'\sin\phi'$, which here would give $46.68 + 16.35 = 63.0$ kPa. That construction assumes the circle can grow about a fixed mean effective stress of 45.63 kPa, and a circle of radius 63.0 kPa centred there would require $\sigma_3' = -17.4$ kPa. Soil cannot sustain that effective tension — it cracks first — so the construction is inadmissible at this stress state, which is a consequence of the unusually large $c'$. The plane strength $\tau_f = c' + \sigma'\tan\phi'$ used above involves no such assumption and is the value to report.

Quantity	Value
Unit weight of the fill	16.677 kN/m$^3$
$\Delta\sigma_1$ for the 3 m lift	50.03 kPa
$\Delta\sigma_3$	25.02 kPa
Excess pore pressure $\Delta u$	29.42 kPa
$\sigma_1'$ and $\sigma_3'$ after the lift	70.64 kPa and 20.61 kPa
Shear strength immediately after the lift	$\tau_f = 77.1$ kPa
Mobilised maximum shear stress	25.02 kPa
Stress history	Over-consolidated ($c' = 50$ kPa, $A = 0.2$)

Topic. Skempton’s pore pressure parameters and undrained strength under a staged fill.

Key relations. $\Delta u = B[\Delta\sigma_3 + A(\Delta\sigma_1 - \Delta\sigma_3)]$; $\sigma' = \sigma - u$; $\tau_f = c' + \sigma'\tan\phi'$; $s' = (\sigma_1' + \sigma_3')/2$, $t = (\sigma_1' - \sigma_3')/2$; $B \to 1$ for a saturated soil, $A_f \approx 0.5$–$1.0$ normally consolidated, $A_f < 0.5$ over-consolidated.

Why this works. Skempton split the pore pressure response into two parts because a soil responds differently to being squeezed all round than to being sheared. The isotropic part, governed by $B$, is essentially complete in a saturated soil because water is far stiffer than the skeleton. The deviatoric part, governed by $A$, is a proxy for the volume change the soil would like to undergo, which is why $A$ is large for a contractive normally consolidated clay and small or negative for a dilatant over-consolidated one. Once $\Delta u$ is known the problem becomes an ordinary effective-stress calculation, and the strength follows from the envelope measured in the laboratory.

Common pitfall. Applying the whole 6 m of fill undrained, which double-counts the first stage and drives the minor effective stress negative. A second trap is using total stresses in the Mohr–Coulomb expression: $c'$ and $\phi'$ are effective-stress parameters and must be paired with an effective normal stress. A third is assuming $B = 1$ when the question supplies $0.98$; here the difference is small, but on a partly saturated fill it is not.

Source. Craig’s Soil Mechanics 8th ed., §4.7 and §5.6 (pore pressure coefficients, embankment loading); Das & Sobhan 9th ed., §12.15–12.17; Terzaghi, Peck & Mesri, Art. 20.

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