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16-Civ-B3 Geotechnical Design · December 2014

Question 6 of 10: Design axial capacity of a driven H-pile in layered clay

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 2014 — 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 five design questions of 24 marks each (answer any three); the examinable total is 4 × 7 + 3 × 24 = 100 marks. All ten 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).

Sources of charts and assumed values (page-1 Note 6). Note 6 of this paper requires the candidate to identify the source of every design chart and every assumed value. Each chart reading and each assumption below is therefore named where it is used, and the values assumed in the absence of data are collected here:

  • Q6 — adhesion factor α from Das, Principles of Foundation Engineering, Table 11.6 (Terzaghi, Peck & Mesri form, α against $c_u/p_a$); $\lambda$ from Vijayvergiya & Focht (1972) as tabulated by Das, Table 11.7.
  • Q7 — overburden correction $C_N$ from Liao & Whitman (1986); $\phi'$ from Wolff (1989) and from Hatanaka & Uchida (1996), both reproduced in Das, Ch. 2; settlement-controlled bearing pressure from Meyerhof (1965) as given by Das, Ch. 5, used only as a serviceability check because the question forbids direct correlations of bearing capacity to penetration index. Table I prints the blow counts as field values $N_f$; with no hammer data they are converted as $N_{60} = N_f$, i.e. a safety hammer at the reference 60 per cent energy ratio with borehole, sampler and rod-length factors of 1 (Das, Ch. 2, hammer-efficiency and correction-factor tables).
  • Q8 — embankment influence factor from Osterberg (1957), reproduced as Das Fig. 6.24; the closed form of that chart is used so the reading carries no chart-scaling error.
  • Q9 — Meyerhof general bearing-capacity equation with the shape factors of De Beer (1970) and the depth factors of Hansen (1970), as set out in Das, Ch. 3.
  • Q10 — Coulomb active earth-pressure coefficient, Das Eq. 13.31; unit weight of the mass-concrete wall assumed $\gamma_c = 24\ \text{kN/m}^3$ (CFEM 4th ed., normal-density concrete), the only value the figure does not supply.

Question 6: Design axial capacity of a driven H-pile in layered clay (24 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. A 0.3 m steel H-pile driven 10 m into a two-layer clay deposit, with the shaft resistance to be taken on the enclosing 0.3 m by 0.3 m box perimeter.

Given data — Question 6
QuantitySymbolValue
Pile section (box dimensions)$d$0.3 m by 0.3 m
Embedded length$L$10 m
Layer 1 (0 to 5 m), normally consolidated clay$c_{u1}$, $\gamma_1$50 kPa, 18 kN/m3
Layer 2 (5 to 10 m), slightly overconsolidated clay$c_{u2}$, $\gamma_2$100 kPa, 18 kN/m3
Factor of safety$FS$2

Find. The allowable (design) axial compressive capacity $Q_{all} = Q_u / FS$, where the ultimate capacity is the sum of the shaft and the toe resistance.

Q (applied) ground surface Layer 1 — NC clay γ = 18 kN/m³, cu = 50 kPa cu/pa = 0.5, α = 0.68 Layer 2 — slightly OC clay γ = 18 kN/m³, cu = 100 kPa cu/pa = 1.0, α = 0.48 Qs1 = 204 kN Qs2 = 288 kN Qp = 9 cu Ap = 81 kN 0 m 5 m 10 m (toe) perimeter p = 4(0.3) = 1.2 m Ap = 0.3 x 0.3 = 0.09 m²
Figure 6.1 — Driven H-pile, box perimeter assumption. Skin friction acts on the enclosing 0.3 m by 0.3 m rectangle (perimeter 1.2 m) and end bearing on the enclosed plan area 0.09 m2, the standard plugged-section idealisation for a driven H-pile in clay.

Approach. Compute the shaft resistance layer by layer with the total stress $\alpha$-method, $f = \alpha c_u$, add the undrained end bearing $Q_p = 9 c_u A_p$, and divide the sum by the factor of safety of 2; then repeat the shaft calculation with the independent $\lambda$-method as the cross-check that Note 6 of the paper effectively demands.

  1. Establish the section constants. The question directs that skin friction be taken on the enclosing rectangle, so the pile is treated as a plugged 0.3 m square box: $$p = 4d = 4(0.3) = 1.2\ \text{m}, \qquad A_p = d^2 = (0.3)^2 = 0.09\ \text{m}^2$$ This is the standard idealisation for an H-pile driven into clay, where the soil trapped between the flanges moves with the pile and the failure surface is the enclosing rectangle rather than the true steel perimeter.
  2. Select the adhesion factor for each layer. The $\alpha$-method writes the unit shaft friction as $f = \alpha c_u$, with $\alpha$ read against the normalised strength $c_u/p_a$, where $p_a = 100\ \text{kPa}$ is atmospheric pressure (Das, Principles of Foundation Engineering, Table 11.6). For Layer 1, $c_u/p_a = 50/100 = 0.5$, which falls between the tabulated points $(0.4,\ 0.74)$ and $(0.6,\ 0.62)$; linear interpolation gives $$\alpha_1 = 0.74 + (0.62 - 0.74)\,\frac{0.5 - 0.4}{0.6 - 0.4} = 0.68$$ For Layer 2, $c_u/p_a = 100/100 = 1.0$ reads directly from the table as $\alpha_2 = 0.48$. The reduction with increasing strength is physical: a stiffer clay remoulds and generates larger excess pore pressures against the shaft during driving, and recovers a smaller proportion of its intact strength.
  3. Shaft resistance of Layer 1 (0 to 5 m). With $f_1 = \alpha_1 c_{u1} = 0.68 (50) = 34.0\ \text{kPa}$ acting over the box perimeter for 5 m, $$Q_{s1} = \alpha_1 c_{u1}\, p\, L_1 = 0.68 (50)(1.2)(5) = 204.0\ \text{kN}$$
  4. Shaft resistance of Layer 2 (5 to 10 m). Here $f_2 = 0.48 (100) = 48.0\ \text{kPa}$, so $$Q_{s2} = \alpha_2 c_{u2}\, p\, L_2 = 0.48 (100)(1.2)(5) = 288.0\ \text{kN}$$ Although the second layer is twice as strong, its shaft contribution is only some 40 per cent larger, because the adhesion factor has fallen from 0.68 to 0.48. Summing the two layers, $$\boxed{Q_s = Q_{s1} + Q_{s2} = 204.0 + 288.0 = 492.0\ \text{kN}}$$
  5. End bearing at the toe. For a pile in saturated clay loaded in the short term the bearing-capacity factor is $N_c^{*} = 9$, and the toe sits in the slightly overconsolidated layer with $c_{u2} = 100\ \text{kPa}$: $$Q_p = 9\, c_{u2}\, A_p = 9 (100)(0.09) = 81.0\ \text{kN}$$ The toe therefore supplies only about 14 per cent of the ultimate capacity, which is the expected result for a small-section friction pile in clay.
  6. Ultimate and allowable capacity. Adding the two components and applying the specified factor of safety, $$Q_u = Q_s + Q_p = 492.0 + 81.0 = 573.0\ \text{kN}$$ $$\boxed{Q_{all} = \frac{Q_u}{FS} = \frac{573.0}{2} = 286.5 \approx 287\ \text{kN}}$$
  7. Independent check by the $\lambda$-method. Vijayvergiya and Focht (1972) express the average unit shaft friction over the whole embedded length as $f_{av} = \lambda\left(\bar{\sigma}'_v + 2\bar{c}_u\right)$. With a uniform $\gamma = 18\ \text{kN/m}^3$ and no water table stated, the mean effective vertical stress over 10 m is $\bar{\sigma}'_v = \gamma L/2 = 18(10)/2 = 90\ \text{kPa}$, and the length-weighted mean strength is $\bar{c}_u = \left[50(5) + 100(5)\right]/10 = 75\ \text{kPa}$. Das, Table 11.7, gives $\lambda = 0.245$ for $L = 10\ \text{m}$, so $$f_{av} = 0.245\left[90 + 2(75)\right] = 0.245(240) = 58.8\ \text{kPa}$$ $$Q_{s,\lambda} = f_{av}\, p\, L = 58.8(1.2)(10) = 705.6\ \text{kN}$$ giving $Q_{u,\lambda} = 705.6 + 81.0 = 786.6\ \text{kN}$ and $Q_{all,\lambda} = 393\ \text{kN}$.

The $\lambda$-method is therefore some 37 per cent less conservative than the $\alpha$-method on this profile. That spread is normal — the two methods were calibrated against different pile databases, the $\lambda$-method largely against long offshore piles — and the correct engineering response is to adopt the lower value for design and to confirm it by load testing, which is exactly what CFEM requires before a resistance factor better than the default may be used.

Final results — Question 6
QuantitySymbolValue
Box perimeter / toe area$p$ / $A_p$1.2 m / 0.09 m2
Adhesion factors (Das Table 11.6)$\alpha_1$ / $\alpha_2$0.68 / 0.48
Shaft resistance, Layer 1 (0 to 5 m)$Q_{s1}$204.0 kN
Shaft resistance, Layer 2 (5 to 10 m)$Q_{s2}$288.0 kN
Total shaft resistance$Q_s$492.0 kN
End bearing$Q_p$81.0 kN
Ultimate axial capacity$Q_u$573.0 kN
Design (allowable) axial capacity, FS = 2$Q_{all}$287 kN
Cross-check, $\lambda$-method$Q_{all,\lambda}$393 kN (not adopted)

Check: no groundwater table is stated. Because the $\alpha$-method is a total-stress method, the shaft and toe results above are unaffected by that omission; only the $\lambda$-method cross-check depends on it, and the calculation assumes a dry profile with $\gamma = 18\ \text{kN/m}^3$ throughout. If the deposit is in fact saturated to the ground surface, $\bar{\sigma}'_v$ falls to about 41 kPa and the $\lambda$-method shaft resistance drops to roughly 561 kN, which brings the two methods much closer together. The adopted design value of 287 kN is unchanged either way.