Deep Foundations (30 marks): 07-Str-B5, May 2015, Q2

Question 2 — Deep Foundations (30 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	Symbol	Value
Dead / live load	DL, LL	7000 kN, 3000 kN
Pile cap level	—	1 m below ground surface
Pile embedment	L	12 m (1 m to 13 m depth)
Pile diameter, spacing	D, s	0.60 m, 2.0 m c/c
Undrained strength at toe / average	c_u,b, c_u,av	175 kPa, 105 kPa
Clay unit weight	γ	18.5 kN/m³
Undrained modulus	E_u	65 MN/m²
Coefficient of volume compressibility	m_v	0.07 m²/MN
Clay thickness over rock	—	25 m
Allowable settlement	s_lim	30 mm

Find. The number and arrangement of 0.6 m piles needed for an overall factor of safety of 2.5 in (a) the short term and (b) the long term; then whether that group satisfies (c) the factored bearing-resistance check and (d) the 30 mm settlement limit.

Approach. Compute single-pile capacity by the total-stress (α) method for the short term and the effective-stress (β) method for the long term, choose the grid from the allowable single-pile load, verify that block failure does not govern, then run the limit-states load/resistance comparison and an equivalent-raft settlement analysis with the Skempton–Bjerrum correction.

Figure 2.1 — Elevation of the adopted 4 × 4 group, with the equivalent raft placed at two-thirds of the pile embedment and the load spreading at 1:4 to bedrock.

Fix the pile geometry. For a 0.6 m circular pile driven from 1 m to 13 m depth, $$A_p = \frac{\pi D^{2}}{4} = 0.2827\ \text{m}^{2}, \qquad p = \pi D = 1.885\ \text{m}, \qquad L = 12\ \text{m}$$ The choice between precast concrete and steel pipe piles does not change these numbers; it changes only the adhesion factor, and for a driven displacement pile in stiff clay the two are taken as equal.
(a) Short term — end bearing by the total-stress method. Undrained end bearing uses $N_c = 9$ for a deep foundation: $$Q_b = 9\,c_{u,b}\,A_p = 9 \times 175 \times 0.2827 = 445.3\ \text{kN}$$
Shaft resistance by the α method. With $c_{u,av} = 105$ kPa in a stiff clay, Tomlinson's adhesion chart for driven piles gives α ≈ 0.5 (the API expression $\alpha = 0.5\psi^{-0.5}$ with $\psi = c_u/\sigma^{\prime}_{v} = 105/129.5 = 0.81$ returns 0.56, so 0.5 is the safer of the two): $$Q_s = \alpha\,c_{u,av}\,p\,L = 0.5 \times 105 \times 1.885 \times 12 = 1187.5\ \text{kN}$$ $$Q_{ult} = Q_b + Q_s = 445.3 + 1187.5 = 1632.8\ \text{kN}, \qquad Q_{all} = \frac{1632.8}{2.5} = 653.1\ \text{kN}$$
Choose the grid. The number of piles required is $$n \ge \frac{10\,000}{653.1} = 15.31 \;\Rightarrow\; \boxed{n = 16 \text{ piles in a } 4 \times 4 \text{ grid at } 2.0\ \text{m c/c}}$$ giving a group plan dimension $B_g = 3(2.0) + 0.6 = 6.60$ m and a pile cap of about 7.2 m square.
Verify that block failure does not govern. A pile group in clay can fail as a single block, so the smaller of (sum of individual capacities) and (block capacity) controls. For the block, $N_c = 5(1 + 0.2B_g/L_g)(1 + 0.2L/B_g) = 5(1.2)(1.364) = 8.18$, which is below Skempton's limit of 9 for a square block, so it is used as calculated: $$Q_{block} = N_c\,c_{u,b}\,B_g^{2} + c_{u,av}\,(4B_g)\,L = 8.18(175)(6.60^{2}) + 105(26.4)(12) = 62\,370 + 33\,264 = 95\,634\ \text{kN}$$ $$\textstyle\sum Q_{ult} = 16 \times 1632.8 = 26\,125\ \text{kN} \;\ll\; 95\,634\ \text{kN}$$ Individual pile failure governs, the group efficiency is 1.0, and the allowable group load is $26\,125/2.5 = 10\,450$ kN > 10 000 kN ✓.
(b) Long term — switch to effective stresses. In the long term the excess pore pressures set up by driving and loading have dissipated, so shaft friction is governed by the effective stress acting on the pile and the drained strength of the remoulded clay at the interface. The β (Burland) method gives the unit shaft friction as $$f_s = \beta\,\sigma^{\prime}_{v}, \qquad \beta = K_0 \tan\delta = (1 - \sin\phi^{\prime})\tan\phi^{\prime}$$ With φ′ = 24° (see the note below), $K_0 = 1 - \sin 24^{\circ} = 0.593$ and $\beta = 0.593 \times 0.445 = 0.264$.
Evaluate the long-term shaft and base resistance. At mid-shaft (7 m) and at the toe (13 m) the vertical effective stresses are $\sigma^{\prime}_{v} = 18.5(7) = 129.5$ kPa and $18.5(13) = 240.5$ kPa. Hence $$Q_s = \beta\,\sigma^{\prime}_{v,mid}\,p\,L = 0.264(129.5)(1.885)(12) = 773.7\ \text{kN}$$ $$Q_b = N_q\,\sigma^{\prime}_{v,toe}\,A_p = 12(240.5)(0.2827) = 816.0\ \text{kN}$$ taking Berezantsev's $N_q = 12$ for φ′ = 24° at L/D = 20. The long-term single-pile capacity is therefore $$Q_{ult} = 773.7 + 816.0 = 1589.7\ \text{kN}, \qquad Q_{all} = 635.9\ \text{kN}$$
Confirm the same grid in the long term. The required number of piles is $$n \ge \frac{10\,000}{635.9} = 15.73 \;\Rightarrow\; \boxed{n = 16 \text{ piles} \;-\; \text{the same } 4 \times 4 \text{ group}}$$ with an allowable group load of $16 \times 635.9 = 10\,174$ kN > 10 000 kN ✓. The two analyses agree closely (1633 kN against 1590 kN per pile), which is typical for a stiff clay: neither drainage condition is decisively critical, so the same group serves both.
(c) Bearing-resistance limit state. The factored load effect and the factored geotechnical resistance are $$Q_f = 1.25(7000) + 1.5(3000) = 8750 + 4500 = 13\,250\ \text{kN}$$ $$R_f = f\,n\,Q_{ult} = 0.7 \times 16 \times 1632.8 = 18\,288\ \text{kN}\ \text{(short term)}, \qquad 0.7 \times 16 \times 1589.7 = 17\,805\ \text{kN}\ \text{(long term)}$$ $$\boxed{R_f = 17\,805\ \text{kN} \;\ge\; Q_f = 13\,250\ \text{kN} \quad \checkmark \ (\text{ratio } 1.34)}$$ The bearing-resistance limit state is comfortably satisfied in both drainage conditions.
(d) Set up the equivalent raft for settlement. For friction piles the load is transferred mainly by the shaft, so the settlement is computed for an imaginary raft at two-thirds of the pile penetration, of the same plan area as the group: $$z_{raft} = 1 + \tfrac{2}{3}(12) = 9.0\ \text{m}, \qquad B_{raft} = 6.60\ \text{m}, \qquad q = \frac{10\,000}{6.60^{2}} = 229.6\ \text{kPa}$$ Below the raft the load is assumed to spread at 1:4, so at a depth z below it the loaded width is $B = 6.60 + z/2$.
Compute the one-dimensional (oedometer) settlement. The compressible clay between the raft and bedrock is 16 m thick; dividing it into four 4 m sub-layers,

Question 2(d) — stress increase and oedometer compression, 1:4 spread below the equivalent raft
Sub-layer (depth, m)	z below raft (m)	Spread width B (m)	Δσ′ = Q/B² (kPa)	m_vΔσ′H (mm)
9 – 13	2.0	7.60	173.1	48.5
13 – 17	6.0	9.60	108.5	30.4
17 – 21	10.0	11.60	74.3	20.8
21 – 25	14.0	13.60	54.1	15.1
Total oedometer settlement s_oed				114.8

Apply the modified one-dimensional (Skempton–Bjerrum) theory. One-dimensional compression over-predicts the consolidation settlement of a finite loaded area because lateral strain relieves part of the excess pore pressure. The correction is $$s_c = \mu\,s_{oed}, \qquad \mu \approx 0.5 \ \text{for an over-consolidated stiff clay}$$ $$s_c = 0.5 \times 114.8 = 57.4\ \text{mm}$$ The immediate (undrained) component follows from the undrained modulus applied to the equivalent raft: $$s_i = q\,B_{raft}\,\frac{1 - \nu_u^{2}}{E_u}\,I_s = 229.6(6.60)\frac{1 - 0.5^{2}}{65\,000}(0.82) = 14.3\ \text{mm}$$
Compare with the serviceability criterion. The total settlement is $$s = s_i + \mu\,s_{oed} = 14.3 + 57.4 = 71.7\ \text{mm}$$ $$\boxed{s = 71.7\ \text{mm} \;>\; 30\ \text{mm} \quad \times \ \text{ SLS is NOT satisfied}}$$
Revise the group so that the serviceability limit state is met. Settlement falls when the equivalent raft is made larger (lower contact pressure) and deeper (less clay left to compress). Increasing the spacing to 3.0 m and the pile length to 20 m gives $B_g = 4(3.0) + 0.6 = 12.60$ m for a 5 × 5 group, an equivalent raft at 14.3 m, $q = 63.0$ kPa, and $$s = 7.5 + 0.5(33.0) = 24.0\ \text{mm} \;<\; 30\ \text{mm} \quad \checkmark$$ $$\boxed{\text{Adopt } 5 \times 5 = 25 \text{ piles at } 3.0\ \text{m c/c}, \ L = 20\ \text{m} \;(s \approx 24\ \text{mm})}$$

Check — parameters assumed in Question 2. (i) Adhesion factor α = 0.5 for the driven piles in stiff clay (Tomlinson); the API correlation gives 0.56, which would raise the single-pile capacity by 7 % but not change the 16-pile grid. (ii) Drained friction angle φ′ = 24° for part (b); the paper supplies no effective-stress parameters, and a long-term analysis is impossible without one. Holding N_q = 12, a value of 22° would need 17 piles and 26° would need 16, so φ′ must be measured before construction. (iii) No groundwater table is given; effective stresses in part (b) are computed from the bulk unit weight 18.5 kN/m³, i.e. assuming the water table lies below the pile toe. If the clay is submerged from the surface, σ′_v falls to 47 % of these values and 34 piles would be required. (iv) Skempton–Bjerrum μ = 0.5, mid-range for a stiff over-consolidated clay; the conclusion that the SLS fails is insensitive to it (μ = 0.7 gives 95 mm, μ = 0.3 gives 49 mm — both above 30 mm). (v) Equivalent-raft convention: the verdict is, however, sensitive to how the raft is sized. Taking the raft as the group outline with a 2:1 spread below it (Das) gives 50.5 mm, still a failure; spreading the load at 1:4 from the pile heads down to the raft (B = 10.6 m, q = 89.0 kPa, 2:1 below) gives 28.4 mm, a marginal pass. On any convention the 16-pile group is at best marginal on settlement, which is why a revised group is proposed.

Question 2 — final results
Item	Result
(a) Short-term single-pile capacity	Q_b = 445.3 kN, Q_s = 1187.5 kN, Q_ult = 1632.8 kN
(a) Allowable single-pile load (FS = 2.5)	653.1 kN → n ≥ 15.31
(a) Design group	4 × 4 = 16 piles at 2.0 m c/c, B_g = 6.60 m
(a) Block check	95 634 kN > 26 125 kN → individual failure governs
(b) Long-term single-pile capacity (β = 0.264)	Q_s = 773.7 kN, Q_b = 816.0 kN, Q_ult = 1589.7 kN
(b) Long-term group	Same 16 piles (allowable 10 174 kN > 10 000 kN)
(c) Factored load / factored resistance	13 250 kN / 17 805 kN → satisfied, ratio 1.34
(d) Settlement of the 16-pile group	s_i = 14.3 mm, μs_oed = 57.4 mm, total 71.7 mm > 30 mm — fails
(d) Revised group meeting the SLS	5 × 5 = 25 piles at 3.0 m c/c, L = 20 m, s ≈ 24 mm

Topic: Axial capacity of driven piles in stiff clay by the total-stress and effective-stress methods, group block failure, and equivalent-raft settlement with the Skempton–Bjerrum correction.

Key relations: $Q_{ult} = 9c_{u,b}A_p + \alpha c_{u,av}pL$ (short term); $Q_{ult} = N_q\sigma^{\prime}_{v,toe}A_p + \beta\sigma^{\prime}_{v,mid}pL$ with $\beta = (1-\sin\phi^{\prime})\tan\phi^{\prime}$ (long term); block capacity $N_c c_{u,b}B_g^{2} + c_{u,av}(4B_g)L$; equivalent raft at $2L/3$ with $s = s_i + \mu\,m_v\!\sum\!\Delta\sigma^{\prime}H$.

Why this works: Immediately after driving, a clay has no time to drain, so its strength is the undrained shear strength and capacity is expressed in total stresses through α and N_c = 9. In the long term the pore pressures have equalised and the same soil is described by effective stresses, so shaft friction becomes proportional to the vertical effective stress through β. The equivalent-raft idealisation recognises that friction piles do not transfer their load to the toe but shed it gradually along the shaft; placing a fictitious raft two-thirds of the way down puts the load where the soil actually feels it. The Skempton–Bjerrum factor then corrects the oedometer result, which assumes zero lateral strain, for the three-dimensional reality beneath a finite-width foundation.

Common pitfall: Stopping at the capacity check. This question is designed so that a group sized purely on strength — and comfortably passing the factored resistance check with a ratio of 1.34 — settles more than twice the allowable amount. For friction piles in clay, serviceability nearly always governs, and the fix is a larger, deeper equivalent raft rather than simply more piles at the same spacing. A second trap is computing the group settlement from the settlement of a single pile; the group's pressure bulb is many times deeper, and single-pile results are irrelevant to it.

Source: Tomlinson & Woodward, Pile Design and Construction Practice, 6th ed., Ch. 4 (single-pile capacity in clay) and Ch. 5 (pile groups, block failure, equivalent raft); Craig's Soil Mechanics, 9th ed., Ch. 11; Canadian Foundation Engineering Manual, 4th ed., Ch. 18 (deep foundations, resistance factors).

Question 2 of 6: Deep Foundations (30 marks)

Question 2 — Deep Foundations (30 marks)