Deep Foundations (30 marks): 07-Str-B5, May 2015, Q5

Question 5 — 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
Undrained shear strength	c_u	50 kPa
Elastic modulus, Poisson's ratio	E, ν	50 MPa, 0.5
Bulk unit weight	γ	20 kN/m³
Coefficient of volume change	m_v	2.3 × 10⁻⁴ m²/kN
Initial void ratio	e₀	1.47
Depth to very dense sand	—	34 m
Piles	n, L, D	25, 18 m, 0.40 m
Pile spacing (from the figure)	s	1.0 m c/c, 5 × 5 grid
Pile cap	—	5 m × 5 m, founded 1 m below ground
Adhesion factor	α	0.8
Applied vertical centric load	Q	10 MN = 10 000 kN

Find. (a) the ultimate capacity of one pile; (b) the ultimate capacity of the group; (c) the factor of safety under 10 MN, with comment; (d) the settlement of the group by the equivalent-raft method.

Approach. Use the α (total-stress) method for the single pile, compare the sum of individual capacities with block failure of the group, divide into the applied load for the factor of safety, and place an equivalent raft at two-thirds of the pile embedment for an elastic settlement calculation.

Figure 5.1 — The 25-pile group. At 1.0 m centres the piles are only 2.5 diameters apart, so the group is close to a solid block; the equivalent raft sits 13 m down with 21 m of clay still to compress beneath it.

(a) Pile geometry. For a 0.4 m diameter pile, $$A_p = \frac{\pi(0.4)^{2}}{4} = 0.1257\ \text{m}^{2}, \qquad p = \pi(0.4) = 1.2566\ \text{m}, \qquad L = 18\ \text{m}$$
Shaft resistance by the α method. Undrained skin friction is a fraction of the undrained shear strength: $$Q_s = \alpha\,c_u\,p\,L = 0.8 \times 50 \times 1.2566 \times 18 = 904.8\ \text{kN}$$ The high adhesion factor (0.8) is consistent with a soft-to-firm clay: the softer the clay, the closer the remoulded interface strength is to the intact strength.
End bearing. With $N_c = 9$ for a deep circular base in clay, $$Q_b = 9\,c_u\,A_p = 9 \times 50 \times 0.1257 = 56.5\ \text{kN}$$ $$\boxed{Q_{ult} = Q_s + Q_b = 904.8 + 56.5 = 961.3\ \text{kN per pile}}$$ End bearing supplies less than 6 % of the total — these are floating friction piles, and that observation governs everything that follows.
(b) Sum the individual capacities. If each pile fails independently, $$\textstyle\sum Q_{ult} = 25 \times 961.3 = 24\,033\ \text{kN}$$
Check block failure of the group. At 1.0 m centres the out-to-out plan dimension is $B_g = 4(1.0) + 0.4 = 4.40$ m. Treating the group and the soil between the piles as one deep block, whose sides are sheared at the full undrained strength, $$Q_{block} = N_c\,c_u\,B_g^{2} + c_u\,(4B_g)\,L$$ Here $N_c = 5(1 + 0.2B/L_g)(1 + 0.2L/B_g) = 10.9$, which exceeds the limiting value of 9, so $N_c = 9$ is used: $$Q_{block} = 9(50)(4.40^{2}) + 50(17.6)(18) = 8712 + 15\,840 = 24\,552\ \text{kN}$$
Take the lesser of the two. The group capacity is $$\boxed{Q_{g,ult} = \min(24\,033;\ 24\,552) = 24\,033\ \text{kN}}$$ with a group efficiency of $24\,033/24\,033 = 1.00$. The two mechanisms are within 2 % of each other, which is exactly what a spacing of 2.5 diameters produces: the group is on the boundary between behaving as 25 separate piles and behaving as one solid block.
(c) Factor of safety under the applied load. $$\boxed{FS = \frac{Q_{g,ult}}{Q} = \frac{24\,033}{10\,000} = 2.40}$$
(d) Set up the equivalent raft. Because more than 94 % of the load is shed through the shaft, the equivalent raft is placed at two-thirds of the pile embedment below the pile heads: $$z_{raft} = 1 + \tfrac{2}{3}(18) = 13.0\ \text{m}, \qquad B_{raft} = B_g = 4.40\ \text{m}, \qquad q = \frac{10\,000}{4.40^{2}} = 516.5\ \text{kPa}$$
Compute the elastic settlement. Using the elastic parameters supplied, with the rigid-square influence factor $I_s = 0.82$ and no depth-embedment reduction (conservative), $$s_e = q\,B_{raft}\,\frac{1-\nu^{2}}{E}\,I_s = 516.5(4.40)\frac{1 - 0.5^{2}}{50\,000}(0.82) = 0.02795\ \text{m}$$ $$\boxed{s_e = 28.0\ \text{mm}}$$ The paper states no allowable settlement for this foundation; against the CFEM guidance of 50 mm total settlement for a foundation on clay, and 25 mm where a structure is settlement-sensitive, 28 mm is acceptable for a normal structure but marginal for a sensitive one. Applying the Fox depth-correction factor for $D/B \approx 3$ ($I_f \approx 0.5$) would halve it to about 14 mm.
Cross-check with the consolidation parameters — and note the contradiction. Poisson's ratio of 0.5 means the elastic calculation above is an undrained, immediate settlement; long-term consolidation must be added, and the paper supplies m_v and e₀ for that purpose. Spreading the load at 1:4 from the equivalent raft through the 21 m of clay remaining above the dense sand gives $s_{oed} = m_v\sum\Delta\sigma^{\prime}H = 710$ mm. That result is not credible, and the reason is visible in the data: the constrained modulus implied by m_v is $$E_{oed} = \frac{1}{m_v} = \frac{1}{2.3\times10^{-4}} = 4350\ \text{kPa} = 4.35\ \text{MPa}$$ which is an order of magnitude softer than the stated E = 50 MPa, although a constrained modulus must be the stiffer of the two. The two parameters cannot describe the same clay.
State the serviceability verdict. On the basis actually requested — the equivalent raft analysed elastically with the given E and ν — $$\boxed{s = 28\ \text{mm} \;<\; 50\ \text{mm} \ (\text{CFEM guidance}) \;\Rightarrow\; \text{SLS satisfied}}$$ but this verdict holds only if E = 50 MPa is the correct stiffness. If the supplied m_v is correct instead, the group settles of the order of 0.7 m and the foundation is unbuildable as designed. Oedometer and stiffness testing must be reconciled before construction.

Comment on the design of this pile group (part c). A factor of safety of 2.40 falls just short of the 2.5 normally required for a pile group under working load, so the foundation is marginally under-designed on capacity alone. More telling is why it is marginal. The piles are spaced at only 2.5 diameters, where the recommended minimum for friction piles in clay is 3 diameters and 3.5 to 4 is preferred; at this spacing the block and individual mechanisms coincide, the shafts overlap each other's stress fields, and the group buys almost nothing for the twenty-fifth pile. At the same time the piles stop 15 m short of the very dense sand at 34 m, so they float in clay and every kilonewton of load is carried into compressible ground rather than into a firm stratum. Three improvements follow directly: increase the spacing to 1.4–1.6 m (3.5–4 D), which raises the block capacity sharply and, far more importantly, enlarges the equivalent raft and cuts settlement; or extend the piles to found in the dense sand at 34 m, converting them to end-bearing piles with a small fraction of the settlement; or reduce the pile count and increase the diameter, which is usually cheaper than driving 25 slender piles. The settlement calculation in part (d) — and the unresolved conflict between E and m_v — should be settled before any of these is chosen.

Check — conflicting stiffness data. The paper supplies both E = 50 MPa and m_v = 2.3 × 10⁻⁴ m²/kN (constrained modulus 4.35 MPa). A constrained modulus is always stiffer than Young's modulus for the same soil, so these two values are mutually inconsistent by more than a factor of ten. Part (d) explicitly asks for an elastic analysis, so E = 50 MPa is used for the boxed answer; the m_v-based result (710 mm) is reported alongside so the inconsistency is visible rather than hidden. No allowable settlement is stated in the question either; the 50 mm CFEM guideline for foundations on clay is adopted as the criterion.

Question 5 — final results
Item	Result
(a) Shaft resistance, single pile	904.8 kN
(a) End bearing, single pile	56.5 kN
(a) Ultimate single-pile capacity	961.3 kN
(b) Sum of individual capacities	24 033 kN
(b) Block capacity (B_g = 4.40 m)	24 552 kN
(b) Ultimate group capacity	24 033 kN (efficiency 1.00)
(c) Factor of safety under 10 MN	2.40 — below the usual 2.5
(d) Equivalent raft	4.40 m square at 13.0 m depth, q = 516.5 kPa
(d) Elastic settlement	28.0 mm — SLS satisfied against the 50 mm guideline
(d) m_v-based consolidation cross-check	710 mm — data conflict, must be resolved

Topic: Undrained capacity of a friction-pile group in uniform clay, block failure, and equivalent-raft settlement by elastic analysis.

Key relations: $Q_{ult} = \alpha c_u p L + 9c_u A_p$; block capacity $Q_{block} = N_c c_u B_g^{2} + c_u(4B_g)L$ with $N_c \le 9$; group efficiency $\eta = Q_{g}/\sum Q_{ult}$; equivalent raft at $2L/3$, $s_e = qB(1-\nu^{2})I_s/E$; constrained modulus $E_{oed} = 1/m_v$.

Why this works: In clay a driven pile's shaft resistance is governed by the strength of a thin remoulded annulus, so it is expressed as a fraction α of the intact undrained strength rather than through effective stresses; and because the base is deep, the bearing-capacity factor is the fully-confined value of 9 rather than the surface value 5.14. The block check exists because closely spaced piles cannot each mobilise their own independent failure surface — the surfaces intersect, and the group ends up shearing the soil only around its outside perimeter. The equivalent-raft idealisation then answers the separate question of deformation: friction piles hand their load to the soil gradually, so the load appears to the deep clay as if it were applied on a raft two-thirds of the way down the shafts.

Common pitfall: Assuming the block check is a formality. Here it returns 24 552 kN against 24 033 kN — a 2 % difference — which is the diagnostic signature of piles spaced too closely. A designer who skips the check learns nothing about why the group is inefficient. The second pitfall is trusting a single stiffness parameter without cross-checking: E and m_v describe the same soil and must be mutually consistent, and when they are not, the settlement answer is worthless until the discrepancy is resolved.

Source: Das, Principles of Foundation Engineering, 9th ed., Ch. 11 (pile foundations — α method, group efficiency, elastic settlement of groups); Tomlinson & Woodward, Pile Design and Construction Practice, 6th ed., Ch. 5; Canadian Foundation Engineering Manual, 4th ed., Ch. 18 and Table 8.1 (tolerable settlements).

Question 5 of 6: Deep Foundations (30 marks)

Question 5 — Deep Foundations (30 marks)