16-Civ-B19 Foundation Engineering · December 2019

Question 5 of 6: Deep Foundations — bored pile group in over-consolidated clay

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

Question 5: Deep Foundations — bored pile group in over-consolidated clay (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.

Find. The undrained ultimate capacity of one pile, a group that satisfies the factored bearing-resistance requirement within the 6 m × 6 m cap, the global factor of safety of that group, and its total settlement by the equivalent-raft method.

Approach. Compute the single-pile capacity from α-method skin friction plus $9c_u$ end bearing, then test how many piles the 6 m × 6 m cap can actually hold and whether that number satisfies the factored resistance requirement; revise the design if it does not, compute the global factor of safety of the revised group against the lesser of individual-sum and block failure, and finally assess consolidation and immediate settlement from an equivalent raft.

1. Part (a) — skin friction on a single pile. With $A_p=0.2827$ m² and $p=1.885$ m as before, and the adhesion factor given directly as 0.8, $$Q_s=\alpha c_upL=(0.8)(60)(1.885)(18)=\boxed{1629\ \text{kN}}$$
2. End bearing in the same clay. For a pile whose embedded length exceeds about four diameters the bearing-capacity factor for undrained conditions has reached its asymptote $N_c^*=9$: $$Q_p=9c_uA_p=9(60)(0.2827)=\boxed{152.7\ \text{kN}}$$ $$Q_u=Q_s+Q_p=1629+153=\boxed{1781\ \text{kN}}$$ End bearing supplies only 8.6 % of the capacity, so this is a friction pile in the fullest sense — a fact that will determine where the equivalent raft is placed in part (d).
3. Part (b) — establish the factored demand. $$Q_f=1.25Q_D+1.5Q_L=1.25(8000)+1.5(2000)=10\,000+3000=13\,000\ \text{kN}$$ Each 18 m pile offers a factored resistance of $0.4Q_u=0.4(1781)=712.5$ kN, so the number of piles required is $$n=\frac{13\,000}{712.5}=18.25\quad\Longrightarrow\quad 19\ \text{piles}$$
4. Test that number against the cap constraint — this is where the design bites. With the piles equally spaced at 1.8 m in a square array, an $m \times m$ group needs a cap of at least $$B_g=(m-1)s+d$$ For $m=4$: $B_g=3(1.8)+0.6=6.0$ m, which exactly fills the permitted cap. For $m=5$: $B_g=4(1.8)+0.6=7.8$ m, which is 1.8 m too wide. The 6 m × 6 m limit therefore caps the group at 4 × 4 = 16 piles, and 19 cannot be provided. Checking the 16 that fit: $$R_f=16(0.4)(1781)=11\,400\ \text{kN}\ \lt\ Q_f=13\,000\ \text{kN}\quad(\text{ratio }0.88)$$ The stated 18 m pile length is insufficient — 16 piles fall 1600 kN short.
5. Revise the design by lengthening the piles. Since neither the pile count nor the spacing can grow, and increasing the diameter would push the piles past the edge of the cap (a 4 × 4 array at 1.8 m already leaves only 0.3 m from the outer pile centres to the cap edge, exactly one 600 mm radius), the free variable is length. Requiring the 16 piles to supply the factored demand, $$16(0.4)\left[\alpha c_upL+Q_p\right]\ge13\,000\quad\Longrightarrow\quad L\ge\frac{13\,000/6.4-152.7}{(0.8)(60)(1.885)}=20.76\ \text{m}$$ Adopt $L=21$ m. The toe then lies 22 m below ground, comfortably above the dense sand at 34 m, so the piles remain wholly in clay and the analysis above still applies: $$Q_u=(0.8)(60)(1.885)(21)+152.7=1900+153=2053\ \text{kN}$$ $$R_f=16(0.4)(2053)=\boxed{13\,137\ \text{kN}\ \ge\ 13\,000\ \text{kN}}\quad\checkmark$$ Design adopted: 16 cast-in-place piles, 600 mm diameter, 21 m long, at 1.8 m centres under a 6.0 m × 6.0 m cap founded 1 m below ground.
6. Part (c) — check the group against block failure before quoting a factor of safety. The footprint is $B_g=L_g=3(1.8)+0.6=6.0$ m and $L/B_g=21/6=3.5 \gt 2.5$, so $N_c^*=9$: $$Q_{block}=2(B_g+L_g)Lc_u+B_gL_gN_c^*c_u=(24)(21)(60)+(36)(9)(60)=30\,240+19\,440=49\,680\ \text{kN}$$ $$\sum Q_u=16(2053)=32\,843\ \text{kN}\ \lt\ 49\,680\ \text{kN}$$ The sum of individual capacities governs, so $Q_g=32\,843$ kN.
7. Form the global factor of safety. The unfactored service load is $Q=8000+2000=10\,000$ kN: $$FS=\frac{Q_g}{Q}=\frac{32\,843}{10\,000}=\boxed{3.28}$$ This sits within the conventional 2.5–3 range for pile groups and is consistent with the limit-states check: the two formats are simply different accounting systems for the same reserve.
8. Part (d) — place the equivalent raft. Because 92 % of the capacity is shaft friction, the load is delivered to the clay progressively down the shaft rather than at the toe, and Tomlinson's rule places the equivalent raft at two thirds of the embedded pile length below the pile heads: $$z_{raft}=1.0+\tfrac{2}{3}(21)=1.0+14.0=15.0\ \text{m below ground level}$$ The raft carries the full service load over the group footprint: $$q=\frac{Q}{B_gL_g}=\frac{10\,000}{36}=277.8\ \text{kPa}$$ No overburden is deducted, because the pile group adds this load to a soil column that was never excavated.
9. Compute the consolidation settlement of the clay beneath the raft. The clay extends 19 m from the raft down to the dense sand; dividing it into five 3.8 m sub-layers and spreading the load at 2 : 1 ($\Delta\sigma=Q/(B_g+z)^2$) with $\sigma'_0=\gamma(z_{raft}+z)$:

   Sub-layer	$z$ to mid-depth (m)	$\Delta\sigma$ (kPa)	$\sigma'_0$ (kPa)	$S_c$ (mm)
   1	1.90	160.2	338.0	72.8
   2	5.70	73.1	414.0	30.5
   3	9.50	41.6	490.0	15.3
   4	13.30	26.9	566.0	8.7
   5	17.10	18.7	642.0	5.4
   Total consolidation settlement				132.6

   using $S_c=\dfrac{C_c}{1+e_0}H\log_{10}\dfrac{\sigma'_0+\Delta\sigma}{\sigma'_0}$ with $C_c/(1+e_0)=0.25/2.20=0.114$.
10. Add the immediate settlement and compare with a tolerable limit. Treating the equivalent raft as rigid under undrained conditions ($\nu_u=0.5$, $E_u=60$ MPa, $I_s=0.82$): $$S_i=\frac{qB_g(1-\nu_u^2)I_s}{E_u}=\frac{(277.8)(6.0)(0.75)(0.82)}{60\,000}=17.1\ \text{mm}$$ $$S_{total}=S_i+S_c=17.1+132.6=\boxed{150\ \text{mm}}$$ No allowable settlement is stated in the question; total settlements of about 25–50 mm are commonly targeted for ordinary pile-supported structures, and 150 mm exceeds even the upper figure threefold. The serviceability limit state is not satisfied.
11. State what the settlement result means for the design. The group passes the ultimate limit state with a global factor of safety of 3.28 and fails the serviceability limit state by a factor of three — a very common outcome for friction piles in a compressible clay, and a reminder that capacity and deformation are independent questions. The remedy follows from where the settlement comes from: 55 % of it is generated in the first 3.8 m below the equivalent raft, where $\Delta\sigma/\sigma'_0$ is largest. Driving or boring the piles the full 33 m to the very dense sand converts the group to end bearing, moves the equivalent raft to the top of the sand and effectively eliminates the clay compression; that is the appropriate design change here. Enlarging the cap to permit more, longer piles would help but is excluded by the stated constraint. One conservatism should also be recorded: the clay is described as slightly over-consolidated, but no preconsolidation pressure or recompression index is given, so the calculation treats it as normally consolidated throughout. If the loading in the upper sub-layers were in fact still on the recompression line, the true settlement could be several times smaller, and determining $\sigma'_c$ by oedometer testing would be the first investigation to commission before committing to the deeper piles.