16-Civ-B19 Foundation Engineering · December 2019

Question 2 of 6: Deep Foundations — driven 4 × 4 pile group in clay over sand

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

Question 2: Deep Foundations — driven 4 × 4 pile group in clay over sand (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 overall factor of safety of the group, the bearing-resistance limit-state check at factored loads, and the group settlement by both the settlement-ratio and equivalent-raft methods.

Approach. Build the single-pile capacity from an α-method shaft resistance in the clay and an $N_q^*$ end bearing in the sand, compare the sum of the individual capacities with the block capacity to obtain the group capacity, then use that capacity in both the working-stress and limit-states checks; settlement follows from Vesic's ratio applied to a three-component single-pile settlement, and independently from Meyerhof's SPT relation applied to an equivalent raft at the pile toes.

1. Part (a) — establish the pile geometry constants. For a 0.6 m circular pile, $$A_p=\frac{\pi d^2}{4}=\frac{\pi(0.6)^2}{4}=0.2827\ \text{m}^2,\qquad p=\pi d=\pi(0.6)=1.885\ \text{m}$$
2. Compute the shaft resistance by the α method. With $c_u/p_a=60/100=0.60$, Das's tabulated adhesion factor (Table 9.4, after Terzaghi, Peck and Mesri) gives $\alpha=0.62$: $$Q_s=\alpha c_u p L=(0.62)(60)(1.885)(15)=\boxed{1052\ \text{kN}}$$
3. Compute the end bearing in the sand. An SPT of 20 in a medium dense sand corresponds to $\phi'\approx35^\circ$, for which Meyerhof's bearing-capacity factor (Das's tabulated $N_q^*$: 56.7 at 30°, 143 at 35°, 346 at 40°) is $N_q^*=143$. The theoretical value $q'N_q^*=(18)(15)(143)=38\,610$ kPa is far above the limiting unit end bearing that a real pile can mobilise, so Das's ceiling governs: $$q_p=\min\left[q'N_q^*,\ 0.5\,p_aN_q^*\tan\phi'\right]=\min\left[38\,610,\ 0.5(100)(143)\tan35^\circ\right]=5006\ \text{kPa}$$ $$Q_p=q_pA_p=(5006)(0.2827)=1415\ \text{kN}$$
4. Assemble the single-pile ultimate capacity. $$Q_u=Q_s+Q_p=1052+1415=\boxed{2467\ \text{kN per pile}}$$ End bearing supplies 57 % of the capacity, so the seating of the toes in the sand is what this design relies on (see step 8).
5. Compare the sum of individual capacities with block failure. The group footprint is $$B_g=L_g=3s+d=3(1.8)+0.6=6.0\ \text{m}$$ Summing the individual piles gives $\sum Q_u=16(2467)=39\,477$ kN, while the block prism mobilises the full undrained strength on its perimeter plus the sand bearing over its base: $$Q_{block}=2(B_g+L_g)Lc_u+B_gL_gq_p=(24)(15)(60)+(36)(5006)=21\,600+180\,216=201\,816\ \text{kN}$$ The block is about five times stronger, so the sum of the individual capacities governs and the group efficiency is effectively unity: $$Q_g=\boxed{39\,477\ \text{kN}}$$
6. Form the total factor of safety. The unfactored working load is $Q=8000+4000=12\,000$ kN, so $$FS=\frac{Q_g}{Q}=\frac{39\,477}{12\,000}=\boxed{3.29}$$ This meets the factor of 2.5–3 conventionally required for pile groups designed from soil parameters alone.
7. Part (b) — apply the load and resistance factors. The factored load effect and the factored geotechnical resistance are $$Q_f=1.25Q_D+1.5Q_L=1.25(8000)+1.5(4000)=10\,000+6000=16\,000\ \text{kN}$$ $$R_f=f_cQ_g=0.6(39\,477)=23\,686\ \text{kN}$$ $$R_f=23\,686\ \text{kN}\ \ge\ Q_f=16\,000\ \text{kN}\quad\Longrightarrow\quad\boxed{\text{ULS satisfied } (R_f/Q_f=1.48)}$$
8. Interpret the margin and the assumption behind it. The factored resistance exceeds the factored demand by 48 %, but more than half of it is end bearing, and Meyerhof's limiting value presumes the toe is seated in the sand: his layered-soil rule builds the unit tip resistance up from the clay value toward the sand limit over roughly ten diameters of penetration. If the toes merely touched the sand, $q_p$ would fall toward $9c_u=540$ kPa, the overall factor of safety would drop to 1.61 and the limit state would fail. The design therefore depends on the piles being driven a real distance into the sand, and the capacity should be confirmed by dynamic monitoring during driving. Adopting $\phi'=35^\circ$ for $N_{60}=20$ sits at the conservative end of the usual correlations, which partly offsets that uncertainty.
9. Part (c)(i) — split the working load on one pile between shaft and toe. Each pile carries $Q_w=12\,000/16=750$ kN, shared in proportion to the components of capacity: $$Q_{wp}=750\frac{1415}{2467}=430.2\ \text{kN},\qquad Q_{ws}=750\frac{1052}{2467}=319.8\ \text{kN}$$
10. Compute the three components of the single-pile settlement. The elastic shortening of the pile shaft, with $\xi=0.5$ for a roughly uniform-to-parabolic friction distribution, is $$S_{e(1)}=\frac{(Q_{wp}+\xi Q_{ws})L}{A_pE_p}=\frac{[430.2+0.5(319.8)](15)}{(0.2827)(200\times10^6)}=0.16\ \text{mm}$$ The settlement caused by the load at the toe, with $I_{wp}=0.85$ and the sand modulus, is $$S_{e(2)}=\frac{q_{wp}d}{E_s}(1-\nu_s^2)I_{wp}=\frac{(1521.6)(0.6)}{24\,000}(1-0.3^2)(0.85)=29.43\ \text{mm}$$ and that caused by the friction transmitted along the shaft, with $I_{ws}=2+0.35\sqrt{L/d}=2+0.35\sqrt{25}=3.75$ and the undrained clay modulus $E_c=400c_u=24\,000$ kPa, is $$S_{e(3)}=\left(\frac{Q_{ws}}{pL}\right)\frac{d}{E_c}(1-\nu_c^2)I_{ws}=\left(\frac{319.8}{28.27}\right)\frac{0.6}{24\,000}(0.75)(3.75)=0.80\ \text{mm}$$ $$S_e=0.16+29.43+0.80=\boxed{30.4\ \text{mm}}$$
11. Scale to the group with Vesic's settlement ratio. A group loads a far larger volume of soil than a single pile, and Vesic's empirical ratio captures that as the square root of the width ratio: $$S_g=S_e\sqrt{\frac{B_g}{d}}=30.4\sqrt{\frac{6.0}{0.6}}=30.4(3.162)=\boxed{96.1\ \text{mm}}$$
12. Part (c)(ii) — recompute by the equivalent-raft method on SPT values. Because the settlement is to be based on the SPT of the sand, the group is replaced by an equivalent 6.0 m × 6.0 m raft at the level of the pile toes, 15 m below ground. No soil was excavated to install the piles, so the gross pressure is the load increment: $$q=\frac{12\,000}{6.0\times6.0}=333.3\ \text{kPa}$$ Meyerhof's (1976) SPT relation for pile groups in sand, in the SI form given by Das, includes an influence factor for the embedment of the group: $$I=1-\frac{L}{8B_g}=1-\frac{15}{8(6.0)}=0.6875\ \ (\ge0.5)$$ $$S_g(\text{mm})=\frac{0.96\,q\sqrt{B_g}\,I}{N_{60}}=\frac{0.96(333.3)\sqrt{6.0}\,(0.6875)}{20}=\boxed{26.9\ \text{mm}}$$
13. Reconcile the two settlement estimates. The two methods differ by a factor of about 3.6 (96.1 mm against 26.9 mm), and the difference comes from the methods, not from an arithmetic error. The settlement-ratio method starts from a single-pile settlement dominated by $S_{e(2)}$ — 430 kN of toe load pressing into a 24 MPa sand over a 0.6 m width — and multiplies it by $\sqrt{B_g/d}=3.16$. It is well known to be conservative for large groups, because a group does not concentrate toe pressure the way a single pile does. Meyerhof's relation instead spreads the whole 12 MN over 36 m² at toe level and takes the sand stiffness from the SPT directly, with a reduction for the depth of embedment. Design practice is to treat them as brackets, here roughly 27–96 mm. Total settlements of about 25–50 mm are commonly targeted for ordinary pile-supported structures: the Meyerhof estimate meets that and the settlement-ratio estimate does not. With the end bearing correctly evaluated, bearing resistance is not the concern for this foundation; deformation is the open question, and a load test on a working pile, which measures the single-pile load–settlement response directly, is the usual way to decide which bound is realistic.