Question 7 of 9: Ultimate capacity of a driven concrete pile by two methods

Question 7: Ultimate capacity of a driven concrete pile by two methods (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.

Find. The ultimate axial capacity $Q_u = Q_p + Q_s$ of the single pile, computed by two independent methods so that the answers can be compared and a design value recommended.

Approach. Both methods split the capacity into shaft and toe. Method 1 is the conventional semi-empirical treatment — the total-stress $\alpha$ method in the clay, the $K\tan\delta^{\prime}$ effective-stress method in the sand, and Meyerhof’s $N_q^{*}$ with its limiting point resistance at the toe. Method 2 replaces the clay shaft with Vijayvergiya & Focht’s $\lambda$ method and the toe with Janbu’s closed-form bearing-capacity factor; because the $\lambda$ method was calibrated only on piles in clay, the sand shaft keeps its $K\tan\delta^{\prime}$ value in both methods.

1. Geometry and effective stresses. With the water table at the surface both layers are submerged:$$\begin{aligned}A_p &= \frac{\pi D^2}{4} = 0.1963\ \text{m}^2 \\ p &= \pi D = 1.5708\ \text{m}\end{aligned}$$$$\begin{aligned}\gamma^{\prime}_{clay} &= 17 - 9.81 = 7.19 \\ \gamma^{\prime}_{sand} &= 20 - 9.81 = 10.19\ \text{kN/m}^3 .\end{aligned}$$The vertical effective stress is therefore $\sigma^{\prime}_v = 7.19(5.0) = 35.95$ kPa at the top of the sand and $35.95 + 10.19(8.0) = 117.47$ kPa at the pile toe.
2. Method 1 — shaft resistance in the clay ($\alpha$ method). For $c_u/p_a = 50/100 = 0.5$, Das’s tabulation of the Terzaghi–Peck–Mesri adhesion factor gives $\alpha = 0.68$. Then$$Q_{s,clay} = \alpha c_u p L_c = 0.68(50)(1.5708)(5.0) = 267.0\ \text{kN}.$$
3. Method 1 — shaft resistance in the sand. The unit friction is $f = K\sigma^{\prime}_v \tan\delta^{\prime}$. For a high-displacement driven pile take $K = 1.4K_0 = 1.4(1-\sin 35^{\circ}) = 0.597$ and $\delta^{\prime} = 0.8\phi^{\prime} = 28^{\circ}$. Field measurements show the friction ceasing to grow below a critical depth $L^{\prime} = 15D = 7.5$ m, where $\sigma^{\prime}_v = 61.42$ kPa, so$$\int \sigma^{\prime}_v\,dz = \underbrace{\tfrac{35.95+61.42}{2}(2.5)}_{121.7} + \underbrace{61.42(5.5)}_{337.8} = 459.6\ \text{kPa}\cdot\text{m},$$$$Q_{s,sand} = p\,K\tan\delta^{\prime}\int\sigma^{\prime}_v\,dz = 1.5708(0.597)(0.5317)(459.6) = 229.1\ \text{kN}.$$
4. Method 1 — end bearing (Meyerhof). Meyerhof’s chart (as tabulated by Das) gives $N_q^{*} = 143$ for $\phi^{\prime} = 35^{\circ}$, and the point resistance is capped by his limiting value:$$\begin{aligned}q_p &= \sigma^{\prime}_v N_q^{*} = 117.47(143) = 16\,798\ \text{kPa} \\ q_{p(\lim)} &= 0.5\,p_a N_q^{*}\tan\phi^{\prime} = 0.5(100)(143)(0.7002) = 5006\ \text{kPa}.\end{aligned}$$The limit governs, as it always does for an embedment ratio of $L/D = 26$, so$$Q_p = A_p q_{p(\lim)} = 0.1963(5006) = 983.0\ \text{kN}.$$
5. Method 1 total. Adding the three contributions:$$\boxed{Q_{u,1} = 983.0 + 267.0 + 229.1 = 1479.2\ \text{kN}}$$
6. Method 2 — clay shaft by the $\lambda$ method. Vijayvergiya and Focht express the average unit friction in clay as $f_{av} = \lambda(\bar{\sigma}^{\prime}_v + 2\bar{c}_u)$, with $\lambda$ read at the full embedded length of the pile: interpolating Das’s tabulation between 0.245 at $L = 10$ m and 0.200 at $L = 15$ m gives $\lambda = 0.218$ at $L = 13$ m. The method was calibrated on piles in clay, so the means are taken over the clay layer only, $\bar{\sigma}^{\prime}_v = 35.95/2 = 17.98$ kPa and $\bar{c}_u = 50$ kPa. Hence$$\begin{aligned}f_{av} &= 0.218\,[17.98 + 2(50)] = 25.72\ \text{kPa} \\ Q_{s,clay} &= f_{av}\,p\,L_c = 25.72(1.5708)(5.0) = 202.0\ \text{kN}.\end{aligned}$$The sand shaft has no $\lambda$ counterpart, so the $K\tan\delta^{\prime}$ value of 229.1 kN is carried into Method 2 unchanged.
7. Method 2 — end bearing by Janbu’s equation. Janbu gives a closed-form factor that needs no chart:$$N_q^{*} = \left(\tan\phi^{\prime} + \sqrt{1+\tan^{2}\phi^{\prime}}\right)^{2} e^{2\eta\tan\phi^{\prime}},$$with $\eta = 105^{\circ} = 1.833$ rad for a dense sand. Substituting $\tan 35^{\circ} = 0.7002$ gives $N_q^{*} = (1.9210)^2 e^{2.567} = 3.690(13.02) = 48.04$, so$$Q_p = A_p \sigma^{\prime}_v N_q^{*} = 0.1963(117.47)(48.04) = 1108.1\ \text{kN}.$$
8. Method 2 total.$$\boxed{Q_{u,2} = 1108.1 + 202.0 + 229.1 = 1539.2\ \text{kN}}$$
9. Compare and recommend. The two methods differ by only 4 %, which is unusually close for pile-capacity predictions and gives real confidence in the answer. They divide the capacity differently — Method 1 puts 66 % of it in the toe and Method 2 puts 72 % there — so a designer who cares about the load–settlement response, not just the ultimate load, should note that mobilising full end bearing needs a toe movement of order $0.1D = 50$ mm whereas the shaft is fully mobilised at 5–10 mm. Taking the lower of the two and a global factor of safety of 2.5,$$Q_{all} = \frac{1479.2}{2.5} = 592\ \text{kN} \quad\text{(say 590 kN per pile).}$$Canadian practice (CFEM) would confirm this with a static load test on about one pile in ten, because the $\alpha$ and $N_q^{*}$ values are empirical.