Ultimate capacity of a driven pile by two methods: 16-Civ-B3 Geotechnical Design, December 2018, Q6

Question 6: Ultimate capacity of a driven 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.

Quantity	Symbol	Value
Pile diameter (circular, precast concrete, driven)	D	0.5 m
Length in clay / in sand / total	L₁ / L₂ / L	5.0 m / 8.0 m / 13.0 m
Clay: undrained strength, bulk unit weight	c_u, γ	50 kPa, 17 kN/m³
Sand: effective cohesion, friction angle, bulk unit weight	c′, φ′, γ	0, 35°, 20 kN/m³
Water table (from the figure symbol)	—	at ground surface
Unit weight of water; atmospheric reference	γ_w, p_a	9.81 kN/m³, 100 kPa

Find. The ultimate axial compressive capacity Q_u = Q_p + Q_s of the single pile, computed by two independent methods, and a recommended allowable load.

[Figure not reproduced: Figure 1 (redrawn) — Precast concrete pile, 0.5 m diameter, driven 13.0 m: 5.0 m through soft-to-firm clay and 8.0 m into dense sand, with the water table at ground level. See the official exam paper.]

Approach. Because the water table stands at the ground surface, every vertical stress used in an effective-stress expression must be an effective stress, so the profile is built first; the capacity is then assembled as end bearing in the sand plus shaft friction over the clay and over the sand, and the whole calculation is repeated with a genuinely different set of models — total stress and Meyerhof for the first, effective stress and Janbu for the second — so that the two answers form a bracket rather than a duplicate.

Establish the pile geometry and the effective stress profile. For a circular pile, $$A_p=\frac{\pi D^2}{4}=\frac{\pi(0.5)^2}{4}=0.19635\ \text{m}^2,\qquad p=\pi D=\pi(0.5)=1.5708\ \text{m}$$ With the water table at the surface both layers are submerged, so $$\gamma'_{\text{clay}}=17-9.81=7.19\ \text{kN/m}^3,\qquad \gamma'_{\text{sand}}=20-9.81=10.19\ \text{kN/m}^3$$ The effective vertical stress at the top of the sand and at the pile toe follows directly: $$\sigma'_{v(5)}=7.19(5.0)=35.95\ \text{kPa},\qquad \sigma'_{v(13)}=35.95+10.19(8.0)=\boxed{117.47\ \text{kPa}}$$
Method 1, shaft in clay: the α method. The adhesion factor is read against the normalised strength c_u/p_a = 50/100 = 0.50, which falls between the tabulated entries α = 0.74 at 0.4 and α = 0.62 at 0.6, giving α = 0.68 by linear interpolation. Hence $$Q_{s(\text{clay})}=\alpha c_u\,p\,L_1=0.68(50)(1.5708)(5.0)=267.0\ \text{kN}$$
Method 1, shaft in sand: the K–δ formulation with a critical depth. A precast concrete pile is a high-displacement driven pile, so the earth pressure coefficient is taken as 1.4 K₀ and the interface friction angle as 0.8φ′: $$K=1.4(1-\sin 35^{\circ})=1.4(0.42642)=0.5970,\qquad \delta'=0.8(35^{\circ})=28^{\circ}$$ Unit friction in sand is taken to grow with effective stress only to the critical depth L_c = 15D = 7.5 m below ground and to remain constant below it; the effective stress at that level is σ′_v = 35.95 + 10.19(2.5) = 61.42 kPa. Integrating over the two portions of the sand, $$\int_{5.0}^{7.5}\sigma'_v\,dz=\tfrac{1}{2}(35.95+61.42)(2.5)=121.72\ \text{kPa}\!\cdot\!\text{m},\qquad \int_{7.5}^{13.0}\sigma'_v\,dz=61.42(5.5)=337.84\ \text{kPa}\!\cdot\!\text{m}$$ so that $$Q_{s(\text{sand})}=p\,K\tan\delta'\left(121.72+337.84\right)=1.5708(0.5970)(0.53171)(459.56)=229.1\ \text{kN}$$
Method 1, end bearing: Meyerhof with the limiting point resistance. For φ′ = 35° the bearing-capacity factor for driven piles is N_q* = 143. The unfactored expression gives an implausible value, $$q_p=\sigma'_{v(13)}N_q^{*}=117.47(143)=16\,798\ \text{kPa}$$ so the limiting form must be applied. With L/D = 13/0.5 = 26, well beyond the depth at which point resistance ceases to grow, $$q_{p(\text{lim})}=0.5\,p_a N_q^{*}\tan\phi'=0.5(100)(143)(0.70021)=5006.5\ \text{kPa}$$ The limiting value governs, and $$Q_p=A_p\,q_{p(\text{lim})}=0.19635(5006.5)=\boxed{983.0\ \text{kN}}$$
Method 1 total. Summing the three contributions, $$Q_{u(1)}=267.0+229.1+983.0=\boxed{1479\ \text{kN}}$$ The base supplies 66 per cent of the capacity, which is what one expects of a pile driven a substantial distance into dense sand.
Method 2, shaft in clay: the λ method. Vijayvergiya and Focht express the average unit friction over the clay embedment as f_av = λ(σ′_v(av) + 2c_u), with λ read against that embedment; for 5 m, λ = 0.336. The mean effective overburden over the clay is σ′_v(av) = 7.19(5.0)/2 = 17.98 kPa, so $$f_{\text{av}}=0.336\left[17.98+2(50)\right]=0.336(117.98)=39.64\ \text{kPa}$$ $$Q_{s(\text{clay})}=f_{\text{av}}\,p\,L_1=39.64(1.5708)(5.0)=311.3\ \text{kN}$$ This is 16.6 per cent above the α-method value — a modest spread for two independent empirical routes, and in the less conservative direction, which is one reason the lower of the two totals is adopted below.
Method 2, shaft in sand: the β method with no critical-depth cut-off. Working entirely in effective stress and taking the at-rest coefficient with full interface friction, $$\beta=K_0\tan\delta'=(1-\sin 35^{\circ})\tan 35^{\circ}=0.42642(0.70021)=0.2986$$ The mean effective stress over the sand is σ′_v(av) = (35.95 + 117.47)/2 = 76.71 kPa, so $$Q_{s(\text{sand})}=\beta\,\sigma'_{v(\text{av})}\,p\,L_2=0.2986(76.71)(1.5708)(8.0)=287.8\ \text{kN}$$ The lower earth-pressure coefficient and the absence of a critical-depth cap very nearly cancel, leaving this within 26 per cent of the Method 1 value.
Method 2, end bearing: Janbu's closed-form bearing-capacity factor. Janbu expresses the point factor as a function of the friction angle and of the angle η′ subtended by the failure surface at the pile toe, taken as 105° = 1.8326 rad for a dense sand: $$N_q^{*}=\left(\tan\phi'+\sqrt{1+\tan^{2}\phi'}\right)^{2}e^{2\eta'\tan\phi'}=(0.70021+1.22078)^{2}e^{2(1.8326)(0.70021)}$$ $$N_q^{*}=3.6902\,e^{2.5661}=3.6902(13.015)=48.04$$ This needs no chart and therefore no chart citation, which is exactly why it is worth carrying alongside the Meyerhof value that page-1 Note 6 obliges us to attribute. Hence $$Q_p=A_p\,\sigma'_{v(13)}N_q^{*}=0.19635(117.47)(48.04)=1108\ \text{kN}$$
Method 2 total, and the design value. Summing, $$Q_{u(2)}=311.3+287.8+1108=\boxed{1707\ \text{kN}}$$ The two independent estimates, 1479 kN and 1707 kN, differ by 14.3 per cent about their mean of 1593 kN — close agreement for pile capacity, where a spread of a factor of two between methods is not unusual. Adopting the lower of the two with a factor of safety of 3, as CFEM recommends for a capacity estimated from soil parameters without a load test, $$Q_{\text{all}}=\frac{Q_{u(1)}}{\text{FS}}=\frac{1479}{3}=\boxed{493\ \text{kN}\ \text{(say 490 kN)}}$$

Quantity	Method 1 (α + K–δ + Meyerhof)	Method 2 (λ + β + Janbu)
Shaft resistance in clay, Q_s (clay)	267.0 kN	311.3 kN
Shaft resistance in sand, Q_s (sand)	229.1 kN	287.8 kN
End bearing, Q_p	983.0 kN (limiting value governs)	1108 kN
Ultimate capacity, Q_u	1479 kN	1707 kN
Allowable load at FS = 3	493 kN	569 kN
Recommended design capacity: Q_all ≈ 490 kN (governed by the lower estimate), to be confirmed by a static load test on one pile in ten.

Check — assumed values and chart sources (page-1 Note 6).

α = 0.68 interpolated from the α-versus-c_u/p_a table in Das, Principles of Foundation Engineering, 9th ed., Ch. 11 (Terzaghi, Peck and Mesri).
N_q* = 143 at φ′ = 35° from the interpolated Meyerhof values for driven piles, Das Ch. 11; the limiting point resistance 0.5p_aN_q*tanφ′ is from the same section.
λ = 0.336 at 5 m embedment from the Vijayvergiya and Focht (1972) table, Das Ch. 11 (the table reads 0.500 at 0 m, 0.336 at 5 m, 0.245 at 10 m, 0.200 at 15 m). Because only 5 m of the pile is in clay, λ has been read against that length; reading it against the full 13 m instead would give λ ≈ 0.218 and reduce Q_s(clay) to about 202 kN, lowering Q_u(2) to roughly 1598 kN — a 6 per cent change that does not alter the conclusion.
K = 1.4K₀ and δ′ = 0.8φ′ are mid-range values for a high-displacement driven pile (Das Ch. 11 quotes K₀ to 1.8K₀); η′ = 105° is Janbu's value for dense sandy soil.
Critical depth L_c = 15D is the conventional value for a medium-dense to dense sand. It is a convenient design device rather than a measured phenomenon; omitting it raises Q_s(sand) in Method 1 to 306 kN and Q_u(1) to 1556 kN.
Section shape. The figure labels the dimension "Diameter", so a circular section is assumed. Precast piles are frequently square; a 0.5 m × 0.5 m square section would give A_p = 0.25 m² and p = 2.0 m, raising the Method 1 capacity to 1883 kN — 27 per cent higher. The section shape should be confirmed before the result is used.
The water table is load-bearing on this answer. Had the profile been dry, σ′_v(13) would be 245 kPa rather than 117 kPa; the Meyerhof point resistance would still be capped at the same limiting value, but the Janbu end bearing would roughly double. The submerged case computed here is the conservative and, per the figure symbol, the correct one.
The pile is driven into dense sand, so the end-of-driving (undrained) clay resistance is used for the α method. Long-term set-up in the clay would increase the shaft resistance; ignoring it is conservative.

Topic: Ultimate axial capacity of a single driven pile in a layered clay-over-sand profile, estimated by independent total-stress and effective-stress routes.

Key relations: Q_u = Q_p + Q_s; Meyerhof point resistance Q_p = A_pσ′_vN_q* ≤ A_p(0.5p_aN_q*tanφ′); Janbu N_q* = (tanφ′ + √(1 + tan²φ′))²e^{2η′tanφ′}; α method f = αc_u; λ method f_av = λ(σ′_v(av) + 2c_u); β method f = βσ′_v with β = Ktanδ′.

Why this works: A pile carries load in two physically distinct ways — by bearing on the soil beneath its toe and by shearing the soil along its shaft — and the two mechanisms mobilise at very different displacements, shaft friction at a few millimetres and end bearing at perhaps ten per cent of the diameter. Summing them at "ultimate" is a design convention, not a statement about simultaneous mobilisation, which is one reason a substantial factor of safety is used. The limiting point resistance exists because the linear growth of q_p with depth predicted by a rigid-plastic bearing-capacity theory is contradicted by every load test on a long pile: the sand around the toe is compressible, so it dilates less and the failure surface shrinks as confining stress rises, and the resistance flattens out. Requiring two methods, as this question does, is not a mathematical exercise but the standard safeguard against a single unrepresentative empirical coefficient carrying the whole design.

Common pitfall: Using total unit weights to compute σ′_v when the water table is at the surface, which more than doubles the effective stress at the toe and inflates the Janbu end bearing accordingly. A second and very common error is quoting the unlimited Meyerhof value — here 3298 kN of end bearing, more than three times the correct figure — because the limiting expression was forgotten. A third is applying the λ method to a thin sub-layer, when it is defined only on averages over the embedment.

Source: Das, Principles of Foundation Engineering, 9th ed., Ch. 11 (pile foundations: Meyerhof's method, Janbu's method, the α, λ and β methods, allowable capacity and load testing); CFEM 4th ed., Ch. 18 (axial capacity of deep foundations and recommended factors of safety).

Question 6 of 9: Ultimate capacity of a driven pile by two methods

Section B — Design Questions (answer any three; all four are answered)

Question 6: Ultimate capacity of a driven pile by two methods (24 marks)