In layered clay the shaft resistance is:

P x sum of alpha c_u L for each layer

End bearing uses c_u of:

The stiff clay at the tip, 90 kPa

Group allowable load is 2984 kN because:

Sum of piles (7459 kN) is smaller, then /2.5

2076 Asoj · 8 marks

Steel pipe pile group 3×4, layered clay

Short answer

Single pile skin friction by α-method: 564.4 kN, end bearing 57.3 kN, ultimate 621.6 kN. Sum for 12 piles 7459 kN; block capacity 14895 kN. Group capacity governed by sum, allowable 2984 kN with FS 2.5.

A 15 m long closed end steel pipe pile group (3×4) consists of 12 piles of a 300 mm diameter and evenly spaced at 900 mm, center-to-center, is driven into layered undrained clay. The top 6 m consists of clay with undrained cohesion of 50 kPa and adhesion factor of 0.74, followed by 6 m of clay with undrained cohesion of 65 kPa and adhesion factor of 0.62, which was underlain by stiff clay with undrained cohesion of 90 kPa and adhesion factor of 0.50. Estimate the allowable load carrying capacity of pile group. Take factor of safety = 2.5.

Given: closed-end steel pipe piles, 3 × 4 group (12 piles), $D = 0.30$ m, $L = 15$ m, spacing $0.9$ m. Layers: 0–6 m ( $c_u = 50$ , $\alpha = 0.74$ ); 6–12 m ( $c_u = 65$ , $\alpha = 0.62$ ); 12–15 m ( $c_u = 90$ , $\alpha = 0.50$ ). FS $= 2.5$ .

1. Single pile

\begin{aligned} P &= \pi(0.3) = 0.9425\ \text{m},\quad A_p = 0.07069\ \text{m}^2 \\ \sum\alpha c_uL &= 0.74(50)(6) + 0.62(65)(6) + 0.50(90)(3) = 222.0 + 241.8 + 135.0 = 598.8\ \text{kN/m} \\ Q_s &= 0.9425 \times 598.8 = 564.4\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 90 \times 0.07069 = 57.3\ \text{kN} \\ Q_u &= 621.6\ \text{kN} \end{aligned}

Sum for 12 piles $= 12 \times 621.6 = 7459.3$ kN.

2. Block failure

\begin{aligned} B_g &= 2(0.9) + 0.3 = 2.1\ \text{m},\quad L_g = 3(0.9) + 0.3 = 3.0\ \text{m},\quad P_g = 2(2.1 + 3.0) = 10.2\ \text{m} \\ Q_{block} &= P_g\sum c_uL + 9c_uB_gL_g \\ &= 10.2(50 \times 6 + 65 \times 6 + 90 \times 3) + 9(90)(2.1)(3.0) \\ &= 9792 + 5103 = 14\,895\ \text{kN} \end{aligned}

3. Group capacity

Group capacity $=$ smaller of $7459.3$ kN and $14\,895$ kN $= 7459.3$ kN (efficiency 1).

Q_{all} = \frac{7459.3}{2.5} = 2983.7\ \text{kN}

Answer: allowable load of the pile group $\approx 2984$ kN.

✓ Answer checked by Claude Sonnet 5.5

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Steel pipe pile group 3×4, layered clay

1. Single pile

2. Block failure

3. Group capacity

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