2078 Kartik · 7 marks

Friction pile length, 350 mm, 1200 kN

Short answer

Shaft load =840 kN (70% of ultimate). Using Q_s = α c_u π D L, L =840/(0.8×40×π×0.35) =23.87 m. Required length ≈24 m.

A friction pile 350 mm diameter is proposed to be driven in a layer of uniform cohesive soil with unconfined compressive strength of 80 kPa. Considering the pile end carry only 30% of the total load, determine the length of the pile required for the ultimate load of 1200 kN. Assume adhesion factor = 0.80.

Given: $D = 0.35$ m, $q_u = 80$ kPa so $c_u = 40$ kPa, $\alpha = 0.80$ , ultimate load $Q_u = 1200$ kN, end bearing is 30% of the total load.

\begin{aligned} Q_p &= 0.30 \times 1200 = 360\ \text{kN} \\ Q_s &= 1200 - 360 = 840\ \text{kN} \end{aligned}

Shaft friction: $Q_s = \alpha c_u\pi DL$

\begin{aligned} 840 &= 0.80 \times 40 \times \pi \times 0.35 \times L = 35.19\,L \\ L &= \frac{840}{35.19} = 23.87\ \text{m} \end{aligned}

Answer: length of pile required $\approx 24$ m.

✓ Answer checked by Claude Sonnet 5.5

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