Question:
The load carrying capacity of an individual friction pile is $300 \text{ kN}$. What is the total load carrying capacity of group of 10 such piles with a group efficiency factor of 0.8?
The load carrying capacity of an individual friction pile is $300 \text{ kN}$. What is the total load carrying capacity of group of 10 such piles with a group efficiency factor of 0.8?
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Group efficiency is usually less than 1 for friction piles in loose sand or clay because pressure bulbs overlap, reducing the effective soil resistance.
Updated On: May 20, 2026
- $240 \text{ kN}$
- $2400 \text{ kN}$
- $3000 \text{ kN}$
- $300 \text{ kN}$
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The Correct Option is B
Solution and Explanation
Concept:
The total capacity of a pile group ($Q_g$) is not necessarily the simple sum of individual capacities. It is adjusted by the group efficiency factor ($\eta_g$), which accounts for the overlap of stress zones between adjacent piles.
Step 1: Apply the group capacity formula.
The formula is: \[ Q_g = n \times Q_i \times \eta_g \] Where:
• $n = 10$ (number of piles)
• $Q_i = 300 \text{ kN}$ (individual capacity)
• $\eta_g = 0.8$ (efficiency factor)
Step 2: Calculation.
\[ Q_g = 10 \times 300 \times 0.8 \] \[ Q_g = 3000 \times 0.8 \] \[ Q_g = 2400 \text{ kN} \]
Step 1: Apply the group capacity formula.
The formula is: \[ Q_g = n \times Q_i \times \eta_g \] Where:
• $n = 10$ (number of piles)
• $Q_i = 300 \text{ kN}$ (individual capacity)
• $\eta_g = 0.8$ (efficiency factor)
Step 2: Calculation.
\[ Q_g = 10 \times 300 \times 0.8 \] \[ Q_g = 3000 \times 0.8 \] \[ Q_g = 2400 \text{ kN} \]
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