Question:
The load carrying capacity of an individual friction pile is \(200\) kN. What is the total load carrying capacity of a group of \(9\) such piles with the group efficiency factor of \(0.8\)?
The load carrying capacity of an individual friction pile is \(200\) kN. What is the total load carrying capacity of a group of \(9\) such piles with the group efficiency factor of \(0.8\)?
Show Hint
For pile groups,
\[
\boxed{
Q_g=\eta nQ_p
}
\]
If the group efficiency is less than \(1\), the pile group capacity is less than the sum of the individual pile capacities.
Updated On: Jul 23, 2026
- \(1800\) kN
- \(1640\) kN
- \(1440\) kN
- \(900\) kN
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
The load carrying capacity of a pile group is obtained by multiplying the capacity of one pile by the number of piles and the group efficiency.
The formula is
\[
\boxed{
Q_g=\eta \times n \times Q_p
}
\]
where
\[
\eta=\text{Group efficiency},
\]
\[
n=\text{Number of piles},
\]
\[
Q_p=\text{Load carrying capacity of one pile}.
\]
Step 1: Write the given data. \[ Q_p=200\text{ kN} \] \[ n=9 \] \[ \eta=0.8 \]
Step 2: Calculate the pile group capacity. \[ Q_g = 0.8\times9\times200 = 1440\text{ kN} \] Hence, \[ \boxed{Q_g=1440\text{ kN}.} \] Therefore, the correct option is \[ \boxed{(C)\;1440\text{ kN}.} \]
Step 1: Write the given data. \[ Q_p=200\text{ kN} \] \[ n=9 \] \[ \eta=0.8 \]
Step 2: Calculate the pile group capacity. \[ Q_g = 0.8\times9\times200 = 1440\text{ kN} \] Hence, \[ \boxed{Q_g=1440\text{ kN}.} \] Therefore, the correct option is \[ \boxed{(C)\;1440\text{ kN}.} \]
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