Question:

A 4 x 4 group pile, with each pile 20 m long and 500 mm in diameter, is installed in a square pattern in a clayey soil, as shown in the figure. The average unconfined compressive strength of the soil is 100 kN/m\(^2\), and the adhesion factor is 0.8. Neglect the bearing at the tip of the piles. For a group efficiency factor of 1.0, the centre to centre spacing(s) of the piles (in m) would be ......... (round off to two decimal places). 

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When designing pile foundations, always consider factors such as group efficiency, the adhesion factor, and the unconfined compressive strength of the soil. Ensure to account for the pile spacing to avoid excessive pile group interaction.
Updated On: Apr 3, 2025
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Solution and Explanation

The group efficiency \( \eta_g \) is calculated as: \[ \eta_g = \frac{Q_{{avg}}}{n Q_u} \] Where:
- \( \eta_g = 1 \) (group efficiency factor)
- \( Q_{{avg}} = \frac{UCS}{2} = \frac{100}{2} = 50 \, {kN/m}^2 \) (average unconfined compressive strength of the soil)
- \( Q_u \) is the ultimate bearing capacity of the pile
The total bearing capacity \( \overline{Q} \) is given by: \[ \overline{Q} = n Q_{{up}} = n \left( \alpha c \times (\pi \times D) \right) \] Where:
- \( n = 16 \) (number of piles)
- \( \alpha = 0.8 \) (adhesion factor)
- \( D = 0.5 \, {m} \) (diameter of the pile)
The pile spacing is calculated as: \[ \overline{Q} = 50 \times 4 \times B \times 20 = 16 \times 0.8 \times (\pi \times 0.5 \times 20) \] Now, solving for the spacing \( B \): \[ B = 5.026 \, {m} \] Thus, the spacing \( s \) of the piles is: \[ s = B + D = 5.026 + 0.5 = 1.51 \, {m} \] Thus, the centre to centre spacing of the piles is \( 1.51 \, {m} \).
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