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A pile group of 25 piles has to be proportioned in a uniform pattern in soft clay with equal spacing in all directions.

Assuming the value of cohesion is to be constant throughout the depth of piles, determine the optimum value of spacing of the piles in the group. Assuming adhesion factor 0.7. Neglect the end bearing effect given the diameter of pile as 0.5 m. Also calculate the efficiency of group using converse Labarre formula.

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Given Data: $$\begin{array}{l} d=0.5 m \\ \alpha=0.7 \end{array}$$

• Case I :-Considening as individeral pile,

$Q_{u}=Q_{f}+Q_{up}$ (neglected) $Q_{u}=\alpha \mathrm{CuAs}_{4}$ $Q_{u}=0.7 \mathrm{Cu} \times \pi \times 0.5 \times \mathrm{L}$

• Qug1 $=25 \times 0.7 \times \pi \times 0.5 \times \mathrm{Cu×L}$
• Qug1 $=27.49 \mathrm{CuL}$

• Case 2 :- Considering as Group Piles

$$\begin{array}{l} Q u g_{2}=\alpha \mathrm{cu} \text { As } \\ Q ug_{2}=1 \times \mathrm{Cu} \times 4 \times(4S+0.5) \times \mathrm{L} \end{array}$$ Thus, $Q u g 1=Q ug_{2}$ $\Rightarrow 27.49 \mathrm{Cu} L=4 \times(4S+0.5) \mathrm{CuL}$ $\Rightarrow \quad 4 \mathrm{~L}+0.5=6.87$

• Hence,S=1.593m

• As Per Is code

Smin $=3 d=1.5 \mathrm{~m}\lt S(OK)$ Thus, Provide,spacing $=1.6 \mathrm{~m}(\mathrm{say})$

• Group Efficiency as per Converse labarres

\begin{aligned} \theta &=\tan ^{-1}\left(\frac{d}{S}\right) \\ \Rightarrow & n g=1-\frac{17.354}{90} \times\left[\frac{(4 \times 5)+(5 \times 4)}{5 \times 5}\right] \\ n_{g} &=0.6915=69.15 \% \end{aligned}

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