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

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written 6.1 years ago by

teamques10 ★ 69k

modified 2.9 years ago by

Chandan15 • 300

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.

advanced geotechnical engineering

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written 2.9 years ago by

Chandan15 • 300

• modified 2.9 years ago

pile 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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