Question

foundation

Pile on rock (resting and socket)

Design pile on rock depth 10 meter and load 1000 ton

Find the diameter of this pile

Answer 1

Socketing the piles in rock is being done, or rather being insisted upon, by cutting the sound rock for a minimum of 1D (for large diameter piles), where D = dia of pile.

Piles on rocks carry the load by point bearing. To ensure the proper contact between base of pile and rock surface, it is needed to socket the pile into hard rock by cutting through weathered and soft rock and by cutting the hard rock for a reasonable depth mainly to get the level surface of rock and to remove the top weathered surface of rock. This reasonable depth may vary from 150 mm to 300 mm.

Most of the times there are differences in recognizing the type of rock–weathered/soft/hard. First, the classification of rock layer is to be done properly. For this purpose we can refer to field test reports like RQD, CRR and SPT results and crushing strength of rock core samples.

Based on Uniaxial Compression Strength of Rock having safe load carrying capacity of the pile socketed in rock, diameter of pile can be calculated by combining the skin friction and end bearing components of the load.

Qs, is given by

Qs = qcNjNdAp + qcπDlsαβ

where qc = uniaxial compression strength of rock, T/m2

Nd = Depth factor = 0.8 + 0.2ls/D, limited to 2

Ap = Area of pile toe, sq m

D = Dia of pile, m

ls = socket length into rock, m

α = rock socket side resistance reduction factor

β = rock socket correction factor

Nj = values as taken from code- also depends on spacing of discontinuities

For sound rock having negligible discontinuities, qc can be taken as equal to safe strength of concrete in T/m2

Nj = 0.3

α = 0.05

β = 1 and

Nd = 1

For given load of 1000 tonne, pile with M40 grade concrete can be taken and socket length into rock = 10m , the dia of socketed pile will be derived.

The final calculation is left on the student as the grade of concrete is not mentioned. All other values have been well explained.