The design and selection of a particular type of foundation is a work of the RCC consultant The construction person must know the background so that the interaction with the designer is easy. Read this article so that you know the basics.

Today we shall have a look at the primary considerations for calculating the basic parameters of a friction pile.

But before we take up the discussions , let us have a look at this picture :

Now ,a look at the Friction piles :

Whenever, the depth of the hard strata is so deep that boring up to that depth is either not economical or it takes too much time , it is suggested to go for Friction Piles.

The load is carried by the friction developed between the concrete surface and the surrounding soil

Now let us study the basic guiding calculations :

To illustrate this point , let us solve a few examples :

Another example :

Before making a sketch of the Pile cap , have a look at this conceptual diagram :

As a construction person , you should also be able to calculate the material required for the construction . Today , we shall learn how to calculate the materials required for the Pile foundations.

Data :

Pile Shape : Circular

Pile Diameter : 600 mm

Pile Depth : 8.35 M.

Mix of concrete : M20

Reinforcement: 6 HYSD bars of 16mm diameter .

Stirrups : 8 mm diameter bars provided at a spacing of 200 mm C/C [ Center to Center ]

As a construction person , we know the following :

1.Unit weight of 16mm dia HYSD bars : 1.58 Kg/m

2.Unit weight of 8mm dia HYSD bars : 0.39 Kg/m

3.Cover for Pile reinforcement : 50 mm

4.Material consumption for Concrete M20 :

Cement : 7.87 Bags/ CuM

Sand : 0.41 CuM.

Aggregate : 0.83 CuM.

4.Formula for volume of the cylinder = (0.785).D^{2}.H where D : Pile diameter in M & H : Pile height in M.

Having thus prepared , we shall now go for the actual calculations as follows .

Part A : REINFORCEMENT :

A.01 : Longitudinal bars :

Cutting Length of bar : 8350 –( 50+50 ) = 8250 mm = 8.25 M.

Weight :

6 Nos X 8.25 m X 1.58 Kg/m = 78.21 Kg.

A.02 : Stirrups :

Stirrup diameter : 600 –( 50+50 ) = 500 mm.

Cutting length : ( 3.14) D +( 2x10d )

D : Stirrup Diameter =500 mm

d: Diameter of bar used for stirrup = 8 mm.

= (3.14 x 500 )+( 2x 10×8) = 1730 mm = 1.73 M.

No of stirrups : ( 8350 / 200 )+1 = 43 Nos

Weight : 43 Nos x 1.73M x 0.39 Kg/m = 29 Kg.

Summary of Reinforcement :

16mm dia HYSD bars : 78 Kg / Pile

8mm dia HYSD bars : 29 Kg / Pile

Part B : M20 Concrete :

Volume of pile :

(0.785).D^{2}.H

=0.785 x 0.6 x 0.6 x 8.35

=2.36 CuM /Pile

Cement required :

2.36 x 7.87 bags = 18.57 bags

Sand Required :

2.36x 0.41 CuM = 0.968 CuM.

Aggregate Required :

2.36x 0.83 CuM = 1.96 CuM.

Final Summary :

Material required per pile of 600 mm dia. having a length of 8.35 M.

According to Karl von Terzaghi, a foundation is shallow if its depth is equal to or less than its width. Foundations constructed by open excavation are also called shallow foundations.

The term footing is used in conjunction with shallow foundations commonly. A footing is a foundation unit constructed in brick work, masonry or concrete under the base of a wall or a column for the purpose of distributing the load over a large area. A footing or a shallow foundation is placed immediately below the lowest part of the Superstructure supported by it.
Some examples of footings are:

Spread footings for columns

Spread footings for walls

Combined footings for columns and walls

Strap footing

The Deep foundations are simply called foundations. The depth of the foundation is more than the width. Some examples are:

Pile foundation

Stages in forming vibro-expanded piles

Well foundation

Coffer dams

The two terms are also defined as : ‘A foundation is the part of the structure which is in direct contact with the ground to which the loads are transmitted, while the lowermost portion of the foundation which is in direct contact with the sub-soil is called the footing.’