Group Action of Piles

By-Prateek Saxena (091541)Rachit Khandelwal (091542)Rupesh Mishra (091543)

Introduction

A structure is never founded on a single pile.

The bearing capacity of a pile group is not necessarily the capacity of the individual pile multiplied by the number of piles in the group.

The phenomenon by virtue of which this difference occurs is known as ‘group action of piles’.

Number of Piles

Piles for walls are commonly installed in a staggered arrangement to both sides of the centre line of the wall.

For a column, at least three piles are used in a triangular pattern, even for small loads.

When more than three piles are required, the arrangement of piles is symmetrical about the point or area of load application.

Pile Group Pattern

Pile CapColumn and wall loads are usually transferred to the pile

group through a pile cap, which is typically a reinforced concrete slab structurally connected to the pile heads to help the group act as a unit

Spacing of Piles

It Depends upon the following factors

Overlapping of stresses of adjacent piles

Cost of foundation

Desired efficiency of the pile group.

Minimum SpacingThe spacing may vary from 2d to 6d for straight uniform

cylindrical piles. For friction piles, the recommended minimum spacing is 3d. For point-bearing piles, the minimum spacing is 2.5d when

the piles rest in compact sand & 3.5d when the piles rest in stiff clay.

The minimum spacing may be 2d for compaction piles.

where d = diameter of the pile.

Group Capacity of PilesThe capacity of a pile group is not necessarily the capacity of

the individual pile multiplied by the number of individual piles in the group.

Disturbance of soil during the installation of the pile and overlap of stresses between the adjacent piles, may cause the group capacity to be less than the sum of the individual capacities.

Determination of Pile Group Capacity

To determine the capacity of a pile group, the sum of the capacities of the individual piles is compared with the capacity of the single large equivalent (block) pile.

The smaller of the two values is taken.

Applying an appropriate factor of safety to this chosen value, the design load of the pile group is obtained.

Qup= Qeb + Qsf = qu Ab + fs As

Where Qup=Ultimate Bearing Capacity of the Pile

Qeb=End Bearing Resistance of the Pile

Qsf=Skin Friction of the Pile

1.Pile Group in SandQu= σ Nq Ap + σav K Tanδ As

2. Pile Group in ClayQu= Cu Nc Ap + α Cu As

Pile Group EfficiencyThe ‘efficiency’, ηg, of a pile group is defined as

the ratio of the group capacity, Qg , to the sum of the capacities of the number of piles, n, in the group:

ηg =Qg/(n.Qp)

Numerical

Q.1 200mm dia 8m long concrete piles are used as foundation for a column in a uniform deposit of (a) medium clay and (b) sand having qu=100 KN/m2 The Spacing between the piles is 500 mm. There are 9 piles in the ground arranged in a square pattern. Calculate the ultimate pile load capacity of the group. Assume Adhesion factor = 0.9Φ=32 , Nq=27

Unit weight of soil = 18 KN/m3

Case-1 -> ClayUnconfined Compressive Strength, qu=100 KN/m2

Cu = qu/2= 50 KN/m2

Ultimate Capacity of a single pile is given byQu= Cu Nc Ap + α Cu As

= (50*9*3.14*(.2)2)/4 + 0.9*50*3.14*0.2*8 = 14.137 + 226.194 = 240.33. KN

Ultimate Pile Load Capacity of pile group= 9*240.333 = 2163 KN

Width of the Pile group= 0.5+0.5+0.1+0.1= 1.2m

Ultimate Load Capacity of the pile group Qug=Cu Nc Ab + Pb L Cu

= 50*9*1.2*1.2 + 4*1.2*8*50

= 2568 KN

Taking the lower Value, the ultimate load capacity is 2163 KN

Case-2 -> Sandδ=(3Φ)/4= 240

K=1 (Concrete Pile)Critical Depth of the pile Lc= 15 D= 15 *0.2=3.0 m

Limiting σ at 3 m depth=18 *3 = 54 KN/m2

From 3m to 8m depth , unit point bearing resistance qpu and unit skin friction resistance remain constant at σ=54 KN/m2

qpu=σ Nq and fs= σav K Tanδ

qpu= 54*27= 1458 KN/m2

Over the Length 3m σav= 54/2 = 27 KN/m2

fs=27 *1.0* Tan 24= 12 KN/m2

Skin Friction Resistance= fs As= 12 * 3.14*.2*3=22.61 KN

Over the length 5m σav= 54 KN/m2

fs=54 *1.0* Tan 24= 24 KN/m2

Skin Friction Resistance= fs As= 24 * 3.14*.2*5=75.39 KN

Qf=22.61 +75.39= 98 KN

Now Qpu= qpu Ab

= 1458*3.14*0.22/4=45.78 KN

Qu= 98+45.78= 143.78 KN

For 9 Piles= 9*143.78= 1294 .03 KN

Thank You