Symmetric pile group is discussed a lot in literature. We have to opt for the option of Asymmetric pile group due to some obstructions or when it is found that some of the piles among the group are defective and new piles are proposed.
First query which comes in this regard is whether we should refer to CG of pile cap or pile group. CG of pile cap is relevant for spread foundations(as we assume that foundation is resting on uniform soil across the foundation plan area) but not for pile foundations. For pile foundations, CG of pile group is governed by the positions of piles.
Now second question comes is how to calculate CG of pile group. Same is explained below with the example :
Pile group of 5 piles is considered. Due to shift of pile (P2) on left side, pile group has become asymmetric and this will also result in shift of CG axis by x which is unknown and need to be determined.
Method 1 – First method which I came across is taking arithmetic average of pile co-ordinates (either x co-ordinates or y co-ordinates).
2.7+2.7+0-2.7-4.45 = -1.75. Divide it by number of piles
x = -1.75 / 5 = 0.35 Shift is 0.35
I tried to investigate mechanics behind this simple method which is explained below.
I tried to develop equation, summation of forces (compressive and tensile) = 0 about new CG axis. Although you have 5 pile reactions, you can express those in terms of P1 only based on the assumption that pile reaction varies linearly.
[(2.7-x)m]+[(4.45-x)m]-mx-[(2.7+x)m]-[(2.7+x)m]=0
m is reaction which develops at 1 m distance from CG axis. m get’s cancelled here and
shift x = 0.35
Method 2 – Load distribution among the pile group is non uniform due to moment load acting on the pile cap. Pile cap is expected to rotate about some point due to moment acting on the pile cap. Obviously, equal and opposite moment need to be provided by pile reactions.
Accordingly equation can be written for condition, moment of compressive pile reactions about CG of pile group is equal to moment tensile pile reactions about CG of pile group.
P2 = P1 X (4.45-x) / (2.7-x)
P3 = (P1 X x)/(2.7-x)
P4 = P1 X (2.7 + x) / (2.7-x) & P4=P5
P1((2.7-x)^2) + P1((4.45-x)^2) = P1(x^2) + 2P1((2.7+x)^2)
x=0.488
It is clear that both methods do not give same answers. They may give same answers for the case of symmetrical pile groups.
However, for asymmetric pile groups, method 1 do not seems right method as it is based on the assumption that summation of compressive reactions is equal to summation of tensile reactions. In this case, since pile cap is rotating, our focus should be on moments due to compressive and tensile reactions as considered in method 2.