Design Example of Pile Cap for Concentric Loading

A pile cap have to support a 18”X18” column which is subjected to live load=170 kips and dead load=160 kips under service loading. The column is reinforced with longitudinal bars of 12 No. 7 bars. Consider fy=60 Ksi and f’C =3 Ksi. The diameter of pile is 12”. The ultimate pile capacity=70 kip/pile and service load capacity=42 kip/pile as conformed by testing.
Complete pile cap design will be illustrated here; except
superimposed load weight of pile cap also exerts some load on piles. Lets
consider cap weight exerts 3 kips load on each pile.
Thus net service load = (42-3) kips/pile


=39 kip/pile


Total unfactored load= (170 + 160) kips


=330 kips
Pile number required, N =330/39
=8.46 or, 9 piles

Let’s arrange piles as per following pattern:

Pile arrangement in pile cap (9 piles)
Factored load = (1.2X160+1.6X170) kips
=464 kips
i.e. Ultimate load on pile cap, Wu = 464 kips


Ultimate load on pile, Pu = 464/9 kips/pile



=51.56 kips/pile <70 kips
(ultimate pile capacity).




Let’s try to check punching shear failure; at first consider
punching shear due to single pile:



Assume an effective depth of cap=18”


Thus punching perimeter, bo = π X
(pile diameter + effective depth of cap)
Punching perimeter around pile



Now, permissible shear around single pile, Vc=
=355910 lb

=356 kips
Here shear force =Nominal reaction of pile=Pn



pile cap design-1








=68.88 kips <356 kips.



This means no pile will be punched when effective depth,



Now what can be punched through pile cap? One is pile
itself; another is column or columns supported by pile cap. Should we check
both cases?


It depends on experience of designer, if he or they has/
have not enough experience, should check both. The basis is to have idea about
punching perimeter relative to load. If a column is heavily loaded and he
selects relatively small size of piles (cross-section) with more pile, the
piles can be punched.

Again if column load is small and relatively greater
cross-section of pile is chosen, pile may not be punched. But calculation for
punching of column through pile foundation must be checked for all situations.

Let’s calculate perimeter shear around 18”X18” column.
Consider No.6 (20 mm) bars will be used in pile cap. A 3” clearance is assumed
above pile butt, the pile is embedded in cap by 6”.
Total depth calculation of pile cap


Thus total depth becomes=18”+3”+6”+6/8




The practical depth provided=28.0”


i.e., effective depth will becomes =28.0”-3”-6”-6″/8




So, b0 =4 X (18.25+18)





ϕVc = ϕ4f’c




=434823 lb


=435 kips

=412.5 kips < ϕVc=435
Therefore column will also not be punched.

As shown in Figure below, the critical section runs by
a distance d/2 from column face. Only one pile falls in side of this perimeter,
thus remaining (9-1) =8 Nos.piles contribute in calculating Vu. If critical
section passes through a cross-section of pile or piles by any fraction,
this/those piles reactions are also contribute (added) in calculating Vu.
But a partial contribution is considered (Pu divided by fraction
outside the critical section).
Critical perimeter for punching around column


Now we calculate beam shear capacity of pile cap. We
know critical section lies at a distance d from face of column (figure below). 

Calculation of width of pile cap
width of pile cap, B= (Pile c/c distance)
face to cap edge distance)


The critical section lies 18.25” from column face. Now
distance between column face to pile face (inside). 
In one word, clear distance
among them = 3’-9”/2-6”/12
If all piles lie inside of critical section for beam shear,
no shear is considered to produce shear failure in foundation.

Thus ϕVc= ϕ2f’cbd
=150 kips


=154.7 kips >ϕVc=150

Thus depth of pile should
be increased, let consider d=19”


ϕVc= ϕ2f’cbd

=156kips >154.7kips (Vu)
This means the effective depth of 19”is enough to avoid
failure due to punching and beam shear. Beam shear is often called flexure

Let’s check bending failure. The critical section is
located at the face of column. For each direction there have three piles to
contribute to bending moment.





=348.03 kip-ft
In normal beam section design, the following procedure
is followed, if depth is not fixed.


[for concrete of f’c≤ 4000 psi]
Equation (1) is used for strength reduction factor=0.9



We need to establish


Mu ≤ ϕ Mn



Minimum depth for bending moment


= 0.9×0.0135x60xbd2(1-0.59×0.0135×60/3)


=0.61x bd2



Consider Mu
or, 348=61d2
or,d2= 348/61=5.70
or, d= 2.38”
In checking moment capacity above procedure based on pmax
is not valid here as effective depth is governed mainly by shear and required
depth (2.38”) in far below provided depth 19”. Here in most cases, minimum steel
area governs. Lets check required steel area for this moment.
Assume, a=1 in
=4.18 in2
Now to check assumed a:


Assume a=0.98 in
= 4.18 in2

no more iteration is required. Thus required As=4.18 in2


Now check minimum steel area:

=5.20≥6.33 in2
The controlling value of minimum reinforcement is 6.33 in2
which is more than bending requirement of steel, 4.18 in2.
i.e. required steel area, As =6.33 in2.
Number of bar required= 6.33/0.44
= 14.4
≈15 (as per previous assumption, No.6 bars are considered)
As ACI code do not allow plain concrete (unreinforced) in pile cap, we have to provide minimum steel as calculated above. Structural engineers sometimes also use shrinkage and temperature steel as minimum steel. 
These are
As (min) =0.002bh for fy=40 ksi
=0.0018bh for fy=60 ksi
Earlier we have assumed cap exerts 3 kips load on each pile. Now
we have to check whether it satisfies by 19” depth or not.
Total depth provided=19”+3”+6”+6”/8
Total weight of pile cap =8.33×8.33×28.75×150/12
=24.93 kips
Now each pile share equal load this is 24.93/9=2.77 kips
This is below our assumed load’ thus 19” depth is satisfactory
in all respect. 
Reinforcement details of pile cap
If share load (2.77 kips) calculated above is more than assumed
load (3 kips), we have to revise total calculation assuming new cap load to
calculate net service load on pile.  
If opposite situation arise, as in this example, the design is safe but to achieve economy we have check rational difference between two values.

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