# Design Example of Pile Cap for Concentric Loading

A

*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***pile***is reinforced with longitudinal bars of 12 No. 7 bars. Consider***column****f**=60 Ksi and_{y}**f’**=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._{C}Complete pile cap design will be illustrated here; except

superimposed load weight of pile cap also exerts some

consider cap weight exerts 3 kips load on each pile.

superimposed load weight of pile cap also exerts some

*. Lets***load on piles**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

Factored load = (1.2X160+1.6X170) kips

=464 kips

i.e. Ultimate load on pile cap, W

_{u }= 464 kips

Ultimate load on pile, P

_{u }= 464/9 kips/pile

=51.56 kips/pile <70 kips

(ultimate

(ultimate

*).***pile capacity**

Let’s try to check punching shear failure; at first consider

punching shear due to single pile:

punching shear due to single pile:

Assume an effective depth of cap=18”

Thus punching perimeter, b

(pile diameter + effective depth of cap)

_{o}= π X(pile diameter + effective depth of cap)

=πX(12+18)

inches

inches

=90.25”

Now, permissible shear around single pile, V

4√f’

b

_{c}=4√f’

_{c}b

_{0}d=4X√3000

X90.25X18

X90.25X18

=355910 lb

=356 kips

Here shear force =Nominal reaction of pile=P

_{n}

Now

=51.56/0.75

=68.88 kips <356 kips.

This means no pile will be punched when effective depth,

d=18”

d=18”

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?

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.

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

cross-section of pile is chosen, pile may not be punched. But calculation for

punching of column through

*must be checked for all situations.***pile foundation**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”.

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”.

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

=27.75”

The practical depth provided=28.0”

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

=18.25”

So, b

_{0}=4 X (18.25+18)=145”

Now

ϕV

b

_{c }= ϕ4√f’_{c}b

_{0}d

=0.75X4X√3000

X145X18.25

X145X18.25

=434823 lb

=435 kips

V

_{u}=P_{u}X8=51.56X8

=412.5 kips < ϕV

kips

_{c}=435kips

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 V

section passes through a cross-section of pile or piles by any fraction,

this/those piles reactions are also contribute (added) in calculating V

But a partial contribution is considered (P

outside the critical section).

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 V

_{u}. If criticalsection passes through a cross-section of pile or piles by any fraction,

this/those piles reactions are also contribute (added) in calculating V

_{u}.But a partial contribution is considered (P

_{u}divided by fractionoutside the critical section).

Now we calculate beam shear capacity of pile cap. We

know critical section lies at a distance d from face of column (figure below).

know critical section lies at a distance d from face of column (figure below).

The

width of pile cap, B= (Pile c/c distance) X2+(Pile

dia)X1/2X2+(pile

face to cap edge distance)X2

width of pile cap, B= (Pile c/c distance) X2+(Pile

dia)X1/2X2+(pile

face to cap edge distance)X2

=3’X2+6”/12X2+8”/12X2

=8’-4”

=100”

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

distance between column face to pile face (inside). In one word, clear distance

among them = 3’-9”/2-6”/12

=1.75”

If all piles lie inside of critical section for beam shear,

no shear is considered to produce shear failure in foundation.

no shear is considered to produce shear failure in foundation.

Thus ϕV

_{c}= ϕ2√f’_{c}bd=0.75X2X√3000

X100X18.25

X100X18.25

=150 kips

V

_{u}=3X51.56=154.7 kips >ϕV

kips

_{c}=150kips

i.e.,V

_{u}>ϕV_{c}Thus depth of pile should

be increased, let consider d=19”

be increased, let consider d=19”

ϕV

_{c}= ϕ2√f’_{c}bd

=0.75X2X√3000

X100X19

X100X19

=156kips >154.7kips (V

_{u})This means the effective depth of 19”is enough to avoid

failure due to punching and beam shear. Beam shear is often called flexure

shear.

[for concrete of f’

failure due to punching and beam shear. Beam shear is often called flexure

shear.

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.

located at the face of column. For each direction there have three piles to

contribute to bending moment.

M

_{n}=3×51.56x(21”+6”)x1/12

=348.03 kip-ft

In normal beam section design, the following procedure

is followed, if depth is not fixed.

is followed, if depth is not fixed.

*β*

_{1}=0.85

[for concrete of f’

_{c}≤ 4000 psi]

Equation (1) is used for strength reduction factor=0.9

We need to establish

Mu ≤ ϕ M

_{n}

= 0.9×0.0135x60xbd

^{2}(1-0.59×0.0135×60/3)

=0.61x bd

^{2}

=0.61x100xd

^{2}^{=61d2 }

Consider Mu

= ϕ

M

= ϕ

M

_{n}or, 348=61d

^{2}or,d

^{2}= 348/61=5.70or, d= 2.38”

In checking moment capacity above procedure based on p

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.

_{max}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 in

^{2}Now to check assumed a:

=0.9835

in

in

Assume a=0.98 in

i.e.

no more iteration is required. Thus required A_{s}=4.18 in^{2}

Now check minimum steel area:

=3x√3000x100x19x1/60000

≥200x100x19x1/60000

≥200x100x19x1/60000

=5.20≥6.33 in

^{2}The controlling value of minimum reinforcement is 6.33 in

which is more than bending requirement of steel, 4.18 in

^{2}which is more than bending requirement of steel, 4.18 in

^{2}.i.e. required steel area, A

_{s}=6.33 in^{2}.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

A

_{s (min) }=0.002bh for f_{y}=40 ksi=0.0018bh for f

_{y}=60 ksiEarlier we have assumed cap exerts 3 kips load on each pile. Now

we have to check whether it satisfies by 19” depth or not.

we have to check whether it satisfies by 19” depth or not.

Total depth provided=19”+3”+6”+6”/8

=28.75

Total weight of pile cap =8.33×8.33×28.75×150/12

=24936.6lbs

=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.

in all respect.

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.

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.