Foundation, Concrete and Earthquake Engineering

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

=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)

X(12+18) inches

=90.25”

Now, permissible shear around single pile, Vc= 4f’c b0d

=4X3000 X90.25X18

=355910 lb

=356 kips

Here shear force =Nominal reaction of pile=Pn

Now

=51.56/0.75

=68.88 kips <356 kips.

This means no pile will be punched when effective depth, 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?

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”.
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, b0 =4 X (18.25+18)

=145”

Now

ϕVc = ϕ4f’c b0d

=0.75X4X3000 X145X18.25

=434823 lb

=435 kips

Vu=PuX8

=51.56X8

=412.5 kips < ϕVc=435 kips

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).

Now we calculate beam shear capacity of pile cap. We 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

=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
=1.75”

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
=0.75X2X3000 X100X18.25
=150 kips

Vu=3X51.56
=154.7 kips >ϕVc=150 kips

i.e.,Vu>ϕVc

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

ϕVc= ϕ2f’cbd

=0.75X2X3000 X100X19

=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 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.

Mn=3x51.56x(21”+6”)x1/12

=348.03 kip-ft

In normal beam section design, the following procedure is followed, if depth is not fixed.

β1=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 Mn

= 0.9x0.0135x60xbd2(1-0.59x0.0135x60/3)

=0.61x bd2

=0.61x100xd2
=61d2

Consider Mu = ϕ Mn

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:

=0.9835 in
Assume a=0.98 in
i.e. no more iteration is required. Thus required As=4.18 in2

Now check minimum steel area:

=3x√3000x100x19x1/60000 ≥200x100x19x1/60000

=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

=28.75

Total weight of pile cap =8.33x8.33x28.75x150/12

=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.
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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