Chapter Eleven

Pile Foundation

11.1 Introduction

Piles are structural members that are made of steel, concrete, or timber. They are used
to build pile foundations (classified as deep foundations) which cost more than
shallow foundations . Despite the cost, the use of piles often is necessary to ensure
structural safety. The most case in which pile foundations are required, is when the
soil supporting the structure is weak soil (expansive soil, or collapsible soil, etc...) we
use piles to transmit the foundation load to the nearest bed rock layer, and if bed rock
is not encountered, we use piles to transmit the load to the nearest stronger soil layer
to ensure the safety for the structure. The following figure clarifies the function of pile
foundation.
11.2 Capacity of Piles

The ultimate load capacity of the pile may be expressed as:

(11.1)

When the pile penetrates weak soil to rest on strong soil or bed rock, the pile will
supported by the bed rock or the strong soil from at the pile end (end point of pile),
So:

=Load carried at the pile end point

In addition, when the pile penetrates the soil, the shearing resistance between the soil
and the pile should be considered in where:

=Load carried by the skin friction developed at the sides of the pile (caused by
shearing resistance between the soil and the pile).
11.3 Types of Piles

1. Point Bearing Piles:

If the soil supporting the structure is weak soil, pile foundation will used to transmit
the load to the strong soil layer or to the bed rock (if encountered), here the pile will
resist the entire load depending on its end point load and the value of (frictional
resistance) is very small in this case, so:

(Point Bearing Piles)

2. Friction Piles:

When no strong layer or rock is present at reasonable depth at a site, point bearing
piles becomes very long (to reach strong layer) and uneconomical. In these type of
soil profiles, piles are driven through the softer (weaker) soil to specified depth, and
here the point bearing load ( ) is very small and can be considered zero, however
the load on the pile will resisted mainly by the frictional resistance between soil and
pile ( ) so:
 (Friction Piles)

In practice, we assume the pile resist the applied loads by its point bearing load and
its frictional resistance to estimate the ultimate load the pile can carry.

11.4 Calculation of Point Bearing Load( )

We will use Meyerhof method to calculate the value of for sand and clay.

Calculation of for sand:

= × ≤ (11.2)

=Cross-sectional area of the end point of the pile (bearing area between pile and
soil).
 (11.3)

= Effective vertical stress at the level of the end of the pile.

=Load capacity Factor (depends only on ϕ−value) as shown in Table 11.1.

Table 11.1: Values of for different soil friction angle

=Limiting value for point resistance

= 0.5 x × X X tan∅

=atmospheric pressure=(100 kN/ or 2000 Ib/ )

So, for sandy soil the value of 𝐢𝐬:

∅ (11.4)
Important Note:

The soil profile may consists of several sand layers, the value of friction angle (ϕ)
which used to calculate as shown in the above equation is the friction angle for the
soil that supporting the pile end (for last soil layer).

Calculation of 𝐏 for Clay:

= × × (11.5)

= Cohesion for the soil supported the pile at its end.

=Bearing capacity factor for clay=9 (when ∅=0.0)

=9× ×

Calculation of for 𝐂−𝛟 𝐒𝐨𝐢𝐥:

If the supporting the pile from its end is C−ϕ soil:

∅ × × (11.6)

Calculation of Frictional Resistance ( 𝐬)

Calculation of 𝐬 for sand:

The general formula for calculating 𝐬 is:

𝐬 𝐏 ∑ 𝐢 𝐢 (11.7)

P=pile perimeter=π×D (if the pile is circular, D=Pile diameter)

=4×D (if the pile is square, D=square dimension)

=unit friction resistance at any depth

=depth of each soil layer

f= ×N (11.8)

Here the value of (f) is vertical, so N must be perpendicular to f (i.e. N must be

=friction coefficient between soil and pile=tan δ

δ=soil−pile friction angle=0.8∅→μs=tan(0.8∅)(for each layer)

N= Horizontal stress from the soil to the pile

→N= ×K (for each soil layer)

=vertical effective stress for each layer

But, to calculate for each soil layer, to be representative, we take the average
value for for each layer.

K=Effective earth pressure coefficient

K=1−sin∅ or K= 0.5+0.008 =relative density (%)

If you are not given the relative density for each layer, use K=1−sinϕ
Now, N= ×K (for each soil layer)

→f=tan(0.8ϕ)× ×K (for each soil layer) (11.9)

Now, how we calculate the value of for each soil layer:

We draw the vertical effective stress along the pile, but the stress will linearly
increase to a depth of (15D), after this depth the stress will be constant and will not
increase. (this is true only if we deal with sandy soil). If there is one soil layer before
reaching 15D:

If there are more than one soil layer before reaching 15D:
Finally we can calculate the value of as following:

∑ ∅ (11.10)

i=each soil layer

Note:

We take soil layer every change in soil properties or every change in slope of vertical
stress.