Pile Foundation Design: A Student Guide

SINGLE PILE DESIGN

5.1 End bearing piles

If a pile is installed in a soil with low bearing capacity but resting on soil beneath with high bearing capacity, most of the load is carried by the end bearing.

In some cases where piles are driven in to the ground using hammer, pile capacity can be estimated by calculating the transfer of potential energy into dynamic energy . When the hammer is lifted and thrown down, with some energy lose while driving the pile, potential energy is transferred into dynamic energy. In the final stage of the pile’s embedment,On the bases of rate of settlement, it is able to calculate the design capacity of the pile.

For standard pile driving hammers and some standard piles with load capacity (F_Rsp,), the working load for the pile can be determined using the relationship between bearing capacity of the pile, the design load capacity of the pile described by: F_Rsp ³ g _n× F_Sd and table 5-1

where: F_Sd = design load for end baring.

The data is valid only if at the final stage, rate of settlement is 10 mm per ten blow. And pile length not more than 20 m and geo-category 2 . for piles with length 20 - 30 m respective 30 - 50 m the bearing capacity should be reduced by 10 res. 25%.

Table 5-1 Baring capacity of piles installed by hammering

Example 5.1

A concrete pile with length 26 m and cross-sectional area (235)× (235) is subjected to a vertical loading of 390 kN (ultimate) load. Determine appropriate condition to halt hammering. Type of hammer Drop hammer activated by rope and friction winch. Class 2, GC 2, pile length 20 m

solution:

F_Rsp³ g _n. F_sd

g _n = 1.1 (table 10-3)

vertical load 390 kN Þ F_Rsp³  (1.1)390/0.9***= 477kN

Pile cross-sectional area Þ 0.235² = 0.055 m² Þ

type of hammer: Drop hammer activated by rope and friction winch Þ

***For piles 20m - 30m length, the bearing capacity should be reduced by 10%

\ Table value (table 5-1): Hammer weight = 4 ton Þ fall height 0.45m (interpolation)

Hammer weight = 3 ton Þ fall height 0.54 m

4 ton hammer with fall height 0.45m is an appropriate choice.

5.2 Friction piles

Load on piles that are driven into friction material, for the most part the weight is carried by friction between the soil and the pile shaft. However considerable additional support is obtained form the bottom part.

In designing piles driven into friction material, the following formulas can be used

where: q_ci = consolidation resistance

*a can be decided using table 10-4

A_b = end cross-sectional area of the pile

A_mi = shaft area of the pile in contact with the soil.

should be ³ 1.5 for piles in friction material

q_cs = end resistance at the bottom of the pile within 4 × pile diameter from the end of the pile

Example 5.2

Pile length 22 m, steel pile, friction pile, external diameter 100 mm, GC2,

Determine the ultimate bearing capacity of the pile

solution:

The values are slightly scattered then the usual while the rest of the condition is favourable.

g _Rd = 1.5 (the lowest value)

g _n = 1.1

 

At the base where condition is unfavourable we get :

a _s = 0.5

a _m = 0.0025

Design bearing capacity of the pile is 62 KN.

5.3 Cohesion piles

Piles installed in clay: The load is carried by cohesion between the soil and the pile shaft. Bearing capacity of the pile can be calculated using the following formula for pile installed in clay.

Where:

a _i = adhesion factor for earth layer

c_udci = undrained shear strength of clay.

A_mi = area of pile shaft in contact with the soil.

The adhesion factor a is taken as 0 for the firs three meters where it is expected hole room and fill material or week strata. For piles with constant cross-sectional area the value of a can be taken as 1.0 and for piles with uniform cross-sectional growth the value of a can be taken as 1.2 .

Example 5.3

18 m wood pile is installed small end down in clay. Pile diameter is 125 mm at the end and 10 mm/m increase in diameter. The undrained shear strength of the soil, measured from the pile cut-off level is: 0-6 m = 12 kP 6-12 m = 16 kPa 12-18 m = 19 kPa. Determine the ultimate load capacity of the pile. Pile cut-off level is 1.5m from the ground level. g _Rd = 1.7

decide the values for a

a = 0 for the first 3.0 meters

a = 1.2 for the rest of the soil layer

divided the pile into 3 parts (each 6.0 m in this case)

calculate Average diameter at the middle of each section:

Average diameter : Bottom (section) = 0.125+3.0× (0.01) = 0.15

Middle (section) = 0.155+6× 0.01 = 0.21

Top (section) = 0.215+(3+2.25)× (0.01) = 0.268

=

\ Ultimate bearing capacity of the pile is 117kN

5.4 Steel piles

Because of the relative strength of steel, steel piles withstand driving pressure well and are usually very reliable end bearing members, although they are found in frequent use as friction piles as well. The comment type of steel piles have rolled H, X or circular cross-section(pipe piles). Pipe piles are normally, not necessarily filled with concrete after driving. Prior to driving the bottom end of the pipe pile usually is capped with a flat or a cone-shaped point welded to the pipe.

Strength, relative ease of splicing and sometimes economy are some of the advantages cited in the selection of steel piles.

The highest draw back of steel piles is corrosion. Corrosive agents such as salt, acid, moisture and oxygen are common enemies of steel. Because of the corrosive effect salt water has on steel, steel piles have restricted use for marine installations. If steel pile is supported by soil with shear strength greater than 7kPa in its entire length then the design bearing capacity of the pile can be calculated using the following formulas. Use both of them and select the lowest value of the two:

Where:m _m = correction factor

E_SC = elasticity module of steel

I = fibre moment

f_yc characteristic strength of steel

A = pile cross-sectional area

C_uc = characteristic undrained shear strength of the soil.

 

 

Example 5.4

Determine the design bearing capacity of a Steel pile of external diameter 100 mm, thickness of 10 mm. Treated against corrosion. pile. Consider failure in the pile material. C_c of the soil is 18 kPa, favourable condition. S2

Steel BS 2172

solution :

g _n = 1.1

m _m = 0.9

 

E_sc = 210 Gpa

for BS 2172 f_yc = 320 MPa

=

= =

The first formula gives us lower value, therefore, the design bearing capacity of the pile is 0.3 MN

If we consider corrosion of 1mm/year Þ

= =

5.5 Concrete piles

Relatively, in comparable circumstances, concrete piles have much more resistance against corrosive elements that can rust steel piles or to the effects that causes decay of wood piles, furthermore concrete is available in most parts of the world than steel.

Concrete piles may be pre-cast or cast-in place. They may be are reinforced, pre-stressed or plain.

5.5.1 Pre-cast concrete piles

These are piles which are formed, cast to specified lengths and shapes and cured at pre casting stations before driven in to the ground. Depending up on project type and specification, their shape and length are regulated at the prefab site. Usually they came in square, octagonal or circular cross-section. The diameter and the length of the piles are mostly governed by handling stresses. In most cases they are limited to less than 25 m in length and 0.5 m in diameter. Some times it is required to cut off and splice to adjust for different length. Where part of pile is above ground level, the pile may serve as column.

If a concrete pile is supported by soil with undrained shear strength greater than 7 MPa in its entire length, the following formula can be used in determining the bearing capacity of the pile :

Where: N_u = bearing capacity of the pile, designed as concrete column

E_sc = characteristic elasticity module of concrete

I_c = fibre moment of the concrete cross-section ignoring the reinforcement

C_uc = characteristic undrained shear strength of the soil in the loose part of the soil within a layer of 4.0 m

 

Example 5.5

Concrete pile (0.235) × (0.235) cross-section installed in clay with characteristic undrained shear strength of 12 kPa. In favourable condition. C50. Determine design load of the pile. Consider failure in the material.

Solution:

j _ef = 1.3

l_c /h = 20

k_c = 0.6, kj = 0.24, k_s = 0.62

f_cc = 35.5 /(1.5× 1.1) = 21.5 MPa

f_st = 410/(1.15× 1.1) = 324 MPa

Effective reinforced area:

F_Rd = m _m× N_U

m _m = 0.9 Þ F_Rd = (0.9)0.769 = 0.692 MN

Failure checking using the second formula:

E_cc = 34 GPa

The lowest value is 0.632 MN Þ Design capacity =0.63 M

5.6 Timber piles (wood piles)

Timber piles are frequently used as cohesion piles and for pilling under embankments. Essentially timber piles are made from tree trunks with the branches and bark removed. Normally wood piles are installed by driving. Typically the pile has a natural taper with top cross-section of twice or more than that of the bottom.

To avoid splitting in the wood, wood piles are sometimes driven with steel bands tied at the top or at the bottom end.

For wood piles installed in soil with undrained shear strength greater than 7kPa the following formula can be used in predicting the bearing capacity of the pile:

Where: = reduced strength of wood

A = cross-sectional area of the pile

If the wood is of sound timber, (e.g. pinewood or spruce wood with a diameter > 0.13m), then (reduced strength) of the pile can be taken as 11MPa.

Increase in load per section of pile is found to be proportional to the diameter of the pile and shear strength of the soil and can be decided using the following formula:

where: A_m, = area of pile at each 3.5 m section mid point of pile

C_m = shear strength at each 3.5m section mid point of pile

d_m = diameter of pile at each 3.5 m section mid point of pile

P_mi = pile load at the middle of each section

 

Example 5.6

Determine the design bearing capacity of a pile 12m pile driven in to clay with characteristic undrained shear strength 10KPa and 1.0kPa increase per metre depth. Piling condition is assumed to be favourable and the safety class 2. The pile is cut at 1.5m below the ground level. Top diameter of the pile is 180mm and growth in diameter is 9mm/m.

*Often it is assumed that cohesive strength of the soil in the fires three meters is half the values at the bottom.

solution:

First decide which part of the pile is heavily loaded. To do so, divide the pile which is in contact with the soil in three parts or sections (see fig.4.1) in this example the pile is divided into three 3.5m parts

Calculate and decide diameter of the pile at the mid point of each 3.5m section (0.180+0.009(y_i) ; y_i growth per meter from the end point.

Calculate the shear strength of the soil at the mid point of each 3.5m section C_mi = (22 - 1(y_i) ). Shear strength at the end of the pile = (10MPa + 1MPa (12m))=22 MPa

Decide the values of the partial coefficients from table (10-1 - 10-4)

P_ti = pile load at the top of each section

= stress at the top of the pile

\ The bearing capacity of the pile is 55.1kN

Now using the equation in (6-7), we will check the pile for failure

f_Red = 11MPa (see section 5.6)

m _n = 0.9

g _n = 1.1

\ =

In consideration of failure in the pile material, the pile can be loaded up to 9.0 MPa

In consideration of cohesion force, the pile can be loaded up to 55 MPa

the bearing capacity of the pile is therefore, 55 MPa

5.6.1 Simplified method of predicting the bearing capacity of timber piles

Consider the previous case and use the following formula :

regarded the pile in its full length

calculate average diameter of the pile Þ

calculate average shear strength of the pileÞ

3. decide the values of g _{Rd, g m} and a (table 10-1 - 10-4) :

g _Rd, = 1.7

g _m = 1.8× (0.8) = 1.44

a = 1.2

\ the bearing capacity of the pile is 56 kN

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Last updated: May 13, 1999.