‫ﻙ ﺍﻟﺧﻠﻳﺞ ﻟﺣﻔﺭ‬

PILE DESIGN

Project : MR. HAMDAN SAEED SALEM ALBADAL ALZAABI

Consultant : FAZAA ENGINNERING CONSSULTANCY

Contractor : WAFA ALHAYAT GEN. CONT. CO.

Piling Contractor : Pile JET Foundation Drilling

Location : Plot No. (69) , Sector : (SE-25)- KHALIFA CITY

ABU DHABI, UNITED ARAB EMIRATES
PILE TYPICAL DETAILS
1)PILE CALCULATION FOR MAJLIS AND GUARD ROOM
1)PILE CALCULATION
FOR MAJLIS AND
GUARD ROOM
PILE DESIGN CALCULATION:
Ref : Site investigation Report .
NO. : 3934REP200S
Dated : 09/08/2008
By : SHADID lab.
Based on the above mentioned soil report the following data shall be used for pile design .
From Borehole

Design Data :
Pile diameter = 500 mm Cover = 75 mm.
Pile length = 12.00 m
Cut off length = 1.00 m
Socket length = 1.00 m
Design load = 100 ton
Test load = 150 ton
Concrete grade = 40 N/ mm2
Main Reinforcement = 6 T 16 of Length 11.50 m
Stirrups = 8 mm @150 mm
q b = 34.93 kg / cm2
q s = 16.34 kg /cm2
RQD = 84 ℅
J = 0.20
a) α = 0.30 Tomlinson ( fig 4.33 ), β = 0.65 Tomlinson (fig 4.34 )
Nφ = 2.75
where :
q b : Unconfined Comp. strength of rock.
q s : Average unconfined comp strength in socket bearing layer
RQD: Rock Quality Designation
J : Mass factor
α : Rock socket reduction factor
β : Rock socket correction factor
φ : Angle of internal friction
Geotechnical Design Calculations : 1 )
1.1) The Ultimate Pile Capacity :
Qult = Ultimate pile capacity
Qbase = Ultimate base capacity
Qshaft = Ultimate shaft capacity
1.1.a ) Using Tomlinson Approach :
Qbase = 2 x Nφ x qB x Ab Eq ( 1)
Qshaft = α x β x q s x As Eq (2)

Where :
Nφ = tan2 (45+φ/2) = 2.75
Ab = Area of pile
As = surface area of pile socket

From Eq(1) :
Qbase = 2 x 2.75x 34.93 x п( 50.0)2 / 4000
= 377 ton
From Eq(2) :
Qshaft = 0.30 x 0.65 x 16.34 x (п x 50.0 x 100 )/1000
= 50 tons
SO,
Qult = 377.0 + 50.0
= 427 tons

F.O.S. =Ultimate Pile Capacity / Design Vertical Load

=427 / 100.0 = 4.27 Ok
1.2 ) Chech of Stress under Pile Tip :
1.2.1 ) Using Boussinesq's equation :
Δσav = q {1- 1/[ (R/Z )2 + 1 ]^(3/2) }
Where :
Δσav : Vertical Stress caused by appoint load
q : Intensity of Pressure on Circular area at radius (R)
= 50.91 kg / cm2
Z : Distance below the center of Pile = 100 cm
R : Radius of Pile = 25 cm
Δσav = q {1- 1/[ (25/100 )2 + 1]^(3/2)
= 50.91 x 0.0869
= 4.42 kg / cm2 OK
1.2.2 ) Using Approximate method :
Δσav = Q / [п x ( D + Z ) ^2 / 4]
where :
Δσav : Vertical Stress caused by appoint load
Q : Design load = 100 ton
D : Pile diameter = 50 cm
Z : Distance below the center of Pile = 100 cm
SO ,
Δσav = 100000 / [п x (150) ^2 / 4 ]
Δσav = 5.65 kg / cm2 OK
 1.3 ) Calculation of Pile Settlement :
1.3.1 ) Settlement of Pile shaft :
S1 = (Qwp + ζ Qws ) L / (Ap xEp)
Where:
Qwp : load carried by end bearing under working load condition
= 549.0 KN
Qws : Load carried by shaft friction under working load
= 451.0 KN
ζ : Skin friction distribution factor = 0.67
L : length of pile from cut off level = 11000 mm.
Ap : Area of pile Tip = 196349 mm2
Ep : Yung's modulus of Pile material = 28 KN / mm2
Get: S1 = 1.67 mm.
1.3.2 ) Settlement of pile by the load at end bearing :
S2 = qwp x D x ( 1-μs2 ) Iwp / Es
Where :
qwp : point load per unit area at pile Tip = 1655 KN / m2
D : pile diameter = 500 mm
ES :Young's modulus of rock at pile tip = 338875 KN / m2
μs : Poisson's ratio of soil = 0.5
Iwp : Influence factor = 0.85 mm2
Get: S2 = 1.86 mm.
1.3.3 ) Settlement of pile by load transmitted along pile shaft :
S3 = (Qws / (p*le)) x (D / ES ) x (1- μS2) x IWS
Where:
P : Perimeter of the pile = 1571 mm
Le : Embeded Length of the pile = 1000.0 mm
Iws : Influence factor = 2 + 0.35 x (le/D)^1/2 = 2.45 mm
S3 = 0.35 mm Get,
SO, Stotal = S1 + S2 + S3 = 3.88 mm.
2 ) Structure Design Calculation :
2.1 ) Allowable Working Stress in Concrete :
According to B.S. 8004 :1986 , page 97 Article 7.4.3.3.1
Maximum working stress on pile concrete = 25 ℅ of Fcu
For 500 mm. diameter pile carring 100 ton
Working load = (Design vertical load / cross section area )
= (100000 /[ п x (500)^2 / 4 ] )
= 5.08 N / mm2
For, 500 mm. diameter pile carrying 150 ton test load
Testing load = (Test load / Cross Section Area )
= 7.63 N/ mm2
For, concrete grade 40
0.25℅ Fcu =0.25 x 40 = 10.00 N / mm2 > working load
> Test load
SO, The cross section of pile can carry working and testing load safety
2.2 ) Reinforcement in pile :
Provide 6 bars 16 mm. diameter as main reinforcement
According to B.S. 8110 : part 1, 1997 , table 3.25,
The minimum percentage of reinforcement for acomp. Member
= 0.4℅ of 460 N/ mm2
Actual area of provide steel = (6 x п x (16 )^2 / 4 )
= 1206 mm2
Area of pile cross section = п x ( 500 )^2 / 4
= 196349 mm2
Actual percentage of steel provided = (1206 / 196349 )
= 0.61 ℅ > 0.4 ok safe
but, The Ultimate load applied on each pile should not exceed the ( N ) value given
below as per B.s. 8110 : part 1 , section 3.8.4.3 , 1997
N = 0.4 Fcu Ac + 0.8 Fy As
Where :
Ac = Area of concrete
As = Area of steel
N = 0.4 x 40 x 195143 + 0.8 x 460 x 1206
= 3566 KN
Safe working load = ( N / safety factor )
= ( 3566 / 1.5 )
= 2377 KN
= 238 ton > 100 ton
SO, The pile section can carry the design load.
2.3 ) Check 0f Deflection , Bending Moment , Soil Reaction :
According to the pile requirements of B.S. 8004 :1986 , article 7.4.2.5.4 ,
Maximum out of plumb tolerance = 1/ 75
We shall consider horizontal load on pile = 5 ℅ of the vertical load
H = 0.05 x 1000 + 1000/ 75
= 50 + 13.33
= 63.33
Hult = 1.4 x 63.33 = 88.00 KN
Maximum out of position tolerance = 75 mm.
Bending moment due to position = 1000 x 0.075
= 75.00KN.m
Ultimate Bending Moment = 1.4 x 75.00
= 105 KN/m
For, fixed pile head we shall use Tomlinson eq…. (6.57) , (6.58) , (6.59)
Defliction ( Yf ) = ( Fy x Hult x T3 ) / (EI)
Bending moment ( Mf) = Fm x Hult x T
Where :
Fy , Fm = Coefficient of Defliction , Bending Moment (fig 6.39)
Hult = The Horizontal Force
T = The stiffness factor
= ( EI / nh )^ (0.2)
E = 140 x √Fcu
= 28000 MN / m2 for Fcu =40 N/mm2
I = ( п x D4 / 64 )
 = 0.00307 m4
nh = Coefficient of modulus
= 15 MN/m3
Get : T = 1.42
The maximum length of pile = 11.00 m from the pile cut off level
SO, Zmax = ( L / T ) = (11.00 / 1.42 ) =7.74
Defliction Bending moment

Yf = Fy H
X (m) Z= X/T Fy T^3/EI Fm Mf = Fm H T

0 0 0.93 2.94 -0.71 -89

1 0.70 0.70 2.22 -0.2 -25

2 1.41 0.40 1.27 0.17 21

3 2.11 0.18 0.57 0.25 31

4 2.82 0.02 0.06 0.19 24

5 3.52 -0.02 -0.06 0.10 13

Fore the above table the bending moment below 5.0 m can be neglected
However, a 11.50 m of steel cages length will be used for all piles as typical pile
details
Maximum moment = 105 +89
= 194 KN.M
From B.S. 8110 for circular columns :
At Fcu = 40 N/ mm2
h = 500 mm
hs = 334 mm
Then ,
hs/h = 0.67
Mu/h3 = 1.55 N/mm2
Nu/h2 = 5.60 N/mm2
From chary 14 Asc = 0.6 ℅
However ,
6 bars 16 mm diameter provided of 0.61 ℅ is ok
2.4 ) Check of Clear Spacing :
According to B.S. 8110: part 1 , 1997 , art 3.12.11.2.4 max clear spacing is 300 mm.

For 6 bar of 16 mm diameter clear spacing = 166 mm ‹ 300 mm.

2.5 ) Check of Bond Length ( using deformed bars ) :
From B.S. 8110- part 1 , 1997 ,table 3.27 , for Fcu 40 N/mm2
Anchorage length = 35 x bar size = 35 x 16
= 560 mm < 800 mm
2.6 ) Check of Spacing Stirrups :
According to B.S. 8110 part 1 , 1997 , table 3.8
When Fcu = 40 N/mm2
Area of steel = 1206 mm2
Max permitted shear stress = Vc
= 0.5 N/mm2 for concrete grade 25
= 0.5 x ( 40 / 25 )^ (1/3) N/mm2 for grade 40
= 0.58 N/mm2
Working shear stress in pile = Vw
= (Hult / [(п x ( D ) 2 / 4 ]
 = ( 88.00 / (п ( 500 )2 /4 )
= 0.48 N/mm2 < Vc ok
According to B.S. 8110 : part 1 , 1997 , table 3.7
Shear reinforcement required = (0.4 x 500 x 150) / ( 0.95 x 460 ) = 69 mm2
8 mm @ 150 mm = ( п x ( 8 )2 / 4 ) x 2

= 101 mm2 > 69 mm ok

REFRANCE :

1 ) British Standard code of practice for Foundation

BS 8004 : 1986

British Standard Institute1986

2 ) British Standard Structure use of Concrete

B.S. 8110 : part 1

British Standard Institute

3 ) Pile Design and Construction practice

M.J.Tomlinson

Viewpoint Publication

3 rd Edition .1987