Page No. 1

1.0 Basic Pile Design Data* Type of Pile : - : Bored Cast in-situ.

* Pile Diameter = 1000 mm

* Working Load in Compression = 10,000 kN

* Minimum Factor of Safety = 3.00

* Reference Point for Level Measurements : Existing Ground Level

* Cut-off Level = -3.00 m

* Bedrock Level = -12.77 m

* Pile Toe Level = -32.50 m

* Concrete = 60

* Cement Type : Sulphate Resisting Cement (S.R.C.)

* Reinforcement : High Tensile [Black] Steel.

* Yield Strength of Steel = 460

* Longitudinal Bars

A] Top 12.00 m : 15 Nos. Φ 32 mm

B] Rest of Pile : 7 Nos. Φ 32 mm

* Helical Links

A] Top 2.00 m : Φ 10 mm @ 90 mm c/c

B] Rest of Pile : Φ 10 mm @ 150 mm c/c

* Cover of Reinforcement = 75 mm

* Number of Piles = 122

N/mm2

N/mm2

Page No. 2

2.0 Pile Capacity Calculations* The soil investigation report has been done by : Al-Hai & Al-Mukaddam for Geotechnical Works

Report No. SR/0710609 dated Nov. 27, 2007

Borehole No. 5

* Reference Point for Level Measurements : Existing Ground Level

* Pile Diamter = 1000 mm

* Pile Toe Level = -32.50 m

* Bedrock Level = -12.77 m

* Socketd Length of Pile in Rock = 19.73 m

: Pile base bearing capacity.

: Pile socket friction bearing capacity.

: Angle of Friction

=

: Unconfined compressive strength of the socket zone

: Area of pile base.

:

b : Correction factor related to the discontinuity spacing in the rock mass.

: Unconfined compressive strength along the shaft area.

: Area of pile shaft.

Ultimate Bearing Capacity of Pile (Q u)

Where Qb

Qs

Where fNf

qub

Ab

Where a Reduction factor relating to qus.

qus

As

Qu=Qb+Qs

tan2( 45o+φ2)

=qu( lab)

5

Q s=αβ qus A s

Qb=2 Nφ qub Ab

Page No. 3

Friction Capacity of Pile

No.

Depth RangeLength RQD

a bFrom To

m m % MPa kN

1 -12.77 -18.05 5.28 50 0.96 0.50 0.65 16.59 5164

2 -18.05 -21.75 3.70 50 1.51 0.40 0.65 11.62 4537

3 -21.75 -23.35 1.60 50 1.78 0.35 0.65 5.03 2050

4 -23.35 -26.15 2.80 50 3.45 0.22 0.65 8.80 4276

5 -26.15 -32.25 6.10 50 4.11 0.21 0.65 19.16 10811

6 -32.25 -32.50 0.25 50 3.85 0.21 0.65 0.79 418

qucs As Qs

m2

* Ultimate Skin Friction Capacity = 27255 kN

Page No. 4

End Bearing Capacity of Pilef = 27

= 2.66

= 0.77 MPa

= 0.79

* Ultimate End Bearing Capacity = 3217 kN

* Ultimate Total Capacity = 30472 kN

* Actual Factor of Safety = 3.05

Thus, the pile can safely carry the applied load

Qs

o

Nf

qub

Ab m2

Qb

Qu

` Page No. 5

3.0 Elastic AnalysisIn granular soil-rock environment where the soil modulus is assumed to increase linearly with depth,

Where E : Modulus of elasticity of concrete

I : Moment of inertia of pile

D : diameter of pile.

: Modulus of subgrade reaction

E = 32.00 GPa

I = 0.05

= 45

* Stiffness Factor T = 2.04 m

* Net Pile Length = 29.50 m

Shear Forces

* Due to 5% of Axial Load = 500.00 kN

* Due to Out-of-Verticality = 133.33 kN

* Total Shear Force H = 633.33 kN

stiffness factor (T) is given by: -

, where Ko = 20

nh

m4

nh MN/m3

H1

H2

T=5√ EInh

= πD4

64

E=Ko+0 .2 f cu

=N

75

* The following evaluation is based on the elastic analysis of laterally-loaded fixed pile head (pile cap)

and linearly increasing soil modulus (Reese & Matlock in Tomlinson 1994).

* Bending Moment

: Moment coefficient

Page No. 6

Depth Z Moment

(x)

m kNm

0 0.00 0.00 -0.92 -1185.77

1 1.00 0.49 -0.46 -590.44

2 2.00 0.98 -0.08 -107.11

3 3.00 1.47 0.17 215.34

4 4.00 1.97 0.25 328.00

5 5.00 2.46 0.24 311.55

6 6.00 2.95 0.18 228.44

7 7.00 3.44 0.11 139.77

8 8.00 3.93 0.05 62.22

9 9.00 4.42 0.01 8.00

10 10.00 4.91 -0.01 -10.67

11 11.00 5.41 0.00 0.00

1213141516

Where Fm

Fm Mf

N.B. (x) is the depth from the cut-off level

-1500 -1000 -500 0 5000.00

2.00

4.00

6.00

8.00

10.00

12.00Moment (kNm)

Dep

th (

m)

M f=Fm HT

xT

171819202122232425262728

Page No. 7

4.0 Longitudinal Reinforcement

= 1185.77 kNm [From the elastic analysis]

* Out-of-Position Moment = 750.00 kNm

* Ultimate Moment = 2903.66 kNm [Factored to 1.50]

* Ultimate Axial Load = 15000 kN [Factored to 1.50]

* Pile Diameter h = 1000 mm

d' = 101.00 mm

= 60

= 460

* Required Steel Percentage

1] From PROKON = 1.45 % ------ (1)

= 0.30 % ------ (2) [BD 74/00, C9.2]

3] Minimum of 6 bars = 0.61 % ------ (3) [BD 74/00, C9.2]

= 0.62 % ------ (4) [BD 74/00, C9.2]

Top 12.00 m

* Maximum of (1), (2), (3) & (4) = 1.45 %

M1

M2 = 0.075 N

Mult

Nult

fcu N/mm2

fy N/mm2

2] 0.30% Ac

4] 0.15N/fy

rreq.

* Provided Steel Diameter = 32 mm

* Required No. of Steel Bars = 15

* Provided No. of Steel Bars = 15

* Spacing Between Bars c/c = 167 mm [Maximum 300 mm]

OK

Rest of Pile

* Maximum of (2), (3) & (4) = 0.62 %

* Provided Steel Diameter = 32 mm

* Required No. of Steel Bars = 7

* Provided No. of Steel Bars = 7

OK

Page No. 8

5.0 Stress in Concrete

* Pile Diameter D = 1000 mm

* Compressive Strength of Concrete = 60

* Working Axial Load N = 10000 kN

* Area of Concrete = 785398

* Permissible service stress should not exceed 25 % of the specified cube strength as per BS8004: 7.4.4.3

= 15.00

* Actual Stress in Concrete,

= 12.73

OK

6.0 Ultimate Vertical Load

rreq.

fcu N/mm2

Ac mm2

s1 N/mm2

s2 N/mm2

σ=NA

* Proposed Vertical Reinforcement : 15 Nos. Φ 32 mm

As per BS8110, Part 1 - 1997, The ultimate axial load should not exceed the value of "N" given by:

* Yield strength of steel. = 460

* Area of steel = 12064

* Compressive strength of concrete. = 60

* Area of concrete = 773334

N = 20125 kN

* Allowable Axial Load = 13416 kN [Divided by 1.50]

* Applied Axial Load = 10000 kN

OK

Page No. 9

7.0 Settlement CalculationsThe settlement of piles under a vertical working load is calculated as follows: -

s : Total settlement.

: Settlement of pile shaft due to elastic shortening.

: Settlement of pile caused by pile point load.

: Settlement of pile caused by pile shaft-transmitted load.

where : Load carried at the pile point under working load condition.

x : Unit skin resistance distribution along pile shaft.

: Load carried by frictional (skin) resistance under working load condition.

L : Length of pile.

: Area of pile cross-section.

fy N/mm2

Asc mm2

fcu N/mm2

Ac mm2

Nall

Napp

s1

s2

s3

Qwp

Qws

Ap

N=0.35 f cu Ac+0 .7 A sc f y

s=s1+s2+s3

s1=(Qwp+ξQws )L

A p E p

: Modulus of elasticity of pile material.

* Ultimate End Bearing Capacity = 3217 kN

* Ultimate Skin Friction Capacity = 27255 kN

* Factor of Safety of End Bearing = 3.05

* Factor of Safety of Skin Friction = 3.05

= 1056 kN

= 8944 kN

x = 0.67

L = 29.50 m

= 0.79

= 32.00 GPa

= 8.27 mm

Page No. 10

Where : Point load per unit area at the pile point =

D : Diameter of pile.

: Modulus of elasticity of soil at or below pile point.

: Poisson's ratio of soil.

:

= 1344

D = 1000 mm

= 1.00 GPa

= 0.30

Ep

Qb

Qs

Qwp

Qws

Ap m2

Ep

s1

qwp

Es

ms

Iwp Influence factor = ar for circular foundations.

qwp kN/m2

Es

ms

s2=qwp D

Es

(1−μs2 )I wp

Qwp

A p

= 0.85

= 1.04 mm

where p : Perimeter of pile.

: Influence factor =

p = 3142 mm

= 3.90

= 0.34 mm

Total Settlement s = 9.66 mm

Page No. 11

8.0 Shear Calculations on PilesAssumptions and Considerations

* Concrete Compressive Strength = 60

* Yield Strength of Stirrups = 460

* Applied Shear Force = 633.33 kN

* Applied Normal Force = 10000 kN

* Factored Shear Force V = 950 kN [Factored to 1.50]

* Factored Normal Force N = 15000 kN [Factored to 1.50]

* Pile Diameter = 1000 mm

Iwp

s2

Iws

Iws

s3

fcu N/mm2

fyv N/mm2

dp

s3=(Qws

pL) D

E s

(1−μ2 ) I ws

2+0 . 35√ LD

* Cross-sectional Area of Pile = 785398

* Main Reinforcement : No. of Bars = 15

Diameter of Bars = 32 mm

* Area of Tension Reinforcement = 6032 [Half the Total Reinforcement].

* Concrete Cover = 75 mm

Shear Stress Calculations

* Width of Pile b = 1000 mm

* Centroid of Tension Zone c = 239 mm

* Effective Depth d = 739 mm

* Shear Stress v = 1.29

* Maximum Shear Stress = 4.75 [BD 74/00, C5.1]

Shear Stress is OK

Page No. 12

* Calculating the factor

= 0.91 (Table 9, BS 5400)

= 1.25 (Table 8, BS 5400)

* Shear Stress in Concrete = 0.69

= 0.63

= 1.95

Ac mm2

As mm2

N/mm2

vmax N/mm2

xs

gm

vc N/mm2

N/mm2

(1+0. 05 N

A c

)

ξs vc

ξs vc

= 1.22

Shear Reinforcement Criterion (BD 74/00)

* The criterion of shear reinforcement is based on the following cases:

Case (1) , No shear reinforcement is required.

Case (2) , Minimum shear reinforcement is required.

Case (3) , Shear reinforcement is to be provided.

v = 1.29

= 1.22

Therefore, case (3) governs

Page No. 13

* Minimum Shear Reinforcement = 1.00

Provided Reinforcement

A] Top 2.00 m

* Number of Legs = 2 legs

N/mm2

N/mm2

N/mm2

mm2/mm

ξs vc∗¿ ¿

ξs vc∗¿ ¿

v<ξs

vc∗¿2

¿

ξs

vc∗¿2

≤v≤ξs vc∗¿ ¿¿

v>ξs vc∗¿ ¿

* Diameter of Stirrups = 10 mm

* Spacing = 90 mm

* Provided Area of Stirrups = 1.75

* Required Area of Stirrups = 1.65

OK

B] Rest of Pile

* Number of Legs = 2 legs

* Diameter of Stirrups = 10 mm

* Spacing = 150 mm

* Provided Area of Stirrups = 1.05

* Required Area of Stirrups = 1.00

OK

Formulas

or 0.70 whichever is greater.

for case (3)

Asv mm2/mm

Asv mm2/mm

Asv mm2/mm

Asv mm2/mm

ξs=4√500

d

vc=0 . 27

γ m

3√100 A s

bd3√ f cu

( A sv

sv)min

= 0 . 4 b0. 87 f yv

A sv

sv

=b( v−ξs vc)

0.87 f yv