1.0 Basic Pile Design Data * * * * * * * * * * * *

Type of Pile : -

:

Bored Cast in-situ.

Pile Diameter

=

Working Load in Compression

=

Minimum Factor of Safety

=

Reference Point for Level Measurements

:

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

A] Top 12.00 m

:

15 Nos. Φ 32 mm

B] Rest of Pile

:

7 Nos. Φ 32 mm

A] Top 2.00 m

:

Φ 10 mm @ 90 mm c/c

B] Rest of Pile

:

Φ 10 mm @ 150 mm c/c

1000

mm

10,000 kN 3.00 Existing Ground Level

N/mm2

N/mm2

* Longitudinal Bars

* Helical Links

* Cover of Reinforcement * Number of Piles

=

75

=

122

mm

1

Page No.

2.0 Pile Capacity Calculations * The soil investigation report has been done by :

2

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

Ultimate Bearing Capacity of Pile (Qu)

Qu = Qb + Qs Where

Qb : Pile base bearing capacity. Qs : Pile socket friction bearing capacity.

Qb = 2 Nφ qub Ab Where

φ : Angle of Friction

Nφ = tan 2 (45 o + φ ) 2

qub : Unconfined compressive strength of the socket zone = qu (lab) 5 Ab : Area of pile base.

Qs = α βqus As Where

α : Reduction factor relating to qus. β : Correction factor related to the discontinuity spacing in the rock mass. qus : Unconfined compressive strength along the shaft area. As : Area of pile shaft.

Page No.

Friction Capacity of Pile Depth Range No.

From

To m

Length RQD qucs m

%

MPa

α

β

As

Qs

m2

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

3

*

Qs

Ultimate Skin Friction Capacity

=

27255 kN

Page No.

End Bearing Capacity of Pile

φ

=

27

Nφ

=

2.66

qub

=

0.77

MPa

Ab

=

0.79

m2

* Ultimate End Bearing Capacity

Qb

=

3217

kN

* Ultimate Total Capacity

Qu

=

30472

kN

=

3.05

* Actual Factor of Safety

o

Thus, the pile can safely carry the applied load

4

`

Page No.

5

3.0 Elastic Analysis In granular soil-rock environment where the soil modulus is assumed to increase linearly with depth, stiffness factor (T) is given by: -

EI nh Where

T =5

E : Modulus of elasticity of concrete E = K o + 0.2 f c u , where Ko = 20 4 I : Moment of inertia of pile = πD 64 D : diameter of pile. nh : Modulus of subgrade reaction

* Stiffness Factor * Net Pile Length

E

=

32.00

GPa

I

=

0.05

m4

nh

=

45

T

=

2.04

m

=

29.50

m

H1

=

500.00 kN

H2

=

133.33 kN

H

=

633.33 kN

MN/m3

Shear Forces

* Due to 5% of Axial Load * Due to Out-of-Verticality * Total Shear Force

=

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 M f = Fm HT Where

Fm : Moment coefficient

Page No.

Z

Moment

(x)

Fm

m

x T

0.00

0.00

-0.92

-1185.77

1.00

0.49

-0.46

-590.44

2.00

0.98

-0.08

-107.11

3.00

1.47

0.17

215.34

4.00

1.97

0.25

328.00

5.00

2.46

0.24

311.55

6.00

2.95

0.18

228.44

7.00

3.44

0.11

139.77

8.00

3.93

0.05

62.22

9.00

4.42

0.01

8.00

10.00

10.00

4.91

-0.01

-10.67

12.00

11.00

5.41

0.00

0.00

Mf kNm

-1500 0.00

-1000

-500

0

2.00

Depth (m)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Depth

6

4.00 6.00 8.00

Moment (kNm)

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

500

17 18 19 20 21 22 23 24 25 26 27 28 Page No.

7

4.0 Longitudinal Reinforcement

* * * *

M1

=

1185.77 kNm

[From the elastic analysis]

Out-of-Position Moment

M2

=

750.00 kNm

= 0.075 N

Ultimate Moment

Mult

=

2903.66 kNm

[Factored to 1.50]

Ultimate Axial Load

Nult

=

15000

kN

Pile Diameter

h

=

1000

mm

d'

=

fcu

=

60

N/mm2

fy

=

460

N/mm2

1] From PROKON

=

1.45

%

------ (1)

2] 0.30% Ac

=

0.30

%

------ (2)

[BD 74/00, C9.2]

3] Minimum of 6 bars

=

0.61

%

------ (3)

[BD 74/00, C9.2]

4] 0.15N/fy

=

0.62

%

------ (4)

[BD 74/00, C9.2]

=

1.45

%

[Factored to 1.50]

101.00 mm

* Required Steel Percentage

Top 12.00 m

* Maximum of (1), (2), (3) & (4)

ρreq.

* * * *

Provided Steel Diameter

=

32

Required No. of Steel Bars

=

15

Provided No. of Steel Bars

=

15

Spacing Between Bars c/c

=

167

mm

mm

[Maximum 300 mm]

OK Rest of Pile

* * * *

ρreq.

Maximum of (2), (3) & (4)

=

0.62

Provided Steel Diameter

=

32

Required No. of Steel Bars

=

7

Provided No. of Steel Bars

=

7

% mm

OK

Page No.

8

5.0 Stress in Concrete

* * * *

Pile Diameter

D

=

1000

mm

Compressive Strength of Concrete

fcu

=

60

Working Axial Load

N

=

10000

Area of Concrete

Ac

=

785398 mm2

N/mm2 kN

* Permissible service stress should not exceed 25 % of the specified cube strength as per BS8004: 7.4.4.3 * Actual Stress in Concrete, σ = N

A

σ1

=

15.00

N/mm2

σ2

=

12.73

N/mm2

OK

6.0 Ultimate Vertical Load

* 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:

N =0.3 5f c uAc 0. 7 As c f y * * * *

Yield strength of steel.

fy

=

460

Area of steel

Asc

=

12064

Compressive strength of concrete.

fcu

=

60

Area of concrete

Ac

=

773334 mm2

N

=

20125

kN

Nall

=

13416

kN

Napp

=

10000

kN

* Allowable Axial Load * Applied Axial Load

N/mm2 mm2 N/mm2

[Divided by 1.50]

OK

Page No.

9

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

s = s1 + s2 + s3 s : Total settlement. s1 : Settlement of pile shaft due to elastic shortening. s2 : Settlement of pile caused by pile point load. s3 : Settlement of pile caused by pile shaft-transmitted load.

s1 =

(Qwp + ξQws ) L

where

Ap E p Qwp

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

ξ

: Unit skin resistance distribution along pile shaft.

Qws

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

L

: Length of pile.

Ap

: Area of pile cross-section.

Ep

* * * *

: Modulus of elasticity of pile material.

Ultimate End Bearing Capacity

Qb

=

3217

kN

Ultimate Skin Friction Capacity

Qs

=

27255

kN

Factor of Safety of End Bearing

=

3.05

Factor of Safety of Skin Friction

=

3.05

Qwp

=

1056

kN

Qws

=

8944

kN

ξ

=

0.67

L

=

29.50

m

Ap

=

0.79

m2

Ep

=

32.00

GPa

s1

=

8.27

mm

Page No.

s2 =

qwp D Es

Where

(1 − µ s2 ) I wp

Qwp

qwp

: Point load per unit area at the pile point =

D

: Diameter of pile.

Es

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

µs

: Poisson's ratio of soil.

Iwp

: Influence factor = αr for circular foundations. qwp

=

1344

kN/m2

D

=

1000

mm

Es

=

1.00

GPa

µs

=

0.30

Ap

10

Q D s3 = ( ws ) (1 − µ 2 ) I ws pL Es where

Iwp

=

0.85

s2

=

1.04

p

: Perimeter of pile.

Iws

: Influence factor =

Total Settlement

2 + 0.35

mm

L D

p

=

3142

mm

Iws

=

3.90

s3

=

0.34

mm

s

=

9.66

mm

Page No.

8.0 Shear Calculations on Piles Assumptions and Considerations

* * * * * * *

Concrete Compressive Strength

fcu

=

60

N/mm2

Yield Strength of Stirrups

fyv

=

460

N/mm2

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

dp

=

1000

mm

11

Ac * Cross-sectional Area of Pile * Main Reinforcement : No. of Bars Diameter of Bars

* Area of Tension Reinforcement * Concrete Cover

As

=

785398 mm2

=

15

=

32

mm

=

6032

mm2

=

75

mm

[Half the Total Reinforcement].

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

N/mm2

Maximum Shear Stress

vmax

=

4.75

N/mm2

[BD 74/00, C5.1]

Shear Stress is OK

Page No.

* Calculating the factor ξ s vc

ξs

* Shear Stress in Concrete

(1 +

=

0.91

(Table 9, BS 5400)

γm

=

1.25

(Table 8, BS 5400)

vc

=

0.69

N/mm2

ξ s vc

=

0.63

N/mm2

0.05N ) Ac

=

1.95

12

ξ s vc * Shear Reinforcement Criterion

=

1.22

N/mm2

(BD 74/00)

* The criterion of shear reinforcement is based on the following cases: Case (1) v <

ξ s vc * 2

, No shear reinforcement is required.

Case (2)

ξ s vc * ≤ v ≤ ξ s vc * 2

, Minimum shear reinforcement is required.

Case (3)

v > ξ s vc *

, Shear reinforcement is to be provided. v

ξ s vc *

=

1.29

N/mm2

=

1.22

N/mm2

Therefore, case (3) governs

Page No.

* Minimum Shear Reinforcement

=

1.00

=

2

mm2/mm

Provided Reinforcement A] Top 2.00 m

* Number of Legs

legs

13

* * * *

Diameter of Stirrups

=

10

mm

Spacing

=

90

mm

Provided Area of Stirrups

Asv

=

1.75

mm2/mm

Required Area of Stirrups

Asv

=

1.65

mm2/mm

OK B] Rest of Pile

* * * * *

Number of Legs

=

2

legs

Diameter of Stirrups

=

10

mm

Spacing

=

150

mm

Provided Area of Stirrups

Asv

=

1.05

mm2/mm

Required Area of Stirrups

Asv

=

1.00

mm2/mm

OK Formulas

ξs = 4

vc =

500 d

0 .27 3 100A s γm bd

A sv sv

or 0.70 whichever is greater.

m in

=

3

f cu

0. 4b 0 . 87 f yv

A sv b v−ξ s vc = sv 0.87 f y v

for case (3)

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