Circular Concrete Pile Design 1t5o3a

LAEMCHABANG

Typical Piles

SCOTT WILSON

DESIGN OF UNBRACED CIRCULAR COLUMN (Compression + Moment + Shear) Typical Piles

Column Mark :

1.

Design Parameters

h

fcu

=

45

N /mm2

fy

=

460

N /mm2

fyv

=

460

N /mm2

Diameter of Column Section,

2.

hs

(1235 = 1270 - 2*17.5) 1235 mm

=

Clear height of Column,

h Lo

=

28850

mm

Concrete cover to links,

c

=

75

mm

3. Value of au and emin

Type of Column

Factor a

The column is unbraced. End condition at top End condition at bottom Factor Effective height, Le / h =

 L e =  Lo 28.0

=

1

= =

1 1.2

=

34620

>

10

=

1/2000 ( Le / h )2

Assume Deflection at ULS, mm

= =

0.393 1

=

485

mm

=

62

mm

=

20

mm

61.3

=

992

mm

=

0.803

say

(1235^2 )

=

1.31

K au = aKh

0.05 h = 0.05 x 1235 Minimum eccentricity, emin

Therefore the column is slender column.

4.

5.

Design Axial Load, Bending Moment and Shear Force from Analysis Design Compression Force,

N

=

Design Bending Moment,

Mo

Design Shear Force,

V

= =

2000

kN

8800

kNm

1500

kN

(at top of pile)

Longitudinal Reinforcement Design Try

( 2 nos.

52

of 2-bar

T

40

bundles

vertical reinforcement

+ 16 nos.

of 3-bar

=

61.3

Equivalent dia. of bar bundles

hs

bundles )

=

hs / h

h - 2 c - 2(dia. of links) -

=

992

/

1235

0.8

mm

1) Compression Force + Maximum Moment Additional Moment Due to Axial Compression Force, Madd(N)

= 2000

Design moment, M = M / h3 =

9770

x

485

/ 1000

8800 x 106

/

N / h2 = =

970

kNm

+ 970

=

9770

kNm

(1235^3 )

=

5.19

=

64687

2000

From BCA's Chart No.

x 103

/

19,

100Asc / Ac and

Asc required

=

5.4

x

Ac / 100

K

=

5.4 =

1

=

64687

mm2

2) Reinforcement Area of vertical reinforcement, As required Vertical reinforcement to be provided As provided

=

As provided / Ac

6.

65345 =

5.5%

= mm2

52 >

T 64687

mm2 40 mm2

OK

Shear Reinforcement Design

327928097.xls

Page 2 of 3

08/27/2016

LAEMCHABANG

Typical Piles

SCOTT WILSON

DESIGN OF UNBRACED CIRCULAR COLUMN (Compression + Moment + Shear) Refer to 'Shear in Piles and Circular Columns' of 'Notes on Structures' by OAP (a copy attached) r

=

rs

=

h/2

=

617.5

mm mm

hs / 2

=

496

sin 

=

2 rs /  r

=

0.511



=

0.537

cos 

=

0.859

bd

=

r2 (/2 +  + sin  cos )

b

=

r [(/2 +  + sin  cos ) / (1 + sin )]

d

=

bd / b

=

=

933

971087.48 mm2 =

=

Mo / N = v

0.6

x

=

=

933 +

1039

=

623

mm

2000.0

=

4400

mm

V / bd

=

1.54

###

=

32673

8800.0

/

As

=

vc =

0.79 (100As / (bd))1/3 (400/he)1/4 / m

V he / Mo

=

0.18

N /mm2

<

=

1.11

Asv/Sv req'd

=

0.6 h

= 1.149

1039

mm

Checking of shear is required. OK

=

3.365

N /mm2 =

1.30

N /mm2

1.04

Required minimum links for containment of compression reinforcement

327928097.xls

>

=

vc' = vc + 0.6 (N V he / AcMo)

< 1

v < (vc'+0.4)

=

+ 75

100As / bd

Shear reinforcement should be provided.

Asv/Sv provided

61.3 / 2

5 N/mm^2

mm2

v > vc'

Shear links to be provided

mm

mm

Depth of equivalent rectangular section, he 0.6 he

1041

1 >

T 1.04

=

T 10 16

-

-

480

(Cl. 3.12.7.1 of BS 8110 : Part 1)

175

OK

Page 3 of 3

08/27/2016