Pile Cap Design Examples

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2-Pile Pilecap Design Example

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Calculation Sheet

Project : Anchor Bolt Design ACI 318-11 Crane Beam Design

Eng : Test

www.civilbay.com Doc No : www.civilbay.com

Chk : Test

Jon No :

Date : 2013-03-06

Subject : Pilecap Design Examples

Rev : 0 1 of 4

DESCRIPTION 2 piles pilecap design based on

Code Abbreviation

A23.3-04 Design of Concrete Structures

A23.3-04

INPUT Pcf = 1500

[kN]

Column width

bc = 330

[mm]

As1 bar size = 25 As1 bar No = 7

Factored column load

Column length

hc = 500

[mm]

Pilecap width

bf = 1000

[mm]

Pilecap length

hf = 1750

[mm]

Pilecap thickness

tf = 620

[mm]

Ver reinft. 500

As1 prov = 3500 Equivalent to

25

@

133

Hor reinft. Pile diameter

dp = 250

[mm]

As2 bar size = 20

Pile spacing

sp = 750

[mm]

As2 bar No = 7

Pile edge distance

ed = 500

[mm]

Concrete strength

f'c = 30

[MPa]

Rebar yield strength

fy = 400

[MPa]


Concrete resistance factor

 c = 0.65

Rebar resistance factor

 s = 0.85

Low density conc factor

 = 1.0

Backfill soil weight above cap

Ws = 0

Pilecap self weight

W f = bf x hf x tf x 23.5

300

20

@

[kN] = 25

[kN]

258


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 2 of 4

Factored total pilecap load Pilecap effective depth Factored pile axial reaction

Pf = Pcf + W s + W f

= 1532

[kN]

d = tf - 1.5db -150

= 440

[mm]

= 766

[kN]

Qpr = Pf / 2

Vertical Direction Factored moment to column face

Mf = Qpr x (0.5sp - 0.5hc)

= 96

[kNm]

Factored shear to column face

Vf = Qpr

= 766

[kN]

CONCLUSION

[Pilecap Reinft by S&T Model Is Adequate]

[Pilecap Reinft by Deep Beam Model Is Adequate]

[Column One Way Shear Is Adequate]


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 3 of 4 Code Reference

S&T AND BEARING STRESS LIMIT MODEL by Adebar Column Bearing Stress Ac = 1.7E+05 [mm 2]

 A2   1  1 . 0  = 0 .33  A   c

 2d

 = 0 . 33 

 Ac

Allowable stress Column stress

fb =

  1  1 . 0 

 c ( 0.6 f 'c  6  f 'c )

fcb = Pcf / (bc x hc)

A2 = 1.8E+06 [mm 2]

= 0.75

= 0.39

= 17.9

[MPa]

= 9.1

[MPa]

< fb

OK

Pile Bearing Stress Ap = 4.9E+04 [mm 2]

  A2  1  1 . 0   Ap

 = 0 .33  

d

  1  1 .0 

 = 0 . 33  

 dp

Allowable stress Pile stress

fb =

 c ( 0.6 f 'c  6  f 'c )

fcb = Qpr / ( *

dp2

/4)

A2 = 7.9E+05 [mm 2]

= 1.00

= 0.25

= 17.1

[MPa]

= 15.6

[MPa]

< fb

OK

Required Reinft Vertical reinft.

l = 0.5sp - 0.25hc

= 250

[mm]

= 435

[kN]

As1 = Fh / ( s x fy)

= 1280

[mm 2]

Asmin = 0.2% x bf x tf

= 1240

[mm 2]

= 1280

[mm 2]

3500

[mm 2]

Fh = l / d x Qpr

As1 = max ( As1, Asmin ) As1 prov =

> As1


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 4 of 4

REDUCED LEVER ARM MODEL by Park and Paulay

Vertical Direction dVf / Mf = 3.52

>1.0 -> pilecap acts as deep beam

dVf / Mf > 2.0

Mf / Vf = 125

[mm]

= 150

[mm]

As1 = Mf / ( s fy z )

= 1877

[mm 2]


= 1240

[mm 2]

= 1877

[mm 2]

= 3500

[mm 2]

z = 0.4(d + Mf / Vf ) for dVf / Mf >=1.0 1.2 Mf / Vf

for dV f / Mf >=2.0


> As1

OK

[Deep Beam Model Is Applicable]


COLUMN ONE WAY SHEAR Qpr = 766

[kN]

bc = 330

[mm]

Pile diameter

dp = 250

[mm]

hc = 500

[mm]

Pile edge distance

ed = 500

[mm]

bf = 1000

[mm]

d = 440

[mm]

hf = 1750

[mm]

dc = 250

[mm]

Vf = r x Qpr

= 0

[kN]

bw = bf

= 1000

[mm]

 = 230 / (1000 + d)

= 0.160

Vc =  c    fc b w d

= 250

Factored single pile load

Pilecap effective depth

Hor Critical Shear Line Dist from col face to pile edge Fraction of pile outside critical line Factored shear force

When no transverse reinft.

r = 0.00

'

> Vf


11.3.6.3 Eq (11-9)

[kN] OK

11.3.4 Eq (11-6)

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Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 1 of 6


Code Abbreviation


A23.3-04

INPUT Ver reinft. in 2xed width range

Pcf = 4500

[kN]

Column width

bc = 400

[mm]

Asver bar size = 25 Asver bar No = 7


Column length

hc = 400

[mm]

Pilecap width

bf = 2450

[mm]

Pilecap length

hf = 2309

[mm]

Pilecap thickness

tf = 980

[mm]

500

Asver prov = 3500 Equivalent to

25

@

233

Hor reinft. in 2xed width range Pile diameter

dp = 350

[mm]

As1 bar size = 25

Pile spacing

sp = 1050

[mm]

As1 bar No = 7

Pile edge distance

ed = 700

[mm]

Concrete strength

f'c = 30

[MPa]


fy = 400

[MPa]



 c = 0.65


 s = 0.85


 = 1.0


Ws = 0

Pilecap self weight


500

25

@

[kN] = 130

[kN]

233


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 2 of 6



= 4663

d = tf - 1.5db -150

= 800

[mm]

= 1554

[kN]

Qpr = Pf / 3

[kN]


Mf = Qpr x (0.577sp - 0.5hc)

= 631

[kNm]


Vf = Qpr

= 1554

[kN]

CONCLUSION


[Pilecap Reinft by Deep Beam Model Is NOT Adequate]

[Pile One and Two Way Shear Is Adequate]



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06




 A2   1  1 . 0  = 0 .33  A   c

 2d

 = 0 . 33 

 Ac


fb =

  1  1 . 0 

 c ( 0.6 f 'c  6  f 'c )


A2 = 5.2E+06 [mm 2]

= 1.00

= 1.00

= 33.1

[MPa]

= 28.1

[MPa]

< fb

OK


  A2  1  1 . 0   Ap

 = 0 .33  

d

  1  1 .0 

 = 0 . 33  

 dp


fb =

 c ( 0.6 f 'c  6  f 'c )

fcb = Qpr / ( *

dp2

/4)

A2 = 1.5E+06 [mm 2]

= 1.00

= 0.43

= 20.9

[MPa]

= 16.2

[MPa]

< fb

OK

Required Reinft Vertical direction

l = 0.577sp - 0.25hc

= 506

[mm]

Fh = l / d x Qpr

= 984

[kN]

Asver = Fh / ( s x fy)

= 2893

[mm 2]

Asmin = 0.2% x 2ed x tf

= 2744

[mm 2]

Asver = max ( Asver, Asmin )

= 2893

[mm 2]

3500

[mm 2]

> Asver

OK

Asver prov =

Horizontal direction b1 = 425

l = sqrt(

[mm] b12

+

h12)

h1 = 203

[mm]

= 471

[mm]

Fh = l / d x Qpr

= 915

[kN]

FT2 = b1 / l x Fh

= 826

[kN]

As1 = FT2 / ( s x fy)

= 2429

[mm 2]


= 2744

[mm 2]

= 2744

[mm 2]

3500

[mm 2]


> As1


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 4 of 6




dVf / Mf < 2.0

Mf / Vf = 406

[mm]

= 482

[mm]

Asver = Mf / ( s fy z )

= 3849

[mm 2]


= 2744

[mm 2]


= 3849

[mm 2]

= 3500

[mm 2]

< Asver

NG


Asver prov =



for dV f / Mf >=2.0


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 5 of 6

CORNER PILE ONE WAY SHEAR

Code Reference A23.3-04

Factored pile load

Vf = 1554

[kN]

Pile diameter

dp = 350

[mm]

Pile edge distance

ed = 700

[mm]

d = 800 dc = 166

Pilecap effective depth Dist from pile edge to col corner

ec = 525

[mm]

[mm]

d/2 = 400

[mm]

[mm]

dc + 50 = 216

[mm]

< d/2 Critical shear line intersect column, one way shear needs NOT to be checked sqrt(f'c ) = min ( sqrt(fc') , 8 ) min( d, 1.414 ec ) = 742 When no transverse reinft.

= 5.477

[mm]

bw = 2058

 = 230 / (1000 + d)

= 0.128

Vc =  c    fc b w d

= 749

'

< Vf

[MPa]

11.3.4

[mm] 11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK

CORNER PILE TWO WAY SHEAR

Factored pile load

Vf = 1554

[kN]

Pile diameter

dp = 350

[mm]

Pile edge distance

ed = 700

[mm]

d = 800 dc = 166


ec = 525

[mm]

[mm]

d/2 = 400

[mm]

[mm]

dc + 50 = 216

[mm]

< d/2 Critical shear line intersect column, one way shear needs NOT to be checked

 = 0.0 Critical perimeter length

bo =  (dp + d) (270-2)/360



vc1 =  1 



2   0 .19   c f 'c  c 

= 2168

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 3.304

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)

 c = width / length = 1 for pile  s d   0 .19  λ  c f 'c  bo 

vc2 = 

 s = 2 for corner pile vc3 = 0 38   c f 'c vc = min( vc1 , vc2 , vc3 )

= 1.353

[MPa]

Vc = vc bo d

= 2346

[kN]

> Vf


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 6 of 6


[kN]

bc = 400

Pile diameter

dp = 350

[mm]

hc = 400

[mm]

Pile edge distance

ed = 700

[mm]

bf = 2450

[mm]

d = 800

[mm]

hf = 2309

[mm]

dc = 278

[mm]

= 0

[kN]

bw = bf

= 2450

[mm]

 = 230 / (1000 + d)

= 0.128

Vc =  c    fc b w d

= 892



[mm]



r = 0.00 Vf = r x 2 x Qpr

'

> Vf


11.3.6.3 Eq (11-9)

[kN] OK

11.3.4 Eq (11-6)

CivilBay



CivilBay



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 1 of 7


Code Abbreviation


A23.3-04

INPUT Pcf = 2000

[kN]

Column width

bc = 800

[mm]



Column length

hc = 400

[mm]

Pilecap width

bf = 1400

[mm]

Pilecap length

hf = 1400

[mm]

Pilecap thickness

tf = 540

[mm]

Vertical reinft. 500


25

@

92

Horizontal reinft. Pile diameter

dp = 200

[mm]

As2 bar size = 25

Pile spacing

sp = 600

[mm]

As2 bar No = 7

Pile edge distance

ed = 400

[mm]

Concrete strength

f'c = 30

[MPa]


fy = 400

[MPa]



 c = 0.65


 s = 0.85


 = 1.0


Ws = 0

Pilecap self weight


500

25

@

[kN] = 25

[kN]

92


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 2 of 7



= 2031

[kN]

d = tf - 1.5db -150

= 360

[mm]

= 508

[kN]

Qpr = Pf / 4


Mf = 2 x Qpr x 0.5(Sp - hc)

= 102

[kNm]


Vf = 2 x Qpr

= 1016

[kN]

Factored moment to column face

Mf = 2 x Qpr x 0.5(Sp - bc)

= -102

[kNm]


Vf = 2 x Qpr

= 1016

[kN]

Horizontal Direction

CONCLUSION




[Column One and Two Way Shear Is Adequate]


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06




 A2   1  1 . 0  = 0 .33  A   c

 2d

 = 0 . 33 

 Ac


fb =

  1  1 . 0 

 c ( 0.6 f 'c  6  f 'c )


A2 = 2.0E+06 [mm 2]

= 0.49

= 0.09

= 12.7

[MPa]

= 6.3

[MPa]

< fb

OK


  A2  1  1 . 0   Ap

 = 0 .33  

d

  1  1 .0 

 = 0 . 33  

 dp


fb =

 c ( 0.6 f 'c  6  f 'c )

fcb = Qpr / ( *

dp2

/4)

A2 = 5.0E+05 [mm 2]

= 1.00

= 0.27

= 17.4

[MPa]

= 16.2

[MPa]

< fb

OK

Required Reinft b1 = 100

l = sqrt(

[mm] b12

h12)

h1 = 200

[mm]

= 224

[mm]

Fh = l / d x Qpr

= 315

[kN]

FT1 = h1 / l x Fh

= 282

[kN]

As1 = FT1 / ( s x fy)

= 830

[mm 2]

Asmin = 0.2% x bf x tf / 2

= 756

[mm 2]

= 830

[mm 2]

+

Vertical direction


3500 > As1

[mm 2] OK

Horizontal direction FT2 = b1 / l x Fh

= 141

[kN]

As2 = FT2 / ( s x fy)

= 415

[mm 2]

Asmin = 0.2% x hf x tf / 2

= 756

[mm 2]

= 756

[mm 2]


3500 > As2


[mm 2] OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 4 of 7




dVf / Mf > 2.0

Mf / Vf = 100

[mm]

= 120

[mm]


= 2489

[mm 2]


= 1512

[mm 2]


= 2489

[mm 2]

= 7000

[mm 2]

> Asver

OK


for dV f / Mf >=2.0

As prov = 2 x As1 prov

Horizontal Direction dVf / Mf = -3.60

<1.0 -> pilecap does NOT act as deep beam

dVf / Mf < 2.0

Mf / Vf = -100

z = 0.4(d + Mf / Vf ) for dVf / Mf >=1.0

[mm]

= 104

[mm]

Ashor = Mf / ( s fy z )

= -2872

[mm 2]

Asmin = 0.2% x hf x tf

= 1512

[mm 2]

Ashor = max ( Ashor, Asmin )

= 1512

[mm 2]

= 7000

[mm 2]

> Ashor

OK

1.2 Mf / Vf

for dV f / Mf >=2.0


[Deep Beam Model Is NOT Applicable. The Result Is Abandoned]



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 5 of 7



Factored pile load

Vf = 508

[kN]

Pile diameter

dp = 200

[mm]

Pile edge distance

ed = 400

[mm]

d = 360

[mm]


dc = 41

[mm]

ec = 300

[mm]

d/2 = 180

[mm]

dc + 50 = 91

[mm]

< d/2 Critical shear line intersect column, one way shear needs NOT to be checked sqrt(f'c ) = min ( sqrt(fc') , 8 ) min( d, 1.414 ec ) = 360 When no transverse reinft.

= 5.477

[mm]

bw = 1563

 = 230 / (1000 + d)

= 0.169

Vc =  c    fc b w d

= 339

'

< Vf

[MPa]

11.3.4

[mm] 11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK


Factored pile load

Vf = 508

[kN]

Pile diameter

dp = 200

[mm]

Pile edge distance

ed = 400

[mm]

d = 360

[mm]


dc = 41

[mm]

ec = 300

[mm]

d/2 = 180

[mm]

dc + 50 = 91

[mm]



bo =  (dp + d) (270-2)/360

= 1319

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 2.619

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)

vc = min( vc1 , vc2 , vc3 )

= 1.353

[MPa]

Vc = vc bo d

= 643

[kN]



vc1 =  1 



2   0 .19   c f 'c  c 


vc2 = 

 s = 2 for corner pile vc3 = 0 38   c f 'c

> Vf


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 6 of 7


[kN]

bc = 800

Pile diameter

dp = 200

[mm]

hc = 400

[mm]

Pile edge distance

ed = 400

[mm]

bf = 1400

[mm]

d = 360

[mm]

hf = 1400

[mm]

Vf = r x 2 x Qpr

= 0

[kN]

bw = hf

= 1400

[mm]

 = 230 / (1000 + d)

= 0.169

Vc =  c    fc b w d

= 303



[mm]

Ver Critical Shear Line Dist from col face to pile edge Fraction of pile outside critical line Factored shear force


dc = 0

[mm]

r = 0.00

'

> Vf

11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK



dc = 200

[mm]

r = 0.00 Vf = r x 2 x Qpr

= 0

[kN]

bw = bf

= 1400

[mm]

 = 230 / (1000 + d)

= 0.169

Vc =  c    fc b w d

= 303

'

> Vf

11.3.6.3 Eq (11-9)

[kN] OK

11.3.4 Eq (11-6)


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 7 of 7

COLUMN TWO WAY SHEAR A23.3-04 Qpr = 508

[kN]

bc = 800

[mm]

Pile diameter

dp = 200

[mm]

hc = 400

[mm]

Pile edge distance

ed = 400

[mm]

bf = 1400

[mm]

[mm]

hf = 1400

[mm]

= 3840

[mm]

= 1.353

[MPa]

13.3.4.1 (13-5)

= 2.012

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 1870

[kN]



d = 360

Ver Critical Shear Line Dist from col face to pile edge

dc = 0

Fraction of pile outside critical line

r1 = 0.00

[mm]

Hor Critical Shear Line Dist from col face to pile edge

dc = 200


r2 = 0.10

[mm]

Factored shear force

Vf = 203

Critical perimeter length

bo = 2x(bc + d + hc + d)



vc1 =  1 



[kN]

2   0 .19   c f 'c  c 

 c = col length / col width  s d   0 .19  λ  c f 'c  bo 

vc2 = 

s = 4

= 2.0

for interior column

vc3 = 0 38   c f 'c

> Vf


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 1 of 7

PILECAP DESIGN 4 piles pilecap design based on

Code Abbreviation


A23.3-04

INPUT Pcf = 700

[kN]

Column width

bc = 900

[mm]

As1 bar size = 20

Column length

hc = 900

[mm]

As1 bar No = 6

Pilecap width

bf = 2200

[mm]

Pilecap length

hf = 3200

[mm]

Pilecap thickness

tf = 900

[mm]

dp = 324

[mm]

As2 bar size = 20

Pile spacing

sp1 = 2200

[mm]

As2 bar No = 8

Pile spacing

sp2 = 1200

[mm]

Pile edge distance

ed = 500

[mm]

Concrete strength

f'c = 30

[MPa]


fy = 400

[MPa]




20

@

200



 c = 0.65


 s = 0.85


 = 1.0


Ws = 0

Pilecap self weight


300


20

@

200

[kN] = 149

[kN]

001


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 2 of 7



= 886

[kN]

d = tf - 1.5db -150

= 720

[mm]

= 222

[kN]

Qpr = Pf / 4


Mf = 2 x Qpr x 0.5(Sp1 - hc)

= 288

[kNm]


Vf = 2 x Qpr

= 443

[kN]


Mf = 2 x Qpr x 0.5(Sp2 - bc)

= 66

[kNm]


Vf = 2 x Qpr

= 443

[kN]


CONCLUSION

[Pilecap Reinft by S&T Model Is NOT Adequate]




002


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06




 A2   1  1 . 0  = 0 .33  A   c

 2d

 = 0 . 33 

 Ac


fb =

  1  1 . 0 

 c ( 0.6 f 'c  6  f 'c )


A2 = 7.0E+06 [mm 2]

= 0.65

= 0.20

= 14.5

[MPa]

= 0.9

[MPa]

< fb

OK


  A2  1  1 . 0   Ap

 = 0 .33  

d

  1  1 .0 

 = 0 . 33  

 dp


fb =

 c ( 0.6 f 'c  6  f 'c )

fcb = Qpr / ( *

dp2

/4)

A2 = 7.9E+05 [mm 2]

= 0.70

= 0.41

= 17.8

[MPa]

= 2.7

[MPa]

< fb

OK


l = sqrt(

[mm] b12

h12)

h1 = 875

[mm]

= 952

[mm]

Fh = l / d x Qpr

= 293

[kN]

FT1 = h1 / l x Fh

= 269

[kN]

As1 = FT1 / ( s x fy)

= 792

[mm 2]


= 1980

[mm 2]

= 1980

[mm 2]

1650

[mm 2]

+

Vertical direction


< As1

NG


= 115

[kN]

As2 = FT2 / ( s x fy)

= 339

[mm 2]


= 2880

[mm 2]

= 2880

[mm 2]

2400

[mm 2]


< As2

NG

[Pilecap Reinft by S&T Model Is NOT Adequate] 003


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 4 of 7




dVf / Mf < 2.0

Mf / Vf = 650

[mm]

= 548

[mm]


= 1546

[mm 2]


= 3960

[mm 2]


= 3960

[mm 2]

= 3300

[mm 2]

< Asver

NG


for dV f / Mf >=2.0


Horizontal Direction dVf / Mf = 4.80


dVf / Mf > 2.0

Mf / Vf = 150

[mm]

= 180

[mm]


= 1086

[mm 2]


= 5760

[mm 2]


= 5760

[mm 2]

= 4800

[mm 2]

< Ashor

NG



for dV f / Mf >=2.0



004


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 5 of 7



Factored pile load

Vf = 222

[kN]

Pile diameter

dp = 324

[mm]

Pile edge distance

ed = 500

[mm]

d = 720 dc = 505


ec = 338

[mm]

[mm]

d/2 = 360

[mm]

[mm]

dc + 50 = 555

[mm]

> d/2 Critical shear line doesn't intersect column, one way shear shall be checked sqrt(f'c ) = min ( sqrt(fc') , 8 ) min( d, 1.414 ec ) = 478 When no transverse reinft.

= 5.477

[mm]

bw = 2458

 = 230 / (1000 + d)

= 0.134

Vc =  c    fc b w d

= 843

'

> Vf

[MPa]

11.3.4

[mm] 11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK


Factored pile load

Vf = 222

[kN]

Pile diameter

dp = 324

[mm]

Pile edge distance

ed = 500

[mm]

d = 720 dc = 505


ec = 338

[mm]

[mm]

d/2 = 360

[mm]

[mm]

dc + 50 = 555

[mm]

> d/2 Critical shear line doesn't intersect column, one way shear shall be checked


bo =  (dp + d) (270-2)/360



vc1 =  1 



2   0 .19   c f 'c  c 

= 2156

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 3.055

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


vc2 = 


= 1.353

[MPa]

Vc = vc bo d

= 2100

[kN]

> Vf

OK

[Pile One and Two Way Shear Is Adequate] 005


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 6 of 7


[kN]

bc = 900

Pile diameter

dp = 324

[mm]

hc = 900

[mm]

Pile edge distance

ed = 500

[mm]

bf = 2200

[mm]

d = 720

[mm]

hf = 3200

[mm]

dc = 312

[mm]

Vf = r x 2 x Qpr

= 0

[kN]

bw = hf

= 3200

[mm]

 = 230 / (1000 + d)

= 0.134

Vc =  c    fc b w d

= 1097



[mm]



r = 0.00

'

> Vf

11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK



dc = 812

[mm]

r = 0.28 Vf = r x 2 x Qpr

= 126

[kN]

bw = bf

= 2200

[mm]

 = 230 / (1000 + d)

= 0.134

Vc =  c    fc b w d

= 754

'

> Vf

11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK

006


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 7 of 7


[kN]

bc = 900

[mm]

Pile diameter

dp = 324

[mm]

hc = 900

[mm]

Pile edge distance

ed = 500

[mm]

bf = 2200

[mm]

d = 720

[mm]

hf = 3200

[mm]

Dist from col face to pile edge

dc = 312

[mm]


r1 = 0.00

= 6480

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 2.259

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 6312

[kN]



Ver Critical Shear Line


dc = 812


r2 = 1.00

[mm]


Vf = 886


bo = 2x(bc + d + hc + d)



vc1 =  1 



[kN]

2   0 .19   c f 'c  c 


vc2 = 

s = 4

= 1.0

for interior column

vc3 = 0 38   c f 'c

> Vf

OK


007

CivilBay



CivilBay



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 1 of 7


Code Abbreviation


A23.3-04

INPUT Pcf = 2500

[kN]

Column width

bc = 880

[mm]



Column length

hc = 440

[mm]

Pilecap width

bf = 2061

[mm]

Pilecap length

hf = 2061

[mm]

Pilecap thickness

tf = 600

[mm]



25

@

143


dp = 250

[mm]

As2 bar size = 25

Pile spacing

sp = 1061

[mm]

As2 bar No = 7

Pile edge distance

ed = 500

[mm]

Concrete strength

f'c = 30

[MPa]


fy = 400

[MPa]



 c = 0.65


 s = 0.85


 = 1.0


Ws = 0

Pilecap self weight


500

25

@

[kN] = 60

[kN]

143


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 2 of 7



= 2575

[kN]

d = tf - 1.5db -150

= 420

[mm]

= 515

[kN]

Qpr = Pf / 5


Mf = 2 x Qpr x 0.5(Sp - hc)

= 320

[kNm]


Vf = 2 x Qpr

= 1030

[kN]


Mf = 2 x Qpr x 0.5(Sp - bc)

= 93

[kNm]


Vf = 2 x Qpr

= 1030

[kN]


CONCLUSION



[Pile One and Two Way Shear Is NOT Adequate]



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06




 A2   1  1 . 0  = 0 .33  A   c

 2d

 = 0 . 33 

 Ac


fb =

  1  1 . 0 

 c ( 0.6 f 'c  6  f 'c )


A2 = 4.2E+06 [mm 2]

= 0.77

= 0.12

= 13.6

[MPa]

= 6.5

[MPa]

< fb

OK


  A2  1  1 . 0   Ap

 = 0 .33  

d

  1  1 .0 

 = 0 . 33  

 dp


fb =

 c ( 0.6 f 'c  6  f 'c )

fcb = Qpr / ( *

dp2

/4)

A2 = 7.9E+05 [mm 2]

= 1.00

= 0.23

= 16.5

[MPa]

= 10.5

[MPa]

< fb

OK


l = sqrt(

[mm] b12

h12)

h1 = 421

[mm]

= 523

[mm]

Fh = l / d x Qpr

= 641

[kN]

FT1 = h1 / l x Fh

= 516

[kN]

As1 = FT1 / ( s x fy)

= 1516

[mm 2]


= 1237

[mm 2]

= 1516

[mm 2]

3500

[mm 2]

+

Vertical direction


> As1

OK


= 381

[kN]

As2 = FT2 / ( s x fy)

= 1120

[mm 2]


= 1237

[mm 2]

= 1237

[mm 2]

3500

[mm 2]


> As2


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 4 of 7




dVf / Mf < 2.0

Mf / Vf = 311

[mm]

= 292

[mm]


= 3219

[mm 2]


= 2473

[mm 2]


= 3219

[mm 2]

= 7000

[mm 2]

> Asver

OK


for dV f / Mf >=2.0




dVf / Mf > 2.0

Mf / Vf = 91

z = 0.4(d + Mf / Vf ) for dVf / Mf >=1.0

[mm]

= 109

[mm]


= 2524

[mm 2]


= 2473

[mm 2]


= 2524

[mm 2]

= 7000

[mm 2]

> Ashor

OK

1.2 Mf / Vf




for dV f / Mf >=2.0


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 5 of 7



Factored pile load

Vf = 515

[kN]

Pile diameter

dp = 250

[mm]

Pile edge distance

ed = 500

[mm]

d = 420 dc = 198


ec = 375

[mm]

[mm]

d/2 = 210

[mm]

[mm]

dc + 50 = 248

[mm]


= 5.477

[mm]

bw = 1864

 = 230 / (1000 + d)

= 0.162

Vc =  c    fc b w d

= 451

'

< Vf

[MPa]

11.3.4

[mm] 11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

NG


Factored pile load

Vf = 515

[kN]

Pile diameter

dp = 250

[mm]

Pile edge distance

ed = 500

[mm]

d = 420 dc = 198


ec = 375

[mm]

[mm]

d/2 = 210

[mm]

[mm]

dc + 50 = 248

[mm]



bo =  (dp + d) (270-2)/360

= 1579

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 2.571

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vc = vc bo d

= 897

[kN]



vc1 =  1 



2   0 .19   c f 'c  c 


vc2 = 

 s = 2 for corner pile vc3 = 0 38   c f 'c

> Vf


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 6 of 7


[kN]

bc = 880

Pile diameter

dp = 250

[mm]

hc = 440

[mm]

Pile edge distance

ed = 500

[mm]

bf = 2061

[mm]

d = 420

[mm]

hf = 2061

[mm]

dc = 216

[mm]

Vf = r x 2 x Qpr

= 0

[kN]

bw = hf

= 2061

[mm]

 = 230 / (1000 + d)

= 0.162

Vc =  c    fc b w d

= 499



[mm]



r = 0.00

'

> Vf

11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK



dc = 436

[mm]

r = 0.06 Vf = r x 2 x Qpr

= 64

[kN]

bw = bf

= 2061

[mm]

 = 230 / (1000 + d)

= 0.162

Vc =  c    fc b w d

= 499

'

> Vf

11.3.6.3 Eq (11-9)

[kN] OK

11.3.4 Eq (11-6)


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 7 of 7


[kN]

bc = 880

[mm]

Pile diameter

dp = 250

[mm]

hc = 440

[mm]

Pile edge distance

ed = 500

[mm]

bf = 2061

[mm]

d = 420

[mm]

hf = 2061

[mm]


dc = 216

[mm]


r1 = 0.02

= 4320

[mm]

= 1.353

[MPa]

13.3.4.1 (13-5)

= 2.061

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 2455

[kN]





dc = 436


r2 = 0.90

[mm]


Vf = 1862


bo = 2x(bc + d + hc + d)



vc1 =  1 



[kN]

2   0 .19   c f 'c  c 


vc2 = 

s = 4

= 2.0

for interior column

vc3 = 0 38   c f 'c

> Vf


OK

CivilBay



CivilBay



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 1 of 8


Code Abbreviation


A23.3-04

INPUT Pcf = 7500

[kN]

Column width

bc = 610

[mm]



Column length

hc = 410

[mm]

Pilecap width

bf = 2100

[mm]

Pilecap length

hf = 3000

[mm]

Pilecap thickness

tf = 1180

[mm]



25

@

173


dp = 300

[mm]

As2 bar size = 25

Pile spacing

sp = 900

[mm]

As2 bar No = 6

Pile edge distance

ed = 600

[mm]

Concrete strength

f'c = 30

[MPa]


fy = 400

[MPa]



 c = 0.65


 s = 0.85


 = 1.0


W s = 10

Pilecap self weight


500

25

@

[kN] = 175

[kN]

181


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 2 of 8



= 7731

[kN]

d = tf - 1.5db -150

= 1000

[mm]

= 1288

[kN]

Qpr = Pf / 6


Mf = 2 x Qpr x (Sp - 0.5hc)

= 1791

[kNm]


Vf = 2 x Qpr

= 2577

[kN]


Mf = 3 x Qpr x (Sp - 0.5bc)

= 560

[kNm]


Vf = 3 x Qpr

= 3865

[kN]


CONCLUSION






Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06




 A2   1  1 . 0  = 0 .33  A   c

 2d

 = 0 . 33 

 Ac


fb =

  1  1 . 0 

 c ( 0.6 f 'c  6  f 'c )


A2 = 6.3E+06 [mm 2]

= 1.00

= 1.00

= 33.1

[MPa]

= 30.0

[MPa]

< fb

OK


  A2  1  1 . 0   Ap

 = 0 .33  

d

  1  1 .0 

 = 0 . 33  

 dp


fb =

 c ( 0.6 f 'c  6  f 'c )

fcb = Qpr / ( *

dp2

/4)

A2 = 1.1E+06 [mm 2]

= 1.00

= 0.78

= 28.3

[MPa]

= 18.2

[MPa]

< fb

OK


l = sqrt(

[mm] b12

h12)

h1 = 798

[mm]

= 851

[mm]

Fh = l / d x Qpr

= 1097

[kN]

FT1 = h1 / l x Fh

= 1028

[kN]

As1 = FT1 / ( s x fy)

= 3022

[mm 2]


= 2478

[mm 2]

= 3022

[mm 2]

3000

[mm 2]

+

Vertical direction


< As1

NG


= 383

[kN]

As2 = FT2 / ( s x fy)

= 1127

[mm 2]


= 2360

[mm 2]

= 2360

[mm 2]

2750

[mm 2]


> As2


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 4 of 8




dVf / Mf < 2.0

Mf / Vf = 695

[mm]

= 678

[mm]


= 7769

[mm 2]


= 4956

[mm 2]


= 7769

[mm 2]

= 6000

[mm 2]

< Asver

NG


for dV f / Mf >=2.0




dVf / Mf > 2.0

Mf / Vf = 145

[mm]

= 174

[mm]


= 9474

[mm 2]


= 7080

[mm 2]


= 9474

[mm 2]

= 8250

[mm 2]

< Ashor

NG





for dV f / Mf >=2.0


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 5 of 8



Factored pile load

Vf = 1288

[kN]

Pile diameter

dp = 300

[mm]

Pile edge distance

ed = 600

[mm]


d = 1000

Dist from pile edge to col corner

dc = 560

ec = 450

[mm]

[mm]

d/2 = 500

[mm]

[mm]

dc + 50 = 610

[mm]


= 5.477

[mm]

bw = 2997

 = 230 / (1000 + d)

= 0.115

Vc =  c    fc b w d

= 1227

'

< Vf

[MPa]

11.3.4

[mm] 11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

NG


Factored pile load

Vf = 1288

[kN]

Pile diameter

dp = 300

[mm]

Pile edge distance

ed = 600

[mm]


d = 1000 dc = 560

ec = 450

[mm]

[mm]

d/2 = 500

[mm]

[mm]

dc + 50 = 610

[mm]



bo =  (dp + d) (270-2)/360



vc1 =  1 



2   0 .19   c f 'c  c 

= 2550

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 3.469

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


vc2 = 


= 1.353

[MPa]

Vc = vc bo d

= 3450

[kN]

> Vf

OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 6 of 8

EDGE PILE TWO WAY SHEAR

Factored pile load

Vf = 1288

[kN]

Pile diameter

dp = 300

[mm]

Pile edge distance

ed = 600

[mm]


d = 1000 dc = -5

[mm] [mm]

ec = 450

[mm]

d/2 = 500

[mm]

dc + 50 = 45

[mm]



bo =  (dp + d) (360-2)/360 vc1

 2   0 .19   c f 'c =  1   c  

= 3571

[mm] A23.3-04

= 2.029

[MPa]

13.3.4.1 (13-5)

= 3.667

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 4831

[kN]


vc2 = 

 s = 3 for edge pile vc3 = 0 38   c f 'c

> Vf


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 7 of 8


[kN]

bc = 610

Pile diameter

dp = 300

[mm]

hc = 410

[mm]

Pile edge distance

ed = 600

[mm]

bf = 2100

[mm]

[mm]

hf = 3000

[mm]

Vf = r x 3 x Qpr

= 0

[kN]

bw = hf

= 3000

[mm]

 = 230 / (1000 + d)

= 0.115

Vc =  c    fc b w d

= 1228



d = 1000

[mm]



dc = 295

[mm]

r = 0.00

'

> Vf

11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK



dc = 845

[mm]

r = 0.00 Vf = r x 2 x Qpr

= 0

[kN]

bw = bf

= 2100

[mm]

 = 230 / (1000 + d)

= 0.115

Vc =  c    fc b w d

= 860

'

> Vf

11.3.6.3 Eq (11-9)

[kN] OK

11.3.4 Eq (11-6)


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 8 of 8


[kN]

bc = 610

[mm]

Pile diameter

dp = 300

[mm]

hc = 410

[mm]

Pile edge distance

ed = 600

[mm]

bf = 2100

[mm]

[mm]

hf = 3000

[mm]

= 6040

[mm]

= 1.586

[MPa]

13.3.4.1 (13-5)

= 3.034

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 8171

[kN]



d = 1000


dc = 295


r1 = 0.00

[mm]


dc = 845


r2 = 1.00

[mm]


Vf = 5154


bo = 2x(bc + d + hc + d)



vc1 =  1 



[kN]

2   0 .19   c f 'c  c 


vc2 = 

s = 4

= 1.49

for interior column

vc3 = 0 38   c f 'c

> Vf


OK

CivilBay



CivilBay



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 1 of 7


Code Abbreviation


A23.3-04

INPUT Ver reinft. in 2xed width range

Pcf = 8750

[kN]

Column width

bc = 740

[mm]



Column length

hc = 740

[mm]

Pilecap width

bf = 3000

[mm]

Pilecap length

hf = 2759

[mm]

Pilecap thickness

tf = 960

[mm]

300


20

@

114

Hor reinft. in 2xed width range Pile diameter

dp = 300

[mm]

As2 bar size = 20

Pile spacing

sp = 900

[mm]

As2 bar No = 12

Pile edge distance

ed = 600

[mm]

Concrete strength

f'c = 30

[MPa]


fy = 400

[MPa]



 c = 0.65


 s = 0.85


 = 1.0


Ws = 0

Pilecap self weight


300

20

@

[kN] = 187

[kN]

114


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 2 of 7



= 8983

d = tf - 1.5db -150

= 780

[mm]

= 1283

[kN]

Qpr = Pf / 7

[kN]


Mf = 2 x Qpr x (0.866Sp - 0.5hc)

= 1051

[kNm]


Vf = 2 x Qpr

= 2567

[kN]


Mf = 2Qpr 0.5(Sp-bc)+Qpr(Sp-0.5bc)

= 886

[kNm]


Vf = 3 x Qpr

= 3850

[kN]


CONCLUSION



[Edge Pile Two Way Shear Is Adequate]



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06




 A2   1  1 . 0  = 0 .33  A   c

 2d

 = 0 . 33 

 Ac


fb =

  1  1 . 0 

 c ( 0.6 f 'c  6  f 'c )


A2 = 8.3E+06 [mm 2]

= 0.96

= 0.37

= 19.3

[MPa]

= 16.0

[MPa]

< fb

OK


  A2  1  1 . 0   Ap

 = 0 .33  

d

 = 0 . 33  

 dp


fb =

  1  1 .0 

 c ( 0.6 f 'c  6  f 'c )

fcb = Qpr / ( *

dp2

/4)

A2 = 1.1E+06 [mm 2]

= 1.00

= 0.53

= 23.1

[MPa]

= 18.2

[MPa]

< fb

OK


[mm]

Fh = b1 / d x Qpr

= 1176

[kN]

As1 = Fh / ( s x fy)

= 3460

[mm 2]

= 2880

[mm 2]

= 3460

[mm 2]

3450

[mm 2]

Asmin = 0.2% x bf x tf / 2 As1 = max ( As1, Asmin ) As1 prov =

< As1


NG


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 4 of 7




dVf / Mf < 2.0

Mf / Vf = 409

[mm]

= 476

[mm]


= 6496

[mm 2]


= 5760

[mm 2]


= 6496

[mm 2]

= 7350

[mm 2]

> Asver

OK


for dV f / Mf >=2.0

As prov =



dVf / Mf > 2.0

Mf / Vf = 230

[mm]

= 276

[mm]


= 9436

[mm 2]


= 5297

[mm 2]


= 9436

[mm 2]

= 6717

[mm 2]

< Ashor

NG


As prov =



for dV f / Mf >=2.0


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 5 of 7



Factored pile load

Vf = 1283

[kN]

Pile diameter

dp = 300

[mm]

Pile edge distance

ed = 600

[mm]

d = 780 dc = 380


ec = 450

[mm]

[mm]

d/2 = 390

[mm]

[mm]

dc + 50 = 430

[mm]



bo =  (dp + d) (360-2)/360

= 3393

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 3.132

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 3580

[kN]



vc1 =  1 



2   0 .19   c f 'c  c 


vc2 = 


> Vf

[Edge Pile Two Way Shear Is Adequate]

OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 6 of 7


[kN]

bc = 740

Pile diameter

dp = 300

[mm]

hc = 740

[mm]

Pile edge distance

ed = 600

[mm]

bf = 3000

[mm]

d = 780

[mm]

hf = 2759

[mm]

dc1 = 230

[mm]



[mm]


dc2 = 680

[mm]

r2 = 0.00


r1 = 0.00


Vf = (2r1 + r2) Qpr

= 0

[kN]

bw = hf

= 2759

[mm]

 = 230 / (1000 + d)

= 0.129

Vc =  c    fc b w d

= 990


'

> Vf

11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK



dc = 559

[mm]

r = 0.00 Vf = r x 2 x Qpr

= 0

[kN]

bw = bf

= 3000

[mm]

 = 230 / (1000 + d)

= 0.129

Vc =  c    fc b w d

= 1076

'

> Vf

11.3.6.3 Eq (11-9)

[kN] OK

11.3.4 Eq (11-6)


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 7 of 7


[kN]

bc = 740

[mm]

Pile diameter

dp = 300

[mm]

hc = 740

[mm]

Pile edge distance

ed = 600

[mm]

bf = 3000

[mm]

d = 780

[mm]

hf = 2759

[mm]


dc = 230

[mm]

dc = 680

[mm]


r1 = 0.00




r2 = 0.97


dc = 559


r3 = 0.56

[mm]


Vf = 5380


bo = 2x(bc + d + hc + d)

[kN] = 6080

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 2.503

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 6416

[kN]



vc1 =  1 



2   0 .19   c f 'c  c 


vc2 = 

s = 4

= 1.0

for interior column

vc3 = 0 38   c f 'c

> Vf


OK

CivilBay



CivilBay



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 1 of 8


Code Abbreviation


A23.3-04

INPUT Pcf = 8000

[kN]

Column width

bc = 930

[mm]



Column length

hc = 620

[mm]

Pilecap width

bf = 2500

[mm]

Pilecap length

hf = 2299

[mm]

Pilecap thickness

tf = 900

[mm]

Ver. reinft. 300


20

@

167

Hor. reinft. Pile diameter

dp = 250

[mm]

As2 bar size = 20

Pile spacing

sp = 750

[mm]

As2 bar No = 8

Pile edge distance

ed = 500

[mm]

Concrete strength

f'c = 30

[MPa]


fy = 400

[MPa]



 c = 0.65


 s = 0.85


 = 1.0


Ws = 0

Pilecap self weight


300

20

@

[kN] = 122

[kN]

143


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 2 of 8



= 8152

d = tf - 1.5db -150

= 720

[mm]

= 1019

[kN]

Qpr = Pf / 8

[kN]


Mf = 3 x Qpr x (0.866Sp - 0.5hc)

= 1038

[kNm]


Vf = 3 x Qpr

= 3057

[kN]


Mf = 2Qpr (Sp-0.5bc)+Qpr0.5(Sp-bc)

= 489

[kNm]


Vf = 3 x Qpr

= 3057

[kN]


CONCLUSION






Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06




 A2   1  1 . 0  = 0 .33  A   c

 2d

 = 0 . 33 

 Ac


fb =

  1  1 . 0 

 c ( 0.6 f 'c  6  f 'c )


A2 = 5.7E+06 [mm 2]

= 0.72

= 0.30

= 16.3

[MPa]

= 13.9

[MPa]

< fb

OK


  A2  1  1 . 0   Ap

 = 0 .33  

d

  1  1 .0 

 = 0 . 33  

 dp


fb =

 c ( 0.6 f 'c  6  f 'c )

fcb = Qpr / ( *

dp2

/4)

A2 = 7.9E+05 [mm 2]

= 1.00

= 0.63

= 25.1

[MPa]

= 20.8

[MPa]

< fb

OK


l = sqrt(

[mm] b12

h12)

h1 = 495

[mm]

= 716

[mm]

Fh = l / d x Qpr

= 1013

[kN]

FT1 = h1 / l x Fh

= 700

[kN]

As1 = FT1 / ( s x fy)

= 2058

[mm 2]


= 1500

[mm 2]

= 2058

[mm 2]

2100

[mm 2]

+

Ver. Reinft.


> As1

OK

Hor. Reinft. FT2 = b1 / l x Fh

= 732

[kN]

As2 = FT2 / ( s x fy)

= 2154

[mm 2]

Asmin = 0.2% x hf x tf / 2.4

= 1724

[mm 2]

As2 = max ( As1, Asmin )

= 2154

[mm 2]

2400

[mm 2]

As2 prov =

> As2


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 4 of 8




dVf / Mf > 2.0

Mf / Vf = 340

[mm]

= 407

[mm]


= 7493

[mm 2]


= 4500

[mm 2]


= 7493

[mm 2]

= 4140

[mm 2]

< Asver

NG


for dV f / Mf >=2.0

As prov =



dVf / Mf > 2.0

Mf / Vf = 160

[mm]

= 192

[mm]


= 7493

[mm 2]


= 4138

[mm 2]


= 7493

[mm 2]

= 4408

[mm 2]

< Ashor

NG


As prov =



for dV f / Mf >=2.0


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 5 of 8



Factored pile load

Vf = 1019

[kN]

Pile diameter

dp = 250

[mm]

Pile edge distance

ed = 500

[mm]

d = 720 dc = 318


ec = 375

[mm]

[mm]

d/2 = 360

[mm]

[mm]

dc + 50 = 368

[mm]


= 5.477

[mm]

bw = 2384

 = 230 / (1000 + d)

= 0.134

Vc =  c    fc b w d

= 817

'

< Vf

[MPa]

11.3.4

[mm] 11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

NG


Factored pile load

Vf = 1019

[kN]

Pile diameter

dp = 250

[mm]

Pile edge distance

ed = 500

[mm]

d = 720 dc = 318


ec = 375

[mm]

[mm]

d/2 = 360

[mm]

[mm]

dc + 50 = 368

[mm]



bo =  (dp + d) (270-2)/360



vc1 =  1 



2   0 .19   c f 'c  c 

= 2286

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 2.920

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


vc2 = 


= 1.353

[MPa]

Vc = vc bo d

= 2226

[kN]

> Vf

OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 6 of 8


Factored pile load

Vf = 1019

[kN]

Pile diameter

dp = 250

[mm]

Pile edge distance

ed = 500

[mm]

d = 720 dc = 215


ec = 375

[mm]

[mm]

d/2 = 360

[mm]

[mm]

dc + 50 = 265

[mm]



bo =  (dp + d) (360-2)/360 vc1

 2   0 .19   c f 'c =  1   c  

= 3047

[mm] A23.3-04

= 2.029

[MPa]

13.3.4.1 (13-5)

= 3.200

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 2968

[kN]


vc2 = 


> Vf


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 7 of 8


[kN]

bc = 930

Pile diameter

dp = 250

[mm]

hc = 620

[mm]

Pile edge distance

ed = 500

[mm]

bf = 2500

[mm]

d = 720

[mm]

hf = 2299

[mm]

dc1 = 410

[mm]



[mm]


dc2 = 35

[mm]

r2 = 0.00


r1 = 0.00


Vf = (2r1+r2) x Qpr

= 0

[kN]

bw = hf

= 2299

[mm]

 = 230 / (1000 + d)

= 0.134

Vc =  c    fc b w d

= 788


'

> Vf

11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK



dc = 465

[mm]

r = 0.00 Vf = r x 3Qpr

= 0

[kN]

bw = bf

= 2500

[mm]

 = 230 / (1000 + d)

= 0.134

Vc =  c    fc b w d

= 857

'

> Vf

11.3.6.3 Eq (11-9)

[kN] OK

11.3.4 Eq (11-6)


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 8 of 8


[kN]

bc = 930

[mm]

Pile diameter

dp = 250

[mm]

hc = 620

[mm]

Pile edge distance

ed = 500

[mm]

bf = 2500

[mm]

d = 720

[mm]

hf = 2299

[mm]


dc = 410

[mm]


r1 = 0.20

= 5980

[mm]

= 1.578

[MPa]

13.3.4.1 (13-5)

= 2.391

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 5825

[kN]





dc = 465


r2 = 0.42

[mm]


Vf = 3030


bo = 2x(bc + d + hc + d)



vc1 =  1 



[kN]

2   0 .19   c f 'c  c 


vc2 = 

s = 4

= 1.50

for interior column

vc3 = 0 38   c f 'c

> Vf


OK

CivilBay



CivilBay



Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 1 of 8


Code Abbreviation


A23.3-04

INPUT Pcf = 9000

[kN]

Column width

bc = 750

[mm]



Column length

hc = 750

[mm]

Pilecap width

bf = 2500

[mm]

Pilecap length

hf = 2500

[mm]

Pilecap thickness

tf = 1060

[mm]

Ver. reinft. 300


20

@

143

Hor. reinft. Pile diameter

dp = 250

[mm]

As2 bar size = 20

Pile spacing

sp = 750

[mm]

As2 bar No = 8

Pile edge distance

ed = 500

[mm]

Concrete strength

f'c = 30

[MPa]


fy = 400

[MPa]



 c = 0.65


 s = 0.85


 = 1.0


Ws = 0

Pilecap self weight


300

20

@

[kN] = 156

[kN]

143


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 2 of 8



= 9195

d = tf - 1.5db -150

= 880

[mm]

= 1022

[kN]

Qpr = Pf / 9

[kN]


Mf = 3Qpr x (Sp - 0.5hc)

= 1149

[kNm]


Vf = 3 x Qpr

= 3065

[kN]


Mf = 3Qpr x (Sp - 0.5bc)

= 1149

[kNm]


Vf = 3 x Qpr

= 3065

[kN]


CONCLUSION






Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06




 A2   1  1 . 0  = 0 .33  A   c

 2d

 = 0 . 33 

 Ac


fb =

  1  1 . 0 

 c ( 0.6 f 'c  6  f 'c )


A2 = 6.3E+06 [mm 2]

= 0.78

= 0.45

= 19.2

[MPa]

= 16.0

[MPa]

< fb

OK


  A2  1  1 . 0   Ap

 = 0 .33  

d

  1  1 .0 

 = 0 . 33  

 dp


fb =

 c ( 0.6 f 'c  6  f 'c )

fcb = Qpr / ( *

dp2

/4)

A2 = 7.9E+05 [mm 2]

= 1.00

= 0.84

= 29.6

[MPa]

= 20.8

[MPa]

< fb

OK


l = sqrt(

[mm] b12

h12)

h1 = 563

[mm]

= 795

[mm]

Fh = l / d x Qpr

= 924

[kN]

FT1 = h1 / l x Fh

= 653

[kN]

As1 = FT1 / ( s x fy)

= 1921

[mm 2]


= 1767

[mm 2]

= 1921

[mm 2]

2400

[mm 2]

+

Ver. Reinft.


> As1

OK

Hor. Reinft. FT2 = b1 / l x Fh

= 653

[kN]

As2 = FT2 / ( s x fy)

= 1921

[mm 2]


= 1767

[mm 2]

= 1921

[mm 2]

2400

[mm 2]


> As2


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 4 of 8




dVf / Mf > 2.0

Mf / Vf = 375

[mm]

= 450

[mm]


= 7512

[mm 2]


= 5300

[mm 2]


= 7512

[mm 2]

= 4830

[mm 2]

< Asver

NG


for dV f / Mf >=2.0

As prov =



dVf / Mf > 2.0

Mf / Vf = 375

[mm]

= 450

[mm]


= 7512

[mm 2]


= 5300

[mm 2]


= 7512

[mm 2]

= 4830

[mm 2]

< Ashor

NG


As prov =



for dV f / Mf >=2.0


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 5 of 8



Factored pile load

Vf = 1022

[kN]

Pile diameter

dp = 250

[mm]

Pile edge distance

ed = 500

[mm]

d = 880 dc = 405


ec = 375

[mm]

[mm]

d/2 = 440

[mm]

[mm]

dc + 50 = 455

[mm]


= 5.477

[mm]

bw = 2544

 = 230 / (1000 + d)

= 0.122

Vc =  c    fc b w d

= 975

'

< Vf

[MPa]

11.3.4

[mm] 11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

NG


Factored pile load

Vf = 1022

[kN]

Pile diameter

dp = 250

[mm]

Pile edge distance

ed = 500

[mm]

d = 880 dc = 405


ec = 375

[mm]

[mm]

d/2 = 440

[mm]

[mm]

dc + 50 = 455

[mm]



bo =  (dp + d) (270-2)/360



vc1 =  1 



2   0 .19   c f 'c  c 

= 2115

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 3.639

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


vc2 = 


= 1.353

[MPa]

Vc = vc bo d

= 2518

[kN]

> Vf

OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 6 of 8


Factored pile load

Vf = 1022

[kN]

Pile diameter

dp = 250

[mm]

Pile edge distance

ed = 500

[mm]

d = 880 dc = 250


ec = 375

[mm]

[mm]

d/2 = 440

[mm]

[mm]

dc + 50 = 300

[mm]



bo =  (dp + d) (360-2)/360 vc1

 2   0 .19   c f 'c =  1   c  

= 3003

[mm] A23.3-04

= 2.029

[MPa]

13.3.4.1 (13-5)

= 3.807

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 3575

[kN]


vc2 = 


> Vf


OK


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 7 of 8


[kN]

bc = 750

Pile diameter

dp = 250

[mm]

hc = 750

[mm]

Pile edge distance

ed = 500

[mm]

bf = 2500

[mm]

d = 880

[mm]

hf = 2500

[mm]

dc1 = 500

[mm]



[mm]

Ver Critical Shear Line Dist from col face to pile edge Fraction of pile outside critical line

r1 = 0.00


Vf = r1 x 3Qpr

= 0

[kN]

bw = hf

= 2500

[mm]

 = 230 / (1000 + d)

= 0.122

Vc =  c    fc b w d

= 958


'

> Vf

11.3.6.3 Eq (11-9)

[kN]

11.3.4 Eq (11-6)

OK



dc = 500

[mm]

r = 0.00 Vf = r x 3Qpr

= 0

[kN]

bw = bf

= 2500

[mm]

 = 230 / (1000 + d)

= 0.122

Vc =  c    fc b w d

= 958

'

> Vf

11.3.6.3 Eq (11-9)

[kN] OK

11.3.4 Eq (11-6)


Calculation Sheet


Eng : Test


Chk : Test

Jon No :

Date : 2013-03-06


Rev : 0 8 of 8


[kN]

bc = 750

[mm]

Pile diameter

dp = 250

[mm]

hc = 750

[mm]

Pile edge distance

ed = 500

[mm]

bf = 2500

[mm]

d = 880

[mm]

hf = 2500

[mm]


dc = 500

[mm]


r1 = 0.24

= 6520

[mm]

= 2.029

[MPa]

13.3.4.1 (13-5)

= 2.599

[MPa]

13.3.4.1 (13-6)

= 1.353

[MPa]

13.3.4.1 (13-7)


= 1.353

[MPa]

Vr = vc bo d

= 7762

[kN]





dc = 500


r2 = 0.24

[mm]


Vf = 2707


bo = 2x(bc + d + hc + d)



vc1 =  1 



[kN]

2   0 .19   c f 'c  c 


vc2 = 

s = 4

= 1.00

for interior column

vc3 = 0 38   c f 'c

> Vf


OK

Pile Cap Design Examples

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