Making Bar Bending Schedule Of Pile Foundation
What is a Pile Cap?
Single pile can be used in construction. Piles also are used together and connected by a reinforced concrete cap.
What is Pile Foundation?
Pile foundation is a type of deep foundation. Pile foundation is made of long columnar elements of steel or reinforced concrete. When the depth of foundation is more than three times its breadth then it is called as piled. Pile foundation is used to transfer the loads from the superstructures. Pile foundation is used for large structures.
Types of Pile Foundation
1. Bored Piling
2. Driven Piling
3. Screw Piling
4. Mini Piling
5. Sheet Piling
BBS Process
Pile cap stands for rectangular concrete slab which can be calculated with the rectangular volume formula.
The pile column must be attached with the inner spacer ring. At the bottom level the anchorage bar is bent into the column. At top of the column the development length is arranged outside. The pile has three parts: Vertical bar, Inner spacer ring and Outer Helical Ring.
Measurement of the Length of Vertical Bar
Cutting length of vertical bar = Pile anchorage length Bottom + Height of Pile + Development Length Upper + Lap Length(50d) – Clear Cover bottom
Vertical Bar Calculation
Development length = 50 d
Steel | |
0 to -12 m | 16-25 ф |
-12 to -21m | 16-20 ф |
-21 to -30.5m | 16-16 ф |
Length of 1 bar 25 ф
= 50 d + 12000 + 50 d
= (50 x 25) + 12000 + (50 x 25)
= 14500 mm
= 14.5 m
Therefor,
Length for 16 no. bar = 14.5 x 16 = 232 m
Length of 1 bar 20 ф
= 9000 + 50 d
= 9000 + (50 x 20)
= 10000 mm
= 10 m
Therefor,
Length for 16 no. bar = 10 x 16 = 160 m
Length of 1 bar 16 ф
= 9500 + 300 – (bend)
= 9500 + 300 – (1 x 2 x d)
= 9500 + 300 – (1 x 2 x 16)
= 9768 mm
= 9.768 m
Therefor,
Length of 16 no. bar = 9.768 x 16 = 156.29 m
Master Ring Calculation
Given:
Dia. Of master ring = 16 mm @ 1500 mm c/c
Length of pile = 30.5 m or 30500 mm
r = Radius of pile – clear cover – spiral ring dia. – vertical bar dia. – half of the master ring dia.
r = 500 – 75 – 8 – 25 – (16/2)
r = 384 mm
r = 0.384 m
Length of 1 master ring = Circumference of circle
= 2 πr
= 2 x 3.14 x 0.384
= 2.4 m
Total no. of master ring = (Length of pile / Spacing) + 1
= (30500 / 1500) + 1
= 21 no.
For 24 nos. of master ring
= 2.4 x 21
= 50.4 m
Spiral Ring Calculation
Given:
Pitch or Spacing = 150 mm
Clear cover = 75 mm
Dia. of pile = 1000 mm or 1 m
Length of pile = 30.5 m or 30500 mm
Dia. of spiral bar = 8 mm ф
Net radius of spiral in caging = radius of pile – clear cover – half of spiral bar dia.
= 500 – 75 – ( 8/2 )
= 500 – 75 – 4
r = 421 mm
Length of one spiral ring = circumference of one spiral ring
= 2 πr
= 2 x 3.14 x 421
= 2645 mm
= 2.645 m
No. of spirals = ( Length of pile / Pitch of pile ) + 1
= ( 30500 / 150 ) + 1
= 204 no.
Total length of spiral ring = 2.645 x 204 = 539.58 m
We know the length of one full length bar is 12 m.
Lap considered 50 d
= 50 x 8
= 400 mm
= 0.4 m
Number of lap = [ Total no. of bar / One full length bar ] – 1
= [ 539.58 / 12 ] – 1
= 44 nos.
Total length of lap = 0.4 x 44 = 17.6 m
Total length of spiral ring = 539.58 + 17.6 = 557.18 m
Total weight of Steel required for Pile | |||
Dia. of bar in mm | Total length in m | Unit weight of steel in kg/m | Total weight in kg |
8 mm | 557.18 m | 0.395 kg/m | 220 kg |
16 mm | 206.69 m | 1.58 kg/m | 326.6 kg |
20 mm | 160 m | 2.47 kg/m | 395.2 kg |
25 mm | 232 m | 3.86 kg/m | 895.5 kg |
Pile Cap Steel Bar Calculation
Explanation for finding a & b value
a = 1000 – 75 – 50 – 12.5 + 150
a = 1012.55 mm
a = 1.0125 m
b = 1000 – 50 – 6 + 150
b = 1094 mm
b = 1.094 m
Pile Cap (4200 x 4200)
Clear cover = 50 mm
Bottom reinforcement 1st layer
ф = Length of pile cap – ( 2 x clear cover ) – ( 2 x half of bar ) + (2 x a) – ( 2 x 2d )
25 ф = 4200 – ( 2 x 50 ) – ( 2 x 25/2 ) + ( 2 x 1012.5 ) – ( 2 x 2d )
= 6000 mm
= 6 m
Number of bar = [ Length of bar / Spacing ] + 1
= [ 4200 / 100 ] + 1
= 43 nos.
For 43 nos,
Total length of bar required = 6 x 43 = 258 m
Top reinforcement
12 ф = Length of pile cap – (2xclear cover) – (2xhalf of bar) + (2xb) – (2x2d)
12 ф = 4200 – ( 2 x 50 ) – ( 2 x 6 ) + ( 2 x 1094 ) – ( 2 x 2 x 12 )
= 6.228 m
Number of bar = [ Length of bar / Spacing ] + 1
= [ 4200 / 125 ] + 1
= 35 nos.
For 35 nos.
Total length of bar required = 6.228 x 35 = 218 m
Pile Cap (4200 x 2000)
Explanation for finding value a^{1} & b^{1}
a^{1} = 1000 – 75 – 50 – 12.5 + 150
a^{1} = 1012.55 mm
a^{1} = 1.0125 m
b^{1} = 1000-50-6+150
b^{1} = 1094 mm
b^{1} = 1.094 m
Pile Cap (4200 x 4200)
Clear cover = 50 mm
Bottom reinforcement 2st layer
25 ф = Length of pile cap–( 2 x clear cover )–( 2 x half of bar )+(2 x a1)–( 2 x 2d)
25 ф = 4200 – ( 2 x 50 ) – ( 2 x 12.5 ) + ( 2 x 987.5 ) – ( 2 x 2 x 25 )
Number of bar = [ Length of bar / Spacing ] + 1
Total length of bar required = 5.95 x 43 = 255.85 m
12 ф = Length of pile cap – (2xclear cover) – (2xhalf of bar) + (2xb1) – (2x2d)
12 ф = 4200 – ( 2 x 50 ) – ( 2 x 6 ) + ( 2 x 1082 ) – ( 2 x 2 x 12 )
Number of bar = [ Length of bar / Spacing ] + 1
Total length of bar required = 6.204 x 35 = 217.14 m
Side Face Reinforcement
Explanation for a value a
a= 2100 – 50 – 6 + 150
= 2194 mm
a= 2.194 m
12 ф = { Length of pile cap – ( 2 x clear cover ) – ( 2 x half of bar ) + (2xa) – (2 x 2d) } x 2
= { 4200 – (2x50) – ( 2 x 6 ) + ( 2 x 2194 ) – ( 2 x 2 x 12 ) } x 2
= 16.856 m
For 7 no. of bars
= 16.856 x 16
= 118 m
Pile Cap Steel Quantity | |||
Dia. in mm | Total length in m | Unit weight in kg | Total weight in kg |
25 mm | 513.85 m | 3.86 kg | 1983.4 kg |
12 mm | 553.14 m | 0.889 kg | 491.7 kg |