In this article, we
explained How to make a bar bending schedule of pile foundation
PILE FOUNDATION DIAGRAM
GIVEN DATA:
Pile cap length =
4000mm
Pile cap breadth=
4000mm
Pile diameter = 1.2m or
1200mm
Pile radius = 0.5m or
500mm
Number of Vertical rod
in piles = 16nos
Length of the pile
driven = 32.5m
Clear cover of the
pile= 75mm
Vertical Rods Diameter and Numbers of Piles
Length |
No of vertical rods are used in piles |
Diameter of steel |
0 to -12
m |
16nos |
25Ã˜ |
-12 to
-21m |
16nos |
20Ã˜ |
-21 to
-32.5m |
16nos |
16Ã˜ |
The length of the vertical rods, divided by 12meter of rod length and
diameter of the rod changes on each segment, which will be decided by the structural consultant. The above table is for calculation purpose
Master Ring Calculation
Dia. Of master ring =
16 mm @ 1500 mm c/c
No. of master ring =
(Length of pile / Spacing) + 1 = 32.5/1.5 = 21.67 = 22nos
Helical Ring Calculation
No. of spirals =
(Length of pile / Pitch of pile) + 1 = 32.5/0.15 = 216.67 = 217nos
MASTER RING CALCULATION
Details
Given for Master Ring
Length of pile = 32.5 m or 32500 mm
Diameter Of master ring = 16 mm @ 1500 mm c/c
Length
of 1 master ring = Circumference of ring/circle = 2 Ï€r
r = R = Radius = diameter/2
Diameter
of The Master Ring = Diameter of pile – (2 x clear cover) – (2 x spiral ring
diameter) – (2 x Vertical bar diameter) – (Master ring dia)
Vertical bar diameter varies in each segment as 16, 20
and 25. We will take 25mm dia for the vertical bar diameter because it is the
highest dia
Dia = 1200 – (2x75) – (2x8) – (2x25) – (16)
Dia = 968mm
r = dia/2 = 968/2 = 484mm = 0.484m
Length of 1 master ring = Circumference of circle = 2 Ï€r
Length of 1 master ring = 2 x 3.14 x 0.484 = 3.039m
Total
number of master ring (N_{r})= (Length of pile / Spacing) + 1 =
(32500 / 1500) + 1 = 22.67 - 23nos.
For 23 nos. of master ring = 3.039 x 23 = 69.897m of 16mm dia
SPIRAL OR HELICAL RING CALCULATION
Details
Given For Spiral or
Helical Ring:
Dia. of pile = 1200 mm or 1.2 m
Length of pile = 32.5 m or 32500 mm
Pitch or Spacing = 150 mm
Clear cover = 75 mm
Dia. of spiral bar = 8 mm Ã˜
Length
of 1 helical ring = Circumference of ring/circle= 2 Ï€r
r = R = Radius = diameter/2
Diameter
of the helical Ring = Diameter of the pile – (2 x clear cover) – (Spiral bar
dia)
Diameter of the helical Ring = 1200 – (2x75) – (8) = 1042mm
Diameter = r/2 = 1042mm
r = dia/2 = 1042/2 = 521mm
r = 521mm or 0.521mm
Length of 1 spiral or helical ring = 2Ï€r = 2 x Ï€ x 0.521
= 3.27mm
No.
of helical or spirals ring (N_{r}) = (Length of pile / Pitch of pile) +
1
= (32500 / 150) + 1
= 217.666666667 – 218
nos.
Total length of spiral or helical ring = 3.27 x 218 = 712.86m
We know that the length of a full-length bar is 12 m.
so, we should find the total length of lap
To
find, Lap considered 50 d
As we know diameter of spiral ring 8mm = 50 x 8 = 400 mm
= 0.4 m
Number
of lap needed = [Total no. of bar / One full length bar] – 1
= (712.86 / 12) – 1 = 58.405 – 59nos
Total length of lap = 0.4 x 59 = 23.6 m
Total
Length of Spiral Ring = 712.86m + 23.6m = 736.46 meters of 8mm dia
VERTICAL BAR CALCULATION
Details
Given on Vertical Bar Calculation
Length
of one bar 25 Ñ„ – 12m
Length |
No vertical rods used in piles |
Diameter of steel |
0 to -12
m |
16nos |
25Ã˜ |
= 50 d +12000 + 50d = (50 x 25) + 12000 + (50 x 25) = 14.5 m - 14500 mm
Therefore,
Length for 16 no. bar = 14.5 x 16 = 232m
Length
of one bar 20 Ñ„ – 12m
Length |
No of vertical rods are used in piles |
Diameter of steel |
-12 to
-21m |
16nos |
20Ã˜ |
= 50 d +12000 = (50 x 25) + 12000 = 13.25 m - 13250 mm
Therefore,
Length for 16 no. bar = 13.25 x 16 = 212m
Length
of one bar 16 Ñ„ – 8.5m
Length |
No of vertical rods are used in piles |
Diameter of steel |
-21 to
-32.5m |
16nos |
16Ã˜ |
= 8500 + 300 – (bend) = 8500 + 300 – (2d) = 8500 + 300 –
(2x16) = 8.768m - 8768mm
Therefore,
Length for 16 no. bar = 13.25 x 8.768 = 116.176m
Dia. of bar in mm |
Spiral ring |
Master ring |
Vertical rod |
Total length in m |
Unit weight of
steel in kg/m |
Total weight in kg |
8 mm |
736.46 m |
- |
- |
736.46 m |
0.395 kg/m |
290.90 kg |
16 mm |
- |
69.897m |
116.176m |
186.07m |
1.58 kg/m |
293.99 kg |
20 mm |
- |
- |
212m |
212 m |
2.47 kg/m |
523.64 kg |
25 mm |
- |
- |
232m |
232 m |
3.86 kg/m |
895.52 kg |
Total Weight in Kg |
2004.05 |
How to Make BBS of
Pile Cap - Pile Cap Steel Bar Calculation
Given Details
Pile Cap size = 4.0m x 4.0m
Bottom Reinforcement 1st layer
Pile Cap Size = 4000 X 4000
Clear cover = 50 mm
Bottom reinforcement dia = 25 Ñ„
Length
of the Bottom reinforcement 1st layer = Length of pile cap – (2 x
clear cover) – (2 x half of dia bar) + (2 x development length (x)) – (2 x bend)
As we
know,
Clear cover = 50mm
Diameter = 25mm
Bend = 2d = 2 x 25 = 50mm
Development length x = 750 – 75 – 50 (clear cover) –
12.5 (half dia) + 150 = 762.5mm – 0.762m
Therefore,
Length of the Bottom reinforcement 1st layer = 4000 – (2
x 50) – (2 x 25/2) + (2 x 762.5) – (2 x 50) = 5300 mm = 5.3m
Number
of bar (25 Ñ„ @ 100c/c) = [Length of bar / Spacing] + 1 =
[4000 / 100] + 1 = 41 nos.
For
41 nos,
Total length of bar required = 5.3 x 41 = 217.3m
Top Reinforcement 1st layer
Length
of the Top reinforcement 1st layer = Length of pile cap – (2 x
clear cover) – (2 x half of dia bar) + (2 x development length (x)) – (2 x bend)
As we
know,
Clear cover = 50mm
Diameter = 12mm
Bend = 2d = 2 x 12 = 24mm
Development length Y = 750 – 50 (clear cover) – 6 (half
dia) + 150 = 844mm – 0.844m
Therefore,
Length
of the Top reinforcement 1st layer = 4000 – (2 x 50) – (2 x 12/2)
+ (2 x 844) – (2 x 24) = 5528mm = 5.5m
Number
of bar (12 Ñ„ @ 125 c/c) = [Length of bar / Spacing] + 1 =
[4000 / 125] + 1 = 33 nos.
For
33 nos,
Total length of bar required = 5.5 x 33 = 181.5m
Bottom Reinforcement 2nd layer
Length
of the Bottom reinforcement 2_{nd} layer 25 Ñ„ =
Length of pile cap – (2 x clear cover) – (2 x half of dia bar) + (2 x
development length (x_{1})) – (2 x bend)
As we
know,
Clear cover = 50mm
Diameter = 25mm
Bend = 2d = 2 x 25 = 50mm
Development length X_{1} = 750 – 75 – 50 (clear cover)
– 12.5 (half dia) – 25 (1^{st} reinforcement dia) + 150 = 737.5mm – 0.737m
Therefore,
Bottom
reinforcement 2_{nd} layer 25 Ñ„ = 4000 – (2 x 50) – (2 x
25/2) + (2 x 737.5) – (2 x 50) = 5250mm –
5.2m
Number
of bar (25 @ 100 c/c) = [Length of bar / Spacing] + 1 =
[4000 / 100] + 1 = 41 nos.
For
41 nos,
Total length of bar required = 5.2 x 41 = 213.2m
Top Reinforcement 2nd layer
Length
of the Top reinforcement 2_{nd} layer 12 Ñ„ =
Length of pile cap – (2 x clear cover) – (2 x half of dia bar) + (2 x
development length (Y_{1})) – (2 x bend)
As we
know,
Clear cover = 50mm
Diameter = 12mm
Bend = 2d = 2 x 12 = 24mm
Development length Y_{1} = 750 – 50 (clear cover)
– 6 (half dia) – 12 (1^{st} reinforcement dia) + 150 = 832mm – 0.832m
Therefore,
Top
reinforcement 2_{nd} layer 12 Ñ„ = 4000 – (2 x 50) – (2 x
12/2) + (2 x 832) – (2 x 24) = 5504mm –
5.5m
Number
of bar (12 Ñ„ @ 125 c/c) = [Length of bar / Spacing] + 1 =
[4000 / 125] + 1 = 33 nos.
For
33 nos,
Total length of bar required = 5.5 x 33 = 181.5m
Side Face Reinforcement
Length
of the Side reinforcement 12 Ñ„ = (Length of pile cap – (2 x clear cover) – (2 x
half of dia bar) + (2 x development length) – (2 x bend)) x 2 (sides)
As we
know,
Clear cover = 50mm
Diameter = 12mm
Bend = 2d = 2 x 12 = 24mm
Development length Y_{1} = 2000 – 50 (clear cover)
– 6 (half dia) –+ 150 = 2094mm – 2.094m
Therefore,
Length
of the Side reinforcement 12 Ñ„ = (4000 – (2 x50) – (2 x 6)
+ (2 x 2094) – (2 x 24)) x 2 = 16056mm -
16m
Number
of bar (12 @ 125 c/c) = [Length of bar / Spacing] + 1 =
[2000 / 125] + 1 = 17 nos.
For
17 nos,
Total length of bar required = 16 x 17 = 272m
Dia.
of bar in mm |
Spiral
ring |
Master
ring |
Vertical
rod |
Side
reinforcement |
Top reinforcement |
Total
length in m |
Unit
weight of steel in kg/m |
Total
weight in kg |
8 mm |
736.46 m |
- |
- |
- |
- |
736.46 m |
0.395 kg/m |
290.90 kg |
12 mm |
- |
- |
- |
272m |
363m |
635 m |
0.889 kg/m |
564.51 kg |
16 mm |
- |
69.897m |
116.176m |
- |
- |
186.07m |
1.58 kg/m |
293.99 kg |
20 mm |
- |
- |
212m |
- |
- |
212 m |
2.47 kg/m |
523.64 kg |
25 mm |
- |
- |
232m |
- |
430.5m |
662.5 m |
3.86 kg/m |
2557.25 kg |
Total
weight of steel required = 4230.29/kg - 4.2ton
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really a good and helpfull article
ReplyDeleteno software to this calculations ?
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