Let us make BBS of the pile in a triangular pile cap foundation as shown below.
Given data:
Length of pile = 8000mm.=8m.
Diameter of pile = 300mm = 0.3m
Clear cover = 50mm
Pitch = 150 mm. = 0.15m
The diameter of helical stirrup bar = 8mm.= 0.008m.
Main reinforcement. = 12mm.∅ -6 nos.
Master ring =12∅ @ 1500 c/c
Development length Ld = 550mm.
Bottom anchorage = 150mm.
Calculation:
A. Cutting length:
1. Main reinforcement:
The cutting length of the reinforcement bar
= [ pile depth + development length + bottom anchorage - clear cover - ( 90° bend + 45° bend)]
= [ 8000mm. + 550mm. + 150mm. - 50mm. - ( 2d + 1d )]
= [ 8650mm.- ( 2 × 12mm. + 1 × 12mm.)]
= 8614mm.
= 8.61m.< 12m.
Hence no lapping is required.
2. Master ring:
The cutting length of the master ring
= [2πr + hook length]
Here,
The radius of the master ring r
= [pile dia. - {2nos. × (clear cover + spiral stirrup bar dia. + rebar dia.+(1/2 × master ring bar dia.)}] ÷2
=[300mm. - {2nos.× (50mm. +8mm. + 12mm.+ ( 1/2 × 12mm.)}] ÷2
= [300mm. - {2nos. × 76mm.}] ÷2
= [148mm] ÷2
= 74mm.
The cutting length of the master ring
= [ (2 × 3.142 × 74mm.) + (10d)]
= [ (465.02mm) + (10 × 12mm)]
= 585.02mm.
= 0.585m.
3. Spiral stirrup or helix:
The cutting length of the spiral stirrup is calculated by the formula
= n√C2 + P2
Where,
n = no. of turns.
C = circumference of the stirrup.
P = pitch.
⧭. Number of turns ( n )
= [{( length of pile - clear cover) ÷ pitch } + no. of closure rings]
= [ {(8000mm. - 50mm.) ÷ 150mm } + 2 ]
= [53 + 2]
= 55 nos.
Note: Here, no 2 i.e. added denotes the closure rings at the top & bottom of the helical stirrup as shown in the below drawing.
⧭. Circumference of the helical stirrup ( C )
= πD or 2πr
Here,
r = [(diameter of pile ) - (2 nos.× clear cover ) - (2nos. ×1/2 × dia. of the spiral stirrup bar )] ÷ 2
r = [(300mm ) - ( 2nos. × 50mm ) - (2nos. × 1/2 × 8mm )] ÷ 2
= [300mm - 100mm -8mm] ÷ 2
= [192mm] ÷ 2
= 96mm.
= 0.096m.
Circumference of the helical stirrup
C = 2πr
= [2 × 3.142 × 0.096 m.]
= 0.603 m.
The cutting length of the helical stirrup or spiral stirrup
= n√C2 + P2
= 55 × √0.6032 + 0.152
= 55 × √ 0.364 + 0.0225
= 55 × √ 0.386
= 55 × 0.621
= 34.17m > 12m.
To get the exact cutting length of the spiral stirrup bars, we have to add lap length, as the length of an individual bar is limited to 12m.
Let us provide a lap length of 50d, where d = diameter of the helical bar.
⧭. Number of lapping required
= [Calculated length of the helical bar ÷ length of a single bar.]
= [34.17m ÷ 12m]
= 2.84 say 3nos.
The total length of the helical stirrup bar required in a pile
= [ calculated cutting length of the bar + ( no. of lapping × lap length )]
= [34.17m. + ( 3 nos. × 50d )]
= [34.17m. + ( 3 × 50 × 0.008m. ) ]
= [34.17m. + 1.2m.]
= 35.37m.
Go through the article👇
B. No of master ring:
No. of master ring required
= [{( length of pile - clear cover) ÷ c/c distance } + 1]
= [ {(8000mm. - 50mm.) ÷ 1500mm } + 1 ]
= [5.3 + 1]
= 6.3 nos. say 7 nos.
Now, let us prepare a BBS table for a single pile.
|
Sl. No. |
Type of Bar |
Dia. in mm. |
Nos. |
Length in m. |
Total length in m. |
Weight in Kg/m. |
Total bar wt. in kg. |
|
1. |
Main reinforcement. |
12 |
6 |
8.61 |
51.66 |
0.888 |
45.87 |
|
2. |
Helical stirrup. |
8 |
1 |
35.37 |
35.37 |
0.395 |
13.97 |
|
3. |
Master ring. |
12 |
7 |
0.585 |
4.095 |
0.888 |
3.64 |
|
4. |
Total weight of steel bars = |
63.48 |
|||||
|
5. |
Add 3% wastage = |
1.90 |
|||||
|
6. |
The grand total wt. of rebar's = |
65.38 |
|||||
Continued 👉 Type-2
Thank you for going through these calculation steps❤. Have a good day 😄.
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