Bar bending schedule of two way slab and How to calculate the cutting length and weight of rebar’s in 2 way slab?
In this Article we will calculate the cutting length and prepare the B.B.S of the 2 way slab as shown below.
Let us redraw the above drawing with the dimensions for the calculation purpose as below.
Given Data.
Dimension of slab
Lx = five m.= 5000-mm.
Ly = three m. = 3000-mm.
Thickness = 125-mm.
Main bar-One = 10mm@ 150 c/c
Main bar-Two or say distribution bar) = 10-mm @ 150 c/c
Top extra bar-One = 8-mm @ 150 c/c
Top extra bar Two 8-mm @ 150 c/c
Clear cover of the slab = 25-mm.
Development length Ld = 45-d.
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Calculation steps.
1 -The no. of the bars
a – Main bars.
The no. of main bars – One
= [Lx ÷ bar spacing ] + One
= [ 5000 ÷ 150] + One
= 33.33 + One
= 43.33 nos.
By rounding off = 44 nos.
b – Main bars – 2.
The no. of the main bars – Two ( or say distribution bar)
= [Ly ÷ bar spacing ] + One
= [ 3000 ÷ 150] + One
= 20 + One
= 21 nos.
c – Top extra bars- One.
The no. of the top extra bars-One on one side
= [{(Ly ÷ Fore)÷ bar spacing} + One]
= [{ (3000 ÷ Fore) ÷ 150} +One]
= [{ 750 ÷ 150} +One]
= [{ 5} +One]
= Six nos.
The total no. of the top extra bars on both sides.
= [2 sides × Six nos.]
= 12 nos.
Note: The top extra bars are provided at the Ly/fore span of the main bar on the both sides.
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c – Top extra bars- Two .
The no. of the top extra bars- Two on one side
= [{(Lx ÷ 4)÷ bar spacing} +One]
= [{ (5000 ÷ 4) ÷ 150} +One]
= [{ 1250 ÷ 150} +One]
= [{ 8.33} +One]
= 9.33 nos.
By rounding off = Ten nos.
The total no. of the top extra bars on both sides
= [Two sides × 10nos.]
= 20 nos.
Note: The top extra bars are provided at the Lx/fore span of the main bar on the both sides.
2 – Cutting length.
a – Main bar- One.
Cutting length of the main bar- One
= [clear span +(Two nos. × development length) + (extra crank length) – (Two nos. × 45° bend)]
= [ 3000mm. + ( Two nos. × Ld) + ( 0.42D ) – ( Two nos. × One d)]
Here, D = [slab thickness – ( top and bottom clear cover) + bar diameter ]
= [125mm – (Two nos × 25 mm.) +Ten mm ]
= [125mm – 60 mm]
D = 65 mm.
Cutting length of the main bar- One.
= [ 3000mm. + ( Two nos. × 45 × Ten mm. ) + ( 0.42× 65mm. ) – ( Two nos. × One × 10mm)]
= [ 3000mm. + 900mm. + 27.3mm. – 20mm. ]
= 3907.30 mm. = 3.91m.
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b – Main bar- Two.
Cutting length of the main bar – Two
= [clear span +(Two nos. × development length) + (extra crank length) – (Two nos. × 45° bend)]
= [ 5000mm. + ( Two nos. × Ld) + ( 0.42D ) – ( Two nos. × One d)]
= [ 5000mm. + ( Two nos. × 45 × 10mm. ) + ( 0.42× 65mm. ) – ( Two nos. × One × 10mm)]
= [ 5000mm. + 900mm. + 27.3mm. – 20mm. ]
= 5907.30 mm. = 5.91m.
c – Top extra bar – One.
Cutting length of the top extra bar- One
= [clear span Lx + (Two nos. × development length)]
= [5000mm. + ( Two nos. × 45d )]
= [ 5000mm. + ( Two nos. × 45 × 8mm. )]
= [ 5000mm. + 720mm.]
= 5720 mm. = 5.72m.
d – Top extra bar- Two.
Cutting length of the top extra bar- Two
= [clear span Ly + (Two nos. × development length)]
= [3000mm. + ( Two nos. × 45d )]
= [ 3000mm. + ( Two nos. × 45 × 8mm. )]
= [ 3000mm. + 720mm.]
= 3720 mm. = 3.72m.
Let us prepare the B.B.S table to calculate the weight of rebars.
Sl. No. |
Type of Bar |
Dia.in mm. |
Nos. |
Lengthin m. |
Total lengthin m. |
WeightKg/m. |
Totalbar wt.in kg. |
1. |
Main bar- One |
10 |
44 |
3.91 |
172.04 |
0.617 |
106.15 |
2. |
Main bar- Two |
10 |
21 |
5.91 |
124.11 |
0.617 |
76.58 |
3. |
Top extra bar- One |
8 |
12 |
5.72 |
68.64 |
0.395 |
27.11 |
4. |
Top extra bar-Two |
8 |
20 |
3.72 |
74.40 |
0.395 |
29.39 |
4. |
Total weight of the bars = |
239.23 |
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5. |
Add three % wastage = |
7.18 |
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6. |
The grand total wt. of the rebar’s = |
246.41 |
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