The bar bending schedule, or bbs, is essential in estimating the steel needed for beams, columns, and slabs. It helps determine bar shape, size, length, weight, bending dimension, etc. I will create a slab bar bending schedule with examples in this article.
Slabs are classified into two types: one-way slabs and two-way slabs. A one-way slab is supported on two opposing sides and only bears load in one direction. The longer span to shorter span ratio equals or exceeds 2. i.e., l/b ≥ 2.
Ly/Lx = Longer span/shorter span = 5000/2000 = 2.5 > 2
It is a one-way slab.
Given,
Length of slab = 5000 mm
Width of slab = 2000 mm
Main bar = 12 mm @ 150 mm c/c
Distribution ber = 8 mm @ 150 mm c/c
Clear cover (Top and bottom) = 25 mm
The Thickness of the slab = is 150 mm.
Development length = 40d
Where d is the dia of the bar.
Determine the number of main and distribution bars required first.
Cutting length of main bar = Clear span of slab + (2 x development length) + Inclined length – (Bend length)
Clear span of slab = 2000 mm
Development length = 40d
Inclined length 0.42d
1d is for every 45 bend
Where d = diameter of the bar.
Now compute D
D = Thickness of slab – Both clear side cover (top & bottom) – dia of bar
150 – (25+25) -12 = 88 mm
Length of main bar = Ly + 2Ld+ (1 x 0.42d) -(1d x 4)
= 2000 + (2 x 40 x 12) + (2 x 0.42 x 88) – (1 x 12 x 4)
= 2838 mm = 2.838 m
Weight of main bars = d^2 x L/162 x 34 = 12^2 x 2.838/162 = 86 kg
Length Of Distribution Bar:
= Clear span of slab + 2 x development length = Lx + 2 Ld
= 5000 + 2 x 40 x 8 = 5640 mm = 5.64 m
Weight of distribution bars = d^2 x L/162 x 14
= 8^2 x 5.64/162 x 14 = 31 kg
No. of top bars = {(Ly/4)/spacing +1} x 2 = {(2000/4)150 +1} x 2 = 9
Length of extra bar = Ly – 2 x Ly/4 + 2 x 100 {Here 2 x 100 is for both side lapping of 100 mm for extra bar}
= 2000 – (2 x 2000/4) +200
= 1200 mm = 1.2 m
Weight of extra bar = d^2 x L/162 x 9 = 8^2 x 1.2/162 x 9 = 4.266 kg
It is a two-way slab.
Given,
Length of longer span = 5000 mm
length of shorter span = 3000 mm
Main bar = 12 mm @ 150 mm c/c
Distribution ber = 8 mm @ 150 mm c/c
Clear cover (Top and bottom) = 25 mm
The Thickness of the slab = is 200 mm.
Development length = 40d
Where d is dia of a bar.
Step 1: For Sections A-A
Calculate the number of bars required for the main bars and distribution bars.
No. of bars = Length of slab/spacing + 1
Total No. of bars = Ly/150 +1 = 4000/150 +1 = 27
No. of main bars = 14
No. of distribution bars = 13
Clear span of slab = 5000 mm
Development length= Ld = 40d
Inclined length 0.42d
1d is or every 45° bend
Where d = diameter of the bar.
Now Calculate D
D = Thickness of slab – Both clear side cover (top & bottom) – dia of bar
200 – (25+25) -12 = 138 mm
Length of main bar = Lx + (2 x Ld)+ (2 x 0.42d) -(1d x 4)
= 5000 + (2 x 40 x 12) + (2 x 0.42 x 138) – (1 x 12 x 4)
= 2838 mm = 6.02 m
For 14 bars total length = 6.02 x 14 = 84.28 m
Weight of main bars = d^2 x L/162 = 12^2 x 84.28/162 = 75 kg
Cutting Length Of Distribution Bar:
= Clear span of slab + 2 x development length = Lx + 2 L
= 5000 + 2 x 40 x 8 = 5640 mm = 5.64 m
Total length of distribution bar = 5.64 x 13 = 73.32
Weight of distribution bars = d^2 x L/162
= 8^2 x 73.32/162 = 29 kg
No. of extra bars = {(Ly/4)/spacing +1} x 2 = {(4000/4)/150 +1} x 2 = 16
Length of extra bar = Lx – 2 x Lx/4 + 2 x 100 {Here 2 x 100 is for both side lapping of 100 mm for extra bar}
= 5000 – (2 x 5000/4) + 200
= 2800 mm = 2.7 m
Total length of extra bar = 2.7 x 16 = 43.8 m
Weight of extra bar = d^L/162 = 8^2 x 43.2/162 = 17 kg
Step 1: For Section B-B
Determine the number of main and distribution bars required.
Total no. of bars = Length of slab/spacing + 1
Total No. of bars = Lx/150 +1 = 5000/150 +1 = 35
No. of main bars = 18
No. of distribution bars = 17
Step 2:
Clear span of slab = 4000 mm
Development length= Ld = 40d
Inclined length 0.42d
1d is or every 45° bend
Where d = diameter of the bar.
Now Calculate D
D = Thickness of slab – Both clear side cover (top & bottom) – dia of bar
= 200 – (25+25) – 12 = 138 mm
Cutting length of main bar = Ly + (2 x Ld)+ (2 x 0.42d) – (1d x 4)
= 4000 + (2 x 40 x 12) + (2 x 0.42 x 138) – (1 x 12 x 4)
= 5028 mm = 5.02 m
For 18 bars total length = 5.02 x 18 = 90.36 m
Weight of main bars = d^2 x L/162 = 12^2 x 90.36/162 = 80 kg
Cutting Length Of Distribution Bar:
= Clear span of slab + 2 x development length = Ly + 2 x Ld
= 4000 + 2 x 40 x 8 = 4640 mm = 4.64 m
Total length of distribution bar = 4.64 x 17 = 78.88
Weight of distribution bars = d^2 x L/162
= 8^2 x 78.88/162 = 31 kg
Step 3:
Length of extra bar = Ly – (2 x Ly/4) + (2 x 100) {Here 2 x 100 is for both side lapping of 100 mm for extra bar}
= 4000 – (2 x 4000/4) + 200
= 2200 mm = 2.2 m
Total length of extra bar = 2.2 x 9 = 19.8 m
Weight of extra bar = d^L/162 = 8^2 x 19.8/162 = 8 kg
Slabs are classified into two types: one-way slabs and two-way slabs. A one-way slab is supported on two opposing sides and only bears load in one direction. The longer span to shorter span ratio equals or exceeds 2. i.e., l/b ≥ 2.
The slab is supported by four sides in a two-way slab, and loads are carried in both directions. The longer to shorter span ratio is less than 2. i/e l/b 2. Read the entire article to get an idea of how to prepare the slab's bar bending schedule.
Bar Bending Schedule Of One Way Slab:
Example 1:
 |
| One way slab |
Ly/Lx = Longer span/shorter span = 5000/2000 = 2.5 > 2
It is a one-way slab.
Given,
Length of slab = 5000 mm
Width of slab = 2000 mm
Main bar = 12 mm @ 150 mm c/c
Distribution ber = 8 mm @ 150 mm c/c
Clear cover (Top and bottom) = 25 mm
The Thickness of the slab = is 150 mm.
Development length = 40d
Where d is the dia of the bar.
Bar Bending Schedule Of Slab (One Way):
Step 1:
Calculate no. of barsDetermine the number of main and distribution bars required first.
No. of bars = Length of slab/spacing + 1
No. of main bars = Lx/Spacing + 1 = 5000/150 = 34
No. of distribution bars = Ly/Spacing + 1 = 2000/150 +1 = 14
No. of main bars = Lx/Spacing + 1 = 5000/150 = 34
No. of distribution bars = Ly/Spacing + 1 = 2000/150 +1 = 14
Step 2:
Determine the cutting length of the main and distribution bars.Cutting length of main bar = Clear span of slab + (2 x development length) + Inclined length – (Bend length)
Clear span of slab = 2000 mm
Development length = 40d
Inclined length 0.42d
1d is for every 45 bend
Where d = diameter of the bar.
Read More
Now compute D
D = Thickness of slab – Both clear side cover (top & bottom) – dia of bar
150 – (25+25) -12 = 88 mm
Length of main bar = Ly + 2Ld+ (1 x 0.42d) -(1d x 4)
= 2000 + (2 x 40 x 12) + (2 x 0.42 x 88) – (1 x 12 x 4)
= 2838 mm = 2.838 m
Weight of main bars = d^2 x L/162 x 34 = 12^2 x 2.838/162 = 86 kg
Length Of Distribution Bar:
= Clear span of slab + 2 x development length = Lx + 2 Ld
= 5000 + 2 x 40 x 8 = 5640 mm = 5.64 m
Weight of distribution bars = d^2 x L/162 x 14
= 8^2 x 5.64/162 x 14 = 31 kg
Step 3:
Calculate the Top bar (Extra bar) at the top of the crucial length (L/4) area.No. of top bars = {(Ly/4)/spacing +1} x 2 = {(2000/4)150 +1} x 2 = 9
Length of extra bar = Ly – 2 x Ly/4 + 2 x 100 {Here 2 x 100 is for both side lapping of 100 mm for extra bar}
= 2000 – (2 x 2000/4) +200
= 1200 mm = 1.2 m
Weight of extra bar = d^2 x L/162 x 9 = 8^2 x 1.2/162 x 9 = 4.266 kg
Bar Bending Schedule Of Slab (Two Way)
Ly/Lx = Longer span/Shorter span = 5000/3000 = 1.66 < 2It is a two-way slab.
Given,
Length of longer span = 5000 mm
length of shorter span = 3000 mm
Main bar = 12 mm @ 150 mm c/c
Distribution ber = 8 mm @ 150 mm c/c
Clear cover (Top and bottom) = 25 mm
The Thickness of the slab = is 200 mm.
Development length = 40d
Where d is dia of a bar.
Bar Bending Schedule Of Two-Way Slab:
Calculate the number of bars required for the main bars and distribution bars.
No. of bars = Length of slab/spacing + 1
Total No. of bars = Ly/150 +1 = 4000/150 +1 = 27
No. of main bars = 14
No. of distribution bars = 13
Step 2:
Determine the cutting length of the primary and distribution bars.Main bar cutting length = clear span of slab + (2 x development length) + inclined length - (Bend length)
Clear span of slab = 5000 mm
Development length= Ld = 40d
Inclined length 0.42d
1d is or every 45° bend
Where d = diameter of the bar.
Now Calculate D
D = Thickness of slab – Both clear side cover (top & bottom) – dia of bar
200 – (25+25) -12 = 138 mm
Length of main bar = Lx + (2 x Ld)+ (2 x 0.42d) -(1d x 4)
= 5000 + (2 x 40 x 12) + (2 x 0.42 x 138) – (1 x 12 x 4)
= 2838 mm = 6.02 m
For 14 bars total length = 6.02 x 14 = 84.28 m
Weight of main bars = d^2 x L/162 = 12^2 x 84.28/162 = 75 kg
Cutting Length Of Distribution Bar:
= Clear span of slab + 2 x development length = Lx + 2 L
= 5000 + 2 x 40 x 8 = 5640 mm = 5.64 m
Total length of distribution bar = 5.64 x 13 = 73.32
Weight of distribution bars = d^2 x L/162
= 8^2 x 73.32/162 = 29 kg
Step 3:
Calculate the Top bar (Extra) provided at the top of the critical length (L/4) area.No. of extra bars = {(Ly/4)/spacing +1} x 2 = {(4000/4)/150 +1} x 2 = 16
Length of extra bar = Lx – 2 x Lx/4 + 2 x 100 {Here 2 x 100 is for both side lapping of 100 mm for extra bar}
= 5000 – (2 x 5000/4) + 200
= 2800 mm = 2.7 m
Total length of extra bar = 2.7 x 16 = 43.8 m
Weight of extra bar = d^L/162 = 8^2 x 43.2/162 = 17 kg
Step 1: For Section B-B
Determine the number of main and distribution bars required.
Total no. of bars = Length of slab/spacing + 1
Total No. of bars = Lx/150 +1 = 5000/150 +1 = 35
No. of main bars = 18
No. of distribution bars = 17
Step 2:
Calculate the cutting length of the main bars
Cutting length of main bar = Clear span of slab + (2 x development length) + (Inclined length) – (Bend length)Clear span of slab = 4000 mm
Development length= Ld = 40d
Inclined length 0.42d
1d is or every 45° bend
Where d = diameter of the bar.
Now Calculate D
D = Thickness of slab – Both clear side cover (top & bottom) – dia of bar
= 200 – (25+25) – 12 = 138 mm
Cutting length of main bar = Ly + (2 x Ld)+ (2 x 0.42d) – (1d x 4)
= 4000 + (2 x 40 x 12) + (2 x 0.42 x 138) – (1 x 12 x 4)
= 5028 mm = 5.02 m
For 18 bars total length = 5.02 x 18 = 90.36 m
Weight of main bars = d^2 x L/162 = 12^2 x 90.36/162 = 80 kg
Cutting Length Of Distribution Bar:
= Clear span of slab + 2 x development length = Ly + 2 x Ld
= 4000 + 2 x 40 x 8 = 4640 mm = 4.64 m
Total length of distribution bar = 4.64 x 17 = 78.88
Weight of distribution bars = d^2 x L/162
= 8^2 x 78.88/162 = 31 kg
Step 3:
Calculate no. of extra bar:
No. of extra bars = {(Lx/4)/spacing +1} x 2 = {(5000/4)150 +1} x 2 = 9Length of extra bar = Ly – (2 x Ly/4) + (2 x 100) {Here 2 x 100 is for both side lapping of 100 mm for extra bar}
= 4000 – (2 x 4000/4) + 200
= 2200 mm = 2.2 m
Total length of extra bar = 2.2 x 9 = 19.8 m
Weight of extra bar = d^L/162 = 8^2 x 19.8/162 = 8 kg