A Vision To Clear Solutions.
First ,find number of rods required for main reinforcement and distribution
Step 2
D = Slab thickness – 2 side clear cover – dia of bar = 150 – 50 -12 = 88 mm
Step 1
Inclined Length = 0.42 D
NOTE
SLAB REINFORCEMENT CALCULATION
One way Slab Reinforcement DetailingThe Biggest misconception is considering beam outer to outer as slab span.Please refer the below diagram.
Let's take an example of the below one way slab diagram.
Given
- Main Bars are 12mm in diameter @150mm centre to centre spacing
- Distribution bars are 8mm in diameter @150mm centre to centre spacing.(Main Bar & Distribution bar difference)
- Top & Bottom clear cover is 25mm
- Consider Development length as 40d
- Thickness of slab -150mm
One way slab Bar bending Schedule Calculation
Step-1
First ,find number of rods required for main reinforcement and distribution
Number of Bars Formula=(Length of slab/spacing)+1
Number of Main Bars=(Ly/spacing)+1=(5000/150)+1=34 nos
Number of Distribution Bars=(Lx/spacing)+1=(2000/150)+1=14 nos
Step 2
Find cutting length of main bars and distribution bars
Cutting Length of Main Bar,
= Clear Span of Slab (Lx) + (2 X Developement Length) +(1xInclined length) – (45°bend x 2)
Crank Length = 0.42 D, We have already discussed this on Cutting Length of Main Bar post
= 2000 + (2 x 40 x 12) + (1 x 0.42 x D) – (1d x 2)
= 2000 + 960 + 0.42D – (1x12x2) = 2960 +0.42D – 24
D = Slab thickness – 2 side clear cover – dia of bar = 150 – 50 -12 = 88 mm
Length of Main Bar = 2960+(0.42 x 88) – 24 = 2973 mm or 2.97 m
Step 3
Find length of distribution bar
= Clear Span (Ly) + (2 x Development Length (Ld))
= 5000 + (2 x 40 x 8) = 5640 mm or 5.64 m
Step 4
Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,
Please refer the drawing section A-A
Number of top bars = (Lx/4) / spacing + 1 = (2000/4) / 150 +1 = 4 Nos x 2 sides = 8Nos
Length of Top Bar (L) = same as distribution bars = 5.64 m
Bar Bending Schedule for One Way Slab
S.no
|
Description
|
No.of Bars
|
Unit Qty
|
Total Qty
|
Diameter of Bar
|
Cutting Length
|
1
|
Main Bar
|
27
|
1
|
27
|
12
|
3.53m
|
2
|
Distribution Bar
|
21
|
1
|
21
|
8
|
4.34m
|
3
|
Top Bar (Ly)
|
5
|
1
|
5
|
8
|
4.96m
|
4
|
Top Bar (Lx)
|
6
|
1
|
6
|
8
|
3.96m
|
Two way slab Reinforcement Detailing
Now Calculate Bar Bending Schedule for Two Way Slab.
Let’s take an example the below two-way slab diagram
- Main bars are 12 mm in diameter @ 150 mm centre to centre spacing
- Distribution bars are 8 mm in diameter @ 150 mm centre to centre spacing.
- Top and Bottom Clear Cover is 25 mm
- Development length – 40 d
- Thickness of Slab – 150 mm
One Way Slab Bar Bending Schedule Calculation
Step 1
- First, find number of rods required for main reinforcement and distribution
- Number of Required Bars Formula = (Length of slab / spacing) + 1
- Number of Main Bars = Ly / spacing + 1 = (4000/150) + 1 = 27 nos
- Number of Distribution Bars = Lx / spacing + 1 = 3000/150 + 1 = 21 nos
Step 2
Find cutting length of main bars and distribution bars
Cutting Length of Main Bar,
= Clear Span of Slab (Ly) + (1 X Developement Length) +(1 x inclined length) – (45°bend x 2)
Inclined Length = 0.42 D
= 3000 + (1 x 40 x 12) + (1 x 0.42 x D) – (1d x 2)
= 3000 + 480 + 0.42D – (1x12x2) = 3480 + 0.42 D -24
D = Slab thickness – 2 side clear cover – dia of bar = 150 – 50 -12 = 88 mm
Length of Main Bar = 3480+(0.42 x 88)-24 = 3492.9 mm or 3.49 m
Step 3
Find cutting length of distribution bar
= Clear Span of Slab (Ly) + (1 X Developement Length) +(1 x inclined length) – (45°bend x 2)
Inclined Length = 0.42 D
= 4000 + (1 x 40 x 8) + (1 x 0.42 x D) – (1d x 2)
= 4000 + 320 + (0.42×88) – (1x8x2) = 4340.96mm or 4.34 m
Step 4
Find Top Bar (Extra) ; Top Bars are provided at the top of critical length (L/4) area,
Please refer the drawing section A-A
Number of Top bars on Ly side = (Lx/5) / spacing + 1 = (3000/5) / 150 +1 = 5 Nos
Length of top bar on Ly side = Clear Span of Slab (Ly) + (2 X Development Length)
= 4000+(2*40*12) = 4000+960 = 4960 mm or 4.96 m
Number of Top bars on Lx side = (Ly/5) / spacing + 1 = (4000/5) / 150 + 1 = 6 Nos
Length of top bar on Ly side = Clear Span of Slab (Ly) + (2 X Development Length)
= 3000+(2*40*12) = 3000+960 = 3960 mm or 3.96 m
Bar Bending Schedule for Two Way Slab
S.no
|
Description
|
No.of
Bars
|
Unit
Qty
|
Total
Qty
|
Diameter
of Bar
|
Cutting
Length
|
1
|
Main
Bar
|
27
|
1
|
27
|
12
|
3.53m
|
2
|
Distribution
Bar
|
21
|
1
|
21
|
8
|
4.34m
|
3
|
Top
Bar (Ly)
|
5
|
1
|
5
|
8
|
4.96m
|
4
|
Top
Bar (Lx)
|
6
|
1
|
6
|
8
|
3.96m
|