Let us calculate the cutting length & prepare the BBS of the two-way slab as shown below.
Let us redraw the above drawing with dimensions for the calculation purpose as below.
Given data:
Dimension of the slab
Lx = 5m.= 5000mm.
Ly = 3m. = 3000mm.
Thickness = 125mm.
Main bar-1 = 10mm@ 150 c/c
Main bar-2 (or say distribution bar) = 10 mm @ 150 c/c
Top extra bar-1= 8mm @ 150 c/c
Top extra bar-2= 8mm @ 150 c/c
Clear cover of slab = 25mm.
Development length Ld = 45d.
Calculation steps:
1. The no. of bars:
a. Main bars:
The no. of main bars-1
= [Lx ÷ bar spacing ] +1
= [ 5000 ÷ 150] +1
= 33.33 +1
= 43.33 nos.
By rounding off = 44 nos.
b. Main bars-2:
The no. of main bars-2 ( or say distribution bar)
= [Ly ÷ bar spacing ] +1
= [ 3000 ÷ 150] +1
= 20 +1
= 21 nos.
c. Top extra bars-1:
The no. of top extra bars-1 on one side
= [{(Ly ÷ 4)÷ bar spacing} +1]
= [{ (3000 ÷ 4) ÷ 150} +1]
= [{ 750 ÷ 150} +1]
= [{ 5} +1]
= 6 nos.
The total no. of top extra bars on both sides
= [2 sides × 6nos.]
= 12 nos.
Note: The top extra bars are provided at the Ly/4 span of the main bar on both sides.
c. Top extra bars-2:
The no. of top extra bars-2 on one side
= [{(Lx ÷ 4)÷ bar spacing} +1]
= [{ (5000 ÷ 4) ÷ 150} +1]
= [{ 1250 ÷ 150} +1]
= [{ 8.33} +1]
= 9.33 nos.
By rounding off = 10 nos.
The total no. of top extra bars on both sides
= [2 sides × 10nos.]
= 20 nos.
Note: The top extra bars are provided at the Lx/4 span of the main bar on both sides.
2. Cutting length:
a. Main bar-1:
Cutting length of the main bar-1
= [clear span +(2nos. × development length) + (extra crank length) - (2nos. × 45° bend)]
= [ 3000mm. + ( 2nos. × Ld) + ( 0.42D ) - ( 2nos. × 1d)]
To know why extra crank length is taken as 0.42D, Go through the article 👇
👀. What is 0.42D in a crank bar (bent-up bar) cutting length formula? - its derivation.
Here, D = [slab thickness - ( top & bottom clear cover) + bar diameter ]
= [125mm - (2nos × 25 mm.) +10mm ]
= [125mm - 60 mm]
D = 65 mm.
Cutting length of the main bar-1
= [ 3000mm. + ( 2nos. × 45 × 10mm. ) + ( 0.42× 65mm. ) - ( 2nos. × 1× 10mm)]
= [ 3000mm. + 900mm. + 27.3mm. - 20mm. ]
= 3907.30 mm. = 3.91m.
To know, why the bend deduction for 45° bend is taken as 1d, Go through the article 👇
👀. What are bend deductions for different angles in reinforcement bars?
b. Main bar-2:
Cutting length of the main bar-2
= [clear span +(2nos. × development length) + (extra crank length) - (2nos. × 45° bend)]
= [ 5000mm. + ( 2nos. × Ld) + ( 0.42D ) - ( 2nos. × 1d)]
= [ 5000mm. + ( 2nos. × 45 × 10mm. ) + ( 0.42× 65mm. ) - ( 2nos. × 1× 10mm)]
= [ 5000mm. + 900mm. + 27.3mm. - 20mm. ]
= 5907.30 mm. = 5.91m.
c. Top extra bar-1:
Cutting length of the top extra bar-1
= [clear span Lx + (2nos. × development length)]
= [5000mm. + ( 2nos. × 45d )]
= [ 5000mm. + ( 2nos. × 45 × 8mm. )]
= [ 5000mm. + 720mm.]
= 5720 mm. = 5.72m.
d. Top extra bar-2:
Cutting length of the top extra bar-2
= [clear span Ly + (2nos. × development length)]
= [3000mm. + ( 2nos. × 45d )]
= [ 3000mm. + ( 2nos. × 45 × 8mm. )]
= [ 3000mm. + 720mm.]
= 3720 mm. = 3.72m.
Let us prepare a BBS table to calculate the weight of the rebars.
|
Sl. No. |
Type of Bar |
Dia. in mm. |
Nos. |
Length in m. |
Total length in m. |
Weight Kg/m. |
Total bar wt. in kg. |
|
1. |
Main bar-1. |
10 |
44 |
3.91 |
172.04 |
0.617 |
106.15 |
|
2. |
Main bar-2 |
10 |
21 |
5.91 |
124.11 |
0.617 |
76.58 |
|
3. |
Top extra bar-1. |
8 |
12 |
5.72 |
68.64 |
0.395 |
27.11 |
|
4. |
Top extra bar-2 |
8 |
20 |
3.72 |
74.40 |
0.395 |
29.39 |
|
4. |
Total weight of bars = |
239.23 |
|||||
|
5. |
Add 3% wastage = |
7.18 |
|||||
|
6. |
The grand total wt. of rebar's = |
246.41 |
|||||
To know, how to calculate the weight of rebars per RMT., Go through the article 👇
❤. How to calculate the weight of reinforcement bars?
Thank you for going through these calculation steps❤. Have a good day 😄.
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