Part- 2 👈 Back
Let us calculate the cutting length & weight of rebars in the RCC continuous beam as shown in the below-given drawing.
Given data:
Beam width = 300mm.
Beam depth = 450mm.
Column width = 300mm.
Clear span L = 4000mm. = 4m.
Top bar= 16mm ∅ - 2nos.
Top bent-up bar = 16mm ∅ - 2nos.
Bottom reinforcement = 25mm ∅ - 2nos.
Stirrup = 8mm∅ @ 150c/c
Clear cover = 30mm.
Development length Ld = 45d
Calculation:
A. Cutting length:
1. Bottom bar:
Cutting length of the bottom bar
= [clear span + (2nos. × development length) - ( 2nos. × 90° bend)]
= [4000mm. + (2nos. × 45d) - ( 2nos. × 2d)]
= [4000mm. + (2nos. × 45 × 25mm.) - ( 2nos. × 2× 25mm.)]
= [ 4000mm. + 2250mm - 100mm.]
= 6150mm.
= 6.15m.
2. Top bar:
Cutting length of top bar
= [clear span + (2nos. × development length) - ( 2nos. × 90° bend)]
= [4000mm. + (2nos. × 45d) - ( 2nos. × 2d)]
= [4000mm. + (2nos. × 45 × 16mm.) - ( 2nos. × 2× 16mm.)]
= [ 4000mm. + 1440mm - 64mm.]
= 5376mm.
= 5.38m.
3. Bent-up bar or crank bar
Cutting length of the bent-up (crank) bar
= [clear span + ( 2nos. × development length ) + (2 nos.× extra crank length) - {(2nos.× 90° bend) + (4nos.× 45° bend)}]
= [4000mm + ( 2nos. × 45d ) + ( 2nos. × 0.42D ) - {(2 nos.× 2d ) + (4nos.× 1d )}]
Note:
1. As the crank bar is bent at 45°, the resulting slope length will be 0.42D extra as it represents the hypotenuse of a triangle.
2. There are 2nos. of 90° bend & 4 nos of 45° bend as shown in the below drawing.
Cutting length of the crank bar
= [4000mm + ( 2 × 45 × 16mm.) + ( 2nos. × 0.42D ) - {(2nos.× 2 × 16mm.) + (4nos.× 1 × 16mm.)}]
Here, D
= [beam depth + bar diameter - {( top & bottom clear cover) + (top & bottom stirrup dia.)}]
= [ 450mm. + 16mm. - {(2nos × 30 mm.) + ( 2nos. × 8mm.)}]
= [ 450mm. + 16mm. - 76mm.]
= 390mm.
Cutting length of the bent-up (crank) bar
= [4000mm + (2 × 45 × 16mm.) + (2nos. × 0.42 × 390mm.) - {(2nos.× 2 × 16mm.) + (4nos.×16mm)}]
= [4000mm + (1440mm.) + (327.6mm.) - {64mm. + 64mm}]
= [ 5767.60 mm. - 128mm.]
= 5639.6mm.
= 5.64m.
Important note: If you are curious to know, why the extra crank length is taken as 0.42D in the above formula, you can click here.
4. Stirrup:
The cutting length of the stirrup bar
Where,
a = {beam width - (2 × cover) - (2 × 1/2 × stirrup dia.)}
= {300mm. - (2 × 30mm.) - ( 2 × 1/2 × 8mm.)}
= {300mm. - (60mm.) - (8mm.)}
= 232mm.
b = {beam depth - (2 × cover) - (2 × 1/2 × stirrup dia.)}
= {450mm. - (2 × 30mm.) - ( 2 × 1/2 × 8mm.)}
= {450mm. - (60mm.) - (8mm.)}
= 382mm.
The cutting length of the stirrup bar
= [{2 nos. × (232mm. + 382mm.)} + (2nos ×10d ) - (3 nos. × 2d ) - (2 nos. × 3d) ]
To know, why we take bend deductions, Go through the article 👇
👀. What are bend deductions for different angles in reinforcement bars?
Note: Here, hook length is taken as 10d.
The cutting length of the stirrup
= [{2 nos. × 614mm.} + (2nos. × 10 × 8mm) - ( 3 nos. × 2 × 8mm ) - ( 2 nos. × 3 × 8 mm.)]
= [1228mm. + 160 mm - 48mm - 48mm.]
= [1388mm. - 96 mm.]
= 1292 mm i.e. 1.292 m.
B. No. of stirrups
No of stirrups in beam
= [{ clear span of the beam ÷ stirrup spacing } + 1]
= [{ 4000mm. ÷ 150mm. } + 1]
= [26.66 +1]
= 27.66 nos.
By rounding off = 28 nos.
Now, let us prepare a BBS table for the simply supported beam.
|
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. |
Bottom bar |
25 |
2 |
6.15 |
12.30 |
3.853 |
47.39 |
|
2. |
Top bar |
16 |
2 |
5.38 |
10.76 |
1.578 |
16.98 |
|
3. |
Bent-up bar |
16 |
2 |
5.64 |
11.28 |
1.578 |
17.80 |
|
4. |
Stirrup bar |
8 |
28 |
1.292 |
36.176 |
0.395 |
14.29 |
|
5. |
Total weight of bars = |
96.46 |
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|
6. |
Add 3% wastage = |
2.89 |
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|
7. |
The grand total wt. of rebar's = |
99.35 |
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Thank you for going through these calculation steps❤. Have a good day 😄.
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