Important Point

## General guidelines to be followed in preparing BBS:

- The bars should be grouped together for each structural unit, e.g., beam, etc.
- In a building structure, the bars should be listed floor by floor
- For cutting and bending purposes, schedules should be provided as separate A4 sheets and not as part of the detailed reinforcement drawings.
- The form of bar and fabric schedule and the shapes of the bar used should be in accordance with BS 8666.
- It is preferable that bars should be listed in the schedule in numerical order.
- It is essential that the bar mark reference on the label attached to a bundle of bars refers uniquely to a particular group or set of bars of defined length, size, shape, and type used on the job.
- This is imperative as a bar mark reference can then point to a class of bar characteristics. Also, this helps steel fixers and laborers keep track of the type and number of bars needed to complete a certain work.

Also, read: House Construction Cost Calculator Excel Sheet

**Bar Bending Schedule Required Ment Data**

Estimation of steel quantity is an essential skill of any civil engineer. Every civil engineer must know the method of calculation of steel quantity from drawing.

Data requires for estimation of steel quantity:

**1. Plan, Elevation & Section **

**2. Structural Detail (Slab & Beam)**

**3. All dimensions must be clear and co-related.**

Also, read :How to Find House Construction Cost

## How to prepare a Slab BBS?

We calculated her two parts of caution of **BBS as below**

**Slab Beam.****Slab.**

**Slab Beam Bar Bending Schedule **

**Slab Beam Location**

Here many beam shows. So we calculate TB1 Beam.

**TB1 Beam.**

**Numbers of bars for Stirrups **

Suppose the spacing of stirrups is 200 c/c and the length along which they are placed is 14100 mm, we can find the number of bars by the formula below

[ (14.100-200) / 200)] + 1 =**70.5 Nos**

**Total Numer of Stirrups = 71 Nos**

#### Stirrups Cutting Length

Cutting length beam stirrup

**90 degree hook:**

Length of stirrup = 2 (A + B) + 20 x diameter

**135 degree hook:**

**Length of stirrup = 2 (A + B) + 24 x diameter**

**Length of stirrup = 2 ( (Length -2 Cover) + (Breadth-2 Cover) + 24 x diameter**

diameter of Stirrups.

Length of stirrup = 2 (0.150 + 0.550) + 24 x 8

**Length of stirrup = 1.592 m**

**Total Weight of Stirrups**

**Total Weight Stirrups = (No of ring x Cutting Length ) x (Weight diameter of bars)**

**Total Weight Stirrups **= ( 71 x 1.592) x (0.395)** = 44.648 kg**

We must remember than steel is ductile in nature and is subject to elongation. Hence, the length of a bar is increased when bends or hooks are introduced. Hence, certain deductions are needed to offset this increase in length.

Also, read: Building Estimation Step by Step In Excel Sheet

**Cutting Length Bar:**

**Cutting Length = True Length of a bar – Deductions**

For 45 degree

**Cutting length = Total length – 1 x Diameter of bar x No. of bends**

Cutting length = (Total Length) +( 2 x L length) + (Lap Length (45 x d) ) – (2 x Bend) – (Cover)

**Bottom Bar **

**2 nos bar 16 mm diameter**

Cutting length = (11.400+2.400+0.200) + ( 2x 0.550) + ( 45 x 16) – ( 2 x 0.016 ) -(2 x 0.025)

Cutting length = (14.100) + (1.100) + (0.720) – (0.032) – (0.050)

Cutting length = 15.838 m

**Total weight of bottom bar = No of Bars x Cutting length x Weight of bars**

**Total weight of bottom bar** = 2 x 15.838 x 1.580 **= 50.048 kg**

**Top Bar**

**2 nos bar 20 mm diameter**

Cutting length = (11.400+2.400+0.200) + ( 2x 0.550) + ( 45 x 20) – ( 2 x 0.020 ) -(2 x 0.025)

Cutting length = (14.100) + (1.100) + (0.900) – (0.040) – (0.050)

Cutting length = 16.010 m

**Total weight of top bar = No of Bars x Cutting length x Weight of bars**

**Total weight of top bar** = 2 x 16.010 x 2.469 **= 79.057 kg**

**Total Weight of Beam = Total Weight Stirrups + Total weight of bottom bar + Total weight of top bar**

**Total Weight of Beam = ****44.648 kg + 50.048 kg + 79.057 kg = 173.753 kg **

**Slab Bar Bending Schedule Excel Sheet**

**Cutting Length Slab Bar**

**Cutting Length of Bar = (Total length) + ( Bentup Bar Length (0.42 x d)) +Lap**

Area ( -1 to 3)/(-A to +B)

**For X Direction**

**8 mm Diameter 200 mm c/c**

**No of Bars** = (7 /0.2) + 1 = 36.0 Nos. **= 36 Nos**

**Cutting Bar** = (0.100+2.400+4.300+1.000+2.800+1.000+0.100) +( 0.42 * (0.125-0.025-0.025)) + 0

Cutting Bar = (11.700) + (0.032) **= 11.732 m**

**Total weight of X direction slab bar = No of Bars x Cutting length x Weight of bars**

**Total weight of X direction slab bar** = 36 x 11.732 x 0.395 **= 166.829 kg**

**For Y Direction**

**8 mm Diameter 200 mm c/c**

**No of Bars** = (11.700/0.200) + 1 = 59.5 Nos. **= 60 Nos**

**Cutting Bar** = (0.100+1.800+3.200+1.800+0.100) +( 0.42 * (0.125-0.025-0.025)) + 0

Cutting Bar = (7.000) + (0.0315) **= 7.032 m**

**Total weight of Y direction slab bar = No of Bars x Cutting length x Weight of bars**

**Total weight of Y direction slab bar** = 60 x 7.032 x 0.395 **= 166.658 kg**

**Extra Bars **

**10 mm Diameter 200 mm c/c**

**No of Bars** = (11.700/0.200) + 1 = 59.5 Nos. **= 60 Nos**

TB2 area A/( – 1 to +3) = (0.500 + 0.200 + 0.900 ) = **1.600 m**

**Total weight of TB1 = No of Bars x Cutting length x Weight of Diameter bar**

**Total weight of TB1** = 60 x 1.600 x 0.616 **= 59.136 kg **

**No of Bars** = (7 /0.2) + 1 = 36.0 Nos. **= 36 Nos**

TB3 area 4 (-A to +B) = (0.700 + 0.200 + 1.250 ) = **2.150 m **

**Total weight of TB3** = 36 x 2.150 x 0.616 **= 47.678 kg **

**No of Bars** = (7 /0.2) + 1 = 36.0 Nos. **= 36 Nos**

TB5 area 1/2 (-A to +B) = (1.250 + 0.200 + 1.400 ) = **2.850 m**

**Total weight of TB5** = 36 x 2.850 x 0.616 **= 63.202 kg **

**Total weight of Slab** = **Total weight of X direction slab bar** + **Total weight of Y direction slab bar + Total weight of TB1 + Total weight of TB3 + Total weight of TB5**

**Total weight of Slab = 166.829 kg + 166.658 kg + 59.136 kg + 47.678 kg + 63.202 kg = 503.03 kg **

## How to Calculate Slab Steel Quantity from Drawing Excel Sheet

How to Calculate Slab Steel Quantity from Drawing Excel Sheet **Click Here**

### Bar Bending Schedule for Slab

S.NO. |
Description |
Shape Of Bar |
Grid |
No of Column |
Nos. of Bar |
Length |
Dia. Of Bar |
Dia Wise Length In Mtr. |
Total Weight In KG |
||||||||

8 mm |
10 mm |
12 mm |
16 mm |
20 mm |
25 mm |
Sub Total |
Cumulative |
||||||||||

0.395 |
0.616 |
0.888 |
1.579 |
2.466 |
3.854 |
||||||||||||

A | Slab Beam |
||||||||||||||||

1 | TB1 ( 200 X 600 ) | ||||||||||||||||

-A / (-1 to +3) | |||||||||||||||||

+B / (-1 to +3) | |||||||||||||||||

Top Bar | 2 | 2 | 15.200 | 20 | – | – | – | – | 60.80 | – | |||||||

Bottom Bar | 2 | 2 | 15.200 | 16 | – | – | – | 60.80 | – | – | |||||||

– | – | – | – | – | – | ||||||||||||

Lap | |||||||||||||||||

Top Bar | 2 | 2 | 1 | 20 | – | – | – | – | 4 | – | |||||||

Bottom Bar | 2 | 2 | 0.8 | 16 | – | – | – | 3.20 | – | – | |||||||

Ring | – | – | – | – | – | – | |||||||||||

200 C/C ( -1 to 3) | 2 | 55 | 1.560 | 8 | 171.60 | – | – | – | – | – | |||||||

150 C/C ( 3 to+ 3) | 2 | 16 | 1.560 | 8 | 49.92 | – | – | – | – | – | |||||||

Total Length as per dia | 221.52 | – | – | 64.00 | 64.80 | – | |||||||||||

Dia in K.g/m | 0.395 | 0.616 | 0.888 | 1.579 | 2.466 | 3.854 | |||||||||||

Total weight as per dia | 87.5 | 0 | 0 | 101.056 | 159.797 | 0 | 348.353 | ||||||||||

Total Qty in K.g | 348.353 kg |

## Steel Calculations

- Weight of bar in kg/meter should be calculated as (
**d**) where^{2}÷ 162**‘d’ is the diameter of bar in mm.** - ‘
**L**‘ for column main**steel**in footing should be considered as minimum**30cm/as specified.** - For bent up bar add
**0.42d**in a straight length of the bar where’d’ is the clear depth of**beam/slab**.

## Steel Quantity Calculation

**(Calculation of bars no’s)**: first, calculate the number of bars required (main and distribution both). Formula = (total length – clear cover)/center to center spacing + 1 main bar, = (5000 – (25+25))/100 + 1, = 4950 divided by 100 + 1, = 51 bars., distribution bar = (2000 – (25+25))/125 + 1, = 1950 divided by 125 + 1, = 17 bars.**(Cutting length) main bar**: formula = (l) + (2 x ld) + (1 x 0.42d) – (2 x 1d), where l = clear span of the slab, ld = development length which is 40 d (where d is diameter of bar), 0.42d = inclined length (bend length), 1d = 45° bends (d is diameter of bar) first calculate the length of “d“. D = (thickness) – 2 (clear cover at top, bottom) – diameter of the bar, = 150 – 2(25) -12, d = 88 mm ans, by putting values: cutting length = 2000 + (2 x 40 x 12) + (1 x 0.42 x 88) – (2 x 1 x 12), cutting length = 2000 + 960 + 36.96 – 24, =2972.96 mm ~ 2973 mm or 2.973 m.**(Cutting length) Distribution bar:**formula = clear span + (2 x development length (ld)), = 5000 + (2 x 40 x 8), = 5640 mm or 5.64 m**(Steel Quantity Calculation) Conclusion:**- Main bar nos = numbers 51, main bar length = length (51 x 2.973 m) = 151.623 m, main bar weight = weight (d^2/162) x length = 134.776 kg.
- Distribution bar nos = numbers 17, distribution bar length= length (17 x 5.64 m) = 95.88 m, distribution bar weight = weight (d^2/162) x length = 37.87 kg.

### How to Calculate Steel Weight?

**Weight of Steel = (l/1000) x (w/1000) x t x η**

**l** = length in mm

**w** = width in mm

**t** = thickness in mm

**η** = Specific material density ( e.g.: steel = 7.85 kg / dm³)

### Steel Quantity Calculation

(Cutting length) main bar: formula **= (l) + (2 x ld) + (1 x 0.42d) – (2 x 1d)**, where l = clear span of the slab, ld = development length which is 40 d (where d is diameter of bar), 0.42d = inclined length (bend length), 1d = 45° bends (d is diameter of bar) first calculate the length of “d“.

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Vishal Sutariya says

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1. Brick work

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Jesse says

How to quantify main bars and distribution bars using R.c detailing of slab thanks.