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Bar bending schedule (BBS) of an concrete footing provides the reinforcement details and the total steel quantity required for the footing construction. The BBS and quantity of steel reinforcement required for a simply isolated footing are calculated and explained by means of a workout example.**Calculation of Steel Reinforcement For Isolated Footing**

The figure-1 below shows the cross-sectional plan and section of an isolated footing. The details of reinforcement used in the footing are calculated from the detailed drawing of the structure prepared by the designer. The structural drawing provides the reinforcement location and their specifications.
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**: Isolated Footing Design Example**

### Details Obtained from Structural Drawing

Following details are obtained from the drawings and specifications:- Length of Footing = X
- The breadth of Footing = Y
- The height of the footing (Thickness) = h
- The diameter of the Main reinforcement bars = d
_{m} - The diameter of Distribution Reinforcement Bars = d
_{d} - The spacing of reinforcement bars = s
- Cover for reinforcement = c

**Formulas for Calculation**

From the figure-1, the following parameters are determined:
- Number of Main Reinforcement Bars ( X-Bars)
- Number of Distribution Reinforcement Bars ( Y-Bars)
- Cutting Length of Main Reinforcement Bars (Cm)
- Cutting Length of Distribution Reinforcement Bars (Cd)
- Steel Quantity Required

**1. Calculation of Number of Main Reinforcement (X- Bars)**From the figure-1(c), the x-bars are distributed along the y-direction. Hence the number of bars is:

**Nm = (Y / Spacing of Main Reinforcement) + 1 Eq.1**

**2. Calculation of Number of Distribution Reinforcement (Y-Bars)**From Figure-1 (c), the y-bars are distributed along X-axis. Hence the number of bars

**Nd = (X/Spacing of Distribution Reinforcement) +1 Eq.2**

**3.**

**Cutting Length of Main Reinforcement (X-Bars)**Cm1 = [Length of Footing – 2(cover)] +2[ Thickness of footing -2(cover)] - 2[Bend] From the figure,

**Cm1 = [X-2C] +2[h-2C]-2[2d _{m}] Eq.3**

**Total Cutting Length of Main Reinforcement (Cm)**Cm = Number of Main Reinforcement x Cutting length of Single Main Bar Cm = Nm x Cm1

**Cm = Nm {[X-2C] +2[h-2C]-2[2d _{m}]} Eq.4**

**4. Cutting Length of Distribution Reinforcement (Y-Bars)**Cd= [Breadth of Footing – 2(cover)] +2[ Thickness of footing -2(cover)] - 2[Bend] From the figure,

**Cd1 = [Y-2C] +2[h-2C]-2[2d _{m}] Eq.5**

**Total Cutting Length of Main Reinforcement**Cd = Number of Distribution Reinforcement x Cutting length of Single Distribution Bar Cd = Nd x Cd1

**Cd = Nd {[X-2C] +2[h-2C]-2[2d _{m}] } Eq.6**

**5. Estimation of Steel Quantity**The steel quantity is determined by the formula:

**W = D ^{2}L/162 Eq.7**

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**Example – ****Footing ****Bar Bending Schedule**

**Estimation of Steel Quantity for Footing**

Given below is an example of an isolated footing with specifications and dimensions.
### Details Obtained from the Drawing

The following details are obtained from the figure-2:- Length of Footing = X = 2m
- Breadth of Footing = Y = 1.6m
- Height of the footing (Thickness) = h =0.3m
- Diameter of the Main reinforcement bars = d
_{m}= 12mm - Diameter of Distribution Reinforcement Bars = d
_{d}=12mm - Spacing of Main reinforcement bars = s
_{m}= 150mm c/c - Spacing of Distribution Reinforcement bars = s
_{d}= 150mm c/c - Cover for reinforcement = c = 50mm

**1. Calculation of Number of Main Reinforcement (X- Bars)**From Eq.1

Nm = (Y / Spacing of Main Reinforcement) + 1

Nm = (1.6/.15)+1

**Nm= 12nos**

**2. Calculation of Number of Distribution Reinforcement (Y-Bars)**From Eq.2

Nd = (X/Spacing of Distribution Reinforcement) +1

Nd = (2/.15)+1

**Nd= 14nos**

**3. Cutting Length of Main Reinforcement (X-Bars)**

Cm1 = [X-2C] +2[h-2C]-2[2d_{m}]

Cm1 = [2-(2x.05)]+2[.3-(2 x .05)]-2 (2 x .012)

Cm1 = 1.9+.4-.048

Cm1=2.252m

From Eq.4,**Total Cutting Length of Main Reinforcement**

Cm= Nm {[X-2C] +2[h-2C]-2[2d_{m}]}

Cm = 11 x 2.252

**Cm = 24.772m**

**4.**

**Cutting Length of Distribution Reinforcement (Y-Bars)**From Eq.5, Cd1 = [Y-2C] +2[h-2C]-2[2d

_{m}] Cd1 = [1.6-(2x.05)]+2[.3-(2 x .05)]-2 (2 x .012) Cd1 = 1.5+.4-.048 Cd1 = 1.852m From Eq.6,

**Total Cutting Length of Main Reinforcement**

Cd = Nd x Cd1

Cd = 14 x 1.852

**Cd = 25.928m**

**5. Estimation of Steel Quantity**From Eq.7, W = D

^{2}L/162;

**Total Steel Quantity for Main Reinforcement,**

Wm = (12 x 12 x 24.772)/162

**Wm= 22kg**

**Total Steel Quantity for Distribution Reinforcement,**

Wd=(12x12x25.928)/162

**Wd =23.05kg**

^{2}/162 gives the weight of single bar

**Footing ****Bar Bending Schedule and Quantity of Steel**

The number of reinforcement bars, their length and the quantity of steel for the given Isolated footing is estimated and consolidated in a tabulated form ( Table-1).
**Table.1: Bar Bending Schedule For Isolated Footing**

SL. No | Bar Type | Dia: of Bar (mm) | Shape of Bar | No: of Bar | Length of Bar(m) | Total Length of Bar(m) | Weight of 1m length Bar (Kg) | Total Weight of Bar Kg |

1 | Main Reinforcement (X-Bars) | 12 | 11 | 2.252 | 24.772 | 0.889 | 22.02 | |

2 | Distribution Reinforcement (Y-Bars) | 12 | 14 | 1.852 | 23.05 | 0.889 | 23.05 | |

TOTAL | 45.07 |