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.

Plan and Cross-Sectional View of an Isolated Footing

Fig.1. Plan and Cross-Sectional View of an Isolated Footing

Read More: Isolated Footing Design Example

Details Obtained from Structural Drawing

Following details are obtained from the drawings and specifications:

  1. Length of Footing = X
  2. The breadth of Footing = Y
  3. The height of the footing (Thickness) = h
  4. The diameter of the Main reinforcement bars = dm
  5. The diameter of Distribution Reinforcement Bars = dd
  6. The spacing of reinforcement bars = s
  7. Cover for reinforcement = c

Formulas for Calculation

From the figure-1, the following parameters are determined:

  1. Number of Main Reinforcement Bars ( X-Bars)
  2. Number of Distribution Reinforcement Bars ( Y-Bars)
  3. Cutting Length of Main Reinforcement Bars (Cm)
  4. Cutting Length of Distribution Reinforcement Bars (Cd)
  5. Steel Quantity Required

From the given details, the following values are calculated:

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[2dm]         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[2dm]}     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[2dm]        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[2dm] }      Eq.6

5. Estimation of Steel Quantity

The steel quantity is determined by the formula:

W = D2L/162                        Eq.7

Note: Equation 7 is obtained by the solving the formula:

Weight of Steel(W) = Volume of the Material(V) x Density of the Material

Where, V= Area of steel x length of the steel; Density of Steel = 7850 kg/m3

Example – Footing Bar Bending Schedule

Estimation of Steel Quantity for Footing

Given below is an example of an isolated footing with specifications and dimensions.

Plan and Cross-Sectional View of an Isolated Footing

Fig.1. Example of Plan and Cross-Sectional View of an Isolated Footing

Details Obtained from the Drawing

The following details are obtained from the figure-2:

  1. Length of Footing = X = 2m
  2. Breadth of Footing = Y = 1.6m
  3. Height of the footing (Thickness) = h =0.3m
  4. Diameter of the Main reinforcement bars = dm= 12mm
  5. Diameter of Distribution Reinforcement Bars = dd =12mm
  6. Spacing of Main reinforcement bars = sm = 150mm c/c
  7. Spacing of Distribution Reinforcement bars = sd = 150mm c/c
  8. Cover for reinforcement = c = 50mm

From the given details, the following values are calculated:

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)

From Eq.3

Cm1     = [X-2C] +2[h-2C]-2[2dm]

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[2dm]}

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[2dm]

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 = D2L/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

Note: W = D2/162 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. NoBar TypeDia:

of Bar

(mm)

Shape of BarNo: of BarLength of Bar(m)Total Length of Bar(m)Weight of 1m length Bar (Kg)Total Weight of Bar Kg
1Main Reinforcement (X-Bars)12112.25224.7720.88922.02
2Distribution Reinforcement

(Y-Bars)

12141.85223.050.88923.05
TOTAL45.07