Long years back when there was no concept of bar bending schedule, certain recommendation given by BN Dutta were used for estimating steel for different components of building.

Description | Percentage of steel |

Beam | 2% of total volume of concrete |

Column | 5% of total volume of concrete |

Slab | 1% of total volume of concrete |

Footing | 0.8% of total volume of concrete |

**Percentage of steel**– Bar Bending Schedule

But these recommendation are outdated and not used now a days as it does not give accurate values if more bars are used in a single member. Hence** ****BBS** gives better estimation of steel.

## What is bar bending schedule ?

Bar bending schedule or schedule of bars is comprehensive list that describes location ( in different component of structure), mark, type, size, length and number of reinforcement bars and bending details required for a given reinforced concrete item.

**Bar bending** schedule is also a way of organizing re-bars for each structural unit which generally describes particulars of bars, shape of bending with sketches, total length and weight of bars. These information help in preparing an estimate of quantities. Fig 1 shows standard form of reinforcement schedule.

## Bar Bending schedule formulas

### 1. Hooks and bends :

A hook commonly provided in stirrups, is an extra length left at the one corner of the stirrup so that the stirrup remains in shape. Unless otherwise specified, semicircular hook or a bend forming an anchorage to a bar is bent with an internal radius of Kd as shown n figure 2a and 2b.

The figure 2a and figure 2b shows the standard hook for 180° and 90° bend.

The hook length of a 180° and 90° bend is calculated as follows

### Length of hook for 180° bend:

Length of hook H = length 1-2 + length 2-3 – (k+1) d

Where d is the diameter of bar and the values of K are given in the table 1

= 4d + πr – (K+1) d

=4d + π (kd + d/2) – (K+1) d

Note: The value of K is taken as 2 and ‘d’ is the diameter of bar.

= 4d + π (2d + 0.5d) – 3d

= 8.85d ≈ **9d**

**Table 1 : Value of K**

Bar Diameter | Values of K | Mild steel confirming to IS : 432 1960 or IS : 1139 – 1959 | Medium tensile steel confirming to IS : 432 -1960 or IS : 1139-1959 | Cold worked steel confirming to IS : 1786 – 1961 (HYSD Bars ) |

≤ 25 mm | K | 2 | 3 | 4 |

≥ 25 mm | K | 3 | 4 | 6 |

### Length of hook for 90° bend:

Length of hook H = length 1-2 + length 2-3 – (K+1) d

Where d is the diameter of bar and the values of K are given in the table 1

= 4d + πr/2 – (K+1) d

= 4d + (π*(Kd + d/2))/2 – (K + 1) d

Note: The value of K is taken as 2

= 4d + (π(2d+0.5d))/2 – 3d

= 4d + (π*2.5d/2) – 3d

=4.92d ≈ **5d**

**Table 2 : Hook and bend allowances**

Value of K | Hook allowance | |

2 | 9d | Rounded of to nearest 5 mm but not less than 75 mm |

3 | 11d | |

4 | 13d | |

6 | 17d |

**Hook**Allowance

Value of K | Bend allowance | |

2 | 5d | Rounded of to nearest 5 mm but not less than 75 mm |

3 | 5.5d | |

4 | 6d | |

6 | 7d |

### 2. Bend length in cranked bars :

**Why cranking of bars important ?**

Fig 3** **shows how crank bars are arranged in the slab. The slab which is supported at two ends exerts positive (sagging) moments in the middle of slab and negative (hogging) moments at the ends. When the member sags, the bottom fibers experience tension and top fibers experience compression.

This is opposite when the member undergoes hogging. To resist this tension due to hogging, the reinforcement is provided at the top face at beam supports. Hence the bars are cranked at the ends so that the reinforcement is provided at the bottom face in the middle portion of slab and is provided at the top face in the edges of slab. In some cases, separate straight bars are used instead of cranked bars (Fig 4 ). In such cases, the bottom rods get curtailed at a distance of one third of the span of the slab.

**Crank length calculation :**

Fig 5 shows how cranking of bars are done at the slab edges. To find total length of bar, an extra length la (Fig 6 ) is calculated as follows,

Extra length la = l1 – l2

From Fig 6,

Tan θ = D/l2

Sin θ = D/l1

Hence, l1 = D/sin θ and l2 = D/tan θ

Therefore, la = (D/sin θ) – (D/tan θ)

For different cranking angles θ, different values of la are obtained as shown in the following table.

Table 3 : Value of la for different values of **θ**

θ° | D/sin θ | D/tan θ | la = (D/sin θ) – (D/tan θ) |

30° | 2D | 1.732D | 0.27D |

45° | 1.414D | D | 0.42D |

60° | 1.1547D | 0.577D | 0.58D |

90° | D | 0 | D |

135° | 1.414D | -D | 2.42D |

Additional length la is added while calculating total length of bar.

Now, total length of bar for angle of 45° is given by (Fig 5),

L + 0.42D *2 + (L/4) + (L/4)

Note :

(I) An extra bar of length (L/4) is given on both sides to keep crank bars in position.

(ii) D =depth of slab – top cover – bottom cover

### 3. Bend length calculation for stirrups :

By referring fig 7 total length of stirrup is calculated as follows

Extra length of 135° hook = length 1-2 + length 2-3

** = **4d + πr

= 4d + (π*(Kd + d/2))

Note : * The value of K is taken as 2

* d is the diameter of bar

= 4d + (π(2d+0.5d))

= 11.85d ≈ **12d**

Since in fig 7a there are two 135° hooks,

Extra length of 135° hook = 12d * 2= 24d

b = B – 2*cover

a = A – 2*cover

Total length of stirrups = 2 (b +a ) + 24d

Sample example for preparation of bar bending schedule

**RCC simply supported beam of size (400*500) is reinforced with 4 No of 20mm bars. Main bars are provided in one row and bent up bars are provided on second. Two anchor bars of 6 mm diameter and 6 mm dia stirrups of 150 mm c/c. span of beam is 4.6 m and bearing is 30 cm. Calculate total quantity of mild steel reinforcement and prepare bar bending schedule.**

Given,

Width B = 400mm

Overall Depth D = 500mm

Span of beam = 4600 mm

Edge bearing = 300 mm

Top anchor bar = 2 # 12

Main straight bar = 2 # 20

Bottom bent up bars = 2 # 20

Stirrups = 6 Ф @ 150 c/c

Assume clear cover = 25 mm

Soln :

Total length = 4600 + 300 + 300 = 5200 mm

**Length of main straight bars**

Length of straight bars = [total length – (2x clear cover) + (2 x 9d) )

Where d is the diameter of bar

Length of straight bars = [5200 – (2 x 25) + (9 x 2 x 20) ] = 5510 mm

**Length of bent up bars**

Length of bent up bars = [ total length – (2 x clear cover) + (2 x 9d) + (2 x 0.42D) ]

Where D = overall depth – top cover – bottom cover

= [ 5200 – (2 x 25 ) + (2 x 9 x 20 ) + (2 x 0.42 x 450) ]

Length of bent up bars = 5888 mm

**Length of anchor bars**

Length of anchor bars = [ total length – (2 x clear cover ) + ( 2 x 9d )]

= [ 5200 – ( 2 x 25 ) + ( 2 x 9 x 12 ) ]

= 5366 mm

**Length of Stirrups :**

Referring fig 7a

a = A – (2 X clear cover )

a = 500 – ( 2 X 25)

a = 450 mm

b = B – ( 2 x clear cover )

b = 400 – ( 2 x 25 )

b = 350 mm

Length of stirrups = 2 ( a + b ) + 2(12d)

Where d is the diameter of stirrup

Length of stirrup = 2 ( 450 + 350 ) + 2 ( 12 x 6 )

= 1744 mm

Number of stirrups = [(TL – 2(Clear cover))/spacing ] + 1

= [ ( 5200 – 2 ( 25 ))/ 150 ] + 1

= 35.33 ≈ 36 Nos

Total quantity of steel required = 80 kg (Refer Xl sheet for the bar bending schedule calculation )

## Uses of bar bending schedule

- BBS gives an idea of how much reinforcement is required for a particular project
- BBS helps in estimating the cost of reinforcement
- As the cutting and bending is done at the factory and transported to site, Wastage of reinforcement ( 5 to 10 % ) at the site can be reduced and it leads to faster execution of work at the site.
- BBS improves quality since the BBS of reinforcement are prepared according to the standard codes.