# Bar Bending Schedule for footings :-

Bar Bending schedule plays a vital role in the construction of High rise buildings.  It is very important to learn Bar Bending Schedule for finding out the quantities of Steel reinforcement required for every component of the building.

For Suppose, consider the case of high rise buildings, It requires tons of steel to complete 10+ floor building. It’s impossible to order all the steel required for whole construction at a time it creates a problem of space and also steel is prone to corrosion by the contact of water (rain). To avoid this, high rise building orders reinforcement (steel) as per requirement. Firstly they find the Estimation of Steel reinforcement in footings (steel quantities)  [Bar Bending Schedule for footings],  required for the construction of footings. After the completion of footings they go for next order and so on.

If you are new to Bar Bending Schedule, Refer the Basics of Bar Bending schedule

if you wanted to know different types of footings check here Different types of footings

## Quantity of Reinforcement (Steel) required for footings / Bar Bending Schedule for footings:-

The procedure to finding out the quantities of steel required for the footings, We are considering the below footing plan.

Observations from the above fig:

• F1, F4, F7 is Plain footing (1.0×1.0×0.8)
• F2 is Stepped Footing (0.9×0.9×1.35)
• F3,  F8  is Isolated footing (0.9×0.9×0.5)
• F5 is Combined Isolated footing (4.2×1.7×0.9)
• F6 is Shoe footing (0.6×0.6×0.4)

True dimensions and shapes of the footings are decided and designed by the structural engineer based on soil history, type of construction, the total expected load of the structure. All the dimensions of the above columns considered only for explanation purpose.

We use different types of mesh (reinforcement) in footings as per design. You can refer different types of Reinforcement used in footings from here.

## Steps involved in calculating the bar bending schedule of a footings:-

Remember, Steel required for construction is ordered in Kgs or Number of Bars. The standard size of each bar is 12m. The final output of BBS calculation is in Kgs or in Number of “12m” Bars.

To make it easier calculation is divided into two parts, X bar Calculations and Y bar calculations.

X Bars are Horizontal bars in X direction and Y Bars are vertical one projected in Y Direction.

1. Deduct the concrete cover to find the dimensions of bars.
2. Find the Length of single X Bars & Y Bars
3. Find the total length of X bars. & Y bars
4. Calculate the weight of steel required per 1m
5. Calculate the total number of 12m bars required
6. Find the total weight of steel required.

## Plain footing Bar Bending schedule:

For the calculation of the total quantity of steel required for the Plain footing, we are adopting these dimensions for bars.

• Dimensions of Footing  are 1.0 × 1.0 × 0.9 (Length × Breadth × Depth)
• Plain mesh is adopted for F1, F4, F7 footings
• Dia of X Bars is 16mm (Dia 16mm @ 100mm  C/C)
• Dia of Y Bars is 12mm (Dia 12mm @ 100mm C/C)
• which means Center to center spacing between X bars & Y bars is 100mm

Remember, Proper Concrete cover should be adopted for the reinforcement  in Footings to resist it from corrosion.

Concrete Cover deduction:

As per condition, concrete cover of  0.1m is deducted from all sides of  mesh. True dimensions post deducting is 0.8×0.8 (length and breadth)

Refer below image for more details:

 Length ofEach X bar = 0.8m Length ofEach Y bar = 0.8m No. of X bars [(Y Bar Length)/Spacing]+1= [0.8/0.1]+1= 9bars No. of Y bars [(X Bar Length)/Spacing]+1= [0.8/0.1]+1= 9bars Total Lengthof X bars = Length of each X bar ×No. of X Bars= 0.8 × 9 =7.2m Total Lengthof Y bars = Length of each Y bar ×No. of Y Bars= 0.8 × 9 =7.2m Total No. of ’12m’X bars = 7.2/12= 0.6bars Total No. of ’12m’Y bars = 7.2/12= 0.6bars Weight of steelrequired for1m of 16mm bar = D2/162= 162/162= 1.58kg/m Total weight of steelrequired for X bars = 1.58 × 7.2= 11.37Kgs Weight of steelrequired for1m of 12mm bar = D2/162= 122/162= 0.88kg/m Total weight of steelrequired for Y bars = 0.88 × 7.2= 6.33Kgs

Total Weight of Plain Mesh

= Weight of steel required for X bars + Weight of steel required for Y bars

= 11.37Kgs+6.33Kgs =17.70Kgs

## Isolated footings Bar Bending Schedule :-

For the calculation of the total quantity of steel required for the Isolated footing, we are adopting below dimensions for bars.

• Hook mesh is adopted for F3,F8 footings
• Dimensions of Footing  are 0.9×0.9×0.5 (Length × Breadth × Depth)
• Dia of X Bars is 16mm (Dia 16mm @ 100mm  C/C)
• Dia of Y Bars is 12mm (Dia 12mm @ 90mm C/C)
• which means Center to center spacing between X bars is 100mm & Y bars is 90mm

Remember Proper Concrete cover should be adopted for the reinforcement  in Footings to resist it from corrosion.

Concrete Cover deduction:

As per condition, concrete cover of  0.1m is deducted from all sides of  mesh. In hook mesh, hook is provided at the end of each bar. Each bar has two ends and therefore, hook length is included in the calculation of length of bar

Hook Length = 9d (d is the dia of bar)

Total Hooks for each bar = 2

True dimensions post deducting concrete cover is length = (0.7m+2x9d) & breadth = (0.7m+2x9d)

Refer below image for more details

 Length ofEach X bar = 0.7+2×9dd = 16mm = 0.016m= 0.7+2×9×0.016= 0.988m Length ofEach Y bar = 0.7+2×9dd = 12mm = 0.012m= 0.7+2×9×0.012= 0.916m No. of X bars [(Y Bar Length)/Spacing]+1= [0.7/0.1]+1= 8bars(Don’t include hooklength in calculating no. of bars) No. of Y bars [(X Bar Length)/Spacing]+1= [0.7/0.09]+1= 9bars Total Lengthof X bars = Length of each X bar ×No. of X Bars= 0.988 × 8 =7.9m Total Lengthof Y bars = Length of each Y bar ×No. of Y Bars= 0.916 × 9 =8.24m Total No. of ’12m’X bars = 7.9/12= 0.65bars Total No. of ’12m’Y bars = 8.24/12= 0.68bars Weight of steelrequired for1m of 16mm bar = D2/162= 162/162= 1.58kg/m Total weight of steelrequired for X bars = 1.58 × 7.9= 12.48Kgs Weight of steelrequired for1m of 12mm bar = D2/162= 122/162= 0.88kg/m Total weight of steelrequired for Y bars = 0.88 × 8.24= 7.25Kgs

Total Weight of Hook Mesh

= Weight of steel required for X bars + Weight of steel required for Y bars

= 12.48Kgs+7.25Kgs =19.73Kgs

## Stepped footings Bar Bending Schedule:

For the calculation of the total quantity of steel required for the Plain footing, we are adopting these dimensions for bars.

• Hook mesh is adopted for F2 footings
• Dimensions of Footing  are 0.9×0.9×1.35 (Length × Breadth × Depth)
• Dia of X Bars is 16mm (Dia 16mm@110mm  C/C)
• Dia of Y Bars is 20mm (Dia 12mm@115mm C/C)
• which means Center to center spacing between X bars is 110mm & Y bars is 115mm

Remember Proper Concrete cover should be adopted for the reinforcement  in Footings to resist it from corrosion.

Concrete Cover deduction:

As per condition, concrete cover of  0.1m is deducted from all sides of  mesh. In hook mesh, hook is provided at the end of each bar. Each bar has two ends and therefore, hook length is included in the calculation of length of bar

Hook Length = 9d (d is the dia of bar)

Total Hooks for each bar = 2

True dimensions post deducting concrete cover is length = (0.7m+2×9d) & breadth = (0.7m+2×9d)

Refer below image for more details

 Length ofEach X bar = 0.7+2×9dd = 16mm = 0.016m= 0.7+2×9×0.016= 0.988m Length ofEach Y bar = 0.7+2×9dd = 12mm = 0.012m= 0.7+2×9×0.020= 1.06m No. of X bars [(Y Bar Length)/Spacing]+1= [0.7/0.11]+1=~7bars(Don’t include hooklength in calculating no. of bars) No. of Y bars [(X Bar Length)/Spacing]+1= [0.7/0.115]+1= ~6bars Total Lengthof X bars = Length of each X bar ×No. of X Bars= 0.988 × 7 =7.9m Total Lengthof Y bars = Length of each Y bar ×No. of Y Bars= 1.06 × 6 =6.36m Total No. of ’12m’X bars = 7.9/12= 0.65bars Total No. of ’12m’Y bars = 6.36/12= 0.53bars Weight of steelrequired for1m of 16mm bar = D2/162= 162/162= 1.58kg/m Total weight of steelrequired for X bars = 1.58 × 7.9= 12.48Kgs Weight of steelrequired for1m of 12mm bar = D2/162= 202/162= 2.46kg/m Total weight of steelrequired for Y bars = 2.46 × 6.36= 15.64Kgs

Total Weight of Hook Mesh

= Weight of steel required for X bars + Weight of steel required for Y bars

= 12.48Kgs+15.64Kgs =28.12Kgs

## Eccentric/Shoe footings Bar Bending Schedule:-

For the calculation of the total quantity of steel required for the Plain footing, we are adopting these dimensions for bars.

• Mesh extended till Depth of Footing is adopted for F5 footings
• Dimensions of Footing  are 0.6×0.6×0.4 (Length × Breadth × Depth)
• Dia of X Bars is 16mm (Dia 16mm@80mm  C/C)
• Dia of Y Bars is 16mm (Dia 16mm@80mm C/C)
• which means Center to center spacing between X bars & Y bars is 80mm

Remember Proper Concrete cover should be adopted for the reinforcement  in Footings to resist it from corrosion.

Concrete Cover deduction:

As per condition, concrete cover of  0.1m is deducted from all sides of  mesh. In Mesh extended till Depth of Footing, extra bar is bent towards the depth.

 Length ofEach X bar = 0.4+0.3+0.3= 1.0m Length ofEach Y bar = 0.4+0.3+0.3= 1.0m No. of X bars [(Y Bar Length)/Spacing]+1= [0.4/0.08]+1=~6bars(Don’t include extra barlength in calculating no. of bars) No. of Y bars [(X Bar Length)/Spacing]+1= [0.4/0.08]+1= ~6bars Total Lengthof X bars = Length of each X bar ×No. of X Bars= 1.0 × 6 =6m Total Lengthof Y bars = Length of each Y bar ×No. of Y Bars= 1.0 × 6 =6m Total No. of ’12m’X bars = 6/12= 0.5bars Total No. of ’12m’Y bars = 6./12= 0.5bars Weight of steelrequired for1m of 16mm bar = D2/162= 162/162= 1.58kg/m Total weight of steelrequired for X bars = 1.58 × 6= 9.48Kgs Weight of steelrequired for1m of 12mm bar = D2/162= 122/162= 1.58kg/m Total weight of steelrequired for Y bars = 1.58 × 6= 9.48Kgs

Total Weight of Mesh extended till Depth of Footing

= Weight of steel required for X bars + Weight of steel required for Y bars

=9.48Kgs+9.48Kgs =18.96Kgs

## Combined Footing Bar Bending Schedule:-

For the calculation of the total quantity of steel required for the combined footing, we are adopting these dimensions for bars.

• Hook Mesh is usually adopted for F5 footing.
• Dimensions of Footing  are 4.2×1.7×0.9(Length × Breadth × Depth)
• Dia of X Bars is 12mm (Dia 12mm@100mm  C/C)
• Dia of Y Bars is 12mm (Dia 12mm@100mm C/C)
• which means Center to center spacing between X bars & Y bars is 100mm

Remember Proper Concrete cover should be adopted for the reinforcement  in Footings to resist it from corrosion.

Concrete Cover deduction:

As per condition, concrete cover of  0.1m is deducted from all sides of  mesh. In Mesh extended till Depth of Footing, extra bar is bent towards the depth.

 Length ofEach X bar = 4.2+2×9dd = 12mm = 0.012m= 4+2×9×0.012= 4.41m Length ofEach Y bar = 1.7+2×9dd = 12mm = 0.012m= 1.7+2×9×0.012= 1.91m No. of X bars [(Y Bar Length)/Spacing]+1= [1.7/0.1]+1=18bars(Don’t include hooklength in calculating no. of bars) No. of Y bars [(X Bar Length)/Spacing]+1= [4.2/0.1]+1= 43bars Total Lengthof X bars = Length of each X bar ×No. of X Bars= 4.41 × 18 =79.38m Total Lengthof Y bars = Length of each Y bar ×No. of Y Bars= 1.91 × 43 =82.13m Total No. of ’12m’X bars = 79.38/12= 6.6bars Total No. of ’12m’Y bars = 82.13/12= 6.84bars Weight of steelrequired for1m of 16mm bar =D2/162=122/162=0.88kg/m Total weight of steelrequired for X bars = 0.88 × 79.38= 69.85Kgs Weight of steelrequired for1m of 12mm bar =D2/162=122/162=0.88kg/m Total weight of steelrequired for Y bars = 0.88 ×82.13= 72.27Kgs

Total Weight of Hook Mesh

= Weight of steel required for X bars + Weight of steel required for Y bars

= 69.85Kgs+72.27Kgs =142.12Kgs

Abstract of above calculation:-

Total Weight of steel required for above plan

= Total weight of steel required for X bars +Total weight of steel required for Y bars

= 150.88Kgs+130.88Kgs =281.76Kgs = 0.28Ton