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### How to calculate rebar cutting length in a beam with lapping? / Calculating the cutting length of rebar's for lapping in a beam.

Let us calculate the cutting length of the top & bottom bar of a beam as shown below.

Given data:

Beam width = 300mm.

Beam depth = 550mm.

Column width = 400mm.

Effective length L1= 6m = 6000mm.

Effective length L2 = 5.5m. = 5500mm.

Top reinforcement = 20mm ∅ - 2nos.

Bottom  reinforcement = 25mm ∅ - 2nos.

Clear cover = 30mm.

Development length Ld = 50d.

Lapping length = 50d.

Calculation:

A. Cutting length:

1. Bottom bar:

Total length of the bottom bar

= [L1 + L2+ (2nos. × development length) - (2nos. × 1/2 × column width) - ( 2nos. × 90° bend)]

= [6000mm. + 5500mm. + (2nos. × 50d) - (2nos. × 1/2 × 400mm.)  - ( 2nos. × 2d)]

= [6000mm. + 5500mm. + (2nos. ×50 × 25mm.) - (2nos. × 1/2 × 400mm.)- ( 2nos. × 2× 25mm.)]

= [ 11500mm. +  (2500mm.) - (400mm.)- ( 100mm.)]

13,500mm.

= 13.50m  > 12m.

The total length (13.50m.) is greater than the available bar length(12m.). Hence there will be lapping in the bar.

Before proceeding further, Go through the article ðŸ‘‡

ðŸ‘€. Where to provide lapping in beam reinforcement?

So that you will understand the correct position of lapping in the beam & the basic rules of lapping.

a. Cutting length of the part-1 of the bottom bar

= [L1 + (development length) + (half of lap length ) - (1/2 × column width) - ( 1no × 90° bend)]

= [ 6000mm. + (50d) + ( 1/2 × 50d) - (1/2 × 400mm.) - ( 2d)]

= [ 6000mm. + (50 × 25mm  ) + ( 1/2 × 50 × 25mm. ) - (1/2 × 400mm.) - ( 2 × 25mm.)]

= [ 6000mm. + (1250mm  ) + ( 625mm. ) - ( 200mm.) - (50mm.)]

= 7625mm.

= 7.625m.

b. Cutting length of the part-2 of the bottom bar

= [L2 + (development length) + (half of lap length ) - (1/2 × column width) - ( 1no × 90° bend)]

= [ 5500mm. + (50d) + ( 1/2 × 50d) - (1/2 × 400mm.) - ( 2d)]

= [ 5500mm. + (50 × 25mm  ) + ( 1/2 × 50 × 25mm. ) - (1/2 × 400mm.) - ( 2 × 25mm.)]

= [ 5500mm. + (1250mm  ) + ( 625mm. ) - ( 200mm.) - (50mm.)]

= 7125mm.

= 7.125m.

Check:

[Total length of bar + lap length ] = [cutting length of (part-1 + part-2)]

[13500mm + 50d] = [7625mm. +7125mm.]

[13500mm. + 50 × 25mm.] = 14750mm.

14750mm = 14750mm.

2. Top bar:

Total length of the top bar

= [L1 + L2+ (2nos. × development length) - (2nos. × 1/2 × column width) - ( 2nos. × 90° bend)]

= [6000mm. + 5500mm. + (2nos. × 50d) - (2nos. × 1/2 × 400mm.)  - ( 2nos. × 2d)]

= [6000mm. + 5500mm. + (2nos. ×50 × 20mm.) - (2nos. × 1/2 × 400mm.)- ( 2nos. × 2× 20mm.)]

= [ 11500mm. +  (2000mm.) - (400mm.)- ( 80mm.)]

13,020mm.

= 13.02m  > 12m

The total length (13.02m.) is greater than the available bar length(12m.). Hence there will be lapping in the bar.

a. Cutting length of the part-1 of the top bar

= [L1 +( L2/2) + (development length) + (half of lap length ) - (1/2 × column width) - (90° bend)]

= [ 6000mm. + (5500mm.÷ 2) + (50d) + ( 1/2 × 50d) - (1/2 × 400mm.) - ( 2d)]

= [ 6000mm. + (2750mm.) + (50 × 20mm  ) + ( 1/2 × 50 × 20mm. ) - (1/2 × 400mm.) - ( 2 × 20mm.)]

= [ 6000mm. + (2750mm.) + (1000mm  ) + ( 500mm. ) - ( 200mm.) - (40mm.)]

= 10,010mm.

= 10.010m.

b. Cutting length of the part-2 of the top bar

= [( L2/2) + (development length) + (half of lap length ) - (1/2 × column width) - (90° bend)]

= [ (5500mm.÷ 2) + (50d) + ( 1/2 × 50d) - (1/2 × 400mm.) - ( 2d)]

= [ (2750mm.) + (50 × 20mm  ) + ( 1/2 × 50 × 20mm. ) - (1/2 × 400mm.) - ( 2 × 20mm.)]

= [(2750mm.) + (1000mm  ) + ( 500mm. ) - ( 200mm.) - (40mm.)]

= 4,010mm.

= 4.010m.

Check:

[Total length of bar + lap length ] = [cutting length of (part-1 + part-2)]

[13020mm + 50d] = [10010mm. +4010mm.]

[13020mm. + 50 × 20mm.] = 14,020mm.

14,020mm = 14,020mm.

Thank you for going through these calculation steps. Have a good day ðŸ˜„.

### Bar bending schedule (BBS) of a continuous beam. (Type-2) / How to calculate the cutting length & weight of rebar's in continuous beam.

ðŸ‘ˆ Back

Let us calculate the cutting length & weight of rebars in the RCC continuous beam as shown in the below-given drawing.

The schedule of the beam is given below.

 Bottom bar Top bar Stirrup Left Mid Right Left Mid Right Left L/3 Mid L/3 Right L/3 2 -25∅ 2-25∅ 2-25∅ 2-20∅ 2-20∅ 2-20∅ 8mm. 8mm. 8mm. ------- 2 -20∅ ------- 2-20∅ ------- 2-20∅ 100c/c 125c/c 100c/c

Given data:

Beam width = 300mm.

Beam depth = 550mm.

Column width = 400mm.

Column length = 480mm.

Effective length L1= 5m = 5000mm.

Effective length L2 = 4.5m. = 4500mm.

Top continuous reinforcement = 20mm ∅ - 2nos.

Top curtail reinforcement (EOS bar) =      20mm ∅ - 2nos.

Bottom continuous reinforcement = 25mm ∅ - 2nos.

Bottom curtail reinforcement     = 20mm ∅ - 2nos.

Stirrup at zone-1 = 8mm∅ @ 100c/c

Stirrup at zone-2 = 8mm∅ @ 125c/c.

Clear cover = 30mm.

Development length Ld = 50d

Calculation:

A. Cutting length:

a. Bottom bar:

1. Cutting length of the continuous bottom bar

= [L1 + L2+ (2nos. × development length) - (2nos. × 1/2 × column width) - ( 2nos. × 90° bend)]

= [5000mm. + 4500mm. + (2nos. × 50d) - (2nos. × 1/2 × 400mm.)  - ( 2nos. × 2d)]

= [5000mm. + 4500mm. + (2nos. ×50 × 25mm.) - (2nos. × 1/2 × 400mm.)- ( 2nos. × 2× 25mm.)]

= [ 9500mm. +  (2500mm.) - (400mm.)- ( 100mm.)]

11,500mm.

= 11.50m. < 12m.

The total length (11.50m.) is less than the available bar length(12m.). Hence there will be no lapping in the bar.

If the total cutting length exceeds 12m., the cutting length should be calculated as given in the below-linked article.

2. Cutting length of the bottom mid curtail bar-1

= [L1 - {0.1L1 + 0.15L1}]

= [5000mm. - {(0.1× 5000mm. ) + ( 0.15 × 5000mm.)}]

= [5000mm. - {(500mm. ) + (750mm.)}]

= [ 5000mm. - 1250mm ]

3750mm.

= 3.75m.

3. Cutting length of the bottom mid curtail bar-2

= [L2 - {0.1L2 + 0.15L2}]

= [4500mm. - {(0.1× 4500mm. ) + ( 0.15 × 4500mm.)}]

= [4500mm. - {(450mm. ) + (675mm.)}]

= [ 4500mm. - 1125mm ]

3375mm.

= 3.375m.

b. Top bar:

4. Cutting length of the continuous top bar

= [L1 + L2+ (2nos. × development length) - (2nos. × 1/2 × column width) - ( 2nos. × 90° bend)]

= [5000mm. + 4500mm. + (2nos. × 50d) - (2nos. × 1/2 × 400mm.)  - ( 2nos. × 2d)]

= [5000mm. + 4500mm. + (2nos. ×50 × 20mm.) - (2nos. × 1/2 × 400mm.)- ( 2nos. × 2× 20mm.)]

= [ 9500mm. +  (2000mm.) - (400mm.)- ( 80mm.)]

11,020mm.

= 11.02m. < 12m.

5. Cutting length of the top left curtail bar

= [(L1/4) + (development length) - ( 90° bend) ]

= [(5000mm.÷ 4) + (50d) - (2d)]

= [1250mm. + (50 × 20mm.) - ( 2 ×20mm. )]

= [ 1250mm. +1000mm.- 40mm ]

= 2210mm.

= 2.21m.

6. Cutting length of the top mid curtail bar

= [L1/4 + column width +L2/4]

= [5000/4 + 400mm. + 4500/4]

= [ 1250mm. + 400mm. +1125mm.]

2775mm.

= 2.775m.

7. Cutting length of the top right curtail bar

= [(L2/4) + (development length) - ( 90° bend) ]

= [(4500mm.÷ 4) + (50d) - (2d)]

= [1250mm. + (50 × 20mm.) - ( 2 ×20mm. )]

= [ 1125mm. +1000mm.- 40mm ]

= 2085mm.

= 2.085m.

c. Stirrup:

8. The cutting length of the stirrup bar

= [{2nos. ×  (a +b )} + (2nos.× hook length) -  ( 3nos. × 90° bend) - (2nos. ×135° bend )]

Where,

a = {beam width - (2 × cover) - (2 × 1/2 × stirrup dia.)}

= {300mm. - (2 × 30mm.) - ( 2 × 1/2 × 8mm.)}

= {300mm. - (60mm.) - (8mm.)}

= 232mm.

b = {beam depth - (2 × cover) - (2 × 1/2 × stirrup dia.)}

= {550mm. - (2 × 30mm.) - ( 2 × 1/2 × 8mm.)}

= {550mm. - (60mm.) - (8mm.)}

= 482mm.

The cutting length of the stirrup

= [{2 nos. × (232mm. + 482mm.)} + (2nos ×10d ) - (3 nos. × 2d ) - (2 nos. × 3d) ]

To know, why we take bend deductions, Go through the article ðŸ‘‡

Note:  Here, hook length is taken as 10d.

The cutting length of the stirrup

= [{2 nos. × 714mm.} + (2nos. ×  10 × 8mm) - ( 3 nos. × 2 × 8mm ) - ( 2 nos. × 3 × 8 mm.)]

= [1428mm. + 160 mm - 48mm - 48mm.]

= [1588mm. - 96 mm.]

= 1492 mm i.e. 1.492 m.

B. No. of  stirrups:

For clear span C1:

1. Stirrups in left span:

No. of stirrups

= [{( C1 /3) ÷ stirrup spacing } + 1]

Where,

C1= [(Effective span L1) - (2nos.× 1/2 of column width)]

= [(5000mm.) - (2 × 1/2 × 400mm.)]

= 4600mm.

No. of stirrups

= [{( 4600mm. /3) ÷ 100mm. } + 1]

= [{1533.33mm. ÷ 100mm. } + 1]

=   [15.33 +1]

= 16.33 nos.

By rounding off = 17 nos.

2. Stirrups in mid-span:

No. of stirrups

= [{( C1 /3) ÷ stirrup spacing } - 1]

= [{( 4600mm. /3) ÷ 125mm. } - 1]

= [{1533.33mm. ÷ 125mm. } - 1]

=   [12.26 -1]

= 11.26 nos.

By rounding off = 11 nos.

3. Stirrups in right span:

Stirrups in right span = stirrups in the left span = 17 nos.

Note:

1. As the spacing is the same, the left & right spans have equal no of stirrups.

2. As we have added 2nos. of stirrups in left & right span, we have to deduct 1no. from mid-span.

Total no. of stirrups in span C1

= Stirrups in [left +mid +right]

= [17nos. + 11nos. + 17nos.]

= 45 nos.

For clear span C2:

4. Stirrups in left span:

No. of stirrups

= [{( C2 /3) ÷ stirrup spacing } + 1]

Where,

C2= [(Effective span L2) - (2nos.× 1/2 of column width)]

= [(4500mm.) - (2 × 1/2 × 400mm.)]

= 4100mm.

No. of stirrups

= [{( 4100mm. /3) ÷ 100mm. } + 1]

= [{1366.66mm. ÷ 100mm. } + 1]

=   [13.66 +1]

= 14.66 nos.

By rounding off = 15 nos.

2. Stirrups in mid-span:

No. of stirrups

= [{( C2 /3) ÷ stirrup spacing } - 1]

= [{( 4100mm. /3) ÷ 125mm. } - 1]

= [{1366.66mm. ÷ 125mm. } - 1]

=   [10.93 -1]

= 9.93 nos.

By rounding off = 10 nos.

3. Stirrups in right span:

Stirrups in right span = stirrups in the left span = 15 nos.

Total no. of stirrups in span C2

= Stirrups in [left +mid +right]

= [15nos. + 10nos. + 15nos.]

40 nos.

The total no. of the stirrups in the beam

= Stirrups in [span C1 + span C2]

= [45 nos. + 40 nos.]

85 nos.

Now, let us prepare a BBS table for the RCC continuous beam.

 Sl. No. Type of Bar Dia. in mm. Nos. Length in m. Total length in m. Weight in Kg/m. Total  bar wt. in kg. 1. Bottom continuous bar 25 2 11.50 23.0 3.854 88.642 2. Bottom curtail bar-1 20 2 3.75 7.50 2.466 18.495 3. Bottom curtail bar-2 20 2 3.375 6.75 2.466 16.646 4. Top continuous bar 20 2 11.02 22.04 2.466 54.35 5. Top left curtail bar. 20 2 2.21 4.42 2.466 10.899 6. Top mid curtail bar. 20 2 2.775 5.55 2.466 13.686 7. Top right  curtail bar. 20 2 2.085 4.17 2.466 10.283 8. Stirrup bar 8 85 1.492 126.82 0.395 50.094 9. Total weight of bars = 263.09 10. Add 3% wastage = 7.89 11. The grand total wt. of rebar's = 270.98

To know, how to calculate the weight of rebars per RMT., Go through the article ðŸ‘‡

Continued ðŸ‘‰ Part-3ðŸ‘‡

ðŸ‘€. Bar bending schedule (BBS) of a simply supported beam. (Type-3)

Thank you for going through these calculation steps. Have a good day ðŸ˜„.

### Bar bending schedule (BBS) of RCC beam. (Type-1) / How to calculate the cutting length & weight of rebar's in RCC beam.

Let us calculate the cutting length & weight of rebars in the RCC beam as shown in the below-given drawing.

Given data:

Beam width = 300mm.

Beam depth = 550mm.

Clear span = 4500mm. = 4.5m.

Top reinforcement = 16mm ∅ - 2nos.

Bottom rebar 1st layer = 20mm ∅ - 2nos.

Bottom rebar 2nd layer = 16mm ∅ - 2nos.

Stirrup at zone-1 = 8mm∅ @ 125c/c

Stirrup at zone-2 = 8mm∅ @ 150c/c.

Clear cover = 30mm.

Spacer bar = 25mm∅

Development length Ld = 50d

Calculation:

A. Cutting length:

1. Bottom layer-1 bar:

Cutting length of the main bar

= [clear span + (2nos. × development length) - ( 2nos. × 90° bend)]

= [4500mm. + (2nos. × 50d) - ( 2nos. × 2d)]

= [4500mm. + (2nos. ×50 × 20mm.) - ( 2nos. × 2× 20mm.)]

= [ 4500mm. + 2000mm - 80mm.]

= 6420mm.

2. Bottom layer-2 bar:

Cutting length of the bar

= [clear span + (2nos. × development length) - ( 2nos. × 90° bend)]

= [4500mm. + (2nos. × 50d) - ( 2nos. × 2d)]

= [4500mm. + (2nos. ×50 × 16mm.) - ( 2nos. × 2× 16mm.)]

= [ 4500mm. + 1600mm - 64mm.]

= 6036mm.

3. Top bar:

As the dia. of top bar = dia of 2nd layer bar,

The cutting length will be the same i.e. = 6036mm.

4. Spacer bar:

The cutting length

= [beam width - (2nos. × clear cover) +10mm]

= [300mm.- ( 2nos. × 30mm.) + 10mm.]

= 250mm.

5. Stirrup bar:

The cutting length of the stirrup bar

= [{2nos. ×  (a +b )} + (2nos.× hook length) -  ( 3nos. × 90° bend) - (2nos. ×135° bend )]

Where,

a = {beam width - (2 × cover)}

= {300mm. - (2 × 30mm.)}

= 240mm.

b = {beam depth - (2 × cover)}

= {550mm. - (2 × 30mm.)}

= 490mm.

The cutting length of the stirrup

= [{2 nos. × (240mm. + 490mm.)} + (2nos ×10d ) - (3 nos. × 2d ) - (2 nos. × 3d) ]

To know, why we take bend deductions, Go through the article ðŸ‘‡

Note:  Here, hook length is taken as 10d.

= [{2 nos.×730mm.} + (2nos. ×10 × 8mm) - ( 3 nos. × 2 × 8mm ) - ( 2 nos. × 3 × 8 mm.)]

= [1460mm. + 160 mm - 48mm - 48mm.]

= [1620mm. - 96 mm.]

= 1524 mm i.e. 1.524 m.

B. No. of bars:

1. Stirrups for zone-1:

No. of stirrups

= 2 sides × [{( clear span of the beam /3) ÷ stirrup spacing } + 1]

= 2 sides ×[{( 4500mm. /3) ÷ 125mm. } + 1]

= 2 sides × [{1500mm. ÷ 125mm. } + 1]

=  2 × [12 +1]

= 2× 13 nos.

= 26 nos.

2. Stirrups for zone-2:

No. of stirrups

= [{( clear span of the beam /3) ÷ stirrup spacing } -1]

= [{( 4500mm. /3) ÷ 150mm. } - 1]

= [{1500mm. ÷ 150mm. } - 1]

= [10 -1]

= 9 nos.

The total no. of the stirrup.

= stirrups in [zone-1 + zone-2]

= [26nos. + 9nos.]

= 35 nos.

Note:

As we have added 2nos. of stirrups in zone-1, we have to deduct 1no. from zone-2.

3. Spacer bar:

The no of spacer bars

= [{clear span ÷ 1000mm. } + 1]

= [ {4500mm ÷ 1000mm. } + 1]

= [4.5 +1]

= 5.5nos.

By rounding off = 6 nos.

To know what is a spacer bar, Go through the article ðŸ‘‡

Now, let us prepare a BBS table for the RCC beam.

 Sl. No. Type of Bar Dia. in mm. Nos. Length in m. Total length in m. Weight in Kg/m. Total  bar wt. in kg. 1. Bottom bar-1. 20 2 6.42 12.84 2.466 31.663 2. Bottom bar-2 16 2 6.036 12.072 1.578 19.049 3. Top bar 16 2 6.036 12.072 1.578 19.049 4. Spacer bar 25 6 0.25 1.5 3.854 5.781 5. Stirrup bar 8 35 1.524 53.34 0.395 21.069 4. Total weight of bars = 96.611 5. Add 3% wastage = 2.898 6. The grand total wt. of rebar's = 99.51

To know, how to calculate the weight of rebars per RMT., Go through the article ðŸ‘‡

Continued ðŸ‘‰ Part-2ðŸ‘‡

Thank you for going through these calculation steps. Have a good day ðŸ˜„.