# bbs of beam reinforcement and steel quantity calculation

bbs of beam reinforcement and steel quantity calculation,Bar bending schedule of beam,in this topic we know about BBS of beam reinforcement and steel quantity calculation and bar bending schedule of beam.

We know that beam is provided in building construction in plinth level ,lintel level and wall as wall beam. Minimum size of RCC beam should be 9 ×9 inch or 225 mm×225mm having 5 steel bar in which 3 bar is of 12mm dia as main reinforcement having one bent Up Bar and 2 bar is of 10 mm dia of reinforcement used at top of beam that is known as anchor bar.

## Types of reinforcement ( rebar) used in beam

There are three types of reinforcement used in beam main bar, bent up bar and anchor bar

1) main bar :- we know that bottom part of beam is in tensile zone, the reinforcement which have higher dia provided at bottom of beam in tensile zone is known as main bar

2) bent up bar :- one main bar used as bottom of beam as bent Up Bar which provide stability to the beam against sagging and hogging bending moment. Bent up bar is also known as crank bar

3) anchor bar :– second reinforcement of lesser dimension provided at top of beam compression zone is known as anchor bar.

## BBS (Bar bending schedule) of beam

We have given following data according to diagram

Beam size = 225 × 300 mm

Cross sectional length= 225 mm

Cross sectional width = 300 mm

Clear span = 4000 mm ( clear span is distance between inner age of two column )

Clear cover at all side = 20 mm

Dia of main bar =12 mm of 2 nos

Dia of bent up bar =12 mm of 1nos

Dia of anchor bar = 10 mm of 2 nos

Dia of stirrups = 8 mm

Spacing between stirrup = 150 mm

For bar bending schedule of beam we have to calculate following

1) cutting length of main bar

2) cutting length of bent Up Bar

3) cutting length of anchor bar

4) cutting length of stirrup

5) number of stirrups

### How to calculate cutting length of main bar in bbs of beam

1) cutting length of main bar

Cutting length = clear span + 2 (cross sectional length of column) _ 2 clear cover

Clear span = 4000 mm

Length of column =225 mm

Clear cover = 20 mm

CL = 4000+(2×225)_(2×20) mm

CL = 4000+450_40 mm

Cutting length = 4410 mm

Cutting length of 2 main bar =2×4410 mm

Cutting length of 2 main bar=8820 mm

### cutting length of bent up bar in bbs of beam

CL = CL of main bar + (2×0.42 H) _bend

Cutting length of main bar = 4410 mm

H = height

H = width _ 2 clear cover _ dia of main bar

H = 300 _2×20_12 mm

Height = 300_40_12 mm

Height = 300_52 mm

Height = 248 mm

Bend elongation = 4×1d ( we have 4 bend angle that is 45° which is equal to 1d and value of d is 12 mm diameter of main bar

Bend = 4×12 mm = 48 mm

Put all the value in given formula

CL = CL of main bar + (2×0.42 H) _bend

CL = 4410 +(2×0.42×248)_48 mm

C L = 4410+208_48 mm

Cutting length of bent Up Bar = 4570 mm

3) total cutting length of 3 main bar

Total cutting length = 2 straight bar + 1 bent up bar

Total cutting length = 8820+4570 mm

Total cutting length = 13390 mm

### weight of main bar and quantity calculation of main bar for bbs of beam

We have one formula for approx calculation of weight of reinforcement

Weight =( D^2/162)×lenght kg/m

Where we have

Dia of main bar = 12mm

Length of main bar = 13390 mm=13.39m

Weight= (12×12/162)×13.39 kg

Weight = 11.9 kg

So total quantity of steel as main bar is needed about 11.9 kilogram

### quantity calculation of anchor reinforcement for bbs of beam

We know that cutting length of main bar is equal to cutting length of anchor bar

So we have cutting length of two anchor bar is equal to 8820 mm

We have length = 8820 mm =8.82 meter

Dia of anchor bar = 10 mm

Weight = ( D^2/162)×lenght kg/m

Weight = (10×10/162)×8.82 kg

Weight = 5.44 kg

So we need 5.44 kg of Steel as anchor bar

### total quantity of steel needed for bbs of beam

Total weight of steel= weight of main bar + weight of anchor bar

Total weight =11.9 +5.44 kg = 17.34 kg

So we have approx 17.34 kilogram of Steel reinforcement required for 1 beam having size 225×300×4000 mm

### we have to calculate cutting length of stirrup

Dia of stirrup = 8mm

Size of beam = 225×300 mm

CL of stirrup = perimeter + hook length _ bend

Perimeter= 2(a+b)

Length of a side = 225_2×20_2×4 ( half dia)

Length a = 225_40_8 mm

Length of a = 177 mm

Length of b = 300_2×20-2×4 mm

Length of b = 252 mm

Perimeter = 2(177+252) mm

Perimeter of stirrup = 858 mm

Hook length = 2×10d

Hook length = 2×10×8= 160 mm

Bend = 4×1d ( four bend of 45°)

Bend = 4×10 mm =40 mm

CL of stirrup = perimeter + hook length _ bend

CL of stirrup= 858+160_40 mm

Cutting length of stirrup = 978 mm =0.978 m

You should visits:-

●How to calculate cutting length of triangular stirrup

● 8) calculate number of stirrup

Length of straight bar = 4410 mm

Spacing =150 mm

No. of stirrup = (length/spacing)+1

No. of stirrup= (4410/150) +1

No. of stirrup= 29.4 +1 =30.4

Rounded it No. of stirrup= 30 nos