Bar Bending Schedule Of Plinth Beam

Raja Junaid Iqbal April 30, 2021 Bar Bending Schedule

Bar Bending Schedule Of Plinth Beam

Let us prepare a Bar Bending Schedule for the simple plinth beam as shown below.

Given data :

Clear span of the beam = 2500 mm., beam size 300mm × 380 mm.

Top & bottom rebar dia. = 12 mm., number of bars = 4 nos.

Stirrups 8 mm.∅ @ 150 mm c/c., clear cover =25 mm. on all the sides.

The cutting length of the main bar

= [(2 times × Ld) + inner distance between columns – (2nos. × 2d ) ]

(Let us take development length Ld as 40d, and we have taken 2d for the 90° bend deductions.)

= [ ( 2 × 40 × 12 mm.)+ 2500 mm. – ( 2 nos. × 2 × 12 mm ) ]

= [ 480 mm + 2500 mm – 48 mm]

= 2932 mm i.e. 2.932 m.

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Cutting length of the stirrup

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

Where a = beam width – 2 × cover, & b = beam depth – 2 × cover

= [2 nos. × { ( 300 mm – 2 × 25mm.) + ( 380 mm – 2 × 25mm )} + (2nos ×10d ) – (3 nos. × 2d ) – (2 nos. × 3d) ]

Here, 10d is taken for hook length.

We have deducted 2d for 90° bend -3nos., & 3d for 135° bend -2nos. as shown in the above drawing.

= [2 nos.×{ ( 250 mm ) + ( 330 mm ) } + (2nos. ×10 × 8mm) – ( 3 nos. × 2 × 8mm ) – ( 2 nos. × 3 × 8 mm.)]

= [2 nos. × { 580 mm } + 160 mm – 48mm – 48mm.]

= [1160 mm + 160 mm – 96 mm.]

= 1224 mm i.e. 1.224 m.

Number of stirrups

= ( plinth beam length ÷ stirrup spacing ) + 1

= (2500 mm. ÷ 150 mm) +1

= 16.66 + 1

= 17.66 nos.

By rounding off, the no. of stirrups required = 18 nos.

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

sl. bar dia. no. length total weight total

no. ( mm) (m.) length kg/m weight

1. main bar 12 4 2.932 22.8 0.89 20.292

2. stirrups 8 18 1.224 22.032 0.395 8.702

Total weight of bars = 28.994 kgs.

Add 3% wastage = 0.869 kgs.

Bar Bending Schedule Of Plinth Beam