Bar Bending Schedule Of Plinth Beam
Let us prepare a Bar Bending ScheduleĀ for the simple plinth beamĀ as shown below.
Given data :
Clear span of the beam = 2500 mm.,Ā beam size 300mm Ć 380 mm.
Top & bottom rebar dia. = 12 mm.,Ā number of bars = 4 nos.
Stirrups 8 mm.ā @ 150 mm c/c.,Ā clear cover =25 mm. on all the sides.
The cutting length of the main bar
=Ā [(2 times Ć Ld) + inner distance between columns – (2nos. Ć 2d ) ]
(Let us take development length Ld as 40d, and we have takenĀ 2d for the 90Ā° bendĀ deductions.)
= [ ( 2 Ć 40 Ć 12 mm.)+ 2500 mm. – ( 2 nos. Ć 2 Ć 12 mm ) ]
= [ 480 mm + 2500 mm – 48 mm]
= 2932 mm i.e.Ā 2.932 m.
Learn MoreĀ
Bar Bending Schedule Of Lintel Beam With Full Detail
Cutting length of the stirrup
= [2nos. Ć { (a +b )} + (hook length) –Ā (90Ā° bend) – (135Ā° bend )]
Where a = beam width – 2Ā Ć cover, & b = beam depth – 2 Ć cover
= [2 nos.Ā Ć { ( 300 mm – 2 Ć 25mm.) + ( 380 mm – 2Ā Ć 25mm )} + (2nosĀ Ć10d ) – (3 nos. Ć 2d ) – (2 nos. Ć 3d) ]
Here, 10d is taken for hook length.
We have deducted 2d for 90Ā° bend 3nos., & 3d for 135Ā° bend 2nos. as shown in the above drawing.
= [2 nos.Ć{ ( 250 mm ) + ( 330 mm ) } + (2nos.Ā Ć10 Ć 8mm) – ( 3 nos. Ć 2 Ć 8mm ) – ( 2 nos. Ć 3 Ć 8 mm.)]
= [2 nos. Ć { 580 mm } + 160 mm – 48mm – 48mm.]
= [1160 mm + 160 mm – 96 mm.]
= 1224 mm i.e.Ā 1.224 m.
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Number of stirrupsĀ
= ( plinth beam length Ć· stirrup spacing ) + 1
= (2500 mm.Ā Ć· 150 mm) +1
= 16.66 + 1
= 17.66 nos.
By rounding off, the no. of stirrups required =Ā 18 nos.Ā
Now, let us prepare aĀ BBSĀ table for the plinth beam.
1.Ā main barĀ Ā Ā Ā Ā Ā Ā 12Ā Ā Ā Ā Ā 4Ā Ā 2.932Ā Ā Ā Ā Ā 22.8Ā Ā Ā Ā Ā Ā Ā 0.89Ā Ā Ā 20.292
2. stirrupsĀ Ā Ā Ā Ā Ā Ā Ā Ā 8Ā Ā Ā Ā Ā 18Ā Ā 1.224Ā Ā Ā 22.032Ā Ā Ā Ā Ā 0.395Ā Ā Ā Ā Ā 8.702
Total weight of barsĀ =Ā 28.994 kgs.
Add 3% wastageĀ Ā Ā Ā Ā = 0.869Ā kgs.
Ā Ā Ā Grand total of rebarsĀ =Ā Ā 29.863Ā kgs.Ā Ā Ā
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