Bar bending schedule of column.

Bar bending schedule of column.

In this article we learn how to calculate bbs or bar bending schedule of column.
to calculate bbs of any component it is very necessary for you to carefully observe the drawing and available data.

KNOWN DATA,

DIA OF LONGITUDNAL BAR=20MM

NO, OF LONGITUDNAL BAR= 4

Ld =DEVELOPMENT LENGTH =40D

DIA OF TIE BAR = 8MM

SPACING BETWEEN TIES = 250MMC/C

HOOK =10D

CLEAR COVER:-   COLIUMN = 40MM

                               FOOTING= 50MM

                                SLAB= 25 MM

CUTTING LENGTH OF LONGITUDNAL BAR:-

  =300+(1000-50-12-12)+300+3000+Ld-BENDS

  =300+926+300+3000+(40X20)-2(2X20)

  =300+926+300+3000+800-80

  =5246mm =5.246M

TOTAL LENGTH OF LONGITUDNAL BARS = 4 X 5.246=20.984

now,

NO. OF TIES:-(LENGTH OF LONGITUDNAL BAR-(Ld+300))/SPACING+1

  =(5246-{(40X20)+300})/250+1

  =17.58=18

CUTTING LENGTH OF TIE= 2{300-(2X CLEAR COVER)- DIA OF TIE BAR}+ 2{350-(2X CLEAR COVER)- DIA OF TIE BAR} + HOOK – BENDS

  =2(300-80-8)+2(350-80-8)+(2X10X8)-2(3D)-3(2D)

   =2(300-80-8)+2(350-80-8)+(2X10X8)-2(3X8)-3(2X8)

  =1012MM=1.012M

TOTAL LENGTH OF TIES=18 X 1.012= 18.216

WEIGHT CALCULATION:-

WEIGHT OF 1 LONGITUDNAL BAR = D^2/162 XLENGTH

  =〖20〗^2/162 X5.246 =13.115KG

TOTAL WEIGHT OF LONGITUDNAL BARS=4X13.115=52.46KG

WEIGHT OF 1 TIE=  D^2/162 X LENGTH

  =8^2/162 X1.012 =0.399KG

TOTAL WEIGHT OF TIES = NO. OF TIES X WEIGHT OF 1 TIE

  = 18 X 0.399= 7.182KG