How to calculate bar bending schedule for column?

Bar Bending Schedule of Column: –

Consider the height of the column is 4.5 m and having a cross-sectional area is 300 x 300 mm and having a 40 mm of clear cover. 4 bars are going to use has a dia. of 16 mm. The 8 mm dia. of stirrups is going to use @ 160mm and www.fbhdtj.blogspot.si - 792210 RUB TINK0FF eny mm at L/3. So, calculate the cutting length (or BBS) of bars and stirrups?

Given Data:

Height = 4.5 m

Cross section = 300 x 300 mm

Clear Cover = 40 mm

No of vertical bar = 4 no’s

Diameter of vertical bar = 16 mm

Diameter of Stirrup = 8 mm

Stirrups centre to centre spacing = @160mm or @ 210 mm

BBS of Column =?

 

Solution: –

The Calculation was taken into two steps.

1. Vertical bar calculation
2. Cutting length of stirrups

Vertical Bar Calculation

Length of 1 bar = Height of column (H) + development length (Ld)

Length of 1 bar

= 4500 mm + 40d

= 4500 + 40 x 16

= 4500 + 640

= 5140 mm or 5.140 m

So, the length of 1 vertical bar is 5.140 m and we have a total 4 number of bars,

Total length

= 4 x 5.140

= 20.56 m long vertical bar is required.

Cutting the length of stirrups

Cross-sectional area column is 300 mm  300mm

Cross-section area of the stirrup

= 300 – 2 x clear cover

= 300 – 2 x 40

= 300 – 80

= 220 mm (A)

Number of stirrups

= 4500/3

= 1500mm or 1.5m

Formula = L/3 / spacing + 1

(No of stirrups in end zone)

= 1500 / 160

= 9.37 no’s say 10 no’s

There is total two zones of 160mm spacing and one zone of 210mm spacing

= 2 x 10

= 20 no’s (at end zones)

At Mid Zones

= 1500 / 210

= 7.14 no’s say 8 no’s

Total no of stirrups

= 20 + 8

= 28 no’s

Cutting Length of one stirrup

Formula: –

= (4 x A) + hook – bend

Cutting length

= (4 x A) + 2 x 10d – 5 x 2d

Where

Hook = 10d

Bend = 5  2d (5 bends in stirrup)

d = dia. of bar

= (4 x 220) + 2 x 10 x 8 – 2 x 5 x 8

= 880 +160 – 80

= 960 mm or 0.96 m

So, we have a total 28 no’s of stirrups are going to use,

Total length

= 28 x 0.96

= 27m long 8 mm bar

 

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