BAR BENDING SCHEDULE FOR COLUMN-LCETED -lceted LCETED INSTITUTE FOR CIVIL ENGINEERS

## BAR BENDING SCHEDULE FOR COLUMN

GIVEN DATA

•Slab Thickness – 125 mm & 100 mm

•Floor height – 3000 mm or 3 m

•Ground Floor Level – 3300 mm

•Footing Height – 300 mm

•Development Length – 50d

•The column has 6 numbers of 20 mm dia bars

•8mm stirrups @ 150 mm C/C

•Footing Clear Cover – 40 mm

•Slab Clear Cover – 25 mm

Step 1 – Find the length of Vertical Bar

Length of Vertical bar = Development length (Ld) + Height of Ground Level + Floor Heights (1,2,3)+ Slab Thicknesses + Overlap Length (Det.B)

= (50×20)+3300+ (3×3000)+(3×125)+100 +(50×20)

= 14775 mm or 14.78 m

Now we know the length of one vertical bar. Normally column reinforcement drawings won’t come with lapping details.

Step 2 – Find out lapping

As we know that lapping length required is 50d = 50 * Diameter of the bar = 50×20 = 1000 mm.

We know that each bar is 12.25 m or 40 feet length (approx).Total Length of Vertical Bar = 14.2 m which is more than 12.25 m so each rod will be lapped at least once to attain the required length.

So we have added the lapping length with the total length = 14775+1000 = 15775 mm or 15.78 m

Step 3 – Cutting Length of Stirrups

Length of One Hook
= 9d (Cutting length of Stirrup = Perimeter of stirrup + Number of Bends +  Number of Hooks)

= 2(a+b) + 3 numbers of 90 degree bends + 2 numbers of hooks

= 2(500+200)+(3 x2d)+(2x9d)= 2×700+3x2x20+2x9x20

Cutting length of Stirrup = 1880 mm

Step 4 – Number of Stirrups

Number of stirrups required = (Total length of Column/spacing of stirrups)+1  = (3300+125+3000+125+3000+125+3000+100)/150 + 1

Number of stirrups= 85 no.s

Step 5 – Bar Bending Schedule

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