How to make bar bending schedule of column in detail?

VERTICAL BAR CALCULATION

Total length of bar = 20 ф

20 ф = 300 + (1500 – clear cover – bottom bar dia.) + 600 + 200 + clear height between floor level to slab bottom + slab thickness + lap length - bend

20 ф= 300 + (1500 – 50 – 25 – 25) + 600 + 200 + (3000 x 4) + (175 x 3) + (50 x 20) – (2 x 2 x 20)

20 ф = 15945 mm

= 15.945 m

Steel bar available in the market is 12 m long but we get the ans is 15.945 m which greater than 12 m bar 

Therefor, add one lap i.e. 50 d

Therefor, 15945 + 50 d

15945 + (50 x 20)

16945 mm

16.845 m

Total no. of 20 ф bar is 4 no.

Therefor total length of bar = 16.845 x 4 = 67.78 m

Total length of bar = 16 ф

16 ф = 300 + (1500 – clear cover – only one bar rested) + 600 + 200 + (3000 x 4) + (175 x 3) + (50 x 16) – (2 x 2 x 16)

= 15786 mm

= 15.786 m

Steel bar available in the market is 12 m long but we get the ans is 15.945 m which greater than 12 m bar 

Therefor, add one lap i.e. 50 d

Therefor 15.786 + 50 d

= 15.786 + (50 x 16)

= 16586 mm

= 16.586 m

Total no. of 16 ф bar = 2 no.

Therefor total length of bar = 16.586 x 2 = 33.172 m

Cutting length of vertical bar

Part 1

300 + (1500 – 50 – 25 – 25) + 600 + 200 + (300 x 2) + (175 x 2) + (b¹) – bend

Explaination for [b¹ = 1500 – 75 = 1425 mm] and bend is (1 x 2 x 2.0)

= 10235 mm

= 10.235 m

Part 2

a¹ + 175 + 3000 + Ld – ( 1 x 2 x d )

Explaination for [ Ld = 50 d = (50 x 20) = 1000 mm

and a¹ = 1500 + 75 + Ld = 1500 + 75 + (50 x 20) = 2575 mm ]

= 2575 + 175 + 3000 + (50 x 20) – (1 x 2 x 20)

= 6710 mm

= 6.71 m

For 16 ф

Part 1

300 + (1500 – 50 – 25) + 600 + 200 + b – bend

Explaination for [ b = 1500 + 75 + (50d) = 2375 mm ]

= 300 + (1500 – 50 – 25) + 600 + 200 + 2375 – (1 x 2 x 16)

= 4868 mm

= 4.868 m

Part 2

a + (175 x 3) + (3000 x 3) + Ld – Bend

Explaination for [a = 1500 – 75 = 1425 mm]

= 1425 + (175 x 3) + (3000 x 3) + (50 x 16) – (1 x 2 x 16)

= 11718 mm

= 11.718 m

Cutting length of stirrups

a = 600 – (2 x 44) = 512 mm

b = 300 – (2 x 44) = 212 mm

cutting length = (2 x a) + (2 x b) + (2 x 10 x d) – (5 x 2 x d)

= (2 x 512) + (2 x 212) + (2 x 10 x 8) – (5 x 2 x 8)

= 1528 mm

= 1.528 m

Number of stirrups

Part A

= [Length/spacing] + 1

= [2700 / 100] + 1

= 28 no.

Part B

= [Length/spacing] + 1

= [2000 / 150] + 1

= 12 no.

12 x 4 = 48 no. ( i.e. Part B is in 4 part)

Part C

= [Length/spacing] + 1

= [1175 / 100] + 1

= 13 no.

13 x 3 = 39 no.

Part D

= [Length/spacing] + 1

= [650 - 30 / 100] + 1…….where 30 is clear cover

= 7 no.

Total nos. = A + B + C

= 122 no.