How to make bar bending schedule of beam reinforcement?

TYPE II

BBS OF BEAM: TYPE – 2

BEAM MKD	BOTTOM BAR			TOP BAR			STIRRUPS
	LEFT	MID	RIGHT	LEFT	MID	RIGHT
B1	2-32ф	2-32ф + 2-25ф	2-32ф	2-25ф   + 2-25ф	2-25ф	2-25ф + 2-25ф	8 ф @ Left 100, Mid 150, Right 100 c.c

BOTTOM BAR

1^st layer

32ф = Total Length + Development Length – Bend + Lapping

= (6000 + 600 + 6000) + (50d x 2) – (2d x 2) + (50d)

= (6000 + 600 + 6000) + (50 x 32 x 2) – (2 x 32 x 2) + (50 x 32)

= 17272 mm

= 17.372 meter

For 2 nos.

= 17.372 x 2

= 34.544 meter

CUTTING LENGTH

a= L + (Ld) + Length of half of column + half of lap length – bend

a= 6000 + (50 x 32 x 1) + (300) + [(50 x 32) / 2] – (2 x 32 x 1)

a= 8.636 meter

b= L + (Ld) + Length of half of column + half of lap length – bend

b= 6000 + (50 x 32 x 1) + (300) + [(50 x 32) / 2] – (2 x 32 x 1)

b= 8.636 meter

2^nd Layer (Bottom Extra Bar)

25ф = L – ( 2 x (L/7) )

= 6000 – ( 2 x (6000/7) )

= 4285.7 mm

= 4.285 meter

For 4 nos.

= 4.285 x 4

= 17.14 meter

TOP BAR

1^st Layer

25ф = Length + Ld – Bend + Lapping

25ф = (6000 + 600 + 6000) + (50 x 25 x 2) – (2 x 25 x 2) + (50 x 32 x 1)

25ф = 16250 mm

25ф = 16.25 meter

For 2 nos.

= 16.25 x 2

= 32.5 meter

CUTTING LENGTH

a= L + (Ld) + Length of column + (L/2) + (Lap/2) – Bend

a= 6000 + (50 x 25 x 1) + (600) + (6000/2) +[ (50 x 25)/2] – (2 x 25 x 1)

a= 11.425 meter

b= L/2 + Ld + (Lap/2) – Bend

b= (6000/2) + (50 x 25 x 1) + [(50 x 25) / 2] – (2 x 25 x 1)

b= 4825 mm

b= 4.825 meter

2^nd Layer ( Extra Over Support Bar)

a= L/3 + Ld – Bend

a= (6000 / 3) + (50 x 25) – (2 x 25 x 1)

a= 3.2 meter

for 4 nos.

= 3.2 x 4

= 12.8 meter

b= column length + [ 2 x (L/3)]

b= 600 + [2 x (6000/3)]

b= 4.6 meter

for 2 nos.

= 4.6 x 2

= 9.2 meter

STIRRUPS

Therefor,

= L/3

= 6000 / 3

= 2000 mm

No. of stirrups in zone : 1

(Length /Spacing) + 1

= (2000 / 100) + 1

= 21 nos.

Total no. of stirrups for zone : 1

Therefor,

= 21 x 4

= 84 nos.

No, of stirrups in zone : 2

(Length /Spacing) - 1

= (2000 / 150) - 1

= 12 nos.

Total no. of stirrups for zone : 2

Therefor,

= 12 x 2

= 24 nos.

Total no. of stirrups

= 84 + 24

= 108 nos.

Therefor,

a= 300 – (2 x clear cover) – (2 x half of dia.)

a= 300 – (2 x 30) – (2 x 4)

a= 232 mm

b= 600 – (2 x clear cover) – (2 x half of dia.)

b=600 – (2 x 30) – (2 x 4)

b= 532 mm

cutting length

= (2 x a) + (2 x b) + hook – bend

= (2 x 232) + (2 x 532) + (10 x 8 x 2) – (3 x 8 x 2) – (2 x 8 x 3)

= 1592 mm

= 1.592 meter

For 108 nos

= 1.592 x 108

= 171.93 meter


NOTE

90 degree = 2d

135 degree = 3d