HOW TO CALCULATE THE QUANTITY OF STEEL IN COLUMN (BBS) -lceted LCETED INSTITUTE FOR CIVIL ENGINEERS

## HOW TO CALCULATE THE QUANTITY OF STEEL IN COLUMN (BBS)

EXAMPLE:
We have a column. The height of the column is 3 m and having a cross-sectional area is 500 x 400 mm. Six bars are going to use having a diameter of 16 mm. The diameter of the stirrup is 8 mm and having a spacing @ 150 mm and @200 mm at L/3 respectively.

GIVEN DATA:

Height = 3 meter

Cross section = 500 x 400 mm

No of vertical bars = 6 nos.

the diameter of the vertical bar = 16 mm.

Diameter of stirrup = 8 mm.

stirrups center to center spacing = @150 or @ 200 mm.

SOLUTION:
The calculation was to proceed in two steps.

·       Longitudinal bars calculation
·       Cutting length of stirrups

STEP 1: LONGITUDINAL BARS CALCULATION

Length of 1 Rod = H + Ld

H = Height

Ld = development length (50d)

Length of 1 Rod

= 3000mm + 50d (where d is dia of the bar)

= 3000 + 50(16)

= 3000 + 800

= 3800mm or 3.8m

Total length

The length one vertical bar is 3.8m. as we have six vertical bars

= 6 x 3.8

Total length = 22.8 m bar of 16mm needed

STEP 2: CUTTING THE LENGTH OF RECTANGULAR STIRRUPS

The cross-sectional area of the column is 500 mm x 400 mm

Take it has,

500mm is side A
400mm is side B

Minimum clear cover to the reinforcement: 15 mm to the bars in slabs, 25 mm to bars in beams and columns. In large columns, say 450 mm in thickness, the cover should be 40 mm.

 SIDE A SIDE B A =500 – 2 Side clear cover A = 500  – 2 x clear cover A = 500  – 2 x 40 A = 500  – 80 A = 420 mm B =400 –  2 x Top & Bottom cover B = 400 – 2 x clear cover B = 400 – 2 x 40 B = 400 – 80 B = 320 mm

NO OF STIRRUPS REQUIRED

As mentioned spacing @ 150 mm and @200 mm at L/3 respectively.

Formula = L/3 +1

L/3= 3000/3

=1000 mm or 1 m

NO OF STIRRUPS NEEDED IN TOP ZONE

=1000 / 150

= 6.67 nos say 7 nos

NO OF STIRRUPS NEEDED IN BOTTOM ZONE

=1000 / 150

= 6.67 nos say 7 nos

NO OF STIRRUPS NEEDED IN MID ZONE

= 1000/200

=5 nos

TOTAL STIRRUPS NEEDED = 7+5+7 =19 nos

Formula = 19 +1=20 nos

NO OF STIRRUPS REQUIRED = 20 nos

CUTTING LENGTH OF ONE STIRRUP

Formula:

Cutting Length of Stirrups = Perimeter of Shape + Total hook length – Total Bend Length

LCETED TRIVIA

Perimeter Of Shape
Perimeter of Rectangle = 2 ( length + breadth)
Perimeter of Square = 4 x side length
Perimeter of circle or Circumference of Circle = 2πr = πd
(r= radius, d= Diameter of Circle)
Total hook Length
1 Hook length = 9d or 75mm
Total Bend Length
45° Bend length = 1d
90° Bend length = 2d
135° Bend length = 3d (Remember, d = Diameter of Bar)

Cutting Length of rectangular Stirrup = 2 ( length + breadth) + 2 numbers of hooks – 3 numbers of 90° bends – 2 numbers of 135° bends

= 2(a+b) + 2(9d) – 3(2d) – 2 (3d)

= 2(500+400) + 2(9x8) – 3(2x8) – 2(3x8)

= 2(900) + 2(72) – 3(16) – 2(24)

= 1800 + 144 – 48 – 48

= 1848mm say as 1.84m

We have a total of 20 nos of stirrups, which are going to use,

Total length

= 20 x 1.84

= 36.8 m long 8 mm bar

CUTTING THE LENGTH OF RECTANGULAR STIRRUPS = 36.8 m

RESULT:

TOTAL LONGITUDINAL BARS (6 nos) = 22.8 m bar of 16mm

CUTTING THE LENGTH OF RECTANGULAR STIRRUPS =
36.8 m of 8mm

As we know the formula for weight calculation of steel bar for 1 meter is = D2/162 To know more click this link

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