## 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 into 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. we have total bar six 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 = 2Ï€r = Ï€d

(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 = D**

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