**Example**

suppose we have a column. The height of the column is 4 m and having a cross-sectional area is 300 x 400 mm and having a 40 mm of the clear cover. 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 = 4 meter

Cross section = 300 x 400 mm

Clear cover = 40 mm.

No of vertical bars = 6 no’s.

The diameter of the vertical bar = 16 mm.

Diameter of stirrup = 8 mm.

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

BBS of Column =?

**Solution**

The calculation was to proceed into two steps.

- vertical bars calculation
- Cutting length of stirrups

**Step 1:Vertical Bar Calculation **

Length of 1 bar = H + L_{d}

#Where

L_{d} = development length

H = Height of column

Length of 1 bar

= 4000 mm + 40d <where d is dia of bar>

= 4000 + 40 x 16

= 4000 + 640

= **4640 mm or 4.640 m Ans..**

The length one vertical bar is 4.640 m. we have total bar six bars,

Total length

= 6 x 4.640

= **27.84 m long vertical bar is ****required**.

**Step 2:Cutting The Length Of Stirrups **

The cross-sectional area of the column is 300 mm x 400 mm

A: is the vertical cross-section area of the stirrup

B: is a horizontal cross-section area of the stirrup

**Calculation of length A **

A = Horizontal Distance – 2 Side clear cover

A = 300 – 2 x clear cover

A = 300 – 2 x 40

A = 300 – 80

A = 220 mm

**The length of B **

B = Vertical distance – 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 **

= 4000/3

=1333.3 mm or 1.33 m

Formula = L/3 / spacing + 1

(no of stirrups in end zone)

=1333.3 / 150

= 8.8 nos say 9 nos

There are total two zones of 150 mm spacing and one zone of 200 mm spacing.

= 2 x 9

= 18 nos (at end zones)

At Mid Zone

=1333.3 / 200

= 6.6 nos say 7 nos

Total no of stirrups

= 18 + 7

= 25 nos

**Cutting length of one stirrup **

**Formula**

= (2 x A) + (2 x B) + hook – bend

Cutting length

= (2 x A) + (2 x B) + 2 x10d – 5 x 2d

# where

hook = 10d

bend = 5 x 2d (we have 5 bends in one stirrup)

d = is diameter of bar

= (2 x 220) + ( 2 x 320) + 2 x 10 x 8 – 2 x 5 x 8

= 440 + 640 + 160 – 80

= **1160 mm or 1.16 m**

We have a total 25 nos of stirrups, which are going to use,

Total length

= 25 x 1.16

= **29 m long 8 mm bar**.

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## 3 comments

Is a very encouraged site and as well assisting researcher to achieved their aims at easy stage.

Quantity of materials required in cc in col.

If you’ll give us more brief description it will be better…