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**How to calculate quantity of steel and concrete in column and how to calculate quantity of cement ,sand and aggregate in column.**

**● **Description :- in column or pillar formation of building we have require Steel bar ,sand ,cement and aggregate. There is a simple question arises how to calculate quantity of material required for column formation like that quantity of cement ,quantity of steel bar, quantity of sand and quantity of aggregate.

Let we try to solve it by simple calculation method.

**What is M20 grade of concrete**

**● **M-20 mix :- for the column formation we have used concr M-20 mix :- for the column formation we have used concrete ete type of M-20 mix, it is mixture of sand ,**cement and aggregate. **

In M20 mix there is a ratio of its constituent( 1 : 1.5 :3 )are you used in which one part is cement 1.5 part is sand and 3 part is aggregate**.**

**How to calculate quantity of concrete in column**

**● Column :- there is a two section of column first one is plan section and second one is cross section**

( 1) Plane section :- plane section have given dimension like that 300 mm length and 300 mm wide

** Wide = 300 mm = 0.3 m**

** Length = 300mm = 0.3 m**

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(2) Cross section :– in cross section there is a 4 number of steel bar each of length 4 metre having Dia 12 mm are used in column formation.

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● **G****ivendimension**:

Wide = 0.3m

Length = 0.3 m

Height = 4 m

Steel bar = 4 nos ,each of 4 m lenght

Density of steel = 7850 kg /m3

Density of cement = 1450 kg/m3

Concrete type = M-20 mix

**Steel calculation for column**

We know that Steel bar calculated in kilogram , weight of Steel bar is equal to multiplication of volume of Steel bar and density of steel bar.

Weight = volume ×density

For steel weight calculation we have used one formula in Kilogram per metre.

Weight = (D)2/162 ×L kg/m

Where D= diameter of Steel bar= 12mm

L = total running length= 4×4 m =16 m

Weight = 12×12/162×16 kg

Weight of Steel for making one column

Weight = 14 .22 kg Ans.

● Wet volume = 0.3×0.3×4 m3

Wet volume=0.36 m3

**● dry volume** :- we know that in wet volume there is a water and it’s bubble it can be remove, hence require more quantity of material ,that’s why volume of material is increased that is dry volume. it is increased by 54% ,that’s why we multiply 1.54 cofactor in wet volume for calculation of dry volume

Dry volume = 1.54×0.36m3

Dry volume = 0.5544 m3

We know that concrete have also Steel bar in column, for the calculation of Net volume of concrete we should subtract volume of Steel bar from dry volume to get net volume.

Volume of steel = weight÷density

V steel = 14.22 kg/7850 kg/m3

V steel = 0.0018 m3

Net volume =V dry _ V steel

V net = 0.5544-0.0018 m3

V net = 0.5526 m3

**quantity of cement in column**

Total R = 1+1.5 +3 =5.5

Weight = Net volume×density

Quantity of cement= 1/5.5×0.5526 m3×1450 kg/m3

Weight =801.27/5.5 kg

W = 145.685 kg

We know 1bag cement = 50 kg

Number of bag of cement=145.685/50

No. of bags = 2.914 Ans.

● **quantity of sand:-**

We should calculate quantity of sand in cubic feet

Q of sand = 1.5/5.5×0.5526 ×35.3147 cft

1 m3 = 35.3147 cft

Quantity of sand = 5.32 cft

●**quantity of aggregate:-**

Also quantity of aggregate is calculated in cubic feet

Q of stone = 3/5.5×0.5526×35.3147 cft

Quantity of aggregate=10.644 cft

**●Results**:- for making one column of building we have require following material:-

Weight of steel = 14.22 kg

No. of bags cement = 2.914

Quantity of sand = 5.32 cft

Quantity of aggregate = 10.644 cft

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This is the perfect calculation, most of the website avoid steel calculation, but you have done this perfectly. Thank you

Very interesting. Thanks for this