Why so many bricks ???

Dear Reader ,

In my experience , the subject of estimation is quite boring to the students as well as field people.

In that context , we are going through a series of articles devoted to the arithmetic involved in construction.

We have already discussed the estimation of materials in concrete

( Read post : Vow..what a concrete )

Now we shall travel in the masonry today.

Just keep your patience when you read.

If you get bored , just Stop and take a break.

IMG-20140213-WA0007

Take a scrap pad and a calculator and do some jugglery yourselves.

That’s how you shall keep aligned .

Re-continue and reflect as many times as you feel.

There is no hurry.

Happy Calculating !

Overview:

We talked about the general aspects of the masonry in the previous post.
We also spoke about  the stone masonry

( Read : Throw the stone..carefully )

Now we shall learn the arithmetic issues involved in counting the bricks.

We shall  be able to find out the :
Number of bricks needed for the masonry work and also the mortar needed for the job.
In other words ,for a given masonry wall we shall be able to compute:

1.No of bricks 

2.No of cement bags
3.Sand

And then we can also find out the total material cost.

This shall help us in:
1.Budgeting for materials
2.Ordering the material

Ok ? so all geared up ?
let us begin..here we go:

The Data:
The brick wall which we need to build is 5 m long, 4 m high and it is 200 mm thick.
It is to be jointed in Cement mortar 1:6.
The bricks of size 200 mm x100 mm x 70 mm have been approved.

CHECK THE ACTUAL DIMENSIONS OF THE BRICKS ON SITE
CHECK THE ACTUAL DIMENSIONS OF THE BRICKS ON SITE

The method:
1.Calculate the volume of one brick :
0.2m x 0.10m x 0.07m = 0.0014 CuM.
2.Find number of bricks per CuM:
= 1/0.0014 = 714 nos.

So 714 bricks shall be needed if there is no mortar joint. Right ?

3. Now,assume that there is a mortar joint of 10 mm thickness between the  bricks.

So, increase each dimension of brick by 10 mm.

Thus,the volume of each brick with mortar shall be :
0.21 x 0.110 x 0.08 =0.001848 CuM.

4..Again, Find number of bricks per CuM:

= 1/0.0011848  = 541 nos.

NORMALLY CEMENT MORTAR CM ( 1:6 ) IS USED FOR THE BRICK WORK
NORMALLY CEMENT MORTAR CM ( 1:6 ) IS USED FOR THE BRICK WORK

5.Thus, number of bricks have reduced from 714 to 541 due to the presence of mortar.

6.Therefore, 714 – 541 = 173 Bricks represent the volume of cement mortar.

173 x 0.0014= 0.2422 CuM.
7.Summary I:

We shall need 541 bricks and 0.24 CuM of cement mortar for 1 CuM of brick work.

TRAY FOR MIXING CEMENT MORTAR at the location of brickwork
TRAY FOR MIXING CEMENT MORTAR at the location of brickwork

8.Now we also need to find out how many cement bags and how much quantity of sand is needed 

for 0.24 CuM of mortar .

Let us assume that the mortar quantity is CuM.

The mix is CM 1:6.
We shall use dry mix method.

Cement: [1.52/(1+6)]x 28.8 =  [0.2171]x 28.8 = 6.25 Bags

Sand: [0.2171] x 6  = 1.302  CuM.

Consumption of materials for 1 CuM of cement mortar (1:6):

Cement: 6.25 Bags
Sand:   1.302 CuM.

A CEMENT BAG WEIGHS 50 KG AND HAS A VOLUME OF 35 LITRES.
A CEMENT BAG WEIGHS 50 KG AND HAS A VOLUME OF 35 LITRES.
Now ,we actually need only 0.24 CuM of mortar for 1 CuM of brick work

Hence for 0.24 CuM of mortar , we shall need :
Cement: 6.25 Bags x 0.24 =1.5 bags
Sand:   1.302 CuM. x 0.24= 0.3125 CuM.

IMG-20140819-WA0010

9.Summary II:
Consumption of materials for 1 CuM of Brickwork :
Cement: 1.5 bags
Sand:     0.3125 CuM.
Bricks: 541 Bricks

10.Now we are at the final lap:
Volume of brick wall:
5 x 4x 0.2 = 4 CuM

Therefore material consumption shall be:

Cement: 1.5 bags x 4 =  bags
Sand:     0.3125 CuM. x 4 = 1.25 CuM.
Bricks: 541 Bricks x 4 = 2164 Bricks

If the unit rates are known we can also find out total cost of the materials.

RCC FRAMED STRUCTURE. [ Brickwork in full swing ]
RCC FRAMED STRUCTURE.
[ Brickwork in full swing ]
11.Assume that :

Bricks : INR 4200 per 1000. [ INR 4.2  per no ]

Sand : INR 4500 per Brass or 100 Cft.

[ INR 45 per cft or 45 x 35.29 = INR 1588 per cuM.]

Cement : INR 300 per bag of 50 Kg.

The material cost shall be as follows :

Cement:  bags@ 300 =       1800
Sand:     1.25 CuM.@ 1588 = 1985
Bricks:  2164 Bricks@ 4.2 =  9088

Total cost :1800 + 1985 + 9088

= INR 12,873 

12. Now we can find out the materials cost per CuM  of  brick work :

12,873 /4 = INR 3218.25 per CuM of brick work 

or 3218.25/35.29 = INR 91.19 per Cubic feet of brick work

BRICKS STORED AT THE SITE
BRICKS STORED AT THE SITE

Author’s note :

Dear Reader ,

Hope you have understood the basics.

1.Remember , you don’t have to do all this calculation every .

Just memorize the unit cost i.e INR 3218.25 per CuM or INR 91.19 per Cubic feet of brick work .

As long as the material rates are the same, the cost shall be the same.

2. For those comfortable with spread sheets , once do this calculation in the spread sheet with proper formulae. Later , when any of the variable changes ,the moment you revise that figure, the entire computations shall be automatically revised.

SAK.PORTRAIT.27.8.2012

Its really a fun.

believe me.

Happy journey on the path of estimations.

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2 thoughts on “Why so many bricks ???”

    1. Yogesh
      Great to see this.
      There is already an article about edge column footing.
      Have u gone through ?
      Pros and Cons ..let me post another article soon.
      Thanks for the fantastic suggestion.

      Like

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