Let us fix the tile of **2' × 2'** size in the room of area **10 ft.× 10 ft. **

Let us use cement mortar of **1:4** ratio having **40 mm**. thickness to fix these tiles.

__ Calculation:__

** 1. The number of tiles required**

= room area ÷ tile area

= [(10ft × 10 ft.)÷ (2 ft. × 2 ft.)]

= [100 sqft. ÷ 4 sqft.]

= **25 nos.**

**The wet volume of cement mortar **

= room area × mortar thickness

= 100 sq. ft. × (40 × 0.00328084) ft.

( 1mm = 0.00328084 ft. )

= 100 sqft.× 0.13123 ft.

= **13.123 cu ft. **

Dry volume of cement mortar

= 1.33 × 13.123 cu ft

= 17.453 cu ft.

=** 0.4942 cum**.

( 1cu ft = 0.0283168 cum. )

**2. The volume of cement required**

= dry volume of floor mortar × (ratio of cement ÷ total ratio)

= 17.453 cu ft × ( 1 ÷ (1+4 ))

= 17.453 × 0.2

= 3.4906 cu ft.

= **0.09884 cum.**

( 1cuft = 0.0283168 cum. )

As you know, the density of cement = 1440kg/cum.

So, the total quantity of cement required in kgs.

= 0.09884 cum. × 1440 kg/cum.

= 142.3296 kg

One bag of cement weighs 50kg

The number of cement bags required

= 142.329 ÷ 50

= 2.84** bags.**

**3. The volume of sand required**

= dry volume of floor mortar × (ratio of sand ÷ total ratio)

= 17.453 cu ft.× ( 4 ÷ (1+4 ))

= 17.453 cu ft.× 0.8

= 13.9624 cu.ft.

= 0.3953 cum.

or

Volume of sand

= volume of cement × 4

= 0.09884 cum.× 4

= 0.3953cum.

or

Volume of sand

= total dry volume- the volume of cement

= 0.49421 cum - 0.09884 cum

= **0.3953 cum.**

Density of dry sand = 1602 kg/cum.

Weight of sand required

= 0.3953 cum × 1602 kg/cum

= **633.387 kg.**

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