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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Thank you for going through these calculation steps❤. Have a good day 😄.
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