Let us consider a wall of 5m × 3m, that should be plastered internally.³
Let us plaster this brick wall in a 1: 4 ratio having 12mm. in thickness.
Here, length L = 5m., height H = 3m., thickness T = 12mm.
Area (A) of the wall that would be plastered
A = [L × H]
= [5m × 3m]
A = 15 sq.m.
Volume of the plaster
= [surface area of wall × plaster thickness]
= [15 sq.m. × 0.012m.]
= 0.18 cum.
Let us take 20% extra material to fill the voids, joints, uneven surfaces & for wastage.
So, the actual wet volume of the plaster
= [0.18 +( 0.18 × 20 ÷ 100)]
= [0.18 + 0.036]
= 0.216 cum.
Dry volume of the plaster mortar
= [1.33 × 0.216]
= 0.2873 cum.
The volume of cement required
= [dry volume of plaster mortar × (ratio of cement ÷ total ratio)]
= [0.2873 cum × ( 1 ÷ (1+4 ))]
= [0.2873 × 0.2]
= 0.05746 cum.
As you know, the density of cement = 1440kg/cum.
So, the total quantity of cement required in kgs.
= [0.05746 × 1440]
= 82.7424 kg
One bag of cement weighs 50kg
The number of cement bags required
= [82.7424 ÷ 50]
= 1.6548 bags.
The volume of sand required
= [dry volume of plaster mortar × (ratio of sand ÷ total ratio)]
= [0.2873 cum × ( 4 ÷ (1+4 ))]
= [0.2873 × 0.8]
= 0.22984 cum.
= 8.116 cu.ft.
( 1cu.m = 35.315 cu.ft. )
or
Volume of sand
= [volume of cement × 4]
= [0.05746 cum.× 4]
= 0.22984 cum.
Quantity of water required
= [weight of cement × w/c ratio]
= [82.7424kg × 0.55]
= 45.51 kg i.e. 45.51 liter.
Now, let us find out the volume of materials required for plastering a 100 sq.m. of the brick wall.
The volume of cement required
= [(100sq.m ÷ 15 sq.m.) × 0.05746 cu.m.]
(As we have already calculated the material requirement for the 15 sq.m. wall surface.)
= [6.6667 × 0.05746]
= 0.383066 cum.
= 551.618 kg. of cement
= 11.03 bags of cement
(By similar procedure as we did it above.)
Volume of sand required
= [0.383066 × 4]
= 1.5346 cum.
= 54.196 cu.ft.
( 1cu.m. = 35.315 cu.ft.)
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