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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