# How to find out the quantity of cement and sand in plastering work

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In this construction video tutorial you will learn some useful tips which will help you to calculate the quantity of cement and sand in a plastering work.

The proportion of mortar is taken as 1:6 i.e. one part cement and six part sand.

The density of plaster is taken as 12 mm i.e. 0.012 m

The density of plaster is generally recommended as follow :-

12 mm – Toward plan surface of brick wall or for interior walls.

15 mm – Toward rough surface of 9” or 4” brick wall.

20 mm – Toward two layer of plaster normally applied for exterior walls.

Density of cement = 1440 kg/m^{3}

1 m3 is equivalent to 35.3147 ft^{3}

The density of the wall is 1 m x 1 m and the thickness of plaster is 12 mm.

Volume of mortar will be determined as 1 x 1 x .012 = 0.012 m^{3} (wet volume)

To covert wet volume to dry volume, the following formula is used :-

0.012 x 1.30 = 0.0156 m^{3}

In dry volume, the mortar is made of cement and sand and it is 30 to 35% more than wet volume

i.e. 1.30 x wet volume

Now, dry volume for concrete is formed with cement, sand and aggregate i.e. 1.54 x wet volume

We already know that dry volume of mortar = 0.0156 m^{3}

Mortar ratio = 1:6 = 1+6 = 7

Density of cement = 1440 kg/m^{3}

The quantity of cement = 1/7 x 0.0156 x 1440 = 3.21 kg

In bags = 3.21/50 = 0.064 bags = 1 bag. (as one cement bag = 50 kg)

The quantity of sand will be determined as follow :-

6/7 x 0.0156 = 0.0133 m^{3}

To covert to cubic feet = 0.0133 x 35.3147 = 0.47 ft^{3} (as 1 m^{3} = 35.3147 ft^{3})

To get online demonstration, watch the following video tutorial.

Video Source TECH@7K

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