All about civil construction knowledge- PARAM VISIONS

How to calculate the quantity of materials required for cement plastering?/ Estimating the volume of cement & sand in plasterwork.

 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 × wet volume

    = [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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