# Find out the quantity of dry material required for 50 m of cement plaster 1.20 cm thick and in 1:5

Share

Get Access to:

- Ask & get answers from experts & other users
- Play Quiz and test your skills
- Free Download eBooks, Notes, Templates, etc.
- Study Materials
- Latest Articles

Get Access to:

- Ask & get answers from experts & other users
- Play Quiz and test your skills
- Free Download eBooks, Notes, Templates, etc.
- Study Materials
- Latest Articles

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Given data: Mix ratio = 1:5 , thickness of plaster = 1.2 cm = 0.012 m Only one side of cement plaster is given i.e. 50m Let us assume another side = 1 m Therefore, the Area of cement plaster = ( 50*1) mÂ² Now, the wet volume of cement plasterÂ = area Ã— thickness of cement plaster = 50mÂ² Ã— 0.012m =0.6Read more

Given data:

Mix ratio = 1:5 , thickness of plaster = 1.2 cm = 0.012 mOnly one side of cement plaster is given i.e. 50m

Let us assume another side = 1 m

Therefore, the Area of cement plaster = ( 50*1) mÂ²

Now, the wet volume of cement plasterÂ = area Ã— thickness of cement plaster

= 50mÂ² Ã— 0.012m =0.6 mÂ³

Adding 20% mortar for unevenness, joint filling, etc.

Therefore, total wet volume of cement plaster = 0.6 +( 20% Ã—0.6)

= 0.72 mÂ³

Now, Dry volume of cement plaster = wet volume Ã— 1.35

[ here 1.35 is the safety factor ]

= 0.72 Ã— 1.35 = 0.972 mÂ³

For cement:Quantity of cementÂ = (total dry volume Ã— cement ratio)/total ratio

= (0.972Ã—1)/ (1+5)= 0.972/6 = 0.162mÂ³

Now, the weight of cement = Volume of cement Ã— density of cement

= 0.162mmÂ³ Ã— 1440Kg/mmÂ³ = 233.28 Kg

As we know,

1 bag of cement = 50Kg

So,

233.28/50 = 4.66 = 5 bags

Here, 5 bags of dry cement is used.

For sand:Quantity of sandÂ = ( Total dry volume Ã— sand ratio) / total ratio

= ( 0.972 Ã—5) /(1+5) = 4.86/6 = 0.81mÂ³

Now, the weight of sand = Volume of sand Ã— density of sand

= 0.81mmÂ³Â Ã— 1920 Kg/mmÂ³ = 1555.2 Kg

[Density of cement and sand are considered from standards]

For water :Quantity of waterÂ = 20% of total dry material

= 20 % Ã— ( wt. of cement + wt. of sand )

= 20 % Ã— ( 233.28 + 1555.2 ) = 357.696 liters

See less