How to Calculate quantity of Materials for 9″ Thick Wall
Quantity of Bricks
To calculate the Quantity of cement & sand required for brickwork,
Let us take the example of
Brick size = 9”x4.5”x3”
The thickness of wall = 9”
The thickness of mortar = 0.39”
Brick size with mortar= (9+0.39)x(4.5+0.39)x(3+0.39) = 9.39”x4.89”x3.39”
No. of bricks = Volume of Brickwork/ Volume of 1 Brick with mortar
Volume of one Brick = (9/12)x(4.5/12)x(3/12)= 0.75’x0.375’x0.25’ = 0.070 CFT
Volume of one Brick with mortar = 0.783’x0.408’x0.283’=0.090 CFT.
Volume of Brickwork = 10’ x 6’ x 0.75’ = 45 Cubic Foot (CFT)
No. of bricks = Volume of Brickwork/ Volume of 1 Brick with Mortar
∴ No. of bricks = 45 CFT/ 0.090 CFT
No of Bricks = 500
Brick wastage–> 5 to 10 %
Consider 10% brick wastage,
∴ No. of bricks= 500 + (500 x 0.1)
= 550 Number
Quantity of cement:-
Quantity of mortar = Quantity of brickwork – Volume of Bricks
Volume of bricks =0.070 CFT x 500 No’s = 35 CFT
Quantity of mortar = 45 – 35 = 10 CFT
Mix Ratio –> 1:5
Dry volume of mortar = Wet volume x (1+0.33) (33% of volume)
Dry Volume of mortar= 10 x 1.33 = 13.3 CFT
Consider 5% wastage of mortar.
Dry volume of mortar = 13.3 + (13.3 x 0.05) = 13.3+0.665 = 13.965 ≊ 14 CFT
Quantity of Cement = {(Dry Volume of mortar x Cement ratio) / (Sum of the ratio)}
∴Quantity of cement = (14 x 1) /(1+5)=2.333 CFT
Density of Cement = 1440 kg/m³
1 m³=35.3147 Cubic Feet (CFT)
∴ Quantity of Cement = 2.333/35.3147 = 0.066 m³
∴ Weight of Cement = 1440 x 0.066 = 95.04 Kg
1 bag of cement contains 50 kg of cement
∴ Number of bags = 95.04 Kg / 50 kg = 1.90 No’s (Say 2 No of bags)
Quantity of Sand:-
Cement : Sand :: 1:5
Quantity of Sand = Quantity of Cement x 5
∴ Quantity of Sand = 2.333 x 5= 11.665 CFT
1 m³=35.3147 Cubic Feet (CFT)
∴ Quantity of sand = 11.665/35.3147 = 0.330 m³
Density of sand = 1600 kg/m³
∴ Weight of the sand= 0.330 x 1600= 528 kg (0.528 tonnes)
Therefore- all materials required for 1 Cubic Meter Brick is,
Numbers of Brick- 500
Cement required –95 kg
Sand Quantity – 0.330 Cubic Meter
