Calculation Of Quantity Of Cement, Sand And Brick In Wall
Assuming
1- Grade of mortar = 1:6 (cement : sand)
2- Class A brick (19 cm x 9 cm x 9 cm)
3- Volume of brick work = 1.0 m^3, (wall size =1*1*1 =1.0 m^3)
4- Thickness of mortar = 10 mm
Step 1 :- Calculation of bricks
No. of bricks = (volume of brick work / volume of one brick with mortar)
Volume of one brick (without mortar) = .19*.09*.09 = 0.001539 m^3
since thickness of mortar = 10 mm (0.01 m)
Volume of brick with mortar = (0.19+0.01) x (0.09+0.1) x (0.09+0.1) = 0.2×0.1×0.1 = 0.002 m^3
Number of brick required in brick work volume = 1.0 m^3
therefore,
No.of bricks = 1.0/ (0.002) = 500 Nos.
consider the percentage of waste as 10 % or 15 %
Total no. of bricks = 500 + (10 x 500 )/100=550
Step 2 :- Calculation for quantity of mortar
Since we need 500 no of bricks
volume occupied by bricks = No of bricks x volume of one brick
Volume of bricks = 500 x 0.001539 = 0.7695 m^3
Volume of mortar = Volume of brick work – Volume of Bricks
therefore,
volume of Mortar = 1.0 – 0.7695 = 0.2305 m^3
Step 3:- Calculation for Quantity of Cement
Cement = (dry volume of mortar x Cement ratio)/ sum of the ratio,s(proportion)
Dry volume of Mortar = 1.33 x 0.2305 = 0.3065 m^3 ( 33% increment due to volume shrink after water addition )
Vol. of cement= (0.3065 x 1) / (1+6) = 0.3065/7= 0.043795 m^3
cement (kg) = 0.043795 x 1440 = 63.0648 kg
No. of bags = 63.0648 / 50 = 1.26 bags approximately 2 bags
Step 4 :- Calculation for Quantity of Sand
Sand = (dry volume of mortar x Sand ratio)/ sum of the ratio,s(proportion)
Sand = 0.3065*6/7 = 0.2627 m^3
or
sand = cement volume x 6 ( since 1:6 grade of mortar)
Sand = 0.043795 * 6 =0.2627 m^3
sand = 0.2627 x 1920 = 504.384 kg /1000 = 0.5043 tonnes or
sand = 0.2627 x 35.3147 = 9.27 cft or
sand = 0.2627 x 0.354 = 0.0930 brass
So, For
- Grade of mortar = 1:6 (cement : sand)
- Class A brick (19 cm x 9 cm x 9 cm)
- Volume of brickwork = 1.0 m^3
- Thickness of mortar = 10 mm
Number of Bricks = 500
Cement = 1.26 bags of 50 Kg
Sand = 0.2627 m^3
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