Q: For a given plan find
(i) No. of bricks
(ii) C:M required for Brick work with ratio 1:6
D = 1.0 X 2.10
Height of Ceiling = 3.0 m
Solution:

For External wall or 20 cm wall
Length of wall (L ) = 10.40 x 2 + 4.0 x 2 = 28.80 m
L = 28.80 m
B = 0.20 m
D = 3.0 m
Volume of brickwork = Length x width x Thickness of wall
= 28.80 x 3.0 x 0.20
= 17.28 m³
 For internal wall or 10 cm wall
Length of wall (L ) = 4.0 m
L = 4 m
B = 0.20 m
D = 3.0 m
Volume of brickwork = Length x width x Thickness of wall
= 4.0 x 3.0 x 0.10
= 1.2 m³
∴ Total volume brickwork = 1.2 + 17.28 = 18.48
Deduction of Door = L x B x D = 1.0 x 2.10 x 0.20 = ‐0.42 m³
Deduction of Internal Door = L x B x D = 1.0 x 2.10 x 0.10 = 0.21 m³
Deduction of Door Lintel = L x B x D = 1.20 x 0.10 x 0.20 = ‐0.024 m³
Since, Breadth of Lintel = thickness of wall = 0.20 m
Deduction of Internal Door Lintel = L x B x D = 1.20 x 0.10 x 0.10 = 0.012 m³
Since, Breadth of Lintel = thickness of wall = 0.10 m
Total Volume of brick work after deductions = 18.48 ‐ 0.42 ‐ 0.024 – 0.21 0.012= 17.814 m³
Assuming:
 Brick size = 19cm x 19cm x 19cm
 Thickness of mortar = 10 mm
Quantity of Bricks:
No. of bricks = Volume of brickwork
volume of 1 brick with mortar
= 17.814/0.002
∴ No. of bricks = 8907 bricks
Volume of 1 Brick with mortar = 0.20×0.10×0.10=0.002 m³
CM required for brickwork:
Volume of total no.of bricks = (0.19×0.09×0.09) x 8907
= 0.001539 x 8907
= 13.7079 m³
Quantity of mortar = Quantity of brickwork – Volume of bricks
Quantity of mortar = 17.814 – 13.7079
= 4.1061 m³
Mix Ratio –> 1:6
Dry volume of mortar = Wet volume x 1.33
∴Dry volume of mortar = 4.1061 m³x 1.33
= 5.4611 m³
Quantity of cement:
Quantity of Cement = Dry Volume of mortar x Cement ratio
Sum of the ratio
∴Quantity of cement = (5.4611 x 1) /(1+6)
= 0.7802 m³
Density of Cement = 1440 kg/m³
∴ Weight of Cement = 1440 x 0.7802
= 1123.488 Kg
1 bag of cement contains 50 kg of cement
∴ Number of bags = 1123.488 Kg / 50 kg
= 22.4697 No’s
Quantity of Sand:
Cement : Sand :: 1:6
Quantity of Sand = Quantity of Cement x 6
∴ Quantity of Sand = 0.7802 m³x 6
= 4.6812 m³
1 m³=35.3147 Cubic Feet (CFT)
∴ Quantity of sand = 4.6812 x 35.3147
= 165.315 CFT
Density of sand = 1650 kg/m³
∴ Weight of the sand = 4.6812 x 1650
= 7723.98 kg
=7.7239 tonnes
Watch the video given in the link below, for better understanding and detailed explanation
Easy method to CALCULATE NO OF BRICKS AND QUANTITY OF CEMENT & SAND in BRICKWORK
Click here to download pdf of “How to Calculate Quantity of Bricks, Cement, Sand for Brickwork in a Building”
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