Q:- For a given plan find (i) No. of bricks (ii) C:M required for Brick work with ratio 1:5.
D = 1.0 X 2.10
Height of Ceiling = 3.0 m
Solution:
Length of wall (L ) = 5.40 x 2 + 4.0 x 2 = 18.80 m
L = 18.80 m
B = 0.20 m
D = 3.0 m
Volume of brickwork = Length x width x Thickness of wall
= 18.80 x 3.0 x 0.20
= 11.28 m³
Deduction of Door = L x B x D = 1.0 x 2.10 x 0.20 = ‐0.42 m 3
Deduction of Door Lintel = L x B x D = 1.20 x 0.10 x 0.20 = ‐0.024 m 3
Since, Breadth of Lintel = thickness of wall = 0.20 m
Total Volume of brick work after deductions = 11.28 ‐ 0.42 ‐ 0.024 = 10.836 m3
Assuming:-
- Brick size = 19cm x 9cm x 9cm
- Thickness of mortar = 10 mm
Quantity of Bricks:-
No. of bricks = Volume of brickwork
volume of 1 brick with mortar
= 10.836/0.002
∴ No. of bricks = 5418 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 5418
= 0.001539 x 5418
= 8.3383 m³
Quantity of mortar = Quantity of brickwork – Volume of bricks
Quantity of mortar = 10.836 – 8.3383
= 2.4977 m³ ≅ 2.5 m³
Mix Ratio –> 1:6
Dry volume of mortar = Wet volume x 1.33
∴Dry volume of mortar = 2.5 m³x 1.33
= 3.325 m³
Quantity of cement:-
Quantity of Cement = Dry Volume of mortar x Cement ratio
Sum of the ratio
∴Quantity of cement = (3.325 x 1) /(1+6)
= 0.475 m³
Density of Cement = 1440 kg/m³
∴ Weight of Cement = 1440 x 0.475
= 684 Kg
1 bag of cement contains 50 kg of cement
∴ Number of bags = 684 Kg / 50 kg
= 13.68 No’s
Quantity of Sand:-
Cement : Sand :: 1:6
Quantity of Sand = Quantity of Cement x 6
∴ Quantity of Sand = 0.475 m³x 6
= 2.85 m³
1 m³=35.3147 Cubic Feet (CFT)
∴ Quantity of sand = 2.85 x 35.3147
= 100.647 CFT
Density of sand = 1650 kg/m³
∴ Weight of the sand = 2.85 x 1650
= 4702.5 kg
=4.7025 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
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