How to Calculate the Number of Bricks, Cement,
and Sand Required for 1 Cum of Brickworks in Nepal?
Brickwork is an
essential component in construction and is widely used in residential,
commercial, and infrastructural projects. Whether constructing a new building
or renovating an existing one, understanding how to calculate the required
quantities of materials like bricks, cement, and sand is crucial. In this blog,
we will guide you through the process of determining these quantities for 1
cubic meter (1 cum) of brickwork, specifically when using two common mortar
ratios: 1:6 and 1:4.
- A
mortar ratio (1:6) is used for 230mm OR 9 inches of wall.
- A
mortar ratio (1:4) is used for 115mm OR 4.5 inches of wall.
Key Details:
- Brick Size: In Nepal,
the standard size of a brick used for construction is typically 23 cm
(Length) x 11 cm (Width) x 7 cm (Height).
- Mortar Ratios: We will be
working with two common ratios:
- 1:6 (one
part cement, six parts sand)
- 1:4 (one
part cement, four parts sand)
 |
| Nos. of bricks |
- Volume
of 1 Brick without mortar: 23*11*7=1771 cm3 = 0.0001771= 0.001771cum
·
Let's
consider the thickness of the mortar is 10mm i.e. 1cm. Therefore, the size of the brick
with mortar will be 24*12*8cm
· Volume of 1 Brick
with mortar:
24*12*8=2304 cm3 = 0.002304 cum
· Numbers of bricks
required in 1cum = (Vol. of wall)/ (Vol. of 1 brick with mortar)
=
1/0002304
= 434.02 i.e. 434 Bricks
Add 5% wastage= 434+5% of 434 = 455.7 =
457 bricks
*Now, the net volume
of bricks will be Nos. of bricks X Vol. of 1 Brick without mortar
=
434*0.001771
=0.768614cum
*Therefore, the net volume of mortar will be:
= Vol.
of 1cum wall- Vol. of total bricks required in 1cum wall
=1-0.768614
=0.231386 cum (Wet volume of mortar)
Therefore, dry
volume of mortar =
1.33* wet vol. of mortar
= 1.33*0.231386
=0.30774338cum (dry volume of
mortar)
Step 2:
Calculating the quantity of Cement and Sand for mortar ratio 1:6 
Cement and Sand
In this ratio, 1 part
is for cement and 6 part is for sand thus the calculation strep will be
Required
cement= 1/ (sum of ratio) x dry vol. of mortar x density of cement
= (1/7) *0.30774338
cum *1440 kg/m3
= 0.044 cum x1440
kg/m3
= 63.31 kg
=1.27 bags (63.31/50, because 1 bag= 50 kg of cement)
*Add 2% wastage = 63.31+ 2% of 63.31 kg
= 64.58 kg = 1.3 bags
Required
sand= 1/ (sum of ratio) x dry vol. of mortar
= (6/7)
*0.30774338
= 0.264 cum
=9.32 cft. (0.264*35.314, because 1 cum = 35.314 cft)
*Add 8% wastage = 0.264+ 8% of 0.264 = 0.285
cum = 10.06 cft
Step 2:
Calculating the quantity of Cement and Sand for mortar ratio 1:4
In this ratio, 1
part is for cement and 4 part is for sand thus the calculation strep will be
Required
cement= 1/ (sum of ratio) x dry vol. of mortar x density of cement
= (1/5)
*0.30774338 cum *1440 kg/m3
= 0.0615 cum x1440
kg/m3
= 88.56 kg
=1.77 bags (88.56/50, because 1 bag= 50 kg of cement)
*Add 2% wastage = 88.56+ 2% of 88.56 kg
= 90.33 kg = 1.81 bags
Required
sand= 1/ (sum of ratio) x dry vol. of mortar
= (4/5)
*0.30774338
= 0.246 cum
=8.69 cft. (0.246*35.314, because 1 cum = 35.314 cft)
*Add 8% wastage =0.246 + 8% of 0.246 =
0.266 cum = 9.39 cft
Summary of
Quantity of Bricks, Cement, and Sand Required for 1 cum of brickwork with
mortar thickness 10mm and mortar ratio 1:6 & 1:4 (without considering the
wastage)
|
Items |
Mortar
Ratio (1:6) |
Mortar
Ratio (1:4) |
Unit |
|
Bricks |
434 |
434 |
Nos. |
|
Cement |
1.27 |
1.77 |
Bags |
|
0.044 |
0.0615 |
cum |
|
|
1.56 |
2.17 |
cft |
|
|
Sand |
0.264 |
0.246 |
cum |
|
9.32 |
8.69 |
cft |
Summary of
Quantity of Bricks, Cement, and Sand Required for 1 cum of brickwork with
mortar thickness 10mm and mortar ratio 1:6 & 1:4 (including 5% wastage in bricks, 2% wastage in
cement,t and 8% wastage in sand)
|
Items |
Mortar
Ratio (1:6) |
Mortar
Ratio (1:4) |
Unit |
|
Bricks (incl. 5% wastage) |
457 |
457 |
Nos. |
|
Cement (incl. 2% wastage) |
1.3 |
1.81 |
Bags |
|
0.0451 |
0.0628 |
cum |
|
|
1.592 |
2.218 |
cft |
|
|
Sand (incl. 8% wastage) |
0.285 |
0.266 |
cum |
|
10.06 |
9.39 |
cft |
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