Calculate the quantity of material for load bearing wall with length 6 m thickness 40 cm and height 3.5 m preparation of mortar (1:6)
Volume of wall = base quantity
= b x d x t = l x h x t
= 6 x 3.5 x 0.4
= 8.4 m3
Standard diamension of brick with mortar = 0.2 x 0.1 x 0.1 = 0.002 m3
Number of bricks = 8.4 / 0.002 = 42.00 nos.
Actual volume of one brick = 0.19 x 0.9 x 0.9 = 1.539 x 10-3
Total volume of bricks = 4200 x 1.539 x 10-3 = 6.46 m3
Volume of mortar = 8.4 – 6.46 = 1.94
Actual volume of water required will 25 % more due to losses
= 1.94 + 25 % = 2.42 m3
Quantity of cement required
= 2.42 / 7 = 0.34 m3
1 m3 = 30 bag
0.35 x 30 = 10.5 bags say 11 bags
Quantity of sand required
= 0.35 x 6 = 2.1 m3
1 truck = 5.87 m3
Therefore 0.08 m3 = 0.35 trucks
Sr. no. | Item | Quantity | Rate | Total |
1 | Bricks | 4200 Nos. | 750/100 nos. | 3150 |
2 | Sand | 1 ½ Brick | 2000/truck | 1000 |
3 | cement | 11 bag | 110/bag | 1210 |
5,360 Rs. |
Partition wall with bricks
Find the estimation of material required for building partition wall with brick work second class, cement mortar 1:6 and base quantity 10 m3
Total volume = base quantity = 10 m3
Volume of one brick with mortar
= 0.2 x 0.1 x 0.1 = 0.002 m3
No. of bricks required = 10 / 0.002 = 5000 nos.
Actual volume of one brick
= 0.10 x 0.9 x 0.9
= 1.539 x 10-3 x 5000
= 7.69 m3 say 7.7 m3
Actual volume of mortar = 10 – 7.7 = 2.3 m3
Considering 25 % more for losses.
Actual volume of mortar required
= 2.3 + 25 % 2.3 = 2.87 m3
Quantity of cement required
=2.87 / 7 = 0.41 m3
= 0.41 x 30 = 12.3 bags say 13 bags
Quantity of sand required
= 0.41 x 6 = 2.46 m3
1 truck = 5.87 m3
2.46 m3 = ½ truck
Sr. no. | Items | Quantity | Rate | Total |
1 | Bricks | 5000 | 750/1000 | 3750 |
2 | Sand | ½ truck | 2000/truck | 1000 |
3 | Cement | 1/13 bag | 110/bag | 1430 |
6,180 Rs. |
Estimate for the material required for load bearing wall of U.C.R. masonry consider base quantity 10m3 and proportion of mortar (1:6)
Quantity of stones required = 125 % of base
Quantity = 12.5 m3
Quantity of mortar required
Considering all stones.
42 % of base quantity = 4.2 m3
Quantity of cement required
= 4.2/7 = 0.6 m3
1m3 = 30 bags
0.6m3 = 18 bags
Quantity of sand required
= 0.6 x 6 = 3.6 m3
1 truck = 5.87 m3
Therefore 3.6 m3 = 0.6 trucks = 1 truck
Sr. no. | Items | Quantity | Rate | Total |
1 | Stone | 13 m3 | 275/m3 | 3575 |
2 | Sand | 1 truck | 2000/truck | 2000 |
3 | Cement | 18 bags | 110/bag | 1980 |
7,555 Rs. |
Estimate the quantity of material required for plastering the area 100 m2. Mix proportion used 1:5. Thickness of plaster is 12 mm.
Volume of mortars = 100 x 12 / 10 x 100 = 1.2 m3
Considering 25 % more due to losses
= 1.2 + 2.5 % . 12 = 1.5 m3
Quantity of cement required
= 1.5 / 6
=0.25 m3
0.25 m3 = 0.25 x 30
7.5 bags = 8 bags
Quantity of sand = 0.25 x 5 = 1.25 m3
1 truck = 5.87 m3
1.25 m3 = 0.21 truck
= ½ truck
Sr. no. | Items | Quantity | Rate | Total |
1 | Cement | 8 bags | 110/bag | 880 |
2 | Sand | ½ truck | 2000/truck | 1000 |
1,880 Rs. |
Calculate the estimate for mosaic tiles flooring over an area of 100 m2 with 20 mm thick lime.
Size of tile = 20 cm x 20 cm
Proportion of mortar = (1:2:4)
Base quantity = area = 100 m2
Size of tile = 20 cm x 20 cm
Thickness of lime base = 20 mm
Area of one tile = 20 cm x 20 cm
= 400 cm2 = 0.004 m2
Total no. of tiles required = 100 /0.04 = 25 00 nos.
Volume of lime base = 100 x (20 / 10 x 100) = 2 m3
Considering 25 % more due to looses = 2 + 25 % .2 = 2.5 m3
Quantity of cement required = 2.5 / 7 = 0.35 m3
1 m3 = 30 bags
0.35 x 30 = 10.5 bags say 11 bags
Consider 6 bags of cement for 100 m2 surface for slurry
= total quantity = 10.5 +6 = 16.5 say 17 bags
Quantity of lime required = 0.35 x 2 = 0.7 m3
1 m3 = 30 bags
7 m3 = 21 bags
Quantity of sand required = 0.35 x 4 = 1.4 m3 = ½ truck
Sr.no. | Items | Quantity | Rate | Total |
1 | Tiles | 2500 | 750/brass | 13020 |
2 | Cement | 17 bags | 110/bag | 1870 |
3 | Sand | ½ truck | 2000/truck | 1000 |
4 | Lime | 21 bags | 840 | |
16730 Rs. |
1 brass = 144 tiles
2500 / 144 = 17.36 brass
