The Dimension Of This Slab
Length – 6m or 20 feet
Breadth – 6m or 20 feet
Depth – 0.150m or 0.5 feet
Grade of concrete – M20
M20 Mix Ratio = 1 : 1.5 : 3
The Mix Ratio Denotes The Following
1 – Cement
1.5 – Sand (Fine aggregate – 1.5 Times of Cement Quantity)
3 – Blue metal (Coarse aggregate – 3 Times of Cement Quantity)
Volume of concrete we need = length x breadth x depth =
L x b x h = 6 x 6 x 0.15 = 5.4 m3
OK
First, we can find the volume of concrete for 1m3 so we can use on different volume
The volume of 1m3 concrete
We find out the volume of concrete in wet condition 1m3 (After the consolidation of Cement + sand + water). So in dry conditions, we have to add 30 to 35% sand bulkage and add 20% for wastage.
Accounting bulkage of sand - 34%
Accounting wastage of material -20%
Hence 1 cum becomes 1.54 (1+0.34+0.2) cum
Total Part of the Concrete = 1+1.5+3 = 5.5 Parts
Calculation of Volume of Cement in 1m3 of Concrete
Cement Quantity= (Cement Part / Concrete Parts) x Concrete Volume
Cement part = 1 (1:1.5:3)
Concrete parts = 1+1.5+3 = 5.5 Parts
Density of cement - 1440 kg/cum
Cement = (1/5.5) x 1.54 = 0.2798
Density of Cement = 1440 kg/ m3 = 0.2798 X 1440 = 403.19 kg
1 Cement Bag Weight = 50 Kg
Hence 403.29/50 = 8.06 bags
LCETED TRIVIA
To be simple
Volume of,
One cement bag = 0.0347 cum
One cement bag = 1.225 cft
Hence Required cement = ((1/5.5) x (1.54))/(0.0347) = 8.06 bags
Calculation of The volume of sand in 1m3 of Concrete
Sand Quantity= (sand Part / Concrete Parts) x Concrete Volume
Sand part = 1.5 (1:1.5:3)
Concrete parts = 1+1.5+3 = 5.5 Parts
Density of sand = 1500–1550 kg/cum
Sand = (1.5/5.5) x 1.54 = 0.42
Density of Cement = 1500 kg/ m3 = 0.42 X 1500 = 630 kg
But we all are buying sand in cft
Therefore,
1 Cum = 35.31 Cft
Sand Quantity = 0.42
= 0.42 x 35.31= 14.83 Cft
100 cft = 1 Unit
2.83 cum = 1 unit
1 load - 4 unit
Generally on lifting for sand, 1 unit sand equals 35 cement bags of sand
1 cft box = 1 x 1 x 1.25 ft
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Calculation of Volume of aggregate in 1m3 of Concrete
Aggregate Quantity = (sand Part / Concrete Parts) x Concrete Volume
Aggregate part = 3 (1:1.5:3)
Concrete parts = 1+1.5+3 = 5.5 Parts
Density of aggregate = 1520 - 1680 kg/cum
Aggregate = (3/5.5) x 1.54 = 0.84
But we all are buying aggregate in cft
Therefore,
1 Cum = 35.31 Cft
Aggregate Quantity = 0.82
= 0.84 x 35.31 = 29.66 Cft
Total Materials Required for 1m3
Cement - 8 bags
Sand - 14.83 Cft
Aggregate - 29.66 Cft
Therefore,
We need the volume of concrete for 5.4m3
Cement = 5.4 x 8 ( 8 bag of cement for 1m3) = 43.2 – 43 bags
Sand = 5.4 x 14.83 (14.83 cft of sand for 1m3) = 80.08 cft
Aggregate = 5.4 x 29.66 (29.66 cft of aggregate for 1m3) = 160.164 cft
Total Materials Required for 5.4m3
Cement - 43 bags
Sand – 80.08 Cft
Aggregate – 160.164 Cft
Smaller size aggregate's price little high than the larger size
Generally, we use 40 mm aggregate instead of 20mm in PCC for cost-effective
Calculation of Volume of water in 1m3 of Concrete
Water cement ratio 0.5 for a bag = 0.5 x 50 ( weight of a bag) = 25 litre
Therefore,
We have 43 bags so 25 litre x 43 = 1075 litre
1 drum = 200 (208) litre
So,
For 43 bags = 1075/208 = 5.16 drums
As LCETED civil engineer, you should know how to use the Excel sheet to find this quantity. so use this sheet for your uses
You Can Change This Value As Per Your Specification
You Can Change This Value As Per Your Specification
Download it by clicking the link
LCETED TRIVIA
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