Answer Posted / guppi chandra sekhar, au colle

Calculation of materials for 1cum of 1:6 Cement Mortar by
Weighing method:
Specific Gravity of Cement = 3.15 ; Density of Cement =
1440kg/cum
Specific Gravity of Sand = 2.6 ; Density of Cement =
1550kg/cum @ 5% moisture
First of all we have to calculate yield of 1 cement bag
(50kg) mixed in 1:6
Weight of Cement = 50kg
Weight of sand mixed = 6 X 50 = 300kg (contains 5%
moisture also)
Approximate volume of sand = 300 / (2.6 X 1000) = 0.12 cum
(contains 5% moisture also)
Weight of only sand = 95% of 300kg = 285 kg
Absolute volume of Cement = 50kg / (3.15X1000) = 0.016 cum
Absolute volume of sand = Weight of sand / (2.6 X 1000) =
285 / (2.6 X 1000) = 0.11 cum
Let volume of water added = 25 itres = 0.025 cum
Moisture present in the sand = 300 - 285 = 15kg = 15
litres = 0.015 cum
Total volume of water = 0.025 + 0.015 = 0.04 cum
Let, Air content in the mortar = 2% of yield
Let, yield = Y
(1-2%)Y = Absolute volumes of (Cement + Sand + Water) =
0.016 + 0.11 + 0.04 = 0.166 cum
Total Yield Y = 0.166 / 0.98 = 0.17 cum
1bag of 50kg Cement produces = 0.17 cum of 1:6 Cement
Mortar with 0.12 cum (equals to 300kg) of Sand @ 5%
Moisture and 25 litres of water.
Therefore, 1 cum of 1:6 Cement Mortar requires :
Cement = 1bag X 1/ 0.17 = 5.88 bags
Sand = 300 X 1 / 0.17 = 1765kg (or) equivalent to 0.71 cum
by volume
Water = 25 X 1 / 0.17 = 147 litres

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