then add 60% extra for voids & cavities in case of plaster

then total volume of work = 1 + 60/100 x1 =1.6 cum
quantity of cement in cum = 1.6 / (1+6)= 1.6/7=0.228 cum
but as we know cement is available in market in 50 kg
bags (volume of 50kg cement is = 0.0347 cum)

thu quantity of cement in bags = 0.228/0.0347= 6.57 bags

quantity of sand in cum= 0.228 x 6 =1.36 cum

eg. if cement mortar ratio is ( 1:4)

suppose 1 cum of cement mortar work is to be done

then add 60% extra for voids & cavities in case of plaster

then total volume of work = 1 + 60/100 x1 =1.6 cum
quantity of cement in cum = 1.6 / (1+4)= 1.6/5=0.32 cum
but as we know cement is available in market in 50 kg
bags (volume of 50kg cement is = 0.0347 cum)

thu quantity of cement in bags = 0.32/0.0347= 9.22 bags

quantity of sand in cum= 0.32 x 4 =1.28 cum

this is exact procedure.you can calculate for any ratio of
cement & mortar

Calculation of materials for 1cum of 1:6 Cement Mortar by
Volumetric 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
Volume of Cement = (50/1440) = 0.03472cum
Volume of Sand mixed = 6 X 0.03472 = 0.208cum (contains 5%
moisture)
Volume of only sand = 95% of 0.208 = 0.198 cum
Absolute volume of Cement = 50kg / (3.15X1000) = 0.016 cum
Absolute volume of sand = Weight of sand / (2.6 X 1000) =
(0.198 X 1550) / (2.6 X 1000) = 0.118 cum
Let volume of water added = 30litres = 0.03 cum
Moisture present in the sand = (0.208 - 0.198) = 0.01 cum
Total volume of water = 0.03 + 0.01 = 0.04 cum
Let, Moisture content in the mortar = 2% of yield
Let, yield = Y
(1-2%)Y = Absolute volumes of (Cement + Sand + Water) =
0.016 + 0.118 + 0.04 = 0.174 cum
Total Yield Y = 0.174 / 0.98 = 0.178 cum
1bag of 50kg Cement produces = 0.178 cum of 1:6 Cement
Mortar with 0.208cum of Sand @ 5% Moisture and 30 litres of
water.
Therefore, 1 cum of 1:6 Cement Mortar requires :
Cement = 1bag X 1/ 0.178 = 5.62 bags
Sand = 0.208 X 1 / 0.178 = 1.17 cum
Water = 30 X 1 / 0.178 = 168 .5 litres

I think Mr. kamlesh is right
The considraion of voides for mortor is 25% and rest method
is same as our conventionel menthod, but for plaster 30 %
should be increase for undulation ( uneven wall surface)
also so total dry volumn is 1 + 25% + 30% = 1.55 Cum

take an example 1:4, total ratio(1+4)is 5. assume total
volume of mortar required is 10cubic metre. this will
contain 1/5of10=2cubic metre of cement and 4/5of10=8cubic
metres of sand. density of cement as 1440kg/m3, mass of
cement will be 2x1440=2880kg and number of bags
2880/50=57.6bags. note
(i) 50kg is mass of 1 bag of cement.
(ii)other irregularities like covering voids, reduction in
volume/shrinkage due to setting of cement etc should be
considered by factoring the total volume by say 1.4.
(iii)for water use about 30% of the raw volume of sand and
cement.

the unit weight of cement is 1440kg/m3.
for 1 m3 of 1:1 cement mortor the reuirement of sand is 1m3
and cement will be 1part of sand that is 1 m3 thus weight
of cement required is 1x1440kg/m3=1440 kg.
for 1 m3 of 1:2 cement mortor the reuirement of sand is 1m3
and cement will be 1/2 part of sand that is 0.5 m3 thus
weight of cement required is 0.5x1440kg/m3=720 kg.
for 1 m3 of 1:4 cement mortor the reuirement of sand is 1m3
and cement will be 1/4 part of sand that is 0.25 m3 thus
weight of cement required is 0.25x1440kg/m3=360 kg.
for 1 m3 of 1:6 cement mortor the reuirement of sand is 1m3
and cement will be 1/6 of part sand that is 0.5 m3 thus
weight of cement required is 1/6x1440kg/m3=240 kg.

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

Please can anybody let me know how to calculate cement,
sand & aggregete content for one meter cube of concrete.
please if possible provide calculations for cement in no.
of bags, sand & aggregate in cu.mt.
Thanks.