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

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.

in cement morter the void between sand and cement is less
and take 25% total wet volume of water.take 10 cu m wet
volume of morter increse 25% than total volume of dry
material 12.5 cu m . if we find the material in 1:6 morter
than 12.5/(1+6)
cement=1.78 cu m sand=1.78*6=10.71cu m

For planter 20mm thickness
area 100m2
= 100*.02= 2m3
20% for wide of brick work & westage.
=2*.2= 2.4m3
Bluk 1.3
=2.4*1.3=2.99m3
Ratio
1:2
=1/3*2.99=.996m3(cement)
For bag .996/.03496=28.45bagscement
for slarry 1sqmtr =2kg cement.
So 100sqmtr 200kg/50 =4 bags
total cement qty =28.45+4= 32.45 bags cement
Sand
=2.99*2=5.98m3 sand
Note: this qty condider for only for plaster.

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

Say cement sand ratio is 1:6
calculate the quantity of mortar in 1m2 of brickwork, say
it is 0.01912.
just devide by the ratio considering 60% for wastage, say
0.01912/1*0.6=0.01147 volume of cement in the mortar.
Divide the total volume of cement by 0.0345 which s the
volume of 1 bag, say 0.01147/0.0345=0.33257/m2*number of
m2, say 100m2=34bags.

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.

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

As per thumb rule cement required per 1 cum of M20 concrete
is 411Kg/cum(1.57/5.5=0.285cum of cement i.e
0.285*1440=411Kg).But as per PWD rate analysis cement
required is 347Kg/cum.the difference is 64Kg.This kind of
difference is coming every grade of concrete.Can you give
perfect suggestion regarding this.

plz tell.. how can we find weight(in kg) of 1 cum of m20
mix..? how can we find the ratio of cement sand and
agregate on site i.e if we given some amount of m20 mix
then hw we find the ratio..?