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
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