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Question
How to calculate cement and sand quantities in cement mortar
As 1:1,1:2,1:4,1:6 ?
 Question Submitted By :: Civil-Engineering
I also faced this Question!!     Answer Posted By  
 
Answer
# 1
the procedure to calculate quanity is given below

eg. if cement mortar ratio is ( 1:6)

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+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
 
Is This Answer Correct ?    154 Yes 41 No
Santosh Singh
 
Answer
# 2
1:4 (or) 1:6 
Is This Answer Correct ?    12 Yes 5 No
Arjun
 
 
 
Answer
# 3
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
 
Is This Answer Correct ?    11 Yes 4 No
Guppi Chandra Sekhar
 
Answer
# 4
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
 
Is This Answer Correct ?    8 Yes 6 No
V.kapil
 
Answer
# 5
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.
 
Is This Answer Correct ?    5 Yes 4 No
Osbert (civ Eng. Student Maker
 
Answer
# 6
Calculation Method
Plastering with CM 1:5 12 mm thick - 10 m2

Area to be Plastered = 10 m2

Thickness of mortar Layer = 12 mm = 0.012 m

Volume of mortar required = 10 x 0.012 = 0.12 m3

Add 15 percent allowance for undualation of surface & wastage = 0.018 m3

so, total volume of Mortar Required = 0.138 m3 say 0.14 m3

sand required for 0.14 m3 of CM 1:5 = 0.14 m3

cement required for 0.14 m3 of CM 1:5 = 0.14/5 = 0.028 m3

wt of cement required = 0.028 x 1440 = 41 kg
 
Is This Answer Correct ?    1 Yes 0 No
Prakash
 
Answer
# 7
how to calculate cement and sand quantities in cement
mortar as 1:3?
 
Is This Answer Correct ?    2 Yes 2 No
Saritha
 
Answer
# 8
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.
 
Is This Answer Correct ?    40 Yes 41 No
T.chandrasekar
 
Answer
# 9
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
 
Is This Answer Correct ?    1 Yes 2 No
Guppi Chandra Sekhar, Au Colle
 
Answer
# 10
ratio: 1:4
for 1 m3,cement= 1/(1+4)=1/5=.2m3,let it be .2+.6*.2=.32
quqntity of cement:


.32*1440(unit wt of cement)*1=460.8

bags rqd= 460.8/50(1 bag)=9.2````let be 10;


sand requred

4*.2*1=.8m3



typical one load==5m3
 
Is This Answer Correct ?    0 Yes 1 No
Dilshad Akthar
 

 
 
 
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