Project on Concrete Mix Design for various grades of Concrete
Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
concrete mix design of M35 grade with fly ash
lets consider,
grade M35
cement OPC 43 grade (IS 8112)
mineral admixture FLY ASH (IS 3812 part 1)
max nominal size of aggregate 20mm
min cement content 340 Kg/m3 ( table 5 IS 456:2000 )
max W/C ratio is 0.45 (table 5 IS 456:2000)
workability 100mm slump value for pumpable concrete
exposure condition ( very severe) for reinforced concrete
crushed angular aggregates
admixture superplasticizer
sp.gravity of cement is 3.15
sp.gravity of fly ash 2.2
sp.gravity of FA 2.74
sp.gravity of CA 2.74
CA and FA is of grade 1
target strength
f'ck=fck+1.65s
=35+1.65(5) , where s = 5 for M35 mix
= 43.25 N/mm2
selection of W/C ratio
from graph 1 of compressive strength vs water cement ratio (IS 10262:2009)
for compressive strength of 43.25 N/mm2 is 0.37
from graph 2
for 43.25N/mm2 is 0.42
and from environmental exposure
for very severe condition is 0.45
the lower value is concidered as per experience the value is taken as 0.4
selection of water content
from table 2 of IS 10262:2009
for nominal size of 20mm aggregate is 186 kgs
for slump value of 100mm
then = 186+(6% of 186)
= 197.16 kg
but here superplasticizer is used so the water content reduced by 29%
so we take
= 197.16 - ( 29% of 197.16)
=139.98 kg of water content is added
calculation of cement content
from table 5 of IS 456:2000
min content of cementatious material for very sever condition is 340 kg/m3
then, W/C=0.4
cementatious material = 139.98/0.4
=349.96 kg/m3
max value is taken i.e 350kg/m3
for fly ash
increse the cementatious material of 10% by experience
then cementatious material content = 350 + (10% of 350)
= 385 kg/m3
so W/C ratio= 140/385 = 0.364
then 30% of flyash @total cemantatious material is
30% of 385 = 115 kg/m3 of fly ash
therefore total opc cement is 385-115= 270kg/m3 of cement
volume of CA and FA
from table 3 IS 10262:2009
the vol of CA is incresed to decrese the FA, as the w/c ratio is lowered by 0.10 the proportion of volume of CA is incresed by 0.02
therefore the vol of CA for w/c of 0.4 is 0.62
hence it is pumpable concrete volume of CA is reduced by 10%
then, 0.62 - ( 10% of 0.62) = 0.56
vol of CA is 0.56
vol of FA is 1-0.56 = 0.44
mix calculation
volume of concrete prepared =1 m3
vol of cement = mass of cement / sp.g of cement *1/1000= 270/3.15*1/1000 =0.086m3
vol of fly ash = mass of fly ash / sp. g of flyash * 1/1000 = 115/2.2 * 1/1000 = 0.052
vol of water = mass of water / sp.g of water * 1 / 1000 = 140/1 * 1 / 1000 = 0.140m3
vol of admixture = mass of admixture / sp.g of admixture * 1 / 1000 =7.7/1.145 * 1/1000= 0.00672 m3
vol of aggregate CA and FA =[ 1 m3 of concrete- ( vol of cement + vol of flyash+ vol of water+vol of admixture)] =[1-(0.086+0.052+0.140+0.00672)] = 0.715 m3
mass of CA = vol of aggregate * vol of CA * sp.g of CA * 1000
= 0.715 * 0.56 * 2.74 * 1000 = 1097.52 kg/m3
mass of FA = vol of aggregate * vol of FA * sp.g of FA * 1000
=0.715 * 0.44 * 2.74 * 1000 = 862.004 kg/m3
mix proportions are :
cement :270 kg/m3
flyash : 115 kg/m3
water : 140 kg/m3
FA : 862.004 kg/m3
CA: 1097.52 kg/m3
admixture : 7.7 kg/m3
w/c ratio is 0.364
mix ratio is : 1 : 2.24 : 2.85
MIX DESIGN OF M50 GRADE without flyash
data required
grade M50
max nominal size of aggregate 20mm
exposure severe
min cement content 320kg/m3
workability 75mm
pumpable concrete
angular aggregates
chemical admixture super plasticizers used
sp.gravity of cement 3.15
sp.gravity of FA 2.74
sp.gravity of CA 2.74
CA and FA is of grade 1
Target strength
f'ck = fck + 1.65s
s= 5 , for M50
f'ck = 50+1.65 * 5
=58.25 N/mm2
selection of water cement ratio
for sever exposure from table 5, IS 456: 2000
i.e 0.45
from experience we take 0.4
selection of water content
Max water content for 20mm aggregate is 186kgs
for slump of 75mm
= 186 + (3% of 186)
= 191 kgs
and use of super plasticizers 29% of water is reduced
= 191 - ( 29% of 191)
= 135.61kgs
calculation of cement content
W/C = 0.4
C = 135.61/0.4
= 339.025 kg/m3
proporationg of Vol of CA and FA
from table 3 IS 10262:2009
the vol of CA is incresed to decrese the FA, as the w/c ratio is lowered by 0.10 the proportion of volume of CA is incresed by 0.02
therefore the vol of CA for w/c of 0.4 is 0.62
hence it is pumpable concrete volume of CA is reduced by 10%
then, 0.62 - ( 10% of 0.62) = 0.56
therefore, vol of FA is = 1- 0.56 = 0.44
mix calculation
volume of concrete prepared =1 m3
vol of cement = mass of cement / sp.g of cement *1/1000= 339.025/3.15*1/1000 =0.107m3
vol of water = mass of water / sp.g of water * 1 / 1000 = 135.61/1 * 1 / 1000 = 0.135m3
vol of admixture = mass of admixture / sp.g of admixture * 1 / 1000 =6.7/1.145 * 1/1000= 0.00592 m3
vol of aggregate CA and FA =[ 1 m3 of concrete- ( vol of cement + vol of flyash+ vol of water+vol of admixture)] =[1-(0.107+0.135+0.00592)] = 0.752m3
mass of CA = vol of aggregate * vol of CA * sp.g of CA * 1000
= 0.752 * 0.56 * 2.74 * 1000 = 1153.99 kg/m3
mass of FA = vol of aggregate * vol of FA * sp.g of FA * 1000
=0.752 * 0.44 * 2.74 * 1000 = 906.707 kg/m3
mix proportions are :
cement :339.025 kg/m3
water : 135.61 kg/m3
FA : 906.707 kg/m3
CA: 1153.86 kg/m3
admixture : 6.7 kg/m3
w/c ratio is 0.4
mix ratio is : 1 : 2.67 : 3.40