Project on Concrete Mix Design for various grades of Concrete
- concrete mix design for M35 using fly ash
1} target mean strength f'ck= fck+ 1.65 * standard deviation
=35 + 1.65* 5
=43.25 N/mm^2
2} selection of W/C ratio = as per table 5 IS 456 MAX w/c ratio is 0.45
3} selection of water content :max water content for 20 mm aggt.= 186 lit.
estimated water content for 100 mm slump = 186+ (6/100)*186 =197 lit.
superplasticizer are used so water content reduced upto 20% and above. w/c ratio reduced 25% is achieved with plasticizer
water content = 197* 0.75= 148 lit.
4}cement and fly ash content: cementitious material (cement+ fly ash) = 148/0.45=329 kg/m^3 > 320 it is ok
fly ash is not active as that of cement,cementitious material increased by 10% based on trial
cementetious material content = 329 *1.1=362 kg/m^3
fly ash as 25%
fly ash content= (25/100)*362=91 kg/m^3
cement=362-91=271kg/m^3
5} mix calculation:
volume of concrete= 1 m^3
volume of cement= (271/3.15)*(1/1000)=0.086 m^3 , specific gravity of cement=3.15
volume of fly ash = (91/2.2)(1/1000)=0.041 m^3 , S.G. of fly ash =2.2
volume of water= 0.148 m^3
volume of chemical admixture, assuming 1.2% by weight=(1.2/100)*362=4.34kg/m^3
=(4.34/1.1)*(1/1000)=0.004 m^3
volume of aggt.={1-[0.086+0.04+0.148+.004]}=0.721m^3
weight of C.A.= 0.721*0.55*2.78* 1000 = 1102 kg/m^3
weight of F.A.= 0721*0.45*2.70*1000 =876kg /m^3
concrete mix design for M50 without fly ash:
1} target mean strength= f'ck=fck+1.65*S
=50+1.65*5=58.25 N/mm^2
2} max water cement ratio = 0.40< 0.45 ok
3}water content for 20 mm aggt. = 186 lit.
estimated water content for 100 mm slump = 186+ (6/100)*186 =197 lit.
superplasticizer used. w/c reduction of 29% is achieved
water content = 197*0.71=140 lit
4} cement content: 140/0.4=350 kg/m^3 > 320kg/m^3 ok.
5} C.A AND F.A content: volume of c.a. corrosponding to 20 mm size aggt. and f.a. (zone 1) water cement ratio = 0.6, here w/c ratio = 0.4, the volume of C.A. needs to increased to decreased the F.A. content.
w/c ratio lower by 0.10 , proportion increased by 0.02
corrected volume of C.A. for water cement ratio of 0.40= 0.62
volume of C.A. = 0.62*0.9 = 0.56 , for pumpable concrete value reduced by 10%
volume of F.A. = 1- 0.56 = 0.44
6} mix calculation:
vol of concrete= 1 m^3
vol. of cement = (350/3.15)*(1/1000)= 0.111 m^3
vol. of water = (140/1)(1/1000) = 0.143 m^3
vol of chemical admixture = (7/1.145)(1/1000) = 0.006 m^3
vol of aggt = {1-(0.111+0.140+0.006) = 0.74 3m^3
mass of C.A.= 0.743*0.56*2.74*1000 = 1140 kg /m^3
mass of F.A. = 0.743*0.44*2.44*1000 = 896 kg /m^3
mix proportion for trial 1
cement =350
water = 140
F.A.= 896
C.A.N=1140
CHM.Admixture =7
w/c ratio = 0.4
trial mix = 1:2.56:3.26