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
the concrete mix design for M35 grade
target mean strength
Fck=fck+1.65S
Where Fck equal to target mean strength and fck equal to characteristic strength and as equal to deviation factor
Fck= 35+1.65(5)
Fck= 43.25 n/mm2
water cement ratio
In is code 456 table number 5 for M35 garde water cement ratio is a 0.45 and minimum cement content is 340 kg/m3
In is code 10262:2009 Table number 2 the maximum size of aggregate is 20 mm than this 20 mm size of aggregate maximum water is 186 litre or kg for 25 to 50 mm slump
But we need 100 mm slump than increase 5% of water
required water is 186*(5/100) + 186
= 9.3+186 = 195 litre or kg
Cement content =195/0.45
=434 kg
Fly ash
in cementitious material content and its percentage may be based on experience and trial
This illustrative example is with increase of 10 percent cementitious material conten
cementitious material conten= 434*1.10 =478 kg/m3
Wate cement ratio= 195/478 =0 .40
Maximum fly ash is 30% use in Cheng with cement than
= 478*.3 = 144 kg/m3 fly ash
And cement is =478-144= 334 kg / m3
Data
Water cement ratio
A) cement=334 kg/m3
B) water=195 lit or kg/m3
C) Water cement ratio= .40
D) fly ash = 144 kg/m3, density of fly ash is 2200kg/m3
Volume for 1 m3
Volume of cement= cement content/density. = 334/3150 =0.10 m3
volume of fly ash= fly ash content/density. =144/2200 =0 .065 m3
volume of water = water content/density.
= 195/1000. =0.195
= 1-(0.10+0.065+0.195) = 0.64 m3
we are using 20 mm size aggregate and zone 4 than value is 0.66
volume of course aggregate =0.64*0.66*2740 = 1158 kg
volume of fine aggregate(sand) =0.64*0.34*2740 = 614 kg
Mix proportion data for m 35
A) cement=334 kg/m3
B) water=195 lit or kg/m3
C) Water cement ratio= .40
D) fly ash = 144 kg/m3,
E) coarse aggregate= 1158 kg/m3
F) fine aggregate= 614 kg/m3
the concrete mix design for M50 grade
target mean strength
Fck=fck+1.65S
Where Fck equal to target mean strength and fck equal to characteristic strength and as equal to deviation factor
Fck= 50+1.65(5)
Fck= 58.25 n/mm2
water cement ratio
assum that w/c= 0.35
maximum size of aggregate is 20 mm
than this 20 mm size of aggregate maximum water is 186 litre or kg for 25 to 50 mm slump
also we are using (2% of cement content) plasticizer and this plasticizer is reducing 20% of water
required water is= 186*(1-0.2)
= 149 litre or kg( for 25 to 50 mm slump)
we need 100 mm slump than increase 5% of water
required water is 149*(5/100) + 149
= 157 litre or kg
Cement content =157/0.35
=448 kg/m3
Plasticizer= 2% of cement content
= 448*(2/100) = 8.9 kg
Data
Water cement ratio
A) cement=448 kg/m3
B) water=157 lit or kg/m3
C) Water cement ratio= 0.35
D) plasticizer = 8.9 kg/ m3
Volume for 1 m3
Volume of cement= cement content/density. = 448/3150 =0.14 m3
volume of water = water content/density.
= 157/1000. =0.157
Volume of plasticizer = 2% of cement content /density
= (448*0.02)/1145 = 8.9/1145 = 0.007m3
= 1-(0.14+0.157+0.007) = 0.696 m3
we are using 20 mm size aggregate and zone 4 than value is 0.66
We are developing a pumpable concrete than reduce 10 % of aggregate than=0.66-{0.66*(10/100)}=0.594m3
volume of course aggregate =0.696*0.6594*2740 = 1132 kg
volume of fine aggregate(sand) =0.696*0.406*2740 = 774 kg
Mix proportion data for m 50
A) cement=448 kg/m3
B) water=157 lit or kg/m3
C) Water cement ratio= .35
D) plasticizer = 8.9 kg/ m3
E) coarse aggregate= 1132 kg/m3
F) fine aggregate= 774 kg/m3