Project on Concrete Mix Design for various grades of Concrete
1)Concrete mix design for M35:
Aim:Concrete meet the strength requirement of the compressive strength,and fullfil durability,ingrediants useage.
Introduction:Mix design based on 3 standards
a)Indian standard
b)British standard
c)American standard.
A)Target mean strength of concrete:f'ck
f'ck = fck + 1.65S
fck = Compressive strength concrete = 35
S = 5 from table 8 IS10262:2009
= 35 + (1.65 x 5)
= 35 + 8.25
= 43.25
B)Calculate water cement ratio:
From graph 1 IS 10262:2009
a)Water cement ratio for f'ck 43.25 = 0.35
From graph 2 IS 10262:2009
b)Water cement ratio for f'ck = 0.42
c)Environmental exposure contition Table 5 IS 456:2000
M35 grade concrete = 0.45
In above a,b&c lower value taken for suitable = 0.35
C)Selection of watercontend: Table 2 IS10262:2009
Max water content for 20mm aggregates size = 186kg
= 186L
D)Selection of cement content: Table 5 IS 456:2000
W = 186
W/C =0.35
C = 186 / 0.35
= 531.5 kg
Cementitious materials content = 531.5 x 1.10
= 584.65
Water content = 186 L
Water cement ratio = 186 / 584.65
= 0.31
Fly ash added 30% of quantity of OPC = 584.65 x (30/100)
= 175.4 kg
Cement contend = 584.65 - 175.4 = 409.25 kg
Fly ash content = 531.50 - 409.25
= 122.25 Kg
E)Volume of coarse aggregade and fine aggregade:Table 3 IS 10262:2009
It's zone 2 = 0.62
F)MIx calculation:
a)Concrete volume = 1m3
b)Volume of cement = (Mass of cement/specific gravity of cement) x (1/1000)
= (531.50 / 3.15) x (1/1000)
= 0.168
c)Volume of fly ash = (122.25 / 2.74) x (1/1000)
= 0.04
d)Volume of water = (Mass of water/Specific gravity of water) x (1/1000)
= (186/1) x (1/1000)
= 0.186
e)Volume of aggregates = a-(b+c+d)
= 1-(0.168+0.04+0.186)
= 1-0.394
= 0.606
f)Mass of coarse aggregates = e x volume of coarse aggregate x specific gravity of CA x 1000
= 0.606 x 0.62 x 2.74 x 1000
= 1029.47
g)Mass of fine aggregates = e x volume of fine aggregate x specific gravity of fA x 1000
= 0.606 x 0.38 x 2.74 x 1000
= 630.97
Result :
Mix propotion:
Cement = 409.25 kg
Fly ash =122.25 kg
Water = 186 L
Fine aggregates = 630.97 kg
Coarse aggregates = 1029.47 kg
water cement ratio = 0.35
Concultion:
Result of mix = 409.25 : 630.97:1029.47
= 1:1.54:2.52
2)Concrete mix design for M50:
Aim:Concrete meet the strength requirement of the compressive strength,and fullfil durability,ingrediants useage.
Introduction:Mix design based on 3 standards
a)Indian standard
b)British standard
c)American standard.
A)Target mean strength of concrete:f'ck
f'ck = fck + 1.65S
fck = Compressive strength concrete = 50
S = 5 from table 8 IS10262:2009
= 50 + (1.65 x 5)
= 50 + 8.25
= 58.25
B)Calculate water cement ratio:
Environmental exposure contition Table 5 IS 456:2000
M50 grade concrete `less then ` 0.40
= 0.35
C)Selection of watercontend: Table 2 IS10262:2009
Max water content for 20mm aggregates size = 186kg
= 186L
D)Selection of cement content: Table 5 IS 456:2000
W = 186
W/C =0.35
C = 186 / 0.35
= 531.5 kg
Water content = 186 L
Water cement ratio = 186 / 531.5
= 0.34
E)Volume of coarse aggregade and fine aggregade:Table 3 IS 10262:2009
It's zone 2 = 0.62
F)MIx calculation:
a)Concrete volume = 1m3
b)Volume of cement = (Mass of cement/specific gravity of cement) x (1/1000)
= (531.50 / 3.15) x (1/1000)
= 0.168
c)Volume of chemical admixture = (10.63 / 1.145) x (1/1000)
= 0.009
d)Volume of water = (Mass of water/Specific gravity of water) x (1/1000)
= (186/1) x (1/1000)
= 0.186
e)Volume of aggregates = a-(b+c+d)
= 1-(0.168+0.009+0.186)
= 1-0.363
= 0.637
f)Mass of coarse aggregates = e x volume of coarse aggregate x specific gravity of CA x 1000
= 0.637 x 0.62 x 2.74 x 1000
= 1082.14
g)Mass of fine aggregates = e x volume of fine aggregate x specific gravity of fA x 1000
= 0.637 x 0.38 x 2.74 x 1000
= 663.24
Result :
Mix propotion:
Cement = 531.50 kg
Chemical admixture =10.63kg
Water = 186 L
Fine aggregates = 663.24 kg
Coarse aggregates = 1082.14 kg
water cement ratio = 0.35
Concultion:
Result of mix = 531.50 : 663.24:1082.14
= 1:1.25:2.03