Project on Concrete Mix Design for various grades of Concrete
For M35 Grade of concrete
Cement is OPC 43 Grade
Severe exposure condition :-
Sp. Gravity of fy ash is 2.2
Sp. Garvity of CA & FA - 2.74
For M35 Grade of concrete :- Standard deviatation is 5.0
A. Cal Target mean strength Fck'
Fck' = Fck + 1.65 S
Fck - compressive strength of concrete after 28 days
S - standard deviatation
= 35 + 1.65 X 5
= 43.25 N/mm2
B. Slelct water cement ratio
From table 5 IS code 456-2000
Severe condition - max water content is 0.45
As per IS code 4031:1958
Table 2 - fck' value is 43.25 n/mm2
check the graph line and determine the water content
= 0.42
As per IS code 10262 : 2009 Table 2
For 20mm size of aggregate :- the water content is 186kg
C. Cal.cement content
W/C ratio = 0.45
cement content = 186/0.45 = 413kg/m3
Add 10% cementations materials content = 413X1.1 = 454kg/m3
Water content = 186kg
WC ratio = 186/454 = 0.40
Fly ash @30% cementation materials = 454X0.30 = 136kg/m3
Cement = 454-136 = 318kg/m3
Saving of cement content = 413-318 = 95kg/m3
Fly ash being ultilized = 136kg/m3
D. Volume of C A = .56
FA = .44
- Mix calculation
a. Volume of concrete = 1 m3
b. volume of cement = mass of cement/sp. gravity of cement X 1 / 1000
= 318/3.15 X 1/1000 = 0.10 cum
c. volume of fly ash = mass of fly ash / sp. garvity of fly ash X 1/1000
= 136/2.2 X 1/1000 = 0.062cum
d. voulme of total aggregate = a-(b+c)
= 1- (0.1+0.062)
= 0.838cum
e. Mass of CA = Voulme of CA X Voulme of Total aggreagte X Sp. Gravit of CA X 1000
= 0.56 X 0.838 X 2.75 X 1000 = 1286kg/m3
f. Mass of FA = Volume of FA X Volume of Total aggregate X Sp. gravity of FA X 1000
= 0.44 X 0.838 X 2.74 X 1000
= 1010kg/m3
Trail ratio is cement : sand : aggregte
= 318 kg : 1010kg : 1286kg
= 1 : 3.18 : 4.05
For M50 Grade of concrete
Cement is OPC 43 Grade
Severe exposure condition :-
Sp. Garvity of CA & FA - 2.74
For M50 Grade of concrete :- Standard deviatation is 5.0
A. Cal Target mean strength Fck'
Fck' = Fck + 1.65 S
Fck - compressive strength of concrete after 28 days
S - standard deviatation
= 50 + 1.65 X 5
= 58.25 N/mm2
B. Slelct water cement ratio
From table 5 IS code 456-2000
Severe condition - max water content is 0.45
As per IS code 4031:1958
Table 2 - fck' value is 58.25 n/mm2
check the graph line and determine the water content
= 0.35
As per IS code 10262 : 2009 Table 2
For 20mm size of aggregate :- the water content is 186kg
C. Cal.cement content
W/C ratio = 0.45
cement content = 186/0.45 = 413kg/m3
Add 10% cementations materials content = 413X1.1 = 454kg/m3
Water content = 186kg
WC ratio = 186/454 = 0.40
D. Volume of C A = .56
FA = .44
- Mix calculation
a. Volume of concrete = 1 m3
b. volume of cement = mass of cement/sp. gravity of cement X 1 / 1000
= 454/3.15 X 1/1000 = 0.144 cum
c. voulme of total aggregate = a-b
= 1- 0.144
= 0.856cum
e. Mass of CA = Voulme of CA X Voulme of Total aggreagte X Sp. Gravit of CA X 1000
= 0.56 X 0.856 X 2.75 X 1000 = 1313kg/m3
f. Mass of FA = Volume of FA X Volume of Total aggregate X Sp. gravity of FA X 1000
= 0.44 X 0.856 X 2.74 X 1000
= 1032kg/m3
Trail ratio is cement : sand : aggregte
= 454 kg : 1032kg : 1313kg
= 1 : 2.27 : 2.89