Project on Concrete Mix Design for various grades of Concrete
1. Mix design of M35 grade of concrete with cement and fly ash.
A. Grade of concrete - M35
B. Type of cement - OPC 53
C. Miniral Admixture - Fly ash
D. Exposure Condition - Very Severe
E. Maximum Nominal Size Aggregate - 20mm
F. Minimum cement content - 340kg (As per IS 456:2000)
G. Maximum Cement content - 450kg (As per IS 456:2000)
H. Fly ash - 30%
I. Maximum W/C ratio - 0.45 - (As per IS 456:2000)
J. Workability (Slump) - 100mm
K. Chemical Admixture - Plasticizer
II. Test data of materials
A. Specific gravity of cement - 3.15
B. Specific gravity of CA - 2.74
C. Specific gravity of FA - 2.74
D. Specific gravity of Fly ash - 2.90 (IS 12089)
E. Specific gravity of Admixture - 1.12
F. Water absorption of CA - 0.5%
G. Water absorption of FA - 1%
H. Chemical admixture quantity - 0.6% of cementenious material
I. Zone of aggrevates (Sieve Analysis) - Zone II
III. Target strength of mix
f'ck = fck+1.65×s
= 35+1.65×5
= 43.25 Mpa
IV. Selection of water cement ratio
Maximum W/C ratio = 0.45
Adopted W/C ratio = 0.45
V. Maximim water content as per IS 10262:2019 = 186 Litre
The above mentioned water content is for the slump range of 25 to 50mm. For every 25mm increment in slump value the water content should be increased by 3%.
Therefore, for 100mm of slump we have to increase the water content by 6%
Hence, water content for 100mm slump = 186+(0.6×186) = 197.16 Litres
Say, 198 L of water
As we are using plasticizer we can reduce the water content by 20%.
Therefore the required water quantity will be'
198 - (0.20×198) = 157.73 L
Adopted water content will be 160 litre. (Based on experience)
VI. Cement and Fly ash content
Adopted water cement ratio = 0.45
Total cementenious material required = water content/w/c ratio
=160/0.45 = 355.55 kg
Say, 356 kg
As we are using fly ash also in cementenious content, it is desirable to increase the quantity of cementenious content by 10% to get good quality concrete. It is totally based on field experience.
Thus the final cementenious content = 356×1.10
= 391.6 kg
Say 392 kg
Minimum cement content as per IS 456:2000 = 340kg
392kg > 340kg, hence ok
Now fly ash content,
Fly ash = 392×30% = 117.6 = 118kg
Cement content = 392 - 118 = 274kg
Content of plasticizer = 0.6% × 392 = 2.35kg
VII. Coarse & Fine Aggregate
CA ratio as per IS 10262:2019 for 0.5 W/C = 0.62
CA ratio for 0.45 W/C ratio = 0.63 (adding 0.01 for every decrease of .05 and vice versa)
FA ratio = 1-0.63 = 0.37
VIII. Mix calculation of M35 grade of concrete,
A. Volume of concrete = 1m3
B. Total volume of cement = Cement content/S.G×1000
=274/3.15×1000 = 0.087m3
C. Total volume of fly ash = Flyash/S.G×1000
= 118/2.9×1000 = 0.041m3
D. Volume of water = Water/S.G×1000
=160/1×1000 = 0.160m3
E. Volume of plasticizer = Plasticizer/S.G×1000
= 2.35/1.12×1000 = 0.002m3
F. Total aggregate required = A-(B+C+D+E)
= 1-(0.087+0.041+0.160+0.002)
= 0.71m3
G. C.A = F×C.A ratio×S.G×1000
= 0.71×0.63×2.74×1000
= 1225.60 kg
H. F.A = F×F.A×S.G×1000
= 0.71×0.37×2.74×1000
= 719.8 kg
Hence, material required for M35 grade of concrete for 1m3 is,
i. Cement = 275 kg
ii. Flyash = 118 kg
iii. F.A = 720 kg
iv. C.A = 1226 kg
v. Water = 160 L
vi. Plasticizer = 2.5 L
Hence ratio will be
Cement:Flyash:FA:CA:Water = 1:0.42:2.6:4.45:0.58
2. Mix design of M50 grade of concrete.
A. Grade of concrete - M50
B. Type of cement - OPC 53
C. Exposure Condition - Very Severe
D. Maximum Nominal Size Aggregate - 20mm
E. Minimum cement content - 340kg (As per IS 456:2000)
F. Maximum Cement content - 450kg (As per IS 456:2000)
G. Maximum W/C ratio - 0.45 - (As per IS 456:2000)
H. Workability (Slump) - 100mm
I. Chemical Admixture - Plasticizer
II. Test data of materials
A. Specific gravity of cement - 3.15
B. Specific gravity of CA - 2.74
C. Specific gravity of FA - 2.74
D. Specific gravity of Admixture - 1.145
E. Water absorption of CA - 0.5%
F. Water absorption of FA - 1%
G. Chemical admixture quantity - 2% of cementenious material
H. Zone of aggrevates (Sieve Analysis) - Zone II
III. Target strength of mix
f'ck = fck+1.65×s
= 50+1.65×5
= 58.25 Mpa
IV. Selection of water cement ratio
Maximum W/C ratio = 0.45
Adopted W/C ratio = 0.45
V. Maximim water content as per IS 10262:2019 = 186 Litre
The above mentioned water content is for the slump range of 25 to 50mm. For every 25mm increment in slump value the water content should be increased by 3%.
Therefore, for 100mm of slump we have to increase the water content by 6%
Hence, water content for 100mm slump = 186+(0.6×186) = 197.16 Litres
Say, 198 L of water
As we are using plasticizer we can reduce the water content by 20%.
Therefore the required water quantity will be'
198 - (0.20×198) = 157.73 L
Adopted water content will be 160 litre. (Based on experience)
VI. Cement content
Adopted water cement ratio = 0.45
Total cementenious material required = water content/w/c ratio
=160/0.45 = 355.55 kg
Corrected quantity = 360 kg
Minimum cement content as per IS 456:2000 = 340kg
360kg > 340kg, hence ok
Content of plasticizer = 2% × 360 = 7.2kg
VII. Coarse & Fine Aggregate
CA ratio as per IS 10262:2019 for 0.5 W/C = 0.62
CA ratio for 0.45 W/C ratio = 0.63 (adding 0.01 for every decrease of .05 and vice versa)
FA ratio = 1-0.63 = 0.37
VIII. Mix calculation of M50 grade of concrete,
A. Volume of concrete = 1m3
B. Total volume of cement = Cement content/S.G×1000
=360/3.15×1000 = 0.114m3
C. Volume of water = Water/S.G×1000
=160/1×1000 = 0.160m3
D. Volume of plasticizer = Plasticizer/S.G×1000
= 7.2/1.145×1000 = 0.006m3
E. Total aggregate required = A-(B+C+D)
= 1-(0.114+0.160+0.006)
= 0.72m3
F. C.A = E×C.A ratio×S.G×1000
= 0.72×0.63×2.74×1000
= 1242.860 kg = 1243 kg
G. F.A = E×F.A×S.G×1000
= 0.72×0.37×2.74×1000
= 730 kg
Hence, material required for M50 grade of concrete for 1m3 is,
i. Cement = 360 kg
ii. F.A = 730 kg
iii. C.A = 1243 kg
iv. Water = 160 L
v. Plasticizer = 7.2 L
Hence ratio will be
Cement:FA:CA:Water = 1:2.02:3.45:0.44