Project on Concrete Mix Design for various grades of Concrete
Aim:- Calculate the Concrete Mix Design for M35 grade concrete with fly ash & M50 grade concrete without Fly ash
Introduction: The object of mix design is to decide the proportions of materials which will produce concrete of required properties. The mix proportions should be selected in such a way that the resulting concrete is of desired workability while fresh and it could be placed and compacted easily for the intended purpose.
A good concrete mix design creates the foundation of a sound infrastructure. Concrete mix design involves a process of preparation in which a mix of ingredients creates the required strength and durability for the concrete structure.
Procedure:- For M35 grade of concrete mix design
1. Target strength of mix proportioning
f'ck= fck + 1.65 S
f'ck = fck + X
Where
f'ck= target mean compressive strength at 28 days in N/mm²
f'ck= characteristic compressive strength at 28 days, in N/mm²
S = standard division in N/mm²
X = factor based on the grade of concrete,
fck for M35 garde is= 35 N/mm²
S = 5 N/mm²
Then
f'ck= 35+ 1.65×5
f'ck = 43.25 N/ mm²
2. Water cement ratio:
w/c ratio for OPC 43 grade = 0.38 ( with the help of graph IS code- 10262 2019)
Calculate water cement ratio from the maximum water cement ratio according to environmental exposure condition
as per table-5 of Is 456-2000
For very severe exposure = 0.45
3. Selection of water content:
water content = 191.5 × 0.7
Water content = 134.05
4. Selection of cement content
w/c = 0.38
134.05/c = 0.38
Then cement content = 352.7kg
5. Calculation of Fly Ash
For cementitious cement content
So 10% increase the value of cement content
= 352.7+ (352.7×10/100)
= 387.97
Then w/c ratio is
w/c = 134.05/387.97
w/c = 0.34
to Fly Ash content
= 30/100× 387.97
= 116.39 kg
For cement content
C + F. A.= 387.97
C+ 116.39 = 387.97
C = 271.5 Kg
Saving of cement = Initial Cement content - final Cement content
= 352.07 - 271.5
Saving of Cement is = 80kg
6. Volume of Aggregate
volume of coarse aggregates= [0.05-0.38 ×0.01]/0.05+0.60
= 0.624
For pumpable concrete reduce the 10% of volume of coarse aggregate
= 0.624-0.624×10/100
= 0.5616
Volume of fine aggregate = 1- Volume of fine aggregate
= 1-0.561
Volume of fine aggregate = 0.438
7. Selection of Volume
Volume of concrete = 1 m³
Volume of entrapped sir
In wet concrete = 0.002
Volume of cement = mass of cement/specific gravity of cement × 1/1000
= 271.5/3.15×0.001
= 0.086
Volume of Fly Ash = mass of fly ash/ specific gravity of F A × 0.001
= 116.39/ 2.2× 0.001
= 0.0529
Volume of water = mass of water/ specific gravity of water × 0.001
= 134.05/1×0.001
= 0.1340
Volume of admixture = mass of chemical admixture/specific gravity of admixture ×0.001
= 7.75/1.145× 0.001
= 0.0067
Volume of aggregate= 1-(0.086+0.0529+0.1340+0.0067)
= 0.7204
Mass of CA =0.7204× 0.5616× 2.2×1000
= 890.06
Mass of FA= 0.7204×0.438×2.2×1000
= 694.17
Mix proportion
- Cement = 271.5kg/ m³
- Fly ash = 116.39 kg/m³
- Water = 134.05 kg/m³
- FA = 694.17 kg/m³
- CA = 890.06
- W/c = 0.38
Mix ratoi = 1:2.55:3.27
Procedure- For M50 grade of concrete mix design
1.Target strength of mix proportioning
. f'ck = fck+ 1.65 S
f'ck = 50+1.65×5
f'ck= 58.25
2. Water Cement ratio
water cement ratio for OPC 53 grade Cement= 0.34
Calculate water cement ratio from the maximum water cement ratio according to environmental exposure condition
as per table-5 of Is 456-2000
For severe exposure = 0.45
3. Selection of water content
Water content = 197.16×0.7
Water content= 138.01kg/m³
4. Selection of cement content
W/c = 0.34
138.01/c = 0.34
C = 405.91 kg/m³
5. Volume of Aggregate
volume of coarse aggregates= [0.05-0.34 ×0.01]/0.05+0.60
= 0.618
For pumpable concrete reduce the 10% of volume of coarse aggregate
= 0.618-0.618×10/100
= 0.5562
Volume of fine aggregate = 1- Volume of fine aggregate
= 1-0.5563
Volume of fine aggregate = 0.4438
6. Selection of Volume
Volume of concrete = 1 m³
Volume of entrapped sir
In wet concrete = 0.002
Volume of cement = mass of cement/specific gravity of cement × 1/1000
= 405.91/3.15×0.001
= 0.1288
Volume of water = mass of water/ specific gravity of water × 0.001
= 138.012/1×0.001
= 0.1380
Volume of admixture = mass of chemical admixture/specific gravity of admixture ×0.001
= 8.11 /1.145× 0.001
= 0.070
Volume of aggregate= 1-(0.1288+0.1370+0.0708)
= 0.6634
Mass of CA =0.6634×0.5562× 2.2×1000
= 811.762
Mass of FA= 0.6634×0.4438×2.2×1000
= 647.717
Mix proportion
- Cement = 405.91kg/m³
- Water. = 138.012 kg/m³
- FA = 647.717 kg/m³
- CA = 811.762kg/m³
- w/c = 0.34
Mix ratio = 1:1.59:1.99